Odds and Ends

1
Odds and Ends

• HP p HC (again)
• Turing reductions
• Strong NP-completeness versus Weak
  NP-completeness
• Vertex Cover to Hamiltonian Cycle

2
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?

3
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.

4
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,vi1)
  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).

5
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,vi1) for 1 i n-1

6
Turing Reducibility

• Consider the following 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.

7
Number Problems

• Problems where the inputs are numbers
• Prime number problem
• Input Integer n
• Yes/No Question Is n prime?
• Partition problem
• Input Set S of n numbers s1, , sn
• Yes/No Question Is there an S subset of S such
  that the sum of numbers in S the sum of
  numbers in S S.
• What is the input size for these problems?

8
Knapsack Problem

• 0-1 Knapsack optimization problem
• Input
• Capacity K
• n items with weights wi and values vi
• Yes/No Question
• Find a set of items S such that
• the sum of weights of items in S is at most K
• the sum of values of items in S is maximized
• We gave a dynamic programming solution for this
  problem
• We showed that Partition p Knapsack on hw 8
• Is this a contradiction?

9
Definining subproblems

• Define P(i,w) to be the problem of choosing a set
  of objects from the first i objects that
  maximizes value subject to weight constraint of
  w.
• Impose an arbitrary ordering on the items
• V(i,w) is the value of this set of items
• Original problem corresponds to V(n, K)

10
Recurrence Relation/Running Time

• V(i,w) max (V(i-1,w-wi) vi, V(i-1, w))
• A maximal solution for P(i,w) either
• uses item i (first term in max)
• or does NOT use item i (second term in max)
• V(0,w) 0 (no items to choose from)
• V(i,0) 0 (no weight allowed)
• What is the running time of this solution?
• Number of table entries
• Time to fill each entry

11
Example
Items
wA 2 vA 40 wB 3 vB 50 wC 1 vC
100 wD 5 vD 95 wE 3 vE 30
Weight
12
Weak NP-completeness

• An NP-complete problem is called weakly
  NP-complete if it has in its description one or
  more integer parameters and the corresponding
  problem where these parameters are represented in
  unary is in P.
• An NP-complete problem is strongly NP-complete if
  the problem is still NP-complete even if integer
  parameters are encoded in unary

13
Vertex Cover to Ham Cycle
Edge Component For every edge in the Minimum
Vertex Cover problem, we create a component in
the Hamiltonian Cycle Problem
v
u
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Observations.
u
v
u
v
u
v
v?
v?
u?
u?
v?
u?
There are only three possible ways that a cycle
can include all of the vertices in this
component. Key property If a path enters v (u),
it leaves on v (u)
15
Node Selection
u
v
All components that represent edges connected to
node u are strung together into a chain. If
there are V vertices, then we will have V of
these chains, all interwoven. Choosing a node u
corresponds to traversing such a chain
u?
v?
u
w
w?
u?
u
x
x?
u?
16
u
w
y
v
u?
w?
y?
v?
u
v
y
w
v
u
x
z
v?
u?
v
u
x
z
Vertex cover (v,u)
x?
u?
z?
v?
17
u
w
y
v
u?
w?
y?
v?
u
v
y
w
v
u
x
z
v?
u?
v
u
x
z
Vertex cover (v,w,x)
x?
u?
z?
v?
18
Tying the Chains Together
If we want to know if its possible to cover the
original graph using only k vertices, this would
be the same as seeing if we can include all of
the vertices using only k chains. How can we
include exactly k chains in the Hamiltonian Cycle
problem? We must add k extra vertices and
connect each of them to the beginning and end of
every chain. Since each vertex can only be
included once, this allows k chains in the final
cycle.
19
Beginning a Transform








20
The Final Transform for k1