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4.7 ANSWERS TO EXERCISES 655 62. The rank is 3, by the method of proof in Theorem W with P = (i i). The border rank cannot be 1, since we cannot have al(zl)bl(u)cl(u) = al(u)b~(zl)cz(u) = ud and al(~)bz(u)ci(u) E ai(u)bi(u)cz(u) G 0 (modulo &l). The border rank is 2 because of the realization (L A), ( f T), (A -A). 63. (a) Let the elements of T(m, n, s) and T(A4, N, S) be denoted by t(2,J~)(3,k~)(k,z~) and T(I,J~)(J,K~)(K,I~), respectively. Each element ~I,J,)(J,K,)(K,I,) of the direct product, where 1 = (6 I), J = (j, J), and K = (b K), is equal to t(,,j,)(j,k,)(k,zl)~(~,~/)(J,KI)(K,r ) by definition, so it is 1 iff I = I and J = J and K = K. (b) We have M(mns) 5 r3, since T(mns, mns, mns) = T(m, n, s) @ T(n, s, m) @ T(s, m, n). If M(P) 5 R we have A4(Ph) 5 Rh for all h, and it follows that M(N) 5 J,f(pk= 1) 2 RhP Nl 2 RN1 gR/logP, [This result appears in Pan s 1972 paper.] (c) We have Md(mns) < r3 for some d, where ?&(n) = rankd(T(n,n,n)). If l&(P) 2 R we have ?&d(Ph) 5 Rh for all h, and the stated formula follows since WPh) 5 2 >Rh by exercise 61. In an infinite field we save a factor of log N. ( hdt2 [This result is due to Bini and Schonhage, 1979.1 (d) Let P = mns; we can perform p 3h independent Ph X Ph matrix multi- plications with at most (hdzfz)p3hr3h noncommutative scalar multiplications. Reduce M(P2h) to A4(Ph) matrix multiplications of size Ph x Ph; thus we have M(Pzh) 5 (hdt2)~3hM(Ph)(l f O(~/T)~~). Iterating this recurrence gives 2 M(N) = O(N p(m,n,s,r)) exp(O(log log n)2). 64. (a) Let a = xij, A = U&z, b = yjk, B = Yt3, c = u&z, C = zjk, so that CZ = O(u ) can be eliminated. [When m = s = 7 and n = 1, this gives M(N) = O(N2e6).] (b) Take cyi = s -1, ~2 = –oc , t23=Q4=-1,C25=1,(-Y6=S-1, and d > 4. [We assume that o-l exists in the field.] (c) Taking the direct product of T(m, n, 2s) $ T(2n, s, m) $ T(s, 2m, n) with itself 3h times gives a tensor whose border rank is at most (2(m + l)n(s + 2))3h. This tensor is the direct sum of 33h terms of the form T(m (2n)3sk, n2s3(2m)k, (2s)%m3nk) where i +j + k = 3h, and (3h)!/h!3 of these have i = j = Ic = h. Thus if we let P = (2mns)h and p = (3h)!/h!3, the border rank of pT(P, P, P) is at most pr, where r = (2(m + l)n(s + 2))3h/p. Exercise 63(d) now implies that M(N) = O(N P(P,PZP, )fe) for all 6 > 0; here P and r are functions of h. We complete the proof by letting h be large in p(P, P, I , r) = log r/log P = (3h log 2(m + l)n(s + 2) -3h log 3 + O(log h))/(3h log Zmns), which equals ,@m, n, 2s, $(m + l)n(s + 2)) + O((log h)/h). [The best value is obtained for m = 5, n = 1, s = 11, p = 31og,,, 52 < 2.522.1
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654 ANSWERS TO EXERCISES 4.6.4 It is easy to verify that at most n extra bits of precision are needed for the intermediate variables in this calculation; i.e., if ]si] 2 M for 0 2 i < 2n at the beginning of the algorithm, then all of the 5 and X variables will be bounded by 2 M throughout. Algorithm N performs A, addition-subtractions, D, halvings, and M, multiplica- tions, where A1 = 5, D1 = 0, MI = 3; for n > 1 we have A, = Ln/2]2 + + 2 n z +1A,+, + ([n/2] + 1)2 + + 2 , D, = 2 n z +1DLn,zJ + ([n/2] + 1)2 +l, and M, = 21n 2J+1 ML+J. The solutions are A, = 11. 2n- +r1gn1 -3.2 + 6. 2nS,, D, = 4. 2 - +bn1 -2.2 + 2. 2 S,, M, = 3 . 2n-1+r gnl; here S, satisfies the recurrence Sr = 0, S, = 2Sl,/zl+ [n/2], and it is not difficult to prove the inequalities +n[lgnl 2 S, 5 inlgn + n. Algorithm C does approximately the same amount of work as Algorithm N. It would be interesting to find a simpler way to carry out the additions and subtractions in step N3 (and the reverse operations in N5), perhaps analogous to Yates s method. The operation Xi + Xi + wkXl sketched above can be done with a procedure that generalizes the data-rotation algorithm of Fletcher and Silver in CACM 9 (1966), 326, but there might be a better way. 60. (a) In Cl, for example, we can group all terms having a common value of j and k into a single trilinear term; this gives y2 trilinear terms when (j, k) E E xE, plus y2 when (j, k) E E X0 and y2 when (j, k) E OX E. When 3 = k we can also include -xjj y3j z?j in Cl, free of charge. [In the case n = 10, the method multiplies 10 by 10 matrices with 710 noncommutative multiplications; this is fewer than required by any other known method, although Winograd s scheme (35) uses only 600 when commutativity is allowed.] (b) Here we simply let S be all the indices (i, j, k) of one problem, S the indices of the other. [When m = n = s = 10, the result is quite surprising: We can multiply two separate 10 X 10 matrices with 1300 noncommutative multiplications, while no scheme is known that would multiply each of them with 650.1 (c) Corresponding to the left-hand side of the stated identity we get the terms summed over (i, j, k) E S and 0 5 E, 5, n 5 1, so we get all the trilinear terms of the form Xz3 yjk ski except when [i/2] = [j/2] = [k/2]; h owever, these missing terms can all be included in Cl, Cz, or Es. The sum Cr turns out to include terms of the form ~i+~,j+< yi+v,j+, times some sum of z s, so it contributes 8v2 terms to the trilinear realization; and CZ, Cs are similar. To verify that the aBC terms cancel out, note that they are z(-l)Sfq xz+r,j+c Yk+c,i+c Zj+c,k+f, so 7 = 1 cancels with 7 = 0. [This technique leads to asymptotic improvements over Strassen s method whenever ijn +6n2 -4n < nlg7, namely when 36 5 n 5 184, and it was the first construction known to break the lg 7 barrier. Reference: SIAM J. Computing 9 (1980), 321-342.1 61. (a) Replace oij(U) by uail(~). (b) Let Q(U) = a+~~, etc., in a polynomial realization of length r = rankd(tijk). Then tijk = xCl+Y+O=d clWe recommend cheap and reliable webhost to host and run your web applications: Coldfusion Web Hosting services.

4.6.4 ANSWERS TO EXERCISES 653 (~0,. . , yzn-1). The algorithms are presented in unoptimized form, for brevity and ease in exposition; the reader who implements them will notice that many things can be streamlined. Cl. [Test for simple case.] If 72 = 1, set so + ~oYo+mYl, 21 + (2o+51)(Yo+Yl)-zo, and terminate. Otherwise set m + 2+-l. C2. [Remainderize.] For 0 2 Ic < m, set (5k, xm+k) c (5k + ZnL+k, Xk -xm+k) and (Yk, ymfk) + (yk + ym+k, yk -ymfk). (Now We have X(U) mod (2~ - 1) = x0 f.. . + xm--lum-l and ~(~1)mod (urn f 1) = xm + ... + ~2~~1; we will compute z(u)y(u) mod (2~~ - 1) and x(u)y(u) mod (urn + l), then we will combine the results by (57).) C3. (Recurse.] Set (20,. , ~~-1) to the cyclic convolution of (zo, . . . , xrn-i) with . , Ym-1). Also set (zm, . . . , ZZ,,-~) to the negacyclic convolution of PXT?%,..., x2,+-1) with (yn, . . . , yz,,+i). C4. [Unremainderize.] For 0 I Ic < m, set (zk, zmfk) e i(zk + zrnfk, Zk -zm+k). Now (~0,. . . , zm--l) is the desired answer. i Nl. [Test for simple case.] If n = 1, set t t xs(yo + yi), zs e t -(x0 + xl)yl, s1 + t + (xi -xs)ys, and terminate. Otherwise set m + 2Ln 2J and r + 2mi21. (The following steps use 2 +l auxiliary variables Xtj for 0 2 i < 2m and 0 < j < r, to represent 2m polynomials Xi(w) = Xi0 +Xilw+. . . +X,(,-l)~ - ; similarly, there are 2nf auxiliary variables Yij.) N2. [Initialize auxiliary polynomials.] Set Xij t X(i+,)j c xmj+i Yij + ~i+,)j + ymj+%, for 0 < i < m and 0 I j < r. (At this point we have x(u) = XO(U ) + UXl(U ) + . . . + u--l Xm-i(um), and a similar formula holds for y(u). Our strategy will be to multiply these polynomials modulo (urn7 + 1) = (u + I), by operating modulo (w f 1) on the polynomials X(W) and Y(W), finding their cyclic correlation of length 2m and thereby obtaining x(z~)y(u) = Zo(zlm) + u,&(u ) + . . + U2m--1Z2m--1(Um).) N3. [Transform.] (Now we will essentially do a fast Fourier transform on the poly- nomials (X0,. . . ,Xm--l, 0,. . . ,0) and (Yo, . . . , Y,-1, 0, . . . ,O), using w as a (2m)th root of unity. This is efficient, because multiplication by a power of w is not really a multiplication at all.) For j = [n/2] -1, . . . , 1, 0 (in this or- der), do the following for all m binary numbers s + t = (~1~12~. . . sj+lO . . .O)z + (0.. . Otj-1 . . . &~)a: Replace (Xs+t(w), X,+t+2j(w)) by the pair of polynomials (Xs+t(w) + ~(~~~)(~~~)X,+~+~j(w),X~+t(w) -w(7 )(s 2)X,+,+23(~)). (See Section 4.3.3 and Eq. 4.3.3-33. The operation Xi(w) + Xi(w) + wkXl(w) means, more precisely, that we set Xi3 +- Xij + X[(j+k) if j + k < r, otherwise Xij t Xi3 -for 0 5 j < r.) Do the same transformation on the Y s. xl(j+k-r), N4. [Recurse.] For 0 < i < 2m, set (Zie, . . . ,2,(,-l)) to the negacyclic convolution of (X0,. . . ,X+-I)) and . . . , Yz(,--I)). (Y,o, N5. [Untransform.] For j=O, 1, . . . , [n/2] (in this order), set (Zs+t(w), Zs+t+2j(w))+ ~(Zs+~(w)+ZS+t+23(w), ~--(~~~)(~~~)(Z,+t(w)--Z~+~+~~(w))), for all m choices of s and t as in step N3. N6. [Repack.] (Now we have accomplished the goal stated at the end of step N2, since it is easy to show that the transform of the Z s is the product of the transforms of the X s and the Y s) Set zi + ZZo -Z(,+i)(,-1) and Z,j+i + Ztj +Z(m+z)()-l) for 0 < j < r, for 0 5 i < m. 1
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652 ANSWERS TO EXERCISES 4.6.4 57. Let N be the smallest power of 2 that exceeds 2n, and let un+l = . . . = UN-1 = vn+1 = . = UN-1 = 0. If u, = COCheck Tomcat Web Hosting services for best quality webspace to host your web application.

4.6.4 ANSWERS TO EXERCISES 651 complex multiplications are all simple since they involve only two real multiplications and no real additions.) We can construct a normal scheme for the two-dimensional m x m case by applying the m scheme to vectors F(t , *) of length m . Each si step becomes m additions; each rnj becomes a Fourier transform on m elements, but with all of the (Y S in this algorithm multiplied by crj; and each tk becomes m additions. Thus the new algorithm has (a m + c a ) complex additions, t t trivial multiplications, and a total of c c complex multiplications. Using these techniques, Winograd has found normal one-dimensional schemes for the following small values of m with the following costs (a, t, c): m = 2 ( 2,2,2) m = 7 (36,1, 9) m = 3 ( 6,1,3) m=8 (2% 6, m = 4 ( 8,4,4) m = 9 (46,1,12) m=5 (17,1,6) m = 16 (74,8,18) By combining these schemes as described above, we obtain methods that use fewer arithmetic operations than the fast Fourier transform (FFT) discussed in exercise 14. For example, when m = 1008 = 7.9.16, the costs come to (17946,8,1944), so we can do a Fourier transform on 1008 complex numbers with 3872 real multiplications and 35892 real additions. It is possible to improve on Winograd s method for combining relatively prime moduli by using multidimensional convolutions, as shown by Nussbaumer and Quandalle in IBM J. Res. and Devel. 22 (1978), 134-144; their ingenious approach reduces the amount of computation needed for 1008-point complex Fourier transforms to 3084 real multiplications and 34668 real additions. By contrast, the FFT on 1024 complex numbers involves 14344 real multiplications and 27652 real additions. If the two-passes-at-once improvement in the answer to exercise 14 is used, however, the FFT on 1024 complex numbers needs only 10936 real multiplications and 25948 additions, and it is not difficult to implement. Therefore the subtler methods are faster only on machines that take significantly longer to multiply than to add. [References: Proc. Nat. Acad. Sci. USA 73 (1976), 1005-1006; Math. Comp. 33 (1978), 175-199; Advances in Math. 32 (1979), 83-117.1 54. max(2eldeg(pl) -1,. . ,2e,deg(p,) -1, q + 1). 55. 2n -q , where n is the degree of the minimum polynomial of P (i.e., the manic polynomial p of least degree such that p(P) is the zero matrix) and r is the number of distinct irreducible factors it has. (Reduce P by similarity transformations.) 56. Let tzjk + t @ = rijk + rjik, for all i, j, k. If A, B, c iS a realization Of (tijk) Of rank 7, then cl m, otherwise ?-ijk = 0. Now rank( r%,k) + rank(?-jik) 2 rank(t,,k), since we obtain a realization of (tijk) by suppressing the last n rows of A and the first m rows of B in a realization A, B, c of (7ijk f $:32k).
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650 ANSWERS TO EXERCISES 4.6.4 (a) Let n(k) = (p-l)pePk- = (p(pePk) for 0 5 k < e, and n(k) = 1 for k 2 e. Represent the numbers { 1, . . . , m} in the form aipk (modulo m), where 0 2 k 2 e and 0 2 i < n(k), and a is a fixed primitive element modulo pe. For example, when m = 9 we can let a = 2; the values are {2 30,,2130, 2 31,2230,2530,2131,2430,2330, 2032}. Then na pk) = COllle COljFrom our experience, we are can tell you that you can find a reliable and cheap webhost service at Java Web Hosting services.

4.6.4 ANSWERS TO EXERCISES 649 48. If A, B, C and A , B , C are realizations of (tzjk) and (t:3k) of respective lengths r and r , then A = A @ A , B = B @ B , C = C @ C , and A = A @ A , B = B @J B , C = C @ C , are realizations of (tyJk) and (tyik) of respective lengths r + r and r . r . Note: Many people have made the natural conjecture that rank((t,,k) @ (tLJk)) = rank(t,,k) f rank($,), but the construction in exercise 60(b) makes this seem much less plausible than it once was. 49. By Lemma T, rank(tijk) 2 rank(ti(jk)). Conversely if M is a matrix of rank r we can transform it by row and column operations, finding nonsingular matrices F and G such that FMG has all entries 0 except for r diagonal elements that are 1; cf. Algorithm 4.6.2N. The tensor rank of FMG is therefore < r; and it is the same as the tensor rank of M, by exercise 44. 50. Let i = (i , i ) where 1 5 i 5 m and 1 2 i 5 n; then t(zr,2/t)3k = &u~&,~, and it iS Clear that rank(ti(jk)) = mn since (i&k)) is a permutation matrix. By Lemma L, rank(tijk) 2 mn. COnVf?rSdy, since (tljk) has only mn nonzero entries, its rank is clearly 2 mn. (There is consequently no normal scheme requiring fewer than the mn obvious multiplications. There is no such abnormal scheme either [Comm. Pure and Appl. Math. 3 (1970), 165-1791. But some savings can be achieved if the same matrix is used with s > 1 different column vectors, since this is equivalent to (m x n) times (n X s) matrix multiplication.) 51. (a) si = y0 + yl, s2 = y0 -yl; ml = $(x0 + ZI)SI, mz = +(x0 -sl)sz; w. = ml +mz, wi = ml -ms. (b) Here are some intermediate steps, using the methodology in the text: ((~0 -~2) + (m -ZZ)U)((YO -YZ) + (YI -~z)u)mod(u~ + u + 1) = ((zo -az)(yo -yz) -(21 -52)(Yl -y2)) + ((x0 - z)(Yo -y2) -(21 -ZO)(Yl -yo))u. The first realization is The second realization is The resulting algorithm computes si = yc + ~1, sz = yc -yl, ss = y2 -yc, s4 = y2 -yl, s5 = so + y2; ml = +(x0 + 21 + 22).5x, m2 = +(x0 + ZI -2z2)s2, m3 = +(x0 -2×1 + x2)53, m4 = &(-220 + zl +x2).%&; tl = ml + m2, t-2 = ml -m2, t3 = ml + m3, w0 = tl -m3, wl = t3 + m4, w2 = t2 -m4. 52. Let i = (i , i ) when i mod n = i and i mod n = i . Then we wish to compute W(k’,k”) = c x(i’,i”)y(~’,j”) summed for i + j = k (modulo n ) and i + j = Ic (modulo n ). This can be done by applying the n algorithm to the 272 vectors X,, and Y$ of length n , obtaining the n vectors wk . Each vector addition becomes n additions, each parameter multiplication becomes n parameter multiplications, and each chain multiplication of vectors is replaced by a cyclic convolution of degree n . [If the subalgorithms use the minimum number of chain multiplications, this algorithm uses 2(n -d(n ))(n -d(n )) more than the minimum, where d(n) is the number of divisors of n.]
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648 ANSWERS TO EXERCISES 4.6.4 /3so = 0. The result set now has at most m + 1 + cl 3. This single canonical form involves m2 + 2m parameters. As the cr s run through all integers and as we run through all chains, the /3 s run through at most 2m2f2m sets of values mod 2, hence the result set does also. In order to obtain all 2n polynomials of degree n with O-l coefficients, we need m2 + 2m > n. (c) Set m + 161 and compute z2, x3, . . . , xm. Let U(X) = ZL~+~(X)X(~+ )~ + . . . + ul(x)xm f u,J(x), where each ~~(5) is a polynomial of degree 5 m with integer coefficients (hence it can be evaluated without any more multiplications). Now evaluate U(X) by rule (2) as a polynomial in xm with known coefficients. (The number of additions used is approximately the sum of the absolute values of the coefficients, so this algorithm is efficient on O-l polynomials. Paterson and Stockmeyer also gave another algorithm that uses about 6 multiplications.) Reference: SIAM J. Computing 2 (1973), 60-66; see also J. E. Savage, SIAh4 J. Computing 3 (1974), 150-158. For analogous results about additions, see Borodin and Cook, SIAM J. Computing 5 (1976), 146-157; Rivest and Van de Wiele, Inf. Proc. Letters 8 (1979), 178-180.1 43. When ai = aj $ ak is a step in some optimal addition chain for n $1, compute xi = z3xk and pi = p& fpj, where pi = x2-l+. . .+x+1; omit the final calculation of xnf . We save one multiplication whenever ak = 1, in particular when i = 1. (Cf. exercise 4.6.3-31 with c = a.) 44. It suffices to show that (TZjk) s rank is at most that of (&k), since we can obtain (tijk) back from (Tijk) by transforming it in the same way with F-l, G-l, H- . If t ajk = c 1 <1<7 azl bjl ckl then it follows immediately that [H. F. de Groote has proved that all normal schemes that yield 2 X 2 matrix products with seven chain multiplications are equivalent, in the sense that they can be obtained from each other by nonsingular matrix multiplication as in this exercise. In this sense Strassen s algorithm is unique.] 45. By exercise 44 we can add any multiple of a row, column, or plane to another one without changing the rank; we can also multiply a row, column, or plane by a nonzero constant, or transpose the tensor. A sequence of such operations can always be found to reduce agive 2 X 2 x 2 tensor to one of the forms ( )( )00 00 9 ( )( )00 7 (lo)(oo),01 (~~)(~~), 00 00 (A y)( z t). The last tensor has rank 3 or 2 according as the polynomial u2 - TU -9 has one or two irreducible factors in the field of interest, by Theorem W (cf. (72)). 46. A general m X n X s tensor has mns degrees of freedom. By exercise 28 it is impossible to express all m X n X s tensors in terms of the (m + n + S)T elements of a realization A, B, C unless (m + n + S)T 2 mns. On the other hand, assume that m 2 n 2 s. The rank of an m X n matrix is at most n, so we can realize any tensor in ns chain multiplications by realizing each matrix plane separately. [Exercise 45 shows that this lower bound on the maximum tensor rank is not best possible, nor is the upper bound. Thomas D. Howell (Ph. D. thesis, Cornell Univ., 1976) has shown that there are tensors of rank 2 [mns/(m + n + s - 2)1 over the complex numbers.]
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4.6.4 ANSWERS TO EXERCISES 647 39. By induction on m. Let W,(Z) = z2m + ~~~~~~~~~~ + . . . + 2~c, WJ~-~(Z) = X 2m-2 +f2m–3x 2m-3 f . . . + VO, a = cyi + yrn, b = om, and let f(r) = Ci,j~o(-l)i+3(i~3)u,+i+2jai~. It follows that v7 = f(r + 2) for r 2 0, and 6, = f(1). If 6, = 0 and a is given, we have a polynomial of degree m -1 in b, with leading coefficient &(21z,,+l -mu) = *(r2 + … +~~-m-d. In Motzkin s unpublished notes he arranged to make 6k = 0 almost always, by choosing y s so that this leading coefficient is # 0 when m is even and = 0 when m is odd; then we almost always can let b be a (real) root of an odd-degree polynomial. 40. No; S. Winograd found a way to compute all polynomials of degree 13 with only 7 (possibly complex) multiplications [Comm. Pure and Applied Math. 25 (1972), 455-4571. L. Revah found schemes that evaluate almost all polynomials of degree n > 9 with ]n/2] + 1 (possibly complex) multiplications [SIAM J. Computing 4 (1975), 381-3921; she also showed that when n = 9 it is possible to achieve [n/2] + 1 multiplications only with at least n+3 additions. By appending sufficiently many additions (cf. exercise 39), the almost all and possibly complex provisos disappear. V. I^a. Pan [Proc. ACM Symp. Theory Comp. 10 (1978), 162-172; IBM Research Report RC7754 (1979)] found schemes with Ln/2]+1 (complex) multiplications and the minimum number n+2+&s of (complex) additions, for all odd n 2 9; his method for n = 9 is V(X) = ((x + a) + D)(x + Y), w(x) = V(X) + x, t(x) = (v(x) + 6)(4x) + 6) -(v(x) + b )(w(x) + E ), 4×1 = (v(x) + c)(e) + rl) + K. The minimum number of real additions necessary, when the minimum number of (real) multiplications is achieved, remains unknown for n 2 9. 41. u(c + d) -(a + b)d + i(u(c + d) + (b-a)~). [B eware numerical instability. Three multiplications are necessary, since complex multiplication is a special case of (69) with p(u) = u2 + 1. Without the restriction on additions there are other possibilities. For example, the symmetric formula UC - bd + i((u + b)(c + d) -UC - bd) was suggested by Peter Ungar in 1963; cf. Eq. 4.3.3-2 with 2 replaced by i. See I. Munro, Proc. ACM Symp. Theory Comp. 3 (1960), 40-44; S. Winograd, Linear Alg. Appl. 4 (1971), 381-388.1 Alternatively, if u2 + b2 = 1 and t = (1 -u)/b = b/(1 + a), the algorithm w = c - td, v = d + bw, u = w -tv for calculating the product (u + bi)(c + di) = u + iv has been suggested by Oscar Buneman [J. Comp. Phys. 12 (1973), 127-1281. In this method if a = cos 6 and b = sin 8, we have t = tan(0/2). [Helmut Alt and Jan van Leeuwen have shown that four real multiplications or divisions are necessary for computing l/(u + b ) a , and four are sufficient for computing u/(b + ci). It is unknown whether (u + bi)/(c + di) can be computed with only five multiplications or divisions.] 42. (a) Let ~1, . . . , 7rTm be the Xi s that correspond to chain multiplications; then 7ri = Pzz-l XPZ~ and U(X) = Pz,,+i, where each Pj has the form @+&x+p~i~i+…+ p37(j)rY(3), where r(j) 5 [j/2] -1 and each of the & and &k is a polynomial in the o s with integer coefficients. We can systematically modify the chain (cf. exercise 30) so that p3 = 0 and ,&l = 1, for 1 5 j 5 2m; furthermore we can assume that
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646 ANSWERS TO EXERCISES 4.6.4 For the given rational function we have QYj Pj w[j (x) & (x) 1 2 x+5 3x+ 19 3 4 1 5 so u[ l(x)/w[o~(x) = 1 + 2/(x + 3 + 4/(x + 5)). Notes: A general rational function of the stated form has 2n + 1 degrees of freedom, in the sense that it can be shown to have 2n + 1 essentially independent parameters. If we generalize polynomial chains to arithmetic chains, which allow division operations as well as addition, subtraction, and multiplication, we can obtain the following results with slight modifications to the proofs of Theorems A and M: An arithmetic chain with q addition-subtraction steps has at most q+l degrees of freedom. An arithmetic chain with m multiplication-division steps has at most 2m + 1 degrees of freedom. Therefore an arithmetic chain that computes almost all rational functions of the stated form must have at least 2n addition-subtractions, and n multiplication- divisions; the method in this exercise is optimal. 38. The theorem is certainly true if n = 0. Assume that n is positive, and that a polynomial chain computing P(x; ~0, . . . , 2~~) is given, where each of the parameters aj has been replaced by a real number. Let Xi = Xj X Xk be the first chain multiplication step that involves one of ~0, . . . , un; such a step must exist because of the rank of A. Without loss of generality, we may assume that X, involves u,; thus, Xj has the form houo +. . . + hnun + f(x), where ho, . . , h, are real, h, # 0, and f(x) is a polynomial with real coefficients. (The h s and the coefficients of f(x) are derived from the values assigned to the cy s.) Now change step i to Xi = oi X Xk, where cr is an arbitrary real number. (We could take QI = 0; general cr is used liere merely to show that there is a certain amount of flexibility available in the proof.) Add further steps to calculate X = (a -f(x) -houo -. . . - h,-lu,-1)/h,; these new steps involve only additions and parameter multiplications (by suitable new parameters). Finally, replace X-,-1 = 2~~ everywhere in the chain by this new element X. The result is a chain that calculates Q(x; 210,. . . , t&.-1) = P(x; uo, . . . , h-1, (a -f(x) -houo -.. . - hn-1%~1)lhn); and this chain has one less chain multiplication. The proof will be complete if we can show that Q satisfies the hypotheses. The quantity (a -f(x))/hn leads to a possibly increased value of m, and a new vector B . If the columns of A are Ao, AI, . . . > A, (these vectors being linearly independent over the reals), the new matrix A corresponding to Q has the column vectors Ao -(ho/h&k , An-1 -(hn-l/hn)An, plus perhaps a few rows of zeros to account for an increased value of m, and these columns are clearly also linearly independent. By induction, the chain that computes Q has at least n -1 chain multiplications, so the original chain has at least n. [Pan showed also that the use of division would give no improvement; cf. Problemy Kibernetiki 7 (1962), 21-30. Generalizations to the computation of several polynomials in several variables, with and without various kinds of preconditioning, have been given by S. Winograd, Comm. Pure and Applied Math. 23 (1970), 165-179.1
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