CODEWITHSUNDEEP
c++algorithm
Marcus Lee
Marcus Lee

Reputation: 43

Aren't both of these conditions the same?

I'm solving this beginner problem on Codeforces, Young Physicist. https://codeforces.com/problemset/problem/69/A

The answer is to find if 3 forces are in equilibrium.

I've submitted two versions of my answers. One passed but one didn't. But I can't figure why it won't pass, aren't both of the condition same?

if (sumx == 0 && sumy == 0 && sumz == 0)
    cout << "YES" << "\n";
else
    cout << "NO" << "\n";

above is the one that passed.

if (sumx + sumy + sumz == 0)
    cout << "YES" << "\n";
else
    cout << "NO" << "\n";

this is the one that didn't pass.

Upvotes: 2

Views: 94

Answers (2)

HTNW
HTNW

Reputation: 29193

Consider sumx = -1, sumy = 1, sumz = 0. There's a 2-dimensional infinity of solutions to x + y + z = 0, and x = y = z = 0 is just a 0-dimensional singularity.

If you graph x + y + z = 0, you get this:

a plane representing the solutions

Every point on this plane is a set of three values that gives you 0 from their sum. The point where all three are zero is just a single point out of the sea. A quick check that tells you you've oversimplified the constraint is that you have 3 variables, but only 1 equation. That means you should still have 3 - 1 = 2 dimensions of freedom, but a single point has 0 dimensions of freedom.

Upvotes: 7

Diogenis Siganos
Diogenis Siganos

Reputation: 797

The answer is no. When you have sumx + sumy + sumz == 0, it doesn't necessarily mean that all these variables are equal to 0. Let's assume that sumx = -1, sumy = 0, symz = 1. This:

if (sumx == 0 && sumy == 0 && sumz == 0)
    cout << "YES" << "\n";
else
    cout << "NO" << "\n";

outputs NO, but this:

if (sumx + sumy + sumz == 0)
    cout << "YES" << "\n";
else
    cout << "NO" << "\n";

outputs YES.

Upvotes: 1

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