http://codeforces.com/contest/1074/submission/45462396

http://codeforces.com/contest/1074/submission/45462405

These two codes are exactly the same, except for this part:

In the Wrong answer on test 14 submission:

```
for (int i : inTree[rootp]) {
inTree[rootq].pb(i);
distToRoot[i] = distToRoot[i] ^ distToRoot[p] ^ edgeWeight ^ distToRoot[q];
root[i] = rootq;
}
```

In the Accepted submission:

```
edgeWeight ^= distToRoot[p] ^ distToRoot[q];
for (int i : inTree[rootp]) {
inTree[rootq].pb(i);
distToRoot[i] ^= edgeWeight;
root[i] = rootq;
}
```

(This is the end of the method, so xor-equals `edgeWeight`

shouldn't change anything later on.)

I thought xor sum is both commutative and associative? What's the difference between these two snippets? Am I missing something really obvious? Thanks in advance.

Not my bug, xor's bugThe for loop also modifies the values of distToRoot[p] and distToRoot[q] hence the WA.In your second submission, the original values of distToRoot[p] and distToRoot[q] are used to update other indices. There is no problem with xor here.

Wow, can't believe I missed that. Thanks!

The xor summing rule is simple: and . In the Accepted submission, the logic value xor summed with

`distToRoot[i]`

in each iteration is computed before the loop. Therefore, the logic value of`distToRoot[p] ^ edgeWeight ^ distToRoot[q]`

is invariant to the iteration variable`i`

, i.e. does not change inside the loop. In the Wrong Answer submission, when the number of negation operations made to`distToRoot[p]`

and`distToRoot[q]`

in previous iterationsof the loop is an odd number, the logic value xor summed with`distToRoot[i]`

is the complement of`edgeWeight`

in the AC code.