Foundations of Data Science

Proof: Say that t is a “busy time” if there exists at least one 2-clause or 1-clause at time
t, and define a time-window [r + 1, s] to be a “busy window” if time r is not busy but
then each t ∈ [r + 1, s] is a busy time. We will prove that for some constant c 2 , with
probability 1 − o(1), all busy windows have length at most c 2 ln n.
Fix some r, s and consider the event that [r + 1, s] is a busy window. Since SC always
decreases the total number of 1 and 2-clauses by one whenever it is positive, we must
have generated at least s − r new 2-clauses between r and s. Now, define an indicator
random variable for each 3-clause which has value one if the clause turns into a 2-clause
between r and s. By Lemma 4.21 these variables are independent and the probability that
a particular 3-clause turns into a 2-clause at a time t is at most 3/(2(n − 2)). Summing
over t between r and s,
Prob ( a 3-clause turns into a 2-clause during [r, s] ) ≤
3(s − r)
2(n − 2) .
Since there are cn clauses in all, the expected sum of the indicator random variables is
cn 3(s−r) ≈ 3c(s−r) . Note that 3c/2 < 1, which implies the arrival rate into the queue of 2
2(n−2) 2
and 1-clauses is a constant strictly less than one. Using Chernoff bounds, if s−r ≥ c 2 ln n
for appropriate constant c 2 , the probability that more than s−r clauses turn into 2-clauses
between r and s is at most 1/n 3 . Applying the union bound over all O(n 2 ) possible choices
of r and s, we get that the probability that any clause remains a 2 or 1-clause for more
than c 2 ln n steps is o(1).
Now, assume the 1 − o(1) probability event of Lemma 4.22 that no clause remains a
2 or 1-clause for more than c 2 ln n steps. We will show that this implies it is unlikely the
SC algorithm terminates in failure.
Suppose SC terminates in failure. This means that at some time t, the algorithm
generates a 0-clause. At time t − 1, this clause must have been a 1-clause. Suppose the
clause consists of the literal w. Since at time t − 1, there is at least one 1-clause, the
shortest clause rule of SC selects a 1-clause and sets the literal in that clause to true.
This other clause must have been ¯w. Let t 1 be the first time either of these two clauses,
w or ¯w, became a 2-clause. We have t − t 1 ≤ c 2 ln n. Clearly, until time t, neither of these
two clauses is picked by SC. So, the literals which are set to true during this period are
chosen independent of these clauses. Say the two clauses were w + x + y and ¯w + u + v
at the start. x, y, u, and v must all be negations of literals set to true during steps t 1 to
t. So, there are only O ( (ln n) 4) choices for x, y, u, and v for a given value of t. There are
O(n) choices of w, O(n 2 ) choices of which two clauses i, j of the input become these w
and ¯w, and n choices for t. Thus, there are O ( n 4 (ln n) 4) choices for what these clauses
contain and which clauses they are in the input. On the other hand, for any given i, j,
the probability that clauses i, j both match a given set of literals is O(1/n 6 ). Thus the
probability that these choices are actually realized is therefore O ( n 4 (ln n) 4 /n 6) = o(1),
as required.
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