1. An Analysis of Lehmer’s Euclidean GCD Algorithm Jonathan Sorenson Proceedings of the 1995 international symposium on Symbolic and algebraic computation Speaker: 張圻毓

2. Outline  An example for Lehmer’s GCD  An example for Algorithm ML  Analysis Algorithm ML  Experiment  Conclusion

3. An example for Lehmer’s GCD x 0 = 768,454,923 y 0 = 542,167,814 b = 10 3 x’y’x”y”ax 0 +by 0 cx 0 +dy 0 q’q” 7695427685431x 0 +0y 0 0x 0 +1y 0 11 5422275432250x 0 +1y 0 1x 0 -1y 0 22 22788225931x 0 -1y 0 -2x 0 +3y 0 22 88519339-2x 0 +3y 0 5x 0 -7y 0 12

4. An example for Lehmer’s GCD x 1 = -2x 0 +3y 0 = 89,593,596 y 1 = 5x 0 -7y 0 = 47,099,917 GCD(x 0,y 0 )= GCD(x 1,y 1 ) x’y’x”y”ax 1 +by 1 cx 1 +dy 1 q’q” 904789481x 1 +0y 1 0x 1 +1y 1 11 474348410x 1 +1y 1 1x 1 -1y 1 11 4344171x 1 -1y 1 -1x 1 +2y 1 105

5. An example for Lehmer’s GCD x 2 = 1x 1 -1y 1 = 42,493,679 y 2 = -1x 1 +2y 1 = 4,606,238 GCD(x 0,y 0 )= GCD(x 1,y 1 )= GCD(x 2,y 2 ) x’y’x”y”ax 2 +by 2 cx 2 +dy 2 q’q” 4344251x 2 +0y 2 0x 2 +1y 2 108

6. An example for Lehmer’s GCD x 3 = y 2 = 4,606,238 y 3 = x 2 mod y 2 = 1,037,537 GCD(x 0,y 0 )= GCD(x 1,y 1 )= GCD(x 2,y 2 )= GCD(x 3,y 3 ) x’y’x”y”ax 3 +by 3 cx 3 +dy 3 q’q” 51421x 3 +0y 3 0x 3 +1y 3 52

7. An example for Lehmer’s GCD x 4 = y 3 = 1,037,537 y 4 = x 3 mod y 3 = 456,090 GCD(x 0,y 0 )= … = GCD(x 4,y 4 ) x’y’x”y”ax 4 +by 4 cx 4 +dy 4 q’q” 20111x 4 +0y 4 0x 4 +1y 4 --

8. An example for Lehmer’s GCD x 5 = y 4 = 456,090 y 5 = x 4 mod y 4 = 125,357 GCD(x 0,y 0 )= … = GCD(x 5,y 5 ) x’y’x”y”ax 5 +by 5 cx 5 +dy 5 q’q” 4571254561261x5+0y51x5+0y5 0x5+1y50x5+1y5 33 12582126780x5+1y50x5+1y5 1x5-3y51x5-3y5 11 824378481x 5 -3y 5 -1x 5 +4y 5 11 43394830-1x 5 +4y 5 2x5-7y52x5-7y5 11 39430182x5-7y52x5-7y5 - 3x 5 +11y 5 91

9. An example for Lehmer’s GCD x 6 = 2x 5 -7y 5 = 34,681 y 6 = -3x 5 +11y 5 = 10,657 GCD(x 0,y 0 )= … = GCD(x 6,y 6 ) x’y’x”y”ax 6 +by 6 cx 6 +dy 6 q’q” 351034111x6+0y61x6+0y6 0x6+1y60x6+1y6 33 1051110x6+1y60x6+1y6 1x 6 -3y 6 211

10. An example for Lehmer’s GCD x 7 = 0x 6 +1y 6 = 10,657 y 7 = 1x 6 -3y 6 = 2,710 GCD(x 0,y 0 )= … =GCD(x 7,y 7 ) x’y’x”y”ax 7 +by 7 cx 7 +dy 7 q’q” 1121031x7+0y71x7+0y7 0x7+1y70x7+1y7 53

11. An example for Lehmer’s GCD x 8 = y 7 = 2,710 y 8 = x 7 mod y 7 = 2,527 GCD(x 0,y 0 )= … =GCD(x 8,y 8 ) x’y’x”y”ax 8 +by 8 cx 8 +dy 8 q’q” 32231x8+0y81x8+0y8 0x8+1y80x8+1y8 10

12. An example for Lehmer’s GCD x 9 = y 8 = 2,527 y 9 = x 8 mod y 8 = 183 GCD(x 0,y 0 )= … =GCD(x 9,y 9 ) Euclid’s Algorithm …

13. An example for Algorithm ML x 0 = 768,454,923 y 0 = 542,167,814 令 k=3 INPUT:Intergers U,V,k U ≧ V ＞ 0 and k ＞ 0 while V ≠ 0 if len(U)-len(V) ≦ k/2 ML(U,V,K); end if; R=U mod V; U=V; V=R; end while; output(U,V);

14. An example for Algorithm ML  If i is even then  done = b ＜ -X i+1 or a – b ＜ Y i+1 -Y i ;  else  done = b ＜ -Y i+1 or a – b ＜ X i+1 -X i; a= 768 b= 542 (x i,y i ) (1,0) (x i+1,y i+1 ) (0,1) (x i+2,y i+2 ) (1,-1) Q= 1 比較 否 542226(0,1)(1,-1)(-2,3)2 否 22690(1,-1)(-2,3)(5,-7)2 否 9046 (-2 ， 3) (5,-7)(-7,10)1 否 4644(5,-7)(-7,10)(12,-17)1 否 442(-7,10)(12,-17)1 2 ＜ 17

15. An example for Algorithm ML  U = 47,099,917 V = 42,493,679 a= 470 b= 424 (x i,y i ) (1,0) (x i+1,y i+1 ) (0,1) (x i+2,y i+2 ) (1,-1) Q= 1 比較 否 42446(0,1)(1,1)(-9,10)9 否 4610(1,-1)(-9,10)(37,-41)4 否 106(-9,10)(37,-41)16<37

16. An example for Algorithm ML  U = 4,606,238 V = 1,037,537 a= 460 b= 103 (x i,y i ) (1,0) (x i+1,y i+1 ) (0,1) (x i+2,y i+2 ) (1,-1) Q= 4 比較 否 10348(0,1)(1,-1)(-2,9)2 否 487(1,-1)(-2,9)(13,-58)6 否 76(-2,9)(13,-58)1 6 ＜ 58

17. An example for Algorithm ML  U = 456,090 V = 125,357 a= 456 b= 125 (x i,y i ) (1,0) (x i+1,y i+1 ) (0,1) (x i+2,y i+2 ) (1,-3) Q= 3 比較 否 12581(0,1)(1,-3)(-1,4)1 否 8144(1,-3)(-1,4)(2,-7)1 否 4437(-1,4)(2,7)(-3,11)1 否 377(2,7)(-3,11)(17,-62)5 否 72(-3,11)(17,-62)3 2 ＜ 62

18. An example for Algorithm ML  U = 34,681 V = 10,657 a= 346 b= 106 (x i,y i ) (1,0) (x i+1,y i+1 ) (0,1) (x i+2,y i+2 ) (1,-3) Q= 3 比較 否 10628(0,1)(1,-3)(-3,10)3 否 2822(1,-3)(-3,10)(4,-13)1 否 226(-3,10)(4,-13)36<13

19. An example for Algorithm ML  U =2,710 V = 2,527 a= 271 b= 252 (x i,y i ) (1,0) (x i+1,y i+1 ) (0,1) (x i+2,y i+2 ) (1,-1) Q= 1 比較 否 25219(0,1)(1,-1)(-13,14)13 否 195(1,-1)(-13,14)35<14

20. An example for Algorithm ML  U = 2527 V = 183  計算完畢即為 183 a= 252 b= 18 (x i,y i ) (1,0) (x i+1,y i+1 ) (0,1) (x i+2,y i+2 ) (1,-14) Q= 14 比較 否 180(0,1)(1,-14)0 否

21. Analysis Algorithm ML  a i+1 ≧ |v i+1 | and a i+1 – a i ≧ |v i+1 +v i |,for all i ≦ k  Algorithm ML computes gcd(u,v), u>v using at most iterations of its main loop.

22. Analysis Algorithm ML  Algorithm ML computes gcd(u,v), u>v using at most bit operations and space.  u,v at most n bits, setting k = obtain an algorithm with a subquadratic runing time of using space

23. Experiment

25. Conclusion  K 值得取決大小適當會直接影響執行的速 率，如果取決大小不好，可能會比原來的 Lehmer’s 的演算法還慢並且多繞了一圈， 因為 Lehmer’s 在同時間已經直接 Mod 解決 不需再判斷。  運算中當 a 與 b 值瞬間拉大時，下一行運算 將會出現比較結果成立。