In an instance of the "Interval Scheduling" problem, assume that i1,…,ikis the list of jobs in A (the output of our greedy algorithm), and j1,…,jk is another optimal solution. Consider these new sequences of jobs:

S1=i1,…,il,jl+1,…,jk(forsomels.t.1

Question

In an instance of the "Interval Scheduling" problem, assume that i1,…,ikis the list of jobs in A (the output of our greedy algorithm), and j1,…,jk is another optimal solution. Consider these new sequences of jobs:

S1=i1,…,il,jl+1,…,jk(forsomels.t.1<l<k)

S2=j1,…,jl,il+1,…,ik(forsomels.t.1<l<k)

Which one(s) are always valid (non-overlapping) solutions?

BlackTom AI · Accepted Answer

Reframe the problem in your own words: we have two sequences of non-overlapping interval jobs, i1, …, ik coming from the greedy interval scheduling algorithm, and j1, …, jk representing some other optimal solution. We form two stitched sequences by taking a prefix from one solution and a suffix from the other, and we want to know which stitched sequences are guaranteed to be non-overlapping.

Option under consideration: S1 is always valid, but S2 can sometimes be invalid.

Analysis of S1 (i1, …, il, jl+1, …, jk):
- Up to i_l, the schedule is non-overlapping by the construction of i1, …, ik from the greedy algorithm.
- After i_l, we switch to j_{l+1} as the next job in the sequence, followed by j_{l+2}, …, j_k. All of these come from the other optimal schedule, so by definition each j_t is non-overlapping with the next in the j-sequence.
- The critical question is whether the transition from i_l to j_{l+1} creates an overlap. Since i_l is some job in the i-sequence and j_{l+1} is the (l+1)th job in the j-sequence, the decision hinges on the end time of i_l versus the start time of j_{l+1}. However, because j1, …, jk is a valid (non-overlapping) schedule and i1, …, il are valid up to i_l, there is a classic exchange-like property: the end time of i_l is not after the end time of j_l, and j_{l+1} starts after j_l ends, which ensures that the switch from i_l to j_{l+1} does not create an overlap. Intuitively, if there were an overlap at the boundary, you could not have both prefixes be part of valid schedules of the same length, contradicting the existence of the two valid sequences. Therefore, S1 preserves non-overlap at the junction, and the entire sequence remains valid.

Analysis of S2 (j1, …, jl, il+1, …, ik):
- Up to j_l, the sequence is valid by the definition of the j-sequence being non-overlapping.
- The potential issue arises at the boundary between j_l and i_{l+1}. While i_{l+1} is the (l+1)th job in the i-sequence, its start time could be earlier than the finish time of j_l, or it could overlap with j_l depending on how the two sequences interleave their end times. Because we are not guaranteed any particular relationship between the end time of j_l and the start time of i_{l+1}, it is possible that i_{l+1} overlaps with j_l. If such an overlap occurs, the concatenated sequence S2 would be invalid. Thus, there is no universal guarantee that S2 stays non-overlapping for all possible i- and j- optimal sequences and all l with 1 < l < k.

Summary of reasoning:
- S1 relies on a safe boundary: switching from i_l to j_{l+1} does not introduce an overlap under the typical exchange-argument structure of interval scheduling proofs, so S1 is consistently valid.
- S2 relies on a boundary between j_l and i_{l+1}, which can fail if i_{l+1} starts before j_l ends. Since such a timing relationship is not guaranteed, S2 can be invalid in some cases.

Therefore, the statement that S1 is always valid while S2 can sometimes be invalid aligns with the known behavior of these stitched constructions in interval scheduling analyses.

CPSC_V 320 201/202/203 2024W2 Reading Quiz #6 (Greedy Algorithms 1)

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