when test most simple case one to one, and many to one these two reasonable cases, it run a very long time without exit the program.

when i test with the example in book Neural Network Design, it can output correctly but only for book example.

restart:
with(ExcelTools):
with(ListTools):
with(plots):
with(LinearAlgebra):
zipplus := proc(mm, pp)
return zip((x,y) -> x+y, mm, pp)
end proc:
zipminus := proc(mm, pp)
return zip((x,y) -> x-y, mm, pp)
end proc:
zipstar := proc(mm, pp)
return zip((x,y) -> x*y, mm, pp)
end proc:

metara := proc(pp,meaght,bb,mapp,deep)
if deep > 0 then
 pp2 := metara([seq(pp[i], i = 1 .. nops(pp))],meaght,bb,mapp,deep-1):
 mp := zip((x,y) -> x*y,pp2,meaght):
 mpsam := sum(mp[m],m=1..nops(pp2)):
 mpb := [seq(0, i = 1 .. nops(pp2))]:
 for ii from 1 to nops(bb) do
  mpb[ii] := evalf(mpsam + bb[ii]);
 od:
 pa := [seq(0, i = 1 .. nops(pp2))]:
 for ii from 1 to nops(bb) do
  pa[ii] := evalf(mapp[deep](mpb[ii])):
 od:
 return pa:
else
 return pp:
end if:
end proc:

perceptronrule1 := proc(p, t1, meaght1, b1, checksum, mapp)
 meaght3 := meaght1:
 b3 := b1:
 checksum2 := checksum;
 print(p[1]):
 while sum(checksum2[jj], jj=1..nops(p)) <> nops(p) do
  #print("sum(checksum2[jj], jj=1..nops(p))");
  #print(sum(checksum2[jj], jj=1..nops(p)));
  for ii from 1 to nops(p) do
   #print("metara(p[ii],meaght3,b3,mapp,1)");
   #print("p[ii]");
   #print(p[ii]);
   #print("b3");
   #print(b3);
   #print("meaght3");
   #print(meaght3);
   #print(metara(p[ii],meaght3,b3,mapp,1));
   #print("t1[ii]");
   #print(t1[ii]);
   e := zipminus(t1[ii], metara(p[ii],meaght3,b3,mapp,1));
   #print("e");
   #print(e);
   meaght2 := meaght3 + zipstar(e,p[ii]);
   #print("meaght2");
   #print(meaght2);
   #print("meaght3");
   #print(meaght3);
   #print("b3");
   #print(b3);
   b2 := b3 + e;
   #print("b2");
   #print(b2);
   #print("b3");
   #print(b3);
   #print("checksum2");
   #print(checksum2);
   diff1 := zipminus(meaght2, meaght3):
   diff2 := zipminus(b2, b3):
   if sum(diff1[m],m=1..nops(diff1)) = 0 and sum(diff2[m],m=1..nops(diff2)) = 0 then
    checksum2[ii] := 1:
   else
    checksum2[ii] := 0:
    b3 := b2:
    meaght3 := meaght2:
   end if:
  od:
 od:
 return [meaght3, b3, checksum];
end proc:

#Example from book
ppp := [[2,2],[1,-2],[-2,2],[-1,1]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[0,0],[1,1],[0,0],[1,1]]:
mmmeaght1 := [seq(0,ii=1..nops(ppp[1]))]:
bbb1 := [seq(0,ii=1..nops(ppp[1]))]:
emap := [(x) -> if x < 0 then 0 else 1 end if, (x) -> evalf(1/(1+exp(x)))]:
#trace(perceptronrule1);
perceptronrule1(ppp,ttt1,mmmeaght1,bbb1,check1,emap);

#my testing for one to one
#after testing, it loop a very long time and not stop
ppp := [[0,0,0,1],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[0,0,0,1],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
mmmeaght1 := [0,0,0,0]:
bbb1 := [0,0,0,0]:
emap := [(x) -> if x < 0 then 0 else 1 end if, (x) -> evalf(1/(1+exp(x)))]:
#trace(perceptronrule1);
perceptronrule1(ppp,ttt1,mmmeaght1,bbb1,check1,emap);

#my testing for many to one
#after testing, it loop a very long time and not stop
ppp := [[1,1,0,0],[0,1,1,0],[0,1,0,1]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[1,0,0,0],[0,1,0,0],[0,0,0,1]]:
mmmeaght1 := [0,0,0,0]:
bbb1 := [0,0,0,0]:
emap := [(x) -> if x < 0 then 0 else 1 end if, (x) -> evalf(1/(1+exp(x)))]:
#trace(perceptronrule1);
perceptronrule1(ppp,ttt1,mmmeaght1,bbb1,check1,emap);

#my testing for one to many
#after testing, it loop a very long time and not stop
ppp := [[1,0,0,0],[0,1,0,0],[0,0,0,1]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[1,1,0,0],[0,1,1,0],[0,1,0,1]]:
mmmeaght1 := [0,0,0,0]:
bbb1 := [0,0,0,0]:
emap := [(x) -> if x < 0 then 0 else 1 end if, (x) -> evalf(1/(1+exp(x)))]:
#trace(perceptronrule1);
perceptronrule1(ppp,ttt1,mmmeaght1,bbb1,check1,emap);

http://www.maplesoft.com/applications/view.aspx?SID=4229

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