On the circuit complexity of the standard and the Karatsuba methods of multiplying integers

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1 On the circit complexity of the standard and the Karatsba methods of mltiplying integers arxiv: v1 [cs.ds] 7 Feb 2016 Igor S. Sergeev The goal of the present paper is to obtain accrate estimates for the complexity of two practical mltiplication methods: standard (school) and Karatsba [1]. Here, complexity is the minimal possible nmber of gates in a logic circit implementing the reqired fnction over the basis {AND, OR, XOR, NAND, NOR, XNOR}. One can find pper estimates for the said methods e.g. in [2]. The standard method has complexity M(n) 6n 2 8n. In the case n = 2 k, the complexity K(n) of the Karatsba method can be dedced from the recrsion K(2n) 3K(n)+49n 8 as K(2 k ) k 49 2 k +4. We intend to show that the above estimates may be improved with the help of the reslt [3] stating that a sm of n bits may be compted via 4.5n operations instead of 5n as in the naive approach. The reslting bonds are M(n) 5.5n 2 6.5n 1+(n mod 2) and K(2 k ) k 38 2 k 81 5 Φ k Φ k+1 +20, (1) where {Φ k } is the Fibonacci seqence: Φ 1 = Φ 2 = 1, Φ k+2 = Φ k+1 +Φ k. Axiliary circits. The circits below are bilt from the following sbcircits: half-adders HA, HA ±, (3,2)-carry save adders (CSA) FA 3, FA 3, FA 0 3, SFA 3, SFA 3, (5,3)-CSA MDFA, MDFA (the MDFA circit was proposed in [3]). Specifically, they implement the fnctions: HA: (x 1, x 2 ) (; v), where x 1 +x 2 = 2+v; HA ± : (x 1, x 2 ) (; v), where x 1 x 2 = 2+v; FA 3 : (x 1, x 2, x 3 ) (; v), where x 1 +x 2 +x 3 = 2+v; FA 3: (x 1, x 2, x 3 ) (; v), where x 1 +x 2 x 3 = 2 v; FA 0 3 : (x 1, x 2, x 3 ) (; v), wherex 1 +x 2 x 3 = 2+v, ifx 1 +x 2 x 3 0; Translated from the Rssian original pblished in: Proc. of XXII Conf. Information means and technology (Moscow, November 18 20, 2014). Vol. 3. Moscow, MPEI, 2014, isserg@gmail.com 1

2 SFA 3 : (x 1, x 1 x 2, x 3 ) (; v), where x 1 +x 2 +x 3 = 2+v; SFA 3 : (x 1, x 1 x 2, x 3 ) (; v), where x 1 x 2 +x 3 = 2 v; MDFA: (x 1, x 1 y 1, x 2, x 2 y 2, z) ( 1, 1 2 ; v), where x 1 +y 1 + x 2 +y 2 +z = 2( )+v; MDFA : (x 1, x 1 y 1, x 2, x 2 y 2, z) ( 1, 1 2 ; v), where x 1 y 1 + x 2 y 2 +z = 2( 1 2 )+v. These circits are shown on Fig. 1. Gates AND,OR,XOR are denoted by symbols,,, respectively. Inverted inpts are marked by small circles. v HA x 2 x 3 x 1 x 1 z x 1 y 1 FA 3 x 2 x 2 y 2 MDFA 1 v v HA ± x 3 x 1 z x 1 y 1 FA 3 x 2 x 2 y 2 MDFA 1 v 1 2 x 1 x 3 x 1 x 2 SFA 3 v FA x 3 v Figre 1: Axiliary circits x 1 x 3 x 1 x 2 SFA 3 v Standard method. The first stage of the standard method of mltiplication of n-bit integers involves n 2 bit mltiplications. The next stage mltiple addition may be performed via smmation of bits in consective colmns (colmn is a set of bits of the same order). The smmation tilizes the aforementioned axiliary sbcircits. The reslt of a colmn smmation 2

3 is a bit of the prodct and a set of carries to the next order. Let s index colmns from 1 in increasing order. Then, after the first phase one has n n k bits in a k-th colmn, k = 1,...,2n 1. Consider the following rle of colmn smmation: if there exist 5 smmand bits, se MDFA; else, if there exist 3 smmand bits, se FA 3 ; else, if there exist 2 smmand bits, se HA. Denote by h(k) nmber of smmand bits in the k-th colmn after completion of smmation in all lower-order colmns. Clearly, h(1) = 1. Let s check by indction that h(k) = 2k 2 for 2 k n. Obviosly, the statement holds for k = 2. Assme, it also holds for k = t and consider smmation in the t-th colmn. By the declared strategy, smmation of 2t 2 bits involves t/2 1 circits MDFA, one circit HA, and in the case of odd t, one more circit FA 3. In total, it prodces t 1 carries to the next order. Hence, h(t+1) = t+1+t 1 = 2(t+1) 2, as reqired. By analogy, we conclde that h(n+1) = 2n 2 and h(2n k) = 2k+1 for 0 k n 2. For smmation in the (n+1)-th colmn one se the same set of circits as for the n-th colmn. For smmation in (2n k)-th colmn we se k/2 circits MDFA, and in the case of odd k 1 an additional circit FA 3. For k = 1 we need a circit SFA 3 instead of FA 3, since smmation in the previos colmn involves MDF A. The se of MDFA reqires a conversion to the special bit encoding (x,y) (x,x y). All MDFA otpts encoded this way may be connected to MDFA inpts of the same encoding, with the exception of the last MDFA, in the (2n 2)-th colmn. Therefore, to execte all smmations we need additionally q+1 XOR gates, where q is a nmber of MDFA. Therefore, if n 4, then the second stage of mltiplication tilizes n circits HA, n 3+2(n mod 2) circits FA 3, one circit SFA 3, q = (n 2 3n)/2+1 (n mod 2) circits MDFA and q +1 XOR gates (the nmber of MDFA is easy to derive from the nmber of (3,2)-CSA, since MDFA redces the total nmber of smmand bits by 2, FA 3 or SFA 3 redces it by 1; the nmber of smmand bits before smmation stage is n 2, and at the end it is 2n). Smming p the complexities of sbcircits we can bond the complexity M(n) of the mltiplication circit as M(n) 5,5n 2 6,5n 1+(n mod 2). The estimate holds also for 2 n 3. Karatsba method. Represent two m-bit mltiplication operands as A 1 2 n +B 1 and A 2 2 n +B 2, where n = m/2, 0 B i < 2 n, 0 A i < 2 m n. Then, the prodct may be compted by the formla: A 1 A 2 2 2n +((A 1 +B 1 )(A 2 +B 2 ) A 1 A 2 B 1 B 2 )2 n +B 1 B 2. 3

4 The implied circit consists of two addition circits compting A 1 +B 1 and A 2 + B 2, three mltiplication sbcircits for (n + 1)-bit, (m n)-bit and n-bit operands, and a sbcircit for the final addition-sbtraction. The strctre of this final addition is shown on the pattern below (see Fig. 2). Symbols + and denote smmand and sbtrahend bits, respectively. Colmns are indexed so that an index i corresponds to a bit with weight 2 i. Pairs of bits in brackets are missing when m is odd. (++) A 1 A 2 2 2n (A 1 +B 1 )(A 2 +B 2 )2 n ( ) A 1 A 2 2 n B 1 B 2 2 n B 1 B 2 4n 1 3n 1 2n 1 n 0 Fig. 2. Pattern of the final addition-sbtraction in the Karatsba method Consider the following colmn smmation rle for the final addition: if there exist 3 smmand bits and 2 sbtrahend bits, se MDFA ; else, if there exist 3 bits, se a sitable circit from the FA 3 family; else, if there exist 2 bits, se an appropriate circit from the HA family. Denote by h + (k), h (k) nmbers of smmand and sbtrahend bits in the k-th colmn after completion of smmation in all lower-order colmns. One can easily verify that h + (n) = h (n) = 2, h + (n+1) = h (n+1) = 3 and h + (k) = 3, h (k) = 4 for n+2 k 2m n 1. We se one FA 3 and one HAinthen-thcolmn, onemdfa andoneha ± inthe(n+1)-thcolmn, one MDFA and one FA 3 in any sbseqent colmn p to (2m n 1)-th. When m is odd, we have h + (k) = h (k) = 3 for k = 3n 1, 3n 2. Therefore, one MDFA and one HA ± shold be sed in the corresponding colmns. Frther, h + (3n) = 3, h (3n) = 2(seMDFA ), h + (3n+1) = 3, h (3n+ 1) = 1 (se SFA 3 and HA± ), h + (3n + 2) = 2, h (3n+2) = 1 (se FA 0 3, since the vale (A 1 +B 1 )(A 2 +B 2 ) A 1 A 2 B 1 B 2 is non-negative). At last, h + (k) = 2, h (k) = 0 for 3n+3 k 2m 1: se HA elsewhere, bt se XOR in the most significant colmn, since no carry is reqired there. As in the standard method, conversion to the special pair-of-bits encoding reqires q + 1 additional XOR gates, where q is the nmber of MDFA sbcircits. Now, we re going to estimate the complexity K(m) of the mltiplication circit. It s a common knowledge, that the complexity of the addition of two n-bit nmbers is 5n 3, the addition of an n-bit nmber and an (n 1)- bit nmber has complexity 5n 6 (see e.g. [2]). If m 10, then the final 4

5 Karatsba adder-sbtractor contains n 1 circits HA or HA ±, 2m 2n 1 circits FA 3, one circit SFA 3, one circit FA0 3, 2n circits MDFA and 2n+2 XOR gates. Note, that one can save some gates. Colmns indexed by n+i and 2n+i, for i = 0,...,n 1, contain identical pairs of bits (from smmands B 1 B 2, A 1 A 2 2 n and B 1 B 2 2 n, A 1 A 2 2 2n, respectively). Arrange the comptation process to pass these bits to the inpts of FA 3 in colmns indexed by n and n+2,...,2m n 1, and to the inpts of MDFA encoded by (x, x y) in other colmns (that is, in (n + 1)-th colmn, and in the case of odd m, also in (3n 2)-th and (3n 1)-th colmns). Ths, for i = 0 and 2 i n 1 2(m mod 2) we can save two gates XOR and ANDNOT via exploiting a CSA from Fig. 3a in a lower-order colmnandacsafromfig. 3binahigher-ordercolmn(CSA saredifferent since the signs of bits in low-order and high-order colmns are rearranged). For i = 1 and n 2(m mod 2) i n 1, one XOR gate is to be saved. x 1 x 3 x 2 Fig. 3a x 3 x 1 x 2 Fig. 3b So, the following recrrent formlae hold: K(2n 1) K(n+1)+K(n)+K(n 1)+38n 16, n 6, K(2n) K(n+1)+2K(n)+38n 2, n 5. (2) A halving iteration of the Karatsba method provides an advantage when m = 16 or m 18. Bonds on the complexity L(m) of mltiplication of m- bit nmbers for m 18 are collected in the Table 1 (symbol indicates vales obtained via Karatsba method). For the convenience of comparison, let s derive the complexity of the Karatsba mltiplication circit in an explicit form for m = 2 k. Denote K(2 k +2) X k = K(2 +1), A = 1 1 1, 38 2 k +36 b k = K(2 k ) k 2 5

6 Table 1: Bonds for the complexity of mltiplication of m-bit nmbers m L(m) m L(m) Recrrences (2) imply X k+1 AX k +b k for k 4. Via common calclations, we obtain (1) as the soltion of the latter ineqality (initial vales of the complexity shold be taken from the Table 1). Research spported in part by RFBR, grant a. References [1] A.A. Karatsba, Y.P. Ofman. Mltiplication of mltidigit nmbers on atomata. DAN USSR 145(2)(1962), (in Rssian). Eng. transl. in Soviet Phys. Dokl. 7 (1963), [2] S.B. Gashkov. Entertaining compter arithmetic. Fast algorithms for operations with nmbers and polynomials. Moscow, Librocom, 2012 (in Rssian). [3] E. Demenkov, A. Kojevnikov, A. Klikov, G. Yaroslavtsev. New pper bonds on the Boolean circit complexity of symmetric fnctions. Inf. Proc. Letters 110(7) (2010),