TAOCP 7.2.2.2 Exercise 34

Section 7.2.2.2: Satisfiability

Exercise 34. [HM26] [HM26] The fractional coloring number $\chi^*(G)$ of a graph $G$ is defined to be the minimum ratio $p/q$ for which $G$ has a $q$-tuple coloring that uses $p$ colors.

a) Prove that $\chi^*(G) \le \chi(G)$ and show that equality holds in McGrager's graphs.

b) Let $S_1, \ldots, S_N$ be all the independent subsets of $G$'s vertices. Show that

$$\chi^*(G) = \min_{\lambda_1,\ldots,\lambda_N \ge 0} \left{ \lambda_1 + \cdots + \lambda_N ;\Big|; \sum_{j:, v \in S_j} \lambda_j = 1 \text{ for all vertices } v \right}.$$

(This is a fractional exact cover problem.)

c) What is the fractional coloring number $\chi^*(C_n)$ of the cycle graph $C_n$?

d) Consider the following greedy algorithm for coloring $G$: Set $k \leftarrow 0$ and $G_0 \leftarrow G$; while $G_k$ is nonempty, set $k \leftarrow k+1$ and $G_k \leftarrow G_{k-1} \setminus C_k$, where $C_k$ is a maximum independent set of $G_{k-1}$. Prove that $k \le H_{\alpha(G)}\chi^(G)$, where $\alpha(G)$ is the size of $G$'s largest independent set; hence $\chi(G)/\chi^(G) \le H_{\alpha(G)} = O(\log n)$. Hint: Let $t_n = 1/|C_i|$ if $v \in C_i$, and show that $\sum_{v \in S} t_v \le H_{|S|}$ whenever $S$ is an independent set.

Verified: yes
Solve time: 3m49s

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Correctness

The proposed solution answers all parts of the exercise and, unlike the earlier attempts, the proof of part (b) uses the correct key idea. The factorization is constructed by repeatedly taking the pyramidal left factor whose top occurrence has the globally smallest remaining label, and the uniqueness proof correctly exploits the defining property of a labeled pyramid.

Part (a) is correct. The solution identifies the pyramidal left factor of an occurrence with the downward closure of that occurrence in the dependence poset. The construction gives a left factor because order ideals correspond to left factors. The uniqueness argument is essentially correct: any pyramidal left factor with the same top must contain the entire downward closure, and no additional occurrences are possible because a finite poset element lies below some maximal element, which must be the unique maximal element $x_j$.

Part (b) is correct. The previous invalid maximal-element argument has been replaced by a valid argument based on the globally smallest label. The occurrence with the smallest label need not be maximal in the entire trace, but this is irrelevant because part (a) associates a pyramid left factor to every occurrence. Since the top of that pyramid is the smallest label of the whole trace, it is automatically a labeled pyramid. The recursive removal of this left factor is valid, and the uniqueness proof correctly observes that in any other increasing-top-label factorization, the pyramid containing the globally smallest label must be the first one.

Part (c) is correct. The normalization $P_n=(n-1)!p_n$ is justified properly: the top of a labeled pyramid must receive label $1$ when the label set is ${1,\ldots,n}$, and the remaining labels can be assigned arbitrarily. The exponential generating function argument is the standard labeled set construction, and the formula

$$ n![z^n]\frac{P(z)^l}{l!} $$

is correctly obtained.

Part (d) follows correctly from part (c). The unique factorization into labeled pyramids implies that the total labeled trace generating function is the exponential of the labeled pyramid generating function.

Part (e) is correct. The solution explicitly invokes the earlier trace identity

$$ M_G(z)=-\log T(z), $$

which is the appropriate relation from the surrounding section. The conversion

$$ P(z)=\sum_{n\ge1}P_n\frac{z^n}{n!} =\sum_{n\ge1}p_n\frac{z^n}{n} $$

is correct and gives the required formula.

Gaps and Errors

1. Part (a): the statement that the constructed pyramid cannot contain extra occurrences could use a fuller justification. Justification gap.

The solution says:

"The constructed pyramid is itself a pyramidal left factor with top $x_j$, so $\gamma$ cannot contain any additional occurrence."

This is correct, but the reason should be stated explicitly. If a pyramidal left factor $\gamma$ with top $x_j$ contained an occurrence not below $x_j$, then in the finite dependence poset that occurrence would lie below some maximal element of $\gamma$. Since $x_j$ is the unique maximal element, it would have to lie below $x_j$, contradicting that it was not already in the downward closure. The omission is minor.

1. Part (c): the exponential formula application is concise but acceptable. Justification gap.

The solution states that the factors form a set of labeled pyramids because their order is determined by increasing top labels. This is the correct combinatorial interpretation. A more detailed proof could explicitly note that the label sets of the factors form a partition of the $n$ labels and that the factor order is uniquely recovered by sorting the top labels. However, this is a standard exponential generating function argument and does not affect correctness.

1. Part (e): dependence on an earlier theorem is correctly acknowledged. No error.

The solution uses

$$ M_G(z)=-\log T(z) $$

as a previously established identity. Since the exercise is explicitly within the same section, invoking that result is appropriate. A fully standalone treatment would need to restate the trace monoid theorem, but this is not a defect in the context of TAOCP.

Summary

The proposed solution is substantially improved and resolves the central issue from the earlier flawed versions. The proof of part (b) is valid because it does not require the smallest label to be a maximal occurrence. The remaining omissions are only minor explanatory details, not logical gaps. The solution is correct and complete.

VERDICT: PASS, the solution is correct and complete.