-- Apr 24 In-Class Exercise
Problem: For each `n\ge0`, give an example graph on `n+1` vertices that has a vertex cover of size 1.
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Solution: Let graph `G` be a graph on `n+1` vertices with a single edge. Let `overline{G}` be the complement of `G`, i.e. `G` will be a nearly complete graph missing just 1 edge. Therefore the `\overline{G}` will have a clique of size `n+1-1 = n`. From our proof of vertex cover, if `\overline{G}` has a clique of size `n`, then G has a vertex cover of size `n+1-n = 1`.
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Problem: a `p`-time isomorphism is a `p`-time reduction which is one-to one and onto and whose inverse is `p`-time computable. Argue the reduction on the last slide is an isomorphism.
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Solution: The complement of graph `G` can be computed in polynomial time and is unique to any specific graph as the complement of a graph is bijective. We can prove this by contradiction by assuming graphs `G1` and `G2` to be distinct with the same complement `\overline{G}`. However, the complement of the complement `G = \overline{\overline{G}}` is distinct and cannot be both `G1` and `G2`, leading to a contraction. Therefore the complement is bijective. Because the complement can be found in polynomial time and is a bijection on the set of all graphs `\mathbb{G}`, the reduction relied on using the complement to find the clique and therefore is a `p`-time isomorphism.
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Problem: What instances of CLIQUE would map to your graphs above?
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Solution: All cliques of size `n` for graphs on `n+1` vertices.
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Edited: 2019-04-24)
'''Problem:''' For each @BT@n\ge0@BT@, give an example graph on @BT@n+1@BT@ vertices that has a vertex cover of size 1.
@BT@@BT@
'''Solution:''' Let graph @BT@G@BT@ be a graph on @BT@n+1@BT@ vertices with a single edge. Let @BT@overline{G}@BT@ be the complement of @BT@G@BT@, i.e. @BT@G@BT@ will be a nearly complete graph missing just 1 edge. Therefore the @BT@\overline{G}@BT@ will have a clique of size @BT@n+1-1 = n@BT@. From our proof of vertex cover, if @BT@\overline{G}@BT@ has a clique of size @BT@n@BT@, then G has a vertex cover of size @BT@n+1-n = 1@BT@.
@BT@@BT@
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@BT@@BT@
'''Problem:''' a @BT@p@BT@-time isomorphism is a @BT@p@BT@-time reduction which is one-to one and onto and whose inverse is @BT@p@BT@-time computable. Argue the reduction on the last slide is an isomorphism.
@BT@@BT@
'''Solution:''' The complement of graph @BT@G@BT@ can be computed in polynomial time and is unique to any specific graph as the complement of a graph is bijective. We can prove this by contradiction by assuming graphs @BT@G1@BT@ and @BT@G2@BT@ to be distinct with the same complement @BT@\overline{G}@BT@. However, the complement of the complement @BT@G = \overline{\overline{G}}@BT@ is distinct and cannot be both @BT@G1@BT@ and @BT@G2@BT@, leading to a contraction. Therefore the complement is bijective. Because the complement can be found in polynomial time and is a bijection on the set of all graphs @BT@\mathbb{G}@BT@, the reduction relied on using the complement to find the clique and therefore is a @BT@p@BT@-time isomorphism.
@BT@@BT@
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@BT@@BT@
'''Problem:''' What instances of CLIQUE would map to your graphs above?
@BT@@BT@
'''Solution:''' All cliques of size @BT@n@BT@ for graphs on @BT@n+1@BT@ vertices.