2005Journal of Combinatorial DesignsOpen access

Ryser's embedding problem for Hadamard matrices

T. S. Michael

Open full text 2 citations

Abstract

Abstract What is the minimum order ${\cal R}\,(a, b)$ of a Hadamard matrix that contains an a by b submatrix of all 1's? Newman showed that where c♯ denotes the smallest order greater than or equal to c for which a Hadamard matrix exists. It follows that if 4 divides both a and b, and if the Hadamard conjecture is true, then ${\cal R}(a,b)=ab$ . We establish the improved bounds for min {a,b} ≥ 2. The Hadamard conjecture therefore implies that if 4 divides both 2ab and ⌈a/2⌉ ⌈b/2⌉, then ${\cal R}$ (a, b) = 2 · max {⌈a/2⌉b, ⌈b/2⌉a}. Our lower bound comes from a counting argument, while our upper bound follows from a sub‐multiplicative property of ${\cal R}$ : Improvements in our upper bound occur when suitable conference matrices or Bush‐type Hadamard matrices exist. We conjecture that any (1,−1)‐matrix of size a by b occurs as a submatrix of some Hadamard matrix of order at most ${\cal R}(a,b)$ . © 2005 Wiley Periodicals, Inc. J Combin Designs

Open-access reader

About this research paper

What this paper is about

Abstract What is the minimum order ${\cal R}\,(a, b)$ of a Hadamard matrix that contains an a by b submatrix of all 1's? Newman showed that where c♯ denotes the smallest order greater than or equal to c for which a Hadamard matrix exists. It follows that if 4 divides both a and b, and if the Hadamard conjecture is true, then ${\cal R}(a,b)=ab$ . We establish the improved bounds for min {a,b} ≥ 2. The Hadamard conjecture therefore implies that if 4 divides both 2ab and ⌈a/2⌉ ⌈b/2⌉, then ${\cal R}$ (a, b) = 2 · max {⌈a/2⌉b, ⌈b/2⌉a}. Our lower bound comes from a counting argument, while our upper bound follows from a sub‐multiplicative property of ${\cal R}$ : Improvements in our upper bound occur when suitable conference matrices or Bush‐type Hadamard matrices exist. We conjecture that any (1,−1)‐matrix of size a by b occurs as a submatrix of some Hadamard matrix of order at most ${\cal R}(a,b)$ . © 2005 Wiley Periodicals, Inc. J Combin Designs

Why it matters

OpenAlex reports 2 citations for this work. Citation counts describe recorded attention and do not establish research quality.

Key contribution

A contribution statement is not available in the OpenAlex record.

Method / approach

Method details are not available in the OpenAlex metadata.

Main findings

Findings are not separately available in the OpenAlex metadata.

Limitations

Limitations are not available in the OpenAlex metadata.

Applications

Application details are not available in the OpenAlex metadata.

Available abstract

Abstract What is the minimum order ${\cal R}\,(a, b)$ of a Hadamard matrix that contains an a by b submatrix of all 1's? Newman showed that where c♯ denotes the smallest order greater than or equal to c for which a Hadamard matrix exists. It follows that if 4 divides both a and b, and if the Hadamard conjecture is true, then ${\cal R}(a,b)=ab$ . We establish the improved bounds for min {a,b} ≥ 2. The Hadamard conjecture therefore implies that if 4 divides both 2ab and ⌈a/2⌉ ⌈b/2⌉, then ${\cal R}$ (a, b) = 2 · max {⌈a/2⌉b, ⌈b/2⌉a}. Our lower bound comes from a counting argument, while our upper bound follows from a sub‐multiplicative property of ${\cal R}$ : Improvements in our upper bound occur when suitable conference matrices or Bush‐type Hadamard matrices exist. We conjecture that any (1,−1)‐matrix of size a by b occurs as a submatrix of some Hadamard matrix of order at most ${\cal R}(a,b)$ . © 2005 Wiley Periodicals, Inc. J Combin Designs

Key concepts: Mathematics, Hadamard matrix, Combinatorics, Hadamard transform, Hadamard's maximal determinant problem, Conjecture, Order (exchange), Upper and lower bounds

Related papers

Back to paper searchBrowse research topicsOriginal source
Ryser's embedding problem for Hadamard matrices — Research Paper | ScholarLens