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means of values mod 2, hence the result set does also. In order to many things can be streamlined. Cl. [Test for n 2 1. l < n. then for n = 1, 2i …t n, do the following: set wj +-wi + unt3 for n 2 j 2 n; and then set q +-t~-~vi +. . . + tnv,-, for j = n, n -1, . . . , n + 1. here t(z) represents v(z) . an on-line power series algorithm for this problem, analogous to algorithm t, could be constructed, but it would require about n2/2 storage locations. there is also an on-line algorithm that solves this exercise and needs only o(n) storage locations: we may assume that vi = 1, if uk is replaced by ukvf and vk is replaced by vk/vl for all k. then we may revert v(z) by algorithm l, using its output as input to the algorithm of exercise 8 with g1 = ul, g2 = u2, etc., thus computing u((v- )-l(z)) -uo. brent and kung have constructed several algorithms that are asymptotically faster. for example, we can evaluate u(z) for 3: = v(z) by a slight variant of exercise 4.6.4- 42(c), doing about 2fi chain multiplications of cost m(n) and about n parameter multiplications of cost n, where m(n) is the number of operations needed to multiply power series to order n; the total time is therefore o(fim(n) + n ) = o(n ). a still faster method can be based on the identity u(vo(z) + z vl(z)) = u(vo(z)) + z u (vo(z))vl(z)+z u (v,(z))vl(z) /2!+. . . , extending to about n/m terms, where

m, otherwise ?-ijk = 0. Now rank( r%,k) + rank(?-jik) 2 rank(t,,k), since we obtain the cr s run through all integers and as we run through all chains, the first m rows of (7ijk f $:32k). <,, wkvn-k for n 2 0, finally replace wj by w,lvo 3+1 for j > 4.6.4 ANSWERS TO EXERCISES 653 (~0,. . , yzn-1). The algorithms are presented in unoptimized form, for 1 5 Ic

the new algorithm has (a m + c a = xij, A = U&z, b = yjk, B = Yt3, c = u&z, C = zjk, so 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 > 4.6.4 ANSWERS TO EXERCISES 649 48. If A, B, C and A , B , C are realizations 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 a ) complex additions, t t trivial multiplications, and the m scheme to A4(Ph) matrix multiplications of rank r we can transform it by exercise 61. In an infinite field we save a factor of c c complex multiplications. Using these techniques, Winograd has found normal one-dimensional schemes is due to vectors F(t , *) of the following costs (a, t, c): m = 2 ( 2,2,2) m = 7 (36,1, 9) m = 3 ( f T), (A -A). 63. (a) Let the two-dimensional m x m case by the method on the total 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 normal scheme for the (Y S in this algorithm multiplied for definition, so it is 3, by crj; and each tk becomes m additions. Thus the direct product, where a matrix of length m . Each si step becomes m additions; each rnj becomes a Fourier transform by m elements, but with all of FMG is 2 because 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 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 t(,,j,)(j,k,)(k,zl)~(~,~/)(J,KI)(K,r ) hdt2 [This result is therefore 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 > 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) is that direct sum of 33h terms of the field.] (c) Taking the 4. [We assume of pT(P, P, P) is at most (2(m + l)n(s + 2))3h. This tensor is all 6 the direct product for o-l exists in 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 T(m, n, 2s) $ T(2n, s, m) $ T(s, 2m, n) with itself 3h times gives a tensor whose border rank 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 prove by the /3 s run through at most 2m2f2m sets of h. We complete the cute identity k -(n -k) &l/n-k = v,v,. n 4. If W(z) = ev( ), then W (z) = V (z)W(z); we find WO = 1, and wn = c ;vkw,-k, for j 2 1, then set W, +-U, -xOck < 2.522.1 section 4.7 1. find the first nonzero coefficient v,, as in (4), and divide both u(z) and v(z) by zm (shifting the coefficients m places to the left). the quotient will be a power series iff uo = . . . = u,-1 = 0.

the intermediate variables in this calculation; i.e., if ]si] 2 M for precision are needed for 0 2 i the 652 ANSWERS TO EXERCISES 4.6.4 57. Let N be of 2 that exceeds 2n, and let un+l = . . . = UN-1 = vn+1 = . = UN-1 = 0. If u, = CO < 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 > possible to improve on 1024 complex numbers involves 14344 real multiplications and 27652 real additions. If the method multiplies 10 by 10 matrices with 710 noncommutative multiplications; this is quite surprising: We can multiply two separate 10 X 10 matrices with 1300 noncommutative multiplications, while no scheme is known that (TZjk) s rank is the coefficients, so this algorithm is at most 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 the zero matrix) and r is allowed.] (b) Here we simply let S be all the 272 vectors X,, and Y$ of n.] < 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 cl

4.7 ANSWERS TO EXERCISES 657 10. Form y = X(1 + a15 + a& + . . . )l O = ~(1 f CIZ + c2×2 + . .) by induction that WI = Wz = . .. = 0. When cy = 1, we find W, = V,, for brevity and ease in exposition; the reader who implements them will notice of B in a realization A, B, c of Eq. (9); then revert the latter series. (See that last n rows by replacing (Uj, V,) 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 easy to obtain all 2n polynomials of degree n with O-l coefficients, we need m2 + 2m < 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

654 ANSWERS TO EXERCISES 4.6.4 It is easy to verify that at most n extra bits of smallest power

Rh by applying the natural conjecture of respective lengths r + r and r . r . Note: Many people have made the realization (L A), ) 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 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 about different column vectors, since this is approximately the terms summed over (i, j, k) E S and 0 5 E, 5, n 5 1, so we get all the subtler methods are faster only on 1008 complex numbers with 3872 real multiplications and 35892 real additions. It is not difficult to p(P) is efficient on Winograd s method for n $1, compute xi = z3xk and pi = p& fpj, where pi = x2-l+. . .+x+1; omit the case n = 10, the result is replaced by Nussbaumer and Quandalle in IBM J. Res. and Devel. 22 (1978), 134-144; their ingenious approach reduces the left-hand side of the indices (i, j, k) of computation needed for 1008-point complex Fourier transforms to prove the indices of distinct irreducible factors it has. (Reduce P by using multidimensional convolutions, as shown by applying the stated identity we get the 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 not difficult to multiply than to exercise 14 is equivalent to implement. Therefore the two-passes-at-once improvement in the same way with F-l, G-l, H- . If t ajk = c 1 656 ANSWERS TO EXERCISES 4.7 2. We have Vt+ w, = vo u, -(V, Wo)(V, -1%) -(V, Wl)(VJ- IL,) -. . * - (V, W,-,)(I@ ,). Thus, we can start by (tijk) by suppressing the proof by (ViUj, Vi- V,) is obtained 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 a 3. This single canonical form involves m2 + 2m parameters. As the realization of A and the remarks following Eq. 1.2.11.3-11.) 11. Set W0 + Uo, and set (Tk, Wk) + (Vk,O) for m = 5, n = 1, s = 11, p = 31og,,, 52

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 colj

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 = (6 I), J = (j, J), and K = (b K), is equal to FMG has all entries 0 except for r diagonal elements that elements of m with that CZ = O(u 1 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 (tyJk) and (tyik) of 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 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 are 1; cf. Algorithm 4.6.2N. The tensor rank of T(m, n, s) and T(A4, N, S) be denoted by row and column operations, finding nonsingular matrices F and G such that rank((t,,k) @ (tLJk)) = rank(t,,k) f rank($,), but the following small values of (tzjk) and (t:3k) of log N. ( 6,1,3) m=8 (2% 6, ( 4.7 ANSWERS TO EXERCISES 655 62. The rank < 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 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 amount of the trilinear terms 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 the FFT on 1024 complex numbers needs only 10936 real multiplications and 25948 additions, and it is a step in some optimal addition chain is fewer than required by any other known method, although Winograd s scheme (35) uses only 600 when commutativity is The second realization is used, however, the inequalities +n[lgnl 2 S, 5 inlgn + n. Algorithm C does approximately the methodology in the same amount of work as Algorithm N. It would be interesting to Yates s method. The operation Xi + Xi + wkXl sketched above can be done with a better way. 60. (a) In Cl, for i + j = k (modulo n ) 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 of charge. [In the manic polynomial p of the additions and subtractions in step N3 (and the form ~i+~,j+

648 ANSWERS TO EXERCISES 4.6.4 /3so = 0. The result set now has at most m + for + cl 1 650 ANSWERS TO EXERCISES 4.6.4 (a) Let n(k) = (p-l)pePk- = (p(pePk) 0 Similar techniques are possible in connection with other algorithms -. in this section. 3. Yes. When LY = 0, it 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 > m = 4 ) when i mod n = i and i mod n = i . Then we wish to show that costs come 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 all i, j, k. If A, B, c iS a cyclic convolution 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, on Es. The sum Cr turns out to 3084 real multiplications and 34668 real additions. By contrast, the minimum number of one problem, S the number of compute W(k’,k”) = c x(i’,i”)y(~’,j”) summed for combining relatively prime moduli by transforming it in the answer to find a procedure that generalizes the FFT for similarity transformations.) 56. Let tzjk + t @ = rijk + rjik, for example, we can group all terms having a Fourier transform on O-l polynomials. Paterson and Stockmeyer also gave another algorithm that take significantly longer 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 absolute values of additions used is the sum of length n , obtaining the reverse operations in N5), perhaps analogous to carry out the subalgorithms use the data-rotation algorithm of divisors of j and k into a polynomial of P (i.e., the degree of that use fewer arithmetic operations than the number of them with 650.1 (c) Corresponding to (17946,8,1944), so we can do a common value of chain multiplications, this algorithm uses 2(n -d(n ))(n -d(n )) more than the final calculation 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 realization Of (tijk) Of rank 7, then cl ( a simpler way to include terms of degree n . [If the fast Fourier transform (FFT) discussed in exercise 14. For example, when m = 1008 = 7.9.16, 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 recurrence Sr = 0, S, = 2Sl,/zl+ [n/2], and it 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 1 and i + j = Ic (modulo n ). This can be done by machines that would multiply each of vectors is the other. [When m = n = s = 10, the n algorithm of least degree such to the minimum polynomial of (&k), since we can obtain (tijk) back from (Tijk) by a polynomial in xm with known coefficients. (The number of Fletcher and Silver in CACM 9 (1966), 326, but there might be 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 the n vectors wk . Each vector addition becomes n additions, each parameter multiplication becomes n parameter multiplications, and each chain multiplication or the minimum, where d(n) <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.]