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Example: HP ≤ p HC. Hamiltonian Path Input: Undirected Graph G = (V,E) Y/N Question: Does G contain a Hamiltonian Path? Hamiltonian Cycle Input: Undirected Graph G = (V,E) Y/N Question: Does G contain a Hamiltonian Cycle?. Specification of R(x).

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Example: HP ≤ p HC

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Example hp p hc l.jpg

Example: HP ≤p HC

  • Hamiltonian Path

    • Input: Undirected Graph G = (V,E)

    • Y/N Question: Does G contain a Hamiltonian Path?

  • Hamiltonian Cycle

    • Input: Undirected Graph G = (V,E)

    • Y/N Question: Does G contain a Hamiltonian Cycle?


Specification of r x l.jpg

Specification of R(x)

  • Consider any undirected graph G = (V,E) as input x

  • R(x) will be a graph G’ = (V’, E’) where

    • V’ = V union {v} where v is not in V

    • E’ = E union {(v,w) | w in V}

  • Argument that R(x) has polynomial size

    • We add exactly 1 node and |V| edges.


X is yes r x is yes l.jpg

x is yes  R(x) is yes

  • Suppose graph G has a Hamiltonian Path

  • Let this path be v1, v2, …, vn

  • We now argue that v1, v2, …, vn, v is a Hamiltonian Cycle in G’

    • First, all nodes in V’ are included exactly once above or else v1, v2, …, vn would not be a HP in G

    • Since G’ has all the edges that G has, (vi,vi+1) is an edge in E’ for 1 ≤ i ≤ n-1

    • Finally, since E’ contains edge (v,w) for all w in V, it must be the case that E’ contains edges (vn, v) and (v,v1).


R x is yes x is yes l.jpg

R(x) is yes x is yes

  • Suppose graph G’ has a Hamiltonian Cycle

  • Let this cycle be v1, v2, …, vn, v

  • We now argue that v1, v2, …, vn is a Hamiltonian Path in G

    • First, all nodes in V are included exactly once above or else v1, v2, …, vn, v would not be a HC in G’

    • Since the only extra edges in E’ compared to E are edges involving node v, it must be the case that E contains edge (vi,vi+1) for 1 ≤ i ≤ n-1


Turing reducibility l.jpg

Turing Reducibility

  • Steve made a suggestion for an alternate reduction.

  • Given graph G, output Q(n2) graphs Gv,w = (V,Ev,w) where

    • Ev,w = E union {(v,w)} where v,w are nodes in V

  • This is not a polynomial-time reduction because we are outputting Q(n2) graphs.

  • However, this idea can be used to show that if HC can be solved in polynomial time, then HP can be solved in polynomial time.

    • Run each graph Gv,w through our procedure that solves HC.

    • If HC says yes for any one of these graphs, return yes.

    • Otherwise return no.

  • This more general reduction is often called a Turing reduction.

  • We allow ourselves to use the procedure that solves HC (or P2) a polynomial number of times rather than just once.


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