Recursion

Functions – reminder A function can call other functions. Return causes the execution of the function to terminate and returns a value to the calling function. The type of the value returned must be the same as the return-type defined for the function. return-type name(argType1 arg1, argType2 arg2, …) { function body; return value; }

A recursive definition C functions can also call themselves!  However, usually not with the same parameters (why?) Some functions can be defined using smaller occurrences of themselves. Such a definition is called a “recursive definition” of a function.

Recursive calling Example: void func(int n){ putchar(`*`); func(n); } Infinite series of * What is the problem ? Look for other problem …

Factorial By definition : n! = 1*2*3*… *(n-1)*n Thus, we can also define factorial the following way:  0! = 1  n! = n*(n-1)! for n>0 (n-1)! *n

Example - factorial int factRec(int n){ if (n==0 || n==1) return 1; return n*factRec(n-1); } int factorial(int n){ int fact = 1; while (n >= 1) { fact *=n; n--; } return fact; }

Conclusions for Recursive calling Every recursive function has a “boundary condition”. The function stops calling itself when it is satisfied.

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… 4*…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns… 3*…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns… FactRec(2) n 2 Returns…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns… FactRec(2) n 2 Returns…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns… FactRec(2) n 2 Returns… 2*…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns… FactRec(2) n 2 Returns… FactRec(1) n 1 Returns…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns… FactRec(2) n 2 Returns… FactRec(1) n 1 Returns…

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns… FactRec(2) n 2 Returns… FactRec(1) n 1 Returns… 1

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns… FactRec(2) n 2 Returns… 2*1

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… FactRec(3) n 3 Returns… 3*2

Recursive factorial – step by step int factRec(int n) { if (n==0 || n==1) return 1; return n*factRec(n-1); } FactRec(4) n 4 Returns… 4*6

הדפסת מספרים - דוגמא 1 #include void print1(int n){ if (n>=0){ printf("%d ",n); print1(n-1); { void main(){ int i = 3; print1(i); putchar('\n'); { 3210

הדפסת מספרים - דוגמא 2 #include void print2(int n){ if (n>=0){ print2(n-1); printf("%d ",n); { void main(){ int i = 3; print2(i); putchar('\n'); { 0123

הדפסת מספרים - דוגמא 3 #include void print3(int n){ if (n>=0){ printf("%d ",n); print3(n-1); printf("%d ",n); { void main(){ int i = 3; print3(i); putchar('\n'); {

הדפסת מספרים - דוגמא 4 #include void print4(int n){ if (n>=0){ print4(n-1); printf("%d ",n); print4(n-1); { void main(){ int i = 3; print4(i); putchar('\n'); {

Another example - power X y = x*x*…*x Recursive definitions (assume non-negative y):  Base: x 0 =1 1. X y = X*(X y-1 ) 2. X y = (X y/2 ) 2 (for even y ’s only) y times

Fibonacci Series Fibonacci definition:  n 0 = 0  n 1 = 1  n n = n n-1 + n n …

Fibonacci Iterative void fibonacci(int n) { int Fn, Fn1, Fn2, ind; Fn2 = 0 ; Fn1 = 1 ; Fn = 0 ; if ( n == 1 ) Fn = 1 ; for (ind=2 ; ind <= n ; ind++){ Fn = Fn1 + Fn2; Fn2 = Fn1; Fn1 = Fn; } printf("F(%d) = %d \n", n, Fn); }

Fibonacci Recursive fibonacci(1) fibonacci(5) fibonacci(4)fibonacci(3) fibonacci(2) fibonacci(1) fibonacci(2)fibonacci(1) fibonacci(0) fibonacci(1)fibonacci(0) int fibonacci(int n) { if (n==0) return 0; if (n==1) return 1; return fibonacci(n-1) + fibonacci(n-2); }

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… 2*… y 5

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… y 4

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… y 4

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… y 4

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… y 2

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… y 2

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… y 2

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2 rec_pow(2, 1) x 2 Returns… y 1

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2 rec_pow(2, 1) x 2 Returns… y 1

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2 rec_pow(2, 1) x 2 Returns… y 1

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2 rec_pow(2, 1) x 2 Returns… y 1

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2 rec_pow(2, 1) x 2 Returns… 2*… y 1

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2 rec_pow(2, 1) x 2 Returns… 2*… y 1 rec_pow(2, 0) x 2 Returns… y 0

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2 rec_pow(2, 1) x 2 Returns… 2*… y 1 rec_pow(2, 0) x 2 Returns… y 0

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2 rec_pow(2, 1) x 2 Returns… 2*… y 1 rec_pow(2, 0) x 2 Returns… 1 y 0

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(…) y 2 rec_pow(2, 1) x 2 Returns… 2*1 y 1

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(…) y 4 rec_pow(2, 2) x 2 Returns… square(2) y 2

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… y 5 rec_pow(2, 4) x 2 Returns… square(4) y 4

rec_pow – step by step int rec_pow(int x, int y) { if (y == 0) return 1; if (y%2 == 0) return square(rec_pow(x,y/2)); else return x*rec_pow(x,y-1); } rec_pow(2, 5) x 2 Returns… 2*16 y 5

Exercise Write a program that receives two non-negative integers and computes their product recursively. Not * multiple operator !! Hint: Notice that the product a*b is actually a+a+…+a (b times).

Solution int recMult( int x, int y ) { if( x == 0) return 0; return y + recMult( x-1,y); {

Exercise Given the following iterative version of sum-of-digits calculation Find the recursive definition of this function (don’t forget the base case!) int sum_digits(int n){ int sum = 0; while (n > 0) { sum += n%10; n = n/10; } return sum; }

Solution int sumOfDigits( int x ) { if( x < 0) x *= -1; if( x == 0 ) return 0; else return x % 10 + sumOfDigits( x / 10 ); }

More uses Recursion is a general approach to programming functions. Its uses are not confined to calculating mathematical expressions! For example : write a function that finds the max member in an array of integer.

Solution int rec_max(int arr[ ], int size){ int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; }

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest … Returns… arr 760 size 2

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest … Returns… arr 760 size 2

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest … Returns… arr 760 size 2

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest … Returns… arr 760 size 2

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest … Returns… arr 760 size 2 rec_max(764, 1) rest … Returns… arr 764 size 1

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest … Returns… arr 760 size 2 rec_max(764, 1) rest … Returns… arr 764 size 1

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest … Returns… arr 760 size 2 rec_max(764, 1) rest … Returns… 3 arr 764 size 1

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest 3 Returns… arr 760 size 2

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest 3 Returns… arr 760 size 2

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest … Returns… arr 756 size 3 rec_max(760, 2) rest 3 Returns… 5 arr 760 size 2

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest 5 Returns… arr 756 size 3

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest 5 Returns… arr 756 size 3

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest 5 Returns… arr 756 size 3

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest … Returns… arr 752 size 4 rec_max(756, 3) rest 5 Returns… 5 arr 756 size 3

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest 5 Returns… arr 752 size 4

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest 5 Returns… arr 752 size 4

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest 5 Returns… arr 752 size 4

max_rec – step by step int rec_max(int arr[ ], int size) { int rest; if (size == 1) return arr[0]; else { rest = rec_max(arr+1, size-1); if (arr[0] > rest) return arr[0]; else return rest; } rec_max(752, 4) rest 5 Returns… 5 arr 752 size 4

חיפוש בינארי int BinarySearch(int arr[],int x, int left, int right) { int middle; if(left>right) return -1; else { middle=(left+right)/2); if(arr[middle]==x) return middle; else if(x < arr[middle]) return BinarySearch(arr,x,left,middle-1); else return BinarySearch(arr,x,middle+1,right); }

Towers of Hanoi The Towers of Hanoi problem consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape. The objective of the puzzle is to move the entire stack to another rod, obeying the following rules:  Only one disk may be moved at a time.  Each move consists of taking the upper disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod.  No disk may be placed on top of a smaller disk.

Towers of Hanoi

Recursive Solution To move n disks from peg A to peg C:  move n−1 disks from A to B. This leaves disk #n alone on peg A  move disk #n from A to C  move n−1 disks from B to C so they sit on disk #n

Recursive Solution - Function void hanoi(int x, char from, char to, char aux){ if(x==1){ printf("Move Disk From %c to %c\n", from, to); } else { hanoi(x-1,from,aux,to); printf("Move Disk From %c to %c\n", from, to); hanoi(x-1,aux,to,from); }

Scope of variables - reminder A variable declared within a function is unrelated to variables declared elsewhere. A function cannot access variables that are declared in other functions.