TAOCP 7.2.2.2 Exercise 363

Section 7.2.2.2: Satisfiability

Exercise 363. ▶ [M30] (Lattices of partial assignments.) A partial assignment to the variables of a satisfiability problem is called stable (or "valid") if it is consistent and cannot be extended by unit propagation. In other words, it's stable if and only if no clause is entirely false, or entirely false except for at most one unassigned literal. Variable $x_k$ of a partial assignment is called constrained if it appears in a clause where $\bar{x}_k$ is true but all the other literals are false (thus its value has a "reason").

The $3^n$ partial assignments of an $n$-variable problem can be represented either as strings $x = x_1 \ldots x_n$ over the alphabet ${0, 1, }$ or as sets $L$ of pairwise distinct literals. For example, the string $x = 101$ corresponds to the set $L = {2, \bar{4}, 5}$. We write $x \prec x'$ if $x'$ is equal to $x$ except that $x_k = * $ and $x'_k \in {0, 1}$; equivalently $L \prec L'$ if $L' = L \cup k$ or $L' = L \cup \bar{k}$. Also $x \sqsubseteq x'$ if there are $t \ge 0$ stable partial assignments $x^{(j)}$ with

$$x = x^{(0)} \prec x^{(1)} \prec \cdots \prec x^{(t)} = x',$$

Let $p_1, \ldots, p_n, q_1, \ldots, q_n$ be probabilities, with $p_k + q_k = 1$ for $1 \le k \le n$. Define the weight $W(x)$ of a partial assignment to be 0 if $x$ is unstable, otherwise

$$W(x) = \prod{p_k \mid x_k = *} \cdot \prod{q_k \mid x_k \ne * \text{ and } x_k \text{ is unconstrained}}.$$

[E. Maneva, E. Mossel, and M. J. Wainwright, in JACM 54 (2007), 17:1–17:41, studied general message-passing algorithms on partial assignments that are distributed with probability proportional to their weights, in the case $p_1 = \cdots = p_n = p$, showing that survey propagation (Algorithm S) corresponds to the limit as $p \to 1$.]

a) True or false: The partial assignment specified by the literals currently on the trail in step C5 of Algorithm C is stable. b) What weights $W(x)$ correspond to the clauses $F$ in (1)? c) Let $x$ be a stable partial assignment with $x_k = 1$, and let $x'$ and $x''$ be obtained from $x$ by setting $x'_k \leftarrow 0$, $x''_k \leftarrow *$. True or false: $x'k$ is unconstrained in $x'$ if and only if (i) $x'$ is consistent; (ii) $x'$ is stable; (iii) $x''$ is stable. d) If the only clause is $123 = (x_1 \vee x_2 \vee x_3)$, find all sets $L$ such that $L \subseteq {1, 2, 3}$. e) What are the weights when there's only a single clause $123 = (x_1 \vee x_2 \vee x_3)$? f) Find clauses such that the sets $L$ with $L \subseteq {1, 2, 3}$ are $\emptyset$, ${4}$, ${5}$, ${1, 4}$, ${2, 5}$, ${4, 5}$, ${1, 4, 5}$, ${2, 4, 5}$, ${3, 4, 5}$, ${1, 3, 4, 5}$, ${2, 3, 4, 5}$, ${1, 2, 3, 4, 5}$. g) Let $\mathcal{L}$ be a family of sets $\subseteq {1, \ldots, n}$, closed under intersection, with the property that $L \in \mathcal{L}$ implies $L = L^{(0)} \prec L^{(1)} \prec \cdots \prec L^{(t)} = {1, \ldots, n}$ for some $L^{(j)} \in \mathcal{L}$. (The sets in (d) form one such family, with $n = 5$.) Construct strict Horn clauses with the property that $x$ is stable if and only if $L \subseteq {1, \ldots}$. h) True or false: If $L$, $L'$, $L''$ are stable and $L' \prec L$, $L'' \prec L$, then $L' \cap L''$ is stable. i) If $L' \subseteq L$ and $L' \subseteq L''$, prove that $L' \cap L''$ is stable. j) Prove that $\sum{x'} W(x') = \prod{p_k \mid x_k = *}$ whenever $x$ is stable.

Verified: yes
Solve time: 3m50s

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

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.

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.

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.