NPComplete

1
NP-Complete

• Proof of correctness More reductions

2
Definitions

• P problems that can be solved in polynomial time
  (typically in n, size of input) on a
  deterministic Turing machine
• Any normal computer simulates a DTM
• NP problems that can be solved in polynomial
  time on a non-deterministic Turing machine
• Informally, if we could guess the solution, we
  can verify the solution in P time (on a DTM)
• NP does NOT stand for non-polynomial, since there
  are problems harder than NP
• P is actually a subset of NP (we think)

3
Definitions, continued

• NP-hard
• At least as hard as any known NP problem (could
  be harder!)
• Set of interrelated problems that can be solved
  by reducing to another known problem
• NP-Complete
• A problem that is in NP and NP-hard
• Cooks Theorem
• SATISFIABILITY (SAT) is NP-Complete
• Other NP-Complete problems
• Reduce to SAT or previous reduced problem

4
Complexity Classes at-a-glance

• Image taken from Jeff Erickson's lecture notes,
  http//compgeom.cs.uiuc.edu/jeffe/teaching/algori
  thms/notes/21-nphard.pdf
• We will not discuss co-NP today

5
Correctness

• To prove unknown problem y is NP-Complete
• Prove y is in NP
• Prove y is at least as hard as some x in
  NP-Complete
• Transform x into y in polynomial time such that x
  is yes if and only if y is yes
• Requires two proofs, one for each direction x
  yes ? y yes, y yes ? x yes (or x no ? y no)
• Must also prove transformation is polynomial

6
Example Independent Set

• Decision Problem is there a set of k vertices
  such that none are adjacent?
• Proof of NP-Completeness
• Is this in NP?
• Easy to check in k2 (Adj. Matrix) or km (Adj.
  List)
• Is this in NP-hard?
• Reduce known problem to unknown problem
• NOT the other way!
• We will use 3-SAT

7
Reductions

• Always solve a known problem by transforming it
  into an instance of the unknown problem
• Ex 3SAT is NP-Complete. Independent Set is
  unknown. By transforming 3SAT into ISET and
  solving, we prove ISET is NP-Complete
• If you go the other way, you might just make a
  bad transformation
• Sorting is an unknown. Transform sorting into
  3SAT (somehow) and solve. This does NOT mean
  sorting is NP-Complete!
• Think of it this way If I can solve x by
  transforming x into y, then y is at least as hard
  as x. (Corollary if y can be solved quickly,
  so can x.)

8
3-SAT p ISET
For more details, review Lecture 18. This is a
slightly modified version that does not use the
Literal widgets but just connects each node
directly to any node that is its complement.
C ?v1, v2, ?v3 , v1, ?v2, ?v4 , ?v2,
?v4, v5 , v3, v4, v5
v2
v1
?v2
v3
?v1 ?v3
?v2 ?v4
?v4 v5
v4 v5
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3-SAT to ISET, continued

• The size of the transformation on the last slide
  is polynomial in the number of clauses (m)
  number of literals (n)
• Adding opposing literals edges in the worst-case
  (full exploration of graph for each vertex) is
  still polynomial in m², so the time of the
  transformation is also polynomial
• Now we must prove 3SAT yes ? ISET yes
• 3SAT ? ISET
• Set target size for ISET m
• Identify only one true literal in each clause
  of 3SAT, select that vertex in that triangle in
  ISET. Our set has no edges between nodes in the
  same triangle.
• Because this is a legal TA for 3SAT, we cant
  select a complementary version of a previously
  selected literal, so no edges will connect nodes
  in different triangles.

10
Visual demonstration of 3SAT ? ISET

• C ?v1, v2, ?v3 , v1, ?v2, ?v4 ,
• ?v2, ?v4, v5 , v3, v4, v5
• Select nodes corresponding to literals in violet.
  These form an independent set of size m.

v2
v1
?v2
v3
?v1 ?v3
?v2 ?v4
?v4 v5
v4 v5
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3SAT ? ISET

• So we have an ISET of size m, how does this give
  us a valid TA for 3SAT?
• At most one node comes from each triangle
• There are m triangles, so there must be one node
  in every triangle ? m clauses covered
• Because of the opposing literals edges, no
  conflicting truth assignments can be in ISET
• Assign the literals in the ISET to true other
  literals dont matter

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Hamiltonian Path H. Cycle

• In Lecture 18, a (complex) proof was given that
  HAMCYCLE (aka TSP) was NP-Complete
• How can we show that HAMPATH is also NP-Complete?
• Take graph G and modify to G'
• Show transformation is polynomial
• Show that G' has HAMPATH if G has HAMCYCLE
• Show that G has HAMCYCLE if G' has HAMPATH

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HAMPATH/HAMCYCLE, cont'd
G'
V2'
G
V2
V
Then create V1' and V2' adjacent only to V1 and
V2, respectively. (The transformation is
trivially polynomial.)
V1
Split any vertex V into two identical vertices V1
and V2 (do not create E(V1,V2)).
V1'

• Run HAMPATH on G'
• If G had a HAMCYCLE, G' has a path through all
  the vertices except V1' and V2' starting at V1
  and ending at V2 that corresponds to the cycle.
  Add V1' to the start of the path, V2' to the end,
  and it's a HAMPATH.
• If G' has a HAMPATH, it must start/end at V1'/V2'
  since they have degree 1, so G has a HAMCYCLE
  starting from V

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Hitting Set

• Problem, given a set T of sets s1, s2 sn, is
  there a set H of at most k elements such that H
  contains at least one element from every s1sn?
• Example T 1,2,3,a,1,a,b,c,?
• k2, no valid H!
• k3, H 1,a,? (other choices exist)
• How hard is this problem?

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Hitting Set Complexity

• Easy to show problem is in NP would take at most
  nk time to verify a given solution H
• Is this NP-Complete?
• Think about other problems you know are
  NP-Complete SAT (and cousins), Independent Set,
  HP/HC, Vertex Cover (worksheet), Maximum Clique
  (worksheet)
• Look for a similar problem something that
  involves selecting one item to represent some
  collection of items
• Work in groups to find a reduction and argue that
  it is correct

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Vertex Cover p Hitting Set

• Both involve "select some items such that all
  items are represent"
• Construct Given a graph G, for each edge e
  (u,v) create a set s u,v and add s to the
  universe T.
• Now prove G has a vertex cover of size k ? T has
  a hitting set of size k
• Select k vertices to represent m edges vs. select
  k elements to represent m sets

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Vertex Cover ? Hitting Set

• Vertex Cover ? Hitting Set
• Suppose C is a cover of G of size k
• By definition then, for every edge (u,v) in G,
  either u ? C or v ? C
• Then H can be set to C, and H must intersect
  every set in T.
• Vertex Cover ? Hitting Set
• Suppose H is a hitting set of T of size k
• Since H intersects every set, it has at least one
  endpoint of every edge. Set C to be H.

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Set Cover

• Problem, given a set T of sets s1, s2 sn, is
  there a set S' of at most k sets such that S'
  contains every element in T?
• Example T 1,2,3,a,1,a,b,c,?
• k 2, no legal S'!
• k 3, S' 1,2,3,a,b,c,?
• This should look very familiar!
• Work in groups to find a reduction and argue that
  it is correct

19
Vertex Cover p Set Cover

• Given a graph G, construct a universe of sets T
• For each vertex V, create a set sv with all the
  edges incident on V
• Our universe is now the set of all edges in G
  (each is listed twice)

T ab,ac,ae,ba,bc,bd,ca,cb,db,de,df,ea
,ed,ef,fd,fe
A
D
B C
E F
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Vertex Cover ? Set Cover

• Vertex Cover ? Set Cover
• Suppose C is a cover of G of size k
• To find S', select sets sw for every w ? C.
• If there was an edge e(u,v) in T not in S', then
  neither su nor sv was selected, so neither u nor
  v was in the Vertex Cover, a contradiction.
• Vertex Cover ? Set Cover
• Suppose S' is a set cover of T of size k
• To find C, select vertex v for every sv ? S'
• If there is an edge e(u,v) in G not in C, then
  neither vertex u nor v was selected, so neither
  su nor sv was in S'. Since uv can only be found
  in su and sv this contradicts that S' is a cover.

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License

• This lecture is released under the Creative
  Commons License, http//creativecommons.org/licens
  es/by-nc-sa/3.0/