Consider the following C functions.int tob(int b, int* arr){ int i; for(i=0; b>0; i++){ if(b%2) arr[i]=1; else arr[i]=0; b = b/2; } return(i); } int pp(int a,int b) { int arr[20]; int i, tot = 1, ex, len; ex = a; len = tob(b,arr); for(i=0; i<len; i++){ if(arr[i]==1) tot = tot * ex; ex = ex * ex; } return(tot); }The value returned by pp(3,4) is _______.

The function pp(a, b) computes $a^b$ using modular/binary exponentiation.

1. tob(4, arr) converts $b = 4$ into its binary representation in reverse order (least significant bit first) and stores it in arr:

   • $i=0$: $b=4 \implies 4\%2 = 0 \implies arr[0] = 0,\; b=2$

   • $i=1$: $b=2 \implies 2\%2 = 0 \implies arr[1] = 0,\; b=1$

   • $i=2$: $b=1 \implies 1\%2 = 1 \implies arr[2] = 1,\; b=0$

   • Loop terminates; tob returns len = 3 with arr = [0, 0, 1].

2. In pp(3, 4):

   • Initial values: ex = 3, tot = 1

   • $i = 0$: arr[0] == 1 is false $\implies tot = 1,\; ex = 3 \times 3 = 9$

   • $i = 1$: arr[1] == 1 is false $\implies tot = 1,\; ex = 9 \times 9 = 81$

   • $i = 2$: arr[2] == 1 is true $\implies tot = 1 \times 81 = 81,\; ex = 81 \times 81 = 6561$

3. The function returns tot = 81.