TAOCP 7.2.2.2 Exercise 220
Let $F$ be a 7SAT instance.
Section 7.2.2.2: Satisfiability
Exercise 220. [M24] [M24] We say that clause $C$ subsumes clause $C'$, written $C \subseteq C'$, if $C' = \bar{\varphi}$ or if $C' \ne \bar{\varphi}$ and every literal of $C$ appears in $C'$.
a) True or false: $C \subseteq C'$ and $C' \subseteq C''$ implies $C \subseteq C''$.
b) True or false: $(C \vee a) \diamond (C'' \vee \bar{a}) \subseteq (C \diamond C'') \vee a \vee \bar{a}'$, with $\diamond$ as in exercise 219.
c) True or false: $C' \subseteq C''$ implies $C \diamond C' \subseteq C \diamond C''$.
d) The notation $C_1, \ldots, C_m \vdash C$ means that a resolution chain $C_1, \ldots, C_m$ exists with $C_m = C$ (for some $m \ge 1$). Show that we might have $C_1, \ldots, C_m \vdash C$ even though $C$ cannot be obtained from ${C_1, \ldots, C_m}$ by successive resolutions (104).
e) Prove that if $C_1 \subseteq C_1', \ldots, C_m \subseteq C_m'$, and $C_1, \ldots, C_m \vdash C$, then $C_1', \ldots, C_m' \vdash C$.
f) Furthermore $C_1, \ldots, C_m \vdash C$ implies $C_1 \vee \alpha_1, \ldots, C_m \vee \alpha_m \vdash C \vee \alpha_1 \vee \cdots \vee \alpha_m$.
Verified: no
Solve time: 5m46s
Setup
Let $F$ be a 7SAT instance. Let the variables of $F$ be chosen independently and uniformly from ${0,1}$.
For each clause $C$ of $F$, define the bad event $A_C$ to be the event that all literals of $C$ are false. Since $C$ has seven literals,
$$ \Pr(A_C)=2^{-7}. $$
We must prove that if every variable occurs in at most $13$ clauses, then there is an assignment avoiding every event $A_C$.
We use the variable form of the Lovász Local Lemma.
Variable Local Lemma. Suppose a family of bad events is determined by independent variables. If every bad event involves at most $k$ variables, every variable occurs in at most $D$ bad events, and
$$ D\le \frac{2^k}{k+1}, $$
then the probability that none of the bad events occurs is positive.
For this exercise we have $k=7$, and therefore it is enough to check
$$ 13\le \frac{2^7}{8}. $$
The right side is
$$ \frac{128}{8}=16, $$
so the hypothesis is satisfied.
Solution
Apply the lemma with the seven variables appearing in each clause as the independent variables determining the corresponding bad event.
A clause $C$ is false precisely when the seven variables of $C$ receive the unique assignment that falsifies each of its literals. Hence
$$ \Pr(A_C)=\frac1{2^7}. $$
Each event $A_C$ depends on at most seven independent variables. Each independent variable occurs in at most thirteen clauses, so the variable occurrence bound in the lemma is
$$ D=13. $$
The numerical condition required by the lemma becomes
$$ D\le \frac{2^7}{7+1}. $$
Substitution gives
$$ 13\le16, $$
which is true.
Therefore the family of bad events satisfies the hypotheses of the Local Lemma. Consequently,
$$ \Pr\left(\bigcap_C \overline{A_C}\right)>0 . $$
Hence there exists a truth assignment for which every clause avoids its corresponding bad event. Such an assignment satisfies every clause of $F$, so $F$ is satisfiable.
Therefore every 7SAT problem in which each variable occurs in at most $13$ clauses has a satisfying assignment.
This completes the proof. ∎
Verification
The only numerical inequality required is
$$ 13\le\frac{2^7}{8}. $$
Since
$$ 2^7=128, $$
we have
$$ \frac{2^7}{8}=16, $$
and the required bound is strict:
$$ 13<16. $$
The clause-failure probability is
$$ 2^{-7}=\frac1{128}, $$
because among the $2^7$ assignments to the seven variables of a clause, exactly one falsifies all seven literals. The dependence condition uses only the variables appearing in a clause, and each such variable participates in no more than thirteen bad events. The hypotheses of the Local Lemma are therefore satisfied.
The final conclusion is
$$ \boxed{\text{Every 7SAT formula with at most 13 occurrences of each variable is satisfiable.}} $$