isoperimetric sequences for infinite complete binary trees ... · isoperimetric sequences for in...

Isoperimetric sequences for infinite completebinary trees and their relation to meta-Fibonacci

sequences and signed almost binary partitions

Frank Ruskey1 Sunil Chandran2 Anita Das2

1Department of Computer Science, U. of Victoria, CANADA.2Department of Computer Science and Automation, Indian Institute of Science

(IISc), Bangalore, INDIA.

CANADAM, Montreal, May 2009

Complete BinaryTrees

Meta-Fibonacci Sequences: Tanny and ConollyThe Tanny sequence:

I Let T (n) = maximum number of leaves possible at the lowestlevel in a binary tree with n nodes. For n = 1, 2, 3, . . .,

1, 1, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 9, . . .

I Recurrence relation (with T (1) = T (2) = 1,T (3) = 2):T (n) = T (n − 1− T (n − 1)) + T (n − 2− T (n − 2)).

The Conolly sequence:

I Integer k occurs a number of times equal to one plus its2-adic evaluation (ruler function). For n = 1, 2, 3, . . .,

1,︸︷︷︸1

2, 2︸︷︷︸2

, 3,︸︷︷︸1

4, 4, 4︸︷︷︸3

, 5,︸︷︷︸1

6, 6︸︷︷︸2

, 7,︸︷︷︸1

8, 8, 8, 8︸︷︷︸4

, 9,︸︷︷︸1

10, 10︸︷︷︸2

, 11,︸︷︷︸1

12, 12, 12︸︷︷︸3

, . . .

I Satisfies the recurrence relation (with C (1) = 1,C (2) = 2):C (n) = C (n − C (n − 1)) + C (n − 1− C (n − 2)).

1, 1, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 9, . . .

1,︸︷︷︸1

2, 2︸︷︷︸2

, 3,︸︷︷︸1

4, 4, 4︸︷︷︸3

, 5,︸︷︷︸1

6, 6︸︷︷︸2

, 7,︸︷︷︸1

8, 8, 8, 8︸︷︷︸4

, 9,︸︷︷︸1

10, 10︸︷︷︸2

, 11,︸︷︷︸1

12, 12, 12︸︷︷︸3

, . . .

1, 1, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 9, . . .

1,︸︷︷︸1

2, 2︸︷︷︸2

, 3,︸︷︷︸1

4, 4, 4︸︷︷︸3

, 5,︸︷︷︸1

6, 6︸︷︷︸2

, 7,︸︷︷︸1

8, 8, 8, 8︸︷︷︸4

, 9,︸︷︷︸1

10, 10︸︷︷︸2

, 11,︸︷︷︸1

12, 12, 12︸︷︷︸3

, . . .

1, 1, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 9, . . .

1,︸︷︷︸1

2, 2︸︷︷︸2

, 3,︸︷︷︸1

4, 4, 4︸︷︷︸3

, 5,︸︷︷︸1

6, 6︸︷︷︸2

, 7,︸︷︷︸1

8, 8, 8, 8︸︷︷︸4

, 9,︸︷︷︸1

10, 10︸︷︷︸2

, 11,︸︷︷︸1

12, 12, 12︸︷︷︸3

, . . .

Tanny numbers again

A sequence of binary trees maximizing the number of leaves at thelowest level.

1 1 1 1 1

T (n) = 1, 1, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, . . .

Each tree is a subtree of the same infinite tree.

The Conolly numbers come from this tree

I Like before except nodes along leftmost path are unlabeled.

I To get C (n) ignore all nodes with labels bigger than n andcount leaves at the bottom level.

I For example C (21) = 12.

I Proofs in paper with Brad Jackson, extensions in paper withformer student Chris Degau.

An “isoperimetric” problem on infinite binary trees

An infinite binary tree T∞ with all leaves at same level..

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A subset S of the tree, |S | = n = 24..

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The size of the cut (red edges): |(S ,S)| = 20.The problem: Given |S |, how small can we make this cut?

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−1 −1

−1 −1 −1−1

−3 −3 −3 −3 −3−3

Label v where (v , par(v)) ∈ (S , S) with f (v) = ±(2` − 1),` = level of v .24 = 31−15+7+7+7+7+7−3−3−3−3−3−3−3−1−1−1−1−1−1.

−1 −1

−3−3−3

−3−3

−1−1

−+ +

+−+−

+−−

Redistribute; sum at v is 1 iff v ∈ S , otherwise 0.24 = 31−15+7+7+7+7+7−3−3−3−3−3−3−3−1−1−1−1−1−1,

the number of parts is 20.

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A better subset, the “greedy” one, with |(S , S)| = 4.24 = 15 + 7 + 1 + 1.

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The best subset, the “connected” one, with |(S ,S)| = 2.24 = 31− 7.

Recap and Preview

For every finite subset of T∞ there is a partition of the number |S |in which every part is ±(2k − 1) and |(S , S̄)| is equal to thenumber of parts in that partition.

Yet to come:

The partition of |S | of the above form with the least number ofparts gives the answer to the isoperimetric problem.

Notation and results

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I Problem: over all subsets S of size n of T∞, what is δ(n), theleast number of edges in the cut (S ,S)?

I Restrictions: S is connected (δC (n)); S consists of“complete” binary trees” (δG (n)).E.g. δ(24) = δC (24) = 2, and δG (24) = 4.

I Notation ν(d) = 2d − 1; ν(P) =∑

p∈P(2p − 1).

I Results (C (n) = Conolly, T (n) = Tanny):I δC (n) = n + 2− 2T (n); δG (n) = 2C (n)− n.I δ(n) = min1≤k≤n{δC (k) + δG (n − k)}.I δ(n) = 1 + min{δ(n − ν(d)), δ(ν(d + 1)− n)},

where d = blg nc.

��

p∈P(2p − 1).

where d = blg nc.

��

p∈P(2p − 1).

where d = blg nc.

��

p∈P(2p − 1).

where d = blg nc.

��

p∈P(2p − 1).

where d = blg nc.

��

p∈P(2p − 1).

where d = blg nc.

One proof:

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Lemma: δC (n) = n + 2− 2T (n).Proof: Let |S | = n be such that |(S ,S)| = δC (n).

I On the one hand, counting edges,∑v∈S deg(v) = 2(n − 1) + |(S , S)| = 2(n − 1) + δC (n).

I On the other,∑

v∈S deg(v) = 3n3 + n1 = 3n − 2n1.

I Thus δC (n) = n + 2− 2n1 = n + 2− 2T (n), because T (n)maximizes n1.

Binary partitions

I Binary partitions (all parts of the form 2j)Example: 12 = 2 + 2 + 4 + 4 = 21 + 21 + 22 + 22.

I Almost binary partitions (all parts of the form νj = 2j − 1).Example 12 = 7 + 3 + 1 + 1.

I We can allow the partitions to be signed.

I We often want to find minimal partitions; those with the leastnumber of parts.

I The greedy algorithm finds (the unique) minimal partitions forbinary and almost binary partitions. Like coin-changing.

I The signed cases are more interesting...

Minimal signed almost binary partitions

I A Signed Almost Binary Partition (SABP) of n is a pair(P,N) where

n =∑i∈P

(2i − 1)−∑j∈N

(2j − 1) = ν(P)− ν(N).

The minimum size of a SABP of n is denotedτ(n) = |P|+ |N|.

I It is an ABP (or GABP) if N = ∅. The minimum size of aABP is τG (n). (G is for “greedy”).

I It is a CABP if |P| = 1. The minimum size of a CABP isτC (n). (C is for “connected”).

I 12 = 15-3 is a mSABP and a mCABP. 12 = 7+3+1+1 is amGAPB.

I Theorem: δ(n) = τ(n), δG (n) = τG (n), and δC (n) = τC (n).

A normal form for a SABP (P ,N)

I Three conditions

(A) P ∩ N = ∅.(B) P = Greedy(ν(P)) and N = Greedy(ν(N)).(C) max(P) ∈ {blg nc, 1 + blg nc}.

I Lemma: For every n there is a minimal SABP (or mGABP ormCABP) that is in normal form.

I Examples:Third condition is not vacuous: 5 = 15− 7− 3.

43 = 31 + 15− 3︸︷︷︸τ=3

= 31 + 7 + 3 + 1 + 1︸︷︷︸τG=5

= 63− 15− 3− 1− 1︸︷︷︸τC=5

A “duality” and “symmetry”

δ(n) =

TheoremLet d = blg nc.

δG (n) = δC (3 · 2d − n − 2) and δ(n) = δ(3 · 2d − n − 2)

Example: 31+15 + 3 + 1↔ 63−15− 3− 1

Note the binary representations:(n)2 1α011 · · · 1

(3 · 2d − n − 2)2 1α011 · · · 1

A “duality” and “symmetry”

δ(n) =

1 311573

TheoremLet d = blg nc.

δG (n) = δC (3 · 2d − n − 2) and δ(n) = δ(3 · 2d − n − 2)

Example: 31+15 + 3 + 1↔ 63−15− 3− 1

Note the binary representations:(n)2 1α011 · · · 1

(3 · 2d − n − 2)2 1α011 · · · 1

How fast does δ(n) grow?

Curves are lg n and (lg n)/2; diamonds are δ(n).

New curves are δG (n) and δC (n).

When is δ(n) = δG (n)?

n = 1023 + 127 + 15 + 1 = (210−1) + (27−1) + (24−1) + (21−1)

TheoremIf the minimum gap is at least 3, then δ(n) = δG (n). As aconsequence, δ(n) ≥ (lg n)/3.

Can 3 be replaced by 2 above? NO!

The smallest known counter-example is n = 46, 912, 496, 118, 419,where 21 = δ(n) = −2 + δG (n). Why? Here n =

∑j∈S ν(j), where

S = {1, 3, . . . , 45}∑j∈S 2j 1010101010101010101010101010101010101010101010∑j∈B 2j 101010101010101010101010101010101010110000000∑j∈A 2j 10000000000000000000000000000000000000000000000

s + b − a 101010

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

m α(m) α(m)− (4m − 1)/(4− 1)1 1 02 2 33 5 164 20 655 83 2586 594 7717 2641 28208 10856 109899 43623 4375810 305766 4375911 1354341 4376012 5548644 4376113 22325859 4376214 89434722 4376315 357870241 4370016 1431612752 4301317 5726580047 43014

Open Problems

I Extend to k-ary trees.

I Is there a faster way to compute δ(n) and the actual mSABP?I.e., something similar to Prodinger’s algorithm for signedbinary partitions.

I We conjecture that the general problem has a nested solution.This means that there is a sequence S1, S2, . . . of subsets ofnodes in T∞ with the following three properties. |Sn| = n,Sn ⊂ Sn+1, and |(Sn,Sn)| = δ(n).

The end

Thanks for coming!Any questions?

Another oddity...

Notation: #1(s) is the number of 1’s in the string s.

δG (n) = mink≥1{#1((n + k)2) ≤ k}.

For example, G (12) = G ((1100)2) = 4 because, takingsuccessively k = 1, 2, 3, 4, we have #1(1101) = 3 > 1,#1(1110) = 3 > 2, #1(1111) = 4 > 3, but #1(10000) = 1 ≤ 4.

isoperimetric sequences for infinite complete binary trees ... · isoperimetric sequences for in...

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