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Which Boolean Functions Are Most Informative?. Gowtham R Kumar and Thomas Courtade. Stanford University, CSoI , AFOSR. Setup. i.i.d. ~ bern ( ). BSC( ). 1-. 1. 1. 0. 0. 1-. Setup. i.i.d. ~ bern ( ). BSC( ). 1-. 1. 1. 0. 0. 1-. Boolean function. Maximize over.
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Which Boolean Functions Are Most Informative? Gowtham R Kumar and Thomas Courtade Stanford University, CSoI, AFOSR
Setup i.i.d. ~bern() BSC() 1- 1 1 0 0 1-
Setup i.i.d. ~bern() BSC() 1- 1 1 0 0 1- Boolean function Maximize over
Guess for that maximizes . If , Can we do better?
Conjecture 0 Trivially achieved by setting
Loose Outer Bound Using Mrs. Gerber’s lemma, Loose outer bound.
Standard Tools don’t work Conjecture 0: Standard Tools from Information Theory? New Tools? Combinatorics?
Standard Tools don’t work Conjecture 0: Standard Tools from Information Theory? New Tools? Combinatorics? Weaker result:
Lexicographic Ordering (Lex) defined • A Boolean function is Lex if
Lexicographic Ordering (Lex) defined • A Boolean function is Lex if • is not Lex if
Discrete Hypercube defined • Vertex set: • Edge between 2 nodes if Hamming distance is 1. 110 111 100 101 010 011 000 001 Discrete Hypercube
Discrete Hypercube defined • Vertex set: • Edge between 2 nodes if Hamming distance is 1. 110 111 100 101 = 0 = 1 010 011 000 001 “Volume”:
Isoperimetric Inequality Of all closed surfaces with a fixed volume, the sphere has the smallest surface area.
Isoperimetric Inequality Of all closed surfaces with a fixed volume, the sphere has the smallest surface area. What about edge boundary?
Edge boundary illustrated • Vertex set: • Edge between 2 nodes if Hamming distance is 1. • : Edge between red (0) and blue (1). • Edge boundary is the number of edges between red and blue. 110 111 100 101 = 0 = 1 010 011 000 001 “Surface Area”: Edge boundary = 7
Lex minimizes Edge boundary 110 110 111 111 100 100 101 101 010 011 010 011 000 001 000 001 Lex: Edge boundary = 5 Edge boundary = 7
Lex is a Sphere Lemma 1: Of all Boolean functions with fixed, the edge boundary is minimized by the that is Lex.
Conjecture 1 Conjecture 1: Of all Boolean functions with fixed, the conditional entropy is minimized by the that is Lex.
Conjecture 1 Conjecture 1: Of all Boolean functions with fixed, the conditional entropy is minimized by the that is Lex. When ,
Split Conjecture 0 split Conjecture 0 Maximize over Structure Inequality Conjecture 1 Maximize over Lex is optimal. Conjecture 2 For Lex,
Split Conjecture 0 split Conjecture 0 Maximize over Structure Inequality Conjecture 1 Maximize over Lex is optimal. Conjecture 2 For Lex,
Representing using a special tabular form. Let be a subset of indices. In the example below, , n=3. Table: Representation of an arbitrary Boolean function .
Representing using a special tabular form Let be a subset of indices. In the example below, , n=3. Table: Representation of an arbitrary Boolean function . =0
Compressions Illustrated Illustrating I-compression. I-compression reduces “Surface Area”! Table: Representation of an arbitrary Boolean function . =0
Representing using a special tabular form Let be a subset of indices. In the example below, , n=5. Table: Representation of an arbitrary Boolean function
Compressions Illustrated Illustrating I-compression. I-compression reduces “Surface Area”! Table: Representation of an arbitrary Boolean function =0
-Compression reduces edge boundary. Lemma: For any , -Compression reduces the number of edges between and .
-Compression reduces conditional entropy for Theorem: For any containing at most 2 elements, -Compression reduces the conditional entropy . Proof: Concavity of entropy and a little work. Unfortunately, we can find counter-examples for where I-Compression increases the conditional entropy.
-Compression reduces conditional entropy for Theorem: For any containing at most 2 elements, I-Compression reduces the conditional entropy . Proof: Concavity of entropy and a little work. Unfortunately, we can find counter-examples for where I-Compression increases the conditional entropy.
Can now verify Conjecture by simulation for up to . Able to reduce the number of candidate Boolean functions for which Conjectures 0,1 needs to be verified.
Plot of For Lex,
Plot of Observation 1: For ,
Plot of Observation 2 (Symmetry):
Plot of Therefore, we need to prove only for .
Plot of for How does vary with ?