Lehmer GCD 五個停止條件

1. Lehmer GCD五個停止條件 張圻毓

2. Outline Lehmer[1938] Collins[1980] Jebelean[1993] Vallee[2004] Wang[2003]

3. Lehmer[1938] q= q’= If q ≠ q’ stop

4. Example U = 768,454,923 V = 542,167,814 b = 103 New U = 89,593,596 V = 47,099,917

5. Collins[1980] & Jebelean[1993] vi < |bi+1| or ui - vi < |bi+1 - bi| If i 為奇數:vi < - bi+1 or ui – vi < ai+1 - ai If i 為偶數:vi < - ai+1 or ui – vi < bi+1 - bi

6. Example U = 768,454,923 V = 542,167,814 q1= 768/542 = 1 (a0,b0) = (1,0) (a1,b1) = (0,1) (a2,b2) = (1,0) – (0,1) = (1,-1) New (u,v) = (542 , 226) 判斷odd vi < - bi+1 or ui – vi < ai+1 – ai 不合 q2= 542/226 = 2 (a3,b3) = (0,1) – 2(1,-1) = (-2,3) New (u,v) = (226 , 90) 判斷even vi < - ai+1 or ui – vi < bi+1 – bi 不合

7. Example q3= 226/90= 2 (a4,b4) = (1,-1) – 2(-2,3) = (5,-7) New (u,v) = (90 , 46) 判斷odd vi < - bi+1 or ui – vi < ai+1 – ai 不合 q4= 90/46 = 1 (a5,b5) = (-2,3) – 1(5,-7) = (-7,10) New (u,v) = (46 , 44) 判斷even vi < - ai+1 or ui – vi < bi+1 – bi合

8. Vallee[2004] If aj > then Qi=qi for all i ≦ j-2 Example: u = 768 v = 542 (a0,b0) = (1,0) (a1,b1) = (0,1) While 542 > √768(≒27) do q1 = u div v = 1 new u = u mod v = 226 a2 = -a1q1+a0 = 1 b2 = -b1q1+b0 = -1 i+1=2

9. Example u = 542 v = 226 While 226 > √768 do q2 = u div v = 2 new u = 90 a3 = -2 b3 = 3 u = 226 v = 90 While 90 > √768 do q3 = u div v = 2 new u = 46 a4 = 5 b4 = -7

10. Example u = 90 v = 46 While 46 > √768 do q4 = u div v = 1 new u = 44 a5 = -7 b5 = 10 u = 46 v = 44 While 44 > √768 do q5 = u div v = 1 new u = 2 a6 = 12 b6 = -17 while 2 < √768 stop

11. Wang[2003] New ui+2 ≧ |qi+1| or New Ui+2 ≧ λ|Qi+1| New u ≧2|qi+2|*|qi+1| or m ≧2λ|Qi+2|*|Qi+1|

12. Wang[2003]