1. Additional NP-complete problems
Giorgi Japaridze
Theory of Computability
Additional NP-complete problems
Section 7.5

2. Why we want to see more NP-completeness proofs
Giorgi Japaridze Theory of Computability
For reasons that are not well understood, most naturally occurring
NP-problems are known to be either in P or to be NP-complete (the
graph isomorphism problem being one notable exception; so was the
COMPOSITES problem until a few years ago).
So, if you seek a polynomial time algorithm for a new NP-problem,
spending part of your effort attempting to prove it NP-complete is
sensible because doing so may prevent you from wasting time.
This explains why we want to exercise more with NP-completeness
proofs.
Such proofs usually proceed by finding a polynomial reduction
from an already known NP-complete language, most typically from
3SAT. When doing so, we look for structures in the language under
question that can simulate the variables and clauses of Boolean
formulas. For example, when reducing 3SAT to CLIQUE, variables
were simulated by vertices, and clauses by triples. Such structures are
often called gadgets.

3. The NP-completeness of CLIQUE
7.5.b
Giorgi Japaridze Theory of Computability
Corollary CLIQUE is NP-complete.
Proof.
As was already (easily) observed, CLIQUE is in NP.
Next, Theorem 7.32 showed that 3SAT is polynomial time reducible
to CLIQUE.
In turn, by Corollary 7.42, 3SAT is NP-complete.
But Theorem 7.36 says that if an NP-complete language is polynomial
time reducible to an NP-language C, then C is NP-complete.
Hence CLIQUE is NP-complete.

4. The VERTEX-COVER problem defined
Giorgi Japaridze Theory of Computability G
If G is an undirected graph, a vertex cover of
G is a subset of the nodes where every edge
of G touches (at least) one of those nodes. 1 2 3 4
VERTEX-COVER =
= {<G,k> | G is an undirected graph that has a k-node vertex cover}
<G,4>  VERTEX-COVER?
<G,3>  VERTEX-COVER?
<G,2>  VERTEX-COVER?
<G,1>  VERTEX-COVER?

5. The NP-completeness of VERTEX-COVER (a)
7.5.d
Giorgi Japaridze Theory of Computability
Theorem VERTEX-COVER is NP-complete.
Proof. Obviously VERTEX-COVERNP (why?).
Next, we are going to set up a polynomial time reduction from 3SAT
to this language. It maps a 3cnf-formula  to a graph G and a value k,
such that  is satisfiable iff G has a vertex cover of size k.
For each variable x in , we produce a horizontal edge connecting two
nodes. We label the two nodes in this gadget x and x. Setting x to be
TRUE is going to correspond to selecting the left node for the vertex
cover, and setting it to be FALSE correspond to selecting the right node.
For example, when
 = (x  x  z)  (x  z  z)  (x  z  z),
the reduction produces the following two variable gadgets: x x z z

6. The NP-completeness of VERTEX-COVER (b)
Giorgi Japaridze Theory of Computability
 = (x  x  z)  (x  z  z)  (x  z  z)
Next, for each clause, we produce a triangle of (interconnected) nodes,
each node of a triangle labeled with the corresponding literal of the
corresponding clause. These triangles are clause gadgets.
Finally, we connect each variable-gadget node to each clause-gadget
node with the identical label. x x z z x x x x z z z z z

7. The NP-completeness of VERTEX-COVER (c)
Giorgi Japaridze Theory of Computability
Thus, G has 2m+3l nodes, where m is the number of variables in 
and l is the number of clauses.
Let k be m+2l. In the present example, k = 2+(23) = 8.
Claim.  is satisfiable iff G has a vertex cover of size k.
 = (x  x  z)  (x  z  z)  (x  z  z) x x z z x x x x z z z z z

8. The NP-completeness of VERTEX-COVER (d)
7.5.g
Giorgi Japaridze Theory of Computability
Suppose  is satisfiable.
Include the variable-gadget nodes corresponding to true
literals in the VC.
Next, for each clause gadget, select a true-literal-labeled node, and
include the other two nodes in the VC.
We got the total of k nodes.
All edges within the variable and clause gadgets are obviously covered.
And every
intergadget edge is covered either by a (true) variable-gadget node, or by one of the
two nodes of the clause gadget that have been included in the vertex cover.
 = (x  x  z)  (x  z  z)  (x  z  z) x x z z x x x x z z z z z

9. The NP-completeness of VERTEX-COVER (e)
Giorgi Japaridze Theory of Computability
Suppose now G has a VC of size k. Clearly it should include at least one node from
each variable gadget and at least two nodes from each clause gadget.
As this number
sums up to k, those nodes are exactly the nodes of the vertex cover.
We take the nodes
of the variable gadgets that are in the VC and assign TRUE to the corresponding literals.
This assignment satisfies  because, if you take any clause gadget, the node which
is not in the VC is (has to be) connected with an edge to a true-literal node of a variable
gadget, for otherwise that intergadget edge would not be covered.
 = (x  x  z)  (x  z  z)  (x  z  z) x x z z x x x x z z z z z

10. The NP-completeness of HAMPATH (a)
7.5.i
Giorgi Japaridze Theory of Computability
Theorem HAMPATH is NP-complete.
Proof. That HAMPATH is in NP has already been shown. Now we set
up a polynomial time reduction from 3SAT to HAMPATH.
Consider an arbitrary 3cnf-formula  with k clauses:
 = (p1  q1  r1)  (p2  q2  r2)  …  (pk  qk  rk)
c c … ck
We assume that  contains l variables x1,…,xl.
Each such variable will be represented with a
diamond-shaped structure with 3k+5 nodes:
And each clause will be represented
with a single node: cj … xi

11. The NP-completeness of HAMPATH (b)
7.5.j
Giorgi Japaridze Theory of Computability s
The overall structure
of G looks like this.
There are additional
edges connecting
clause nodes with
internal nodes of
diamonds, explained
on the next slide. … x1 c1 … x2 c2 c3 . . ck … xl t

12. The NP-completeness of HAMPATH (c)
7.5.k
Giorgi Japaridze Theory of Computability
The horizontal nodes in a diamond structure are grouped as follows: … xi
c c c ck cj
Additional edges when
clause cj contains xi : … … xi cj cj
Additional edges when
clause cj contains xi : … … xi cj

13. The NP-completeness of HAMPATH (d)
Giorgi Japaridze Theory of Computability s
Suppose  is satisfiable.
Consider the path that
zig-zags through true
diamonds, and zag-zigs
through false diamonds. … x1
TRUE
zig-zag c1 … x2
It covers all the nodes
except the clause
nodes c1,…,ck. We can
easily include them by
selecting one true literal
in each clause and then
adding a detour from the
corresponding pair of
horizontal nodes of the
of the corresponding
diamond. E.g.,when c1
includes x1 and x1 is
Selected for c1, we have: c2
FALSE
zag-zig c3 . . ck … xl
TRUE
zig-zag t

14. The NP-completeness of HAMPATH (e)
Giorgi Japaridze Theory of Computability s
On the other hand, if
c1 included x2 and x2
was selected for c1,
we would set up the
detour in the opposite
direction as you see.
The same story for c2,
only the detour will be
from the next pair of
horizontal nodes, i.e.
nodes #6-7
(rather than #3-4).
And so on.
The resulting path with
detours is obviously
Hamiltonian. … x1
TRUE
zig-zag c1 … x2 c2
FALSE
zag-zig c3 . . ck … xl
TRUE
zig-zag t

15. The NP-completeness of HAMPATH (e)
Giorgi Japaridze Theory of Computability s
For the reverse direction,
assume G has a Hamiltonian
path from s to t.
With a little thought,
one can see that the path
should be normal, meaning
that it goes through the
diamonds in order from
the top one to the bottom
one, except for the
detours to the clause nodes.
From such a normal path
we can easily obtain a
satisfying assignment: if it
zig-zags through the diamond,
assign TRUE to the
corresponding variable.
And if it zag-zigs, then
assign FALSE. … x1
TRUE
zig-zag c1 … x2 c2
FALSE
zag-zig c3 . . ck … xl
TRUE
zig-zag t

16. The NP-completeness of UHAMPATH
Giorgi Japaridze Theory of Computability
UHAMPATH is the undirected version of HAMPATH.
Theorem UHAMPATH is NP-complete.
Proof: By reducing HAMPATH to UHAMPATH. Given a directed graph G, we
construct its undirected equivalent G’ by replacing each node u of G by three nodes: uin,
umid and uout; except the node s which gets replaced by sout, and the node t which gets
replaced by tin.
G’ has two types of edges. One type connects each umid with uin and uout. The other
type connects each uout with vin, as long as there is an edge from u to v in G.
Now, suppose G has a Hamiltonian path
s, u1, u2, …, uk, t
Then obviously the following is a Hamiltonian path in G’:
sout, u1in, u1mid, u1out, u2in, u2mid, u2out, …, ukin, ukmid, ukout, tin
For the reverse direction, suppose G’ has a Hamiltonian path. With a little thought we
find that it has to look like
But then obviously is a Hamiltonian path in G.
sout, u1in, u1mid, u1out, u2in, u2mid, u2out, …, ukin, ukmid, ukout, tin
s, u1, u2, …, uk, t

17. The NP-completeness of SUBSET-SUM (a)
Giorgi Japaridze Theory of Computability
Theorem SUBSET-SUM is NP-complete.
Proof: That SUBSET-SUMNP was shown in Theorem Now we construct a
polynomial time reduction from 3SAT to SUBSET-SUM.
Let  be a 3-cnf Boolean formula with variables x1,…,xl and clauses c1,…,ck.
We construct an (l+k)(2l+2k+1) table of decimal digits.
The columns are labeled with x1,…,xl, c1,…,cl, and the rows are labeled with
y1,z1, y2,z2, …, yl,zl, g1,h1, g2,h2, …, gk,hk, t.
The rows of this table are read as decimal numbers. The first 2l+2k rows, as numbers,
comprise the set S, and the last row is the target number t. It will be seen later that S
has a subset R summing up to t iff  is satisfiable.
Specifically, row/number t consists of l 1s followed by k 3s.
Each row yi has a 1 in the xi column, and also in each cj column such that clause cj
contains xi.
Each row zi has a 1 in the xi column, and also in each cj column such that clause cj
Finally, each row gi and hi has a 1 in the ci column.
Digits not specified above are all 0s.

18. The NP-completeness of SUBSET-SUM (b)
7.5.q
Giorgi Japaridze Theory of Computability x1 x2 x3 x4 … xl c1 c2 … ck y1 1 1 z1 1
(x1 x2  x3) 
(x2 x3  …)
(x3 …  …) … y2 1 1 z2 1 1 y3 1 1 1 z3 1 1 … yl 1 zl 1 g1 1 h1 1 g2 1 h2 1 … gk 1 hk 1 t
… … 3

19. The NP-completeness of SUBSET-SUM (c)
7.5.r
Giorgi Japaridze Theory of Computability
Suppose  has a satisfying assignment. We
construct a subset R of S as follows:
Select yi if xi is assigned TRUE, and select
zi if xi is assigned FALSE.
If we add up what we have selected so far,
we obtain a 1 in each of the first l digits
because we have selected either yi or zi for
each i.
Furthermore, each of the last k digits is a
number between 1 and 3 because each
clause is satisfied and so contains between
1 and 3 true literals.
Now we further select enough of the g and h
numbers to bring each of the last k digits
up to 3, thus hitting the target. x1 x2 x3 x4 … xl c1 c2 … ck y1 1 1 z1 1 y2 1 1 z2 1 1 y3 1 1 1 z3 1 1 … yl 1 zl 1 g1 1 h1 1 g2 1 h2 1 … gk 1 hk 1 t
… … 3

20. The NP-completeness of SUBSET-SUM (d)
Giorgi Japaridze Theory of Computability
Now suppose a subset R of S sums to t.
Observe that each column in the table
describing S has at most five 1s. So, a “carry”
into the next column never occurs when a
subset of S is added.
To get a 1 in each of the first l columns of
t, we must have either yi or zi for each i, but
not both. If it is yi, we assign xi TRUE,
otherwise assign FALSE.
This assignment must satisfy  because
in each of the final k columns the sum is always 3. In column cj, at most 2 can come
from gj and hj, so at least 1 in this column
should come from some yi or zi in the subset.
If it is yi, then xi appears in cj and is
assigned TRUE, so cj is satisfied. If it is zi,
then xi appears in cj and xi is assigned FALSE,
so cj is satisfied. Therefore,  is satisfied. x1 x2 x3 x4 … xl c1 c2 … ck y1 1 1 z1 1 y2 1 1 z2 1 1 y3 1 1 1 z3 1 1 … yl 1 zl 1 g1 1 h1 1 g2 1 h2 1 … gk 1 hk 1 t
… … 3