an analysis of lehmer’s euclidean gcd algorithm jonathan sorenson proceedings of the 1995...
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An Analysis of Lehmer’s Euclidean GCD Algorithm
Jonathan SorensonProceedings of the 1995 international symposium on Symboli
c and algebraic computation
Speaker: 張圻毓
Outline
An example for Lehmer’s GCD An example for Algorithm ML Analysis Algorithm ML Experiment Conclusion
An example for Lehmer’s GCD
x0= 768,454,923 y0= 542,167,814 b = 103
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
An example for Lehmer’s GCD
x1= -2x0+3y0= 89,593,596 y1= 5x0-7y0= 47,099,917
GCD(x0,y0)= GCD(x1,y1)
x’ y’ x” y” ax1+by1 cx1+dy1 q’ q”
90 47 89 48 1x1+0y1 0x1+1y1 1 1
47 43 48 41 0x1+1y1 1x1-1y1 1 1
43 4 41 7 1x1-1y1 -1x1+2y1 10 5
An example for Lehmer’s GCD
x2= 1x1-1y1= 42,493,679 y2= -1x1+2y1 = 4,606,238
GCD(x0,y0)= GCD(x1,y1)= GCD(x2,y2)
x’ y’ x” y” ax2+by2 cx2+dy2 q’ q”
43 4 42 5 1x2+0y2 0x2+1y2 10 8
An example for Lehmer’s GCD
x3= y2= 4,606,238 y3= x2 mod y2 = 1,037,537
GCD(x0,y0)= GCD(x1,y1)= GCD(x2,y2)= GCD(x3,y3)
x’ y’ x” y” ax3+by3 cx3+dy3 q’ q”
5 1 4 2 1x3+0y3 0x3+1y3 5 2
An example for Lehmer’s GCD
x4= y3= 1,037,537 y4= x3 mod y3 = 456,090
GCD(x0,y0)= … = GCD(x4,y4)
x’ y’ x” y” ax4+by4 cx4+dy4 q’ q”
2 0 1 1 1x4+0y4 0x4+1y4 - -
An example for Lehmer’s GCD
x5= y4= 456,090 y5= x4 mod y4 = 125,357
GCD(x0,y0)= … = GCD(x5,y5)
x’ y’ x” y” ax5+by5 cx5+dy5 q’ q”
457 125 456 126 1x5+0y5 0x5+1y5 3 3
125 82 126 78 0x5+1y5 1x5-3y5 1 1
82 43 78 48 1x5-3y5 -1x5+4y5 1 1
43 39 48 30 -1x5+4y5 2x5-7y5 1 1
39 4 30 18 2x5-7y5 -3x5+11y5
9 1
An example for Lehmer’s GCD
x6= 2x5-7y5 = 34,681 y6= -3x5+11y5 = 10,657
GCD(x0,y0)= … = GCD(x6,y6)
x’ y’ x” y” ax6+by6 cx6+dy6 q’ q”
35 10 34 11 1x6+0y6 0x6+1y6 3 3
10 5 11 1 0x6+1y6 1x6-3y6 2 11
An example for Lehmer’s GCD
x7= 0x6+1y6 = 10,657 y7= 1x6-3y6 = 2,710
GCD(x0,y0)= … =GCD(x7,y7)
x’ y’ x” y” ax7+by7 cx7+dy7 q’ q”
11 2 10 3 1x7+0y7 0x7+1y7 5 3
An example for Lehmer’s GCD
x8= y7 = 2,710 y8= x7 mod y7 = 2,527
GCD(x0,y0)= … =GCD(x8,y8)
x’ y’ x” y” ax8+by8 cx8+dy8 q’ q”
3 2 2 3 1x8+0y8 0x8+1y8 1 0
An example for Lehmer’s GCD
x9= y8 = 2,527 y9= x8 mod y8 = 183
GCD(x0,y0)= … =GCD(x9,y9)
Euclid’s Algorithm
…
An example for Algorithm ML
x0= 768,454,923 y0= 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);
An example for Algorithm ML If i is even then done = b ＜ -Xi+1 or a – b ＜ Yi+1-Yi ;
else done = b ＜ -Yi+1 or a – b ＜ Xi+1-Xi;
a=
768
b=
542
(xi,yi)
(1,0)
(xi+1,yi+1)
(0,1)
(xi+2,yi+2)
(1,-1)
Q=
1
比較否
542 226 (0,1) (1,-1) (-2,3) 2 否
226 90 (1,-1) (-2,3) (5,-7) 2 否
90 46 (-2， 3) (5,-7) (-7,10) 1 否
46 44 (5,-7) (-7,10) (12,-17) 1 否
44 2 (-7,10) (12,-17) 1 2＜ 17
An example for Algorithm ML
U = 47,099,917 V = 42,493,679
a=
470
b=
424
(xi,yi)
(1,0)
(xi+1,yi+1)
(0,1)
(xi+2,yi+2)
(1,-1)
Q=
1
比較否
424 46 (0,1) (1,1) (-9,10) 9 否46 10 (1,-1) (-9,10) (37,-41) 4 否10 6 (-9,10) (37,-41) 1 6<37
An example for Algorithm ML
U = 4,606,238 V = 1,037,537
a=
460
b=
103
(xi,yi)
(1,0)
(xi+1,yi+1)
(0,1)
(xi+2,yi+2)
(1,-1)
Q=
4
比較否
103 48 (0,1) (1,-1) (-2,9) 2 否48 7 (1,-1) (-2,9) (13,-58) 6 否7 6 (-2,9) (13,-58) 1 6 ＜ 5
8
An example for Algorithm ML
U = 456,090 V = 125,357
a=
456
b=
125
(xi,yi)
(1,0)
(xi+1,yi+1)
(0,1)
(xi+2,yi+2)
(1,-3)
Q=
3
比較否
125 81 (0,1) (1,-3) (-1,4) 1 否81 44 (1,-3) (-1,4) (2,-7) 1 否44 37 (-1,4) (2,7) (-3,11) 1 否37 7 (2,7) (-3,11) (17,-62) 5 否7 2 (-3,11) (17,-62) 3 2 ＜ 6
2
An example for Algorithm ML
U = 34,681 V = 10,657
a=
346
b=
106
(xi,yi)
(1,0)
(xi+1,yi+1)
(0,1)
(xi+2,yi+2)
(1,-3)
Q=
3
比較否
106 28 (0,1) (1,-3) (-3,10) 3 否28 22 (1,-3) (-3,10) (4,-13) 1 否22 6 (-3,10) (4,-13) 3 6<13
An example for Algorithm ML
U =2,710 V = 2,527
a=
271
b=
252
(xi,yi)
(1,0)
(xi+1,yi+1)
(0,1)
(xi+2,yi+2)
(1,-1)
Q=
1
比較否
252 19 (0,1) (1,-1) (-13,14) 13 否19 5 (1,-1) (-13,14) 3 5<14
An example for Algorithm ML
U = 2527 V = 183 計算完畢即為 183
a=
252
b=
18
(xi,yi)
(1,0)
(xi+1,yi+1)
(0,1)
(xi+2,yi+2)
(1,-14)
Q=
14
比較否
18 0 (0,1) (1,-14) 0 否
Analysis Algorithm ML
ai+1 |v≧ i+1| and ai+1 – ai |v≧ i+1+vi| ,for all i k≦
Algorithm ML computes gcd(u,v), u>v using at most
iterations of its main loop.
)log
( kk
vO
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
)loglogloglog
( 2kuvkk
vuO
)2(log 2kkuO
4)(log n
)log( 2 nnO )(nO
Experiment
Experiment
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。