### Flagpoles

My solution during the contest was essentially the same as the official analysis. Afterwards I realised a potential slight simplification: if one starts by computing the second-order differences (i.e., the differences of the differences), then one is looking for the longest run of zeros, rather than the longest run of the same value. That removes the need to communicate the value used in the runs at the start and end of each section.

### Number Bases

I missed the trick of being able to uniquely determine the base from the

*first*point at which X[i] + Y[i] ≠ Z[i]. Instead, at every point where X[i] + Y[i] ≠ Z[i], I determine two candidate bases (depending on whether there is a carry of not). Then I collect the candidates and test each of them. If more than three candidates are found, then the test case is impossible, since there must be two disjoint candidate pairs.### Broken Memory

My approach was slightly different. Each node binary searches for its broken value, using two other nodes to help (and simultaneously helping two other nodes). Let's say we know the broken value is in a particular interval. Split that interval in half, and compute hashes for each half on the node (h1 and h2) and on two other nodes (p1 and p2, q1 and q2). If h1 equals p1 or q1, then the broken value must be in interval 2, or vice versa. If neither applies, then nodes p and q both have broken values, in the opposite interval to that of the current node. We can tell which by checking whether p1 = q1 or p2 = q2.

This does rely on not having collisions in the hash function. In the contest I relied on the contest organisers not breaking my exact choice of hash function, but it is actually possible to write a solution that works on all test data. Let P be a prime greater than \(10^{18}\). To hash an interval, compute the sums \(\sum m_i\) and \(\sum i m_i\), both mod P, giving a 128-bit hash. Suppose two sequences p and q collide, but differ in at most two positions. The sums are the same, so they must differ in exactly two positions j and k, with \(p_j - q_j = q_k - p_k\) (all mod P). But then the second sums will differ by

\(jp_j + kp_k - jq_j - kq_k = (j - k)(p_j - q_j)\), and since P is prime and each factor is less than P, this will be non-zero.

## 1 comment:

nice

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