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The Walsh-Hadamard transform is very fast to compute.

Can it be used to compute the convolution of two functions as it can be done with Fourier transform ?

edited Oct 21 at 15:10

asked Oct 21 at 14:22

chmike
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Qiaochu Yuan · Accepted Answer · 2009-10-21 15:39:58Z UTC

Not in the sense I think you mean it. First of all, the Walsh-Hadamard transform is a Fourier transform - but on the group (Z/2Z)^n instead of on the group Z/NZ. That means you can use it to compute convolutions with respect to the space of functions (Z/2Z)^n -> C. Unfortunately, unlike the case with Z/NZ you can't use this to approximate a compactly supported convolution on Z, at least not directly.

Drazick · Answer 2 · 2009-12-24 22:51:04Z UTC

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@ Qiaochu Yuan, Where could I find more info on Walsh-Hadamard Transform in the meaning you presented it (As private case of Fourier Transform).

Thanks.

answered Dec 24 at 22:51

Drazick
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	You might be interested in the Wikipedia article: en.wikipedia.org/wiki/… . – Qiaochu Yuan Dec 24 at 23:23
	Is there a place elaborating some of those properties of Walsh-Hadamard Transform in a manner easy to understand for someone with background in EE - Signal Processing and not Math? Thanks. – Drazick Jan 14 at 23:16

Can Walsh-Hadmard transform be used for convolution ?

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