Final Exam Review Final exam will have the similar format and requirements as Mid-term exam: Closed book, no computer, no smartphone Calculator is Ok Final.

Final Exam Review

Final exam will have the similar format and requirements as Mid-

term exam:•Closed book, no computer, no smartphone•Calculator is Ok

Final exam questions are contained in:

•Questions in Homework 2 and Programming Assignment 2

•Content listed in the following slides

2

String Similarity

How similar are two strings? ocurrance occurrence

o c u r r a n c e

c c u r r e n c eo

-

o c u r r n c e

c c u r r n c eo

- - a

e -

o c u r r a n c e

c c u r r e n c eo

-

6 mismatches, 1 gap

1 mismatch, 1 gap

0 mismatches, 3 gaps

3

Applications. Basis for Unix diff. Speech recognition. Computational biology.

Edit distance. [Levenshtein 1966, Needleman-Wunsch 1970] Gap penalty ; mismatch penalty pq. Cost = sum of gap and mismatch penalties.

2 + CA

C G A C C T A C C T

C T G A C T A C A T

T G A C C T A C C T

C T G A C T A C A T

-T

C

C

C

TC + GT + AG+ 2CA

-

Edit Distance

4

Goal: Given two strings X = x1 x2 . . . xm and Y = y1 y2 . . . yn find

alignment of minimum cost.

Def. An alignment M is a set of ordered pairs xi-yj such that each

item occurs in at most one pair and no crossings.

Def. The pair xi-yj and xi'-yj' cross if i < i', but j > j'.

Ex: CTACCG vs. TACATG.Sol: M = x2-y1, x3-y2, x4-y3, x5-y4, x6-y6.

Sequence Alignment

C T A C C -

T A C A T-

G

G

y1 y2 y3 y4 y5 y6

x2 x3 x4 x5x1 x6

5

Def. An s-t cut is a partition (A, B) of V with s A and t B.

Def. The capacity of a cut (A, B) is:

Cuts

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10 15 4

4

Capacity = 10 + 5 + 15 = 30

A

cap( A, B) c(e)e out of A

6

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10 15 4

4 A

Cuts

Def. An s-t cut is a partition (A, B) of V with s A and t B.

Def. The capacity of a cut (A, B) is:

cap( A, B) c(e)e out of A

Capacity = 9 + 15 + 8 + 30 = 62

7

Residual Graph

Original edge: e = (u, v) E. Flow f(e), capacity c(e).

Residual edge. "Undo" flow sent. e = (u, v) and eR = (v, u). Residual capacity:

Residual graph: Gf = (V, Ef ). Residual edges with positive residual capacity. Ef = {e : f(e) < c(e)} {eR : f(e) > 0}.

u v 17

6

capacity

u v 11

residual capacity

6

residual capacity

flow

c f (e) c(e) f (e) if e E

f (e) if eR E

8

Ford-Fulkerson Algorithm

s

2

3

4

5 t 10

10

9

8

4

10

10 6 2

G :capacity

9

Augmenting Path Algorithm

Augment(f, c, P) { b bottleneck(P) foreach e P { if (e E) f(e) f(e) + b else f(eR) f(eR) - b } return f}

Ford-Fulkerson(G, s, t, c) { foreach e E f(e) 0 Gf residual graph

while (there exists augmenting path P) { f Augment(f, c, P) update Gf

} return f}

forward edge

reverse edge

10

Certifiers and Certificates: 3-Satisfiability (3-SAT)

SAT. Given a CNF formula , is there a satisfying assignment?

Certificate. An assignment of truth values to the n boolean variables.

Certifier. Check that each clause in has at least one true literal.

Ex.

Conclusion. SAT is in NP.

instance s

certificate t

11

Subset Sum

SUBSET-SUM. Given natural numbers w1, …, wn and an integer W,

is there a subset that adds up to exactly W?

Ex: { 1, 4, 16, 64, 256, 1040, 1041, 1093, 1284, 1344 }, W = 3754.Yes. 1 + 16 + 64 + 256 + 1040 + 1093 + 1284 = 3754.

Remark. With arithmetic problems, input integers are encoded in binary. Polynomial reduction must be polynomial in binary encoding.

Claim. 3-SAT P SUBSET-SUM.

Pf. Given an instance of 3-SAT, we construct an instance of SUBSET-SUM that has solution iff is satisfiable.

12

Subset Sum

Construction. Given 3-SAT instance with n variables and k clauses, form 2n + 2k decimal integers, each of n+k digits, as illustrated below.

Claim. is satisfiable iff there exists a subset that sums to W.Pf. No carries possible.

dummies to get clausecolumns to sum to 4

y

x

z

0 0 0 0 1 0

0 0 0 2 0 0

0 0 0 1 0 0

0 0 1 0 0 1

0 1 0 0 1 1

0 1 0 1 0 0

1 0 0 1 0 1

1 0 0 0 1 0

0 0 1 1 1 0

x y z C1 C2 C3

0 0 0 0 0 2

0 0 0 0 0 1

0 0 0 0 2 0

1 1 1 4 4 4

x

y

z

W

10

200

100

1,001

10,011

10,100

100,101

100,010

1,110

2

1

20

111,444

13

Weighted Vertex Cover

Definition. Given a graph G = (V, E), a vertex cover is a set S V such that each edge in E has at least one end in S.

Weighted vertex cover. Given a graph G with vertex weights, find a vertex cover of minimum weight. (NP hard problem) all nodes with weight of 1 reduces the problem to standard vertex cover problem.

4

9

2

2

4

9

2

2

weight = 2 + 2 + 4 weight = 11

14

Pricing Method

Pricing method. Set prices and find vertex cover simultaneously.

Why S is a vertex cover set? (use contradiction to prove)

Weighted-Vertex-Cover-Approx(G, w) { foreach e in E pe = 0

while ( edge e=(i,j) such that neither i nor j are tight) select such an edge e increase pe as much as possible until i or j tight }

S set of all tight nodes return S}

15

Approximation method: Pricing Method

Pricing method. Each edge must be covered by some vertex. Edge e = (i, j) pays price pe 0 to use vertex i and j.

Fairness. Edges incident to vertex i should pay wi in total.

Lemma. For any vertex cover S and any fair prices pe: e pe

w(S).

Pf. ▪

4

9

2

2

sum fairness inequalitiesfor each node in S

each edge e covered byat least one node in S

16

Pricing Method

vertex weight

Figure 11.8

price of edge a-b

Example shows the pricing method does not provide the optimal weighted vertex cover solution

17

Weighted Vertex Cover: IP Formulation

Weighted vertex cover. Given an undirected graph G = (V, E) with vertex weights wi 0, find a minimum weight subset of

nodes S such that every edge is incident to at least one vertex in S.

Integer programming formulation. Model inclusion of each vertex i using a 0/1 variable xi.

Vertex covers in 1-1 correspondence with 0/1 assignments: S = {i V : xi = 1}

Objective function: minimize i wi xi.

– Constraints:….. Must take either i or j: xi + xj 1.

18

Weighted Vertex Cover: IP Formulation

Weighted vertex cover. Integer programming formulation.

Task: Show the concrete ILP equation set for an example graph.

19

Weighted Vertex Cover

Weighted vertex cover. Given an undirected graph G = (V, E) with vertex weights wi 0, find a minimum weight subset of

nodes S such that every edge is incident to at least one vertex in S.

3

6

10

7

A

E

H

B

D I

C

F

J

G

6

16

10

7

23

9

10

9

33

total weight = 55

32