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Some Problems in Computer Science and Elementary Number Theory

Some Problems in Computer Science and Elementary Number Theory. Elwyn Berlekamp. Among most important unsolved problems in mathematics/ computer science. Does P = NP ?. =. Does there exist a polynomial time algorithm to solve the Traveling Salesman Problem?. The Traveling Salesman Problem.

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Some Problems in Computer Science and Elementary Number Theory

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  1. Some Problems in Computer Science and Elementary Number Theory Elwyn Berlekamp

  2. Among most important unsolved problems in mathematics/ computer science Does P = NP ? = Does there exist a polynomial time algorithm to solve the Traveling Salesman Problem?

  3. The Traveling Salesman Problem Given a graph (with n nodes), find a path which runs through all the nodes without any repeats.

  4. Does this graph have a Hamiltonian Path? NO (Proof coming later)

  5. What about this graph? YES

  6. The Traveling Salesman Problem (All P- equivalent) Version 1: Given a graph (with n nodes), find a path which runs through all the nodes without any repeats. Version 1′: Determine whether or not such a path exists. Version 2: Same as 1, except starting and ending points are given. Version 3: Given a graph, find a Hamiltonian cycle which runs through each node once. Version 4: Given the complete graph of n nodes, and a table that specifies a cost to each of its n(n-1)/2 branches. Find the Hamiltonian cycle with least cost. Version 5: Given a set of n integers: N={a1, a2, a3…an} and a set of pair sums; S = {s1, s2, ...sk}, find a Hamiltonian path for the graph G whose nodes are N, and there is a branch between aiand aj iff ai + ajεS.

  7. Interesting Special Case of the Traveling Salesman Problem: Nodes = interval of j + 1- i consecutive integers: [ i , j ] Permissible pairsums= S= {s1, s2…} We say [ i , j ] can be chained by Siff a Hamiltonian path exist. 18 16 25 36 12 4 21 15 25 25 25 25 16 9 25 36 16 9 7 2 23 13 36 10 16 36 16 16 24 16 20 5 11 14 22 3 25 6 25 16 25 36 25 4 25 1 8 17 19 9 25 36 9 9

  8. Problems: (wide range of difficulty) For what value of n can [1, n] be chained by squares? by cubes? by kth powers? What is the smallest n such that [1, n] can be chained by squares? …? Is there a largest n such that [1, n] cannot be chained by squares? …? If so, what is it?

  9. S= {1, 4, 9, 16, 25, 36, 49, …} 12 4 5 11 13 8 1 3 6 10 2 7 9

  10. S= {1, 4, 9, 16, 25, 36, 49, …} 12 4 5 11 14 13 8 1 3 6 10 2 7 9

  11. S= {1, 4, 9, 16, 25, 36, 49, …} 12 4 5 11 14 2 7 9 13 18 16 8 1 3 6 10 20 17 15 19

  12. Let’s now prove this graph has no Hamiltonian Path: 18 12 4 21 15 16 9 9 7 2 2 23 13 10 24 20 5 11 11 14 22 22 3 6 1 8 17 19 • If branch 2-14 is not used, then use of 18-7 forces an endpoint at 2 or 9. • If branch 2-14 is used, then there is an endpoint at 11 or 22. • So one endpoint is at 18; the other is among {2,9,11,22} • Branch 4-5 third endpoint at 20 or 11 • Branch 3-6 third endpoint at 10 or 19 • Branch 1-15 third endpoint at 21 or 10 Note: these reductions also work if nodes 24 and/or 23 are absent

  13. 18 12 4 21 15 16 9 7 2 23 13 10 24 20 5 11 14 22 3 6 1 8 17 19 • Since 4 cannot be an endpoint, branch 12-4 must be used • Since 8 cannot be an endpoint, branch 1-8 must be used. • Since 24 cannot be an endpoint, branches 12-24 and 24-1 must be used • But now [24,1,8,17,19,6,10,15,21,4,12] is a disjoint cycle • So [1,24] cannot be chained by squares, QED

  14. Can [1,22] be chained by squares? 18 12 4 21 15 16 9 7 2 23 13 10 24 20 5 11 14 22 3 6 1 8 17 19

  15. Can [1,22] be chained by squares? 18 12 4 21 15 16 9 7 2 13 10 20 5 11 14 22 3 6 1 8 17 19 NO

  16. What are all solutions of chaining [1,23] by squares? 18 12 4 21 15 16 9 7 2 23 13 10 20 5 11 14 22 3 6 1 8 17 19 • In [1,23], branch 13-3 would force a third endpoint at 12 or 23. • So it cannot be used.

  17. What are all solutions of chaining [1,23] by squares? 18 12 4 21 15 16 9 7 2 23 13 10 20 5 11 14 22 3 6 1 8 17 19 • [1,23] can be chained by squares in exactly three different ways, with endpoints {18,9}, {18,2}, or {18,22}. Dotted lines cannot be used.

  18. 18 16 25 36 12 4 21 15 25 25 25 25 16 25 36 16 9 7 2 23 13 10 16 20 5 11 14 22 3 6 25 16 25 36 25 4 25 1 8 17 19 9 25 36 [1,23] chained by squares Conclusions: [1,22] cannot be chained by squares [1,23] CAN be chained by squares [1,24] cannot be chained by squares

  19. Squares can chain [1,n] for n= 15, 16, and 17 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9 17, , 16 And 23: 18, 7, 9, 16, 20, 5, 11, 14, 22, 3, 1, 8, 17, 19, 6, 10, 15, 21, 4, 12, 13, 23, 2. And 25: 18, 7, 9, 16, 20, 5, 11, 25, 24, 12, 4, 21, 15, 10, 6, 19, 17, 8, 1, 3, 22, 14, 2, 23, 13. And 26: 18, 7, 9, 16, 20, 5, 11, 25, 24, 12, 13, 3, 22, 14, 2, 23, 26, 10, 6, 19, 17, 8, 1, 15, 21, 4. And 27: 18, 7, 2, 14, 22, 27, 9, 16, 20, 5, 11, 25, 24, 12, 4, 21, 15, 10, 26, 23, 13, 3, 1, 8, 17, 19, 6.

  20. And 28: 18, 7; 2, 23, 26, 10, 6, 19, 17, 8, 28, 21, 15, 1, 24, 25, 11; 14, 22, 27, 9, 16, 20; 5, 4, 12, 13, 3. And 29: 18, 7, (29), 20, 16, 9, 27, 22, 14; 2, 23, 26, 10, 6, 19, 17, 8, 28, 21, 15, 1, 24, 25, 11; 5, 4, 12, 13, 3. And (now trivially) 30 and 31: (31), 18, 7, 29, 20, 16, 9, 27, 22, 14, 2, 23, 26, 10; 6, (30), 19, 17, 8, 28, 21; 15, 1, 24, 25, 11, 5; 4, 12, because {6,19, 30} is the first triangle in the infinite graph. Here is another solution of 29, 30, and 31: (31), 18, 7, 29, 20, 16, 9, 27, 22, 14, 2, 23, 26, 10; 15, 1, 3, 6, (30), 19, 17, 8, 28, 21, 4; 5, 11, 25, 24, 12, 13 which extends to a solution of 31 and 32: 13, 12, 24, 25, 11, 5; 31, 18, 7, 29, 20, 16, 9, 27, 22, 14, 2, 23, 26, 10, 15, 1, 3, 6, 30, 19, 17, 8, 28, 21, 4, (32).

  21. Problems: (wide range of difficulty) For what value of n can [1, n] be chained by squares? by cubes? by kth powers? What is the smallest n such that [1, n] can be chained by squares? …? Is there a largest n such that [1, n] cannot be chained by squares? …? If so, what is it? [Vague?] How fast can the elements of Sgrow such that questions about chaining [1, n] remain interesting?

  22. [RKG’s Conjecture] Fibonacci numbers, F grow exponentially as fast as any interesting set S.

  23. Fibonacci # = {1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144…} 21 11 10 13 13 13 5 8 3 5 9 4 1 2 3 5 8 8 13 21 13 13 21 12 7 6 8 13 RKG: F chains [1, n] for n = 2, 3, 4, 5, 9, 12, 13 7, 8, 11, Fdoesn’t chain [1, n]if n= 6, 10

  24. Joint unpublished result of ERB and RKG [2003]: Fibonacci plays Pool! Fibonacci plays Billiards! [1,34] is chained by {21,34,55} [1, Fk] is chained by {Fk-1, Fk, Fk+1}

  25. Joint unpublished result of ERB and RKG [2003]: Fibonacci plays Pool! [1,34] is chained by {21,34,55} [1, Fk] is chained by {Fk-1, Fk, Fk+1}

  26. Pythagoras plays Billiards, Too! n = 15 is the smallest n such that [1, n] is chained by squares If a, b, c, is a primitive Pythagorean triplet, with a <b <c and a²=b²=c², then [1, b²] is chained by squares If n < 23 and [1, n] is chained by squares, then it is chained by squares without using 2² = 4 †Small elements of Saren’t of much use

  27. Conditions for 4 elements of S to form the corners of a billiard table: B C A D A, B, C, D εS. (A > B > C > D) Corners are at A/2, B/2, C/2, D/2 Perimeter = n = A – C = B – D Height = B – A = C – D Width = B – C

  28. Conditions for 4 elements of S to form the corners of a billiard table: B C A D A, B, C, D εS. (A > B > C > D) Corners are at A/2, B/2, C/2, D/2 Perimeter = n = A – C = B – D Height = B – A = C – D Width = B – C If all corners are integers and if gcd(height, width) > 2, then path is degenerate. If this gcd = 1, path is complete

  29. If S= {s1 , s2 , …sk , …} Where s1<s2 < … < sk-1 < sk < … And if sk + 2 ≤ n < sk+2 – (sk+2 ) Then S cannot chain [1, n] Proof: sk+ 2 sk+ 1 x = sk y = x + 1 z = x + 2 sk 1 n Corollaries: Fibs cannot chain [1, n] unless Fk – 2 ≤ n ≤ Fk + 1 Squares cannot chain [1, n] unless n ≥ 15 Cubes cannot chain [1, n] unless n ≥ 295

  30. F chains [1, n] if nεF Fchains [1, n] if nεF- 1 Fcannot chain [1, n] if Fk-1+ 1< n <Fk- 1 Theorem F chains only 9 εF+ 1 and only 11 εF - 2

  31. 377 233 127 161 216 89 233 17 72 144

  32. 89 55 38 51 21 55 4 17 34

  33. Fk+2 Fk+1 - Fk 2 Fk+1 38 51 21 Fk-1 3Fk 2 Fk-1 - Fk 2 Fk 2 Fk+1 55 4 17 Fk

  34. 9 12 1 4

  35. If sk+2 > sk+1 + sk + 1 and {s1,s2 , …, sk+2} chains [1,n] then so does {s1,s2 , …, sk+1} What is the fastest growing sequence such that for all k, there exists n(k), such that {s1,s2 , …, sk} chains [1, n] but {s1,s2 , …, sk-1} does not? Answer: Super- Fibonaccis: xn = xn-1 +xn-2 + 1 0, 1, 1, 3, 5, 9, 15, 25, 41, 68…

  36. 25 15 9

  37. Engineering of Modified Pool Tables

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