Given a sequence A = {a_1, ..., a_n} of points on the real line. The task is to

Question

Given a sequence A = {a_1, ..., a_n} of points on the real line. The task is to determine the smallest set of unit-length (closed) intervals that contains all of the input points. Consider the following two greedy approaches: (a) Let I be an interval that covers the most points in A. Add I to the solution, remove the points covered by I from A, and recurse/continue. (b) Add the interval I = [a_1, a_1 + 1] to the solution, remove the points covered by I from A, and recurse/continue. One of these approaches is correct, the other one is not. Show which of the approaches is not correct by finding a counter-example. (The counter example consists of an example input and two "solutions" - one is the actual optimal solution, the other is the solution computed by the greedy algorithm, which is not as good as the optimal solution.)

Explanation / Answer

solution (a) will not yeild to optimal solution

let us take

A={0.1, 0.9,   1.1, 1.3, 1.9, 2.3}

as per solution using algorithm (a)

we will first take

I=[ 0.9, 1.9 ]

the we will add

[ 0.1, 1.1 ] to soltuion followed by [ 2.3, 3.3 ]

hence solution = { [ 0.1, 1.1 ], [ 0.9, 1.9 ], [ 2.3, 3.3 ] }

but the actual optimal solution is

{ [ 0.1, 1.1 ], [ 1.3, 2.3 ]   }

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