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

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

What about this graph? YES

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. S N Version 5: Given a set of n integers: N={a 1, a 2, a 3 …a n } and a set of pair sums; S = {s 1, s 2,...s k }, find a Hamiltonian path for the graph G whose nodes are N, and there is a branch between a i and a j iff a i + a j ε S.

Interesting Special Case of the Traveling Salesman Problem: Nodes = interval of j + 1- i consecutive integers: [ i, j ] S Permissible pairsums= S = { s 1, s 2 … } S We say [ i, j ] can be chained by S iff a Hamiltonian path exist

Problems: (wide range of difficulty) For what value of n can [1, n ] be chained by squares? by cubes? by k th 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?

S= {1, 4, 9, 16, 25, 36, 49, …}

S= {1, 4, 9, 16, 25, 36, 49, …}

S= {1, 4, 9, 16, 25, 36, 49, …}

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 Let’s now prove this graph has no Hamiltonian Path: 22

Since 8 cannot be an endpoint, branch 1-8 must be used. Since 4 cannot be an endpoint, branch 12-4 must be used Since 24 cannot be an endpoint, branches 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

Can [1,22] be chained by squares?

NO Can [1,22] be chained by squares?

In [1,23], branch 13-3 would force a third endpoint at 12 or 23. So it cannot be used What are all solutions of chaining [1,23] by squares?

[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. What are all solutions of chaining [1,23] by squares?

[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

8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9 17,, 16 Squares can chain [1,n] for n= 15, 16, and 17 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.

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).

Problems: (wide range of difficulty) For what value of n can [1, n ] be chained by squares? by cubes? by k th 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 S grow such that questions about chaining [1, n ] remain interesting?

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

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

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 Joint unpublished result of ERB and RKG [2003]: [1, F k ] is chained by { F k-1, F k, F k+1 } Fibonacci plays Pool!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

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

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

S Conditions for 4 elements of S to form the corners of a billiard table: B A C D S. A, B, C, D ε S. (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

S Conditions for 4 elements of S to form the corners of a billiard table: B A C D S. A, B, C, D ε S. (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

If S = {s 1, s 2, …s k, …} Where s 1 < s 2 < … < s k-1 < s k < … And if s k + 2 ≤ n < s k+2 – (s k +2 ) Then S cannot chain [1, n] Proof: Corollaries: Fibs cannot chain [1, n ] unless F k – 2 ≤ n ≤ F k + 1 Squares cannot chain [1, n ] unless n ≥ 15 Cubes cannot chain [1, n ] unless n ≥ sksk s k+ 1 s k+ 2 n x = s k y = x + 1 z = x + 2

FF F chains [1, n ] if n ε F F F F chains [1, n ] if n ε F - 1 F F F F cannot chain [1, n ] if F k < n < F k - 1 Theorem FF F chains only 9 ε F + 1 F and only 11 ε F - 2

F k +2 F k +1 FkFk 3F k 2 F k+1 F k -1 F k +1 - F k 2 F k -1 - F k 2 F k 2

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

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Engineering of Modified Pool Tables

Can we make a useful pool table whose corners are CUBES?