The Solution to the Partition Reconstruction Problem Maria Monks - - PowerPoint PPT Presentation

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The Solution to the Partition Reconstruction Problem Maria Monks - - PowerPoint PPT Presentation

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The Solution to the Partition Reconstruction Problem Maria Monks monks@mit.edu AMS/MAA Joint Mathematics Meetings - San Diego, CA p.1/14 Definitions and Notation A partition of a positive integer n is an sequence [ 1 , 2 , . . . ,

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The Solution to the Partition Reconstruction Problem

Maria Monks

monks@mit.edu

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.1/14

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Definitions and Notation

A partition λ of a positive integer n is an sequence [λ1, λ2, . . . , λm] of positive integers which satisfy λ1 ≥ λ2 ≥ · · · ≥ λm and m

i=1 λi = n.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.2/14

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Definitions and Notation

A partition λ of a positive integer n is an sequence [λ1, λ2, . . . , λm] of positive integers which satisfy λ1 ≥ λ2 ≥ · · · ≥ λm and m

i=1 λi = n.

5 + 2 + 2 + 1 = 10

5 2 2 1

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.2/14

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Definitions and Notation

Let λ be a partition of n, and let µ be a partition of n − k. Then µ is a k-minor of λ if µi ≤ λi for all i. [3, 2, 1, 1] is a 3-minor of [5, 2, 2, 1].

3 2 1 1

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.3/14

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Definitions and Notation

We write Mk(λ) to denote the set of all k-minors of λ.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.4/14

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Definitions and Notation

We write Mk(λ) to denote the set of all k-minors of λ. M3([5, 2, 2, 1]) = {[5, 2], [5, 1, 1], [4, 2, 1], [4, 1, 1, 1], [3, 2, 2], [3, 2, 1, 1], [2, 2, 2, 1]} M3

  • =

             

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.4/14

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The Problem

The Partition Reconstruction Problem: For which positive integers n ≥ 2 and k can we always reconstruct a given partition of n from its set of k-minors?

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.5/14

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The Problem

The Partition Reconstruction Problem: For which positive integers n ≥ 2 and k can we always reconstruct a given partition of n from its set of k-minors? Formally, for which n and k does Mk(λ) = Mk(µ) imply λ = µ for any two partitions λ and µ of n?

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.5/14

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The Problem

The Partition Reconstruction Problem: For which positive integers n ≥ 2 and k can we always reconstruct a given partition of n from its set of k-minors? Formally, for which n and k does Mk(λ) = Mk(µ) imply λ = µ for any two partitions λ and µ of n? If this property holds, we say reconstructibility holds.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.5/14

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The Problem

For example, M9([5, 2, 2, 1]) = M9([6, 3, 1]) = {[1]}. We cannot reconstruct partitions of 10 from their sets of 9-minors.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.6/14

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The Problem

For example, M9([5, 2, 2, 1]) = M9([6, 3, 1]) = {[1]}. We cannot reconstruct partitions of 10 from their sets of 9-minors. We can reconstruct them from their 1-minors, by taking “unions” of distinct 1-minors:

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.6/14

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The Problem

For example, M9([5, 2, 2, 1]) = M9([6, 3, 1]) = {[1]}. We cannot reconstruct partitions of 10 from their sets of 9-minors. We can reconstruct them from their 1-minors, by taking “unions” of distinct 1-minors:

∪ =

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.6/14

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Initial Observations

Clearly, reconstructibility fails for k = n − 1 and holds for k = 0.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.7/14

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Initial Observations

Clearly, reconstructibility fails for k = n − 1 and holds for k = 0. The set of all 1-minors of all (k − 1)-minors of a partition is the same as the set of k-minors of that partition.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.7/14

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Initial Observations

Clearly, reconstructibility fails for k = n − 1 and holds for k = 0. The set of all 1-minors of all (k − 1)-minors of a partition is the same as the set of k-minors of that partition. Hence, if reconstructibility holds for n and k, it holds for n and k − 1.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.7/14

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Initial Observations

Clearly, reconstructibility fails for k = n − 1 and holds for k = 0. The set of all 1-minors of all (k − 1)-minors of a partition is the same as the set of k-minors of that partition. Hence, if reconstructibility holds for n and k, it holds for n and k − 1. This implies that there is a function g(n), defined for n ≥ 2, such that reconstructibility holds for n and k if and only if k ≤ g(n).

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.7/14

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Initial Observations

Clearly, reconstructibility fails for k = n − 1 and holds for k = 0. The set of all 1-minors of all (k − 1)-minors of a partition is the same as the set of k-minors of that partition. Hence, if reconstructibility holds for n and k, it holds for n and k − 1. This implies that there is a function g(n), defined for n ≥ 2, such that reconstructibility holds for n and k if and only if k ≤ g(n). What is g?

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.7/14

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Properties of g

We can show g(n) ≤ g(n − 1) + 1.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.8/14

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Properties of g

We can show g(n) ≤ g(n − 1) + 1. If a partition of the form below appears for n, then g(n) ≤ m − 2.

m l

{

{

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.8/14

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Properties of g

Define ρ(a) to be the smallest divisor d of a for which d ≥ √a.

m l

{

{

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.8/14

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Properties of g

Define ρ(a) to be the smallest divisor d of a for which d ≥ √a. We can show g(n) ≤ ρ(n + 2) − 2.

m l

{

{

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.8/14

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Properties of g

Define ρ(a) to be the smallest divisor d of a for which d ≥ √a. We can show g(n) ≤ ρ(n + 2) − 2. Recall that g(n) ≤ g(n − 1) + 1.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.9/14

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Properties of g

Define ρ(a) to be the smallest divisor d of a for which d ≥ √a. We can show g(n) ≤ ρ(n + 2) − 2. Recall that g(n) ≤ g(n − 1) + 1. In fact, the recursion g(n) = min{g(n − 1) + 1, ρ(n + 2) − 2} holds for most positive integers n. Some counterexamples are: n = 5, 12, 21, 32, . . .

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.9/14

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Recursive formula for g(n)

What is this sequence 5, 12, 21, 32, . . .?

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.10/14

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Recursive formula for g(n)

What is this sequence 5, 12, 21, 32, . . .? Just 5, 12, 21, 32. There are no other counterexamples!

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.10/14

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Recursive formula for g(n)

What is this sequence 5, 12, 21, 32, . . .? Just 5, 12, 21, 32. There are no other counterexamples!

  • Theorem. Let n > 2 be a positive integer other than 5,

12, 21, and 32. Then g(n) = min{ρ(n + 2) − 2, g(n − 1) + 1}.

Direct calculation shows that g(2) = 0, g(5) = 1, g(12) = 3, g(21) = 5, g(32) = 7.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.10/14

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Bounds on g

We can use the recursion to obtain bounds on g: √ n + 2 − 2 ≤ g(n) ≤ √ n + 2 + 3

4

√ n + 2.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.11/14

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Bounds on g

We can use the recursion to obtain bounds on g: √ n + 2 − 2 ≤ g(n) ≤ √ n + 2 + 3

4

√ n + 2. Equality holds for the lower bound whenever n + 2 is a perfect square.

n 2 7 14 23 34 47 62 79 98 119 142 5 10 15 g(n)

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.11/14

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The Solution!

We can solve the recursion to obtain an explicit formula for g.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.12/14

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The Solution!

We can solve the recursion to obtain an explicit formula for g.

  • Theorem. Let n and k be positive integers with k < n.

Then partitions of n can be reconstructed from their sets of k-minors if and only if k ≤ g(n), where

g(n) = min

0≤t≤n ρ(n + 2 − t) − 2 + t

for n ∈ {5, 12, 21, 32}, and g(5) = 1, g(12) = 3,

g(21) = 5, g(32) = 7.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.12/14

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Applications

The partition reconstruction problem was posed by J. Siemons and O. Pretzel, in an attempt to answer the following: For which n and k can we always reconstruct an irreducible character of Sn from the irreducible components of its restriction to the stabilizer of {1, 2, . . . , k}?

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.13/14

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Applications

The partition reconstruction problem was posed by J. Siemons and O. Pretzel, in an attempt to answer the following: For which n and k can we always reconstruct an irreducible character of Sn from the irreducible components of its restriction to the stabilizer of {1, 2, . . . , k}? Our main theorem also completely answers this question: we can do so if and only if k ≤ g(n).

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.13/14

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Acknowledgments

This research was done at the University of Minnesota Duluth with the financial support of the National Science Foundation (grant number DMS-0447070-001) and the National Security Agency (grant number H98230-06-1-0013). I would like to thank Reid Barton and Ricky Liu for their suggestions throughout this research project. I would also like to thank Joe Gallian for introducing me to reconstruction problems and for his helpful encouragement. Finally, thanks to my father, Ken Monks, and to the rest of my family for their help and feedback.

AMS/MAA Joint Mathematics Meetings - San Diego, CA – p.14/14