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Tag Archives: 3Partition
Protected: JobShop Scheduling
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Posted in Appendix: Sequencing and Scheduling
Tagged 3Partition, Difficulty 2, Difficulty 5, FlowShop Scheduling, JobShop Scheduling, SS18
Protected: FlowShop Scheduling
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Posted in Appendix: Sequencing and Scheduling
Tagged 3Partition, Difficulty 5, FlowShop Scheduling, OpenShop scheduling, SS15
Protected: Sequencing to Minimize Weighted Tardiness
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Posted in Appendix: Sequencing and Scheduling
Tagged 3Partition, Scheduling to Minimize Weighted Tardiness, SS5
Protected: Dynamic Storage Allocation
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Posted in Appendix: Storage and Retrieval
Tagged 3Partition, Difficulty 7, Dynamic Storage Allocation, No G&J reference, Partition Into Cliques, SR2
Protected: Bin Packing Take 2
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Posted in Appendix: Storage and Retrieval
Tagged 3Partition, Bin Packing, Difficulty 3, No G&J reference, SR1, uncited reduction
Protected: Numerical 3Dimensional Matching
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Posted in Appendix Sets and Partitions
Tagged 3Partition, 3DM, 4Partition, Difficulty 6, Difficulty 8, Numerical 3Dimensional Matching, SP15, SP16
Protected: Intersection Graph For Segments On A Grid
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Posted in Appendix Network Design
Tagged 3Partition, Bin Packing, Difficulty 7, Intersection Graph For Segments on a Grid, ND46
Weighted Diameter
It took over a year (the first problem strictly from the Appendix was Domatic Number, last August), but we’re finally at the end of the Graph Theory section! And the last problem is one that’s actually good for students to solve.
The problem: Weighted Diameter. This is problem GT65 in the appendix
The description: Given a graph G=(V,E) a collection C of E not necessarily distinct, nonnegative integers, and a positive integer K. Can we find a onetoone function f mapping each edge in E to an element of C such that if f(e) is the length of edge e, then G has a diameter of K or less.
In other words, given a set of edge weights C, can we give each edge in E a (distinct) weight from C such that the resulting weighted graph has a path between any two vertices of length ≤ K?
Example: Suppose I have a graph:
And C= {1,2,2,2,3,5}. I think the best was you can label this is:
The Diameter here is 7 (The length of the path from AD).
The reduction: G&J say to use 3Partition, so we’ll go with that. We’re given a set A, with 3m elements, a bound B, and want to split the elements of B into sets of size 3 so that each set adds up to m. We know several things about A, but the important thing for our purposes is that all of the elements in A add up to m*B.
We’ll also assume that A is at least 9. If it’s smaller than that, we can just bruteforce the answer.
What we’re going to do is build a graph that is a tree with 3m+1 vertices. We have a root, and the root has m chains of length 3 extending from it. This gives us exactly 3*m edges.
We set K = 2*B, and set C = A
If there exists a 3parition of A, then each of the sets of 3 elements can map onto a different chain in the graph. This makes the longest path in the graph be between any 2 leaves. Since the length from a leaf to a root is exactly B, the diameter of the graph is 2B.
If there exists a weighted diameter of the graph of cost 2*B, then we need to show the the cost of each chain is exactly B. Suppose it wasn’t, and the cost from the root to some leaf v is > B, let’s say B+x. Then , since there are least 3 chains (since A >= 9) and since the sum of all of the weights is m*B exactly), there must exist some leaf w, with the cost of the chain from the root to w > Bx. The cost of the path from v to w is now > 2B, a contradiction.
If the cost from the root to some leaf is < B, then there must be some other leaf u with the cost from the root to u > B (since the cost of all of the edges add up to m*B) , and we can do the above on u.
Since each chain costs exactly B, we can use the edge weights of each chain as the sets of 3 elements that make our 3Partition.
Difficulty: 4. G&J does say in the comments that this problem is NPComplete even for trees, so that may have been a hint. The proof is a little tricky (getting from “diameter ≤ K” to “set adding up to exactly K” requires some work, and there may be a more elegant way than what I did). But I think this would make a good homework problem.
Posted in AppendixGraph Theory
Tagged 3Partition, Difficulty 4, GT65, Weighted Diameter
Protected: Directed Bandwidth
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Posted in AppendixGraph Theory, Uncategorized
Tagged 3Partition, Bandwidth, Difficulty 10, Directed Bandwidth
Protected: 3Partition
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Posted in Appendix Sets and Partitions
Tagged 3Partition, 4Partition, Difficulty 8, SP15