hadamard seems unnecessarily slow

[Alan Edelman]

the usual seems so much faster

function had(n) begin
        z=1
        for i=1:log2(n)
           z= kron(z,[1 1;1 -1])
        end
    end
    z
    end

n=1024
@time hadamard(n);
@time had(n);
elapsed time: 0.026615681 seconds (16779744 bytes allocated)
elapsed time: 0.006890752 seconds (11186592 bytes allocated)  

While on the subject of hadamard(n), it would be nice if a matrix would
come for n=1,2 and every multiple of 4 , say up to 1000, times a power of 2
(except of course 668 and such)

[Steven G. Johnson]

Fair enough. The hadamard function just takes the FWHT of the columns of the identity matrix, which is certainly not the most efficient way to do things. (Does it matter? I thought that performance-sensitive cases would mainly use the FWHT.)

Regarding non powers of two, I could certainly have a table of Hadamard matrices for various factors, and then build other sizes from Kronecker products of these. However, I have three concerns:

• Is this actually useful? Who needs Hadamard matrices of non-power-of-two sizes?
• Does it matter what ordering I choose? Ordering the factors in different ways seems like it will produce different matrices (albeit equivalent modulo reordering).
• Does it matter which inequivalent Hadamard matrix I choose, in the common case where there are inequivalent choices for a given factor?

[Steven G. Johnson]

Note that it is even faster on my machine to do a "power-by-squaring" algorithm, which keeps the size of the matrices smaller until the last Kronecker product:

function hadsq(m)
    m == 0 && return reshape([1],1,1)
    m == 1 && return H2
    z = hadsq(m >> 1)
    z = kron(z, z)
    iseven(m) ? z : kron(H2, z)
end

[Alan Edelman]

Is this actually useful? Who needs Hadamard matrices of non-power-of-two sizes?

People use them from time to time for experimental design. Not saying it's the world's highest priority.

Does it matter what ordering I choose? Ordering the factors in different ways seems like it will produce different matrices (albeit equivalent modulo reordering).

Does it matter which inequivalent Hadamard matrix I choose, in the common case where there are inequivalent choices for a given factor?

I would say for starters, any one is fine, in part because I'm not sure it's practical to get an exhaustive list of non-Hadamard equivalent matrices for all n anyway, and generating Hadamard
equivalent matrices is easy enough for the user.

[Alan Edelman]

cute

On Mon, Jul 7, 2014 at 9:28 AM, Steven G. Johnson [e-mail]
wrote:

Note that it is even faster on my machine to do a "power-by-squaring"
algorithm, which keeps the size of the matrices smaller until the last
Kronecker product:

function hadsq(m)
m == 0 && return reshape([1],1,1)
m == 1 && return H2
z = hadsq(m >> 1)
z = kron(z, z)
iseven(m) ? z : kron(H2, z)
end

—
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#2 (comment).

[Steven G. Johnson]

Should be mostly fixed. There are a few known orders of Hadamard matrices that aren't supported because I can't yet find the matrices online, but hopefully that will be fixed once a bug in the webhadamard site is fixed (their matrices are currently truncated).