1 CS21 Decidability and Tractability Lecture 21 February 27 2019 February 27 2019 CS21 Lecture 21 Outline NPcomplete problems Hamilton path and cycleTraveling Salesperson Problem ID: 760752
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Slide1
February 27, 2019
CS21 Lecture 21
1
CS21 Decidability and Tractability
Lecture
21
February 27, 2019
Slide2February 27, 2019
CS21 Lecture 21
Outline
NP-complete problems: Hamilton path and cycle,Traveling Salesperson ProblemNP-complete problems: Subset SumNP-complete problems: NAE-3-SAT, max cut
2
Slide3February 27, 2019
CS21 Lecture 21
3
Hamilton Path
Definition: given a directed graph G = (V, E), a
Hamilton path
in G is a directed path that touches every node exactly once.
A language
(decision problem):
HAMPATH = {(G, s, t) : G has a Hamilton path
from s to t
}
Slide4February 27, 2019
CS21 Lecture 21
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HAMPATH is NP-complete
Theorem: the following language is NP-complete:HAMPATH = {(G, s, t) : G has a Hamilton path from s to t}Proof:Part 1: HAMPATH NP. Proof?Part 2: HAMPATH is NP-hard.reduce from?
Slide5February 27, 2019
CS21 Lecture 21
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HAMPATH is NP-complete
We are reducing
from the language
:
3SAT = {
φ
:
φ
is a 3-CNF formula that has a satisfying assignment }
to the language:
HAMPATH = {(G, s, t) : G has a Hamilton path from s to t}
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HAMPATH is NP-complete
We want to construct a graph from
φ
with the following properties:
a satisfying assignment to
φ
translates into a Hamilton Path from s to t
a Hamilton Path from s to t can be translated into a satisfying assignment for
φ
We will build the graph up from pieces called
gadgets
that “simulate” the clauses and variables of
φ
.
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HAMPATH is NP-complete
The variable gadget (one for each xi):
x
i
true:
x
i
false:
…
…
…
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HAMPATH is NP-complete
…
…
…
:
“x
1
”
“x
2
”
“x
m
”
s
t
path from s to t translates into a truth assignment to x
1
…x
m
why must the path be of this form?
Slide9February 27, 2019
CS21 Lecture 21
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HAMPATH is NP-complete
φ = (x1x2 x3)(x1 x4 x3) … (…)How to ensure that all k clauses are satisfied?need to add nodes can be visited in path if the clause is satisfiedif visited in path, implies clause is satisfied by the assignment given by path through variable gadgets
Slide10February 27, 2019
CS21 Lecture 21
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HAMPATH is NP-complete
φ = (x1x2 x3)(x1 x4 x3) … (…)Clause gadget allows “detour” from “assignment path” for each true literal in clause
…
…
…
“x
1
”
“x
2
”
“x
3
”
“C
1
”
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CS21 Lecture 21
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HAMPATH is NP-complete
One clause gadget for each of k clauses:
…
“x
1
”
…
“x
2
”
…
“x
m
”
:
:
:
“C
1
”
“C
2
”
“C
k
”
:
for clause 1
for clause 2
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CS21 Lecture 21
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HAMPATH is NP-complete
…
…
…
:
“x
1
”
“x
2
”
“x
m
”
s
t
“C
1
”
“C
2
”
“C
k
”
:
φ
= (x
1
x
2
x
3
)
(
x
1
x
4
x
3
)
f(φ) is this graph (edges to/from clause nodes not pictured)
f poly-time computable?
# nodes = O(km)
Slide13February 27, 2019
CS21 Lecture 21
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HAMPATH is NP-complete
…
…
…
:
“x
1
”
“x
2
”
“x
m
”
s
t
“C
1
”
“C
2
”
“C
k
”
:
φ
= (x
1
x
2
x
3
)
(
x
1
x
4
x
3
)
YES maps to YES?
first form path from satisfying assign.
pick true literal in each clause and add detour
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CS21 Lecture 21
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HAMPATH is NP-complete
…
…
…
:
“x
1
”
“x
2
”
“x
m
”
s
t
“C
1
”
“C
2
”
“C
k
”
:
φ
= (x
1
x
2
x
3
)
(
x
1
x
4
x
3
)
NO maps to NO?
try to translate path into satisfying assignment
if path has “intended” form, OK.
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HAMPATH is NP-complete
What can go wrong?path has “intended form” unless return from clause gadget to different variable gadget
…
…
…
“x
i
”
“x
j
”
“x
h
”
“
C
i
”
we will argue that this cannot happen:
Slide16February 27, 2019
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HAMPATH is NP-complete
…
…
:
Case 1 (positive occurrence of v in clause)
:
c
x
y
z
path must visit y
must enter from x, z, or c
must exit to z (x is taken)
x, c are taken. can’t happen
“v”
Slide17February 27, 2019
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HAMPATH is NP-complete
…
…
:
Case 2 (negative occurrence of v in clause):
c
x
y
path must visit y
must enter from x or z
must exit to z (x is taken)
x is taken. can’t happen
“v”
z
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Undirected Hamilton Path
HAMPATH refers to a directed graph.
Is it easier on an undirected graph?
A language
(decision problem):
U
HAMPATH = {(G, s, t) :
undirected
G has a Hamilton path from s to t}
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CS21 Lecture 21
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UHAMPATH is NP-complete
Theorem: the following language is NP-complete:UHAMPATH = {(G, s, t) : undirected graph G has a Hamilton path from s to t}Proof:Part 1: UHAMPATH NP. Proof?Part 2: UHAMPATH is NP-hard.reduce from?
Slide20February 27, 2019
CS21 Lecture 21
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UHAMPATH is NP-complete
We are reducing
from the language
:
HAMPATH = {(G, s, t) : directed graph G has a Hamilton path from s to t}
to the language:
UHAMPATH = {(G, s, t) : undirected graph G has a Hamilton path from s to t}
Slide21February 27, 2019
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UHAMPATH is NP-complete
The reduction:
s
t
G
s
out
t
in
G’
u
v
u
in
u
mid
u
out
v
in
v
mid
v
out
replace each node with three (except s, t)
(u
in
, u
mid
)
(u
mid
, u
out
)
(u
out
, v
in
) iff G has (u,v)
Slide22February 27, 2019
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UHAMPATH is NP-complete
Does the reduction run in poly-time?
YES maps to YES?
Hamilton path in G:
s, u
1
, u
2
, u
3
, …, u
k
, t
Hamilton path in G’:
s
out
, (u
1
)
in
, (u
1
)
mid
, (u
1
)
out
, (u
2
)
in
, (u
2
)
mid
, (u
2
)
out
, … (u
k
)
in
, (u
k
)
mid
, (u
k
)
out
, t
in
Slide23February 27, 2019
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UHAMPATH is NP-complete
NO maps to NO?
Hamilton path in G’:
s
out
, v
1
, v
2
, v
3
, v
4
, v
5
, v
6
, …, v
k-2
, v
k-1
, v
k
, t
in
v
1
= (u
i1
)
in
for some i
1
(only edges to ins)
v
2
= (u
i1
)
mid
for some i
1
(only way to enter mid)
v
3
= (u
i1
)
out
for some i
1
(only way to exit mid)
v
4
= (u
i2
)
in
for some i
2
(only edges to ins)
...
Hamilton path in G:
s, u
i1
, u
i2
, u
i3
, …, u
ik
, t
Slide24February 27, 2019
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Undirected Hamilton Cycle
Definition: given a undirected graph G = (V, E), a
Hamilton cycle
in G is a
cycle
in G that touches every node exactly once.
Is finding one easier than finding a Hamilton path?
A language
(decision problem):
U
HAMCYCLE = {G : G has a Hamilton cycle}
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UHAMCYCLE is NP-complete
Theorem: the following language is NP-complete:UHAMCYCLE = {G: G has a Hamilton cycle}Proof:Part 1: UHAMCYCLE NP. Proof?Part 2: UHAMCYCLE is NP-hard.reduce from?
Slide26February 27, 2019
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UHAMCYCLE is NP-complete
The reduction (from UHAMPATH):
s
t
G
s
t
G’
H. path from s to t implies H. cycle in G’
H. cycle in G’ must visit u via red edges
removing red edges gives H. path from s to t in G
u
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Traveling Salesperson Problem
Definition: given n cities v1, v2, …, vn and inter-city distances di,j a TSP tour in G is a permutation of {1…n}. The tour’s length is Σi = 1…n d(i),(i+1) (where n+1 means 1).A search problem:given the {di,j}, find the shortest TSP tourcorresponding language (decision problem):TSP = {({di,j : 1 ≤ i<j ≤ n}, k) : these cities have a TSP tour of length ≤ k}
Slide28February 27, 2019
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TSP is NP-complete
Theorem: the following language is NP-complete:TSP = {({di,j : 1 ≤ i<j ≤ n}, k) : these cities have a TSP tour of length ≤ k}Proof:Part 1: TSP NP. Proof?Part 2: TSP is NP-hard.reduce from?
Slide29February 27, 2019
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TSP is NP-complete
We are reducing
from the language
:
UHAMCYCLE = {G : G has a Hamilton cycle}
to the language:
TSP = {({d
i
,
j
: 1 ≤ i<j ≤ n}, k) : these cities have a TSP tour of length
≤ k
}
Slide30February 27, 2019
CS21 Lecture 21
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TSP is NP-complete
The reduction:given G = (V, E) with n nodes produce:n cities corresponding to the n nodesdu,v = 1 if (u, v) Edu,v = 2 if (u, v) Eset k = n
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TSP is NP-complete
YES maps to YES?
if G has a Hamilton cycle, then visiting cities in that order gives TSP tour of length n
NO maps to NO?
if TSP tour of length ≤ n, it must have length exactly n.
all distances in tour are 1. Must be edges between every successive pair of cities in tour.
Slide32February 27, 2019
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Subset Sum
A language
(decision problem):
SUBSET-SUM = {(S = {a
1
, a
2
, a
3
, …, a
k
}, B) : there is a subset of S that sums to B}
example:
S = {1, 7, 28, 3, 2, 5, 9, 32, 41, 11, 8}
B = 30
30 = 7 + 3 + 9 + 11. yes.
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Subset Sum
SUBSET-SUM = {(S = {a1, a2, a3, …, ak}, B) : there is a subset of S that sums to B}Is this problem NP-complete? in P?Problem set: in TIME(Bpoly(k))
Slide34February 27, 2019
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SUBSET-SUM is NP-complete
Theorem: the following language is NP-complete:SUBSET-SUM = {(S = {a1, a2, a3, …, ak}, B) : there is a subset of S that sums to B}Proof:Part 1: SUBSET-SUM NP. Proof?Part 2: SUBSET-SUM is NP-hard.reduce from?
our reduction had better produce super-
polynomially
large B (unless we want to prove P=NP)