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- /* Non-recursive solution to Towers of Hanoi */
-
- void main();
-
- #include <stdio.h>
- #include <system.h>
-
- #define width (rings+1)
-
- void main()
- {
- int rings, last, next, x, z[500], s[3];
-
- printf("how many rings? "); scanf("%d",&rings);
-
- for(x=1; x<=rings; x++) /* put rings on first peg */
- z[x]=width-x;
- for(x=0; x<=2*width; x+=width) /* set base for each peg */
- z[x]=1000;
-
- /* if even number of rings, put first ring on second peg; if odd, on third */
-
- if(rings%2==0)
- {
- last=1; s[2]=0; s[1]=1;
- z[width+1]=z[rings];
- }
- else
- {
- last=2; s[1]=0; s[2]=1;
- z[2*width+1]=z[rings];
- }
-
- printf("from 1 to %d\n",last+1); s[0]=rings-1;
-
- while(s[0]+s[1]) /* while first and second pegs aren't empty */
- {
- /* next ring to move is smaller of rings on the two pegs not moved onto last */
-
- if(last==0) next=z[width+s[1]]<z[2*width+s[2]]?1:2;
- if(last==1) next=z[s[0]]<z[2*width+s[2]]?0:2;
- if(last==2) next=z[s[0]]<z[width+s[1]]?0:1;
-
- /* top ring of 'to' peg must be larger and an even 'distance' away */
-
- if((z[next*width+s[next]]>z[last*width+s[last]])||
- ((z[last*width+s[last]]-z[next*width+s[next]])%2==0)) last=3-next-last;
-
- printf("from %d to %d\n",next+1,last+1);
-
- s[next]=s[next]-1; s[last]=s[last]+1; /* move from 'next' to 'last' peg */
- z[last*width+s[last]]=z[next*width+s[next]+1];
- }
- }
-