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CSE 246: Computer Arithmetic Algorithms and Hardware Design

CSE 246: Computer Arithmetic Algorithms and Hardware Design. Lecture 6.1 Multiplication Arithmetic. Instructor: Prof. Chung-Kuan Cheng. Topics:. Karatsuba ’ s Method (1962) Toom ’ s Method (1963) Modular Method FFT. Karatsuba ’ s Method. U=2 n U 1 +U 0 , V=2 n V 1 +V 0

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CSE 246: Computer Arithmetic Algorithms and Hardware Design

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  1. CSE 246: Computer Arithmetic Algorithms and Hardware Design Lecture 6.1 Multiplication Arithmetic Instructor: Prof. Chung-Kuan Cheng

  2. Topics: • Karatsuba’s Method (1962) • Toom’s Method (1963) • Modular Method • FFT

  3. Karatsuba’s Method • U=2nU1+U0, V=2nV1+V0 • UV= 22nU1V1+2n(U1V0+U0V1)+U0V0 = (22n+2n)U1V1+2n(U1-U0)(V0-V1)+(2n+1)U0V0 T(2n)<= 3T(n)+cn T(2k)<=c(3k-2k) T(n)=T(2lgn)<=c(3lgn-2lgn)<3cnlg3 lg3=1.585

  4. Toom’s Method • U=2rnUr+…+2nU1+U0 • V=2rnVr+…+2nV1+V0 • U(x)= xrUr+…+xU1+U0 • V(x)= xrVr+…+xV1+V0 • U(x)V(x)=W(x)= x2rW2r+…+xW1+W0 Set 2r+1 equations: W(0)=U(0)V(0) W(1)=U(1)V(1) W(2r)=U(2r)V(2r)

  5. Toom’s Method • T((r+1)n)<= (2r+1)T(n)+cn • T(n)<=cnlogr+1(2r+1)<cn1+logr+12 Theorem: Given e> 0, there exists a multiplication algorithm such that the number of elementary operation T(n) needed to multiply two n-bit numbers satisfies for some constant c(e) independent of n T(n)<c(e)n1+e

  6. Toom’s Method • U=(4,13,2)16, V=(9,2,5)16 • U(x)=4x2+13x+2, V=9x2+2x+5 • W(x)=U(x)V(x) • W(0)=10, W(1)=304,W(2)=1980 • W(3)=7084,W(4)=18526 • W(x)= x2rW2r+…+xW1+W0

  7. Toom’s Method • W(x)= x2rW2r+…+xW1+W0 • Rewrite W(x)= a2rx2r+…+a1x1+a0 where xk=x(x-1)…(x-k+1) W(x+1)-W(x)= 2ra2rx2r-1+(2r-1)a2r-1x2r-2…+a1 (W(x+2)-W(x+1))-(W(x+1)-W(x))= 2r(2r-1)a2rx2r-2+(2r-1)(2r-2)a2r-1x2r-3…+2a2

  8. Toom’s Method • W(*)=10, 304, 1980, 7084, 18526 • W’(*)=294, 1676, 5104, 11442 • W’’(*)=1382, 3428, 6338 • W’’(*)/2= 691, 1714, 3169 • W’’’(*)/2= 1023, 1455 • W’’’(*)/6= 341, 485 • W’’’’(*)/6= 144 • W’’’’(*)/24= 36 • W(x)= 36x4+341x3+691x2+294x1+10 =(((36(x-3)+341)(x-2)+691)(x-1)+294)x+10 = 36x4+125x3+64x2+69x+10

  9. Toom’s Method

  10. Toom and Cook’s Method • Theorem: There is a constant c such that the execution time of Toom and Cook’s method is less than cn23.5sqrt(lgn) cycles

  11. Modular Method (Schonhage) • Recursive formula: q0=1, qk+1=3qk-1 • Thus, we have qk=1/2(3k+1) • Relatively prime pi 6qk-1,6qk+1,6qk+2,6qk+3,6qk+5,6qk+7 • Set six moduli mi=2pi-1

  12. Modular Method • Given U and V, Find W=UxV • Compute ui=Umodmi vi=Vmodmi • Compute wi=uixvimodmi • Recover W T(n)=O(nlog36)=O(n1.631)

  13. FFT Given U(t)=(u0,u1,…uK-1),V(t)=(v0,v1,…vK-1) Find P(t)=(p0,p1,…,pK-1), where pt=sum(i+j=t modK) uivj • Set w=exp(2pi/K), i.e. wK=1 • us= sum(0<=t<K) wstut • vs= sum(0<=t<K) wstvt • U(s)V(s)=(u0v0,u1v1,…,uK-1vK-1) • P(s)=U(s)V(s), ps=usvs • ps= sum(0<=t<K) wstpt

  14. FFT • K>= 2n-1, un=un+1=…=uK-1=0 • vn=vn+1=…=vK-1=0 • pt=sum(i+j=t modK)uivj =utv0+ut-1v1+…+u0vt

  15. FFT (K=2k ,t=(tk-1,…,t0)) • Set A0(tk-1,…,t0)=ut ,i.e. A0(t)=ut • Set A1(sk-1,tk-2,…,t0)= A0(0,tk-2,…,t0)+w2k-1sk-1A0(1,tk-2,…,t0) • Set A2(sk-1,sk-2,tk-3,…,t0)= A1(sk-1,0,tk-3,…,t0)+ w2k-2(sk-2sk-1)2A1(sk-1,1,tk-3,…,t0) • Set Ak(sk-1,sk-2,sk-3,…,s0)= Ak-1(sk-1,…,s1,0)+ w(s0s1…sk-1)2Ak-1(sk-1,…,s1,1)

  16. FFT (K=2k ,t=(tk-1,…,t0)) • Replace tk-1 with sk-1 • sk-1 determinesw2k-1sk-1 • Replace tk-2 with sk-2 • sk-1,sk-2 determines w2k-2(sk-2sk-1)2 • Replace t0 with s0 • sk-1,sk-2,…,s0 determines w(s0s1…sk-1)2 • Binary s=(s0,s1,…,sk-1)2

  17. FFT (K=2k ,t=(tk-1,…,t0)) By induction, we have Aj(sk-1,…,sk-j,tk-j-1,…,t0)= sum(tk-1,…,tk-j)w2k-j (sk-j,…,sk-1)2 (tk-1,…,tk-j)2ut Ak(sk-1,…,s0)= sum(tk-1,…,t0) w(s0,…,sk-1)2(tk-1,…,t0)2ut =us

  18. FFT: k=2 =

  19. FFT: k=2 =

  20. FFT: k=2 =

  21. FFT: k=2 =

  22. FFT: k=3

  23. FFT: k=3

  24. FFT: k=3

  25. FFT: k=3 =

  26. FFT • us=u0+u1s+u2s2+…+u2k-1s2k-1 • us=u0+u2s2+…+u2k-2s2k-2 +u1s+u3s3+…+u2k-1s2k-1 • us= Fe(s2) + sFd(s2) Fe(s2)=u0+u2s2+…+u2k-2s2k-2 Fd(s2)=u1+u3s2+…+u2k-1s2k-1 • us= Fee(s4)+s2Fed(s4) + s[Fde(s4) +s2Fdd(s4)]

  27. FFT • us=u0+u1s+u2s2+…+u2k-1s2k-1 • us= Fee(s4)+s2Fed(s4) + s[Fde(s4) +s2Fdd(s4)] • us= Feee(s8)+ s4Feed(s8) + s2[Fede(s8)+ s4Fedd(s8)] + s{[Fdee(s8)+s4Fded(s8)]+s2[Fdde(s8)+ s4Fddd(s8)]} • Fx…x(s2k-1)= Fx…xe(s2k) + s2k-1Fx…xd(s2k)

  28. FFT • us=u0+u1s+u2s2+u3s3+u4s4+u5s5+u6s6+u7s7 • us= Fe(s2) + sFd(s2) Fe(s2)=u0+u2s2+u4s4+u6s6 Fd(s2)=u1+u3s2+u5s4+u7s6 • Fe(s2)=Fee(s4) + s2Fed(s4) Fee(s4)=u0+u4s4, Fed(s4)=u2+u6s6 • Fd(s2)=Fde(s4) + s2Fdd(s4) Fde(s4)=u1+u5s4, Fdd(s4)=u3+u7s4 • Fx(s=w0)=Fx(s=w4), Fx(s=w2)=Fx(s=w6), Fx(s=w)=Fx(s=w5), Fx(s=w3)=Fx(s=w7) x=e,d (s0,s1,s2)=(-,0,0),(-,0,1),(-,1,0),(-,1,1) • Fxx(s=w0)=Fxx(s=w2)=Fxx(s=w4)=Fxx(s=w6),Fxx(s=w)=Fxx(s=w3)=Fxx(s=w5)=Fxx(s=w7), xx=ee,ed,de,dd, (s0,s1,s2)=(-,-,0),(-,-,1)

  29. FFT (Inversion) • ur== sum(0<=s<K)wrsus = sum(0<=s,t<K)wrswstut = sum(0<=t<K)utsum(0<=s<K)ws(t+r) =Ku(-r)modK sum(0<=s<K)wsj=K if jmodK=0, 0 otherwise.

  30. FFT • 2n<=2k g< 4n, K=2k • Precision m= 6k • Let M= time of m-bit multiplication • Total time to multiply n-bit numbers O(n)+O(Mnk/g)

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