Is a square matrix square palindromic?
Implementation of various properties presented in a paper by Arthur T. Benjamin and K. Yasuda
is.square.palindromic(m, base=10, give.answers=FALSE) is.centrosymmetric(m) is.persymmetric(m)
m |
The square to be tested |
base |
Base of number expansion, defaulting to 10; not relevant for the “sufficient” part of the test |
give.answers |
Boolean, with |
The following tests apply to a general square matrix m of size
n*n.
A centrosymmetric square is one in which
a[i,j]=a[n+1-i,n+1-j]; use is.centrosymmetric() to test
for this (compare an associative square). Note that this
definition extends naturally to hypercubes: a hypercube a is
centrosymmetric if all(a==arev(a)).
A persymmetric square is one in which
a[i,j]=a[n+1-j,n+1-i]; use is.persymmetric() to test for
this.
A matrix is square palindromic if it satisfies the rather complicated conditions set out by Benjamin and Yasuda (see refs).
These functions return a list of Boolean variables whose value depends
on whether or not m has the property in question.
If argument give.answers takes the default value of
FALSE, a Boolean value is returned that shows whether the
sufficient conditions are met.
If argument give.answers is TRUE, a detailed list is
given that shows the status of each individual test, both for the
necessary and sufficient conditions. The value of the second element
(named necessary) is the status of their Theorem 1 on page 154.
Note that the necessary conditions do not depend on the base b
(technically, neither do the sufficient conditions, for being a square
palindrome requires the sums to match for every base b.
In this implementation, “sufficient” is defined only with
respect to a particular base).
Every associative square is square palindromic, according to Benjamin and Yasuda.
Function is.square.palindromic() does not yet take a
give.answers argument as does, say, is.magic().
Robin K. S. Hankin
Arthur T. Benjamin and K. Yasuda. Magic “Squares” Indeed!, American Mathematical Monthly, vol 106(2), pp152-156, Feb 1999
is.square.palindromic(magic(3)) is.persymmetric(matrix(c(1,0,0,1),2,2)) #now try a circulant: a <- matrix(0,5,5) is.square.palindromic(circulant(10)) #should be TRUE
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