TAOCP 7.2.2.2 Exercise 283
Exercise 7.
Section 7.2.2.2: Satisfiability
Exercise 283. [HM46] Algorithm C solves the flower snark problem in linear time? More precisely, let $p_k(M)$ be the probability that the algorithm refutes $\text{fnmark}(q)$ while making at most $M$ flushes, as $q \to \infty$. Is $p_k(N_0) \to 1$ or $p_k(N_0) > \frac{1}{2}$ for all $q$?
Verified: no
Solve time: 2m09s
Setup
Exercise 7.2.2.2.283 asks about the asymptotic behavior of Algorithm C on the unsatisfiable instances $\operatorname{fnmark}(q)$ introduced in exercise 176. Let $p_k(M)$ denote the probability that Algorithm C refutes $\operatorname{fnmark}(q)$ while making at most $M$ flushes, as $q\to\infty$.
The question asks whether a fixed flush bound, specifically $N_0$, is sufficient with probability tending to $1$, or at least with probability exceeding $\frac12$ for every $q$.
A complete answer would require a rigorous probabilistic analysis of Algorithm C's random choices, including the distribution of its clause-generation process, the effect of purging and flushing, and the interaction with the structure of the flower snark clauses.
Known results
The preceding exercises establish that the flower snark instances $\operatorname{fnmark}(q)$ have compact resolution certificates. In particular, exercise 282 constructs a certificate using $O(q)$ clauses, all of length at most $4$. This proves that the unsatisfiability of $\operatorname{fnmark}(q)$ has short resolution proofs.
Algorithm C, however, is not a deterministic proof-search procedure that is guaranteed to discover a shortest certificate. Its behavior depends on the ordering and retention of generated clauses. A short certificate does not imply that Algorithm C finds a refutation before a prescribed number of flushes.
The definition of $p_k(M)$ therefore concerns the average behavior of a particular SAT-solving strategy, not merely the existence of a short proof.
No theorem in the supplied material establishes either
$$ \lim_{q\to\infty}p_k(N_0)=1 $$
or
$$ p_k(N_0)>\frac12 $$
for all $q$.
The problem is formulated as an asymptotic performance question about Algorithm C, and the required conclusion depends on properties of Algorithm C that are not determined by the general resolution-certificate arguments in Section 7.2.2.2.
Partial argument
The certificate from exercise 282 shows that there exists a sequence of clauses of linear length leading to the empty clause. Hence Algorithm C has a possible successful trajectory whose number of generated useful clauses is $O(q)$.
This observation gives an upper bound on the length of some refutations, but it gives no lower bound on the probability that Algorithm C follows such a trajectory. A randomized search procedure can fail to find a short proof even when one exists.
To prove
$$ p_k(N_0)\to1, $$
one would need to show that the probability of avoiding all successful clause combinations decreases to $0$ as $q$ grows. To prove
$$ p_k(N_0)>\frac12, $$
one would need a uniform lower bound on the probability that the search enters a successful region of the proof space before the first $N_0$ flushes.
Neither conclusion follows from the linear-size certificate alone. The certificate only supplies existence of a successful derivation.
Status
The supplied information does not contain a proof resolving the question. The problem is a performance analysis question about Algorithm C rather than a direct consequence of the resolution theory developed in Section 7.2.2.2.
A rigorous solution would require additional analysis of Algorithm C's stochastic behavior on $\operatorname{fnmark}(q)$, including details beyond the given section context. No complete proof of either proposed asymptotic statement can be supplied without those results.
Therefore the problem remains unresolved from the available material. Any assertion that $p_k(N_0)\to1$ or that $p_k(N_0)>\frac12$ for all $q$ would require a theorem not provided here.