M is a TM with start state s defined over the alphabet {#,(,),0,1,a,q,,}:

Transcription

1 M is a TM with start state s defined over the alphabet {#,(,),0,1,a,q,,}: State Sym. Next State Head s # s L s ) t L t 0 h R t 1 t 1 1. What is M? Use the table given to number the states and head instructions. Num State Head 0 h L 1 s R 2 t # 3 ( 4 ) a 8 q 9, 2. Does M halt on input M? 1

2 Final exam tutorial: Monday April 7, 2pm ECS 116. Remaining regular tutorials: Tutorial #7: Tues. April 1. Assignment 5 has been posted: Due Friday April 4, beginning of class. 2

3 1. Which of these sequences correspond to Hamilton cycles in the graph? (a) (b) (c) (d) (e) Give pseudocode for an algorithm to check if a seqence S[0..(n-1)] for a graph G stored in an adjacency matrix A[0..(n-1)][0..(n-1)] gives a Hamilton cycle. 3

4 Introduction to NP-completeness. The class of NP-complete problems is defined. Satisfiability, the most famous NP-complete problem is introduced. To prove new problems are undecidable, we use halting problem reductions (coming soon). To prove new problems are NP-complete, reductions are also used but the transformation must preserve polynomial running time complexity. 4

5 Table 1: Comparing polynomial and exponential time complexity. Assume a problem of size one takes seconds (1 microsecond). Size n n second second second second second second n second second second second second second n second second second second second second n second 3.2 second 24.3 second 1.7 minutes 5.2 minutes 13.2 minutes 2 n second 1.0 second 17.9 minutes 12.7 days 35.7 years 366 centuries 3 n second 58 minutes 6.5 years 3855 centuries 2*10 8 centuries 1.3*10 13 centuries (from M. R. Garey and D. S. Johnson, Computers and Intractability: A Guide to the Theory of NP-completeness, W. H. Freeman, New York, 1979.) 5

6 Table 2: Effect of improved technology on several polynomial and exponential time algorithms. The following table represents the size of the largest problem instance solvable in 1 hour. Time Complexity function With present computer With computer 100 times faster With computer 1000 times faster n N1 100 N N1 n 2 N2 10 N N2 n 3 N N3 10 N3 n 5 N4 2.5 N N4 2 n N5 N N n N6 N N (from M. R. Garey and D. S. Johnson, Computers and Intractability: A Guide to the Theory of NP-completeness, W. H. Freeman, New York, 1979.) 6

7 Class P A decision problem (yes/no question) is in the class P if there is a polynomial time algorithm for solving it. Polynomial time: O(n c ) for some constant c. If a problem is solvable in polynomial time for some sensible encoding of the input some reasonable machine (TM/RAM/PC) it can be solved in polynomial time for all other sensible encodings/reasonable machines. 7

8 A program that runs on a Random Access Machine (such as our modern day computers) in time T(n) can be simulated on a single tape Turing machine in time O(T 3 (n)). This distinction is made between polynomial time and anything else (sometimes referred to as exponential time) because of the blowup in computation times which appear for exponential algorithms. 8

9 Class NP A decision problem (yes/no question) is in the class NP if it has a nondeterministic polynomial time algorithm. Informally, such an algorithm: 1.Guesses a solution (nondeterministically). 2. Checks deterministically in polynomial time that the answer is correct. Or equivalently, when the answer is "yes", there is a certificate (a solution meeting the criteria) that can be verified in polynomial time (deterministically). 9

10 Example problem which is in P and NP: Minimum Weight Spanning Tree (CSC 225). Input: Graph G, integer k. Question: Does G have a spanning tree of weight at most k? If you are provided with a tree with weight at most k as part of the solution, the answer can be verified in O(n) time. 10

11 Matching is in NP: Given a graph G and integer k, does G have a k-edge matching? Matching: disjoint edges. Certificate, k=5: (a,f) (b,g) (c,h) (d,i)(e,j) 11

12 Hamilton Cycle is in NP: Input graph G. Does G have a Ham. cycle? Certificate: 1, 2, 3, 5, 6, 7, 11, 12, 10, 8, 9, 4 Picture from: 12

13 Does P= NP? the Clay Mathematics Institute has offered a $1 million US prize for the first correct proof. Some problems in NP not known to be in P: Hamilton Path/Cycle Independent Set Satisfiability Note: Matching is in P. Learn more in a graph algorithms class. 13

14 NP-completeness I can't find an efficient algorithm, I guess I'm just too dumb. 14

15 I can't find an efficient algorithm, because no such algorithm is possible. 15

16 I can't find an efficient algorithm, but neither can all these famous people. 16

17 NP-complete Problems The class of problems in NP which are the "hardest" are called the NP-complete problems. A problem Q in NP is NP-complete if the existence of a polynomial time algorithm for Q implies the existence of a polynomial time algorithm for all problems in NP. Steve Cook in 1971 proved that SAT is NPcomplete. Proof: will be given in our last class. Other problems: use reductions. 17

18 Bible for NPcompleteness: M. R. Garey and D. S. Johnson, Computers and Intractability: A Guide to the Theory of NP- Completness, W. H. Freeman, 1st ed. (1979). 18

19 A problem Q in NP is NP-complete if the existence of a polynomial time algorithm for Q implies the existence of a polynomial time algorithm for all problems in NP. 19

20 open: Does every fullerene have a Hamilton cycle? Certificate: 0, 1, 2, 11, 10, 9, 8, 7, 6, 5, 14, 15, 16, 17, 18, 19, 12, 13, 3, 4 20

21 Min Number of Hamilton Cycles 21

22 Conjecture For all fullerenes except isomer 38:2, every edge is in at least one Hamilton cycle Isomer 38:2 22

23 Conjecture For all isomers except 38:2 and 40:8, no edge is in every Hamilton cycle. e Isomer 40:8 23

24 Incorrect Conjecture Every fullerene has some Hamilton cycle with no 6-cycles like this: Isomer 74:

25 A dominating set of a graph G is a subset D of the vertices of G such that every vertex v of G is either in the set D or v has at least one neighbour that is in D. 25

26 Only fullerene isomer C_56:649 has dominating set order

27 n LB #LB #LB+1 #LB+2 Adj. list

28 Independent Set is in NP: Given a graph G and integer k, does G have an independent set of order k? Vertices u and v are independent if edge (u, v) is not in G. Certificate, k=3: 2, 3, 8 28

29 Finding a Maximum Independent Set in the 120-cell Sean Debroni, Erin Delisle, Michel Deza, Patrick Fowler, Wendy Myrvold, Amit Sethi, Benoit de La Vaissiere, Joe Whitney, Jenni Woodcock, 29

30 Start with a dodecahedron: Pictures from: 30

31 Glue 12 more on, one per face. After 6: After 12: total 13 31

32 Add 20 more dodecahedra into the 20 dimples (total 33): Keep going to get the 120-cell: 600 vertices, 4-regular, girth 5, cycles, vertex transitive. 32

33 Notices of the AMS: Jan Vol. 48 (1) 33

34 Finishing the problem: LP gives an upper bound of 221 on 120-cell. The resulting solutions indicate that if 221 is possible then there must be at least 25: B4 B1 Also: B1 + 2 B4 7 We planted 7 in all ways up to isomorphism then tried to extend to 25: not possible. 34

35 The End to the Keller Conjecture Jen Debroni The three amigos and their Kraken computer: John Eblen, Mike Langston, and Dinesh Weerapurage. Peter Shor 35

36 If you tile the plane with unit squares, then some squares have an entire side in common 36

37 Conjecture [1930, Ott-Heinrich Keller]: If you tile d-dimensional space with d-dimensional cubes of equal size then some cubes must have a (d-1)-dimensional side in common. 37

38 History: Oskar Perron [1940]: True for d 6 Lagarias and Shor[1992]: False for d 10 John Mackay [2002]: False for d 8 New: True for d=7. 38

39 Keller graph with dimension d: vertices which are numbered with each of the 4 d d-digit numbers, digits are to 0, 1, 2, or 3. Two vertices are adjacent if their labels differ in at least two positions, and in at least one position the difference in the labels is 2 mod 4. Examples (Dimension 5): Adjacent NOT adjacent NOT adjacent

40 In a sense, these cases require only patience- and maybe a high speed computer the size of a major galaxy. No one in his right mind, and no mathematicians either, would set our to sort through possible tilings to check the case of d=7. p. 24, Barry Cipra and Paul Zorn 40

41 SAT (Satisfiability) Variables: u 1, u 2, u 3,... u k. A literal is a variable u i or the negation of a variable u i. If u is set to true then u is false and if u is set to false then u is true. A clause is a set of literals. A clause is true if at least one of the literals in the clause is true. The input to SAT is a collection of clauses. 41

42 This SAT problem has solution u 1 =T, u 2 =F, u 3 = T, u 4 =F (u 1 OR u 2 OR u 4 ) AND ( u 2 OR u 4 ) AND ( u 1 OR u 3 ) AND ( u 4 OR u 1 ) Does this SAT problem have a solution? ( u 1 OR u 2 ) AND ( u 2 OR u 3 ) AND ( u 3 OR u 1 ) AND ( u 2 OR u 3 ) AND ( u 3 OR u 1 ) 42

43 SAT (Satisfiability) The output is the answer to: Is there an assignment of true/false to the variables so that every clause is satisfied (satisfied means the clause is true)? If the answer is yes, such an assignment of the variables is called a truth assignment. SAT is in NP: Certificate is true/false value for each variable in satisfying assignment. 43

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