Foundations of Data Science

we decrease the residual length of the clause by one.
At a general stage, suppose the fates of i variables have already been decided and n−i
remain. The residual length of each clause is known. Among the clauses that are not yet
satisfied, choose a random shortest length clause. Among the n − i variables remaining,
pick one uniformly at random, then pick it or its negation as the new literal. Include
this literal in the clause thereby satisfying it. Since the clause is satisfied, the algorithm
deletes it. For each other clause, do the following. If its residual length is l, decide with
probability l/(n − i) to include the new variable in the clause and if so with probability
1/2 each, include it or its negation. If literal that was set to true is included in a clause,
delete the clause as it is now satisfied. If its negation is included in a clause, then just
delete the literal and decrease the residual length of the clause by one.
Why does this yield the same distribution as the original one? First, observe that the
order in which the variables are picked by the method of deferred decisions is independent
of the clauses; it is just a random permutation of the n variables. Look at any one clause.
For a clause, we decide in order whether each variable or its negation is in the clause. So
for a particular clause and a particular triple i, j, and k with i < j < k, the probability
that the clause contains the i th , the j th , and k th literal (or their negations) in the order
determined by deferred decisions is:
( ) ( ) ( )
1 −
3 1 −
3 · · · 1 −
3 3
n
(
1 −
2
n−i
(
1 − 1
n−j
n−1
) (
1 −
2
n−i−1
) (
1 − 1
n−j−1
)
· · ·
(
n−i+2
n−i+1
1 − 2
n−j+2
)
)
· · · (1
− 1
n−k+2
2
n−j+1
)
1
n−k+1 = 3
n(n−1)(n−2) ,
where the (1 − · · · ) factors are for not picking the current variable or negation to be included
and the others are for including the current variable or its negation. Independence
among clauses follows from the fact that we have never let the occurrence or nonoccurrence
of any variable in any clause influence our decisions on other clauses.
Now, we prove the lemma by appealing to the method of deferred decisions to generate
the formula. T i = t if and only if the method of deferred decisions does not put the current
literal at steps 1, 2, . . . , t − 1 into the i th clause, but puts the negation of the literal at
step t into it. Thus, the probability is precisely
1
2
(
1 −
3
n
) (
1 −
3
n−1
)
· · ·
(
1 −
3
n−t+2
)
3
n−t+1 ≤ 3
2(n−2) ,
as claimed. Clearly the T i are independent since again deferred decisions deal with different
clauses independently.
Lemma 4.22 There exists a constant c 2 such that with probability 1 − o(1), no clause
remains a 2 or 1-clause for more than c 2 ln n steps. I.e., once a 3-clause becomes a
2-clause, it is either satisfied or reduced to a 0-clause in O(ln n) steps.
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