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5.4 T-joins and Postman Problems
Assume πβ₯0 and πΊ is connected. Thm 5.23: A connected graph πΊ has an Euler tour if and only if every node of πΊ has even degree. Let π₯π be the number of extra traversals of edge π in a postman tour. Construct the graph πΊπ₯ by making 1+π₯π copies of π for each π. 1 1 1 1 πΊ, x, and πΊπ₯ Combinatorial Optimization 2016
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Postman Problem is equivalent to the problem:
Minimize (πππ₯π:πβπΈ) (5.30) subject to π₯(πΏ(π£))β‘|πΏ(π£)| (mod 2), for all π£βπ π₯πβ₯0, for all πβπΈ. π₯π integer, for all πβπΈ. There is an optimal solution for which π₯ is {0, 1}-valued. (since πβ₯0) We call a set π½βπΈ a postman set of πΊ if, for every π£βπ, π£ is incident with an odd number of edges from π½ iff π£ has odd degree in πΊ. Postman Problem Given: A graph πΊ=(π, πΈ) and πβπ
πΈ such that πβ₯0. Objective: To find a postman set π½ such that π(π½) is minimum. Combinatorial Optimization 2016
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Optimal T-Join Problem
(T-Joins) Let πΊ=(π, πΈ) be a graph, and let πβπ such that |π| is even. A T-join of πΊ is a set π½ of edges such that |π½β©πΏ(π£)|β‘|πβ©{π£}| (mod 2), for all π£βπ π½ is a π-join iff the odd-degree nodes of the subgraph (π, π½) are exactly the elements of π. Optimal T-Join Problem Given: A graph πΊ=(π, πΈ), a set πβπ such that |π| is even, and a cost vector πβπ
πΈ. Objective: Find a π-join π½ of πΊ such that π(π½) is minimum. Combinatorial Optimization 2016
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Postman sets: Let π={π£βπ: |πΏ(π£)| is odd} ( need πβ₯0) Even set:
Examples: Postman sets: Let π={π£βπ: |πΏ(π£)| is odd} ( need πβ₯0) Even set: Even set is a set π΄βπΈ such that every node of (π, π΄) has even degree. Let π=β
. If costs are nonnegative, β
is optimal. A set is even iff it can be decomposed into edge-sets of edge-disjoint circuits. Hence β
is optimal iff πΊ has no negative-cost circuit. (can find a negative-cost circuit or determine that none exists.) (π, π )-paths: Let π, π βπ, and π={π, π }. Every π-join π½ contains the edge-set of an (π, π )-path (otherwise, the component of the subgraph (π, π½) containing π has only one node of odd degree, which is impossible.). So minimal π-joins are edge-sets of simple (π, π )-paths. Combinatorial Optimization 2016
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(Pf) ( β) Suppose π½ is a π-join and π½β² is a πβ²-join. Let π£βπ.
Deleting a simple (π, π )-path from a (not minimal) π-join π½, we obtain even sets. Hence if the graph contains no negative-cost circuit, optimal π-join (and minimal) is minimum cost simple (π, π )-path. (can solve the shortest simple (π, π )-path problem for undirected graphs when negative edge costs are allowed, but negative cost circuits are not.) Prop 5.24: Let π½β² be a πβ²-join of πΊ. Then π½ is a π-join of πΊ if and only if π½β³π½β² is a (πβ³πβ²)-join of πΊ. (Pf) ( β) Suppose π½ is a π-join and π½β² is a πβ²-join. Let π£βπ. Then |(π½β³π½β²)β©πΏ(π£)| is even β |π½β©πΏ(π£)|β‘| π½β²β©πΏ(π£)| (mod 2) β π£ is an element of neither or both of π and πβ². β π£βπβ³πβ². (β) apply "only if" part with π½ replaced by π½β³π½β² and π replaced by πβ³πβ². ο Combinatorial Optimization 2016
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Solving the Optimal T-Join Problem
First, obtain an algorithm for the case πβ₯0. Then transform a problem with negative costs into one with nonnegative costs. When πβ₯0, there is always an optimal π-join that is minimal. The minimal π-join can be characterized, as follows. Prop 5.25: Every minimal π-join is the union of the edge-sets of |π|/2 edge-disjoint simple paths, which join the nodes in π in pairs. (Pf) {βͺ{ π π }} is a π-join. Now, enough to show that any π-join contains such a set of edge-disjoint paths. Let π’βπ, and let π» be the component of (π, π½) that contains π’. Then there is a node π£β π’ in πβ©π». (otherwise π’ is the only node in π» of odd degree) So there is a simple (π’, π£) path π such that πΈ(π)βπ½. Now π½\E(π) is a πβ²-join, where πβ²=πβ{π’, π£} by Proposition Repeat the argument. ο Combinatorial Optimization 2016
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π(π½\E(π))+π(πΈ(πβ²)β2π( (π½\E(π))β©πΈ(πβ²) ) β€π(π½)βπ(πΈ(π))+π(πΈ(πβ²))<π(π½),
Suppose (πβ₯0 and) an optimal π-join is expressed as union of edge-sets of paths as in Proposition 5.25, and π is one of these paths, with π joining π’, π£βπ. Then π is a minimum cost (π’, π£)-path. Suppose there is a (π’, π£)-path πβ² in πΊ that has smaller cost than π. By Proposition 5.24, π½β³πΈ(π)β³πΈ(πβ²) is a π-join. Since πΈ(π)βπ½, its cost is π(π½\E(π))+π(πΈ(πβ²)β2π( (π½\E(π))β©πΈ(πβ²) ) β€π(π½)βπ(πΈ(π))+π(πΈ(πβ²))<π(π½), a contradiction. Hence Prop 5.26: Suppose that πβ₯0. Then there is an optimal π-join that is the union of |π|/2 edge-disjoint shortest paths joining the nodes of π in pairs. Combinatorial Optimization 2016
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s.t. π’1π£1, β¦, π’ππ£π is a pairing of the elements of π.
For any pair π’, π£ of nodes in π, let π(π’, π£) be the cost of a least cost (π’, π£)-path in πΊ. Let |π|=2π. The minimum cost π-join is (assuming πβ₯0) minimize π π’ π , π£ π :π=1,β¦,π s.t π’1π£1, β¦, π’ππ£π is a pairing of the elements of π. Form a complete graph πΊ = π, πΈ , give edge π’π£ weight π(π’, π£), and find a minimum-weight perfect matching of πΊ . Join the selected pairs in πΊ using shortest paths. If some edges overlap (since ππ=0 allowed), take symmetric difference. π 8 2 π π π£ π 1 1 1 4 4 3 2 2 π 4 π’ π 1 3 9 π‘ Combinatorial Optimization 2016
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π π 8 2 5 2 6 5 7 π π π 3 π π£ 1 π 6 1 1 4 4 5 4 3 4 3 3 2 2 π 4 π’ π π’ π 7 1 3 9 3 π π‘ π‘ 8 2 π π π£ π 1 1 1 4 3 4 2 2 π 4 π’ π Optimal π-join 1 3 9 π‘ Combinatorial Optimization 2016
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Negative Costs Given πβπ
πΈ, let π={πβπΈ: ππ<0}. Let πβ² be the set of nodes of πΊ that have odd degree in the subgraph (π, π). Then π is a πβ²-join. By proposition 5.24, π½ is a π-join β π½β³π is a (πβ³πβ²)-join. π(π½)=π(π½\N)+π(π½β©π) =π(π½\N)βπ(πβJ)+π(π\J)+π(π½βπ) =|π|(π½β³π)+π(π). ( |π| is the vector defined by |π|π=|ππ|.) π(π) is a constant that does not depend on π½. Hence π½ is an optimal π-join w.r.t. cost vector π β π½β³π is an optimal (πβ³πβ²)-join w.r.t. cost vector |π|. Combinatorial Optimization 2016
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Optimal T-Join Algorithm
Step 1. Identify the set π of edges having negative cost, and the set πβ² of nodes incident with an odd number of edges from π. Replace π by |π| and π by πβ³πβ². Step 2. Find a least-cost (π’, π£)-path ππ’π£ w.r.t. cost vector π for each pair π’, π£ of nodes from π. Let π(π’, π£) be the cost of ππ’π£. Step 3. Form a complete graph πΊ = π, πΈ with π’π£ having weight π(π’, π£) for each π’π£β πΈ . Find a minimum-weight perfect matching π in πΊ . Step 4. Let π½ be the symmetric difference of the edge-sets of paths ππ’π£ for π’π£βπ. Step 5. Replace π½ by π½β³π. Combinatorial Optimization 2016
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|π|-optimal (πβ³πβ²)-join
π 2 π π 2 π -2 2 2 2 5 5 -3 3 π π π π π π 1 6 1 6 1 2 1 2 2 2 π π π π πΊ, π, π πΊ, πβ³πβ², |π| π π π π π π π π π π π π π π |π|-optimal (πβ³πβ²)-join π-optimal π-join Combinatorial Optimization 2016
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