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The activity selection problem

Given a set S = {1, 2, . . . , n} of activities that wish to use aresource, which can be used only by one activity at the time.

Activity i has a start time si and a finishing time fi, fi > si.Activities i and j are comparable if [sifi) (sjfj] = .The goal is to maximize the set of mutually compatible activities.Notice, F = {A|A S, A is compatible } and f(A) = |A|.
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Example.

Activity (A): 1 2 3 4 5 6 7 8Start (s): 3 2 2 1 8 6 4 7Finish (f): 5 5 3 5 9 9 5 8

4

41 5 6 7 82 3 9 10

1 5

2 6

3 7 8

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The Activity Selection problem.

Given a set A of activities, wish to maximize the number ofcompatible activities.

Activity selection A

Sort A by increasing order of fiS := {1}j := 1

from i = 2 to n doif si > fjS := S {i} and j := i

return S.

A : 3 1 2 7 8 5 6f : 3 5 5 5 8 9 9SOL: 3 1 8 5

4

1 5

2 6

3 7 8

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TheoremThe previous algorithm produces an optimal solution to the activity

selection problem.

Proof.Given A = {1, . . . , n} sorted by finishing time, we must show thereis an optimal solution that begins with activity 1. Lest

S = {k, . . . , m} be a solution. If k = 1 done.Otherwise, define B = (S {k}) {1}. As f1 fk the activities inB are disjoint. As |B| = |S|, B is also an optimal solution.If S is an optimal solution to A, then S = A {1} is an optimalsolution for A = {i A|si fi}.Therefore, after each greedy choice we are left with an optimizationproblem of the same form as the original. Induction on the numberof choices, the gredy strategy produces an optimal solution

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The satisfiability problem (SAT)

A boolean variable x can take values 0,1. A litteral is a is aboolean variable x or x. A clause is a disjunction of litterals.An instance of SAT: X = {x1, . . . , xn}, a set of clauses {Ci}

mi=1.

and a boolen formula F in Conjuntive Normal Form (CNF). The

problem consists in obtaining a truth assignment of X:( : X {0, 1}) s.t.

(F) =m

i=1

(Ci) = 1.

Notice: for F = 1, i, (Ci) = 1 x Ci, (x) = 1.

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Horn formulae

A SAT instance, where each Ci has at most one positive literal.Example:

(x1x2)(x1x3x4)(x1x2x3x4)(x1)(x1x2x4)(x3)(x1x2x

Recall that clauses with one positive literal can be rewritten asimplications:

(x y z w) (y z w) x

So the previous example can be rewritten as(x1 x2), (x1 x3 x4), (x2 x3 x4 x1), ( x1),(x1 x2 x4), (x3), (x1 x2 x4).

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Give boolean formulaes A and B, A B is false iff A = 1 andB = 0 . Therefore, if we have a defined from a clause with nonegations in a Horn formulae, for the implication to be false, allvariables at the left must be 1 and the variable to the right mustbe 0

(x2 x3 x4 x1)

This fact can be used to develop a greedy algorithm for Hornformulae

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Linear time Greedy algorithm for Horn formulae

Given a Horn formulae F:

Greedy Horn F

for all i [n] do T(xi) = 0

while there is false implication yy := 1

if all clauses with negatives are 1return F = 1

else F = 0.

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Example:

(x1 x2), (x1 x3 x4), (x2 x3 x4 x1), ( x1),(x1 x2 x4), (x3), (x1 x2 x4)Greedy-Horn will make (x1) = (x2) = (x3) = (x4)

then (x1) = 1 because ( x1),then (x2) = 1 because (x1 x2),then (x4) = 1 because (x1 x2 x4),But then (x1 x2 x4) = 1.Therefore, the original formula has no truth assignment.

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