1. SUBSET-SUM Instance: A set of numbers denoted S and a target number t.
Problem: To decide if there exists a subset YS, s.t yYy=t.
Problem definition: Subset Sum - Given a (multi)set A of integer numbers and an integer number t, does there exist a subset of A such that the sum of its elements is equal to t?

2. SUBSET-SUM is in NP On input S,t: Guess YS Accept iff yYy=t.
The length of the certificate: O(n) (n=|S|)
Time complexity: O(n)

3. SUBSET-SUM is NP-Complete
Proof: We’ll show 3SAT  SUBSET-SUM. t

4. Examples SUBSET-SUM … because 2+8=10
… because 11 cannot be made out of {2,4,8}

5. Reducing 3SAT to SubSet Sum
Proof idea
Choosing the subset numbers from the set S corresponds to choosing the assignments of the variables in the 3CNF formula.
The different digits of the sum correspond to the different clauses of the formula.
If the target t is reached, a valid and satisfying assignment is found.

6. Satisfying Clauses c1 c2 …… ck yi zi digit per clause
number per variable xi assigned true: yi
number per variable xi assigned false: zi
1 if xi is in cj
0 otherwise
1 if xi is in cj
0 otherwise

7. Subset Sum + clause: 3CNF formula: Make the number table,1 2 3 4
+x1
–x1
+x2
–x2
+x3
–x3
+x4
–x4 1
3CNF formula:
Make the number table,
and the ‘target sum’ t
dummies + 1 3

8. Reducing 3SAT to SubSet Sum
Let 3CNF with k clauses and  variables x1,…,x.
Create a Subset-Sum instance <S,t> by: 2+2k elements of S = {y1,z1,…,y,z,g1,h1,…,gk,hk}
yj indicates positive xj literals in clauses
zj indicates negated xj literals in clauses
gj and hj are dummies
and

9. Subset Sum + clause: 3CNF formula: Make the number table,1 2 3 4
+x1
–x1
+x2
–x2
+x3
–x3
+x4
–x4 1 y1 z1 y2 z2 y3 z3 y4 z4
3CNF formula:
Make the number table,
and the ‘target sum’ t
dummies + 1 3

10. Subset Sum Note 1: The “1111” in the target forces a proper assignment
+x1
–x1
+x2
–x2
+x3
–x3
+x4
–x4 1
Note 1: The “1111”
in the target forces
a proper assignment
of the xi variables.
Note 2: The target
“3333” is only possible
if each clause is satisfied.
(The dummies can add
maximally 2 extra.) 1 3

11. Subset Sum + is a satisfying assignment +x1 –x1 +x2 –x2 +x3 –x3 +x41 3 +
is a satisfying assignment 1 3

12. Subset Sum + is not a satisfying assignment +x1 –x1 +x2 –x2 +x3 –x31 ? 2 +
is not a
satisfying assignment 1 3

13. Proof 3SAT  Subset Sum
For every 3CNF , take target t=1…13…3 and the corresponding set S.
If 3SAT, then the satisfying assignment defines a subset that reaches the target.
Also, the target can only be obtained via a set that gives a satisfying assignment for .

14. Directed Hamiltonian Path
Given a directed graph G, does there exist a Hamiltonian path, which visits all nodes exactly once?
Two Examples:
but this graph has no
Hamiltonian path
this graph has a
Hamiltonian path

15. 3SAT to Hamiltonian Path
Proof idea: Given a 3CNF , make a graph G such that
… a Hamiltonian path is possible if and only if there is a zig-zag path through G,
… where the directions (zig or zag) determine the assignments (true or false) of the variables x1,…,xL of .

16. “Zig-Zag Graphs” s t
There are 2L different Hamiltonian paths from s to the target t.
For every i, “Zig” means xi=True
… while “Zag” means xi=False

17. Accessing the Clauses stands for If  has K clauses,  = C1 Ck,
then xi has K “hubs”:
Idea: Make K extra
nodes Cj that can only
be visited via the hub- path that defines an
assignment of xi which
satisfies the clause
(using zig/zag = T/F).
stands for

18. Connecting the Clauses
Let the clause Cj contain xi: If
then xi=True=“zig”
reaches node Cj If
then xi=False=“zag”
reaches node Cj

19. Proof 3SAT P HamPath Given a 3CNF (x1,…,xL) with K clauses
Make graph G with a zig/zag levels for every xi
Connect the clauses Cj via the proper “hubs”
If SAT, then the satisfying assignment defines a Hamiltonian path, and vice versa.

20. Example
4 variables…
4 clauses…
Clauses
connected
via zig-zag
“hubs”

21. Example
assignment…
…satifies all four clauses; hence it defines a Hamiltonian Path

22. Example assignment… …does not satify
first clause; hence the path misses the C1 node

23. More on Hamiltonian Path
Useful for proving NP-completeness of “optimal routing problems”
Typical example: NP-completeness of “Traveling Salesman Problem”
Issue of directed versus undirected graphs

24. Forcing Directions Given a directed graph with s,x1,…,xk,t
Replace s with sout, the t with tin ,and
every xi with the triplet “xi,in—xi,mid — xi,out”
Redraw the original directed edges with
edges going from out-nodes to in-nodes.
If and only if the directed graph has a
HamPath from s to t, then so has this graph.

25. Example: Directions becomes becomesy y
xin
sout
xmid
xout
becomes
ymid
yout
yin
xin
sout
xmid
xout
becomes
ymid
yout
yin
“Undirected HamPath” is NP-complete