Tuesday, 28 October 2014

Collatz function

Difficulty: Easy

The Collatz function is defined for a positive integer n as follows.
f(n) = 3n+1 if n is odd
        n/2     if n is even

We consider the repeated application of the Collatz function starting with a given integer n, as follows:
f(n), f(f(n)), f(f(f(n))), …

It is conjectured that no matter which positive integer n you start from, this sequence eventually will have 1 in it. It has been verified to hold for numbers up to 5 × 260  [Wikipedia: Collatz Conjecture].



e.g. If n=7, the sequence is
  1. f(7) = 22
  2. f(f(7)) = f(22) = 11
  3. f(11) = 34
  4. f(34) = 17
  5. f(17) = 52
  6. f(52) = 26
  7. f(26) = 13
  8. f(13) = 40
  9. f(40) = 20
  10. f(20) = 10
  11. f(10) = 5
  12. f(5) = 16
  13. f(16) = 8
  14. f(8) = 4
  15. f(4) = 2
  16. f(2) = 1
Thus if you start from n=7, you need to apply f 16 times in order to first get 1.

In this question, you will be given a positive number <= 32,000. You have to output how many times f has to be applied repeatedly in order to first reach 1.




#include<stdio.h>
int collatz();
int d=0;
int main()
{int n;
scanf("%d",&n);
collatz(n);printf("%d",d);
return 0;}

int collatz(int a)
{
if(a==1)
return 0;
else if(a%2==0)
{collatz(a/2);d++;}
else if(a%2==1)
{
collatz(3*a+1);d++;}
}
at

5 comments:

  1. Unknown27 August 2019 at 22:03

    Consider a matrix M of integers. Divide M into 4 sub-matrices. These sub-matrices are called as Quadrants. Report the Quadrant number which has the smallest minimum-element. If two or more quadrants have same smallest minimum, report the smallest quadrant index.

    The matrix M is divided into four quadrants by halving the rows and columns. If row/column is an odd number, divide them in such a way that the first half of the row/column should be one smaller than the second half

    ReplyDelete
  2. Unknown27 August 2019 at 22:03

    plz send me the solution

    ReplyDelete
  3. Anonymous31 August 2019 at 18:15

    It is showing an error
    Segmentation fault (core dumped)

    ReplyDelete
  4. Unknown12 October 2020 at 04:03

    #include
    int func(int n)
    { int a;
    if (n%2 == 0)
    {
    a=n/2;
    if(a!=1){
    printf("%d ",a);
    return (func(a));}
    else
    return (func(a));
    }
    else if (n == 1)
    {
    printf("%d",n);
    return 1;
    }
    else if (n%2 == 1 && n!=1)
    {
    printf("%d ",3*n+1);
    return (func(3*n + 1));
    }
    }
    int main()
    {
    int n;
    scanf("%d",&n);
    func(n);
    return 0;
    }

    ReplyDelete

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