Re: sudoku solver

• To: mathgroup at smc.vnet.net
• Subject: [mg128216] Re: sudoku solver
• From: Sseziwa Mukasa <mukasa at gmail.com>
• Date: Wed, 26 Sep 2012 04:12:06 -0400 (EDT)
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• References: <20120925083815.DE57C6843@smc.vnet.net>

Raise the recursion limit (apparently your algorithm is recursive and requires more than 256 recursive calls).  To raise the recursion limit set $RecursionLimit = Infinity (or some large number). It's probably a good idea to set it back to 256 after running your routine to prevent infinite recursion. On Sep 25, 2012, at 4:38 AM, lapajne.jure at gmail.com wrote: > Hello, I'm trying to solve euler problem 96 - solve 50 sudokus. I have already written a solution in phyton and it works, but I have a problem in Mathematica. > The code works for some sudokus while for others it gives me an error just before it finishes.The error message is: >$RecursionLimit::reclim: Recursion depth of 256 exceeded. >>
> General::stop: Further output of $RecursionLimit::reclim will be suppressed during this calculation. >> > > examples: > \left( > \begin{array}{ccccccccc} > 0 & 0 & 3 & 0 & 2 & 0 & 6 & 0 & 0 \\ > 9 & 0 & 0 & 3 & 0 & 5 & 0 & 0 & 1 \\ > 0 & 0 & 1 & 8 & 0 & 6 & 4 & 0 & 0 \\ > 0 & 0 & 8 & 1 & 0 & 2 & 9 & 0 & 0 \\ > 7 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 \\ > 0 & 0 & 6 & 7 & 0 & 8 & 2 & 0 & 0 \\ > 0 & 0 & 2 & 6 & 0 & 9 & 5 & 0 & 0 \\ > 8 & 0 & 0 & 2 & 0 & 3 & 0 & 0 & 9 \\ > 0 & 0 & 5 & 0 & 1 & 0 & 3 & 0 & 0 > \end{array} > \right) > > but the next one doesn't finish: > \left( > \begin{array}{ccccccccc} > 2 & 0 & 0 & 0 & 8 & 0 & 3 & 0 & 0 \\ > 0 & 6 & 0 & 0 & 7 & 0 & 0 & 8 & 4 \\ > 0 & 3 & 0 & 5 & 0 & 0 & 2 & 0 & 9 \\ > 0 & 0 & 0 & 1 & 0 & 5 & 4 & 0 & 8 \\ > 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ > 4 & 0 & 2 & 7 & 0 & 6 & 0 & 0 & 0 \\ > 3 & 0 & 1 & 0 & 0 & 7 & 0 & 4 & 0 \\ > 7 & 2 & 0 & 0 & 4 & 0 & 0 & 6 & 0 \\ > 0 & 0 & 4 & 0 & 1 & 0 & 0 & 0 & 3 > \end{array} > \right) > > This one gives an error when it tries to solve the last zero (position = {9,8}). The weird thing is that if I try to solve position {9,8} first = and then everything else in normal order, it just stops one earlier = (position {9,7}. > > My code is written below. I checked where it stops and it stops before = the two for loops in sestopaj[..] module. > code: > main function: > > = \text{sestopaj}[\text{delno0$\_$}]\text{:=}\text{Module}[\{\text{delno}== \text{delno0},i=1,j=1,\text{k1},\text{z1},\text{mozne},d=0,\text{i1}= =0,\text{j1}=0,a=0\},\{\text{i1}=0;\text{j1}=0;\text{If}[\text{r= esen}[\text{delno}],a=2;\text{Return}[]];\text{For}[i=1,i\leq = 9,i\text{++},\text{For}[j=1,j\leq = 9,j\text{++},\text{If}[\text{delno}[[i,j]]==0,\text{i1}=i;\text{j1}== j;\text{Break}[]];];\text{If}[\text{i1}\neq 0 \&\&\text{j1}\neq = 0,\text{Break}[]];];\text{mozne}=\text{moznecifre}[\text{i1},\text{j1},\= text{delno}];d=\text{Length}[\text{mozne}];\text{For}[\text{k1}=1,\tex= t{k1}\leq = 1d,\text{k1}\text{++},\text{delno}[[\text{i1},\text{j1}]]=\text{mozne}[[= \text{k1}]];\text{Print}[\text{delno}];\text{z1}=\text{sestopaj}[\text{d= elno}];\text{If} [\text{z1}\neq \text{None},a=1;\text{Return}[] ]; = ];\text{delno}[[\text{i1},\text{j1}]]=0;\};\text{If}[a==2,\text{Retu= rn}[\text{delno}]];\text{If}[a==1,\text{Return}[\text{z1}],\text{Retur= n}[\text{None}]];] > > auxiliary functions: > > = \text{cifre}[\text{a0$\_$},\text{b0$\_$},\text{delno3$\_$}]\text{:=}\tex= t{Module}[\{a=\text{a0},b=\text{b0},\text{vse},c,e,l,o\},\{c=1;e=1= ;\text{If}[a>10 = \|b>10,\text{Print}[a,b]];\text{vse}=\{1,2,3,4,5,6,7,8,9\};\text{For}[c== 1,c<10,c\text{++},\text{If}[\text{delno3}[[c,b]]\neq 0 = ,\text{vse}=\text{Complement}[\text{vse},\{\text{delno3}[[c,b]] \}] = ];\text{If}[\text{delno3}[[a,c]]\neq 0, = \text{vse}=\text{Complement}[\text{vse},\{\text{delno3}[[a,c]]\} ] = ];];e=\text{IntegerPart}[(a-1)/3]*3+1;o=\text{IntegerPart}[(b-1)/3]*3+= 1;\text{For}[c=e,c<e+3,c\text{++},\text{For}[l=o,l<o+3,l\text{++},\tex= t{vse}=\text{Complement}[\text{vse},\{\text{delno3}[[c,l]]\} ] = ;]]\};\text{vse}]\text{moznecifre}[\text{f0$\_$},\text{g0$\_$},\text{delno= 1$\_$}]\text{:=}\text{Which}[\text{delno1}[[\text{f0},\text{g0}]]\neq = 0,\{\},\text{delno1}[[\text{f0},\text{g0}]]==0, = \text{cifre}[\text{f0},\text{g0},\text{delno1}]]\text{resen}[\text{delno2$=
\_\$}]\text{:=}\text{Which}[\text{MemberQ}[\text{Flatten}[\text{delno2}],=
0
> =
],\text{False},\text{MemberQ}[\text{Flatten}[\text{delno2}],0]==\text{=
False},\text{True}];
>
> I have tried changing a lot of things with no success. I'll appreciate =
any help. Thank you.
>



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