LIBRARY OF THE 
 
 UNIVERSITY OF ILLINOIS 
 
 AT URBANA-OHAMPAIGN 
 
 too. 2. 
 
Digitized by the Internet Archive 
 in 2013 
 
 http://archive.org/details/parallelmethodsb437maru 
 
^Jl^A? Report No. kjj 
 
 JtyUy&{ 
 
 PARALLEL METHODS AND BOUNDS 
 OF EVALUATING POLYNOMIALS 
 
 by 
 
 Kiyoshi Maruyama 
 
 March, 1971 
 
 DEPARTMENT OF COMPUTER SCIENCE 
 UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN 
 
 URBANA, ILLINOIS 
 
Report No. 1*37 
 
 PARALLEL METHODS AND BOUNDS 
 OF EVALUATING POLYNOMIALS 
 
 by 
 Kiyoshi Maruyama 
 
 March, 1971 
 
 Department of Computer Science 
 University of Illinois at Urb ana- Champaign 
 Urbana, Illinois 6l801 
 
 This work was supported in part by the Department of Computer Science 
 at the University of Illinois at Urbana-Champaign, Urbana, Illinois; 
 the National Science Foundation under NSF Grant GJ-328. 
 
Ill 
 
 Abstract a lower bound of the maximum evaluable degree of polynomi? 
 
 each step s is given, and is N(s)>2 " , J ~ ( 2 /s) ' . An upper 1 
 is derived for the number of steps required to evaluate polynomials of 
 n, and is given by T(P m ) < (l + e)log p n, which approaches theoretical I 
 
 n 
 
 bound Eog 2 (2n + 1^ as n— *>&* , where e ^ (2/log n) 
 
 1/2 
 
 ■U~--*A ~-P -l-Vw 
 
 ~-p ~,,~ — ±a — 3 required to achie\ 
 ;e polynomials of d« 
 
 COkrtKMIONS 
 
 1. In Theorem 3 (primal property) on page 15 should he 
 ^/log^) 1 2 . 
 
 ?. In Theorem 7 on page 32, the statement "The minimum 
 
 number of " should he "The upper bound of number 
 
 of ".. 
 
 Better upper bound of number of operations required to 
 
 evaluate polynomials of degree n by I.uraoka's folding 
 
 method is given by 2n + V - 2 , where ^ is an integer ^-il 
 
 such that ^-th Fibonacci number ^ n and (^+ 1) st Fibonacci 
 
 number > n. For example 
 
 f : 123*5678 — 
 
 Fibonacci No.: 112358 13 21 ... 
 
 if n=7 then *« 5 thus the number of operations required to 
 
 evaluate a polynomial of degree 7 is 2»7 + 5 - 2 « 17 t 
 
 which is better than Theorem 7. 
 
 lonstruction of com] 
 
 lods, which achieve; 
 
 better results thai 
 
 2 of a simple schedi 
 
 primal problem, du; 
 LOF, consistency, 
 
Ill 
 
 Abstract a lower bound of the maximum evaluable degree of polynomial for 
 
 each step s is given, and is N(s)>2 S ^ " , J =s (2/s) ' . An upper bound 
 is derived for the number of steps required to evaluate polynomials of degree 
 n, and is given by T(P ) < (l + e)log n, which approaches theoretical lower 
 bound Eog p (2n + 1)1 as n— *oq, where e % (2/log n) 
 
 l/2 
 
 An upper bound of the number of operations required to achieve the 
 theoretical lower bound of steps required to evaluate polynomials of degree 
 n is given as Cn, C > 2. Furthermore a systematic construction of computation 
 trees, mult i- folding and modified multi-folding methods, which achieves the 
 bound is given. Our dual problem approach leads to better results than Brent's 
 (1] primal problem approach, and shows the exi stance of a simple scheduling 
 algorithm to evaluate polynomials within the bound. 
 
 Index terms evaluation, n-th degree polynomial, primal problem, dual problem, 
 
 steps, computation tree, binary, balanced, LHT, RHT, LOF, consistency, segment, 
 subpolynomial, x-term, amplification factor. 
 
IV 
 
 ACKNOWLEDGMENT 
 The author wishes to thank Dr. D. Kuck, Professor of Computer Science at 
 University of Illinois, for suggesting the problem and his comments. He also 
 acknowledges the assistance of Dr. Y. Muraoka and his helpful discussion. 
 Thanks are extended to Miss Sue Cook who helped in getting this report finished. 
 
V 
 
 TABLE OF CONTENTS 
 
 page 
 
 1. INTRODUCTION 1 
 
 2 . COMPUTATION TREES ^ 
 
 3 • ALGORITHMS 11 
 
 k . COMPUTATION RESULTS AND THE BOUNDS 13 
 
 5 • CONCLUSION 22 
 
 LIST OF REFERENCES .- 2k 
 
 APPENDICES 
 
 A. PROOFS OF MAIN PROPERTIES 25 
 
 B. SOME OTHER PROPERTIES 29 
 
 C. PL/l PROGRAMS 35 
 
1. INTRODUCTION 
 
 The computation of polynomials has "been studied for many years. It 
 has "been showm by Pan[5] that 2n operations are required to evaluate a 
 polynomial of degree n. Thus, for a serial machine, Horner's rule is optimal. 
 However, it is easy to see that it requires only l^ogpn - ! steps to evaluate all 
 powers, one step to multiply by the coefficients, and 1 + flog nl steps to sum 
 the terms. Thus by introducing some "redundant" operations one can obtain the 
 result in 2(1 + ^log p r? ) time steps (assuming multiplication and addition times 
 are equal) . This is a crude bound for a multiarithmetic unit machine because 
 some additions can be performed before the final multiplications are performed. 
 Some of the well known methods are Estrin's[3] and the k-th order Horner's 
 rule[2] . And two other methods called a tree method and a folding method have 
 been developed by Muraoka[U] . Hereafter, we assume that arbitrarily many 
 processors are available to evaluate polynomials. 
 
 We write an n-th degree polynomial as 
 
 ■r, / \ n n-1 / n n 
 
 P (x) = a x + a ..x + + a.x + a_ (1) 
 
 n n n-1 10 K ' 
 
 where we assume a. X for < i < n. P may be used to denote P (x) for 
 l - ~~ n n v ' 
 
 simplicity. The lower bound for the number of steps, or time, required to 
 evaluate a polynomial of degree n is 'log p (2n + l)' , or 'log n" + 1, since 2n 
 
 operations are required to evaluate P (x) while a k level balanced binary tree 
 
 k-1 
 contains 2 nodes which correspond to binary operators. 
 
 The dual problem to a problem of finding the minimum number of steps 
 
 s required to evaluate P (x) may be stated as a problem of finding the maximum 
 
 N for which the polynomial P, T (x) may be evaluated for a particular given method 
 
 in s steps. We consider this dual problem hereafter. 
 
Let P (x) and P (x) "be two polynomials and let s, and s^ "be the 
 n K n p 
 
 minimum number of steps required to evaluate P and P , respectively. Clearly 
 
 1 n 2 
 then n n > n implies s.. > s (the converse is not true) and s > s implies 
 
 n.. > n (see appendix B) . 
 
 P (x) may be divided into q segments where each segment consists of 
 n m. 
 
 a subpolynomial and an x-term, i.e., Q, (x) and x . Thus we may write 
 
 n. 
 
 1 
 
 q-1 m. 
 
 p n w = .z \ oo* 1 + p m ( x) ( 2 ) 
 
 i=l i 
 
 q-1 
 
 where n = m + JJ (n. + l) and m. =m+i+ XI n -- Such a segmentatin of a 
 
 i=l 1 X d<i J 
 
 polynomial is called a q-cut. To compute a segment in s' steps, both 
 
 subpolynomial and x-term should be evaluated in (s T - l) steps, i.e., 
 
 s' - 1 > tCQ. ) and s' - 1 > flog m? , where T is a function from polynomials 
 
 i 
 p to the non negative integers I. Such a segment is said to be consistent if 
 n 
 
 both conditions are satisfied. Unless otherwise specified, for each Q (x) we 
 
 i 
 
 choose n. = N such that N = max \ ml s - 1 = T(P )] . 
 i li m J 
 
 The symbol N(s, q), where s is non negative integer and q is integer 
 greater than or equal to 2, is used to denote the maximum degree of polynomial 
 P / v which can be evaluated in s steps by q-cut. It is convenient to consider 
 the formula (2) as a computation tree to evaluate P (x) . The sub computation 
 tree to evaluate the first term of (2) is called the LHT while the sub computation 
 tree for the second term is called the KHT . 
 
 As an example, Muraoka's folding method, q = 2, may be stated as 
 follows: let N(s-l) and N(s) be the maximum degrees of polynomials which can be 
 evaluated in (s - l) and s steps, respectively. Then using his method, 
 N(s+l) = N(s-l) + 1 + N(s), where N(s-l) + 1 and N(s) correspond to the LHT and 
 RHT of the computation tree of P , y and N(s+l) approaches 1.62 times N(s) 
 for relatively large s since the sequence [n(s) + 1 | s>oJ forms a Fibonacci 
 
sequence (reference Table l) . 
 
2. COMMUTATION TEEES 
 
 k k+1 
 
 In general, for such a computation tree, if 2 < q - 1 < 2 , k > 0, 
 
 then the maximum degree of polynomial -which can be evaluated in (s + l) steps, 
 
 N(s+1, q), is given by: 
 
 N(s+1, q) = N(s) + A Q (N(s-k-l+M) + l) + A (N(s-k-2+M) + l) + 
 
 + A M (N(s-k-l) +1) + A M+1 (N(s-k-2) +1) + 
 
 + Vl 1 ^ 8 * 1 ^ +1) J 
 
 M+L 
 = N(s) + V A. (N(s-k-l+M-i) + l) (3) 
 
 i=0 X 
 
 M+L 
 where q - 1 = Yl A. , A. is the number of segments on the level (k-M+i) and 
 
 i=0 1 X 
 A^ T is multiple of 2. L is the level difference between k and the highest 
 
 level of segment in the LHT. M is the level difference between k and the lowest 
 
 level of segment in the LHT. 
 
 A computation tree is said to be q-consistent if each segment in the 
 LHT is consistent. Thus formula (3) is valid if its computation tree is 
 q-consistent, and is illustrated in Figure 1. It is easy to see that a 
 computation tree is q-consistent if and only if the x-term of the leftmost 
 segment in iL is consistent. 
 
 A balanced computation tree with respect toq,2<q-l<2 , 
 is a binary tree such that M = and < L < 1 (or q - 1 = A + A and A = 
 for i t 0, l) . A balanced computation tree is said to be completely balanced 
 if M = and L = l(or q - 1 = 2 k = A and A. = for i X l) ■ A LOF (lower order 
 first) computation tree is a binary balanced tree such that the 
 
r k+L 
 
 k 
 
 M 
 
 k-M 
 
 '« • 
 
 \ / v 
 
 Vo V 
 
 t » 
 
 s-k-L 
 
 s-k 
 
 s-k+M 
 
 
 level 
 
 RHT 
 
 s-1 
 
 s+1 
 step 
 
 Fig 1. a computation tree, j^n denotes a segment 
 
 and @ denotes a binary addition operator 
 
minimum degree of an x-term of segments in A is greater than the maximum degree 
 of an x-term of segments in A, for 2 < q - .1 < 2 , and if q - 1 = 2 then a 
 LOF computation tree is just a completely balanced tree (see Figure 2). It is 
 important to notice that for given s and q, the LOF computation tree is unique. 
 
 Lemma 1 
 
 k k+1 
 
 Let us assume 2 < q - 1 < 2 and LHT be arranged in any non-balanced 
 
 computation tree. If a pair of segments is t levels moved up then it increases 
 the maximum degree which can be evaluated in (s + l) steps by at least 
 
 (0( i - 1)(2 -OON(s-k-l-j) + 1 > 0, < j < L, 
 where (X is an amplification factor defined by the ratio of N(s-k-j) + 1 and 
 N(s-k-l-j) + 1, and 1 < 0((s) < 2. 
 
 Proof: 
 
 Let us assume the computation tree of formula (3) is q-consistent 
 and a pair of segments in A^ . are I levels moved up, and let •*- denote an 
 assignment, 
 
 Vm *" Vm + 2 and Vj-m - Vj.i-i " 1 - 
 
 Thus the improvement achieved by this rearrangement is given by 
 
 2(N(s-k-l-j+£) + 1) + (N(s-k-l-j+l) + 1) - 2(N(s-k-l-j) + l) - (N(s-k-l-j+ +l)+l) 
 
 > (2o/ + CX- of +1 - 2)(N(s-k-l-j) + 1) 
 
 = (d - 1)(2 -O0(N(s-k-l-d) + 1) > 
 where 0U#(s-k-l-j) = (N(s-k-j) + l) / (N(s-k-j-l) + l) and in general 
 0((s+l)> 0((s) for relatively large s. 
 
2(q - 1—2 ) segments 
 
 k+1 
 
 k-1 
 
 k+1 
 (2 - q+l) segments 
 
 DD s - k -! 
 
 s-k 
 
 s-k+1 
 
 
 level 
 
 RHT 
 
 s-1 
 
 s+1 
 steps 
 
 Fig 2. LOF computation tree 
 
Lemma 1 implies that the maximum degree of polynomial evaluated in 
 (s + l) steps by a balanced tree is greater than that achieved by a computation 
 tree which is non-balanced with respect to q. 
 
 Lemma 2 
 
 k k+1 
 For given q, 2 < q-1 < 2 , if a balanced computation tree is 
 
 q-consistent then the LOF computation tree is also q-consistent (the converse 
 is not true). 
 
 Proof : 
 
 Let us assume the LOF computation tree is not q-consistent. Then 
 m. 
 the x-term, x J , of the leftmost segment in A_ is not consistent. Since m. 
 
 is not greater than the power m. of the leftmost segment A_ of any balanced 
 
 computation tree, then such a balanced computation tree is not q-consistent. 
 
 (Q.E.D. ) 
 Lemma 2 implies that the LOF computation tree can be q-consistent 
 while the balanced computation tree is not q-consistent. In other words, the 
 LOF computation tree is better than the regular balanced tree to solve our 
 dual problem. 
 
 Lemma 3 
 
 k k+1 
 The q-consistency condition, for q > 3 such that 2 < q-1 < 2 , 
 
 of an LOF computation tree at step (s + l) is given by 
 
 s - k - 2 > flog (N(s) + (2(q - 1 - 2 k ) - l)(N(s-k-2) + l) + lj 
 
 for s > 7- It is easy to see that 2-consistency always holds. 
 
 Proof: 
 
 See appendix A. 
 
Lemma k 
 
 Let us assume that the LHT of a computation tree at step (s + l) is 
 arranged as a non-LOF computation tree (can be balanced), and let the maximum 
 legree of polynomial evaluated by such a computation tree be denoted by N(s+1, q). 
 
 Then there exist at least one LOF computation tree of q- consistency, q > q and 
 
 N(s+1, q) > N(s+l,q.) . 
 
 Proof: 
 
 It is clear by the definition of q-consistency and the proof of Lemma 3 
 
 (Q.E.D.) 
 Oheorem 1 
 
 For q > 3 such that 2 < q-1 < 2 , k > 0, if the q-consistency 
 condition for the LOF computation tree holds at step (s + l), then the maximum 
 legree of polynomial evaluated in (s + l) steps is given by 
 T(s+l,q) = N(s) + 2(q - 1 - 2 )(N(s-k-2) + l) + (2 - q + l)(N(s-k-l) + l). 
 
 ^or q =2, i.e., the folding method gives the maximum degree. 
 
 N(s+1,2) = N(s) + N(s-l) + 1. 
 
 ^Toof: 
 
 Lemmas 1 through \ lead us to the result that the maximum degree of 
 )olynomial evaluated at each step, s > 0, which is achieved by the LOF 
 :omputation tree is greater than any other computation trees in LHT. By the 
 lefinition of the LOF computation tree and the proof of Lemma 3, q - 1 = A_+ A.. , 
 
 J < q - 1 < 2 k+1 and 2 k = A Q + A-/2, thus we get A Q = 2 + - q + 1 and 
 
 Lj_ =2(q - 1 - 2 k ). 
 
 (Q.E.D.) 
 
10 
 
 Corollary 1 
 
 If a LOF computation tree is q-consistent then N(s+1, q) > N(s+1, q-i) 
 for 1 $ i < q - 3 • 
 
 Proof: 
 
 It Is easy to see that a q-consistent LOF computation tree is (q - i)- 
 
 consistent for 1 < i < q - 2. It is enough to show that W(s+1, q) > N(s+1, q-l) 
 
 k k+1 
 
 Case 1. (for 2 = q - 2 < 2 ' ) 
 
 k k+1 
 
 N(s+1, q) = N(s) + 2(q - 1 - 2 )(N(s-k-2) + l) + (2 ' - q + l)(N(s-k-l) + l), 
 
 N(s+l,q-l) = N(s) + 2 k (N(s-k-l) + l) . Thus 
 
 IT(s+l, q) - N(s+1, q-l) = 2(N(s-k-2) + l) - N(s-k-l) - 1 
 
 = (2 - a )(N(s-k-2) + 1) > 0. 
 
 k k+1 
 
 Case 2. (for 2<q-2<q-1^2 ) 
 
 k k+1 
 
 N(s+l,q) = N(s) + 2(q - 1 - 2 )(N(s-k-2) + l) + (2 - q + l)(N(s-k-l) + l), 
 
 N(s+l,q-l) = N(s) +2(q-2- 2 k )(N(s-k-2) + l) + (2 k+1 - q + 2)(N(s-k-l) + l) . 
 
 Thus N(s+l,q) - N(s+1, q-l) = 2(N(s-k-2) + l) - N(s-k-l) - 1 
 
 = (2 - a)(N(s-k-2) + l) > 0. 
 
 (Q.E.D) 
 
11 
 
 5- ALGORITHMS 
 
 Corollary 1 implies that to maximize N(s+l) we should choose the 
 maximum q such that the LOF computation tree is q-consistent. From this, we 
 get the following algorithm to generate N(s+l) for s > 2. 
 
 Algorithm 1 (multi- folding method). 
 
 Step 1: Set N(o) +- N(l) <- 0, N(2) <- 1, q *- 2 and s «- 2. 
 
 Step 2: If the LOF computation tree is 3-consistent then go to Step 3. Set 
 
 q* *~ q, N(s+l,q*) *- N(s) + N(s-l) + 1, s *- s + 1 and if s > M, where 
 M is an integer used to limit the number of steps then stop, otherwise 
 go back to Step 2. 
 
 Step 3« Let q* be the maximum q such that the LOF computation tree is q-consistent, 
 where 2 < q - 1 < 2 
 
 Step k: Set 
 
 N(s+l,q*) +- N(s) + 2(q* - 1 - 2 k )(N(s-k-2) + l) + (2 k+1 - q* + l) 
 
 (N(s-k-l) + 1) 
 and s ■«- s + 1. If s > M then stop, otherwise go back to Step 3. 
 
 So far, %( s _k_i)( x ) and P N(s-k-2)^ X ^ the maxim ' um degrees of 
 polynomials evaluated in (s-k-l) and (s-k-2) steps, respectively, were assumed 
 
 for each subpolynomial of the segments in A and A on LHT, respectively. 
 
 The q-consistency condition for a LOF computation tree given in Lemma 3 ' 
 
 s - k - 2 > log u 1 , where u = N(s) + (2q* - 2 k+1 - 3)(w(s-k-2) + l) + 1, 
 
 s-k-2 
 so we have information to increase N(s+l). Let f3 = 2 - u > then we may 
 
 expect Y improvement on N(s+l,q*) given by Algorithm 1 by forcing (3 to be zero 
 
 and achieving (q* +1) -consistency. Such ft improvement is defined as 
 
 If = \(N(s-k-2) + p - N(s-k-l)) for 2 k < q* - 1 < 2 k+1 , 
 
 
12 
 
 where A is a function such that A(y) = for y ^0 and A(y) = y for y > 0. This 
 leads to Algorithm 2. 
 
 J 
 
 Algorithm 2 (modified multi-folding method) . 
 
 Step 1: Set N(0) «- N(l) +■ 0, N(2) *" I, q 4- 2 and s«- 2. 
 
 Step 2: If the LOF computation tree is 3-consistent then go to Step 3- Set 
 
 q <- q, N(s+1, q ) *- N(s) + N(s-l) + 1, s<- s + 1 and if s > M then stop, 
 
 otherwise go back to Step 2. 
 
 -x- 
 Step 3: Let q he the maximum q such that the LOF computation tree is q-consi- 
 
 stent, where 2 <c q - 1 < 2 
 Step k: If q* - 1 = 2 k+1 then go to Step 5- Set *- 2 S ~ k ~ 5 - N(s) - 1, 
 
 If*- N(s-k-3) + p - N(s-k-2) and if lf> then set A+- 1, otherwise 
 
 A>**Q. Go to Step 6. 
 Step 5: Set |3<- 2 S " k " 2 - N(s) - 2(q* - 1 - 2 k )(N(s-k-2) + l) - 1. and 
 
 flT<- N(s-k-2) + p - N(s-k-l). If JT«- then set A *- 1, otherwise /I ♦" 0. 
 Step 6: Set N(s+l,q +A) «- N(s) + 2(q - 1 - 2 )(N(s-k-2) + l) + 
 
 (2 - q* + l)(N(s-k-l) + 1) + Xf and s — s + 1. If s > M then stop, 
 
 otherwise go back to Step 3- 
 
13 
 
 
 
 k. COMPUTATION RESULTS AND THE BOUNDS 
 
 Table 1 shows the computation results of the maximum degree of 
 polynomials evaluated by Muraoka's folding method, Algorithm 1, Algorithm 2 
 and the theoretical lower bound of the minimum required steps to evaluate 
 polynomials of degree given by Algorithm 2. The amplification factor (x(s) , 
 
 defined by the ratio between N(s) and N(s-l), approaches 2 as s increases. 
 
 k(k-l) /2 
 Our results agree with Brent's result that polynomials of degree 2 ' 
 
 are evaluated in k(k+l)/2 steps, e.g., for k = 5 the degree is 1021*. and s is 
 
 15 steps. Table 1 shows a polynomial of degree 1024 can be evaluated in 15 
 
 steps by our LOE computation tree methods (multi-folding and modified 
 
 multi- folding) because by Algorithm 1 we can evaluate a polynomial of degree 
 
 1728 in 15 steps. 
 
 The following results from our algorithms may be more interesting 
 
 than those of Brent. Remember that N(s) denotes the maximum degree of 
 
 polynomial evaluated in s steps by the LOF computation tree (multi -folding 
 
 or modified multi-folding methods). The details of derivations of the 
 
 following results are in Appendix A. Any s > 2 can be denoted by either 
 
 k(k+l)/2 - 1 for k > 2 or k(k+l)/2 for k > 2 or k(k+l)/2 + i for k > i + 1, i > 1. 
 
 (1) For k > 2 and k is an integer 
 
 g k(k-l)/2 > N(k(k+1)/2 . 1} > 2 k(k-l)/2 _ x> 
 
 (2) for k > 2 
 
 g k(k-l)/2 + 1 ^ N(k(k+1)/2) > 2 k(k-l)/2 ) 
 
 (3) and for k>i+l, i > 1 
 
 2 k(k-l)/2 + i + 1 N(k(k+l)/2 + > 2 k(k-l)/2 + 
 
 1 
 
Ik 
 
 The above three lead to: 
 
 (1). for s = k(k+l)/2 - 1, k> 2, 
 
 2 s + 3/2 - (2(s + l)) l/2 > N(g) > 2 s + 1/2 - (2(s + l)) 1 /^ 
 
 (2). for s = k(k+l)/2, k £ 2, 
 
 2 s + 1/2 - (2 S )^ > N(s) > 2 s + 1/2 - (2s)^ ^ 
 
 (3) . for s = k(k+l)/2 + i, i £ 1, k > i + 1, 
 
 2 s + 3/2 - (2(s-i)) l/2 > N(g)> 2 s + 1/2 - (2(s-i)) 1 / 2 < 
 
 And we get the following. 
 
 Theorem 2 (dual property) 
 
 For any given (J > 0, N (s) ^ N(s) > 2 ■ "" , where N (s) denotes the 
 
 maximum degree of polynomial which can be evaluated in s steps, (i.e., theoretical 
 
 l/2 
 maximum degree), and £ ^ (2/s) ' for all sufficiently large s. 
 
 Proof: 
 
 It is easily derived from dual properties. 
 
 (Q.E.D.) 
 
 We are now ready to talk about the primal problem, i.e., the minimum 
 number of steps required to evaluate a polynomial of degree n. 
 (1) . For log 2 n = k(k-l)/2 - 1, k > 2, 
 
 -I /o 
 
 T(P n ) < log 2 n(l + (2/(log 2 n + l)) x/ ) + 1/2, 
 (2) . for log 2 n = k(k-l)/2, k > 2, 
 
 1 /? 
 T(P n ) < log 2 n(l + (2/log 2 n)" L/ ) + 1/2 and 
 
 (3) - for log 2 n = k(k-l)/2 + i, k > i + 1, i > 1, 
 T(P n ) < log 2 n(l + (2/(log 2 n + i)) 1 ' 2 ) + l/2. 
 
 : 
 
 
15 
 
 Thus we get Theorem 3 which is identical to Brent's Theorem 2. 
 
 Theorem 3 (primal property) 
 
 0, T(P ) 
 v n 
 
 ,1/2 
 
 For any given e > 0, T(P ) < (l + e)log n, for all sufficiently large 
 
 n, where T: P — » s and e at (21og n)' ' . 
 
 Proof: 
 
 It is derived from primal properties. 
 
 (Q.E.D.) 
 
 Theorem k 
 
 The number of operations required to evaluate a polynomial of degree 
 n , where n is large, in less than or equal to (l + e)log p n steps is given by 
 #(P ) < Cn, where C » 2 €l ° S 2 n > 2. 
 
 Proof: 
 
 From Theorem 3, #(P n ) < 2 S - 1 « 2 (l + e)log 2 n . n2 €log 2 n . 
 
 (Q.E.D.) 
 
16 
 
 r r 
 
 s 
 
 N(s) "by folding 
 
 N(s) by Algorithm 1 
 
 N( 
 
 3) by Algori' 
 
 bhm 2 
 
 lower 
 
 
 method, q = 2 
 
 -X- 
 
 1 
 
 
 
 q +A 
 
 bound 
 
 1 
 
 
 
 
 
 - 
 
 
 
 
 - 
 
 
 2 
 
 1 
 
 1 
 
 2 
 
 
 1 
 
 2 
 
 2 
 
 3 
 
 2 
 
 2 
 
 2 
 
 
 2 
 
 2 
 
 3 
 
 l+ 
 
 1+ 
 
 1+ 
 
 2 
 
 
 k 
 
 2 
 
 k 
 
 5 
 
 7 
 
 7 
 
 2 
 
 
 7 
 
 2 
 
 k 
 
 6 
 
 12 
 
 12 
 
 2 
 
 
 12 
 
 2 
 
 5 
 
 7 
 
 20 
 
 20 
 
 2 
 
 
 20 
 
 2 
 
 6 
 
 8 
 
 33 
 
 36 
 
 3 
 
 
 36 
 
 3 
 
 7 
 
 9 
 
 5^ 
 
 62 
 
 3 
 
 
 62 
 
 3 
 
 7 
 
 10 
 
 88 
 
 101+ 
 
 3 
 
 
 104 
 
 3 
 
 8 
 
 1.1 
 
 ll+3 
 
 183 
 
 1+ 
 
 
 l83 
 
 1+ 
 
 9 
 
 12 
 
 232 
 
 320 
 
 k 
 
 
 320 
 
 k 
 
 10 
 
 13 
 
 376 
 
 572 
 
 5 
 
 
 572 
 
 5 
 
 11 
 
 ll+ 
 
 609 
 
 992 
 
 5 
 
 
 992 
 
 5 
 
 1.1 
 
 15 
 
 986 
 
 1728 
 
 5 
 
 
 I728 
 
 5 
 
 12 
 
 16 
 
 1596 
 
 3059 
 
 6 
 
 
 3059 
 
 6 
 
 13 
 
 IT 
 
 2583 
 
 5^89 
 
 7 
 
 
 5489 
 
 7 
 
 11+ 
 
 18 
 
 1+180 
 
 9767 
 
 7 
 
 
 9767 
 
 7 
 
 15 
 
 19 
 
 676^ 
 
 17^ 
 
 8 
 
 
 Y]k5k 
 
 8 
 
 16 
 
 20 
 
 109^5 
 
 31286 
 
 9 
 
 
 31286 
 
 9 
 
 16 
 
 21 
 
 17710 
 
 55766 
 
 9 
 
 
 55915 
 
 10 
 
 17 
 
 22 
 
 28656 
 
 100914-6 
 
 11 
 
 
 101095 
 
 11 
 
 18 
 
 23 
 
 1+6367 
 
 I82726 
 
 12 
 
 
 182875 
 
 12 
 
 19 
 
 2k 
 
 75021+ 
 
 330690 
 
 13 
 
 
 330839 
 
 13 
 
 20 
 
 25 
 
 12.1392 
 
 6027214 
 
 15 
 
 
 602873 
 
 15 
 
 21 
 
 26 
 
 1961+17 
 
 IO96509 
 
 16 
 
 
 1096807 
 
 16 
 
 22 
 
 27 
 
 317810 
 
 I988781 
 
 17 
 
 
 19911+63 
 
 17 
 
 22 
 
 28 
 
 511+228 
 
 36.II4-52O 
 
 18 
 
 
 3619735 
 
 18 
 
 23 
 
 29 
 
 832039 
 
 6595653 
 
 20 
 
 
 6603699 
 
 20 
 
 2k 
 
 30 
 
 131+6268 
 
 120605211- 
 
 22 
 
 
 12071699 
 
 22 
 
 25 ■ 
 
 31 
 
 2178308 
 
 2211^723 
 
 2k 
 
 
 22129325 
 
 21+ 
 
 26 
 
 32 
 
 3521+577 
 
 14.06393^3 
 
 26 
 
 
 1+0738153 
 
 27 
 
 27 
 
 33 
 
 5702886 
 
 7I4-9.IO7II 
 
 29 
 
 
 7502714-01 
 
 29 
 
 28 
 
 3h 
 
 92271+61+ 
 
 I38188692 
 
 32 
 
 
 138391057 
 
 32 
 
 29 
 
 35 
 
 1^-930331 
 
 ! 2538533614- 
 
 33 
 
 
 2514-222609 
 
 33 
 
 29 
 
 
 Table 1. 
 
 Maximum degrees c 
 
 >f polj 
 
 momi 
 
 .als at each 
 
 step . 
 
 

 IT 
 
 s 
 
 N(s) by folding 
 
 N(s) by Algorithm 1 
 
 N(s) by Algorithm 2 
 
 lower 
 
 
 method, q = 2 
 
 q 
 
 q. +A 
 
 bound 
 
 36 
 
 2lU578l6 
 
 14.66181068 
 
 35 
 
 1+668.12558 
 
 35 
 
 30 
 
 37 
 
 39088.168 
 
 857771783 
 
 38 
 
 .858785I+53 
 
 38 
 
 31 
 
 38 
 
 632^5985 
 
 15831+99885 
 
 1+2 
 
 1585050551 
 
 1+2 
 
 32 
 
 Table 1. 
 
 Maximum degrees of polynomials st each step . 
 
18 
 
 / 12 v 21 ' ( 20 N 
 
 U„x + — ■ + a 00 x + a on ) x (a on x + + a] x + a Q ) 
 
 33 
 
 22 21' 
 
 6 steps 5 steps 
 
 20 
 
 7 steps 
 
 8 steps 
 
 (a). The computation tree for Muraoka's folding method at step 
 (which evaluates a polynomial of degree 33) • 
 
 i 7 \ 29 , 7 \ 21 , 20 N 
 
 ^ a 36 X + "" + a 29^ X + ^ a 28 X + -" + a 21^ x + ( a 20 X + "- + a x x + a ) 
 
 5 steps 5 steps 5 steps 5 steps 7 steps 
 
 v ®' ^ step 
 
 8 steps 
 
 (b) . The computation tree for multi- folding method at step 8 
 (which evaluates a polynomial of degree 36) . 
 
 Fig. 3 Examples of computation trees at steps 8, 
 
19 
 
 
 o 
 
 in 
 
 o 
 
 -a- 
 
 l-l 
 
 o 
 m 
 
 o 
 
 w 
 -d 
 o 
 
 -P 
 
 o 
 
 •H 
 -P 
 CO 
 -P 
 
 ! 
 
 o 
 
 ■a 
 
 o 
 
 H 
 O 
 ft 
 
 H 
 
 H 
 
 cd 
 ft 
 
 <l> 
 
 Ch 
 
 o 
 
 «h 
 
 Oh 
 
 O 
 
 <M 
 
 fl 
 
 o 
 
 O 
 
 
 w 
 
 (D 
 
 •H 
 
 u 
 
 cd 
 
 i 
 
 G 
 
 o 
 
 
 o 
 
 
 H 
 
 
 ,d 
 
 
 ft 
 
 
 cd 
 
 
 M 
 
 
 C5 
 
 T 
 
 sda^s jo jaqrariM 
 
20 
 
 ko 
 
 35 
 
 30 
 
 25 
 
 20 " 
 
 15 
 
 w 
 
 ft 
 
 (D 
 
 02 
 
 10 
 
 Muraoka ' s folding method 
 
 multi- folding 
 method 
 
 theoretical lower bound 
 
 X : given "by Brent . 
 
 (kCk+l)^ 3 ^' 1 ^ 2 ) 
 
 10 
 
 15 
 
 20 
 
 25 
 
 log n n degree of polynomial 
 
 30 
 
 Graph 2 . Steps, s v. s. Polynomial degree n 
 
21 
 
 o 
 
 
 o 
 
 a 
 
 •H 
 H 
 
 o 
 
 O 
 
 
 O 
 
 CVI 
 
 m 
 ft 
 
 CD 
 -P 
 
 ra 
 
 LT\ 
 
 O 
 
 H 
 
 UA 
 
 o 
 
 o 
 
 CM 
 
 O 
 CM 
 
 ITS 
 H 
 
 O 
 
 H 
 
 LTN 
 
 i!ou9q.sxsuoo 
 
 o 
 
 Graph J> . Consistency q v.s. steps s. 
 
22 
 
 5 • CONCLUSION 
 
 Graph 2 shows the maximum degree of polynomial which can be evaluated 
 in a given number of steps "by the LOF computation tree. Brent's result is also 
 plotted. 
 
 Let us compare Brent's result (his Theorem l) which says that a polynomial 
 
 k(k-l) /2 
 of degree 2 ' can be evaluated in less than or equal to k(k+l)/2 steps, 
 
 k(k-l) /2 
 k > 2, with our result that a polynomial of degree greater than 2 ' can be 
 
 evaluated in k(k+l)/2 steps. These two results are slightly different since the 
 
 former statement does not say that there exists at least one polynomial of degree 
 
 k(k-l)/2 
 higher than 2 ' whose evaluation can be accomplished in k(k+l)/2 steps. 
 
 This statement is also clear from Graphs 1 and 2. Our dual problem approach 
 
 becomes more reasonable than the primal problem approach because a set of 
 
 polynomials can be evaluated in s steps and such a set is ordered with respect to 
 
 polynomial degree. Moreover it is easy to see that the dual property implies 
 
 the primal property, while the converse is not true. 
 
 Some of the consequences of this paper are: 
 
 (l) . We found the maximum possiblly evaluable degree of polynomial 
 for each number of steps, i.e., the dual property for not only a particular s but 
 for all s > 1, while Brent's primal property is applied only for descrete degrees 
 or at descrete steps. 
 
 (2) . Dual property implies primal property. 
 
 (3) • We found simple computation trees called LOF, or multi-folding, 
 to achieve Brent's upper bound and to approach the theoretical lower bound. This 
 means that we have found a simple scheduling algorithm to evaluate a polynomial 
 
 of degree n whose evaluation requires a number of steps close to the lower bound. 
 
 l/2 
 
 (k) . We found e as a function of (2/log n) ' for the upper bound 
 
 . 
 
23 
 
 (l + c)log n of the primal problem. 
 
 l/2 
 (5) . We found <f as a function of (2/s) ' for the lower bound 
 
 2 s(l-o) of the dual problem# 
 
 (6) . The upper bound of the number of operations required to evaluate 
 a polynomial of degree n in (l + e)log 2 n steps is given as Cn, C > 2. 
 
 Furthermore, by considering Brent's statement "Theorem 1 is used to 
 obtain an upper bound for t(n), even if n is not a power of 2", we get the 
 
 following . 
 
 o + i /p Cp=? > 1 1 ' 2 
 N^s)^ 2 ' " K J , for integer s> 2. 
 
 Notice that the equality should be included, i.e., > , while our dual results 
 
 do not, i.e.,> (see dual properties developped ealier) . The maximum degree of 
 
 polynomials derived by the above dualization of Brent's Theorem 2, IL^s), for 
 
 each step s is plotted in the Graph 2. By the comparison of our results and his, 
 
 we have 
 
 (7) • the upper bound of the steps required to evaluate polynomials by 
 
 our multi-folding method is lower than or equal to the Brent's, and the lower 
 
 bound of the degree of polynomials evaluated at each step s, i.e., the maximum 
 
 evaluable degree of polynomials, is .higher than Brent's by a factor of \i, 
 
 where 1.5 < u < 2. 
 
LIST OF REFERENCES 
 
 1. Brent , R. , "On the addition of binary numbers", IEEE Transaction on Computers , 
 
 August 1970, pp. 758-759- - - - 
 
 2. Dorn, W. S. , "Generalizations of Horner's Rule for Polynomial Evaluation", 
 
 IBM Journal of Research and Development , 6, April 1962, pp. 239-2^5. 
 
 3. Estrin, G. , "Organization of Computer Systems -- the Fixed plus Variable Structure 
 
 Computer", Proc. of Western Joint Computer Conference , May i960, 
 pp. 33-V). 
 
 \. Muraoka, Y. , "Parallelism Exposure and Exploitation in Programs", Ph.D. Thesis, 
 University of Illinois, 1971? PP- 33- W- 
 
 5. Pan, V. Ya. , "Methods of Computing Values of Polynomials", Russian 
 
 Mathematical Surveys , 21, January- February 1966, pp. IO5-I36. 
 
 6. Winograd, S. , "On the Time Required to Perform Addition", JACM , 12, April 1965, 
 
 pp. 277-285. 
 
 
25 
 
 APPENDIX A 
 PROOFS OF MAIN PROPERTIES 
 
2b 
 
 Proof of Lemma 5« 
 
 If s - 1 2s flog 2 (N(s) + l7 then N(s+1,2) = N(s) + N(s-l) + 1. 
 
 If 8 - 2 > flogg(N(s) + N(s-2) + 2~)1 then 
 
 N(s+1,3) = N(s) + 2(N(s-2) + 1). 
 
 If s - 2 > Eog 2 (N(s) + 2(N(s-3) + 1) + 1)1 and if 
 
 s - 3 ^ log (N(s) + (N(s-3) + 1) +ll then 
 N(s+1,^) = N(s) + 2(N(s-3) + 1) + (N(s-2) + l) . 
 
 If s - 3 > log 2 (N(s) + 3(N(s-3) + 1) + l7 then 
 W(s+1,5) = N(s) + I<-(N(s-3) + 1). 
 
 If s - 3 2 1og 2 (N(s) + 2(N(s-i|-) + 1) + 2(N(s-3) + l) + lj and if 
 S - X ^ Eog 2 (N(s) + (W(s-if) + 1) + 1)1 then 
 N(s+1,6) = N(s) + 2(N(s-if) + 1) +3(N(s-3) + l) • 
 
 It is important to see that "If condition" of the very first case 
 always hold. In general we get the following properties. 
 
 Property 1, 
 
 If q - 1 = 2 k and if s - k - 1 =? Eog 2 (N(s) + (2 k - l)(N(s-k-l) + l) 
 + 1)1 then N(s+l,q) = N(s) + 2 k (N(s-k-l) + l) . 
 
 
 Property 2. 
 
 If 2 k < q - 1 < 2 k+1 , 
 
 s - k - 1 St Eog 2 (N(s) + 2(q - 1 - 2 k )(N(s-k-2) + l) + (2 +l - q)(N(s-k-l) + l)+j 
 and if s - k - 2 ^ Eog 2 (N(s) + (2(q - 1 - 2 k ) - l) (N(s-k-2) + l) + 1)1 then 
 N(s+l,q) =N(s)+2(q-l- 2 k )(N(s-k-2) + l) + (2 k+1 - q + l) (N(s-k-l) + l) . 
 
27 
 
 Property 3. 
 
 If the second condition of Property 2 holds then the first condition 
 
 always holds. 
 
 Proof: 
 
 N(s) + 2(q - 1 - 2 k )(N(s-k-2) + l) + (2 k+1 - q)(N(s-k-l) +l) +1 
 
 < 2N(s) < 2(N(s) + (2(q - 1 - 2 k ) - l)(N(s-k-2) + l) + l). 
 
 Thus from the above arguments, the q-consistency condition 
 (q > 3) for the LOF computation tree is given "by 
 
 
 s - k - 2 > flog 2 (N(s) + (2(q - 1 - 2 k ) - l) (N(s-k-2) + l) + 1? 
 
 ~k _, _ ,, _k+l 
 where 2 < q - 1 < 2 
 
 (Q.E.D. ) 
 
 Dual Properties . 
 
 (1) At steps s = (k(k + l))/2 - 1, for k > 2, 
 
 log 2 N(s) + 2 > log^s] 1 + 1 = (k(k - l))/2 + 1 > log 2 N(s) + 1. 
 
 By rearrangement, we get 
 
 (k(k - l))/2 > log 2 N(s) > (k(k - l))/2 -1 
 and this yields to: 
 
 2 (k(k - l))/2 > N(g) > 2 (k(k - l))/2 - 1. 
 
 (2) At steps s = (k(k +l))/2, for k > 2 (i.e., at Brent's points), 
 log N(s) + 2 > 1og 2 N(s7 + 1 = (k(k - l))/2 + 2 > log 2 N(s) + 1. 
 
 By rearrangement, we get 
 
 (k(k - l))/2 + 1 > log 2 N(s) > (k(k - l))/2 
 and this yields to: 
 
 2 (k(k - l))/2 + l >N(g) >2 (k(k - l))/2. 
 
28 
 
 (3) At steps s = (k(k + l))/2 + i, for k > i + 1, i > 1, 
 
 log 2 N(s) + 2 > ^log 2 N(s")l + 1 = (k(k - l))/2 + i + 2 > log 2 N(s) +] 
 
 By rearrangement, we get 
 
 (k(k - l))/2 + I + 1 > log N(s) > (k(k - l))/2 + i 
 and this yields: 
 
 2 (k(k - l))/2 + i + 1 > N(s) > 2 (k(k - l))/2 + i. 
 
 Primal Properties . 
 
 These properties are derived directly from those dual properties 
 discussed above. 
 
29 
 
 APPENDIX B 
 SOME OTHER PROPERTIES 
 
30 
 
 The following two properties are important and are given without 
 
 proofs . 
 
 Lemma 5 
 
 Let P (x) and P (x) he two polynomials, where n n > n_. If s n and 
 n n p 12 1 
 
 s are the minimum steps required to evaluate P and P , respectively, then 
 s > s (the converse is not true). 
 
 Lemma 6 
 
 Let s and s be the minimum steps required to evaluate polynomials 
 
 P (x) and P (x), respectively. If s > s then n > n (the converse is not 
 n x n 2 12 12 
 
 true ) . 
 
 Theorem 5 
 
 Muraoka's folding method and the minimum steps required to evaluate 
 polynomials achieved by exhaustive division of P (x) into 2 segments, called 2-cut, 
 are equivalent, i.e., both methods give the same maximum degree of polynomials 
 that can be evaluated in a given number of steps. 
 
 
 
 Proof : 
 
 Let s„ and s denote the minimum number of steps required to evaluate 
 
 a polynomial of degree n by the folding method and exhaustive 2-cut method, 
 respectively, and assume T is the function T: P (x) -* s. If s < s we have 
 
 Q (x)x + P r _ 1 (x) € |o ii _.(x)x 1 + P ± _ 1 (x) | 1 < i < n j, 
 
 therefore, s f > s = min lT(Q (x)x 1 + P. 1 (x)) | 1 < i < n k where f is 
 given by the folding method. 
 
31 
 
 Now consider the case s _ > s, . Let us assume s _ < s, , then there exists k such 
 
 f — b f b 
 
 that s > T(Q'(x)x + P 1 (x)). It is also obvious that s - 2 > T(P ,(x)) 
 and s - 3 > log k for k < f (k > ^ is not valid from Lemma k) . But for such k 
 
 s f - 3 > T(Q n _ k (x))must hold, while s f - 2 > T(Q^_ (x)), since k < J" implies 
 
 deg(Q v.( x )) > ^^(Qy, ( x )) and by Lemma k we get 
 
 T(Q k ( x )) ^ T (Q Vl ( x )) which leads us to a contradiction. 
 
 (Q.E.D. ) 
 
 Theorem 6 
 
 The minimum steps s required to evaluate a polynomial of degree 
 n by Muraoka's folding method is upper bounded by (l + C, )log n + C for 
 relatively large n, where C (^ 0.1^) and C (~ 0) are constants. 
 
 Proof: 
 
 Because the sequence (N(s) + 1 | s > Oj forms a Fibonacci 
 sequence, the amplification factor a(s) for relatively large s approaches 
 
 1/2 
 (1+5 )/2 (« 1.62), and then we get the result. 
 
 (Q.E.D. ) 
 
 For example the minimum' steps s required to evaluate P (x) by the 
 folding method for 2 < n < 200 are bounded by 
 
 1(1 + 0.58)log nl + 1 < s < 1(1 + 0.1^)log nl + 1. 
 
32 
 
 Theorem 7 
 
 The •minimum number of operations required to evaluate a polynomial 
 of degree n, N(s) < n < N(s) + N(s-l) + 1, by Muraoka's folding method is given 
 by the following recurrence formula; 
 
 # (P H(S) + 1 + 1 (X)) " #(P 1 (X)) + 5 + # (P H(S) (X)) 
 for i =0, 1, ..., N(s-l), and for s = 1, 2, ".". '. , where #: P (x) -» I,, 
 
 and #(P ) > C'n where C =s 2.6. 
 " n — 
 
 Proof: 
 
 It is obvious from the computation tree of Muraoka's folding 
 
 method, or from our 2- consistent LOF computation tree. 
 
 (Q.E.D. ) 
 
 In table 2, the computation result given by Theorem 7 is listed. 
 Lemma 7 
 
 Let us assume P / >.(x) denotes the polynomial of the maximum 
 
 degree whose evaluation requires at least s steps by "any" computation tree, 
 
 then 
 
 T(P, T , n(x)x X ) > s + 1 for i > 1. 
 N(s) — — 
 
 Proof: 
 
 It is enough to show that T(xP, T / \(x)) > s + 1. Assume xP, t/ k 
 
 N(s) — N(s) 
 
 can be evaluated in s steps. If xP / \ can be evaluated by our multi-folding 
 method then P / •, should be evaluated by (s - l) steps, which leads to 
 contradiction. If the polynomial can be evaluated as 
 
 2 
 
 ...+...+...+ ax + ax 
 
 
33 
 
 2 
 
 since the evaluation of ax and ax require 2 and 1 steps, respectively, it is 
 
 clear to see that we can evaluate P / >.x + c, c is constant, in s steps. 
 Furthermore, P , \ is, by assumption, the polynomial of the maximum degree whose 
 evaluation requires s steps, thus we have a contradiction. Assume the 
 polynomial is evaluated as 
 
 ••• + + + ( )y? + (a,x + a )x 
 
 we need 3 steps to evaluate the last term, but we can also evaluate 
 
 2 
 
 a_x + ax + c in 3 steps, which means we can evaluate P / x + c in s steps 
 
 and leads us to a contradiction in our assumption. 
 
 (Q.E.D. ) 
 
% 
 
 
 n 
 
 W 
 
 n 
 
 #(P n ) 
 
 n 
 
 #(P n ) 
 
 n 
 
 #(P ) 
 n 
 
 
 
 
 
 
 35 
 
 91 
 
 70 
 
 183 
 
 105 
 
 273 
 
 
 1 
 
 2 
 
 36 
 
 9k 
 
 71 
 
 186 
 
 106 
 
 275 
 
 
 2 
 
 5 
 
 37 
 
 97 
 
 72 
 
 188 
 
 107 
 
 278 
 
 
 3 
 
 8 
 
 38 
 
 99 
 
 73 
 
 191 
 
 108 
 
 280 
 
 
 k 
 
 10 
 
 39 
 
 102 
 
 7^ 
 
 193 
 
 109 
 
 283 
 
 
 5 
 
 13 
 
 ko 
 
 1(A- 
 
 75 
 
 196 
 
 110 
 
 286 
 
 
 6 
 
 15 
 
 kl 
 
 107 
 
 76 
 
 199 
 
 111 
 
 288 
 
 
 7 
 
 18 
 
 k2 
 
 110 
 
 77 
 
 201 
 
 112 
 
 291 
 
 
 8 
 
 21 
 
 k3 
 
 112 
 
 78 
 
 203 
 
 113 
 
 29^ 
 
 
 9 
 
 23 
 
 kk 
 
 115 
 
 79 
 
 205 
 
 Hit-- 
 
 296 
 
 
 10 
 
 26 
 
 k5 
 
 1.18 
 
 80 
 
 207 
 
 115 
 
 299 
 
 
 11 
 
 29 
 
 k6 
 
 120 
 
 81 
 
 210 
 
 116 
 
 301 
 
 
 12 
 
 31 
 
 kl 
 
 123 
 
 82 
 
 212 
 
 117 
 
 304 
 
 
 13 
 
 3^ 
 
 k8 
 
 125 
 
 83 
 
 215 
 
 118 
 
 307 
 
 
 Ik 
 
 36 
 
 k9 
 
 128 
 
 84 
 
 218 
 
 119 
 
 309 
 
 
 15 
 
 39 
 
 50 
 
 131 
 
 85 
 
 220 
 
 120 
 
 312 
 
 
 16 
 
 k2 
 
 51 
 
 133 
 
 86 
 
 223 
 
 121 
 
 315 
 
 
 17 
 
 kk 
 
 52 
 
 136 
 
 87 
 
 226 
 
 122 
 
 317 
 
 
 18 
 
 kl 
 
 53 
 
 138 
 
 88 
 
 228 
 
 123 
 
 320 
 
 
 19 
 
 k9 
 
 5k 
 
 llH 
 
 89 
 
 231 
 
 124 
 
 322 
 
 
 20 
 
 52 
 
 55 
 
 Ikk 
 
 90 
 
 233 
 
 125 
 
 325 
 
 
 21 
 
 55 
 
 56 
 
 lk6 
 
 91 
 
 236 
 
 126 
 
 328 
 
 
 22 
 
 57 
 
 57 
 
 lk9 
 
 92 
 
 239 
 
 127 
 
 330 
 
 
 23 
 
 60 
 
 58 
 
 152 
 
 93 
 
 2kl 
 
 128 
 
 333 
 
 
 2k 
 
 63 
 
 59 
 
 l^k 
 
 9k 
 
 2kk 
 
 129 
 
 335 
 
 
 25 
 
 65 
 
 60 
 
 157 
 
 95 
 
 2kG 
 
 130 
 
 338 
 
 
 26 
 
 68 
 
 . 6l 
 
 159 
 
 96 
 
 2^9 
 
 131 
 
 34l 
 
 
 27 
 
 70 
 
 62 
 
 162 
 
 97 
 
 252 
 
 132 
 
 343 
 
 
 28 
 
 73 
 
 63 
 
 165 
 
 98 
 
 25^ 
 
 133 
 
 346 
 
 
 29 
 
 76 
 
 6k 
 
 167 
 
 99 
 
 257 
 
 13^ 
 
 349 
 
 
 i 30 
 
 1 
 
 78 
 
 65 
 
 170 
 
 100 
 
 260 
 
 135 
 
 351 
 
 
 51 
 
 8i 
 
 66 
 
 173 
 
 101 
 
 262 
 
 136 
 
 35^ 
 
 
 32 
 
 8k 
 
 67 
 
 175 
 
 102 
 
 265 
 
 137 
 
 356 
 
 
 33 
 
 86 
 
 68 
 
 178 
 
 103 
 
 267 
 
 138 
 
 359 
 
 
 - 3k 
 
 89 
 
 69 
 
 180 
 
 HA- 
 
 270 
 
 139 
 
 J. 
 
 362 
 
 Table 2. The number of operations required to evaluate 
 polynomial of degree n by the folding method. 
 

 35 
 
 APPENDIX C 
 PL/l PROGRAMS 
 
36 
 
 OFGRFF 
 CC 
 
 : PROC 
 
 L N(0_: 
 
 S BI 
 
 m a i 
 
 DELTA 
 TOWK 
 
 EDLRE GPTICNSIMAIN); 
 5 1 ) _ _R_I N FIXED (31!j_ 
 N FIXED, 
 N FI>FD( 3 1) t 
 
 BIN FIX EC (31), 
 (0:35) BIN FIXFDOl); 
 
 M( 
 N< 
 N( 
 
 N( 
 M( 
 N< 
 
 0)=0? 
 l)=0; 
 
 2) = l: 
 
 3) = ?; 
 4)=a; 
 c ) = 7; 
 
 N( 
 
 M 
 
 TO 
 CO 
 
 _EM 
 
 NC 
 
 DO 
 
 L: 
 
 ._.EN 
 
 OUT:M= 
 IF 
 
 l; 
 
 c 30; 
 
 I }=2*T0fcK( 1-1) ; 
 
 6)=1?; 
 7 ) = ? ; 
 i*K ((.))== 
 
 1 = 1 T 
 
 towu 
 
 DJ 
 
 - -3 • 
 
 - ~- » 
 
 s = 7 t r so; 
 
 r i=o tc 10; 
 
 TF ( TOVK( I XMC-1J £(NC- K=TC WK( I + l ) ) 
 
 THEN CG ; J=l; GO TO OUT; END; 
 
 C_; 
 
 N<S) + ( 2=MN0-1-TCUK( J))-l)*(Nt S-J-2 J+l) + l; 
 
 T W K ( 5 - J- 2 ) >- M 
 
 __PU 
 
 NC 
 
 EN 
 
 END OF 
 
 THEN DC: NCC=NC; NONC+1? K = J; GO TO L: END 
 M( S.+ 1J-N (S )+2*(NCO-l-IQWKlKJ )*< N < S-K-21 + 1 ) 
 ♦ (TOWMK + 1 l-NCO+1 )*(N(S-K-1)+1) ; 
 
 CATMN( S + l) ,NC.D) ; 
 
 T SKIP 
 
 =NC-l; 
 c; 
 GPEE ; 
 
57 
 
 PRINT: 
 PLT SKIP DATA(NCS+1).NCD): 
 
 NC=NC-l; 
 
 end ; 
 
 end degree; 
 
58 
 
 DEGREE: PROCEDURE OPTI GNS< NAI N) ; 
 DC L N(0;51) BIN FIXEO(31), 
 S BIN FIXED, 
 
 M BIN FIXED( 31) t 
 
 OELTA B IN FIXECM31 I, 
 THWK(0:35) PIN FIXED(31); 
 
 N(0)=0; 
 N ( 1 ) =o; 
 
 M I 2 ) = 1 ; 
 M( 3_) = ?j_ 
 N ( 4 ) =4 ; 
 M ( 5 ) = 7 ; 
 
 N ( h ) = 1 2 ; 
 N( 7)=2Q: 
 
 rnwKto )=i; 
 
 no i = i tc 30_; 
 
 TOWM I ) =?*TOWK( 1-1 ) ; 
 
 FNn; 
 
 NC = 3; 
 
 on S=7 TO 50j_ 
 
 L: DH 1=3 TO io; 
 
 IF ( TOWK ( I XNC-DS tNC-K-TDH K ( I +1 1 ) 
 THEN HO: J=l; GG TO OUT; FND; 
 
 END; 
 
 0UT:M=N( S) + < 2* (NO 1-T(*V»K( J ) ) -1 )*(N( S-J-2H-1J+1 ; 
 TF TOWK (S-J-2 )>=M 
 
 THEN DCl: NCD=NC: NC=NC+l; K = J; GO TO L ; END; 
 JF TOWKCK+ U=NCD-1 THEN GO TO O UT1 ; 
 
 DELTA=T0WK(S-K-2 )-N < S )-2* ( NC D- 1-TOWK { K ) )*(N< S-K-21 + ll-l; 
 IF M( S-K-l )>-N(S-K-?)+DELTA __ 
 
 T F FN nn ; 
 
 N I S+ 1 ) =N ( S ) + 2.*_( NC C-l-TCWK(K) )*(N(S -K-2 ) t jj_ 
 +(T0WK(K+1)-NCD+1 )*(N( S-K-l )+l); 
 
 GO TC PR INT L 
 
 FND; 
 FLSF DN; 
 
 N( S+1)=N(S )-M2*( NfC- 1-TOWK (K ) ) + l )*(N(S-K-2)+l) 
 + i TO WK'('k + 1)-NC 1) *<N( S-K-l ) +l) + DELTA: 
 
 NC0=NCD+1 ; 
 
 on to print; . 
 
 END ; 
 
 lll 
 
 DELTA=T0^K(S-K-3 )-N(S) -1; 
 
 f F N ( S -K - 2 )>= N(S-K-3)+DELTA 
 
 twfn on; 
 
 M ( s +_ 1 ) = N I Si + TOWK{K+l)*lNtS-K- 2)+l ); 
 
 GO TO PR INT; 
 
 end; 
 
 FLSF DO; 
 
 N ( S + 1 ) = NM S)-H TOWK(K»l )-l ) *( N ( S-K-2 ) + 1 > + N( S-K-3 ) + DELTA; 
 
 NCD=NCn+] : 
 
 F N D : 
 
39 
 
 BISECT: PROCEDURE OPT IONS( MAIN ) ; 
 
 BISECT: PROCEDURE OPTIONS (MAIN ) ; 
 OCL DPN BIN FIXED; 
 
 get list(dpn); 
 begin; 
 
 DCL ( A,B,C,S1,S2,MINISTEP) BIN FIXED, 
 STEP(Q:DPN) BIN FIXED; 
 
 STEPCO)=0; STEP(1)=2;STEP(2)=3; 
 DO N=3 TO DPN; 
 
 MINISTEP=100; 
 DO 1=1 TO N; 
 
 S1=STEP( 1-1); 
 A=CEIL(LQG2(I )); 
 
 B=STEP(N-I ); 
 S2=MAX(A,B)+1; 
 
 END; 
 END; 
 
 C=MAX(Sl,S2)+l; 
 
 I F C < MINI STEP THEN D O; M INISTEP=C; M l^U END; 
 END; 
 
 STEP(N)=MINISTEP; 
 
 LB=CEIL(LOG2(2*N+l) ) ; 
 PUT SK I P D ATA ( N,MIN1STE P, MI, LB) ; 
 
 END BISECT; 
 
 
ko 
 
 T R I^S EC ; P R OCEDURE OPT IONS ( MA IN ) ; 
 
 fRI_SEC: PROCEDURE OPTIONS(MAIN ) ; 
 
 DCL NN BIN FIXED; 
 
 GET LIST(NN); 
 
 BEGIN; 
 
 DCL (A f B,CtOtSltS2»S3J BIN FIXED, 
 
 STEP(OtNN) BIN FIXED; 
 
 
 STEP =100; 
 
 
 
 STEP(0)=0; STEP(1)=2; 
 
 
 
 DO N=2 TO NN; 
 
 
 
 K=N-i; 
 
 
 
 DO 1=1 TO k; 
 
 S1 = STFP(I-1 ) ; 
 
 
 
 DO J=I+1 TO n; 
 
 
 
 A=STEP(J-I-1) ; 
 
 
 B=CEIL<L0G2(I )) ; 
 S2=MAX( A,B)+l; 
 
 a=sTep(n-j) ; 
 b=ceil(l0g2( j)) 
 
 S3=MAX( A, B)*l; 
 A=MAX(Si,S2,S3); 
 
 IF A=S1 THEN DO; C=MAX ( S2 , S3 )+l ; D=MAX ( SI , C ) +1 ; 
 
 If _P <=STEP tN I TH E N Oj_ S TEP(N ) =D ; MI = 1 ; MJ=J; END; 
 GO TO OUT; "end; 
 
 IF A=S2 THEN DO; C=MAX ( SI , S3 )+l ; D=MAX ( S2 , C ) + 1 1 
 
 IF D<=STEP(N) THEN DO; STEP(N)=D; MI=I; MJ=J; END; 
 
 GO TO OUT; END; 
 
 C=MAX(S1,S2)+1; D=MAX(S3,C)+l; 
 IF D<=STEP(N) THEN DO; STEP(N)=0; MI=I; MJ=j; END; 
 
 OUT: 
 
 END; 
 
 END; 
 
 PUT .SKIP DAT A(STEP(N),MI f MJ); 
 
 END; 
 
 fnd; 
 
 END TRI_SEC; 
 
 . 
 
u 
 
 EXHAUST: PROCEDURE CPT IONS (MAIN > ; 
 
 DCL NK50) BIN FIXED, /* INDEX REGS */ 
 
 -Hi o: iocu am fixed* 
 
 STEP(0:50) BIN FIXED, 
 _PCSXU BIN- FIXED, -73fe--CJUT POS1JION */ 
 (N,NN,NC,A,B,SR,COUNT) BIN FIXED; 
 
 UP_DATE_H: procedure; 
 
 Ih=o; 
 sr=step(ni (i)-i); 
 H4-SRJ-U — - - 
 
 IF NC>=3 THEN 
 DO 4=2 TO NC_1; 
 
 A=STEP(NI(I)-Nl{I-l)-l»; 
 B=CEIL(LOG2(NIU-l) II; 
 SR=MAX(A,8)+l; 
 
 -H4SR4=HISR4-±X4- -■ 
 
 END; 
 
 A=STE P ( N-N4 IHC_ 1 1 1 ; 
 
 S=CEIL(LGG2(NI(NC_1) )>; 
 
 SR=MAX(A,B)+1; 
 
 H(SR)=H(SR)+l; 
 
 COUNT=0; 1=0; 
 
 L: IF H(I)>0 THEN 
 
 DO; IF H(U>1 THEN DO; H CI + L ) = H< 1+1 )+l ; HfI)=H(I)-2; COUNT=COUNT+ 1; 
 
 IF COUNT<NC_l THEN GO TO L; 
 M=I+l; GO TO FINE; 
 END; 
 
 DO 4=1+1 TQ NO; - 
 
 IF H(JI>0 THEN DO; H ( J+l 1=H( J + l ) + 1 ; H(I)=H(I)-1; 
 
 H(J)=H(J)-1; COUNT=CGUNT+l; 
 IF COUNT<NC_l THEN GO TO L ; 
 M=J+l; GO TO FINE; 
 END; 
 EN04 
 END; 
 1=1+1; 
 go to l; 
 
 FINF: 
 
 IF M<=STEP(N) THEN DO; STEP(N)=M; MC=NC; 
 
 DO K=l TO NC_l; P(K)=NI(K); END; END; 
 
 RETURN; 
 
 end up_date_h; 
 get listinn); 
 
 STEP=lOO; Nl=o; 
 
 STEP(0)=0; STEPll)=2; STEP(2)=3; 
 
 DO N=3 TC NN; 
 ND-N+N; 
 p=o; 
 
 IF N>27 THEN HNC=9; ELSE MNC=N/3+l; 
 LO :DO NC=2 TO MNC; 
 
42 
 
 nc_ 
 
 .1=NC- 
 
 i; 
 
 
 LI :D0 NK1 
 
 )=1 TO N-NC+2; 
 
 
 IF NC=2 
 
 THEN DO; CALL UP_DATE_ 
 
 .h; 
 
 
 GO TO OUT1 ; 
 
 end; 
 
 L2 :DQ JIU2 
 
 )=NI(1 )+l JQ N; 
 
 
 IF NC=3 
 
 THEN DO; CALL UP_DATE_ 
 
 .h; 
 
 
 GO TO OUT2 ; 
 
 end; 
 
 L3 :D0 NK3 
 
 )=NI(2 1 + 1 TO n; 
 
 
 IF NOV 
 
 THEN DO; CALL UP_DATE_ 
 
 .h; 
 
 
 GO TO 0UT3 ; 
 
 end; 
 
 L4 :DQ lUli 
 
 ) = NU3_ 1 + 1 TO n; 
 
 
 IF NC=5 
 
 THEN DO; CALL UP-DATE. 
 
 .h; 
 
 
 GO TO OUT4 ; 
 
 END; 
 
 L5 :D0 NU5 
 
 )=NI(4 )+l TO n; 
 
 
 IF NC=6 
 
 THEN DO; CALL UP-DATE. 
 
 .h; 
 
 
 GO TO OUT 5 ; 
 
 end; 
 
 L6 :D0 M(6 
 
 1-NK5 1+1 TO N; 
 
 
 IF NC=7 
 
 THEN DO; CALL UP_DATE_ 
 
 _h; 
 
 
 GO TO 0UT6 ; 
 
 end; 
 
 L7 :D0 NI(7 
 
 )=NI (6 )+l TO N; 
 
 
 IF NC = 8 
 
 THEN DO; CALL UP-DATE. 
 
 _h; 
 
 
 GO TO 0UT7 ; 
 
 end; 
 
 L8 :D0 NH8 
 
 )=NI(7 1+1 TO N; 
 
 
 IF NC=9 
 
 THEN DO; CALL UP-DATE. 
 
 _h; 
 
 
 GO TO CUT8 ; 
 
 end; 
 
 0UT8 : 
 
 END 
 
 L8 ; 
 
 
 GUT7 : 
 
 : END 
 
 L7 ; 
 
 
 0UT6 ' 
 
 : END 
 
 L6 ; 
 
 
 0UT5 : 
 
 END 
 
 L5 ; 
 
 
 0UT4 ' 
 
 : END 
 
 L4 ; 
 
 
 0UT3 : 
 
 : END 
 
 L3 ; 
 
 
 0UT2 
 
 : END 
 
 12 ; 
 
 
 0UT1 . 
 
 : END 
 
 LIS 
 
 
 END LO; 
 
 
 
 PUT SKIP DATA(STEPIN) ,MC); 
 
 
 DO 1=1 
 
 TO MC-l; PUT EDIT(P(I))(F( 5) 1 ; 
 
 END; 
 
 
 
 END E 
 
 KHAUST 
 
 • 
 
 
 end; 
 
%