09/09/2005
Lecture on Computational Complexity
Overview
Chapter 1 introduces problems, algorithms and their classifications,
problem classification, P, NP, NP-Complete problems. Chapter 2
introduces the Theory of NP-Completeness. Chapter 3 discusses
methods of proving NP completeness. Chapter 4 shows the way
problems can be analyzed using NP-Completeness theory. Chapter 5
show the techniques for proving NP-Completeness. Chapter 6 discusses
other intractable problems. Chapter 7 discusses other relevant issues.
1.1 Computers, Complexity, and Intractability
Consider the following situation:
After exhaustive brainstorming the recently recruited algorithm
designer says to the boss “ I can’t find an efficient algorithm, I guess
I’m just too dumb”
“ I can’t find an efficient algorithm, because no such algorithm is
possible!”
“I can’t find an efficient algorithm, but neither can all these famous
people.”- so recruiting somebody else will not save the employer.
Theory of computational complexity will enable you to assert the last
statement if the problem you have been asked really falls into the
category of difficult problems that well-know experts in the filed of
algorithm could not solved it, and therefore, with a high probability you
are not going to be fired.
1.2 Problems, Algorithms, and Complexity
A general question to be answered usually possessing several
parameters, or free variables, whose values are left unspecified. A
problem is described by giving: 1) a general description of all its
parameters, and 2) a statement of what properties the answer or
solution is required to satisfy.
Example of TSP
Parameters: a finite set C={c1, c2,…, cm} of cities, and for each pair of
cities ci and c j the distance d (ci , c j ) between them. A solution is an
ordering <c  (1), c  (2), …,c  (m)>  of the given cities in order to minimize
sum of distances of consecutive cities when they are arranged in a cycle
in this order.
Notion of efficiency is related to use of computing resources- computing
speed and memory. Time requirement is expressed in terms of a single
variable called size of the problem determined by the amount of bits
required to represent the problem in a computer. For TSP m can be
treated as size, again m(m-1) + could also be treated as size. It depends
upon he encoding scheme used.
Time complexity function for an algorithm expresses its time
requirement by giving for each possible input length the largest amount
of time needed by the algorithm to solve a problem instance of that size.
1.3 Polynomial Time algorithms and Intractable Problems
f(n) = O(g(n))  |f(n)|  c|g(n)|
A polynomial time algorithm is bounded by a polynomial. If an
algorithm is not polynomial then it is exponential.
Comparison of several polynomial and exponential time algorithms
show fast growing nature of expo algorithms.
Time
Size n
complexity
10
20
30
40
50
60
functions
n
n2
n3
n5
2n
5n
.00001 s
.00002 s
.00003 s
.00004 s
.00005 s
.00006 s
.0001 s
.0004 s
.0009 s
.0016 s
.0025 s
.0036 s
.001 s
.008 s
.027 s
.064 s
.125 s
.216 s
.1 s
3.2 s
24.3 s
1.7 m
5.2 m
13.0 m
.001 s
1.0 s
17.9 m
12.7 d
35.7 y
366 cen
.059 s
58 m
6.5 y
3855cen
2X108 cen
1.3X1013cen
Fig. 1.2: Comparison of polynomial and exponential algorithms
Size of the largest problem instance solvable in 1 hour
Time complexity
function
n
n2
n3
n5
2n
5n
With present computer
With computer 100
times faster
with computer 1000
times faster
100N1
1000N1
N2
10N 2
31.6N2
N3
4.64N3
10N3
N4
2.5N 4
3.98N4
N5
N5  6.64
N5  9.97
N6
N6  4.19
N6  6.29
N1
Fig. 1.3 Effect of improved technology on polynomial and exponential
algorithms
Figure 1.3 shows that even if our computers speed is multiplied the
largest size of the problem that can be handled by the advanced
computers will not be a multiple of the size of the problem the current
computer can handle. This is why Jack Edmonds equated polynomial
algorithms to good algorithms and problems not solvable by poly
algorithms as not well-solved problems or intractable.
There are exceptions: for small sized problems some exponential
algorithm may be better than poly as n5 is slower than 2n for
n  20 Simplex algorithm although exponential performs on the average
very well in practice although Klee and Zadeh were too critical to find
examples where its exponential nature was revealed. Usually we do not
get algorithms requiring n100or1099 n2 time complexity .
Reasonable encoding
Encoding scheme
string
length
v[1]v[2]v[3]v[4](v[1]v[2])(v[2]v[3])
V/E list
(v[2])(v[]v3])(v[2])()
0100 /1010 / 0010 / 0000
4v  10e // 4v  10e  (v  e) log10 v
2v  8e // 2v  8e  2e log10 v
v 2  v  1// v 2  v  1
16/9/2005
1.4 Provably Intractable Problems
Two types of intractability- exponential time, exponential space. Later
implies former. So we deal with time complexity. TSP with a bound B is
an example of later. Possibly such problems are not well posed.
About 70 years back Alan Turing demonstrated that some problems are
really undecidable- no algorithms exist for solving them.
It is impossible to specify an algorithm which given an arbitrary
computer program and an arbitrary input to that program, can decide
whether or not the program will eventually halt when applied to that
input[1936]
Hilbert’s 10th problem(solvability of polynomial equations in integers)
are examples.
The first results of intractable decidable problems were really
artificially created. In the early 70s real life examples were produced by
Meyer, Fisher, Rabin. These problems were from finite automata theory,
formal language theory, mathematical logic.
1.5 NP-Complete Problems
The principal technique used for demonstrating that two problems are
related is that of reducing one problem into another by giving a
constructive transformation.
Many examples are there. Prof Dantzig[1960] reduced combinatorial
optimization problems to 0-1 integer linear programming problem.
Edmonds reduced the graph theoretic problems of covering all edges
with minimum number of vertices and finding a maximum independent
set of vertices to the general set covering problem. Reduction of TSP to
shortest paths with negative edge weights allowed by Dantzig and others.
These early reductions were really isolated. The foundation for the
theory of NP-Completeness were laid down in a paper of Stephen Cook,
presented in 1971 entitled “The Complexity of Theorem proving
procedures”[1971]
First he emphasized the significance of polynomial time reducibility that
is reduction for which the required transformations can be executed
using polynomial time algorithms. Polynomial time reducibility ensures
that a polynomial time algo of one solves the other also in polynomial
time.
Second, he focused attention on the class NP of decision problems that
can be solved in polynomial time by a nondeterministic computer.
Third, he proved that one particular problem in NP,c alled the
satisfiability problem has the property that every other problem in NP
can be polynomially reduced to it. If the satisfiability problem can be
solved in polynomial time then so can be every problem in NP, and if
any problem in NP is intractable then the satisfiability problem must
also be intractable.
Finally Cook suggested that there might be some other problems in NP
which share with the satisfiability problem this property of being the
hardest member in NP.
He immediately showed that the problem “ Does a given graph G
contain a complete subgraph on a given number of vertices. Then Karp
supplied a list of 6 very important problems that formed the core of NP
complete class.
23/9/2005
Chapter 2 The Theory of NP-Completeness
2.1 Decision Problems, Languages and Encoding Scheme
A Decision problem  consists of a set D of instances and a subset
Y  D of yes-instances.
SUBGRAPH ISOMORPHISM PROBLEM
INSTANCE: Two graphs G1  (V1 , E1 ), G2  (V2 , E2 )
QUESTION: Does G1 contain a subgraph isomorphic to G2 ?
TRAVELLING SALESMAN
INSTANCE: A finite set of cities and non negative distances between
them and a bound B
QUESTION: Is there a tour of length <=B?
While we study decision problems optimization problems can also be
transformed into a series of decision problems. So long as cost functions
are easy to evaluate decision problems are no harder than the
optimization problems. Many decision problems can also be shown to be
not easier than their corresponding optimization problems.
For any finite set of symbols  we denote by   the set of all finite
strings of the symbol set. If L is a subset of   then we call it a
language. Every problem is encoded into a string of symbols. These
strings can be partitioned into 3 subsets- those that are not encoding of
the problem, that are examples of yes instances and that are examples of
no instances.
Standard encoding scheme will map instances into structured strings.
a) The binary representation of an integer is a structured string
b) If x is a structured string so is [x]
c) If x1,x2,…, xn are structured strings representing the objects then
(x1,x2,…,xn) is a structured string representing the sequence
2.2 Deterministic Turing Machines and the Class P
DTM consists of a finite state control, a read-write head, and a tape
made up of a two-way infinite sequence of tape squares -2,-1,0,1,2,…
A program for a DTM specifies the following information
a) A finite set
of tape symbols, including a subset
of input symbols
and a distinguished blank symbol
b) a finite set Q of states, including a distinguished start state q0 and
two distinguished halt states qy and qn.
 : (Q  (qn, qy ))    Q    {1, 1}
c) a transition function
Date: 30/9/2005
An NDTM program is specified in exactly the same way as a DTM
programming, including the tape alphabet  , input alphabet  , blank
symbol b, state set Q, initial state q, halt states qy,qn, and transition
function  : (Q  (qn, qy))    Q    {1, 1} Computation takes in 2 steps:
The first stage is the guessing stage. Input string is written in squares 1
through x and read write head is at 1 the write only head at -1 and finite
state control inactive. The guessing module directs write only head one
step at a time either to write something on the tape square being
scanned and move one sq left or right or stop at which point the
guessing module is inactive and the finite state control is activated at q0.
Notice that NDTM program M will have an infinite # of possible
computations for a given input string x, one for each possible guessed
string from * . We say that NDTM program accepts x if at least one of
these is an accepting computation.
LM  {x  * : M accepts x}
NP  {L : there is a polynomial time NDTM program M for which LM  L}
2.4 The Relationship Between P and NP
P  NP
Theorem 2.1 If   NP, then there is a polynomial p such that  can be
solved by a deterministic algorithm having time complexity O(2 p ( n) )
q (n)  c1n c2
If A is an algorithm requiring q(n) time then an NDTM need only guess
strings of length at most q(n) to get to the accepted state. Number of
guesses can be as large at most as k q ( n )
2.5 Polynomial Transformations and NP-Completeness
P and NP-P is meaningful if P  NP
A polynomial transformation from a language L1  1* to a language L 2  2*
is a function f : 1*   2* that satisfies the following two conditions:
a) There is a polynomial time DTM program that computes f
b) x  1* , x  L1  f ( x)  L2
Lemma 2.1 If L1L2 , then L2  P implies L1  P
HAMILTONIAN CIRCUIT
INSTANCE: A graph G=(V,E)
QUESTION: Does G contain a Hamiltonian Circuit?
Lemma 2.2 If L1L2 , L2L3  L1L3
Lemma 2.3 If L1 and L2 belong to NP, L1 is No-Complete, and L1 poly
reducible to L2, then L2 id NP-Complete.
Theorem 2.1 (Cook’s Theorem) SATISFIABILITY is NP-Complete.
In NP is obvious. Now every language in NP can be described by a
polynomial time NDTM program that recognizes it.
3. Proving NP-Completeness Results
1) Showing that   NP
2) selecting a know NP-Complete problem 
3) constructing a polynomial transformation f from  to 
4) proving that f is a (polynomial) transformation.
3.2 Six Basic NP-Complete Problems
3-SATISFIABILITY (3-SAT)
INSTANCE: Collection C={c1,c2,…,cm} of clauses on a finite set U of
variables such that |ci|=3 for all i
QUESTION: Is there a truth assignment for U that satisfies all the
clauses in C?
3-DIMENSIONAL MATCHING (3DM)
INSTANCE: M  W  X  Y , where W,X,Y are disjoint sets having the
same number q of elements.
QUESTION: Does M contain a matching, that is, a subset M’ in M such
that |M’|=q and no 2 elements of M’ agree in any coordinate?
VERTEX COVER (VC)
INSTANCE: A graph G=(V,E) and a positive integer K<=|V|
QUESTION: Is there a vertex cover of size K or less for G, that is a
subset V’ of V such that |V’|<=K and for each edge (u,v) in E, at least
one of u and v belongs to V’?
CLIQUE
INSTANCE: A graph G=(V,E) and a positive integer J<=|V|
QUESTION: Does G contain a clique of size J or more, that is a subset
of vertices V’ in V such that {V’|>=J and every 2 vertices of V’ are
joined by an edge in E?
HAMILTONIAN CIRCUIT (HC)
INSTANCE: A graph G=(V,E)
QUESTION: Does G contain a HC, that is, an ordering<v1,v2,…,vn> of
the vertices such that all consecutive edges are there including from the
last to the first vertex in G?
PARTITION
INSTANCE: A finite set A and a size s(a) +ve for each a in A.
QUESTION: Is there a subset A’ in A such that sum of elements in both
the partitions are the same?
SATISFIABILITY TO 3SAT TO (3DM AND VC), 3DM TO
PARTITION VC TO (HC AND CLIQUE)
14102005
Proving NP-Completeness Results
3-SATISFIABILITY is a restricted version of satisfiability in which all
instances have exactly three literals per clause.
Theorem 3.1 3-SATISFAIABILITY is NP-Complete.
Proof: it is easy to show that 3SAT  NP since a NDTM need only guess a
truth assignment for the variables and check in polynomial time
whether that truth setting satisfies all the given 3-literal clauses.
We transform SAT to 3SAT. Let U  {u1 , u2 ,..., un } be a set of variables and
C  {c1 , c2, ..., cm } be a set of clauses making up an arbitrary instance of
SAT. We shall construct a collection C of 3-literal clauses on a set U  of
variables such that C is satisfiable iff C is satisfiable. Now we have
U U 


U j and C   C j
Thus we need only to show how C j and U j can be constructed from c j . Let
c j be given by {z1 , z2 ,..., zk } where the zi ’s are all literals derived from the
variables in U . The way these variables are clauses are formed depends
on the value of k
Case 1.
k  1,U j  { y j1 , y j 2 }
C j  {z1 , y j1 , y j 2 },{z1 , y j1 , y j 2 },{z1 , y j1 , y j 2 },{z1, y j1 , y j 2 }
Case 2. k  2,U j  { y j1}, Cj  {{z1, z2 , y j1},{z1, z2 , y j1}
Case 3. k  3, U j  , C j  {{c j }}
Case 4. k  3, k  3, U j  { y j i :1  i  k  3}
Cj  {{z1 , z2 , y j1}} {{ y j i , zi 2 , y j i 1}:1  i  k  4}
{{ y j k 3 , zk 1 , zk }}
To prove that this is indeed a transformation, we must show that the set C is satisfiable
iff C is satisfiable. It is very easy to show that transformations are polynomial and that if
the sets of clauses are simultaneously satisfied.
3.1.2 3-DIMENSIONAL MATCHING
INSTANCE: M  W  X  Y , where W,X,Y are disjoint sets having the
same number q of elements.
QUESTION: Does M contain a matching, that is, a subset M  in M such
that | M  | q | and no 2 elements of M  agree in any coordinate?
Theorem 3.2 3-DIMENSIONAL MATCHING is NP-Complete.
It is easy to show that 3DM  NP since a nondeterministic algorithm
need only guess a subset q | W || X || Y | triples from M and check in
polynomial time that no two of the guessed triples agree in any
coordinate.
We will transform 3SAT to 3DM. Let U {u1 , u2 ,..., un } be the set of variables
and C  {c1 , c2 ,..., cm} be the set of clauses in an arbitrary instance of 3SAT.
We must construct disjoint sets W,X and Y, with | W || X || Y | and a set
M  W  X  Y such that M contains a matching iff C is satisfiable.
The set of ordered triples can be partitioned into 3 separate classes,
grouped according to their intended functions: “truth-setting and fanout(TSFO)”, satisfaction-testing(SF)”, or “garbage collection(GC)”.
Each TSFO corresponds to a single variable u U , and its structure
depends on the total number m of clauses in C. In general TSFO
component for a variable ui involves “internal” elements
ai j [ X ]and bi j [Y ],1  j  m , which will not occur in any triples outside of this
component, and “external” elements ui [ j ], ui [ j ] W ,1  j  m , which will
occur in other triples. The triples making up this component can be
divided into 2 sets:
Ti t  {ui [ j ], ai [ j ], bi [ j ]:1  j  m}
Ti f  {ui [ j ], ai [ j ], bi [ j ]:1  j  m} Ti t  {(ui [m], ai [1], bi [m])}
Since none of the internal elements ai j [ X ]and bi j [Y ],1  j  m will appear in
nay triples outside of Ti  Ti t  Ti f , it is easy to see that any matching
M  will have to include exactly m triples from Ti , either all from
Ti t or all from Ti f . So this will be forcing a matching to make a choice
between setting ui true or false.
Each ST component in M corresponds to a single clause c j  C . It
involves only 2 internal elements s1[ j ]  X and s2 [ j ]  Y and external
elements from ui [ j ], ui [ j ] W ,1  i  n , determined by which literals occur
in clause c j . The set of triples making this component is defined as
follows:
C j  {(ui [ j ], s1[ j ], s2 [ j ]) : ui  c j } {(ui [ j ], s1[ j ], s2 [ j ]) : ui  c j }
Thus any matching M   M will have to choose exactly one triple from
C j . This can only be done, however, if some ui [ j ]or ui [ j ] does not occur in
triples Ti  M  , which will be the case if and only if truth setting
determined by M  satisfies clause c j .
The construction is completed by means of one large GC component G,
involving internal elements g1[k ]  X and g 2 [k ]  Y ,1  k  m(n 1) and
external elements of the form ui [ j ]or ui [ j ] W . It consists of the following
triples:
G  {(ui [ j ], g1[k ], g2 [k ]),(ui [ j ], g1[k ], g 2[k ]) :1  k  m(n 1),1  i  n,1  j  m}
Thus each pair g1[k ]  X and g2 [k ]  Y must be matched with a unique
ui [ j ]or ui [ j ] that does not occur in any triples M   G. There are exactly
m( n  1) such uncovered external elements.
To summarize we set
W  {ui [ j ], ui [ j ]:1  i  n,1  j  m}
X  A  S1  G1 , where A  {ai [ j ]:1  i  n,1  j  m}
S1  {s1[ j ]:1  j  m}, G1  {g1[ j ],1  j  m(n  1)}
Y  B  S2  G2 , where B  {bi [ j ]:1  i  n,1  j  m}
S2  {s2 [ j ],1  j  m}, G2  {g 2 [ j ]:1  j  m(n  1)}
and M  ( Ti )
( Cj ) G
Notice that M  W  X  Y as required. Furthermore since M contains only
2mn  3m  2m2 n(n  1) triples the construction is polynomial.
It is easily verifiable that M cannot contain a matching unless C is
satisfiable. We now show that the existence of a satisfying truth
assignment for C implies that M contains a matching.
Let t : U  {T , F } be any satisfying truth assignment for C. We construct
a matching M   M as follows: For each clause c j  C , let
z j  {ui , ui ,1  i  n}  c j be a literal that is set true by t(one must exist). We
then set
M   (T t i ) (T f i ) ({( z j [ j ], s1[ j ], s2[ j ])}) G, where G is appropriately
chosen subcollection of G that includes all g1[k ], g2[k ] and remaining u
variables.
EXACT COVER BY 3-SETS(X3C)
INSTANCE: A finite set X with |X|=3q and a collection of 3-element
subsets of X/
QUESTION: Does C contain an exact cover for X, that is, a
subcollection C  C such that every element of X occurs in exactly one
member of C ?
Note that every instance of 3DM can be viewed as an instance of X3C
simply by regarding it as an unordered subset of W  X Y . Thus 3DM is
just a restricted version of X3C.
21102005
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VERTEX COVER and CLIQUE
Despite the fact that VERTEX COVER and CLIQUE are
independently useful for proving NP-Completeness results, they are just
really different ways of looking at the same problem. To see this, it is
convenient to consider them in conjunction with a third problem, called
INDEPENDENT SET.
An independent set in a graph G=(V,E) is a subset
V   V | u, v V  (u, v)  E . The INDEPENDENT SET problem asks, for a
given graph G=(V,E), and a positive integer J | V | whether G contains
an independent set V  having | V  | J . The following relationships between
independent sets, cliques, and vertex covers are easy to verify,
Lemma 3.1 For any graph G=(V,E), and subset V   V , the following
statements are equivalent:
a) V  is a vertex cover for G
b) V  V  is an independent set for G.
c) V  V  is a clique in the complement G c of G, where
G c  (V , E c ) with E c  {{u, v}: u, v V and {u, v}  E}
Thus we see that in a strong sense these three problems might be
regarded as “different versions” of one another.
Theorem 3.3 VERTEX COVER is NP-Complete.
Proof: It is easy to see that VC  NP since a nondeterministic algorithm
need only guess a subset of vertices and check in polynomial time
whether that subset contains at least one endpoint of every edge and has
the appropriate size.
We
transform
3SAT
to
VERTEX
COVER.
Let
U  {u1 , u2 ,..., un }and C  {c1 , c2 ,..., cm} be any instance of 3SAT. We must
construct a graph G=(V,E) and a positive integer K | V | such that G has
a vertex cover of size K or less iff C is satisfiable.
The construction is as follows:
For each variable ui U , there is a truth setting component
Ti  (Vi , Ei ) with Vi  {ui , ui } and Ei  {{ui , ui }} Note that any vertex cover will
have to contain at least one of the vertices ui , ui
For each clause c j  C , there is a satisfaction testing component
S j  (V j, E j ) , consisting of 3 vertices and 3 edges joining them to form a
triangle:
V j  {a1[ j ], b1[ j ], a3[ j ]}
Ej  {{a1[ j ], a2 [ j ]},{a1[ j ], a3[ j ]},{a2 [ j ], a3[ j ]}}
Note that any vertex cover will have to contain at least 2 vertices from
V j in order to cover all the edges of the triangle.
The only part of the construction that depends on which literals occur in
which clauses is the collection of communication edges. These are best
viewed from the vantage point of satisfaction testing components. For
each clause c j  C , let the 3 literals be denoted by x j , y j , and z j . Then the
communication edges emanating from S j are given by:
E j  {{a1[ j ], x j },{a2 [ j ], y j },{a3[ j ], z j }}
The construction of our instance of VC is completed by setting
K  n  2m and G  (V , E ), where
V  ( Vi ) ( V j)
E  ( Ei ) ( E j ) ( E j )
It is easy to see how the construction can be accomplished in polynomial
time. All that remains to be shown is that C is satisfiable iff G has a
vertex cover of size K or less.
First, suppose that V   V is a vertex cover of G with | V  | K . By our
previous remarks, V  must contain at least one vertex from each Ti and
at least 2 vertices from each S j . Since this gives a total of at least
n  2m  K vertices, V  must in fact contain exactly one vertex from each
Ti and exactly 2 vertices from each S j . Thus we can use the way in
which V  intersects each truth setting component to obtain a truth
t : U  {T , F }
assignment
.
We
merely
set
t (ui )  T if ui V  and t (ui )  F if ui V .
To see that this truth assignment satisfies each of the clause c j  C ,
consider the 3 edges in E j . Only 2 of these edges can be covered by
vertices in V j  V  , so one of them must be covered by a vertex from
Vi that belongs to V  . But that implies that the corresponding literal
ui or ui from clause c j is true under the truth assignment t, it follows that
t is satisfying truth assignment for C.
Conversely suppose that t : U  {T , F } is a satisfying truth assignment for
C. The corresponding vertex cover V  includes one vertex from each Ti
and 2 vertices from each S j . The vertex from Ti in V  is ui if
t (ui )  T and ui if t (ui )  F . This ensures that at least one of the 3 edges from
each set E j is covered, because t satisfies each clause c j . Therefore we
need only include in V  the endpoints from S j of the other 2 edges in E j .
HAMILTONIAN CIRCUIT(HC)
For convenience whenever  v1 , v2 ,..., vn  is a Hamiltonian circuit, we
shall refer to {vi , vi 1},1  i  n,and {vn , v1} as the edges in the circuit.
Theorem 3.4 HAMILTONIAN CIRCUIT is NP-Complete.
Proof: it is easy to see that HC  NP , because a nondeterministic
algorithm need only guess an ordering of the vertices and check in
polynomial time that all the required edges belong to the edge set of a
given graph.
We transform VERTEX COVER to HC. Let an arbitrary instance of
VC be given by the graph G=(V,E) and the positive integer K | V | . We
must construct a graph G  (V , E ) such that G has a HC if G has a
vertex cover of size K or less.
Once more our construction can be viewed in terms of components
connected together by communication links. First, the graph G has K
selector vertices a1 , a2 ,..., ak which will be used to select K vertices from
the vertex set V of G. Second for each edge in E, G contains a cover
testing component that will be used to ensure that at least one endpoint
of that edges is among the selected K vertices. The component for
e  {u , v}  E has 12 vertices Ve  {(u, e, i),(v, e, i):1  i  6} and 14 edges
Ee  {{(u, e, i),(u, e, i  1)},{(v, e, i),(v, e, i  1)}:1  i  5}
{{(u, e,3),(v, e,1)},{(v, e,3),(u, e,1)}} {{(u, e,6),(v, e, 4)},{(v, e,6),(u, e, 4)}}
In the completed construction, the only vertices from this cover-testing
component that will be involved in any additional edges are
(u,e,1),(v,e,1), (u,e,6) and (v,e,6).
It may be verified that any HC will have to meet the edges in Ee in
exactly one of 3 configurations. This for example, if the circuit enters
this component at (u,e,1) it will have to exit at (u,e,6) and visit wither all
12 vertices in the component or just the 6 vertices (u,e,i), 1<=i<=6, or
(u,e,1)-(u,e,2)-(u,e,3)- (v,e,1)-(v,e,2)-(v,e,3)- (v,e,4)-(v,e,5)-(v,e,6)- (u,e,4)(u,e,5)-(u,e,6)
Additional edges in our construction will serve to join pairs of covertesting components or to join a cover-testing component to a selector
vertex. For each vertex v of V, let the edges incident on v be ordered as
ev[1] , ev[2] ,..., ev[deg( v )]
Ev  {{(v, ev[i ] , 6), (v, ev[i 1] ,1)}:1  i  deg(v)}
The final connecting edges in G join the first and last vertices from
each of these paths to every one of the selector vertices a1 , a2 ,..., ak . These
edges are specified as follows:
E   {{ai , (v, ev[1] ,1},{ai , (v, ev[deg( v )] , 6)}:1  i  K , v V } The completed graph
G  (V , E ) has
V   {ai :1  i  K } (Ve) and
E   ( Ee ) ( Ev )  E 
It is not hard to see that G can be constructed from G and K in
polynomial time.
We claim that G has a HC iff G has a vertex cover of size K or less.
3.1.5 PARTITION
Theorem 3.5 PARTITION is NP-Complete
Proof: It is easy to see that PARTITION∈NP, since a non-deterministic
algorithm need only guess a subset A of A and check in polynomial
time that the sum of the sizes of the elements in A is the same as that fro
the elements remaining in A  A
We transform 3DM to PARTITION. Let the sets W,X,Y, with , and
M  W  X  Y be an arbitrary instance of 3DM. Let the elements of these
sets be denoted by
W  {w1 , w2 ,..., wq }
X  {x1 , x2 ,..., xq }
Y  { y1 , y2 ,..., yq }
And
M  {m1 , m2 ,..., mk }
Where |M|=k. We must construct a set A, and a size s(a)  Z  for each
a  A such that A contains a subset A satisfying
 s(a)   s(a) iff M contains a matching.
aA
aA A
The set A will contain a total of k+2 elements and will be constructed in
two steps. The first k elements of A are {ai :1  i  k} , where the element
ai is associated with the triple mi  M . The size s(ai ) of ai will be specified
by giving its binary representation in terms of strings of 0’s and 1’s
divided into 3q “zones” of p  log2 (k  1) each of these zones is labeled by
an element of W  X Y
The representation for s(ai ) depends on the corresponding triple. It has
a 1 in the rightmost bit position of the zones labeled by w f (i ) , xg (i ) , yh (i ) and
0’s everywhere else. Alternatively we can write
s(ai )  2 p (3q  f (i ))  2 p (2 q  g (i ))  2 p ( q h (i ))
Since each s(ai ) can be expressed in binary with no more than 3pq bits,
it is clear that s(a) can be constructed from the given 3DM instance in
polynomial time.
The important thing to observe about this part of construction is that, if
we sum up all the entries in any zone, over all elements of {ai :1  i  k} ,
thee total number never exceed k  2 p  1 . Hence in adding up  s(a) for
aA
any subset A  {ai :1  i  k} there will be never any carries from one zone
to the other. It follows that if we let
3q 1
B   2 pj then any subset A  {ai :1  i  k} will satisfy
j 0
 s(a)  B .
aA
The final step of the construction specifies the last 2 elements of A.
These are denoted by b1 and b2 and have sizes defined by
k
s(b1 )  2( s(ai ))  B
i 1
k
s(b2 )  ( s(ai ))  B
i 1
Both of these can be specified in binary with no more that 3pq+1 bits
and thus can be constructed in polynomial time of the size of 3DM.