Function to do an operation as much as possible

Question

I'm new to programming and I wanted to see if A is equal to B after applying the following operations toward A.`

If A=1, then A=A.
If A is even, then A = A/2.
If A is odd, then A = A*3+1.

output Yes if A is equal to B, No if it's not. I use the following code:

#include <stdio.h>

int main() {
    int a, b;
    scanf("%d %d", &a, &b);
    getchar();

    if (a == 1) {
        a = a;
    }
    else if (a % 2 == 0) {
        a = a / 2;
    }
    else if (a % 2 == 1) {
        a = a * 3 + 1;
    }

    if (a == b) {
        printf("Yes\n");
    }
    else {
        printf("No\n");
    }

    return 0;
}

This works just fine, but I wanted to do the operation as much as possible. e.g. If A=12 and B=5 -> 12/2=6 ->6/2=3->3+2=5. thus A=B. What function do I use to do just that?

Did you mean to rely on integer truncation for the odd case, or did you forget a `+ 1` or `- 1` somewhere? — Mad Physicist, Dec 13 '21 at 04:18
Specifically, the Collatz conjecture uses `3 * A + 1` for the odd case. — Mad Physicist, Dec 13 '21 at 04:21
@MadPhysicist Because A is an even number, I wanted to divide it by 2 as much as possible until its an odd number. — Maz Diaz, Dec 13 '21 at 04:22
Your third bullet point is a typo, if the code is to be believed. Please [edit] to fix — Mad Physicist, Dec 13 '21 at 04:23
Style note: if `a` is _not_ even, then it must be odd, so `else if (a % 2 == 1) {` can just be `else {`. — Chris, Dec 13 '21 at 04:37

Mad Physicist · Accepted Answer · 2021-12-14T05:11:21.543

This would be a good application for a while loop:

while(a != 1 && a != b) {
    if(a % 2 == 0) {
        a = a / 2;
    } else if(a % 2 == 1) {
        a = a * 3 + 1;
    }
}
if(a == b) {
    printf("Yes\n");
} else {
    printf("No\n");
}

Make sure that you understand the condition of the loop, which determines what "as much as possible" means, and why the final check is done outside the body of the loop.

The numbers 4 and 2 are implicitly part of any convergent sequence, since 4 = 3 * 1 + 1 and 2 = 4 / 2. A complicated was to include them would be to let a reach 1 twice before ending the loop. A simpler check would be to change the final if to

if(a == b || b == 4 || b == 2) {

Since the Collatz conjecture is still an unsolved problem, there is no guarantee that the loop will converge in all cases. At the same time, it also implies that no one has been able to find a case for which a doesn't at least reach unity.

All numbers up to around 2^68 have been shown to converge to 1. That implies that on machines which support long double with 80 bits of mantissa precision, or 128 bit integers, you could conceivably get a non-convergent input, or an input that converges to something other than one. So perhaps testing for re-visited numbers is a better technique than testing for convergence to 1.

We do know the loop terminates for all `int` values even for 64-bit integers. Of course, there is the possibility of `a=2`, `b=4`, where the Collatz sequence does include `b` but you get to `1` first. — Davis Herring, Dec 13 '21 at 04:44
@DavisHerring. I would have been shocked if anything a computer can check hadn't been checked already, way past 64 it even 128 bits. Interesting point about continuing the sequence though. I'll add in a test for that. — Mad Physicist, Dec 13 '21 at 04:50
@DavisHerring. Added. BTW, it's fewer than I initially guesstimated. Apparently we're only up to 2^68 or so, so technically a `long double` or on some machines `long long` *could* break the loop. — Mad Physicist, Dec 13 '21 at 04:56

Function to do an operation as much as possible

1 Answers1