beta-0.89.2

[luatex.git] / source / texk / web2c / triptrap / mptrap.log

This is MetaPost, version 1.203 (kpathsea version 3.5.7dev) (mem=trap 2009.06.02)  2 JUN 2009 22:19
**  trap 
(./trap.mp
{if}
{known(0)}
{not(true)}
{false}
{known("")}
{true}

/*\`'@#->begingroup.message(SUFFIX1)&str(SUFFIX0)&jobname&char.ASCII'`&str(SUFF
IX2)!endgroup
(SUFFIX0)<-/*\
(SUFFIX1)<-`'
(SUFFIX2)<-pass2
{begingroup}
{message}

`'->begingroup'`endgroup
(SUFFIX0)<-
(SUFFIX1)<-`'
{begingroup}

'`->"\*/"
{endgroup}
{("\*/")&("/*\")}
{("\*//*\")&("trap")}

'`->"\*/"
{ASCII("\*/")}
{char(92)}
{("\*//*\trap")&("\")}
{("\*//*\trap\")&("pass2!")}

\*//*\trap\pass2!
{endgroup}
{nonstopmode}

{outer}
{let}
{delimiters}
! Missing symbolic token inserted.
<inserted text> 
                 INACCESSIBLE
l.22 ...opmode; outer\; let next=\; delimiters ^~7
                                                  ! fi
Sorry: You can't redefine a number, string, or expr.
I've inserted an inaccessible symbol so that your
definition will be completed without mixing me up too badly.

! Extra tokens will be flushed.
<to be read again> 
                   !
l.22 ...pmode; outer\; let next=\; delimiters ^~7!
                                                   fi
I've just read as much of that statement as I could fathom,
so a semicolon should have been next. It's very puzzling...
but I'll try to get myself back together, by ignoring
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

! Forbidden token found while scanning to the end of the statement.
<inserted text> 
                ;
<to be read again> 
                   \
l.23 next\
          ; % the second pass will now compute silently; the first pass will...
A previous error seems to have propagated,
causing me to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.

{\}
{batchmode}

! An expression can't begin with `endgroup'.
<inserted text> 
                0
<to be read again> 
                   endgroup
l.24 batchmode; ^~7,endgroup
                             pausing:=1; exitif p exitif boolean pen pencirc...
I'm afraid I need some sort of value in order to continue,
so I've tentatively inserted `0'. You may want to
delete this zero and insert something else;
see Chapter 27 of The METAFONTbook for an example.

! Missing ` INACCESSIBLE' has been inserted.
<to be read again> 
                   endgroup
l.24 batchmode; ^~7,endgroup
                             pausing:=1; exitif p exitif boolean pen pencirc...
I found no right delimiter to match a left one. So I've
put one in, behind the scenes; this may fix the problem.

! Extra `endgroup'.
<recently read> endgroup
                        
l.24 batchmode; ^~7,endgroup
                             pausing:=1; exitif p exitif boolean pen pencirc...
I'm not currently working on a `begingroup',
so I had better not try to end anything.

{pausing:=1}
{exitif}
{exitif}
{pencircle}
{endfor}
! Extra `endfor'.
l.24 ...itif p exitif boolean pen pencircle endfor
                                                  
I'm not currently working on a for loop,
so I had better not try to end anything.

{scantokens}
{begingroup}
{message}
{char(0)}
{("^^@")&("watch this")}
^^@watch this
{-(1)}
{char(-1)}
{("pair p[],';")&("ÿ")}
{endgroup}
{pen(pen)}
{boolean(true)}
{true}
! No loop is in progress.
<to be read again> 
                   pair
<scantokens> pair
                  p[],';ÿ
<to be read again> 
                   path
l.26 path
          p[][]p,w,qw; qw=(1,-2)..(2,-1)..(2.5,0.5)..(1,2)..(turningnumber',...
Why say `exitif' when there's nothing to exit from?

>> p
! Undefined condition will be treated as `false'.
<to be read again> 
                   pair
<scantokens> pair
                  p[],';ÿ
<to be read again> 
                   path
l.26 path
          p[][]p,w,qw; qw=(1,-2)..(2,-1)..(2.5,0.5)..(1,2)..(turningnumber',...
The expression shown above should have had a definite
true-or-false value. I'm changing it to `false'.

{false}
! Missing `;' has been inserted.
<to be read again> 
                   pair
<scantokens> pair
                  p[],';ÿ
<to be read again> 
                   path
l.26 path
          p[][]p,w,qw; qw=(1,-2)..(2,-1)..(2.5,0.5)..(1,2)..(turningnumber',...
After `exitif <boolean exp>' I expect to see a semicolon.
I shall pretend that one was there.

{pair}
! Text line contains an invalid character.
<scantokens> pair p[],';ÿ
                         
<to be read again> 
                   path
l.26 path
          p[][]p,w,qw; qw=(1,-2)..(2,-1)..(2.5,0.5)..(1,2)..(turningnumber',...
A funny symbol that I can't read has just been input.
Continue, and I'll forget that it ever happened.

{path}
{-(2)}
{-(1)}
{turningnumber((xpart ',ypart '))}
Path at line 26, before choices:
(1,-2){curl 1}
 ..(2,-1)
 ..(2.5,0.5)
 ..(1,2)
 ..{curl 1}(0,2.5)

Path at line 26, after choices:
(1,-2)..controls (1.37755,-1.71404) and (1.71404,-1.37755)
 ..(2,-1)..controls (2.33353,-0.55965) and (2.59729,-0.04124)
 ..(2.5,0.5)..controls (2.36812,1.23369) and (1.6712,1.65662)
 ..(1,2)..controls (0.66821,2.16974) and (0.33485,2.33641)
 ..(0,2.5)

{(unknown path qw)=(path)}
{numeric}
! Enormous number has been reduced.
l.27 numeric p[][]; p[[ [$] ]]=10000000000000000
                                                ; "this string constant is i...
I can't handle numbers bigger than 32767.99998;
so I've changed your constant to that maximum amount.

{(p[[ [-1] ]])=(32767.99998)}
## p[[ [-1] ]]=32767.99998
! Value is too large (32767.99998).
l.27 numeric p[][]; p[[ [$] ]]=10000000000000000;
                                                  "this string constant is i...
The equation I just processed has given some variable
a value of 4096 or more. Continue and I'll try to cope
with that big value; but it might be dangerous.
(Set warningcheck:=0 to suppress this message.)

! Incomplete string token has been flushed.
l.27 ...00000; "this string constant is incomplete
                                                  
Strings should finish on the same line as they began.
I've deleted the partial string; you might want to
insert another by typing, e.g., `I"new string"'.

{string}
! Declared variable conflicts with previous vardef.
<to be read again> 
                   ,
l.28 string foo[]p,
                   p~if true:[];
You can't use, e.g., `numeric foo[]' after `vardef foo'.
Proceed, and I'll ignore the illegal redeclaration.

{if}
{true}
{true}
{boolean}
{fi}
{showvariable}
boolean.boolean=unknown boolean
! OK.
l.29 ...,boolean fi.boolean; showvariable boolean;
                                                  

{inner}
{def}
{outer}
{picture}
{pen}
! Illegal suffix of declared variable will be flushed.
<to be read again> 
                   [
<to be read again> 
                   "a"
l.31 pen p~[]~,q["a"
                    ,qq; p~1~=q=pencircle scaled mexp(-3016.57654);
Variables in declarations must consist entirely of
names and collective subscripts, e.g., `x[]a'.
Are you trying to use a reserved word in a variable name?
I'm going to discard the junk I found here,
up to the next comma or the end of the declaration.

{pencircle}
{-(3016.57654)}
{mexp(-3016.57654)}
{(pen)scaled(0)}
{(unknown pen q)=(pen)}
{(unknown pen p~1~)=(pen)}
{transform}
! Illegal suffix of declared variable will be flushed.
<to be read again> 
                   0
l.32 transform p,pp0
                    ; if p=p:qq=makepen((7,0)..cycle) xscaled hex "1000";sho...
Variables in declarations must consist entirely of
names and collective subscripts, e.g., `x[]a'.
Explicit subscripts like `x15a' aren't permitted.
I'm going to discard the junk I found here,
up to the next comma or the end of the declaration.

{if}
{((xpart p,ypart p,xxpart p,xypart p,yxpart p,yypart p))=((xpart p,ypart p,xxpa
rt p,xypart p,yxpart p,yypart p))}
{true}
Path at line 32, before choices:
(7,0)
 ..cycle

Path at line 32, after choices:
(7,0)..controls (7,0) and (7,0)
 ..cycle

{makepen(path)}
{hex("1000")}
! Number too large (4096).
<to be read again> 
                   ;
l.32 ...=makepen((7,0)..cycle) xscaled hex "1000";
                                                  show qq;fi
I have trouble with numbers greater than 4095; watch out.
(Set warningcheck:=0 to suppress this message.)

{(pen)xscaled(4096)}
{(unknown pen qq)=(pen)}
{show}
>> Pen at line 32:
pencircle transformed (28672,0,0,0,0,0)

! OK.
l.32 ...((7,0)..cycle) xscaled hex "1000";show qq;
                                                  fi

{fi}
Path at line 33, before choices:
(4,1)
 ..(1,2)
 ..(5,3)
 ..(3,2)
 ..(1,1)
 ..(0,1)
 ..(1,0)
 ..cycle

Path at line 33, after choices:
(4,1)..controls (3.72015,2.36482) and (1.84834,1.36037)
 ..(1,2)..controls (-5.16994,6.65202) and (8.52596,9.85889)
 ..(5,3)..controls (4.6365,2.29291) and (3.69041,2.34465)
 ..(3,2)..controls (2.30983,1.65547) and (1.78375,0.86862)
 ..(1,1)..controls (0.63611,1.06099) and (0.23091,1.28844)
 ..(0,1)..controls (-0.3216,0.5983) and (0.24828,0.12357)
 ..(1,0)..controls (2.58165,-0.25998) and (4.23677,-0.15471)
 ..cycle

{makepen(path)}
{qq:=pen}
{show}
>> Pen at line 33:
(0,1)
 .. (1,0)
 .. (4,1)
 .. (5,3)
 .. (1,2)
 .. cycle

! OK.
l.33 ...3,2)..(1,1)..(0,1)..(1,0)..cycle);show qq;
                                                  

{vardef}
! Missing parameter type; `expr' will be assumed.
<to be read again> 
                   )
l.34 vardef p[][]p~ begintext suffix a,b endtext()
                                                  )suffix@=show #@; p.a.b() ...
You should've had `expr' or `suffix' or `text' here.

{expandafter}
{\}
{let}

\\->=show
{outer}
{((6,12))-((xpart p7,ypart p7))}
{((0,1))transformed((xpart p,ypart p,xxpart p,xypart p,yxpart p,yypart p))}
{(x)-(x)}
{(2)/(0)}
>> 2
! Division by zero.
<to be read again> 
                   ,
l.36 (6,12)-p7=(0,1)transformed p=(2/(x-x),
                                           3/0)transformed p;
You're trying to divide the quantity shown above the error
message by zero. I'm going to divide it by one instead.

! Division by zero.
l.36 (6,12)-p7=(0,1)transformed p=(2/(x-x),3/0
                                              )transformed p;
I'll pretend that you meant to divide by 1.

{((2,3))transformed((xpart p,ypart p,xxpart p,xypart p,yxpart p,yypart p))}
{((linearform,linearform))=((linearform,linearform))}
## yxpart p=-yypart p
## xxpart p=-xypart p
{((-xpart p7+6,-ypart p7+12))=((linearform,linearform))}
## ypart p7=-ypart p-yypart p+12
## xpart p7=-xpart p-xypart p+6
{\}
{if}
{string(unknown string p~[-1])}
{true}
{(p0.1 0.2)-(p0.1 0.2)}
! The token `endtext' is no longer a right delimiter.
l.37 ...f string p~[$]: p.1.2-p.1.199999,1 endtext
                                                   transformed p;
Strange: This token has lost its former meaning!
I'll read it as a right delimiter this time;
but watch out, I'll probably miss it later.

{((0,1))transformed((xpart p,ypart p,-xypart p,xypart p,-yypart p,yypart p))}
{((linearform,linearform))=((linearform,linearform))}
## ypart p=-yypart p+6
#### ypart p7=6
## xpart p=-xypart p+3
#### xpart p7=3
{(unknown path p1 2p)=((3,6))}
{showstopping:=0}
{showvariable}
p=(-xypart p+3,-yypart p+6,-xypart p,xypart p,-yypart p,yypart p)
p[]=pair
p[][]=numeric
p[][]p=unknown path
p[][]p~=macro:(SUFFIX2)(SUFFIX3)(EXPR4)<suffix>->begingroup.show ETC.
p[]~=unknown boolean
p~=path
p~[]=unknown string
p~[]~=unknown pen
p~[]~[][]=unknown picture
p~[-1]=unknown string p~[-1]
p~1~=pen
p[[ [] ]]=numeric
p[[ [-1] ]]=32767.99998
p0.1 0.2=p0.1 0.2
p1 2p=path
p7=(3,6)
{((-xypart p+3,-yypart p+6,-xypart p,xypart p,-yypart p,yypart p))=((-xypart p+
3,-yypart p+6,-xypart p,xypart p,-yypart p,yypart p))}
{let}
{let}
{xxpart((-xypart p+3,-yypart p+6,-xypart p,xypart p,-yypart p,yypart p))}
{(-xypart p)+(0.002)}
{yxpart((-xypart p+3,-yypart p+6,-xypart p,xypart p,-yypart p,yypart p))}
{((xpart p2,ypart p2))-((xpart p1,ypart p1))}
{(1)*((linearform,linearform))}
{((xpart p1,ypart p1))+((linearform,linearform))}
{(y)+(0.00002)}
{yypart((-xypart p+3,-yypart p+6,-xypart p,xypart p,-yypart p,yypart p))}
{xypart((-xypart p+3,-yypart p+6,-xypart p,xypart p,-yypart p,yypart p))}
{((5,y))=((yypart p,xypart p))}
## y=xypart p
## yypart p=5
#### ypart p=1
#### yxpart p=-5
{((5,xypart p+0.00002))=((5,xypart p))}
{((xpart p2,ypart p2))=((5,xypart p))}
## ypart p2=xypart p
## xpart p2=5
{((-xypart p+0.002,-5))=((5,xypart p))}
## xypart p=-5
#### ypart p2=-5
#### y=-5
#### xpart p=8
#### xxpart p=5
! Inconsistent equation (off by -0.002).
<to be read again> 
                   ;
l.39 ...2]=(5,y+.00001)=(5,y)=(yypart p,xypart p);
                                                  
The equation I just read contradicts what was said before.
But don't worry; continue and I'll just ignore it.

Path at line 40, before choices:
(0,0)..controls (15,4) and (-15,-12)
 ..(4,0)
 ..cycle

Path at line 40, after choices:
(0,0)..controls (15,4) and (-15,-12)
 ..(4,0)..controls (17.52783,8.54388) and (-15.45978,-4.12262)
 ..cycle

{reverse(path)}
{(path)transformed((8,1,5,-5,-5,5))}
{(path)=(unknown path p2 3p)}

p1 2p~(SUFFIX2)(SUFFIX3)(EXPR4)<suffix>->begingroup.show(SUFFIX0);p(SUFFIX2)(SU
FFIX3)((EXPR4)endgroup
(SUFFIX0)<-p1 2p
(SUFFIX1)<-~
(SUFFIX2)<-
(SUFFIX3)<-2 3p~
! A primary expression can't begin with `right delimiter that matches ('.
<inserted text> 
                0
<to be read again> 
                   )
l.41 p[000000000001]2p~(,[2]3p~,-)
                                  =p~1~2[pausing];
I'm afraid I need some sort of value in order to continue,
so I've tentatively inserted `0'. You may want to
delete this zero and insert something else;
see Chapter 27 of The METAFONTbook for an example.

{-(0)}
(EXPR4)<-0
(SUFFIX5)<-
{begingroup}
{show}
>> Path at line 41:
(3,6)


p2 3p~(SUFFIX2)(SUFFIX3)(EXPR4)<suffix>->begingroup.show(SUFFIX0);p(SUFFIX2)(SU
FFIX3)((EXPR4)endgroup
(SUFFIX0)<-p2 3p
(SUFFIX1)<-~
! Missing `,' has been inserted.
<to be read again> 
                   (0)
p1 2p~->...ow(SUFFIX0);p(SUFFIX2)(SUFFIX3)((EXPR4)
                                                  endgroup
<to be read again> 
                   =
l.41 p[000000000001]2p~(,[2]3p~,-)=
                                   p~1~2[pausing];
I've finished reading a macro argument and am about to
read another; the arguments weren't delimited correctly.
You might want to delete some tokens before continuing.

(SUFFIX2)<-
! Missing `,' has been inserted.
<to be read again> 
                   (0)
p1 2p~->...ow(SUFFIX0);p(SUFFIX2)(SUFFIX3)((EXPR4)
                                                  endgroup
<to be read again> 
                   =
l.41 p[000000000001]2p~(,[2]3p~,-)=
                                   p~1~2[pausing];
I've finished reading a macro argument and am about to
read another; the arguments weren't delimited correctly.
You might want to delete some tokens before continuing.

(SUFFIX3)<-
! Missing `)' has been inserted.
<to be read again> 
                   endgroup
<to be read again> 
                   =
l.41 p[000000000001]2p~(,[2]3p~,-)=
                                   p~1~2[pausing];
I've gotten to the end of the macro parameter list.
You might want to delete some tokens before continuing.

(EXPR4)<-0
(SUFFIX5)<-
{begingroup}
{show}
>> Path at line 41:
(8,1)..controls (-48.68579,57.68579) and (52.91974,-43.91974)
 ..(28,-19)..controls (-7,16) and (63,-54)
 ..cycle

>> (8,1,5,-5,-5,5)
! Isolated expression.
<to be read again> 
                   (
p2 3p~->...roup.show(SUFFIX0);p(SUFFIX2)(SUFFIX3)(
                                                  (EXPR4)endgroup
<to be read again> 
                   endgroup
<to be read again> 
                   =
l.41 p[000000000001]2p~(,[2]3p~,-)=
                                   p~1~2[pausing];
I couldn't find an `=' or `:=' after the
expression that is shown above this error message,
so I guess I'll just ignore it and carry on.

! Extra tokens will be flushed.
<to be read again> 
                   (
p2 3p~->...roup.show(SUFFIX0);p(SUFFIX2)(SUFFIX3)(
                                                  (EXPR4)endgroup
<to be read again> 
                   endgroup
<to be read again> 
                   =
l.41 p[000000000001]2p~(,[2]3p~,-)=
                                   p~1~2[pausing];
I've just read as much of that statement as I could fathom,
so a semicolon should have been next. It's very puzzling...
but I'll try to get myself back together, by ignoring
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

{endgroup}
{endgroup}
{(vacuous)=(unknown picture p~1~2 1)}
>> vacuous
>> unknown picture p~1~2 1
! Equation cannot be performed (vacuous=unknown picture).
<to be read again> 
                   ;
l.41 p[000000000001]2p~(,[2]3p~,-)=p~1~2[pausing];
                                                  
I'm sorry, but I don't know how to make such things equal.
(See the two expressions just above the error message.)

{vardef}
Runaway definition?
if.p(SUFFIX1)(SUFFIX1)=(SUFFIX1)(SUFFIX1)p.fi
! Forbidden token found while scanning the definition of p~[].
<inserted text> 
                enddef
<to be read again> 
                   ;
l.42 vardef p~[]@# tertiary t:=if p@ @=@ @p fi;
                                                vardef p[][]p~[]=BAD; inner ;;
I suspect you have forgotten an `enddef',
causing me to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.

{vardef}
! This variable already starts with a macro.
l.42 ...iary t:=if p@ @=@ @p fi; vardef p[][]p~[]=
                                                  BAD; inner ;;
After `vardef a' you can't say `vardef a.b'.
So I'll have to discard this definition.

Runaway definition?
BAD
! Forbidden token found while scanning the definition of a bad variable.
<inserted text> 
                enddef
<to be read again> 
                   ;
l.42 ... t:=if p@ @=@ @p fi; vardef p[][]p~[]=BAD;
                                                   inner ;;
I suspect you have forgotten an `enddef',
causing me to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.

{inner}
{show}
{-(2)}

p~[-2]@#<tertiary>->begingroup.if.p(SUFFIX1)(SUFFIX1)=(SUFFIX1)(SUFFIX1)p.fi.en
dgroup
(SUFFIX0)<-p~
(SUFFIX1)<-[-2]
(SUFFIX2)<-~
! A tertiary expression can't begin with `['.
<inserted text> 
                0
<to be read again> 
                   [
<to be read again> 
                   (3000)
<to be read again> 
                   ,
l.43 show p~[-2]~[3000,
                       x]++4000>path p3; showvariable p,P;
I'm afraid I need some sort of value in order to continue,
so I've tentatively inserted `0'. You may want to
delete this zero and insert something else;
see Chapter 27 of The METAFONTbook for an example.

{(x)-(3000)}
{(0)*(x-3000)}
{(3000)+(0)}
{(3000)++(4000)}
(EXPR3)<-4999.99998
{begingroup}
{if}
{(p[-2][-2])=(-2)}
>> p[-2][-2]+2
! Unknown relation will be considered false.
<to be read again> 
                   [-2]
p~[-2]->....p(SUFFIX1)(SUFFIX1)=(SUFFIX1)(SUFFIX1)
                                                  p.fi.endgroup
<to be read again> 
                   >
l.43 show p~[-2]~[3000,x]++4000>
                                path p3; showvariable p,P;
Oh dear. I can't decide if the expression above is positive,
negative, or zero. So this comparison test won't be `true'.

{false}
! Missing `:' has been inserted.
<to be read again> 
                   [-2]
p~[-2]->....p(SUFFIX1)(SUFFIX1)=(SUFFIX1)(SUFFIX1)
                                                  p.fi.endgroup
<to be read again> 
                   >
l.43 show p~[-2]~[3000,x]++4000>
                                path p3; showvariable p,P;
There should've been a colon after the condition.
I shall pretend that one was there.

{endgroup}
{path((xpart p3,ypart p3))}
{(vacuous)>(false)}
>> vacuous
>> false
! Not implemented: (vacuous)>(boolean).
<to be read again> 
                   ;
l.43 show p~[-2]~[3000,x]++4000>path p3;
                                         showvariable p,P;
I'm afraid I don't know how to apply that operation to that
combination of types. Continue, and I'll return the second
argument (see above) as the result of the operation.

>> false
{showvariable}
p=(8,1,5,-5,-5,5)
p[]=pair
p[][]=numeric
p[][]p=unknown path
p[][]p~=macro:(SUFFIX2)(SUFFIX3)(EXPR4)<suffix>->begingroup.show ETC.
p[]~=unknown boolean
p~=path
p~[]@#=macro:<tertiary>->begingroup.if.p(SUFFIX1)(SUFFIX1)=(SUFFIX1) ETC.
p[[ [] ]]=numeric
p[[ [-1] ]]=32767.99998
p[-2][-2]=p[-2][-2]
p0.1 0.2=p0.1 0.2
p1=(xpart p1,ypart p1)
p1 2p=path
p2=(5,-5)
p2 3p=path
p3=(xpart p3,ypart p3)
p7=(3,6)
> P=tag
{numeric}
{(2)*(alpha)}
{(p3~)=(2alpha)}
## alpha=0.5p3~
{(1)/(-1)}
{(3)*(beta)}
{(p[-1]~)=(3beta)}
## beta=0.33333p[-1]~
{begingroup}
{save}
{showvariable}
> p=tag
{(3)*(0.33333(SAVED)p[-1]~)}
{((SAVED)p[-1]~)=(1)}
## (SAVED)p[-1]~=1
#### beta=0.33333
{restoring p}
{endgroup}
{showvariable}
p=(8,1,5,-5,-5,5)
p[]=pair
p[][]=numeric
p[][]p=unknown path
p[][]p~=macro:(SUFFIX2)(SUFFIX3)(EXPR4)<suffix>->begingroup.show ETC.
p[]~=numeric
p~=path
p~[]@#=macro:<tertiary>->begingroup.if.p(SUFFIX1)(SUFFIX1)=(SUFFIX1) ETC.
p[[ [] ]]=numeric
p[[ [-1] ]]=32767.99998
p[-2][-2]=p[-2][-2]
p[-1]~=1
p0.1 0.2=p0.1 0.2
p1=(xpart p1,ypart p1)
p1 2p=path
p2=(5,-5)
p2 3p=path
p3=(xpart p3,ypart p3)
p3~=p3~
p7=(3,6)
{def}
{def}
! Missing `=' has been inserted.
<to be read again> 
                   false
l.46 def//expr;=enddef;def!primary!false
                                        ):!fi enddef;
The next thing in this `def' should have been `=',
because I've already looked at the definition heading.
But don't worry; I'll pretend that an equals sign
was present. Everything from here to `enddef'
will be the replacement text of this macro.

{def}
{(path)scaled(-1)}
Path at line 47, before choices:
(1,-2)..controls (1.37755,-1.71404) and (1.71404,-1.37755)
 ..(2,-1)..controls (2.33353,-0.55965) and (2.59729,-0.04124)
 ..(2.5,0.5)..controls (2.36812,1.23369) and (1.6712,1.65662)
 ..(1,2)..controls (0.66821,2.16974) and (0.33485,2.33641)
 ..(0,2.5)
 ..(-1,2)..controls (-1.37755,1.71404) and (-1.71404,1.37755)
 ..(-2,1)..controls (-2.33353,0.55965) and (-2.59729,0.04124)
 ..(-2.5,-0.5)..controls (-2.36812,-1.23369) and (-1.6712,-1.65662)
 ..(-1,-2)..controls (-0.66821,-2.16974) and (-0.33485,-2.33641)
 ..(0,-2.5)
 ..cycle

Path at line 47, after choices:
(1,-2)..controls (1.37755,-1.71404) and (1.71404,-1.37755)
 ..(2,-1)..controls (2.33353,-0.55965) and (2.59729,-0.04124)
 ..(2.5,0.5)..controls (2.36812,1.23369) and (1.6712,1.65662)
 ..(1,2)..controls (0.66821,2.16974) and (0.33485,2.33641)
 ..(0,2.5)..controls (-0.37186,2.68167) and (-0.668,2.25146)
 ..(-1,2)..controls (-1.37755,1.71404) and (-1.71404,1.37755)
 ..(-2,1)..controls (-2.33353,0.55965) and (-2.59729,0.04124)
 ..(-2.5,-0.5)..controls (-2.36812,-1.23369) and (-1.6712,-1.65662)
 ..(-1,-2)..controls (-0.66821,-2.16974) and (-0.33485,-2.33641)
 ..(0,-2.5)..controls (0.37186,-2.68167) and (0.668,-2.25146)
 ..cycle

{makepen(path)}
{qq:=pen}
{show}
>> Pen at line 48:
(-2.5,-0.5)
 .. (-1,-2)
 .. (0,-2.5)
 .. (1,-2)
 .. (2,-1)
 .. (2.5,0.5)
 .. (1,2)
 .. (0,2.5)
 .. (-1,2)
 .. (-2,1)
 .. cycle

{primarydef}
{secondarydef}

//<expr>->

//<expr>->
{pencircle}
{length(path)}
{(pen)slanted(1)}
{((3,6))-((5,-5))}

_aa__<secondary>->if(true
{(0.1)*(15)}
{odd(1.50009)}
{not(false)}
{known((8,1,5,-5,-5,5))}
{(true)and(true)}
(EXPR0)<-true
{if}
{true}

!<primary>->false):(EXPR0)fi
(EXPR0)<-(5,-5)
{false}
{(true)or(false)}
{true}
{fi}
{-((5,-5))}
{-((5,-5))}
{+(1)}
{-(1)}
>> -1
! Improper curl has been replaced by 1.
<to be read again> 
                   )
l.52  {curl- +1)
                ..tension atleast1..cycle sqrt2++sqrt2***[[]];
A curl must be a known, nonnegative number.

! Missing `}' has been inserted.
<to be read again> 
                   )
l.52  {curl- +1)
                ..tension atleast1..cycle sqrt2++sqrt2***[[]];
I've scanned a direction spec for part of a path,
so a right brace should have come next.
I shall pretend that one was there.

Path at line 52, before choices:
(-5,5)

Path at line 52, after choices:
(-5,5)

Path at line 52, before choices:
(3,6)..controls (5,-5) and (-5,5)
 ..(-5,5)..tension atleast1
 ..{2896.30943,-2896.30934}cycle

Path at line 52, after choices:
(3,6)..controls (5,-5) and (-5,5)
 ..(-5,5)..controls (-3.29726,7.86205) and (0.64516,8.35484)
 ..cycle

! Missing `)' has been inserted.
<to be read again> 
                   sqrt
l.52  {curl- +1)..tension atleast1..cycle sqrt
                                              2++sqrt2***[[]];
I found no right delimiter to match a left one. So I've
put one in, behind the scenes; this may fix the problem.

{((-2,11))subpath(path)}
{reverse(path)}
{makepen(path)}

**->[[show(EXPR0)*(EXPR1)]]
(EXPR0)<-pen
(EXPR1)<-pen
{begingroup}
{show}
{(pen)*(pen)}
>> pen
>> pen
! Not implemented: (pen)*(pen).
<to be read again> 
                   ]]
<to be read again> 
                   sqrt
l.52  {curl- +1)..tension atleast1..cycle sqrt
                                              2++sqrt2***[[]];
I'm afraid I don't know how to apply that operation to that
combination of types. Continue, and I'll return the second
argument (see above) as the result of the operation.

>> Pen at line 52:
(-5,5)
 .. (3,6)
 .. cycle

{endgroup}
(EXPR0)<-vacuous
{sqrt(2)}
{sqrt(2)}
{(1.41422)++(1.41422)}
{begingroup}
{endgroup}

***->expandafter(EXPR1)scantokens"**oct"(EXPR0)
(EXPR0)<-2
(EXPR1)<-vacuous
{expandafter}
{scantokens}
{oct(2)}
>> 2
! Not implemented: oct(known numeric).
<to be read again> 
                   ;
l.52 ...nsion atleast1..cycle sqrt2++sqrt2***[[]];
                                                  
I'm afraid I don't know how to apply that operation to that
particular type. Continue, and I'll simply return the
argument (shown above) as the result of the operation.


**->[[show(EXPR0)*(EXPR1)]]
(EXPR0)<-vacuous
(EXPR1)<-2
{begingroup}
{show}
{(vacuous)*(2)}
>> vacuous
>> 2
! Not implemented: (vacuous)*(known numeric).
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.52 ...nsion atleast1..cycle sqrt2++sqrt2***[[]];
                                                  
I'm afraid I don't know how to apply that operation to that
combination of types. Continue, and I'll return the second
argument (see above) as the result of the operation.

>> 2
{endgroup}
(EXPR0)<-vacuous
{-(20.5)}
{charcode:=-20.5}
{-(2048)}
{chardp:=-2048}
{shipout}
{nullpicture}
! Enormous chardp has been reduced.
<to be read again> 
                   ;
l.53 ...:=-20.5;chardp:=-2048;shipout nullpicture;
                                                  
Font metric dimensions must be less than 2048pt.

[]
Edge structure at line 53 (just shipped out):
End edges

{if}
{-(275.50002)}
{charexists(-275.50002)}
{known(unknown path p0 0p)}
{(true)>(false)}
{known(path)}
{(true)=(true)}
{true}
{randomseed}
! Missing `:=' has been inserted.
<to be read again> 
                   charcode
l.54 ...>known p0 0p=known p~: randomseed charcode
                                                  ; fi
Always say `randomseed:=<numeric expression>'.

{randomseed:=-20.5}
{fi}
{randomseed}
>> "goof"
! Unknown value will be ignored.
<to be read again> 
                   ;
l.55 randomseed:="goof";
                         a[($,18++1+-+18),(2,3)]=b[(3,2),(1,$);
Your expression was too random for me to handle,
so I won't change the random seed just now.

{(18)++(1)}
{(18.02776)+-+(18)}
{((2,3))-((-1,1))}
{(a)*((3,2))}
{((-1,1))+((3a,2a))}
! Missing `]' has been inserted.
<to be read again> 
                   ;
l.55 ...f"; a[($,18++1+-+18),(2,3)]=b[(3,2),(1,$);
                                                  
I've scanned an expression of the form `a[b,c',
so a right bracket should have come next.
I shall pretend that one was there.

{((1,-1))-((3,2))}
{(b)*((-2,-3))}
{((3,2))+((-2b,-3b))}
{((3a-1,2a+1))=((-2b+3,-3b+2))}
## b=-0.66667a+0.33333
## a=2
#### b=-1
{show}
{(^)+(1)}
{(~)+(2)}
{-(1)}
{((^+1,~+2))slanted(-1)}
{-(2)}
{((linearform,~+2))yscaled(-2)}
{-((3,4))}
{((linearform,-2~-4))zscaled((-3,-4))}
>> (-5~-3^-13,10~-4^+16)
{((xpart pp,ypart pp,xxpart pp,xypart pp,yxpart pp,yypart pp))xscaled(9)}
{((8,1,5,-5,-5,5))transformed((9xpart pp,ypart pp,9xxpart pp,9xypart pp,yxpart 
pp,yypart pp))}
>> (9xpart pp+72xxpart pp+9xypart pp,ypart pp+8yxpart pp+yypart pp,45xxpart pp-
45xypart pp,-45xxpart pp+45xypart pp,5yxpart pp-5yypart pp,-5yxpart pp+5yypart 
pp)
{((xpart pp,ypart pp,xxpart pp,xypart pp,yxpart pp,yypart pp))shifted((1,2))}
{((8,1,5,-5,-5,5))transformed((8,1,5,-5,-5,5))}
{((xpart pp+1,ypart pp+2,xxpart pp,xypart pp,yxpart pp,yypart pp))transformed((
43,-34,50,-50,-50,50))}
>> (50xpart pp-50ypart pp-7,-50xpart pp+50ypart pp+16,50xxpart pp-50yxpart pp,5
0xypart pp-50yypart pp,-50xxpart pp+50yxpart pp,-50xypart pp+50yypart pp)
{(2)-(1)}
{(_0)*(1)}
{(1)+(_0)}
{-(_0+1)}
>> -_0-1
{show}
{normaldeviate}
{-(-2048)}
{uniformdeviate(2048)}
{angle((-0.41438,761.67789))}
{cosd(90.03117)}
{(200)*(-0.00055)}
{mexp(-0.10986)}
{sqrt(0.99957)}
{mlog(0.99979)}
{sind(-0.05469)}
{floor(-0.00096)}
>> -1
{string}
{(unknown string s2)=(unknown string s4)}
{(unknown string s1)=(unknown string s2)}
{(unknown string s3)=(unknown string s5)}
{(unknown string s2)=(unknown string s4)}
! Redundant equation.
<to be read again> 
                   ;
l.59 string s[]; s1=s2=s4; s3=s5; s1=s2;
                                         if s1<=s4<>(s1<>s3):show[[char34=s2...
I already knew that this equation was true.
But perhaps no harm has been done; let's continue.

{if}
{(unknown string s2)<=(unknown string s1)}
{(unknown string s2)<>(unknown string s5)}
>> unknown string s2
>> unknown string s5
! Unknown relation will be considered false.
<to be read again> 
                   )
l.59 ...1=s2=s4; s3=s5; s1=s2; if s1<=s4<>(s1<>s3)
                                                  :show[[char34=s2:=s3]]fi;
The quantities shown above have not been equated.

{(true)<>(false)}
{true}
{show}
{begingroup}
{char(34)}
{s2:=unknown string s5}
{(""")=(unknown string s2)}
{endgroup}
{fi}
>> vacuous
{(0.1)point(path)}
{begingroup}
{pencircle}
{endgroup}
{((3.24413,0.64801))penoffset(pen)}
{((0.09795,-0.49031))rotated(540)}
{((-0.09795,0.49031))/(0.33333)}
{-(1.5)}
! Paths don't touch; `&' will be changed to `..'.
<to be read again> 
                   )
l.61  directiontime postcontrol-1.5of(p~&cycle)
                                               -precontrol1/2of p~(p~)=s1:=s...
When you join paths `p&q', the ending point of p
must be exactly equal to the starting point of q.
So I'm going to pretend that you said `p..q' instead.

Path at line 61, before choices:
(0,0)..controls (15,4) and (-15,-12)
 ..(4,0)
 ..cycle

Path at line 61, after choices:
(0,0)..controls (15,4) and (-15,-12)
 ..(4,0)..controls (17.52783,8.54388) and (-15.45978,-4.12262)
 ..cycle

{(-1.5)postcontrol(path)}
{(0.5)precontrol(path)}
{((-2.75,-5))-((3.75,-1))}
! Missing `of' has been inserted for directiontime.
<to be read again> 
                   (
l.61 ...ontrol-1.5of(p~&cycle)-precontrol1/2of p~(
                                                  p~)=s1:=s4:=s4;
I've got the first argument; will look now for the other.

{((-6.5,-4))directiontime(path)}
{decimal(0.5)}
{((-0.29384,1.47095))substring("0.5")}
{s4:=unknown string s1}
{s1:=unknown string s4}
{("0")=(unknown string s1)}
{path}
{length("  ")}
! Missing `,' has been inserted.
<to be read again> 
                   }
l.62 path p~[]; p~1=p2{length"  "}
                                  &cycle; p~1=p2=p~0; p2..controls-p2..cycle...
I've got the x coordinate of a path direction;
will look for the y coordinate next.

! An expression can't begin with `}'.
<inserted text> 
                0
<to be read again> 
                   }
l.62 path p~[]; p~1=p2{length"  "}
                                  &cycle; p~1=p2=p~0; p2..controls-p2..cycle...
I'm afraid I need some sort of value in order to continue,
so I've tentatively inserted `0'. You may want to
delete this zero and insert something else;
see Chapter 27 of The METAFONTbook for an example.

Path at line 62, before choices:
(5,-5){4096,0}
 ..{4096,0}cycle

Path at line 62, after choices:
(5,-5)..controls (5,-5) and (5,-5)
 ..cycle

{(unknown path p~1)=(path)}
{((5,-5))=(unknown path p~0)}
{(path)=(path)}
! Redundant or inconsistent equation.
<to be read again> 
                   ;
l.62 ...~[]; p~1=p2{length"  "}&cycle; p~1=p2=p~0;
                                                   p2..controls-p2..cycle=p~...
An equation between already-known quantities can't help.
But don't worry; continue and I'll just ignore it.

{-((5,-5))}
Path at line 62, before choices:
(5,-5)..controls (-5,5) and (-5,5)
 ..cycle

Path at line 62, after choices:
(5,-5)..controls (-5,5) and (-5,5)
 ..cycle

{(path)=(unknown path p~2)}
{length(path)}
Path at line 63, before choices:
(3,6)..tension 1.2
 ..(5,-5)..controls (5,-5) and (5,-5)
 ..(5,-5)
 ..(5,-5)..controls (-5,5) and (-5,5)
 ..(5,-5){0,4096}..tension 1 and atleast1
 ..cycle

Path at line 63, after choices:
(3,6)..controls (-3.01212,4.82085) and (-1.0424,-6.01257)
 ..(5,-5)..controls (5,-5) and (5,-5)
 ..(5,-5)..controls (5,-5) and (5,-5)
 ..(5,-5)..controls (-5,5) and (-5,5)
 ..(5,-5)..controls (5,0.02791) and (4.99951,6.39217)
 ..cycle

>> x
! Improper tension has been set to 1.
<to be read again> 
                   ..
l.64  x..
         {curl1}-p7{curl hex "IsBad"}..tension.75and.74999..p2{0,1}&p2{_,'}....
The expression above should have been a number >=3/4.

{-((3,6))}
{hex("IsBad")}
>> "IsBad"
! String contains illegal digits.
<to be read again> 
                   }
l.64  x..{curl1}-p7{curl hex "IsBad"}
                                     ..tension.75and.74999..p2{0,1}&p2{_,'}....
I zeroed out characters that weren't hex digits.

>> 0.74998
! Improper tension has been set to 1.
<to be read again> 
                   ..
l.64 ...7{curl hex "IsBad"}..tension.75and.74999..
                                                  p2{0,1}&p2{_,'}..cycle:=p
The expression above should have been a number >=3/4.

>> _
! Undefined x coordinate has been replaced by 0.
<to be read again> 
                   ,
l.64 ...sBad"}..tension.75and.74999..p2{0,1}&p2{_,
                                                  '}..cycle:=p
I need a `known' x value for this part of the path.
The value I found (see above) was no good;
so I'll try to keep going by using zero instead.
(Chapter 27 of The METAFONTbook explains that
you might want to type `I ???' now.)

>> (xpart ',ypart ')
! Undefined y coordinate has been replaced by 0.
<to be read again> 
                   }
l.64 ...ad"}..tension.75and.74999..p2{0,1}&p2{_,'}
                                                  ..cycle:=p
I need a `known' y value for this part of the path.
The value I found (see above) was no good;
so I'll try to keep going by using zero instead.
(Chapter 27 of The METAFONTbook explains that
you might want to type `I ???' now.)

Path at line 64, before choices:
(3,6)..controls (-3.01212,4.82085) and (-1.0424,-6.01257)
 ..(5,-5)..controls (5,-5) and (5,-5)
 ..(5,-5)..controls (5,-5) and (5,-5)
 ..(5,-5)..controls (-5,5) and (-5,5)
 ..(5,-5)..controls (5,0.02791) and (4.99951,6.39217)
 ..(3,6)
 ..{curl 1}(-3,-6){curl 2989}..tension 0.75 and 1
 ..{0,4096}(5,-5){curl 1}
 ..cycle

Path at line 64, after choices:
(3,6)..controls (-3.01212,4.82085) and (-1.0424,-6.01257)
 ..(5,-5)..controls (5,-5) and (5,-5)
 ..(5,-5)..controls (5,-5) and (5,-5)
 ..(5,-5)..controls (-5,5) and (-5,5)
 ..(5,-5)..controls (5,0.02791) and (4.99951,6.39217)
 ..(3,6)..controls (-2.44821,4.93144) and (-5.41408,-1.00029)
 ..(-3,-6)..controls (-3.70045,-12.19485) and (5,-11.61679)
 ..(5,-5)..controls (12.45406,-3.75087) and (10.41669,7.45464)
 ..cycle

>> path
! Improper `:=' will be changed to `='.
l.64 ...sion.75and.74999..p2{0,1}&p2{_,'}..cycle:=
                                                  p
I didn't find a variable name at the left of the `:=',
so I'm going to pretend that you said `=' instead.

{(path)=(unknown path p~4)}
{-(9)}
{((3.5001,7.00002))subpath(path)}
{((-9,9))subpath(path)}
{(path)=(unknown path p~6)}
{show}
>> Path at line 66:
(-2.5,2.5)..controls (-2.49893,2.49893) and (0.00107,-0.00107)
 ..(5,-5)..controls (5,0.02791) and (4.99951,6.39217)
 ..(3,6)..controls (-2.44821,4.93144) and (-5.41408,-1.00029)
 ..(-3,-6)..controls (-3.70045,-12.19485) and (5,-11.61679)
 ..(5,-5)..controls (5.0001,-4.99998) and (5.00021,-4.99997)
 ..(5.00032,-4.99995)

{((1,2))directiontime(path)}
>> 1
{-(1.00002)}
{((1,-1.00002))directiontime(path)}
>> 3.24937
Path at line 67, before choices:
(0,0)..controls (1,1) and (0,1)
 ..(1,0)

Path at line 67, after choices:
(0,0)..controls (1,1) and (0,1)
 ..(1,0)

{(unknown path p~3)=(path)}
{show}
Path at line 67, before choices:
(5,-5){-2896.30943,-2896.30934}
 ..{1831.78674,3663.57385}cycle

Path at line 67, after choices:
(5,-5)..controls (5,-5) and (5,-5)
 ..cycle

>> Path at line 67:
(5,-5)..controls (5,-5) and (5,-5)
 ..cycle

{((0.31416,1))subpath(path)}
{((1,1))directiontime(path)}
{(1)-(0.31416)}
{(0.27097)*(0.68584)}
{(0.31416)+(0.18584)}
>> 0.5
{((0,0.25))subpath(path)}
{((0.25,1))subpath(path)}
Path at line 69, before choices:
(0,0)..controls (0.25,0.25) and (0.375,0.4375)
 ..(0.4375,0.5625)..controls (0.625,0.9375) and (0.25,0.75)
 ..(1,0)

Path at line 69, after choices:
(0,0)..controls (0.25,0.25) and (0.375,0.4375)
 ..(0.4375,0.5625)..controls (0.625,0.9375) and (0.25,0.75)
 ..(1,0)

! Missing `endtext' has been inserted.
<to be read again> 
                   ;
l.69 ...subpath(.25,1)of p~3)shifted begintext1,0;
                                                  
I found no right delimiter to match a left one. So I've
put one in, behind the scenes; this may fix the problem.

{(path)shifted((1,0))}
{(unknown path p~5)=(path)}
{(2/3)*((xpart ',ypart '))}
{((0.66667xpart ',0.66667ypart '))zscaled((xpart ',ypart '))}
>> (0,0,xpart ',-ypart ',ypart ',xpart ')
! Transform components aren't all known.
<to be read again> 
                   {
l.70 p~3:=2/3'zscaled'{
                       p~3}..controls(2,2/3(3))and penoffset(1/2x,y)of(0,1)(...
I'm unable to apply a partially specified transformation
except to a fully known pair or transform.
Proceed, and I'll omit the transformation.

>> 0.66667xpart '
! Undefined x coordinate has been replaced by 0.
<to be read again> 
                   {
l.70 p~3:=2/3'zscaled'{
                       p~3}..controls(2,2/3(3))and penoffset(1/2x,y)of(0,1)(...
I need a `known' x value for this part of the path.
The value I found (see above) was no good;
so I'll try to keep going by using zero instead.
(Chapter 27 of The METAFONTbook explains that
you might want to type `I ???' now.)

>> 0.66667ypart '
! Undefined y coordinate has been replaced by 0.
<to be read again> 
                   {
l.70 p~3:=2/3'zscaled'{
                       p~3}..controls(2,2/3(3))and penoffset(1/2x,y)of(0,1)(...
I need a `known' y value for this part of the path.
The value I found (see above) was no good;
so I'll try to keep going by using zero instead.
(Chapter 27 of The METAFONTbook explains that
you might want to type `I ???' now.)

>> path
! Undefined coordinates have been replaced by (0,0).
<to be read again> 
                   }
l.70 p~3:=2/3'zscaled'{p~3}
                           ..controls(2,2/3(3))and penoffset(1/2x,y)of(0,1)(...
I need x and y numbers for this part of the path.
The value I found (see above) was no good;
so I'll try to keep going by using zero instead.
(Chapter 27 of The METAFONTbook explains that
you might want to type `I ???' now.)

{(2/3)*(3)}
{(1/2)*(x)}
{((0.5x,-5))penoffset((0,1))}
>> (0.5x,-5)
>> (0,1)
! Not implemented: penoffset(unknown pair)of(pair).
<to be read again> 
                   (
l.70 ...ols(2,2/3(3))and penoffset(1/2x,y)of(0,1)(
                                                  1,0);
I'm afraid I don't know how to apply that operation to that
combination of types. Continue, and I'll return the second
argument (see above) as the result of the operation.

! Missing `..' has been inserted.
<to be read again> 
                   (
l.70 ...ols(2,2/3(3))and penoffset(1/2x,y)of(0,1)(
                                                  1,0);
A path join command should end with two dots.

Path at line 70, before choices:
(0,0)..controls (2,2) and (0,1)
 ..(1,0)

Path at line 70, after choices:
(0,0)..controls (2,2) and (0,1)
 ..(1,0)

{p~3:=path}
{show}
{reverse(path)}
{(path)intersectiontimes(path)}
>> (0.17227,0.28339)
{(0.17227)point(path)}
>> (0.71329,0.78188)
{(1)-(0.28339)}
{(0.71661)point(path)}
>> (0.7133,0.78189)
{show}
{(path)shifted((0.01,0))}
{(path)intersectiontimes(path)}
{xpart((1.47693,1.18973))}
{(1.47693)point(path)}
{(path)shifted((0.01,0))}
{(path)intersectiontimes(path)}
{ypart((1.47693,1.18973))}
{(1.18973)point(path)}
{((1.50499,0.71521))-((1.495,0.71521))}
>> (0.01,0)
{begingroup}
{-(1)}
{-(1.00002)}
{(-1)+-+(-1.00002)}
! Pythagorean subtraction 1+-+1.00002 has been replaced by 0.
l.74 [[ e[-1+-+ -1.00001]
                         =nullpicture; addto e1 also[[
Since I don't take square roots of negative numbers,
I'm zeroing this one. Proceed, with fingers crossed.

{nullpicture}
{(unknown picture e0)=(picture)}
{addto}
{begingroup}
{addto}
{(path)scaled(3)}
{(unknown picture e1)=(unknown picture e2)}
{(picture)=(unknown picture e1)}
{nullpicture}
{endgroup}
{show}
>> Edge structure at line 76:
Filled pen stroke :
(3,0)..controls (3.75,0.75) and (4.125,1.3125)
 ..(4.3125,1.6875)..controls (4.875,2.8125) and (3.75,2.25)
 ..(6,0)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
End edges

{-(3)}
{(picture)shifted((-3,0))}
>> Edge structure at line 76:
Filled pen stroke :
(0,0)..controls (0.75,0.75) and (1.125,1.3125)
 ..(1.3125,1.6875)..controls (1.875,2.8125) and (0.75,2.25)
 ..(3,0)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
End edges

{endgroup}
{addto}
{-(2)}
{(picture)shifted((-2,-1))}
{-(4)}
{(picture)shifted((-4,-1))}
{e1:=picture}
{addto}
{turningnumber(path)}
{oct("180")}
>> "180"
! String contains illegal digits.
<to be read again> 
                   ++
l.79  withpen qq xscaled(oct"180"++
                                   1) rotated-angle(64,$) shifted (9,8);
I zeroed out characters that weren't in the range 0..7.

{(64)++(1)}
{(pen)xscaled(64.00781)}
{angle((64,-1))}
{-(-0.89517)}
{(pen)rotated(0.89517)}
{(pen)shifted((9,8))}
{show}
>> Edge structure at line 80:
Filled pen stroke :
(-1,-1)..controls (-0.25,-0.25) and (0.125,0.3125)
 ..(0.3125,0.6875)..controls (0.875,1.8125) and (-0.25,1.25)
 ..(2,-1)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
Filled pen stroke :
(-3,-2)..controls (-2.25,-1.25) and (-1.875,-0.6875)
 ..(-1.6875,-0.3125)..controls (-1.125,0.8125) and (-2.25,0.25)
 ..(0,-2)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
Filled pen stroke colored (0.4,1,0):
(0,9)
butt ends, mitered joins limited 1 with pen
(-150.99219,4.99976)
 .. (-54.96875,5.00012)
 .. (9.03906,5.5003)
 .. (73.03125,7.00037)
 .. (137.01563,9.00037)
 .. (168.99219,11.00024)
 .. (72.96875,10.99988)
 .. (8.96094,10.4997)
 .. (-55.03125,8.99963)
 .. (-119.01563,6.99963)
 .. cycle
End edges

{charcode:=5}
{chardp:=5}
{-(1.5)}
{xoffset:=-1.5}
## xoffset=-1.5
{shipout}
[5
Envelope spec at line 80:
(0,9) % beginning with offset (-150.99219,4.99976)
   ..controls (0,9) and (0,9)
 ..(0,9) % counterclockwise to offset (-150.99219,4.99976)
 & cycle

]
Edge structure at line 80 (just shipped out):
Filled pen stroke :
(-1,-1)..controls (-0.25,-0.25) and (0.125,0.3125)
 ..(0.3125,0.6875)..controls (0.875,1.8125) and (-0.25,1.25)
 ..(2,-1)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
Filled pen stroke :
(-3,-2)..controls (-2.25,-1.25) and (-1.875,-0.6875)
 ..(-1.6875,-0.3125)..controls (-1.125,0.8125) and (-2.25,0.25)
 ..(0,-2)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
Filled pen stroke colored (0.4,1,0):
(0,9)
butt ends, mitered joins limited 1 with pen
(-150.99219,4.99976)
 .. (-54.96875,5.00012)
 .. (9.03906,5.5003)
 .. (73.03125,7.00037)
 .. (137.01563,9.00037)
 .. (168.99219,11.00024)
 .. (72.96875,10.99988)
 .. (8.96094,10.4997)
 .. (-55.03125,8.99963)
 .. (-119.01563,6.99963)
 .. cycle
End edges

{nullpicture}
{(unknown picture e3)=(picture)}
{showstats}
Memory usage 1247&115 (2853 still untouched)
String usage 18&58 (43&7564 now untouched)
{addto}
Path at line 81, before choices:
(0.5,0){curl 1}
 ..{curl 1}(3.5,1.5)

Path at line 81, after choices:
(0.5,0)..controls (1.5,0.5) and (2.5,1)
 ..(3.5,1.5)

Path at line 82, before choices:
(1,1)
 ..cycle

Path at line 82, after choices:
(1,1)..controls (1,1) and (1,1)
 ..cycle

{makepen(path)}
{(pen)yscaled(1.5)}
{q:=pen}
{-(14.5)}
{charext:=-14.5}
{chardp:=-14.5}
{chardx:=-14.5}
## chardx=-14.5
{charcode:=6}
{shipout}
[6]
Edge structure at line 83 (just shipped out):
Filled pen stroke :
(3,0)..controls (3.75,0.75) and (4.125,1.3125)
 ..(4.3125,1.6875)..controls (4.875,2.8125) and (3.75,2.25)
 ..(6,0)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
End edges

Path at line 84, before choices:
(0,0){curl 1}
 ..(1,0.5)
 ..(5,1.5)
 ..(7,2.5)
 ..(12,3.5)
 ..{curl 1}(13,4)

Path at line 84, after choices:
(0,0)..controls (0.3153,0.20053) and (0.6504,0.36807)
 ..(1,0.5)..controls (2.29114,0.98723) and (3.72412,0.97192)
 ..(5,1.5)..controls (5.68967,1.78545) and (6.30771,2.22081)
 ..(7,2.5)..controls (8.58867,3.1407) and (10.3946,2.90645)
 ..(12,3.5)..controls (12.3506,3.62962) and (12.68594,3.7973)
 ..(13,4)

{(unknown path p~9)=(path)}
{addto}
{addto}
! A primary expression can't begin with `doublepath'.
<inserted text> 
                0
<to be read again> 
                   doublepath
l.85 addto doublepath
                      (-4095,0)..tension
I'm afraid I need some sort of value in order to continue,
so I've tentatively inserted `0'. You may want to
delete this zero and insert something else;
see Chapter 27 of The METAFONTbook for an example.

>> 0
! Not a suitable variable.
<to be read again> 
                   doublepath
l.85 addto doublepath
                      (-4095,0)..tension
At this point I needed to see the name of a picture variable.
(Or perhaps you have indeed presented me with one; I might
have missed it, if it wasn't followed by the proper token.)
So I'll not change anything just now.

! Extra tokens will be flushed.
<to be read again> 
                   doublepath
l.85 addto doublepath
                      (-4095,0)..tension
I've just read as much of that statement as I could fathom,
so a semicolon should have been next. It's very puzzling...
but I'll try to get myself back together, by ignoring
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

{show}
{-(90)}
{(picture)rotated(-90)}
>> Edge structure at line 86:
Filled pen stroke :
(0,-0.5)..controls (0.5,-1.5) and (1,-2.5)
 ..(1.5,-3.5)
dashed (on 3 off 0) shifted 0
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
Filled pen stroke :
(0,0)..controls (0.20053,-0.3153) and (0.36807,-0.6504)
 ..(0.5,-1)..controls (0.98723,-2.29114) and (0.97192,-3.72412)
 ..(1.5,-5)..controls (1.78545,-5.68967) and (2.22081,-6.30771)
 ..(2.5,-7)..controls (3.1407,-8.58867) and (2.90645,-10.3946)
 ..(3.5,-12)..controls (3.62962,-12.3506) and (3.7973,-12.68594)
 ..(4,-13)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
End edges

{if}
{if}
{elseif}
! Missing `:' has been inserted.
<inserted text> 
                :
<to be read again> 
                   elseif
l.87 if "a" if "ab">"b" elseif
                               path reverse (3,4): >="aa":foo elseif fi "bar"

{("ab")>("b")}
{false}
{reverse((3,4))}
{path(path)}
{true}
{("a")>=("aa")}
{false}
{if}
{((xpart ',ypart '))-((1,yy))}
{((xpart '-1,linearform))<((xpart ',ypart '))}
{true}
{fi}
{else}
! Extra else.
l.88 else if '-(1,yy)<': :fi else
                                  def dup text t=[[t;save enddef;def|suffix$=,$
I'm ignoring this; it doesn't match any if.

{def}
{def}
{def}

||<tertiary>->show.substring(EXPR0)of("a"
(EXPR0)<-(2,-1)
{show}
{("a")&("bc")}
{((2,-1))substring("abc")}
>> "ba"
{tertiarydef}
{def}
{show}
{((23.3,4.5))subpath(path)}
Path at line 95, before choices:
(0,0){curl 2}
 ..(13,4)..controls (12.84297,3.89865) and (12.68063,3.80606)
 ..(12.51372,3.72261){curl 3}
 ..{curl 4}(-1,-1){curl 4}
 ..{curl 2}cycle

Path at line 95, after choices:
(0,0)..controls (-40.15552,-36.7088) and (58.71173,33.50317)
 ..(13,4)..controls (12.84297,3.89865) and (12.68063,3.80606)
 ..(12.51372,3.72261)..controls (8.00914,2.1484) and (3.50458,0.5742)
 ..(-1,-1)..controls (-0.66667,-0.66667) and (-0.33333,-0.33333)
 ..cycle

>> Path at line 95:
(0,0)..controls (-40.15552,-36.7088) and (58.71173,33.50317)
 ..(13,4)..controls (12.84297,3.89865) and (12.68063,3.80606)
 ..(12.51372,3.72261)..controls (8.00914,2.1484) and (3.50458,0.5742)
 ..(-1,-1)..controls (-0.66667,-0.66667) and (-0.33333,-0.33333)
 ..cycle

{(2)+(3)}

++->[[dup.showtoken(EXPR0);]];def.x.expr.z.of(EXPR1)=z.enddef;texts(x=((EXPR1)+
0)(EXPR1)+(EXPR1))("xx",foo((EXPR0)))=0]]
(EXPR0)<-path
(EXPR1)<-5
{begingroup}

dup<text>->[[(TEXT0);save
(TEXT0)<-showtoken(path)
{begingroup}
{showtoken}
> (path)
{save}
{restoring ;}
{endgroup}
{def}
! Missing symbolic token inserted.
<inserted text> 
                 INACCESSIBLE
++->....showtoken(EXPR0);]];def.x.expr.z.of(EXPR1)
                                                  =z.enddef;texts(x=((EXPR1)...
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
Sorry: You can't redefine a number, string, or expr.
I've inserted an inaccessible symbol so that your
definition will be completed without mixing me up too badly.


texts(TEXT0)(TEXT1)<expr>->for.n:=,for.n"yy":n,length.if.false:endfor(TEXT1),(T
EXT0),:if.string.n:forsuffixes.n=foo1,[foo(n)],':show(TEXT0),(TEXT1)|(n;exitif.
not('<='+((EXPR2),yy))endfor.for.m= :+endfor.for.m=alpha.step-1.1 3$:+m.endfor.
fi.endfor
(TEXT0)<-x=((5)+0)(5)+(5)
(TEXT1)<-"xx",foo((path))
(EXPR2)<-0
{for}
{for}
! Missing `=' has been inserted.
<to be read again> 
                   "yy"
texts->for.n:=,for.n"yy"
                        :n,length.if.false:endfor(TEXT1),(TEXT0),:if.string....
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
The next thing in this loop should have been `=' or `:='.
But don't worry; I'll pretend that an equals sign
was present, and I'll look for the values next.

{loop value="yy"}
{if}
{false}
{false}
! Incomplete if; all text was ignored after line 95.
<inserted text> 
                fi
<to be read again> 
                    ENDFOR
<for("yy")> (EXPR0),length.if.false: ENDFOR
                                           
texts->for.n:=,for.n"yy":n,length.if.false:endfor
                                                 (TEXT1),(TEXT0),:if.string....
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
A forbidden `outer' token occurred in skipped text.
This kind of error happens when you say `if...' and forget
the matching `fi'. I've inserted a `fi'; this might work.

{[repeat the loop]}
{length("xx")}

foo(TEXT2)->begingroup(TEXT2)endgroup
(SUFFIX0)<-
(SUFFIX1)<-foo
(TEXT2)<-(path)
{begingroup}
{endgroup}

x<expr>of<primary>->(EXPR0)
{(5)+(0)}
(EXPR0)<-5
! Missing `of' has been inserted for x.
<to be read again> 
                   (5)
<argument> x=((5)+0)(5)
                       +(5)
texts->...:n,length.if.false:endfor(TEXT1),(TEXT0)
                                                  ,:if.string.n:forsuffixes....
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
I've got the first argument; will look now for the other.

(EXPR1)<-5
{(5)+(5)}
{loop value="yy"}
{if}
{string("yy")}
{true}
{forsuffixes}
! Missing symbolic token inserted.
<inserted text> 
                 INACCESSIBLE
<for("yy")> if.string(EXPR0):forsuffixes(EXPR0)
                                               =foo1,[foo((EXPR0))],':show.x...

texts->....m=alpha.step-1.1 3$:+m.endfor.fi.endfor
                                                  
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
Sorry: You can't redefine a number, string, or expr.
I've inserted an inaccessible symbol so that your
definition will be completed without mixing me up too badly.


foo(TEXT2)->begingroup(TEXT2)endgroup
(SUFFIX0)<-
(SUFFIX1)<-foo
(TEXT2)<-("yy")
{begingroup}
{endgroup}
>> "yy"
! Improper subscript has been replaced by zero.
<for("yy")> ...suffixes(EXPR0)=foo1,[foo((EXPR0))]
                                                  ,':show.x=((5)+0)(5)+(5),"...

texts->....m=alpha.step-1.1 3$:+m.endfor.fi.endfor
                                                  
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
A bracketed subscript must have a known numeric value;
unfortunately, what I found was the value that appears just
above this error message. So I'll try a zero subscript.

{loop value=foo1}
{show}

x<expr>of<primary>->(EXPR0)
{(5)+(0)}
(EXPR0)<-5
! Missing `of' has been inserted for x.
<to be read again> 
                   (5)
<for(foo1)> show.x=((5)+0)(5)
                             +(5),"xx",foo((path))|(("yy");exitif.not('<='+(...
<for("yy")> ...R0);exitif.not('<='+((0),yy))endfor
                                                  .for.m= :+endfor.for.m=alp...

texts->....m=alpha.step-1.1 3$:+m.endfor.fi.endfor
                                                  
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
I've got the first argument; will look now for the other.

(EXPR1)<-5
{(5)+(5)}
>> 10
>> "xx"
foo(TEXT2)->begingroup(TEXT2)endgroup
(SUFFIX0)<-
(SUFFIX1)<-foo
(TEXT2)<-(path)
{begingroup}
{endgroup}

|<suffix>->,(SUFFIX0)
! Missing `)' has been inserted.
<to be read again> 
                   ("yy")
<for(foo1)> ...+0)(5)+(5),"xx",foo((path))|(("yy")
                                                  ;exitif.not('<='+((0),yy))...
<for("yy")> ...R0);exitif.not('<='+((0),yy))endfor
                                                  .for.m= :+endfor.for.m=alp...

texts->....m=alpha.step-1.1 3$:+m.endfor.fi.endfor
                                                  
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
I've gotten to the end of the macro parameter list.
You might want to delete some tokens before continuing.

(SUFFIX0)<-
>> Path at line 95:
(0,0)..controls (15,4) and (-15,-12)
 ..(4,0)

>> "yy"
{exitif}
{((xpart ',ypart '))+((0,yy))}
{((xpart ',ypart '))<=((xpart ',linearform))}
>> -yy
! Unknown relation will be considered false.
<to be read again> 
                   )
<for(foo1)> ...)|(("yy");exitif.not('<='+((0),yy))
                                                   ENDFOR
<for("yy")> ...R0);exitif.not('<='+((0),yy))endfor
                                                  .for.m= :+endfor.for.m=alp...

texts->....m=alpha.step-1.1 3$:+m.endfor.fi.endfor
                                                  
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
Oh dear. I can't decide if the expression above is positive,
negative, or zero. So this comparison test won't be `true'.

{[repeat the loop]}
{loop value=0}
{not(false)}
{true}
{for}
{for}
>> 0.5p3~
! Improper initial value has been replaced by 0.
<to be read again> 
                   step
<for("yy")> ...or.for.m= :+endfor.for.m=alpha.step
                                                  -1.1 3$:+m.endfor.fi ENDFOR
texts->....m=alpha.step-1.1 3$:+m.endfor.fi.endfor
                                                  
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
When you say `for x=a step b until c',
the initial value `a' and the step size `b'
and the final value `c' must have known numeric values.
I'm zeroing this one. Proceed, with fingers crossed.

{-(1.1)}
! Missing `until' has been inserted.
<to be read again> 
                   3
<for("yy")> ....m= :+endfor.for.m=alpha.step-1.1 3
                                                  $:+m.endfor.fi ENDFOR
texts->....m=alpha.step-1.1 3$:+m.endfor.fi.endfor
                                                  
<to be read again> 
                   ]]
<to be read again> 
                   ;
l.95 ...9{curl3}..($,$){curl4}..cycle;zzz=p~++2+3;
                                                  [[
I assume you meant to say `until' after `step'.
So I'll look for the final value and colon next.

{(3)*(-1)}
{loop value=0}
{[repeat the loop]}
{loop value=-1.1}
{+(0)}
{[repeat the loop]}
{loop value=-2.20001}
{(0)+(-1.1)}
{[repeat the loop]}
{fi}
{[repeat the loop]}
{loop value=2}
{if}
{string(2)}
{false}
{[repeat the loop]}
{loop value=path}
{if}
{string(path)}
{false}
{[repeat the loop]}
{loop value=10}
{if}
{string(10)}
{false}
{[repeat the loop]}
{(-1.1)+(-2.20001)}
{endgroup}
{(zzz)=(-3.30002)}
## zzz=-3.30002
{begingroup}
{let}
{save}
{def}
{def}

texts->input
{input}
(./trap.mp
if->endinput?
{endinput}
{if}
{known(0)}
{not(true)}
{false}
)
! Incomplete if; all text was ignored after line 2.
<inserted text> 
                fi
l.96 ...t?enddef;def texts=input enddef;texts trap
                                                   ]];
The file ended while I was skipping conditional text.
This kind of error happens when you say `if...' and forget
the matching `fi'. I've inserted a `fi'; this might work.

{restoring \}
{restoring if}
{endgroup}

dup<text>->[[(TEXT0);save
Runaway text?
[[def.texts.secondary.x=primarydef.y++y=x@y.enddef;showtoken++;x.enddef ETC.
! Forbidden token found while scanning a text argument.
<inserted text> 
                endgroup
<to be read again> 
                   \
l.97 ...f y++y=x@y enddef; showtoken++;x enddef]]\
                                                  ;
It seems that a right delimiter was left out,
causing me to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.

(TEXT0)<-[[def.texts.secondary.x=primarydef.y++y=x@y.enddef;showtoken++;x.endde
f]]
{begingroup}
{begingroup}
{def}
{endgroup}
{save}
! Missing symbolic token inserted.
<inserted text> 
                 INACCESSIBLE
<to be read again> 
                   \
l.97 ...f y++y=x@y enddef; showtoken++;x enddef]]\
                                                  ;
Sorry: You can't redefine my error-recovery tokens.
I've inserted an inaccessible symbol so that your
definition will be completed without mixing me up too badly.

{\}

texts<secondary>->primarydef.y++y=(EXPR0)@y.enddef;showtoken++;(EXPR0)
(EXPR0)<-"a"
{primarydef}
{showtoken}
> ++=primarydef'd macro:
("a")@(EXPR1)
{("a")&("b")}
ab
texts<secondary>->primarydef.y++y=(EXPR0)@y.enddef;showtoken++;(EXPR0)
{(1.00002)*(a1)}
{-(1.00002a1)}
(EXPR0)<--1.00002a1
{primarydef}
{showtoken}
> ++=primarydef'd macro:
(-1.00002a1)@(EXPR1)
{(a2)+(a3)}
{(linearform)+(a4)}
{(linearform)+(a5)}
{(linearform)+(a6)}
{-(linearform)}
{(-1.00002a1)=(linearform)}
## a1=0.99998a6+0.99998a5+0.99998a4+0.99998a3+0.99998a2
{-((a3,a2))}
{((-a3,-a2))/(0.99998)}
{(a4)+(a5)}
{begingroup}
{showdependencies}
a1=0.99998a6+0.99998a5+0.99998a4+0.99998a3+0.99998a2
alpha=0.5p3~
{endgroup}
{(linearform)+(a6)}
{(a3)+(a4)}
{(linearform)+(a5)}
{(linearform)+(a6)}
{-((linearform,linearform))}
{((-1.00002a3,-1.00002a2))=((linearform,linearform))}
## a2=0.99998a6+0.99998a5+0.99998a4+0.99998a3
## a3=0.99998a6+0.99998a5+0.99998a4
{restoring  INACCESSIBLE}
{endgroup}
{(1.00002)*(0.25a4*4)}
{(0.25a4*4)+(1)}
{(0.25a5*4)+(0.25a6*4)}
{(0.25a4*4+1)=(linearform)}
## a4*4=0.99998a6*4+0.99998a5*4-3.99994
{(0.9)*(0.5p3~)}
{(0.45p3~)+(7)}
{alpha:=0.45p3~+7}
## alpha=0.45p3~+7
{showdependencies}
alpha=0.45p3~+7
a4=0.25a6*4+0.25a5*4-0.99998
a3=0.49998a6*4+0.49998a5*4-0.99997
a2=0.99997a6*4+0.99997a5*4-1.99992
a1=1.99992a6*4+1.99992a5*4-3.99982
{(0.66667)-(0.25a6*4)}
{(0.25a6*4)=(-0.25a6*4+0.66667)}
## a6*4=1.33334
{(0.25a5*4)=(0.33333)}
## a5*4=1.33331
#### a4=-0.33333
#### a3=0.33333
#### a2=0.66664
#### a1=1.33328
{chardp:=27}
{charcode:=27}
{(3.1)=(aaz)}
## aaz=3.1
{show}
>> 3.1
{ligtable}
{show}
{begingroup}
{interim}
{tracingcommands:=0}

lig(TEXT0)(TEXT1)->ligtable0::for*=1step1until60:0kern.boundarychar+*,endfor.sk
ipto0;ligtable(TEXT0):(TEXT0)(TEXT1)0,skipto255;boundarychar:=boundarychar+51.2
9999
(TEXT0)<-"g"
(TEXT1)<-=:|

lig(TEXT0)(TEXT1)->ligtable0::for*=1step1until60:0kern.boundarychar+*,endfor.sk
ipto0;ligtable(TEXT0):(TEXT0)(TEXT1)0,skipto255;boundarychar:=boundarychar+51.2
9999
(TEXT0)<-"h":"i"
(TEXT1)<-|=:
! Character h is already in a ligtable.
<to be read again> 
                   :
<argument> "h":
               "i"
lig->...r+*,endfor.skipto0;ligtable(TEXT0):(TEXT0)
                                                  (TEXT1)0,skipto255;boundar...
l.102 ...ands:=0; lig("g")(=:|); lig("h":"i")(|=:)
                                                  ;
It's not legal to label a character more than once.
So I'll not change anything just now.


lig(TEXT0)(TEXT1)->ligtable0::for*=1step1until60:0kern.boundarychar+*,endfor.sk
ipto0;ligtable(TEXT0):(TEXT0)(TEXT1)0,skipto255;boundarychar:=boundarychar+51.2
9999
(TEXT0)<-"j"
(TEXT1)<-|=:>

lig(TEXT0)(TEXT1)->ligtable0::for*=1step1until60:0kern.boundarychar+*,endfor.sk
ipto0;ligtable(TEXT0):(TEXT0)(TEXT1)0,skipto255;boundarychar:=boundarychar+51.2
9999
(TEXT0)<-"k"
(TEXT1)<-=:|>

lig(TEXT0)(TEXT1)->ligtable0::for*=1step1until60:0kern.boundarychar+*,endfor.sk
ipto0;ligtable(TEXT0):(TEXT0)(TEXT1)0,skipto255;boundarychar:=boundarychar+51.2
9999
(TEXT0)<-"l"
(TEXT1)<-|=:|>
{restoring tracingcommands=2.1}
{endgroup}
>> vacuous
{begingroup}
{let}
{vardef}
{endgroup}
! Variable c.a1 has been obliterated.
<to be read again> 
                   ;
l.104 b1:=c.a[ [[let c=++;vardef b=enddef;1]] ];
                                                 ligtable"m":0=:0,skipto5;
It seems you did a nasty thing---probably by accident,
but nevertheless you nearly hornswoggled me...
While I was evaluating the suffix of this variable,
something was redefined, and it's no longer a variable!
In order to get back on my feet, I've inserted `0' instead.

{b1:=0}
! Variable b1 has been obliterated.
<to be read again> 
                   ;
l.104 b1:=c.a[ [[let c=++;vardef b=enddef;1]] ];
                                                 ligtable"m":0=:0,skipto5;
It seems you did a nasty thing---probably by accident,
but nevertheless you nearly hornswoggled me...
While I was evaluating the right-hand side of this
command, something happened, and the left-hand side
is no longer a variable! So I won't change anything.

{ligtable}
! A statement can't begin with `skipto'.
<to be read again> 
                   !!
l.105 !!
        ; errhelp 0; errmessage "Be like Jane";
I was looking for the beginning of a new statement.
If you just proceed without changing anything, I'll ignore
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

! Extra tokens will be flushed.
<to be read again> 
                   !!
l.105 !!
        ; errhelp 0; errmessage "Be like Jane";
I've just read as much of that statement as I could fathom,
so a semicolon should have been next. It's very puzzling...
but I'll try to get myself back together, by ignoring
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

{errhelp}
>> 0
! Not a string.
<to be read again> 
                   ;
l.105 !!; errhelp 0;
                     errmessage "Be like Jane";
A message should be a known string expression.

{errmessage}
! Be like Jane.
<to be read again> 
                   ;
l.105 !!; errhelp 0; errmessage "Be like Jane";
                                               
This error message was generated by an `errmessage'
command, so I can't give any explicit help.
Pretend that you're Miss Marple: Examine all clues,
and deduce the truth by inspired guesses.

{errhelp}
{errmessage}
! .
<to be read again> 
                   ;
l.106 errhelp "He%%%lp%"; errmessage"";
                                        errhelp ""; errmessage "Another";
He%
lp


{errhelp}
{errmessage}
! Another.
<to be read again> 
                   ;
l.106 ...sage""; errhelp ""; errmessage "Another";
                                                  
(That was another `errmessage'.)

{headerbyte}
>> 0
! Improper location.
<to be read again> 
                   ;
l.107 headerbyte 0;
                    headerbyte(48.5)substring(-9,9)of"long"; for\=0:\
I was looking for a known, positive number.
For safety's sake I'll ignore the present command.

{headerbyte}
! Missing `:' has been inserted.
<to be read again> 
                   substring
l.107 headerbyte 0; headerbyte(48.5)substring
                                             (-9,9)of"long"; for\=0:\
A colon should follow a headerbyte or fontinfo location.

{-(9)}
{((-9,9))substring("long")}
>> "long"
! Invalid code has been replaced by 0.
<to be read again> 
                   ;
l.107 ... headerbyte(48.5)substring(-9,9)of"long";
                                                   for\=0:\
I was looking for a number between 0 and 255, or for a
string of length 1. Didn't find it; will use 0 instead.

{for}
Runaway loop?
! Forbidden token found while scanning the text of a for loop.
<inserted text> 
                endfor
<to be read again> 
                   \
l.107 ...te(48.5)substring(-9,9)of"long"; for\=0:\
                                                  
I suspect you have forgotten an `endfor',
causing me to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.

{loop value=0}
{[repeat the loop]}
{\}
{headerbyte}
{(2)*(0.33333)}
{fontdimen}
{(2)*(0.33333)}
>> "q"
! Improper font parameter.
<to be read again> 
                   ;
l.108 headerbyte 9:2a6,"q"; fontdimen 9:2a6,"q";
                                                 fontdimen 1:2048;
I'm zeroing this one. Proceed, with fingers crossed.

{fontdimen}
{fontmaking:=1}
{extensible}
! Missing `:' has been inserted.
<to be read again> 
                   5
l.109 fontmaking:=1; extensible 5 5
                                   ,"c"255.5,"d"; charlist 0:5:"a":"d";
I'm processing `extensible c: t,m,b,r'.

! Missing `,' has been inserted.
<to be read again> 
                   255.5
l.109 fontmaking:=1; extensible 5 5,"c"255.5
                                            ,"d"; charlist 0:5:"a":"d";
I'm processing `extensible c: t,m,b,r'.

>> 255.5
! Invalid code has been replaced by 0.
<to be read again> 
                   ,
l.109 fontmaking:=1; extensible 5 5,"c"255.5,
                                             "d"; charlist 0:5:"a":"d";
I was looking for a number between 0 and 255, or for a
string of length 1. Didn't find it; will use 0 instead.

{charlist}
! Character code 5 is already extensible.
<to be read again> 
                   :
l.109 ...sible 5 5,"c"255.5,"d"; charlist 0:5:"a":
                                                  "d";
It's not legal to label a character more than once.
So I'll not change anything just now.

{ligtable}
! Too far to skip.
l.110 ligtable255:255::
                       "a"=:"b","d" kern -2048,"c":0:99.5:"e"|=:|"f",0kern';
At most 127 lig/kern steps can separate skipto1 from 1::.

{-(2048)}
! Character code 0 is already in a charlist.
<to be read again> 
                   :
l.110 ...le255:255::"a"=:"b","d" kern -2048,"c":0:
                                                  99.5:"e"|=:|"f",0kern';
It's not legal to label a character more than once.
So I'll not change anything just now.

>> (xpart ',ypart ')
! Improper kern.
<to be read again> 
                   ;
l.110 ... kern -2048,"c":0:99.5:"e"|=:|"f",0kern';
                                                  
The amount of kern should be a known numeric value.
I'm zeroing this one. Proceed, with fingers crossed.

{ligtable}
! Character code 5 is already extensible.
<to be read again> 
                   :
l.111 ligtable 5:
                 0; def clear(text x)=interim x:=$ enddef; clear(hppp); vppp=0;
It's not legal to label a character more than once.
So I'll not change anything just now.

! Illegal ligtable step.
<to be read again> 
                   ;
l.111 ligtable 5:0;
                    def clear(text x)=interim x:=$ enddef; clear(hppp); vppp=0;
I was looking for `=:' or `kern' here.

{def}

clear(TEXT0)->interim(TEXT0):=$
(TEXT0)<-hppp
{interim}
! The token `hppp' isn't an internal quantity.
<to be read again> 
                   hppp
clear->interim(TEXT0)
                     :=$
l.111 ...(text x)=interim x:=$ enddef; clear(hppp)
                                                  ; vppp=0;
Something like `tracingonline' should follow `interim'.

{hppp:=-1}
## hppp=-1
{(vppp)=(0)}
## vppp=0
{begingroup}

clear(TEXT0)->interim(TEXT0):=$
(TEXT0)<-tracingmacros
{interim}
{tracingmacros:=-1}
{interim}
{tracingcommands:=-1}
Runaway text?
tracingoutput
! Forbidden token found while scanning a text argument.
<inserted text> 
                )
<to be read again> 
                   \
l.112 ...ar(tracingcommands); clear(tracingoutput\
                                                  ;
It seems that a right delimiter was left out,
causing me to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.

[95] [96] [97] [98] [99] [100] [101] [102] [103] [104] [105] [106] [107]
## chardy=6
[108] [109]
{restoring tracingoutput=1}
{restoring tracingcommands=2.1}
{restoring tracingmacros=1}
{endgroup}
{def}
{def}
{addto}
Path at line 122, before choices:
(0,0){4096,0}
 ..{4096,0}(1,0){0,-4096}
 ..{0,4096}cycle

Path at line 122, after choices:
(0,0)..controls (0.33333,0) and (0.66667,0)
 ..(1,0)..controls (1,-0.66667) and (0,-0.66667)
 ..cycle


f(SUFFIX0)(EXPR1)(EXPR2)(TEXT3)->numeric.w;show(EXPR1);addto(SUFFIX0)contour(0,
0)..(2,0)..(1,$)..(1,1)..cycle.withpen.qq;addto(SUFFIX0)doublepath(0,0){1,1}..{
2,1}(2,1)withpen.qq;addto(SUFFIX0)doublepath(($,$){1,0}..(1,1){1,0})scaled0.5wi
thpen.nullpen;
! Missing `]' has been inserted.
<to be read again> 
                   ,
l.123 f(e[3,
            w); g(e3,transformed p,penoffset-(1,1.3)of(pencircle scaled20 ys...
I've seen a `[' and a subscript value, in a suffix,
so a right bracket should have come next.
I shall pretend that one was there.

(SUFFIX0)<-e3
(EXPR1)<-unknown path w
! Missing argument to f.
<to be read again> 
                   ;
l.123 f(e[3,w);
                g(e3,transformed p,penoffset-(1,1.3)of(pencircle scaled20 ys...
That macro has more parameters than you thought.
I'll continue by pretending that each missing argument
is either zero or null.

(EXPR2)<-0
! Missing argument to f.
<to be read again> 
                   ;
l.123 f(e[3,w);
                g(e3,transformed p,penoffset-(1,1.3)of(pencircle scaled20 ys...
That macro has more parameters than you thought.
I'll continue by pretending that each missing argument
is either zero or null.

(TEXT3)<-
{numeric}
{show}
>> unknown path %CAPSULE1133
{addto}
Path at line 123, before choices:
(0,0)
 ..(2,0)
 ..(1,-1)
 ..(1,1)
 ..cycle

Path at line 123, after choices:
(0,0)..controls (0.56189,-0.5286) and (1.43811,0.5286)
 ..(2,0)..controls (2.70831,-0.66634) and (1.66634,-1.70831)
 ..(1,-1)..controls (0.4714,-0.43811) and (1.5286,0.43811)
 ..(1,1)..controls (0.33366,1.70831) and (-0.70831,0.66634)
 ..cycle

{addto}
Path at line 123, before choices:
(0,0){2896.30943,2896.30934}
 ..{3663.57385,1831.78674}(2,1)

Path at line 123, after choices:
(0,0)..controls (0.53566,0.53566) and (1.32689,0.66344)
 ..(2,1)

{addto}
Path at line 123, before choices:
(-1,-1){4096,0}
 ..{4096,0}(1,1)

Path at line 123, after choices:
(-1,-1)..controls (0.10457,-1) and (-0.10457,1)
 ..(1,1)

{(path)scaled(0.5)}
{nullpen}

g(SUFFIX0)->show(SUFFIX0)
(SUFFIX0)<-e3
! Too many arguments to g;
  Missing `)' has been inserted.
l.123 f(e[3,w); g(e3,
                     transformed p,penoffset-(1,1.3)of(pencircle scaled20 ys...
I'm going to assume that the comma I just read was a
right delimiter, and then I'll begin expanding the macro.
You might want to delete some tokens before continuing.

{show}
{(picture)transformed((8,1,5,-5,-5,5))}
>> Edge structure at line 123:
Filled pen stroke :
(10.5,-1.5)..controls (13,-4) and (15.5,-6.5)
 ..(18,-9)
dashed (on 3 off 0) shifted 0
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
Filled pen stroke :
(8,1)..controls (8.57388,0.42612) and (9.41167,-0.41167)
 ..(10.5,-1.5)..controls (14.51955,-5.51955) and (21.76099,-12.76099)
 ..(25.5,-16.5)..controls (27.5211,-18.5211) and (28.4345,-19.4345)
 ..(30.5,-21.5)..controls (35.23984,-26.23984) and (45.4408,-36.4408)
 ..(50.5,-41.5)..controls (51.60489,-42.60489) and (52.4432,-43.4432)
 ..(53,-44)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
Filled contour :
(8,1)..controls (13.45242,-4.45242) and (12.54758,-3.54758)
 ..(18,-9)..controls (24.87325,-15.87325) and (24.87325,-15.87325)
 ..(18,-9)..controls (12.54758,-3.54758) and (13.45242,-4.45242)
 ..(8,1)..controls (1.12675,7.87325) and (1.12675,7.87325)
 ..cycle
mitered joins limited 1 with pen
(-10,10)
 .. (5,-5)
 .. (12.5,-12.5)
 .. (15,-15)
 .. (15,-15)
 .. (10,-10)
 .. (-5,5)
 .. (-12.5,12.5)
 .. (-15,15)
 .. (-15,15)
 .. cycle
Filled pen stroke :
(8,1)..controls (8,1) and (11.31726,-2.31726)
 ..(13,-4)
butt ends, mitered joins limited 1 with pen
(-10,10)
 .. (5,-5)
 .. (12.5,-12.5)
 .. (15,-15)
 .. (15,-15)
 .. (10,-10)
 .. (-5,5)
 .. (-12.5,12.5)
 .. (-15,15)
 .. (-15,15)
 .. cycle
Filled pen stroke :
(8,1)..controls (10.76146,-1.76146) and (5.23854,3.76146)
 ..(8,1)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
End edges

{-((1,1.3))}
{pencircle}
{(pen)scaled(20)}
{-(0.5)}
{(pen)yscaled(-0.5)}
{((-1,-1.3))penoffset(pen)}
>> (-9.33345,1.79489)
Path at line 124, before choices:
(0,0)..controls (1,1) and (-1,1)
 ..(3,0)

Path at line 124, after choices:
(0,0)..controls (1,1) and (-1,1)
 ..(3,0)

{((0,1))directiontime(path)}
>> 0.33333
{(3.14159)point(path)}
{((3.14159,4))subpath(path)}
{((7.68668,2.72437))intersectiontimes(path)}
>> (0,0)
Path at line 126, before choices:
(-1,1.1){curl 1}
 ..{curl 1}(1,-1)

Path at line 126, after choices:
(-1,1.1)..controls (-0.33333,0.40001) and (0.33333,-0.3)
 ..(1,-1)

{(-1)precontrol((0,0))}
{(path)intersectiontimes((0,0))}
>> (-1,-1)
{addto}
{-(4094.99998)}
{-(0.00002)}
Path at line 127, before choices:
(-4094.99998,0){curl 1}
 ..{curl 1}(4094.99998,-0.00002)

Path at line 127, after choices:
(-4094.99998,0)..controls (-1365,0) and (1365,-0.00002)
 ..(4094.99998,-0.00002)

{pencircle}
{(0)penoffset(pen)}
>> 0
>> pen
! Not implemented: penoffset(known numeric)of(pen).
<to be read again> 
                   ;
l.128 pencircle;
                 addto e3 also e3 shifted (0,257); ,"flushed with pride"; nu...
I'm afraid I don't know how to apply that operation to that
combination of types. Continue, and I'll return the second
argument (see above) as the result of the operation.

{addto}
{(picture)shifted((0,257))}
! A statement can't begin with `,'.
<to be read again> 
                   ,
l.128 ...rcle; addto e3 also e3 shifted (0,257); ,
                                                  "flushed with pride"; nume...
I was looking for the beginning of a new statement.
If you just proceed without changing anything, I'll ignore
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

! Extra tokens will be flushed.
<to be read again> 
                   ,
l.128 ...rcle; addto e3 also e3 shifted (0,257); ,
                                                  "flushed with pride"; nume...
I've just read as much of that statement as I could fathom,
so a semicolon should have been next. It's very puzzling...
but I'll try to get myself back together, by ignoring
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

{numeric}
{def}

f(EXPR0)(EXPR1)(EXPR2)->showdependencies;tracingcapsules:=1;showdependencies;sh
ow1/3(3,6)*(((EXPR0)+(EXPR1))+((EXPR1)-(EXPR0))),(1,1)/sqrt2zscaled((EXPR0)+1,(
EXPR0)+2)-((EXPR0)+1,(EXPR0)+2)rotated45,(0,1)zscaled(1,(EXPR1)+2)-(1,(EXPR1)+2
)rotated90
{(xx)+(1)}
{(xx+1)/(0.3)}
(EXPR0)<-3.3333xx+3.3333
{(yy)-(1)}
{(yy-1)/(0.5)}
(EXPR1)<-2yy-2
(EXPR2)<-(xx,0)
{showdependencies}
alpha=0.45p3~+7
{tracingcapsules:=1}
{showdependencies}
xpart %CAPSULE2306=xx
%CAPSULE2314=2yy-2
%CAPSULE2322 = 3.3333xx+3.3333
alpha=0.45p3~+7
{show}
{(1/3)*((3,6))}
{(3.3333xx+3.3333)+(2yy-2)}
{(2yy-2)-(3.3333xx+3.3333)}
{(linearform)+(linearform)}
{((1,2))*(4yy-4)}
>> (4yy-4,8yy-8)
{sqrt(2)}
{((1,1))/(1.41422)}
{(3.3333xx+3.3333)+(1)}
{(3.3333xx+3.3333)+(2)}
{((0.7071,0.7071))zscaled((3.3333xx+4.3333,3.3333xx+5.3333))}
{(3.3333xx+3.3333)+(1)}
{(3.3333xx+3.3333)+(2)}
{((3.3333xx+4.3333,3.3333xx+5.3333))rotated(45)}
{((-0.7071,4.71399xx+6.83531))-((-0.7071,4.71399xx+6.83531))}
>> (0,0)
{(2yy-2)+(2)}
{((0,1))zscaled((1,2yy))}
{(2yy-2)+(2)}
{((1,2yy))rotated(90)}
{((-2yy,1))-((-2yy,1))}
>> (0,0)
{(1000)*(o3)}
{(o1)-(o2)}
{(4000)*(linearform)}
{(1000o3)-(linearform)}
{(4000)*(o2)}
{(linearform)+(4000o2)}
{(linearform)+(9)}
{(0.01)*(o3)}
{-(0.01o3)}
{(3)*(ooo)}
{(-0.01o3)+(3ooo)}
{begingroup}
{(o2)+(o4)}
{(linearform)+(o5)}
{(linearform)-(20)}
{(9/10)*(linearform)}
{(oo)=(linearform)}
## oo=0.9o5+0.9o4+0.9o2-18
{(0.5)*(o2)}
{(o1)+(0.5o2)}
{(8/9)*(linearform)}
{(o5)=(linearform)}
## o5=0.44444o2+0.88889o1
{(o4)=(linearform)}
## o4=0.44444o2+0.88889o1
{(0.0001)*(o2)}
{-(0.0001o2)}
{(o6)=(-0.0001o2)}
## o6=-0.0001o2
{showdependencies}
o6=-0.0001o2
o4=0.44444o2+0.88889o1
o5=0.44444o2+0.88889o1
oo=1.7o2+1.6o1-18
%CAPSULE2234 = 3ooo-0.01o3
%CAPSULE2254 = 8000o2-4000o1+1000o3+9
xpart %CAPSULE2330=xpart '
ypart %CAPSULE2330=ypart '
alpha=0.45p3~+7
{numeric}
### 4000o1 = -%CAPSULE2254+8000o2+1000o3+9
### -4.87383o2 = -oo-0.0004%CAPSULE2254+0.39673o3-17.99643
#### o6=-0.00027
### -0.04366o3 = -o4+0.46689oo+8.40439
### -o4 = -o5
### 0.22894o5 = -%CAPSULE2234+0.10689oo+3ooo+1.92412
{begingroup}
{pair}
### -alfa=-%CAPSULE2222
{endgroup}
### -%CAPSULE2222=-xpart %CAPSULE2220
{xpart((xpart %CAPSULE2220,0))}
### -xpart %CAPSULE9=-%CAPSULE2210
{endgroup}
{(%CAPSULE2234)+(%CAPSULE2210)}
### -%CAPSULE2234=-%CAPSULE2198+%CAPSULE2210
### -%CAPSULE2198=-ypart %CAPSULE2196
### -%CAPSULE2254=-xpart %CAPSULE2196
{begingroup}
{save}
{(1)-(p$)}
{restoring p}
### p$=-ypart %CAPSULE2162+1
{endgroup}
{(2/3)*((-ypart %CAPSULE2162+1,ypart %CAPSULE2162))}
### ypart %CAPSULE2162=-xpart %CAPSULE2162+1
### -0.66667xpart %CAPSULE2162=-xpart %CAPSULE2154
{-((xpart %CAPSULE2154,-xpart %CAPSULE2154+0.66667))}
### xpart %CAPSULE2154=-xpart %CAPSULE2134
{((xpart %CAPSULE2196,ypart %CAPSULE2196))=((xpart %CAPSULE2134,-xpart %CAPSULE
2134-0.66667))}
## xpart %CAPSULE2134=-ypart %CAPSULE2196-0.66667
## xpart %CAPSULE2196=-ypart %CAPSULE2196-0.66667
### ypart %CAPSULE2196=-xpart %CAPSULE2196-0.66667
### -xpart %CAPSULE2196=-xpart %CAPSULE2134
{((xpart ',ypart '))=((xpart %CAPSULE2134,-xpart %CAPSULE2134-0.66667))}
## xpart %CAPSULE2134=-ypart '-0.66667
## xpart '=-ypart '-0.66667
{begingroup}
{(0.5)*(ooo)}
{begingroup}
{numeric}
### -ooo=-%CAPSULE2096
{endgroup}
{(%CAPSULE2096)+(1)}
### -%CAPSULE2096=-%CAPSULE2084+1
{(1/2)*(%CAPSULE2084)}
### -0.5%CAPSULE2084=-%CAPSULE2076
{-(%CAPSULE2076)}
### %CAPSULE2076=-%CAPSULE2068
{(2)*(%CAPSULE2068)}
### -2%CAPSULE2068=-%CAPSULE2060
{(-0.5%CAPSULE2060-0.5)=(%CAPSULE2060)}
## %CAPSULE2060=-0.33333
{(oo)=(-0.33333)}
## oo=-0.33333
{endgroup}
{(4000)*(-0.33333)}
{yoffset:=-1333.31299}
## yoffset=-1333.31299
{xoffset:=-1333.31299}
## xoffset=-1333.31299
{addto}

foo(TEXT2)->begingroup(TEXT2)endgroup
(SUFFIX0)<-
(SUFFIX1)<-foo
! Missing argument to foo.
<to be read again> 
                   contour
l.136 addto foo contour
                        begingroup vardef foo=enddef; (0,0)..cycle endgroup;
That macro has more parameters than you thought.
I'll continue by pretending that each missing argument
is either zero or null.

(TEXT2)<-
{begingroup}
{endgroup}
>> vacuous
! Not a suitable variable.
<to be read again> 
                   contour
l.136 addto foo contour
                        begingroup vardef foo=enddef; (0,0)..cycle endgroup;
At this point I needed to see the name of a picture variable.
(Or perhaps you have indeed presented me with one; I might
have missed it, if it wasn't followed by the proper token.)
So I'll not change anything just now.

! Extra tokens will be flushed.
<to be read again> 
                   contour
l.136 addto foo contour
                        begingroup vardef foo=enddef; (0,0)..cycle endgroup;
I've just read as much of that statement as I could fathom,
so a semicolon should have been next. It's very puzzling...
but I'll try to get myself back together, by ignoring
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

Path at line 136, before choices:
(0,0)
 ..cycle

Path at line 136, after choices:
(0,0)..controls (0,0) and (0,0)
 ..cycle

! Extra `endgroup'.
l.136 ... vardef foo=enddef; (0,0)..cycle endgroup
                                                  ;
I'm not currently working on a `begingroup',
so I had better not try to end anything.

{(unknown path p~7)=(unknown path p~8)}
{-(1)}
Path at line 137, before choices:
(0,0){-2896.30943,2896.30934}
 ..{curl 1}(30,0)

Path at line 137, after choices:
(0,0)..controls (-48.28429,48.28426) and (78.28429,48.28426)
 ..(30,0)

{p~8:=path}
{show}
>> Path at line 137:
(0,0)..controls (-48.28429,48.28426) and (78.28429,48.28426)
 ..(30,0)

{-(1)}
{((0,-1))directiontime(path)}
>> 0.83456
{show}
{((1,0))directiontime(path)}
>> -1
{-(1)}
{-(1)}
! Missing `}' has been inserted.
<to be read again> 
                   )
l.139 of ((0,0){1,0}..(0,0){0,-1)
                                 ..cycle);
I've scanned a direction spec for part of a path,
so a right brace should have come next.
I shall pretend that one was there.

Path at line 139, before choices:
(0,0){4096,0}
 ..{0,-4096}(0,0)

Path at line 139, after choices:
(0,0)..controls (0,0) and (0,0)
 ..(0,0)

{((1,-1))directiontime(path)}
>> 0
! Extra tokens will be flushed.
<to be read again> 
                   ..
l.139 of ((0,0){1,0}..(0,0){0,-1)..
                                   cycle);
I've just read as much of that statement as I could fathom,
so a semicolon should have been next. It's very puzzling...
but I'll try to get myself back together, by ignoring
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

{nullpicture}
{(unknown picture e4)=(picture)}
{(unknown picture e9)=(picture)}
{addto}
{addto}
! Picture is too complicated to use as a dash pattern.
<to be read again> 
                   ;
l.141 addto e9 doublepath p~8 dashed e4;
                                         e9:=nullpicture;
When you say `dashed p', every path in p should be monotone
in x and there must be no overlapping.  This failed
so I'll just make it a solid line instead.

{nullpicture}
{e9:=picture}
{nullpicture}
{e4:=picture}
{addto}
Path at line 142, before choices:
(1,1){curl 1}
 ..{curl 1}(4,1)

Path at line 142, after choices:
(1,1)..controls (2,1) and (3,1)
 ..(4,1)

{addto}
{nullpicture}
{e9:=picture}
{addto}
{pencircle}
{makepath(pen)}
{-(2)}
{(picture)scaled(-2)}
{show}
>> Edge structure at line 144:
Filled pen stroke :
(1,1)..controls (2,1) and (3,1)
 ..(4,1)
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
Filled pen stroke :
(0.5,0)..controls (0.5,0.13261) and (0.44731,0.25978)
 ..(0.35355,0.35355)..controls (0.25978,0.44731) and (0.13261,0.5)
 ..(0,0.5)..controls (-0.13261,0.5) and (-0.25978,0.44731)
 ..(-0.35355,0.35355)..controls (-0.44731,0.25978) and (-0.5,0.13261)
 ..(-0.5,0)..controls (-0.5,-0.13261) and (-0.44731,-0.25978)
 ..(-0.35355,-0.35355)..controls (-0.25978,-0.44731) and (-0.13261,-0.5)
 ..(0,-0.5)..controls (0.13261,-0.5) and (0.25978,-0.44731)
 ..(0.35355,-0.35355)..controls (0.44731,-0.25978) and (0.5,-0.13261)
 ..cycle
dashed (on 6 off 0) shifted -2
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,0,0,0,0)
End edges

{nullpicture}
{(unknown picture e5)=(picture)}
{e4:=picture}
{addto}
{-(4)}
{-(3)}
{-(2)}
{-(2.7)}
Path at line 145, before choices:
(-4,-3){curl 1}..tension 2
 ..{curl 1}(-2,-2.7)

Path at line 145, after choices:
(-4,-3)..controls (-3.66667,-2.95) and (-2.33333,-2.75)
 ..(-2,-2.7)

{addto}
{(path)shifted((0,4))}
{(path)scaled(3)}
{show}
{llcorner(picture)}
>> (0.5,9.5)
{nullpicture}
{e6:=picture}
{addto}
Path at line 149, before choices:
(0,0){curl 1}
 ..{curl 1}(30,0)

Path at line 149, after choices:
(0,0)..controls (10,0) and (20,0)
 ..(30,0)

{pencircle}
{makepath(pen)}
>> path
! Improper type.
<to be read again> 
                   dashed
l.150 dashed
             e5 withpen pencircle scaled2 dashed e5shifted(0,-11)
Next time say `dashed <known picture expression>';
I'll ignore the bad `with' clause and look for another.

{pencircle}
{(pen)scaled(2)}
{-(11)}
{(picture)shifted((0,-11))}
{pencircle}
{show}
>> Edge structure at line 151:
Filled pen stroke :
(0,0)..controls (10,0) and (20,0)
 ..(30,0)
dashed (on 2 off 1) shifted -1
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,1,0,0,1)
End edges

{def}
{(path)rotated(90)}
{-(1)}
{(path)xscaled(-1)}
Path at line 152, before choices:
(0,0)..controls (48.28426,-48.28429) and (48.28426,78.28429)
 ..(0,30)
 ..cycle

Path at line 152, after choices:
(0,0)..controls (48.28426,-48.28429) and (48.28426,78.28429)
 ..(0,30)..controls (-8.28427,21.71573) and (-8.28427,8.28427)
 ..cycle

{p~7:=path}
{linecap:=2}
{linejoin:=1}
{nullpicture}
{(picture)=(picture)}
! Redundant or inconsistent equation.
<to be read again> 
                   ;
l.153 e6=nullpicture;
                      addto e6 doublepath p~8 dashed e5 withpen pencircle;
An equation between already-known quantities can't help.
But don't worry; continue and I'll just ignore it.

{addto}
{pencircle}
{clip}
{addto}
{begingroup}
{save}
{picture}
{nullpicture}
{(unknown picture _p)=(picture)}
{string}
{vardef}
{(unknown string _n0)=("trapf")}

_s(EXPR2)(EXPR3)(EXPR4)(EXPR5)(EXPR6)->begingroup.addto._p.also(EXPR2)infont(EX
PR3)scaled(EXPR4)shifted((EXPR5),(EXPR6));endgroup
(SUFFIX0)<-
(SUFFIX1)<-_s
(EXPR2)<-"Aqua"
(EXPR3)<-"trapf"
(EXPR4)<-1
(EXPR5)<-0
(EXPR6)<-0
{begingroup}
{addto}
{("Aqua")infont("trapf")}
{(picture)scaled(1)}
{(picture)shifted((0,0))}
{endgroup}
{interim}
{linecap:=0}
{vardef}

_r(EXPR2)(EXPR3)->begingroup.addto._p.doublepath(EXPR2)withpen.pencircle.scaled
(EXPR3)endgroup
(SUFFIX0)<-
(SUFFIX1)<-_r
{-(3.9851)}
Path at line 11, before choices:
(8.1693,-3.9851){curl 1}
 ..{curl 1}(8.1693,15.9402)

Path at line 11, after choices:
(8.1693,-3.9851)..controls (8.1693,2.65666) and (8.1693,9.29843)
 ..(8.1693,15.9402)

(EXPR2)<-path
(EXPR3)<-0.3985
{begingroup}
{addto}
{pencircle}
{(pen)scaled(0.3985)}
{endgroup}
{(unknown string _n1)=("trapf")}

_s(EXPR2)(EXPR3)(EXPR4)(EXPR5)(EXPR6)->begingroup.addto._p.also(EXPR2)infont(EX
PR3)scaled(EXPR4)shifted((EXPR5),(EXPR6));endgroup
(SUFFIX0)<-
(SUFFIX1)<-_s
(EXPR2)<-"Brass"
(EXPR3)<-"trapf"
(EXPR4)<-2
(EXPR5)<-8.3686
(EXPR6)<-0
{begingroup}
{addto}
{("Brass")infont("trapf")}
{(picture)scaled(2)}
{(picture)shifted((8.3686,0))}
{endgroup}
{setbounds}
{-(3.9851)}

--->{curl1}..{curl1}
{-(3.9851)}

--->{curl1}..{curl1}

--->{curl1}..{curl1}

--->{curl1}..{curl1}
Path at line 15, before choices:
(0,-3.9851){curl 1}
 ..{curl 1}(18.3313,-3.9851){curl 1}
 ..{curl 1}(18.3313,15.9402){curl 1}
 ..{curl 1}(0,15.9402){curl 1}
 ..{curl 1}cycle

Path at line 15, after choices:
(0,-3.9851)..controls (6.11043,-3.9851) and (12.22087,-3.9851)
 ..(18.3313,-3.9851)..controls (18.3313,2.65666) and (18.3313,9.29843)
 ..(18.3313,15.9402)..controls (12.22087,15.9402) and (6.11043,15.9402)
 ..(0,15.9402)..controls (0,9.29843) and (0,2.65666)
 ..cycle

{restoring linecap=2}
{restoring _n}
{restoring _s}
{restoring _r}
{restoring _p}
{endgroup}
{-(2)}
{(picture)shifted((0,-2))}
{addto}
Path at line 155, before choices:
(1,0){curl 1}
 ..{curl 1}(2,1)

Path at line 155, after choices:
(1,0)..controls (1.33333,0.33333) and (1.66667,0.66667)
 ..(2,1)

{addto}
{(0.5)*((1,1,1))}
{pencircle}
{(pen)xscaled(2)}
{(pen)rotated(45)}
{addto}
{begingroup}
{save}
{picture}
{nullpicture}
{(unknown picture _p)=(picture)}
{string}
{vardef}
{(unknown string _n0)=("trapf")}

_s(EXPR2)(EXPR3)(EXPR4)(EXPR5)(EXPR6)->begingroup.addto._p.also(EXPR2)infont(EX
PR3)scaled(EXPR4)shifted((EXPR5),(EXPR6));endgroup
(SUFFIX0)<-
(SUFFIX1)<-_s
(EXPR2)<-"ast"
(EXPR3)<-"trapf"
(EXPR4)<-1
(EXPR5)<-0
(EXPR6)<-0
{begingroup}
{addto}
{("ast")infont("trapf")}
{(picture)scaled(1)}
{(picture)shifted((0,0))}
{endgroup}
{setbounds}
{-(0.9963)}

--->{curl1}..{curl1}
{-(0.9963)}

--->{curl1}..{curl1}

--->{curl1}..{curl1}

--->{curl1}..{curl1}
Path at line 25, before choices:
(0,-0.9963){curl 1}
 ..{curl 1}(1.9925,-0.9963){curl 1}
 ..{curl 1}(1.9925,6.9738){curl 1}
 ..{curl 1}(0,6.9738){curl 1}
 ..{curl 1}cycle

Path at line 25, after choices:
(0,-0.9963)..controls (0.66417,-0.9963) and (1.32832,-0.9963)
 ..(1.9925,-0.9963)..controls (1.9925,1.6604) and (1.9925,4.3171)
 ..(1.9925,6.9738)..controls (1.32832,6.9738) and (0.66417,6.9738)
 ..(0,6.9738)..controls (0,4.3171) and (0,1.6604)
 ..cycle

{restoring _n}
{restoring _s}
{restoring _r}
{restoring _p}
{endgroup}
{(picture)shifted((0,10))}
{show}
>> Edge structure at line 158:
clipping path:
(0,0)..controls (48.28426,-48.28429) and (48.28426,78.28429)
 ..(0,30)..controls (-8.28427,21.71573) and (-8.28427,8.28427)
 ..cycle
Filled pen stroke :
(0,0)..controls (10,0) and (20,0)
 ..(30,0)
dashed (on 2 off 1) shifted -1
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,1,0,0,1)
Filled pen stroke :
(0,0)..controls (-48.28429,48.28426) and (78.28429,48.28426)
 ..(30,0)
dashed (on 1 off 1 on 1 off 9) shifted -9
square ends, round joins with pen
pencircle transformed (0,0,1,0,0,1)
stop clipping
setbounds path:
(0,-5.9851)..controls (6.11043,-5.9851) and (12.22087,-5.9851)
 ..(18.3313,-5.9851)..controls (18.3313,0.65666) and (18.3313,7.29843)
 ..(18.3313,13.9402)..controls (12.22087,13.9402) and (6.11043,13.9402)
 ..(0,13.9402)..controls (0,7.29843) and (0,0.65666)
 ..cycle
"Aqua" infont "trapf"
transformed (0,-2,1,0,0,1)
Filled pen stroke :
(8.1693,-5.9851)..controls (8.1693,0.65666) and (8.1693,7.29843)
 ..(8.1693,13.9402)
butt ends, round joins with pen
pencircle transformed (0,0,0.3985,0,0,0.3985)
"Brass" infont "trapf"
transformed (8.3686,-2,2,0,0,2)
end of setbounds
Filled pen stroke :
(1,0)..controls (1.33333,0.33333) and (1.66667,0.66667)
 ..(2,1)
square ends, round joins with pen
pencircle transformed (0,0,0,0,0,0)
End edges

{lrcorner(picture)}
>> (36.21318,-5.9851)
{ulcorner(picture)}
>> (-6.2132,37.00375)
{charcode:=200}
{shipout}
[200
Envelope spec at line 158:
(3,12) % beginning with offset (2,-1)
   ..controls (3.75,12.75) and (4.125,13.3125)
 ..(4.3125,13.6875)
   ..controls (4.375,13.8125) and (4.41667,13.91667)
 ..(4.44444,14) % counterclockwise to offset (2.5,0.5)
   ..controls (4.5,14.16667) and (4.5,14.25)
 ..(4.5,14.25) % clockwise to offset (-2.5,-0.5)
   ..controls (4.5,14.25) and (4.5,13.5)
 ..(6,12) % counterclockwise to offset (2.5,0.5)
   ..controls (4.5,13.5) and (4.5,14.25)
 ..(4.5,14.25) % counterclockwise to offset (-2.5,-0.5)
   ..controls (4.5,14.25) and (4.5,14.16667)
 ..(4.44444,14) % clockwise to offset (-2,1)
   ..controls (4.41667,13.91667) and (4.375,13.8125)
 ..(4.3125,13.6875)
   ..controls (4.125,13.3125) and (3.75,12.75)
 ..(3,12) % counterclockwise to offset (2,-1)
 & cycle

]
Edge structure at line 158 (just shipped out):
Filled pen stroke :
(3,12)..controls (3.75,12.75) and (4.125,13.3125)
 ..(4.3125,13.6875)..controls (4.875,14.8125) and (3.75,14.25)
 ..(6,12)
dashed (on 2 off 1) shifted -1 (this will be ignored)
butt ends, mitered joins limited 1 with pen
(-2.5,-0.5)
 .. (-1,-2)
 .. (0,-2.5)
 .. (1,-2)
 .. (2,-1)
 .. (2.5,0.5)
 .. (1,2)
 .. (0,2.5)
 .. (-1,2)
 .. (-2,1)
 .. cycle
clipping path:
(0,0)..controls (48.28426,-48.28429) and (48.28426,78.28429)
 ..(0,30)..controls (-8.28427,21.71573) and (-8.28427,8.28427)
 ..cycle
Filled pen stroke colored (0.5,0.5,0.5):
(0,0)..controls (10,0) and (20,0)
 ..(30,0)
dashed (on 2 off 1) shifted -1
butt ends, mitered joins limited 1 with pen
pencircle transformed (0,0,1.41422,-0.7071,1.41422,0.7071)
Filled pen stroke colored (0.5,0.5,0.5):
(0,0)..controls (-48.28429,48.28426) and (78.28429,48.28426)
 ..(30,0)
dashed (on 2 off 1) shifted -1
square ends, round joins with pen
pencircle transformed (0,0,1.41422,-0.7071,1.41422,0.7071)
stop clipping
setbounds path:
(0,-5.9851)..controls (6.11043,-5.9851) and (12.22087,-5.9851)
 ..(18.3313,-5.9851)..controls (18.3313,0.65666) and (18.3313,7.29843)
 ..(18.3313,13.9402)..controls (12.22087,13.9402) and (6.11043,13.9402)
 ..(0,13.9402)..controls (0,7.29843) and (0,0.65666)
 ..cycle
"Aqua" infont "trapf"
colored (0.5,0.5,0.5)transformed (0,-2,1,0,0,1)
Filled pen stroke colored (0.5,0.5,0.5):
(8.1693,-5.9851)..controls (8.1693,0.65666) and (8.1693,7.29843)
 ..(8.1693,13.9402)
dashed (on 2 off 1) shifted -1
butt ends, round joins with pen
pencircle transformed (0,0,1.41422,-0.7071,1.41422,0.7071)
"Brass" infont "trapf"
colored (0.5,0.5,0.5)transformed (8.3686,-2,2,0,0,2)
end of setbounds
Filled pen stroke colored (0.5,0.5,0.5):
(1,0)..controls (1.33333,0.33333) and (1.66667,0.66667)
 ..(2,1)
dashed (on 2 off 1) shifted -1
square ends, round joins with pen
pencircle transformed (0,0,1.41422,-0.7071,1.41422,0.7071)
setbounds path:
(0,9.0037)..controls (0.66417,9.0037) and (1.32832,9.0037)
 ..(1.9925,9.0037)..controls (1.9925,11.6604) and (1.9925,14.3171)
 ..(1.9925,16.9738)..controls (1.32832,16.9738) and (0.66417,16.9738)
 ..(0,16.9738)..controls (0,14.3171) and (0,11.6604)
 ..cycle
"ast" infont "trapf"
transformed (0,10,1,0,0,1)
end of setbounds
End edges

{showstats}
Memory usage 3371&264 (401 still untouched)
String usage 63&214 (8&7476 now untouched)
{("curb")infont("trapf")}
Missing character: There is no c in font trapf!
{-(0.5)}
{(picture)slanted(-0.5)}
{-(1)}
{(picture)scaled(-1)}
{e6:=picture}
{e5:=picture}
{showstats}
Memory usage 2497&264 (401 still untouched)
String usage 61&206 (7&7472 now untouched)
{show}
{pencircle}
{((0,0))penoffset(pen)}
>> (0,0)
{llcorner(picture)}
>> (-3.98508,-7.97011)
{urcorner(picture)}
>> (3.98506,1.99254)
{(picture)rotated(90)}
{(picture)shifted("oops")}
>> "oops"
! Improper transformation argument.
<to be read again> 
                   ;
l.161 e5:=e5 rotated 90 shifted "oops";
                                        show llcorner e5, urcorner e5;
The expression shown above has the wrong type,
so I can't transform anything using it.
Proceed, and I'll omit the transformation.

{e5:=picture}
{show}
{llcorner(picture)}
>> (-1.99254,-3.98508)
{urcorner(picture)}
>> (7.97011,3.98506)
{show}
{char(220)}
{("Ü")infont("trapf")}
Missing character: There is no Ü in font trapf!
{urcorner(picture)}
>> (0,0)
{makepath(pen)}
>> Path at line 162:
(-2.5,-0.5)..controls (-2.5,-0.5) and (-1,-2)
 ..(-1,-2)..controls (-1,-2) and (0,-2.5)
 ..(0,-2.5)..controls (0,-2.5) and (1,-2)
 ..(1,-2)..controls (1,-2) and (2,-1)
 ..(2,-1)..controls (2,-1) and (2.5,0.5)
 ..(2.5,0.5)..controls (2.5,0.5) and (1,2)
 ..(1,2)..controls (1,2) and (0,2.5)
 ..(0,2.5)..controls (0,2.5) and (-1,2)
 ..(-1,2)..controls (-1,2) and (-2,1)
 ..(-2,1)..controls (-2,1) and (-2.5,-0.5)
 ..cycle

{show}
{-(2)}
{-(1)}
{((-2,-1))subpath(path)}
>> Path at line 163:
(0,0)

{pencircle}
{makepath(pen)}
{((9,10))subpath(path)}
>> Path at line 163:
(0.35355,0.35355)..controls (0.25978,0.44731) and (0.13261,0.5)
 ..(0,0.5)

{("hi")infont("qw99z")}
! Font qw99z not usable: TFM file not found.
l.164 "hi" infont "qw99z";
                           special "ij"="ij"; show substring(-2,-1)of"abc";
I wasn't able to read the size data for this font so this
`infont' operation won't produce anything. If the font name
is right, you might ask an expert to make a TFM file

Missing character: There is no h in font nullfont!
Missing character: There is no i in font nullfont!
>> picture
! Isolated expression.
<to be read again> 
                   ;
l.164 "hi" infont "qw99z";
                           special "ij"="ij"; show substring(-2,-1)of"abc";
I couldn't find an `=' or `:=' after the
expression that is shown above this error message,
so I guess I'll just ignore it and carry on.

{special}
{("ij")=("ij")}
>> true
! Unsuitable expression.
<to be read again> 
                   ;
l.164 "hi" infont "qw99z"; special "ij"="ij";
                                              show substring(-2,-1)of"abc";
Only known strings are allowed for output as specials.

{show}
{-(2)}
{-(1)}
{((-2,-1))substring("abc")}
>> ""
{addto}
{begingroup}
{endgroup}
>> vacuous
! Improper `addto'.
<to be read again> 
                   ;
l.165 addto e5 contour [[ ]];
                              addto e5 contour true; pausing:=0;prologues:=1;
This expression should have specified a known path.
So I'll not change anything just now.

{addto}
{true}
>> true
! Improper `addto'.
<to be read again> 
                   ;
l.165 ... e5 contour [[ ]]; addto e5 contour true;
                                                   pausing:=0;prologues:=1;
This expression should have specified a known path.
So I'll not change anything just now.

{pausing:=0}
{prologues:=1}
{charcode:=197}
{shipout}
[197
Warning: cannot open font map file psfonts.map
]
Edge structure at line 166 (just shipped out):
"curb" infont "trapf"
transformed (0,0,0,1,-1,0.5)
End edges

{-(1)}
{-(1)}
Path at line 167, before choices:
(0,0){2881.87209,2910.67511}..tension atleast1
 ..{4096,0}(1,1){-2896.30943,2896.30934}
 ..{2896.30943,-2896.30934}(2,0)

Path at line 167, after choices:
(0,0)..controls (0.3493,0.3528) and (0.99011,1)
 ..(1,1)..controls (0.12732,1.87268) and (1.46066,0.53934)
 ..(2,0)

{(unknown path p~10)=(path)}
{prologues:=0}
{for}
{-(1)}
{-(1)}
{loop value=(-1,1)}
{show}
{((-1,1))directiontime(path)}
>> 1
{[repeat the loop]}
{loop value=(1,-1)}
{show}
{((1,-1))directiontime(path)}
>> 1.2355
{[repeat the loop]}
{loop value=(0,0)}
{show}
{((0,0))directiontime(path)}
>> 0
{[repeat the loop]}
{linecap:=1}
{linejoin:=1}
{nullpicture}
{e5:=picture}
{addto}
{pencircle}
{(pen)scaled(10)}
{makepath(pen)}
{pencircle}
{(pen)shifted((2,0))}
{addto}
Path at line 171, before choices:
(0,0){curl 1}
 ..{curl 1}(12,0)

Path at line 171, after choices:
(0,0)..controls (4,0) and (8,0)
 ..(12,0)

{pencircle}
{(pen)yscaled(3)}
{charcode:=148}
{shipout}
[148]
Edge structure at line 171 (just shipped out):
Filled contour :
(5,0)..controls (5,1.32608) and (4.47322,2.59785)
 ..(3.53554,3.53554)..controls (2.59785,4.47322) and (1.32608,5)
 ..(0,5)..controls (-1.32608,5) and (-2.59785,4.47322)
 ..(-3.53554,3.53554)..controls (-4.47322,2.59785) and (-5,1.32608)
 ..(-5,0)..controls (-5,-1.32608) and (-4.47322,-2.59785)
 ..(-3.53554,-3.53554)..controls (-2.59785,-4.47322) and (-1.32608,-5)
 ..(0,-5)..controls (1.32608,-5) and (2.59785,-4.47322)
 ..(3.53554,-3.53554)..controls (4.47322,-2.59785) and (5,-1.32608)
 ..cycle
round joins with pen
pencircle transformed (2,0,1,0,0,1)
Filled pen stroke :
(0,0)..controls (4,0) and (8,0)
 ..(12,0)
round ends, round joins with pen
pencircle transformed (0,0,1,0,0,3)
End edges

{nullpicture}
{e5:=picture}
{clip}
! An expression can't begin with `;'.
<inserted text> 
                0
<to be read again> 
                   ;
l.172 e5:=nullpicture; clip p~8 to;
                                    clip e5 to e4; clip e5 to p~8; clip e5 t...
I'm afraid I need some sort of value in order to continue,
so I've tentatively inserted `0'. You may want to
delete this zero and insert something else;
see Chapter 27 of The METAFONTbook for an example.

! Variable p~8 is the wrong type (path).
<to be read again> 
                   ;
l.172 e5:=nullpicture; clip p~8 to;
                                    clip e5 to e4; clip e5 to p~8; clip e5 t...
I was looking for a "known" picture variable.
So I'll not change anything just now.

{clip}
>> picture
! Improper `clip'.
<to be read again> 
                   ;
l.172 e5:=nullpicture; clip p~8 to; clip e5 to e4;
                                                   clip e5 to p~8; clip e5 t...
This expression should have specified a known path.
So I'll not change anything just now.

{clip}
! Not a cycle.
<to be read again> 
                   ;
l.172 ...ip p~8 to; clip e5 to e4; clip e5 to p~8;
                                                   clip e5 to p~7;
That contour should have ended with `..cycle' or `&cycle'.
So I'll not change anything just now.

{clip}
{addto}
{pencircle}
{(pen)scaled(50)}
{makepath(pen)}
{addto}
{pencircle}
{(pen)xscaled(2)}
{addto}
{pencircle}
{addto}
{(path)shifted((10,0))}
{(picture)scaled(1)}
{pencircle}
{charcode:=149}
{shipout}
[149
Envelope spec at line 177:
(25,0) % beginning with offset (2.5,0.5)
   ..controls (25,6.63042) and (22.36609,12.98926)
 ..(17.67767,17.67767) % counterclockwise to offset (1,2)
   ..controls (15.7597,19.59564) and (13.56218,21.16978)
 ..(11.18088,22.36044) % counterclockwise to offset (0,2.5)
   ..controls (7.74117,24.08029) and (3.918,25)
 ..(0,25)
   ..controls (-3.918,25) and (-7.74117,24.08029)
 ..(-11.18088,22.36044) % counterclockwise to offset (-1,2)
   ..controls (-13.56218,21.16978) and (-15.7597,19.59564)
 ..(-17.67767,17.67767) % counterclockwise to offset (-2,1)
   ..controls (-20.44812,14.90723) and (-22.50117,11.55353)
 ..(-23.71729,7.90517) % counterclockwise to offset (-2.5,-0.5)
   ..controls (-24.5592,5.37941) and (-25,2.71243)
 ..(-25,0)
   ..controls (-25,-6.63042) and (-22.36609,-12.98926)
 ..(-17.67767,-17.67767) % counterclockwise to offset (-1,-2)
   ..controls (-15.7597,-19.59564) and (-13.56218,-21.16978)
 ..(-11.18088,-22.36044) % counterclockwise to offset (0,-2.5)
   ..controls (-7.74117,-24.08029) and (-3.918,-25)
 ..(0,-25)
   ..controls (3.918,-25) and (7.74117,-24.08029)
 ..(11.18088,-22.36044) % counterclockwise to offset (1,-2)
   ..controls (13.56218,-21.16978) and (15.7597,-19.59564)
 ..(17.67767,-17.67767) % counterclockwise to offset (2,-1)
   ..controls (20.44812,-14.90723) and (22.50117,-11.55353)
 ..(23.71729,-7.90517) % counterclockwise to offset (2.5,0.5)
   ..controls (24.5592,-5.37941) and (25,-2.71243)
 ..(25,0) % counterclockwise to offset (-2.5,-0.5)
   ..controls (25,-2.71243) and (24.5592,-5.37941)
 ..(23.71729,-7.90517) % clockwise to offset (-2,1)
   ..controls (22.50117,-11.55353) and (20.44812,-14.90723)
 ..(17.67767,-17.67767) % clockwise to offset (-1,2)
   ..controls (15.7597,-19.59564) and (13.56218,-21.16978)
 ..(11.18088,-22.36044) % clockwise to offset (0,2.5)
   ..controls (7.74117,-24.08029) and (3.918,-25)
 ..(0,-25)
   ..controls (-3.918,-25) and (-7.74117,-24.08029)
 ..(-11.18088,-22.36044) % clockwise to offset (1,2)
   ..controls (-13.56218,-21.16978) and (-15.7597,-19.59564)
 ..(-17.67767,-17.67767) % clockwise to offset (2.5,0.5)
   ..controls (-22.36609,-12.98926) and (-25,-6.63042)
 ..(-25,0)
   ..controls (-25,2.71243) and (-24.5592,5.37941)
 ..(-23.71729,7.90517) % clockwise to offset (2,-1)
   ..controls (-22.50117,11.55353) and (-20.44812,14.90723)
 ..(-17.67767,17.67767) % clockwise to offset (1,-2)
   ..controls (-15.7597,19.59564) and (-13.56218,21.16978)
 ..(-11.18088,22.36044) % clockwise to offset (0,-2.5)
   ..controls (-7.74117,24.08029) and (-3.918,25)
 ..(0,25)
   ..controls (3.918,25) and (7.74117,24.08029)
 ..(11.18088,22.36044) % clockwise to offset (-1,-2)
   ..controls (13.56218,21.16978) and (15.7597,19.59564)
 ..(17.67767,17.67767) % clockwise to offset (-2.5,-0.5)
   ..controls (22.36609,12.98926) and (25,6.63042)
 ..(25,0) % counterclockwise to offset (2.5,0.5)
 & cycle

]
Edge structure at line 177 (just shipped out):
clipping path:
(0,0)..controls (48.28426,-48.28429) and (48.28426,78.28429)
 ..(0,30)..controls (-8.28427,21.71573) and (-8.28427,8.28427)
 ..cycle
stop clipping
Filled pen stroke :
(25,0)..controls (25,6.63042) and (22.36609,12.98926)
 ..(17.67767,17.67767)..controls (12.98926,22.36609) and (6.63042,25)
 ..(0,25)..controls (-6.63042,25) and (-12.98926,22.36609)
 ..(-17.67767,17.67767)..controls (-22.36609,12.98926) and (-25,6.63042)
 ..(-25,0)..controls (-25,-6.63042) and (-22.36609,-12.98926)
 ..(-17.67767,-17.67767)..controls (-12.98926,-22.36609) and (-6.63042,-25)
 ..(0,-25)..controls (6.63042,-25) and (12.98926,-22.36609)
 ..(17.67767,-17.67767)..controls (22.36609,-12.98926) and (25,-6.63042)
 ..cycle
round ends, round joins with pen
(-2.5,-0.5)
 .. (-1,-2)
 .. (0,-2.5)
 .. (1,-2)
 .. (2,-1)
 .. (2.5,0.5)
 .. (1,2)
 .. (0,2.5)
 .. (-1,2)
 .. (-2,1)
 .. cycle
Filled pen stroke :
(0,0)..controls (-48.28429,48.28426) and (78.28429,48.28426)
 ..(30,0)
dashed (on 2 off 1) shifted -1
round ends, round joins with pen
pencircle transformed (0,0,2,0,0,1)
Filled pen stroke :
(0,0)..controls (-48.28429,48.28426) and (78.28429,48.28426)
 ..(30,0)
dashed (on 2 off 1) shifted -1
round ends, round joins with pen
pencircle transformed (0,0,1,0,0,1)
Filled pen stroke :
(10,0)..controls (-38.28429,48.28426) and (88.28429,48.28426)
 ..(40,0)
dashed (on 2 off 1) shifted -1
round ends, round joins with pen
pencircle transformed (0,0,1,0,0,1)
End edges

{addto}
>> 3
! Not a suitable variable.
<to be read again> 
                   ;
l.177 charcode:=149; shipout e5;  addto 3;
                                           addto p~8 contour p~7;
At this point I needed to see the name of a picture variable.
(Or perhaps you have indeed presented me with one; I might
have missed it, if it wasn't followed by the proper token.)
So I'll not change anything just now.

{addto}
! Variable p~8 is the wrong type (path).
<to be read again> 
                   ;
l.177 ...pout e5;  addto 3; addto p~8 contour p~7;
                                                  
I was looking for a "known" picture variable.
So I'll not change anything just now.

{nullpicture}
{e5:=picture}
{linecap:=2}
{linejoin:=0}
{miterlimit:=10}
{addto}
Path at line 179, before choices:
(10,5)
 ..cycle

Path at line 179, after choices:
(10,5)..controls (10,5) and (10,5)
 ..cycle

{addto}
Path at line 180, before choices:
(5,10){curl 1}
 ..{curl 1}(5,15)

Path at line 180, after choices:
(5,10)..controls (5,11.66667) and (5,13.33333)
 ..(5,15)

{addto}
Path at line 182, before choices:
(0,0)..controls (10,10) and (10,10)
 ..(10,10)
 ..{curl 1}(0,20){curl 1}
 ..{curl 1}(30,10)

Path at line 182, after choices:
(0,0)..controls (10,10) and (10,10)
 ..(10,10)..controls (6.66667,13.33333) and (3.33333,16.66667)
 ..(0,20)..controls (10,16.66667) and (20,13.33333)
 ..(30,10)

{-(5)}
Path at line 182, before choices:
(0,-5)
 ..(0,5)
 ..cycle

Path at line 182, after choices:
(0,-5)..controls (6.66667,-5) and (6.66667,5)
 ..(0,5)..controls (-6.66667,5) and (-6.66667,-5)
 ..cycle

{makepen(path)}
{charcode:=150}
{prologues:=1}
{shipout}
[150
Envelope spec at line 182:
(10,5) % beginning with offset (-2.5,-0.5)
   ..controls (10,5) and (10,5)
 ..(10,5) % counterclockwise to offset (-2.5,-0.5)
 & cycle

Envelope spec at line 182:
(10,5) % beginning with offset (-2.5,-0.5)
   ..controls (10,5) and (10,5)
 ..(10,5) % counterclockwise to offset (-2.5,-0.5)
 & cycle

Envelope spec at line 182:
(5,10) % beginning with offset (2.5,0.5)
   ..controls (5,11.66667) and (5,13.33333)
 ..(5,15) % counterclockwise to offset (-2.5,-0.5)
   ..controls (5,13.33333) and (5,11.66667)
 ..(5,10) % counterclockwise to offset (2.5,0.5)
 & cycle

Envelope spec at line 182:
(0,0) % beginning with offset (0,-5)
   ..controls (10,10) and (10,10)
 ..(10,10) % counterclockwise to offset (0,5)
   ..controls (6.66667,13.33333) and (3.33333,16.66667)
 ..(0,20) % clockwise to offset (0,-5)
   ..controls (10,16.66667) and (20,13.33333)
 ..(30,10) % counterclockwise to offset (0,5)
   ..controls (20,13.33333) and (10,16.66667)
 ..(0,20) % counterclockwise to offset (0,-5)
   ..controls (3.33333,16.66667) and (6.66667,13.33333)
 ..(10,10) % clockwise to offset (0,5)
   ..controls (10,10) and (10,10)
 ..(0,0) % counterclockwise to offset (0,-5)
 & cycle

]
Edge structure at line 182 (just shipped out):
Filled contour :
(10,5)..controls (10,5) and (10,5)
 ..cycle
mitered joins limited 10 with pen
(-2.5,-0.5)
 .. (-1,-2)
 .. (0,-2.5)
 .. (1,-2)
 .. (2,-1)
 .. (2.5,0.5)
 .. (1,2)
 .. (0,2.5)
 .. (-1,2)
 .. (-2,1)
 .. cycle
Filled pen stroke :
(5,10)..controls (5,11.66667) and (5,13.33333)
 ..(5,15)
square ends, mitered joins limited 10 with pen
(-2.5,-0.5)
 .. (-1,-2)
 .. (0,-2.5)
 .. (1,-2)
 .. (2,-1)
 .. (2.5,0.5)
 .. (1,2)
 .. (0,2.5)
 .. (-1,2)
 .. (-2,1)
 .. cycle
Filled pen stroke :
(0,0)..controls (10,10) and (10,10)
 ..(10,10)..controls (6.66667,13.33333) and (3.33333,16.66667)
 ..(0,20)..controls (10,16.66667) and (20,13.33333)
 ..(30,10)
square ends, mitered joins limited 10 with pen
(0,-5)
 .. (0,5)
 .. cycle
End edges

{for}
{sqrt(-1)}
! Square root of -1 has been replaced by 0.
l.183 for @=angle(sqrt$,
                        mlog$):charext:=uniformdeviate$;charht:=2048;
Since I don't take square roots of negative numbers,
I'm zeroing this one. Proceed, with fingers crossed.

{mlog(-1)}
! Logarithm of -1 has been replaced by 0.
l.183 for @=angle(sqrt$,mlog$)
                              :charext:=uniformdeviate$;charht:=2048;
Since I don't take logs of non-positive numbers,
I'm zeroing this one. Proceed, with fingers crossed.

{angle((0,0))}
! angle(0,0) is taken as zero.
l.183 for @=angle(sqrt$,mlog$):
                               charext:=uniformdeviate$;charht:=2048;
The `angle' between two identical points is undefined.
I'm zeroing this one. Proceed, with fingers crossed.

Runaway loop?
charext:=uniformdeviate$;charht:=2048;addto.e3contour(0,-100)..tension ETC.
! Forbidden token found while scanning the text of a for loop.
<inserted text> 
                endfor
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
I suspect you have forgotten an `endfor',
causing me to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.

{loop value=0}
{uniformdeviate(-1)}
{charext:=-0.00507}
{charht:=2048}
{addto}
{-(100)}
{-(99)}
Path at line 206, before choices:
(0,-100)..tension 500
 ..(100,-99)..tension 3000
 ..cycle

Path at line 206, after choices:
(0,-100)..controls (-36.92659,-103.51175) and (136.98943,-101.7726)
 ..(100,-99)..controls (99.9889,-98.99916) and (0.01108,-99.99895)
 ..cycle

{tracingoutput:=0}
{special}
{charcode:=151}
{shipout}
! Enormous charht has been reduced.
<to be read again> 
                   ;
<for(0)> ...special"bye";charcode:=151;shipout.e3;
                                                  interim.char99="c";true=fa...
<inserted text> endfor
                      
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
Font metric dimensions must be less than 2048pt.

[151
Envelope spec at line 206:
(0,0) % beginning with offset (-1,-2)
   ..controls (0.04324,-0.04068) and (0.08835,-0.07198)
 ..(0.13504,-0.09532) % counterclockwise to offset (0,-2.5)
   ..controls (0.64822,-0.3519) and (1.35178,0.35191)
 ..(1.86496,0.09532) % clockwise to offset (-1,-2)
   ..controls (1.91165,0.07198) and (1.95676,0.04068)
 ..(2,0)
   ..controls (2.00706,-0.00664) and (2.01395,-0.01332)
 ..(2.02066,-0.02003) % clockwise to offset (-2.5,-0.5)
   ..controls (2.2398,-0.23918) and (2.27596,-0.49583)
 ..(2.2023,-0.71681) % clockwise to offset (-2,1)
   ..controls (2.16501,-0.82869) and (2.09958,-0.93141)
 ..(2.01549,-1.0155) % clockwise to offset (-1,2)
   ..controls (1.95589,-1.0751) and (1.88692,-1.12534)
 ..(1.81195,-1.16281) % clockwise to offset (0,2.5)
   ..controls (1.63397,-1.2518) and (1.42226,-1.26889)
 ..(1.22209,-1.16881) % clockwise to offset (1,2)
   ..controls (1.15273,-1.13412) and (1.08475,-1.08539)
 ..(1.02003,-1.02068) % clockwise to offset (2.5,0.5)
   ..controls (1.01332,-1.01396) and (1.00664,-1.00706)
 ..(1,-1)
   ..controls (0.78389,-0.77028) and (0.83284,-0.488)
 ..(0.93013,-0.19617) % clockwise to offset (2,-1)
   ..controls (0.97351,-0.06602) and (1.0265,0.06602)
 ..(1.06989,0.19617) % counterclockwise to offset (2.5,0.5)
   ..controls (1.16716,0.488) and (1.21611,0.77028)
 ..(1,1)
   ..controls (0.99336,1.00706) and (0.98668,1.01395)
 ..(0.97997,1.02066) % counterclockwise to offset (1,2)
   ..controls (0.91525,1.08537) and (0.84727,1.13412)
 ..(0.77791,1.16881) % counterclockwise to offset (0,2.5)
   ..controls (0.57773,1.26889) and (0.36601,1.2518)
 ..(0.18803,1.16281) % counterclockwise to offset (-1,2)
   ..controls (0.11307,1.12534) and (0.0441,1.0751)
 ..(-0.0155,1.0155) % counterclockwise to offset (-2,1)
   ..controls (-0.09958,0.93141) and (-0.16502,0.82869)
 ..(-0.20232,0.71681) % counterclockwise to offset (-2.5,-0.5)
   ..controls (-0.27597,0.49583) and (-0.2398,0.2392)
 ..(-0.02066,0.02003) % counterclockwise to offset (-1,-2)
   ..controls (-0.01395,0.01332) and (-0.00706,0.00664)
 ..(0,0)
 & cycle

Envelope spec at line 206:
(0,0) % beginning with offset (1,2)
   ..controls (-0.00706,0.00664) and (-0.01395,0.01332)
 ..(-0.02066,0.02003) % clockwise to offset (2.5,0.5)
   ..controls (-0.2398,0.23918) and (-0.27596,0.49583)
 ..(-0.2023,0.71681) % clockwise to offset (2,-1)
   ..controls (-0.16501,0.82869) and (-0.09958,0.93141)
 ..(-0.01549,1.0155) % clockwise to offset (1,-2)
   ..controls (0.04411,1.0751) and (0.11308,1.12534)
 ..(0.18805,1.16281) % clockwise to offset (0,-2.5)
   ..controls (0.36603,1.2518) and (0.57774,1.26889)
 ..(0.77791,1.16881) % clockwise to offset (-1,-2)
   ..controls (0.84727,1.13412) and (0.91525,1.08539)
 ..(0.97997,1.02068) % clockwise to offset (-2.5,-0.5)
   ..controls (0.98668,1.01396) and (0.99336,1.00706)
 ..(1,1)
   ..controls (1.21611,0.77028) and (1.16716,0.488)
 ..(1.06987,0.19617) % clockwise to offset (-2,1)
   ..controls (1.02649,0.06602) and (0.9735,-0.06602)
 ..(0.93011,-0.19617) % counterclockwise to offset (-2.5,-0.5)
   ..controls (0.83284,-0.488) and (0.78389,-0.77028)
 ..(1,-1)
   ..controls (1.00664,-1.00706) and (1.01332,-1.01395)
 ..(1.02003,-1.02066) % counterclockwise to offset (-1,-2)
   ..controls (1.08475,-1.08537) and (1.15273,-1.13412)
 ..(1.22209,-1.16881) % counterclockwise to offset (0,-2.5)
   ..controls (1.42227,-1.26889) and (1.63399,-1.2518)
 ..(1.81197,-1.16281) % counterclockwise to offset (1,-2)
   ..controls (1.88693,-1.12534) and (1.9559,-1.0751)
 ..(2.0155,-1.0155) % counterclockwise to offset (2,-1)
   ..controls (2.09958,-0.93141) and (2.16502,-0.82869)
 ..(2.20232,-0.71681) % counterclockwise to offset (2.5,0.5)
   ..controls (2.27597,-0.49583) and (2.2398,-0.2392)
 ..(2.02066,-0.02003) % counterclockwise to offset (1,2)
   ..controls (2.01395,-0.01332) and (2.00706,-0.00664)
 ..(2,0)
   ..controls (1.95676,0.04068) and (1.91165,0.07198)
 ..(1.86496,0.09532) % counterclockwise to offset (0,2.5)
   ..controls (1.35178,0.3519) and (0.64822,-0.35191)
 ..(0.13504,-0.09532) % clockwise to offset (1,2)
   ..controls (0.08835,-0.07198) and (0.04324,-0.04068)
 ..(0,0)
 & cycle

Envelope spec at line 206:
(0,0) % beginning with offset (1,-2)
   ..controls (0.17854,0.17854) and (0.38548,0.31178)
 ..(0.60698,0.42253) % clockwise to offset (0,-2.5)
   ..controls (1.04999,0.64404) and (1.55122,0.77562)
 ..(1.99995,1) % counterclockwise to offset (1,-2)
   ..controls (1.99997,1) and (1.99998,1)
 ..(2,1) % counterclockwise to offset (-1,2)
   ..controls (1.99998,1) and (1.99997,1)
 ..(1.99995,1) % clockwise to offset (0,2.5)
   ..controls (1.55122,0.77562) and (1.05,0.64404)
 ..(0.607,0.42253) % counterclockwise to offset (-1,2)
   ..controls (0.3855,0.31178) and (0.17856,0.17856)
 ..(0,0) % counterclockwise to offset (1,-2)
 & cycle

Envelope spec at line 206:
(0,257) % beginning with offset (-1,-2)
   ..controls (0.04324,256.95932) and (0.08835,256.92802)
 ..(0.13504,256.90468) % counterclockwise to offset (0,-2.5)
   ..controls (0.64822,256.6481) and (1.35178,257.35191)
 ..(1.86496,257.09532) % clockwise to offset (-1,-2)
   ..controls (1.91165,257.07198) and (1.95676,257.04068)
 ..(2,257)
   ..controls (2.00706,256.99336) and (2.01395,256.98668)
 ..(2.02066,256.97997) % clockwise to offset (-2.5,-0.5)
   ..controls (2.2398,256.76082) and (2.27596,256.50417)
 ..(2.2023,256.28319) % clockwise to offset (-2,1)
   ..controls (2.16501,256.17131) and (2.09958,256.06859)
 ..(2.01549,255.9845) % clockwise to offset (-1,2)
   ..controls (1.95589,255.9249) and (1.88692,255.87466)
 ..(1.81195,255.83719) % clockwise to offset (0,2.5)
   ..controls (1.63397,255.7482) and (1.42226,255.73111)
 ..(1.22209,255.83119) % clockwise to offset (1,2)
   ..controls (1.15273,255.86588) and (1.08475,255.91461)
 ..(1.02003,255.97932) % clockwise to offset (2.5,0.5)
   ..controls (1.01332,255.98604) and (1.00664,255.99294)
 ..(1,256)
   ..controls (0.78389,256.22972) and (0.83284,256.512)
 ..(0.93013,256.80383) % clockwise to offset (2,-1)
   ..controls (0.97351,256.93398) and (1.0265,257.06602)
 ..(1.06989,257.19617) % counterclockwise to offset (2.5,0.5)
   ..controls (1.16716,257.488) and (1.21611,257.77028)
 ..(1,258)
   ..controls (0.99336,258.00706) and (0.98668,258.01395)
 ..(0.97997,258.02066) % counterclockwise to offset (1,2)
   ..controls (0.91525,258.08537) and (0.84727,258.13412)
 ..(0.77791,258.16881) % counterclockwise to offset (0,2.5)
   ..controls (0.57773,258.26889) and (0.36601,258.2518)
 ..(0.18803,258.16281) % counterclockwise to offset (-1,2)
   ..controls (0.11307,258.12534) and (0.0441,258.0751)
 ..(-0.0155,258.0155) % counterclockwise to offset (-2,1)
   ..controls (-0.09958,257.93141) and (-0.16502,257.82869)
 ..(-0.20232,257.71681) % counterclockwise to offset (-2.5,-0.5)
   ..controls (-0.27597,257.49583) and (-0.2398,257.2392)
 ..(-0.02066,257.02003) % counterclockwise to offset (-1,-2)
   ..controls (-0.01395,257.01332) and (-0.00706,257.00664)
 ..(0,257)
 & cycle

Envelope spec at line 206:
(0,257) % beginning with offset (1,2)
   ..controls (-0.00706,257.00664) and (-0.01395,257.01332)
 ..(-0.02066,257.02003) % clockwise to offset (2.5,0.5)
   ..controls (-0.2398,257.23918) and (-0.27596,257.49583)
 ..(-0.2023,257.71681) % clockwise to offset (2,-1)
   ..controls (-0.16501,257.82869) and (-0.09958,257.93141)
 ..(-0.01549,258.0155) % clockwise to offset (1,-2)
   ..controls (0.04411,258.0751) and (0.11308,258.12534)
 ..(0.18805,258.16281) % clockwise to offset (0,-2.5)
   ..controls (0.36603,258.2518) and (0.57774,258.26889)
 ..(0.77791,258.16881) % clockwise to offset (-1,-2)
   ..controls (0.84727,258.13412) and (0.91525,258.08539)
 ..(0.97997,258.02068) % clockwise to offset (-2.5,-0.5)
   ..controls (0.98668,258.01396) and (0.99336,258.00706)
 ..(1,258)
   ..controls (1.21611,257.77028) and (1.16716,257.488)
 ..(1.06987,257.19617) % clockwise to offset (-2,1)
   ..controls (1.02649,257.06602) and (0.9735,256.93398)
 ..(0.93011,256.80383) % counterclockwise to offset (-2.5,-0.5)
   ..controls (0.83284,256.512) and (0.78389,256.22972)
 ..(1,256)
   ..controls (1.00664,255.99294) and (1.01332,255.98605)
 ..(1.02003,255.97934) % counterclockwise to offset (-1,-2)
   ..controls (1.08475,255.91463) and (1.15273,255.86588)
 ..(1.22209,255.83119) % counterclockwise to offset (0,-2.5)
   ..controls (1.42227,255.73111) and (1.63399,255.7482)
 ..(1.81197,255.83719) % counterclockwise to offset (1,-2)
   ..controls (1.88693,255.87466) and (1.9559,255.9249)
 ..(2.0155,255.9845) % counterclockwise to offset (2,-1)
   ..controls (2.09958,256.06859) and (2.16502,256.17131)
 ..(2.20232,256.28319) % counterclockwise to offset (2.5,0.5)
   ..controls (2.27597,256.50417) and (2.2398,256.7608)
 ..(2.02066,256.97997) % counterclockwise to offset (1,2)
   ..controls (2.01395,256.98668) and (2.00706,256.99336)
 ..(2,257)
   ..controls (1.95676,257.04068) and (1.91165,257.07198)
 ..(1.86496,257.09532) % counterclockwise to offset (0,2.5)
   ..controls (1.35178,257.3519) and (0.64822,256.64809)
 ..(0.13504,256.90468) % clockwise to offset (1,2)
   ..controls (0.08835,256.92802) and (0.04324,256.95932)
 ..(0,257)
 & cycle

Envelope spec at line 206:
(0,257) % beginning with offset (1,-2)
   ..controls (0.17854,257.17854) and (0.38548,257.31178)
 ..(0.60698,257.42253) % clockwise to offset (0,-2.5)
   ..controls (1.04999,257.64404) and (1.55122,257.77562)
 ..(1.99995,258) % counterclockwise to offset (1,-2)
   ..controls (1.99997,258) and (1.99998,258)
 ..(2,258) % counterclockwise to offset (-1,2)
   ..controls (1.99998,258) and (1.99997,258)
 ..(1.99995,258) % clockwise to offset (0,2.5)
   ..controls (1.55122,257.77562) and (1.05,257.64404)
 ..(0.607,257.42253) % counterclockwise to offset (-1,2)
   ..controls (0.3855,257.31178) and (0.17856,257.17856)
 ..(0,257) % counterclockwise to offset (1,-2)
 & cycle

]
{interim}
! The token `char' isn't an internal quantity.
<to be read again> 
                   char
<for(0)> ...;charcode:=151;shipout.e3;interim.char
                                                  99="c";true=false;show.pen...
<inserted text> endfor
                      
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
Something like `tracingonline' should follow `interim'.

{char(99)}
{("c")=("c")}
! Redundant equation.
<to be read again> 
                   ;
<for(0)> ...de:=151;shipout.e3;interim.char99="c";
                                                  true=false;show.penoffset(...
<inserted text> endfor
                      
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
I already knew that this equation was true.
But perhaps no harm has been done; let's continue.

{true}
{false}
{(true)=(false)}
! Inconsistent equation.
<to be read again> 
                   ;
<for(0)> ...pout.e3;interim.char99="c";true=false;
                                                  show.penoffset(2,1)of.qq;s...
<inserted text> endfor
                      
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
The equation I just read contradicts what was said before.
But don't worry; continue and I'll just ignore it.

{show}
{((2,1))penoffset(pen)}
>> (0,-2.5)
{showstats}
Memory usage 3022&607 (167 still untouched)
String usage 69&244 (2&7446 now untouched)
{let}
{showtoken}
> f=tag
{showstats}
Memory usage 2982&536 (167 still untouched)
String usage 69&244 (2&7446 now untouched)
{nullpicture}
{(unknown picture e17)=(picture)}
{addto}
{-(4)}
{-(5)}
Path at line 206, before choices:
(-4,-5){curl 1}
 ..{curl 1}(6,5)

Path at line 206, after choices:
(-4,-5)..controls (-0.66667,-1.66667) and (2.66667,1.66667)
 ..(6,5)

{setbounds}
{pencircle}
{(pen)scaled(9)}
{makepath(pen)}
{show}
{llcorner(picture)}
>> (-4.5,-4.5)
{urcorner(picture)}
>> (4.5,4.5)
{addto}
{-(5)}
{-(5)}
{(picture)shifted((-5,-5))}
{show}
{llcorner(picture)}
>> (-9.5,-9.5)
{truecorners:=2}
{show}
{llcorner(picture)}
>> (-9,-10)
{-(1)}
{truecorners:=-1}
{setbounds}
{pencircle}
{(pen)xscaled(19)}
{makepath(pen)}
{addto}
{-(5)}
{-(10)}
{(picture)shifted((-5,-10))}
{show}
{llcorner(picture)}
>> (-14.5,-10.5)
>> Edge structure at line 206:
Filled contour :
(10,5)..controls (10,5) and (10,5)
 ..cycle
mitered joins limited 10 with pen
(-2.5,-0.5)
 .. (-1,-2)
 .. (0,-2.5)
 .. (1,-2)
 .. (2,-1)
 .. (2.5,0.5)
 .. (1,2)
 .. (0,2.5)
 .. (-1,2)
 .. (-2,1)
 .. cycle
Filled pen stroke :
(5,10)..controls (5,11.66667) and (5,13.33333)
 ..(5,15)
square ends, mitered joins limited 10 with pen
(-2.5,-0.5)
 .. (-1,-2)
 .. (0,-2.5)
 .. (1,-2)
 .. (2,-1)
 .. (2.5,0.5)
 .. (1,2)
 .. (0,2.5)
 .. (-1,2)
 .. (-2,1)
 .. cycle
Filled pen stroke :
(0,0)..controls (10,10) and (10,10)
 ..(10,10)..controls (6.66667,13.33333) and (3.33333,16.66667)
 ..(0,20)..controls (10,16.66667) and (20,13.33333)
 ..(30,10)
square ends, mitered joins limited 10 with pen
(0,-5)
 .. (0,5)
 .. cycle
setbounds path:
(-0.5,-5)..controls (-0.5,-3.80652) and (-0.9741,-2.66194)
 ..(-1.81802,-1.81802)..controls (-2.66194,-0.9741) and (-3.80652,-0.5)
 ..(-5,-0.5)..controls (-6.19348,-0.5) and (-7.33806,-0.9741)
 ..(-8.18198,-1.81802)..controls (-9.0259,-2.66194) and (-9.5,-3.80652)
 ..(-9.5,-5)..controls (-9.5,-6.19348) and (-9.0259,-7.33806)
 ..(-8.18198,-8.18198)..controls (-7.33806,-9.0259) and (-6.19348,-9.5)
 ..(-5,-9.5)..controls (-3.80652,-9.5) and (-2.66194,-9.0259)
 ..(-1.81802,-8.18198)..controls (-0.9741,-7.33806) and (-0.5,-6.19348)
 ..cycle
Filled pen stroke :
(-9,-10)..controls (-5.66667,-6.66667) and (-2.33333,-3.33333)
 ..(1,0)
square ends, mitered joins limited 10 with pen
pencircle transformed (0,0,0,0,0,0)
end of setbounds
setbounds path:
(4.5,-10)..controls (4.5,-9.86739) and (3.49911,-9.74022)
 ..(1.71751,-9.64645)..controls (-0.06409,-9.55269) and (-2.48044,-9.5)
 ..(-5,-9.5)..controls (-7.51956,-9.5) and (-9.93591,-9.55269)
 ..(-11.71751,-9.64645)..controls (-13.49911,-9.74022) and (-14.5,-9.86739)
 ..(-14.5,-10)..controls (-14.5,-10.13261) and (-13.49911,-10.25978)
 ..(-11.71751,-10.35355)..controls (-9.93591,-10.44731) and (-7.51956,-10.5)
 ..(-5,-10.5)..controls (-2.48044,-10.5) and (-0.06409,-10.44731)
 ..(1.71751,-10.35355)..controls (3.49911,-10.25978) and (4.5,-10.13261)
 ..cycle
setbounds path:
(-0.5,-10)..controls (-0.5,-8.80652) and (-0.9741,-7.66194)
 ..(-1.81802,-6.81802)..controls (-2.66194,-5.9741) and (-3.80652,-5.5)
 ..(-5,-5.5)..controls (-6.19348,-5.5) and (-7.33806,-5.9741)
 ..(-8.18198,-6.81802)..controls (-9.0259,-7.66194) and (-9.5,-8.80652)
 ..(-9.5,-10)..controls (-9.5,-11.19348) and (-9.0259,-12.33806)
 ..(-8.18198,-13.18198)..controls (-7.33806,-14.0259) and (-6.19348,-14.5)
 ..(-5,-14.5)..controls (-3.80652,-14.5) and (-2.66194,-14.0259)
 ..(-1.81802,-13.18198)..controls (-0.9741,-12.33806) and (-0.5,-11.19348)
 ..cycle
Filled pen stroke :
(-9,-15)..controls (-5.66667,-11.66667) and (-2.33333,-8.33333)
 ..(1,-5)
square ends, mitered joins limited 10 with pen
pencircle transformed (0,0,0,0,0,0)
end of setbounds
end of setbounds
End edges

{pencircle}
{urcorner(pen)}
>> (0.5,0.5)
{nullpicture}
{e5:=picture}
{path}
{pencircle}
{(pen)scaled(100)}
{makepath(pen)}
{(unknown path ap1)=(path)}
{-(14)}
{-(7)}
Path at line 206, before choices:
(2,-14)..controls (14,-7) and (14,7)
 ..(2,14)

Path at line 206, after choices:
(2,-14)..controls (14,-7) and (14,7)
 ..(2,14)

{(unknown path ap2)=(path)}
{-(50)}
{-(50)}
Path at line 206, before choices:
(-50,0)..controls (-50,0) and (-50,0)
 ..(50,0)

Path at line 206, after choices:
(-50,0)..controls (-50,0) and (-50,0)
 ..(50,0)

{(unknown path ap3)=(path)}
{show}
{(path)scaled(1000)}
{arclength(path)}
! Arithmetic overflow.
<for(0)> ....(50,0);show.arclength(ap2scaled1000),
                                                  arclength(ap3scaled200);sh...
<inserted text> endfor
                      
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
Uh, oh. A little while ago one of the quantities that I was
computing got too large, so I'm afraid your answers will be
somewhat askew. You'll probably have to adopt different
tactics next time. But I shall try to carry on anyway.

>> 32767.99998
{(path)scaled(200)}
{arclength(path)}
! Arithmetic overflow.
<for(0)> ...p2scaled1000),arclength(ap3scaled200);
                                                  show.arclength.ap1,arcleng...
<inserted text> endfor
                      
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
Uh, oh. A little while ago one of the quantities that I was
computing got too large, so I'm afraid your answers will be
somewhat askew. You'll probably have to adopt different
tactics next time. But I shall try to carry on anyway.

>> 32767.99998
{show}
{arclength(path)}
>> 314.15979
{arclength(path)}
>> 34.88
{show}
{(6.4)arctime(path)}
>> 0.4
{(37)arctime(path)}
{((0,0.7179))subpath(path)}
{arclength(path)}
>> 36.99847
{show}
{-(1000)}
{(-1000)arctime(path)}
>> -25.46454
{-(3)}
{(-3)arctime(path)}
>> 0
{(path)scaled(0.00002)}
{(314)arctime(path)}
! Arithmetic overflow.
<for(0)> ...of.ap2,arctime314of(ap1scaled0.00002);
                                                  [[clear(tracingcommands);c...
<inserted text> endfor
                      
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
Uh, oh. A little while ago one of the quantities that I was
computing got too large, so I'm afraid your answers will be
somewhat askew. You'll probably have to adopt different
tactics next time. But I shall try to carry on anyway.

>> 32767.99998
{begingroup}

clear(TEXT0)->interim(TEXT0):=$
(TEXT0)<-tracingcommands
{interim}
{tracingcommands:=-1}
! Enormous charht has been reduced.
<to be read again> 
                   ;
<for(0)> ...=ASCII.char269-13;shipout.nullpicture;
                                                  "careful"for.for=(EXPR0)st...
<inserted text> endfor
                      
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
Font metric dimensions must be less than 2048pt.

[0]
careful METAFONT METAFONT METAFONT METAFONT METAFONT METAFONT METAFONT METAFONT
 METAFONT METAFONT METAFONT METAFONT METAFONT METAFONT METAFONT METAFONT METAFO
NT METAFONT METAFONT METAFONT METAFONT METAFONT METAFONT METAFONT METAFONT META
FONT METAFONT
{restoring tracingcommands=2.1}
{endgroup}
{scrollmode}

{char(31)}
{("hello again")&("^^_")}
hello again^^_
{save}
### -0.45p3~=-alpha+7
{def}
{begingroup}
{tracingonline:=1}

{tracingonline:=1}

f<expr>->let)=];let[=(;show._
(EXPR0)<-xx
{let}
{let}
{show}
>> _1
{showdependencies}
xpart '=-ypart '-0.66667
{qq:=pen}
{showstats}
Memory usage 2372&536 (167 still untouched)
String usage 69&244 (1&6758 now untouched)
{[repeat the loop]}
>> bye
! Isolated expression.
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
I couldn't find an `=' or `:=' after the
expression that is shown above this error message,
so I guess I'll just ignore it and carry on.

! Extra tokens will be flushed.
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
I've just read as much of that statement as I could fathom,
so a semicolon should have been next. It's very puzzling...
but I'll try to get myself back together, by ignoring
everything up to the next `;'. Please insert a semicolon
now in front of anything that you don't want me to delete.
(See Chapter 27 of The METAFONTbook for an example.)

! Forbidden token found while scanning to the end of the statement.
<inserted text> 
                ;
<to be read again> 
                   endtext
l.206 ...pendencies; qq:=q; showstats; bye endtext
                                                  
A previous error seems to have propagated,
causing me to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.

{show}
)
*end
! An expression can't begin with `end'.
<inserted text> 
                0
<to be read again> 
                   end
<*> end
       
I'm afraid I need some sort of value in order to continue,
so I've tentatively inserted `0'. You may want to
delete this zero and insert something else;
see Chapter 27 of The METAFONTbook for an example.

>> 0
! A group begun on line 206 never ended.
<to be read again> 
                   end
<*> end
       
I saw a `begingroup' back there that hasn't been matched
by `endgroup'. So I've inserted `endgroup' now.

{endgroup}
(end occurred when else on line 88 was incomplete)
(end occurred when if on line 37 was incomplete)
(end occurred when elseif on line 22 was incomplete)
 
Here is how much of MetaPost's memory you used:
 71 strings out of 73
 735 string characters out of 7698
 4834 words of memory out of 5000
 366 symbolic tokens out of 16384
 8i,44n,8p,162b,2f stack positions out of 300i,84n,1500p,200b,25f
 13 string compactions (moved 5119 characters, 856 strings)
(illegal design size has been changed to 128pt)
(some chardp values had to be adjusted by as much as 0.5pt)
(local label 0:: was missing)
(local label 5:: was missing)
(4 font metric dimensions had to be decreased)
 
(You used 3w,2h,16d,2i,312l,301k,1e,10p metric file positions)
Font metrics written on trap.tfm.
25 output files written: trap.ps .. trap.200