1 files changed, 0 insertions, 122 deletions
diff --git a/tinyc/tests/tests2/30_hanoi.c b/tinyc/tests/tests2/30_hanoi.c
deleted file mode 100644
index 7c0893b1a..000000000
--- a/tinyc/tests/tests2/30_hanoi.c
+++ /dev/null
@@ -1,122 +0,0 @@
-/* example from http://barnyard.syr.edu/quickies/hanoi.c */
-
-/* hanoi.c: solves the tower of hanoi problem. (Programming exercise.) */
-/* By Terry R. McConnell (12/2/97) */
-/* Compile: cc -o hanoi hanoi.c */
-
-/* This program does no error checking. But then, if it's right, 
-   it's right ... right ? */
-
-
-/* The original towers of hanoi problem seems to have been originally posed
-   by one M. Claus in 1883. There is a popular legend that goes along with
-   it that has been often repeated and paraphrased. It goes something like this:
-   In the great temple at Benares there are 3 golden spikes. On one of them,
-   God placed 64 disks increasing in size from bottom to top, at the beginning
-   of time. Since then, and to this day, the priest on duty constantly transfers
-   disks, one at a time, in such a way that no larger disk is ever put on top
-   of a smaller one. When the disks have been transferred entirely to another
-   spike the Universe will come to an end in a large thunderclap.
-
-   This paraphrases the original legend due to DeParville, La Nature, Paris 1884,
-   Part I, 285-286. For this and further information see: Mathematical 
-   Recreations & Essays, W.W. Rouse Ball, MacMillan, NewYork, 11th Ed. 1967,
-   303-305.
- *
- *
- */
-
-#include <stdio.h>
-#include <stdlib.h>
-
-#define TRUE 1
-#define FALSE 0
-
-/* This is the number of "disks" on tower A initially. Taken to be 64 in the
- * legend. The number of moves required, in general, is 2^N - 1. For N = 64,
- * this is 18,446,744,073,709,551,615 */
-#define N 4
-
-/* These are the three towers. For example if the state of A is 0,1,3,4, that
- * means that there are three discs on A of sizes 1, 3, and 4. (Think of right
- * as being the "down" direction.) */
-int A[N], B[N], C[N]; 
-
-void Hanoi(int,int*,int*,int*);
-
-/* Print the current configuration of A, B, and C to the screen */
-void PrintAll()
-{
-   int i;
-
-   printf("A: ");
-   for(i=0;i<N;i++)printf(" %d ",A[i]);
-   printf("\n");
-
-   printf("B: ");
-   for(i=0;i<N;i++)printf(" %d ",B[i]);
-   printf("\n");
-
-   printf("C: ");
-   for(i=0;i<N;i++)printf(" %d ",C[i]);
-   printf("\n");
-   printf("------------------------------------------\n");
-   return;
-}
-
-/* Move the leftmost nonzero element of source to dest, leave behind 0. */
-/* Returns the value moved (not used.) */
-int Move(int *source, int *dest)
-{
-   int i = 0, j = 0;
-
-   while (i<N && (source[i])==0) i++;
-   while (j<N && (dest[j])==0) j++;
-
-   dest[j-1] = source[i];
-   source[i] = 0;
-   PrintAll();       /* Print configuration after each move. */
-   return dest[j-1];
-}
-
-
-/* Moves first n nonzero numbers from source to dest using the rules of Hanoi.
-   Calls itself recursively.
-   */
-void Hanoi(int n,int *source, int *dest, int *spare)
-{
-   int i;
-   if(n==1){
-      Move(source,dest);
-      return;
-   }
-
-   Hanoi(n-1,source,spare,dest);
-   Move(source,dest);
-   Hanoi(n-1,spare,dest,source);	
-   return;
-}
-
-int main()
-{
-   int i;
-
-   /* initialize the towers */
-   for(i=0;i<N;i++)A[i]=i+1;
-   for(i=0;i<N;i++)B[i]=0;
-   for(i=0;i<N;i++)C[i]=0;
-
-   printf("Solution of Tower of Hanoi Problem with %d Disks\n\n",N);
-
-   /* Print the starting state */
-   printf("Starting state:\n");
-   PrintAll();
-   printf("\n\nSubsequent states:\n\n");
-
-   /* Do it! Use A = Source, B = Destination, C = Spare */
-   Hanoi(N,A,B,C);
-
-   return 0;
-}
-
-/* vim: set expandtab ts=4 sw=3 sts=3 tw=80 :*/