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Foundations of Data Science

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simple.<br />

We supply some intuition before going to the pro<strong>of</strong>. Imagine maintaining a queue <strong>of</strong><br />

1 and 2-clauses. A 3-clause enters the queue when one <strong>of</strong> its literals is set to false and<br />

it becomes a 2-clause. SC always picks a 1 or 2-clause if there is one and sets one <strong>of</strong> its<br />

literals to true. At any step when the total number <strong>of</strong> 1 and 2-clauses is positive, one <strong>of</strong><br />

the clauses is removed from the queue. Consider the arrival rate, that is, the expected<br />

number <strong>of</strong> arrivals into the queue at a given time t. For a particular clause to arrive into<br />

the queue at time t to become a 2-clause, it must contain the negation <strong>of</strong> the literal being<br />

set to true at time t. It can contain any two other literals not yet set. The number <strong>of</strong><br />

such clauses is ( )<br />

n−t 2 2 . So, the probability that a particular clause arrives in the queue at<br />

time t is at most<br />

2<br />

( n−t<br />

(<br />

2<br />

n<br />

3<br />

)<br />

2<br />

2<br />

)<br />

2<br />

3<br />

≤<br />

3<br />

2(n − 2) .<br />

Since there are cn clauses in total, the arrival rate is 3c , which for c < 2/3 is a constant<br />

2<br />

strictly less than one. The arrivals into the queue <strong>of</strong> different clauses occur independently<br />

(Lemma 4.21), the queue has arrival rate strictly less than one, and the queue loses one<br />

or more clauses whenever it is nonempty. This implies that the queue never has too many<br />

clauses in it. A slightly more complicated argument will show that no clause remains as<br />

a 1 or 2-clause for ω(ln n) steps (Lemma 4.22). This implies that the probability <strong>of</strong> two<br />

contradictory 1-length clauses, which is a precursor to a 0-length clause, is very small.<br />

Lemma 4.21 Let T i be the first time that clause i turns into a 2-clause. T i is ∞ if clause<br />

i gets satisfied before turning into a 2-clause. The T i are mutually independent over the<br />

randomness in constructing the formula and the randomness in SC, and for any t,<br />

Prob(T i = t) ≤<br />

3<br />

2(n − 2) .<br />

Pro<strong>of</strong>: For the pro<strong>of</strong>, generate the clauses in a different way. The important thing is<br />

that the new method <strong>of</strong> generation, called the method <strong>of</strong> “deferred decisions”, results in<br />

the same distribution <strong>of</strong> input formulae as the original. The method <strong>of</strong> deferred decisions<br />

is tied in with the SC algorithm and works as follows. At any time, the length <strong>of</strong> each<br />

clause (number <strong>of</strong> literals) is all that we know; we have not yet picked which literals are<br />

in each clause. At the start, every clause has length three and SC picks one <strong>of</strong> the clauses<br />

uniformly at random. Now, SC wants to pick one <strong>of</strong> the three literals in that clause to<br />

set to true, but we do not know which literals are in the clause. At this point, we pick<br />

uniformly at random one <strong>of</strong> the 2n possible literals. Say for illustration, we picked ¯x 102 .<br />

The literal ¯x 102 is placed in the clause and set to true. The literal x 102 is set to false. We<br />

must also deal with occurrences <strong>of</strong> the literal or its negation in all other clauses, but again,<br />

we do not know which clauses have such an occurrence. We decide that now. For each<br />

clause, independently, with probability 3/n include either the literal ¯x 102 or its negation<br />

x 102 , each with probability 1/2. In the case that we included ¯x 102 (the literal we had set<br />

to true), the clause is now deleted, and if we included x 102 (the literal we had set to false),<br />

111

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