Isoperimetric sequences for infinite complete binary trees and their relation to meta-fibonacci sequences and signed almost binary partitions

Transcription

1 Isoperimetric sequences for infinite complete binary trees and their relation to meta-fibonacci sequences and signed almost binary partitions Frank Ruskey Sunil Chandran 2 Anita Das 2 Department of Computer Science, U. of Victoria, CANADA. 2 Department of Computer Science and Automation, Indian Institute of Science (IISc), Bangalore, INDIA. CANADAM, Montreal, May 2009

2 Complete BinaryTrees

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6 Meta-Fibonacci Sequences: Tanny and Conolly The Tanny sequence: Let T (n) = maximum number of leaves possible at the lowest level in a binary tree with n nodes. For n =, 2, 3,...,,, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 9,... Recurrence relation (with T () = T (2) =, T (3) = 2): T (n) = T (n T (n )) + T (n 2 T (n 2)). The Conolly sequence: Integer k occurs a number of times equal to one plus its 2-adic evaluation (ruler function). For n =, 2, 3,...,, 2, 2 2, 3, 4, 4, 4 3, 5, 6, 6 2, 7, 8, 8, 8, 8 4, 9, 0, 0,, 2, 2, 2, 2 3 Satisfies the recurrence relation (with C() =, C(2) = 2): C(n) = C(n C(n )) + C(n C(n 2)).

7 Meta-Fibonacci Sequences: Tanny and Conolly The Tanny sequence: Let T (n) = maximum number of leaves possible at the lowest level in a binary tree with n nodes. For n =, 2, 3,...,,, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 9,... Recurrence relation (with T () = T (2) =, T (3) = 2): T (n) = T (n T (n )) + T (n 2 T (n 2)). The Conolly sequence: Integer k occurs a number of times equal to one plus its 2-adic evaluation (ruler function). For n =, 2, 3,...,, 2, 2 2, 3, 4, 4, 4 3, 5, 6, 6 2, 7, 8, 8, 8, 8 4, 9, 0, 0,, 2, 2, 2, 2 3 Satisfies the recurrence relation (with C() =, C(2) = 2): C(n) = C(n C(n )) + C(n C(n 2)).

8 Meta-Fibonacci Sequences: Tanny and Conolly The Tanny sequence: Let T (n) = maximum number of leaves possible at the lowest level in a binary tree with n nodes. For n =, 2, 3,...,,, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 9,... Recurrence relation (with T () = T (2) =, T (3) = 2): T (n) = T (n T (n )) + T (n 2 T (n 2)). The Conolly sequence: Integer k occurs a number of times equal to one plus its 2-adic evaluation (ruler function). For n =, 2, 3,...,, 2, 2 2, 3, 4, 4, 4 3, 5, 6, 6 2, 7, 8, 8, 8, 8 4, 9, 0, 0,, 2, 2, 2, 2 3 Satisfies the recurrence relation (with C() =, C(2) = 2): C(n) = C(n C(n )) + C(n C(n 2)).

9 Meta-Fibonacci Sequences: Tanny and Conolly The Tanny sequence: Let T (n) = maximum number of leaves possible at the lowest level in a binary tree with n nodes. For n =, 2, 3,...,,, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 9,... Recurrence relation (with T () = T (2) =, T (3) = 2): T (n) = T (n T (n )) + T (n 2 T (n 2)). The Conolly sequence: Integer k occurs a number of times equal to one plus its 2-adic evaluation (ruler function). For n =, 2, 3,...,, 2, 2 2, 3, 4, 4, 4 3, 5, 6, 6 2, 7, 8, 8, 8, 8 4, 9, 0, 0,, 2, 2, 2, 2 3 Satisfies the recurrence relation (with C() =, C(2) = 2): C(n) = C(n C(n )) + C(n C(n 2)).

10 Tanny numbers again A sequence of binary trees maximizing the number of leaves at the lowest level T (n) =,, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8,... Each tree is a subtree of the same infinite tree.

11 The Conolly numbers come from this tree Like before except nodes along leftmost path are unlabeled. To get C(n) ignore all nodes with labels bigger than n and count leaves at the bottom level. For example C(2) = 2. Proofs in paper with Brad Jackson, extensions in paper with former student Chris Degau.

12 The Conolly numbers come from this tree Like before except nodes along leftmost path are unlabeled. To get C(n) ignore all nodes with labels bigger than n and count leaves at the bottom level. For example C(2) = 2. Proofs in paper with Brad Jackson, extensions in paper with former student Chris Degau.

13 The Conolly numbers come from this tree Like before except nodes along leftmost path are unlabeled. To get C(n) ignore all nodes with labels bigger than n and count leaves at the bottom level. For example C(2) = 2. Proofs in paper with Brad Jackson, extensions in paper with former student Chris Degau.

14 The Conolly numbers come from this tree Like before except nodes along leftmost path are unlabeled. To get C(n) ignore all nodes with labels bigger than n and count leaves at the bottom level. For example C(2) = 2. Proofs in paper with Brad Jackson, extensions in paper with former student Chris Degau.

15 An isoperimetric problem on infinite binary trees An infinite binary tree T with all leaves at same level..

16 An isoperimetric problem on infinite binary trees A subset S of the tree, S = n = 24..

17 An isoperimetric problem on infinite binary trees The size of the cut (red edges): (S, S) = 20. The problem: Given S, how small can we make this cut?

18 An isoperimetric problem on infinite binary trees Label v where (v, par(v)) (S, S) with f (v) = ±(2 l ), l = level of v. 24 =

19 An isoperimetric problem on infinite binary trees Redistribute; sum at v is iff v S, otherwise = , the number of parts is 20.

20 An isoperimetric problem on infinite binary trees A better subset, the greedy one, with (S, S) = =

21 An isoperimetric problem on infinite binary trees The best subset, the connected one, with (S, S) = = 3 7.

22 Recap and Preview For every finite subset of T there is a partition of the number S in which every part is ±(2 k ) and (S, S) is equal to the number of parts in that partition. Yet to come: The partition of S of the above form with the least number of parts gives the answer to the isoperimetric problem.

23 Notation and results Problem: over all subsets S of size n of T, what is δ(n), the least number of edges in the cut (S, S)? Restrictions: S is connected (δ C (n)); S consists of complete binary trees (δ G (n)). E.g. δ(24) = δ C (24) = 2, and δ G (24) = 4. Notation ν(d) = 2 d ; ν(p) = p P (2p ). Results (C(n) = Conolly, T (n) = Tanny): δc (n) = n + 2 2T (n); δ G (n) = 2C(n) n. δ(n) = min k n {δ C (k) + δ G (n k)}. δ(n) = + min{δ(n ν(d)), δ(ν(d + ) n)}, where d = lg n.

24 Notation and results Problem: over all subsets S of size n of T, what is δ(n), the least number of edges in the cut (S, S)? Restrictions: S is connected (δ C (n)); S consists of complete binary trees (δ G (n)). E.g. δ(24) = δ C (24) = 2, and δ G (24) = 4. Notation ν(d) = 2 d ; ν(p) = p P (2p ). Results (C(n) = Conolly, T (n) = Tanny): δc (n) = n + 2 2T (n); δ G (n) = 2C(n) n. δ(n) = min k n {δ C (k) + δ G (n k)}. δ(n) = + min{δ(n ν(d)), δ(ν(d + ) n)}, where d = lg n.

25 Notation and results Problem: over all subsets S of size n of T, what is δ(n), the least number of edges in the cut (S, S)? Restrictions: S is connected (δ C (n)); S consists of complete binary trees (δ G (n)). E.g. δ(24) = δ C (24) = 2, and δ G (24) = 4. Notation ν(d) = 2 d ; ν(p) = p P (2p ). Results (C(n) = Conolly, T (n) = Tanny): δc (n) = n + 2 2T (n); δ G (n) = 2C(n) n. δ(n) = min k n {δ C (k) + δ G (n k)}. δ(n) = + min{δ(n ν(d)), δ(ν(d + ) n)}, where d = lg n.

26 Notation and results Problem: over all subsets S of size n of T, what is δ(n), the least number of edges in the cut (S, S)? Restrictions: S is connected (δ C (n)); S consists of complete binary trees (δ G (n)). E.g. δ(24) = δ C (24) = 2, and δ G (24) = 4. Notation ν(d) = 2 d ; ν(p) = p P (2p ). Results (C(n) = Conolly, T (n) = Tanny): δc (n) = n + 2 2T (n); δ G (n) = 2C(n) n. δ(n) = min k n {δ C (k) + δ G (n k)}. δ(n) = + min{δ(n ν(d)), δ(ν(d + ) n)}, where d = lg n.

27 Notation and results Problem: over all subsets S of size n of T, what is δ(n), the least number of edges in the cut (S, S)? Restrictions: S is connected (δ C (n)); S consists of complete binary trees (δ G (n)). E.g. δ(24) = δ C (24) = 2, and δ G (24) = 4. Notation ν(d) = 2 d ; ν(p) = p P (2p ). Results (C(n) = Conolly, T (n) = Tanny): δc (n) = n + 2 2T (n); δ G (n) = 2C(n) n. δ(n) = min k n {δ C (k) + δ G (n k)}. δ(n) = + min{δ(n ν(d)), δ(ν(d + ) n)}, where d = lg n.

28 Notation and results Problem: over all subsets S of size n of T, what is δ(n), the least number of edges in the cut (S, S)? Restrictions: S is connected (δ C (n)); S consists of complete binary trees (δ G (n)). E.g. δ(24) = δ C (24) = 2, and δ G (24) = 4. Notation ν(d) = 2 d ; ν(p) = p P (2p ). Results (C(n) = Conolly, T (n) = Tanny): δc (n) = n + 2 2T (n); δ G (n) = 2C(n) n. δ(n) = min k n {δ C (k) + δ G (n k)}. δ(n) = + min{δ(n ν(d)), δ(ν(d + ) n)}, where d = lg n.

29 One proof: Lemma: δ C (n) = n + 2 2T (n). Proof: Let S = n be such that (S, S) = δ C (n). On the one hand, counting edges, v S deg(v) = 2(n ) + (S, S) = 2(n ) + δ C (n). On the other, v S deg(v) = 3n 3 + n = 3n 2n. Thus δ C (n) = n + 2 2n = n + 2 2T (n), because T (n) maximizes n.

30 Binary partitions Binary partitions (all parts of the form 2 j ) Example: 2 = = Almost binary partitions (all parts of the form ν j = 2 j ). Example 2 = We can allow the partitions to be signed. We often want to find minimal partitions; those with the least number of parts. The greedy algorithm finds (the unique) minimal partitions for binary and almost binary partitions. Like coin-changing. The signed cases are more interesting...

31 Minimal signed almost binary partitions A Signed Almost Binary Partition (SABP) of n is a pair (P, N) where n = (2 i ) i P j N(2 j ) = ν(p) ν(n). The minimum size of a SABP of n is denoted τ(n) = P + N. It is an ABP (or GABP) if N =. The minimum size of a ABP is τ G (n). (G is for greedy ). It is a CABP if P =. The minimum size of a CABP is τ C (n). (C is for connected ). 2 = 5-3 is a msabp and a mcabp. 2 = is a mgapb. Theorem: δ(n) = τ(n), δ G (n) = τ G (n), and δ C (n) = τ C (n).

32 A normal form for a SABP (P, N) Three conditions (A) P N =. (B) P = Greedy(ν(P)) and N = Greedy(ν(N)). (C) max(p) { lg n, + lg n }. Lemma: For every n there is a minimal SABP (or mgabp or mcabp) that is in normal form. Examples: Third condition is not vacuous: 5 = = τ=3 = τ G =5 = } 63 5 {{ 3 }. τ C =5

33 A duality and symmetry δ(n) = Theorem Let d = lg n. δ G (n) = δ C (3 2 d n 2) and δ(n) = δ(3 2 d n 2) Example: Note the binary representations: (n) 2 α0 (3 2 d n 2) 2 α0

34 A duality and symmetry δ(n) = Theorem Let d = lg n. δ G (n) = δ C (3 2 d n 2) and δ(n) = δ(3 2 d n 2) Example: Note the binary representations: (n) 2 α0 (3 2 d n 2) 2 α0

35 How fast does δ(n) grow? Curves are lg n and (lg n)/2; diamonds are δ(n). New curves are δ G (n) and δ C (n).

36 When is δ(n) = δ G (n)? n = = (2 0 ) + (2 7 ) + (2 4 ) + (2 ) Theorem If the minimum gap is at least 3, then δ(n) = δ G (n). As a consequence, δ(n) (lg n)/3. Can 3 be replaced by 2 above? NO! The smallest known counter-example is n = 46, 92, 496, 8, 49, where 2 = δ(n) = 2 + δ G (n). Why? Here n = j S ν(j), where S = {, 3,..., 45} j S 2j j B 2j j A 2j s + b a 000

37 Growth Rate of α(m) = min{n : δ(n) = m} m α(m) α(m) (4 m )/(4 )

38 Open Problems Extend to k-ary trees. Is there a faster way to compute δ(n) and the actual msabp? I.e., something similar to Prodinger s algorithm for signed binary partitions. We conjecture that the general problem has a nested solution. This means that there is a sequence S, S 2,... of subsets of nodes in T with the following three properties. S n = n, S n S n+, and (S n, S n ) = δ(n).

39 The end Thanks for coming! Any questions?

40 Another oddity... Notation: # (s) is the number of s in the string s. Lemma δ G (n) = min k {# ((n + k) 2 ) k}. For example, G(2) = G((00) 2 ) = 4 because, taking successively k =, 2, 3, 4, we have # (0) = 3 >, # (0) = 3 > 2, # () = 4 > 3, but # (0000) = 4.