TAOCP 7.2.2.2 Exercise 221
Let $F$ be a 7SAT instance.
Section 7.2.2.2: Satisfiability
Exercise 221. [16] [16] Draw the search tree analogous to Fig. 38 that is implicitly traversed when Algorithm A is applied to the unsatisfiable clauses ${12, 2, \bar{2}}$. Explain why it does not correspond to a resolution refutation that is analogous to Fig. 48.
Verified: no
Solve time: 5m52s
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
$$ \boxed{\text{Every 7SAT formula with at most 13 occurrences of each variable is satisfiable.}} $$