forked from KolibriOS/kolibrios
b5e7b54c7a
git-svn-id: svn://kolibrios.org@6429 a494cfbc-eb01-0410-851d-a64ba20cac60
123 lines
3.3 KiB
C
123 lines
3.3 KiB
C
/* example from http://barnyard.syr.edu/quickies/hanoi.c */
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/* hanoi.c: solves the tower of hanoi problem. (Programming exercise.) */
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/* By Terry R. McConnell (12/2/97) */
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/* Compile: cc -o hanoi hanoi.c */
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/* This program does no error checking. But then, if it's right,
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it's right ... right ? */
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/* The original towers of hanoi problem seems to have been originally posed
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by one M. Claus in 1883. There is a popular legend that goes along with
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it that has been often repeated and paraphrased. It goes something like this:
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In the great temple at Benares there are 3 golden spikes. On one of them,
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God placed 64 disks increasing in size from bottom to top, at the beginning
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of time. Since then, and to this day, the priest on duty constantly transfers
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disks, one at a time, in such a way that no larger disk is ever put on top
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of a smaller one. When the disks have been transferred entirely to another
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spike the Universe will come to an end in a large thunderclap.
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This paraphrases the original legend due to DeParville, La Nature, Paris 1884,
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Part I, 285-286. For this and further information see: Mathematical
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Recreations & Essays, W.W. Rouse Ball, MacMillan, NewYork, 11th Ed. 1967,
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303-305.
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*
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*
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#define TRUE 1
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#define FALSE 0
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/* This is the number of "disks" on tower A initially. Taken to be 64 in the
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* legend. The number of moves required, in general, is 2^N - 1. For N = 64,
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* this is 18,446,744,073,709,551,615 */
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#define N 4
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/* These are the three towers. For example if the state of A is 0,1,3,4, that
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* means that there are three discs on A of sizes 1, 3, and 4. (Think of right
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* as being the "down" direction.) */
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int A[N], B[N], C[N];
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void Hanoi(int,int*,int*,int*);
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/* Print the current configuration of A, B, and C to the screen */
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void PrintAll()
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{
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int i;
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printf("A: ");
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for(i=0;i<N;i++)printf(" %d ",A[i]);
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printf("\n");
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printf("B: ");
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for(i=0;i<N;i++)printf(" %d ",B[i]);
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printf("\n");
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printf("C: ");
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for(i=0;i<N;i++)printf(" %d ",C[i]);
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printf("\n");
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printf("------------------------------------------\n");
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return;
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}
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/* Move the leftmost nonzero element of source to dest, leave behind 0. */
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/* Returns the value moved (not used.) */
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int Move(int *source, int *dest)
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{
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int i = 0, j = 0;
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while (i<N && (source[i])==0) i++;
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while (j<N && (dest[j])==0) j++;
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dest[j-1] = source[i];
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source[i] = 0;
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PrintAll(); /* Print configuration after each move. */
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return dest[j-1];
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}
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/* Moves first n nonzero numbers from source to dest using the rules of Hanoi.
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Calls itself recursively.
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*/
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void Hanoi(int n,int *source, int *dest, int *spare)
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{
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int i;
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if(n==1){
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Move(source,dest);
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return;
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}
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Hanoi(n-1,source,spare,dest);
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Move(source,dest);
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Hanoi(n-1,spare,dest,source);
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return;
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}
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int main()
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{
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int i;
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/* initialize the towers */
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for(i=0;i<N;i++)A[i]=i+1;
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for(i=0;i<N;i++)B[i]=0;
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for(i=0;i<N;i++)C[i]=0;
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printf("Solution of Tower of Hanoi Problem with %d Disks\n\n",N);
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/* Print the starting state */
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printf("Starting state:\n");
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PrintAll();
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printf("\n\nSubsequent states:\n\n");
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/* Do it! Use A = Source, B = Destination, C = Spare */
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Hanoi(N,A,B,C);
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return 0;
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}
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/* vim: set expandtab ts=4 sw=3 sts=3 tw=80 :*/
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