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Lehmer GCD 五個停止條件

張圻毓

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Outline

Lehmer[1938]

Collins[1980]

Jebelean[1993]

Vallee[2004]

Wang[2003]

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Lehmer[1938]

q=

q’=

If q ≠ q’ stop

)/( 1 ii avau

1/ ii bvbu

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Example

U = 768,454,923 V = 542,167,814 b = 103

New U = 89,593,596 V = 47,099,917

x’ y’ x” y” ax0+by0 cx0+dy0 q’ q”

769 542 768 543 1x0+0y0 0x0+1y0 1 1

542 227 543 225 0x0+1y0 1x0-1y0 2 2

227 88 225 93 1x0-1y0 -2x0+3y0 2 2

88 51 93 39 -2x0+3y0 5x0-7y0 1 2

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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

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ExampleU = 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

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Exampleq3= 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

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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

0a

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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

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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

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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|

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Wang[2003]