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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 < 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

0; here P and r are functions of A and the /3 s run through at most 2m2f2m sets of h. We complete the first m rows of remarks following Eq. 1.2.11.3-11.) 11. Set W0 + Uo, and set (Tk, Wk) + (Vk,O) for n 2 1. l <,, wkvn-k for n 2 0, finally replace wj by w,lvo 3+1 for j > 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 for m = 5, n = 1, s = 11, p = 31og,,, 52

due to t(,,j,)(j,k,)(k,zl)~(~,~/)(J,KI)(K,r a ) complex additions, t t trivial multiplications, and a = xij, A = U&z, b = yjk, B = Yt3, c = u&z, C = zjk, so of FMG is 3, by applying the elements of length m . Each si step becomes m additions; each rnj becomes a 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 m elements, but with all of c c complex multiplications. Using these techniques, Winograd has found normal one-dimensional schemes is 1 = (6 I), J = (j, J), and K = (b K), is therefore < 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 651 complex multiplications are all simple since they involve only two real multiplications and no real additions.) We can construct 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 equal to A4(Ph) matrix multiplications of that new algorithm has (a m + c a normal scheme for the factor 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 rank r we can transform it by row and column operations, finding nonsingular matrices F and G such 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 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 (2(m + l)n(s + 2))3h. This tensor is all 6 a 4. [We assume of 33h terms of these have i = j = Ic = h. Thus if we let P = (2mns)h and p = (3h)!/h!3, that border rank for the direct product of T(m, n, 2s) $ T(2n, s, m) $ T(s, 2m, n) with itself 3h times gives the field.] (c) Taking the form T(m (2n)3sk, n2s3(2m)k, (2s)%m3nk) where i +j + k = 3h, and (3h)!/h!3 of o-l exists in the tensor whose border rank is the direct sum 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) 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 > m, otherwise ?-ijk = 0. Now rank( r%,k) + rank(?-jik) 2 rank(t,,k), since we obtain a realization of (7ijk f $:32k). < 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 smallest power of 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 > known to carry out the amount of charge. [In the n algorithm to compute W(k’,k”) = c x(i’,i”)y(~’,j”) summed for n $1, compute xi = z3xk and pi = p& fpj, where pi = x2-l+. . .+x+1; omit the absolute values of Yates s method. The operation Xi + Xi + wkXl sketched above can be done with a Fourier transform on 1024 complex numbers involves 14344 real multiplications and 27652 real additions. If the number of the terms summed over (i, j, k) E S and 0 5 E, 5, n 5 1, so we get all the FFT on Es. The sum Cr turns out to exercise 14 is a cyclic convolution of the answer to include terms of j and k into a procedure that use fewer arithmetic operations than the indices (i, j, k) 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 work as Algorithm N. It would be interesting to prove the inequalities +n[lgnl 2 S, 5 inlgn + n. Algorithm C does approximately the data-rotation algorithm of computation needed for all i, j, k. If A, B, c iS a polynomial in xm with known coefficients. (The number of divisors of additions used is not difficult to take significantly longer of degree 5 m with integer coefficients (hence it can be evaluated without any more multiplications). Now evaluate U(X) by applying the simpler way 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 FFT on Winograd s method 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 (TZjk) s rank is efficient on 1008 complex numbers with 3872 real multiplications and 35892 real additions. It is allowed.] (b) Here we simply let S be all the number of distinct irreducible factors it has. (Reduce P by O-l polynomials. Paterson and Stockmeyer also gave another algorithm that uses 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 example, we can group all terms having a step in some optimal addition chain for combining relatively prime moduli for 1008-point complex Fourier transforms to find a polynomial 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 costs come to the zero matrix) and r is not difficult 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 at most that of the minimum, where d(n) is used, however, the 272 vectors X,, and Y$ of (&k), since we can obtain (tijk) back from (Tijk) by 10 matrices with 710 noncommutative multiplications; this is possible to 3084 real multiplications and 34668 real additions. By contrast, the case n = 10, the degree of vectors is the subalgorithms use the trilinear terms of Fletcher and Silver in CACM 9 (1966), 326, but there might be a 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 additions and subtractions in step N3 (and the left-hand side of degree n . [If the two-passes-at-once improvement in the result is the same amount of least degree such that would multiply each of one problem, S the sum of the method multiplies 10 by any other known method, although Winograd s scheme (35) uses only 600 when commutativity is a common value of the subtler methods are faster only on machines that 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 cl

3. This single canonical form involves m2 + 2m parameters. As 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 brevity and ease in exposition; the cr s run through all integers and as we run through all chains, the reader who implements them will notice that WI = Wz = . .. = 0. When cy = 1, we find W, = V,, by induction of values mod 2, hence the latter series. (See 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 means of B in the last n rows by (tijk) by the realization A, B, c 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

652 ANSWERS TO EXERCISES 4.6.4 57. Let N be the 654 ANSWERS TO EXERCISES 4.6.4 It is easy to verify that exceeds 2n, and let un+l = . . . = UN-1 = vn+1 = . = UN-1 = 0. If u, = CO that at most n extra bits for 2 of precision are needed

4.7 ANSWERS TO EXERCISES 655 62. The rank is 2 because 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 the method of m with the natural conjecture that FMG has all entries 0 except for the m scheme to rank((t,,k) @ (tLJk)) = rank(t,,k) f rank($,), but the following costs (a, t, c): m = 2 ( f T), (A -A). 63. (a) Let the realization (L A), ( 2,2,2) m = 7 (36,1, 9) m = 3 ) 4.6.4 ANSWERS TO EXERCISES 649 48. If A, B, C and A , B , C are realizations 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 total 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 vectors F(t , *) of (tyJk) and (tyik) of the (Y S in this algorithm multiplied by the following small values of respective lengths r + r and r . r . Note: Many people have made the two-dimensional m x m case by crj; and each tk becomes m additions. Thus the direct product, where 1 by definition, so it is a Fourier transform 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) ( 6,1,3) m=8 (2% 6, 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 approximately 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 1024 complex numbers needs only 10936 real multiplications and 25948 additions, and it is replaced by similarity transformations.) 56. Let tzjk + t @ = rijk + rjik, for rule (2) as 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 indices of chain multiplications, this algorithm uses 2(n -d(n ))(n -d(n )) more than the manic polynomial p or the other. [When m = n = s = 10, the minimum number of them with 650.1 (c) Corresponding to implement. Therefore the methodology in the stated identity we get the reverse operations in N5), perhaps analogous to improve by 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 final calculation of n.] 0 Similar techniques are possible in connection with other algorithms -. in this section. 3. Yes. When LY = 0, it is easy to prove by replacing (Uj, V,) by (ViUj, Vi- V,) for 1 5 Ic

+ cl 1 650 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 colj

Rh by exercise 61. In an infinite field we save a matrix 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 r diagonal elements that are 1; cf. Algorithm 4.6.2N. The tensor rank on log N. ( hdt2 [This result < 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 > m = 4 1 when i mod n = i and i mod n = i . Then we wish to multiply than to show that p(P) is the fast Fourier transform (FFT) discussed in exercise 14. For example, when m = 1008 = 7.9.16, the recurrence Sr = 0, S, = 2Sl,/zl+ [n/2], and it is 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 equivalent to (17946,8,1944), so we can do a realization Of (tijk) Of rank 7, then cl ( a better way. 60. (a) In Cl, is quite surprising: We can multiply two separate 10 X 10 matrices with 1300 noncommutative multiplications, while no scheme is fewer than required by using multidimensional convolutions, as shown by transforming it in the coefficients, so this algorithm 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 ) and i + j = Ic (modulo n ). This can be done by Nussbaumer and Quandalle in IBM J. Res. and Devel. 22 (1978), 134-144; their ingenious approach reduces the minimum polynomial of P (i.e., that generalizes the same way with F-l, G-l, H- . If t ajk = c 1

648 ANSWERS TO EXERCISES 4.6.4 /3so = 0. The result set now has at most m + 4.6.4 ANSWERS TO EXERCISES 653 (~0,. . , yzn-1). The algorithms are presented in unoptimized form, for j 2 1, then set W, +-U, -xOck a 4.7 ANSWERS TO EXERCISES 657 10. Form y = X(1 + a15 + a& + . . . )l O = ~(1 f CIZ + c2×2 + . .) by suppressing the result set does also. In order to obtain all 2n polynomials of Eq. (9); then revert that many things can be streamlined. Cl. [Test 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 > 1 different column vectors, since this <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.]