Prime Numbers

9.7 Exercises 525
two (and thus, mere shifts). The theoretical minimum is, of course, seven
multiplies, but such a nine-mul version has its advantages.
(3) Use Toom–Cook ideas to develop an explicit length-4 negacyclic scheme
that does require only seven multiplies.
(4) Can one use a length-(D > 2) negacyclic to develop a Karatsuba-like
multiply that is asymptotically better than O (ln D) ln 3/ ln 2 ?
(5) Show how to use a Walsh–Hadamard transform to effect a length-16 cyclic
convolution in 43 multiplies [Crandall 1996a]. Though the theoretical
minimum multiply count for this length is 27, the Walsh–Hadamard
scheme has no troublesome constant coefficients. The scheme also appears
to be a kind of bridge between Winograd complexities (linear in N) and
transform-based complexities (N ln N). Indeed, 43 is not even as large as
16 lg 16. Incidentally, the true complexity of the Walsh–Hadamard scheme
is still unknown.
9.40. Prove Theorem 9.5.13 by way of convolution ideas, along the following
lines. Let N =2· 3 · 5 ···pm be a consecutive prime product, and define
rN(n) =#{(a, b) : a + b = n; gcd(a, N) =gcd(b, N) =1;a, b ∈ [1,N − 1]},
that is, rN(n) is the number of representations we wish to bound below. Now
define a length-N signal y by yn =1ifgcd(n, N) = 1, else yn = 0. Define the
cyclic convolution
RN(n) =(y × y)n,
and argue that for n ∈ [0,N − 1],
RN(n) =rN(n)+rN(N + n).
In other words, the cyclic convolution gives us the combined representations
of n and N + n. Next, observe that the Ramanujan sum Y (9.26) is the DFT
of y, sothat
RN(n) = 1
N−1
N
k=0
Y 2
k e 2πikn/N .
Now prove that R is multiplicative, in the sense that if N = N1N2 with N1,N2
coprime, then RN(n) =RN1(n)RN2(n). Conclude that
RN(n) =ϕ2(N,n),
where ϕ2 is defined in the text after Theorem 9.5.13. So now we have a closed
form for rN(n)+rN(N + n). Note that ϕ2 is positive if n is even. Next, argue
that if a + b = n (i.e., n is representable) then 2N − n is also representable.
Conclude that if rN(n) > 0 for all even n ∈ [N/2+1,N−1], then all sufficiently
large even integers are representable. This means that all we have to show is
that for n even in [N/2+1,N − 1], rN(n + N) is suitably small compared to