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Cc] Vice versa: Let T be the operator corresponding to the fht_fft_conversion, T is its own inverse: T = T −1 , or, equivalently T · T = 1. 16) CHAPTER 3. n-1] input,result { fft(c[], n) fht_fft_conversion(c[], n, is) } or the same thing with swapped lines. Of course the same ideas also work for separate real- and imaginaryparts. n-1] input,result { fht(a[], n) for i:=1 to n/2-1 { t := n - i u := a[i] v := a[t] a[i] := 1/2 * (u+v) a[t] := 1/2 * (u-v) } } At the end of this procedure the ordering of the output data c ∈ C is a[0] a[1] a[2] = = = c0 c1 c2 a[n/2] a[n/2 + 1] a[n/2 + 2] a[n/2 + 3] ...

N-1] input, modified CHAPTER 3. spr] It is assumed that the procedure fht() does no normalization. cc] Cf. n-1] input, result { // transform data: fht(x[], n) // convolution in transformed domain: j := n-1 for i:=1 to n/2-1 { ci := x[i] cj := x[j] t1 := ci*cj // = cj*ci t2 := 1/2*(ci*ci-cj*cj) // = -1/2*(cj*cj-ci*ci) x[i] := t1 + t2 x[j] := t1 - t2 j := j-1 } x[0] := x[0]*x[0] if n>1 then x[n/2] := x[n/2]*x[n/2] // transform back: fht(x[], n) CHAPTER 3. spr] For odd n replace the line for i:=1 to n/2-1 by for i:=1 to (n-1)/2 and omit the line if n>1 then x[n/2] := x[n/2]*x[n/2] in both procedures above.

One might hope to find an algorithm that computes h(0) and uses only half the memory compared to the linear convolution or that needs half the work, possibly both. It may be a surprise that no such algorithm seems to be known currently5 . Here is a clumsy attempt to find h(0) alone: Use the weighted transform with the weight sequence vx = V x where V n is very small. Then h(1) will in the result be multiplied with a small number and we hope to make it almost disappear. 000}. At least for integer sequences one could choose V n (more than two times) bigger than biggest possible value in h(1) and use rounding to nearest integer to isolate h(0) .

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