CF 1455A - Strange Functions
The proposed solution correctly identifies the probability requested: the probability that a single subtract-and-shift cycle produces an odd value $w$ in the range $[2^n,2^{n+1})$ while the other argument remains in $Omega$.
Rating: 800
Tags: math, number theory
Solve time: 2m 34s
Verified: no
Solution
Correctness
The proposed solution correctly identifies the probability requested: the probability that a single subtract-and-shift cycle produces an odd value $w$ in the range $[2^n,2^{n+1})$ while the other argument remains in $\Omega$. It enumerates all odd values $w$ in the target range, considers all admissible powers of two $t$ such that $|u-v|=2^t w < M$, counts all pairs $(u,v)$ giving rise to each $w$, and divides by the total number of pairs $N^2$. The reasoning explicitly shows that each pair producing a valid $w$ appears exactly once in the sum and that no other pair can produce a $w$ in the desired interval. Therefore, the solution both enumerates all contributing elements and excludes all non-contributing ones.
Gaps and Errors
- The step asserting that the number of pairs $(u,v)$ with $|u-v|=d$ equals $2(N-d/2)$ relies on the arithmetic progression structure of $\Omega$. While plausible, it is stated without detailed proof. This is a Justification gap, not a critical error, because the claim can be rigorously justified by indexing the odd numbers in $\Omega$ and counting pairs systematically.
- All other steps, including the summation over admissible $t$ and $w$, use standard combinatorial reasoning and are justified.
No critical errors are present. The approach correctly answers the exercise, and all manipulations of powers of two and logarithms are consistent.
Summary
The solution is mathematically sound, complete, and provides an exact formula for the requested probability. Minor gaps in justification do not affect correctness.
VERDICT: PASS - the solution is correct and complete.