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I've been self-studying the book Algorithms by Papadimitriou, Dasgupta and Vazirani. I am having a hard time with a question about local search involving the traveling salesman problem (TSP).

We'll say a local search algorithm is exact if it always returns a globally optimal solution.

Consider a local search algorithm for TSP that uses neighborhoods defined by $k$-change: two tours $T_0$ and $T_1$ are neighbors if one can delete $j \leqslant k$ edges from $T_0$ and add back another $j$ edges to obtain $T_1$. This is known as the $k$-Opt algorithm.

It's easy to see and the book itself discusses how for low values of $k$ (relative to the number $n$ of vertices), $k$-Opt may get stuck on locally optimal solutions that are not globally optimal. In other words, $k$-Opt is not exact for these values of $k$.

The exercise I'm interested in claims that

$k$-Opt with $k = \lceil n/2 \rceil$ is not exact.

I've tried my hand and searched around but couldn't make it.

I found the paper "Some Examples of Difficult Traveling Salesman Problems", available here (and Papadimitriou is one of the authors by the way). It contains a family of examples for which $k$-Opt with $k < (3n/8)$ is not exact, and uses a particular kind of subgraph (which it calls the diamond) to impose a structure on feasible tours.

I've tried to replicate and extend something like this with little success.

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  • $\begingroup$ @InuyashaYagami I guess in theory any one counter example suffices (so you can pick $n$). A family would be nice, though, specifically in the sense that it holds for arbitrarily large $n$ (rather than it being true as a quirk of some specific value of $n$). $\endgroup$ Feb 16, 2022 at 14:26

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Answer to the first question:

Please see the following figure for $n = 8$. All the red and black edges have weight $1$. The green edge $(C,H)$ has weight $2$. And, the blue edges have weight $0$. The edges that are not there have weight $\infty$; I have not drawn these edges for simplicity.

Now, note that the cycle $(A,B,C,D,E,F,G,H)$ is the optimal tour with weight $5$. Also, the tour formed by the colored edges (red, green, and blue) has weight $6$. Note that the tour formed by these colored edges is locally optimal to $\lceil n/2 \rceil$ change but it is not globally optimal.

enter image description here

Why colored tour is locally optimal?

Proof: For the sake of contradiction assume that it is not locally optimal. It means that there exist at most four edges in the tour that can be replaced to obtain a smaller weight tour. Observe that to obtain a smaller weight tour, the edge $(C,H)$ must be replaced and the blue edges can not to be replaced. If we replace edge $(C,H)$, then we must add the edges $(C,B)$ and $(A,H)$; otherwise, the weight of the tour will become $\infty$.

After adding these edges, we can not keep edges $(B,D)$ and $(A,G)$ in the tour. Therefore, we must add edges $(D,E)$ and $(G,F)$; otherwise, the weight of the tour will become $\infty$. Now, we must remove edges $(B,E)$ and $(A,F)$.

To complete the tour, we need to add edge $(A,B)$; however, we can not add any more edges since we have already added $4 = n/2$ new edges. Therefore, a tour can only be formed by adding edges $(B,F)$ and $(E,A)$. It makes the weight of the tour $\infty$.

This proves that the tour formed by colored edges is locally optimal.


The technique can be easily generalized to the general value of $n$. For example, see the following figure for $n = 14$. Here, again, the tour formed by the colored edges is locally optimal to $\lceil n/2 \rceil$ change but it is not globally optimal. enter image description here

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  • $\begingroup$ I like this a lot! Ingenious. $\endgroup$ Feb 16, 2022 at 23:32
  • $\begingroup$ @Fimpellizzeri Thanks! Hope I have not made any mistake. Also, I would recommend you to create a separate question post for the second question if possible. The community generally does not advice to ask more than two questions in the same post 😅. $\endgroup$
    – D G
    Feb 16, 2022 at 23:39
  • $\begingroup$ You've made a typo in saying we couldn't keep $(A, H)$, you most definitely meant $(A, G)$, but I could follow along the parts that felt less clear. After removing $(C, H)$, $(A, H)$ must be added because blue edges must be kept and it's the only cost-feasible way to keep the tour going from $H$ (each vertex in the tour must have degree $2$). Something similar holds for $(C, B)$ with vertex $C$. Now, since we've established that $(A, H)$ and $(H, G)$ must be on the tour, $(G, A)$ can't be on the tour, for it would form a too-short cycle. (Continued...) $\endgroup$ Feb 16, 2022 at 23:50
  • $\begingroup$ (... continued) Then, $(G, F)$ must be added because it's the only cost-feasible way to keep the tour going from $G$. Something similar holds for the upper part of the graph. But we've already spent all our budget on edge additions, and can only remove one additional edge: either $(A, F)$ or $(B, E)$. Regardless of our choice, this does not lead to a valid tour. $\endgroup$ Feb 16, 2022 at 23:52
  • $\begingroup$ Do you feel like it would be better to edit this question to be only about $(1)$, and open a second question solely about $(2)$? I could certainly do that, I just worry about hogging the front page. $\endgroup$ Feb 16, 2022 at 23:53

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