Tutorial I : Solutions

Solution of Exercise 1

1. a = 0
2. a = -10 (literal substitution gives a = 10 - 5 - 10 - 5;)
3. a = 5 (literal substitution gives a = 10 - 5; - 10 - 5;;)

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Solution of Exercise 2

1.    #include <stdio.h>
   
      void binrep ( unsigned int n )
      {
         if (n <= 1) { printf("%d",n); return; }
         binrep(n/2);
         printf("%d",n%2);
      }
   
      int main ()
      {
         unsigned int n;
         printf("n = "); scanf("%u",&n);
         printf("Binary representation: ");
         binrep(n);
         printf("\n");
      }
   

2.    #include <stdio.h>
   
      void twocrep ( int n , int A[] )
      {
         int i;
   
         if (n == -2147483648) {
            printf("10000000000000000000000000000000\n");
            return;
         }
         if (n < 0) { A[31] = 1; n = -n; } else { A[31] = 0; }
         for (i=0; i<=30; ++i) {
            if (A[31]==0) A[i] = n % 2; else A[i] = 1 - n % 2;
            n = n / 2;
         }
         if (A[31] == 1) {
            i = 0;
            while (A[i] == 1) { A[i] = 0; ++i; }
            A[i] = 1;
         }
         for (i=31; i>=0; --i) printf("%d",A[i]);
         printf("\n");
      }
   
      int main ()
      {
         int n, A[32];
         printf("n = "); scanf("%d",&n);
         printf("2's complement representation: ");
         twocrep(n,A);
      }
   

3.    #include <stdio.h>
   
      void leftrep ( unsigned int n )
      {
         if (n <= 1) { printf("%d",n); return; }
         leftrep(n/2);
         printf("%d",n%2);
      }
   
      void rightrep ( float x )
      {
         int i, precision = 10;
   
         for (i=0; i<precision; ++i) {
            x = x * 2;
            printf("%d",(int)x);
            x = x - (int)x;
         }
      }
   
      int main ()
      {
         float x;
         printf("x = "); scanf("%f",&x);
         printf("Binary representation: ");
         leftrep((int)x);
         printf(".");
         rightrep(x - (int)x);
         printf("\n");
      }
   

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Solution of Exercise 3

The product x * y remains constant (up to floating point approximation).

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Solution of Exercise 4

1. The function computes and returns the standard deviation of the elements A[0],A[1],...,A[n-1].
2. The function computes and returns n³ upon input n.

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Solution of Exercise 5

   int isPalindrome ( char A[] , int start, int n )
   {
      if (n <= 1) return 1;
      if (A[start] != A[start+n-1]) return 0;
      return isPalindrome(A,start+1,n-2);
   }

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Solution of Exercise 6

   A[n] = 5

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Solution of Exercise 7

   int frog ( int n )
   {
      if (n <= 0) return 0;
      if (n == 1) return 1;   /* Only one possibility: (1) */
      if (n == 2) return 2;   /* Two possibilities: (1,1) and (2) */
      if (n == 3) return 4;   /* Four possibilities: (1,1,1), (1,2), (2,1) and (3) */
      return frog(n-1) + frog(n-2) + frog(n-3);
   }

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Solution of Exercise 8

We have

     35
   = 32 + 2 + 1
   = 25 + 21 + 20
   = (100011)2.

     0.375
   = 0.25 + 0.125
   = (1/4) + (1/8)
   = 2-2 + 2-3
   = (0.011)2.

Therefore,

     35.375
   = (100011.011)2
   = (1.00011011)2 x 25
   = (1.00011011)2 x 2[132 - 127]
   = (1.00011011)2 x 2[(128 + 4) - 127]
   = (1.00011011)2 x 2[(10000100)2 - 127]
   = (1.00011011000000000000000)2 x 2[(10000100)2 - 127]

Consequently, the 32-bit floating point representation of +35.375 is:

   0 10000100 00011011 00000000 0000000

and that for -35.375 is:

   1 10000100 00011011 00000000 0000000

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