1. 1 Two Fast GCD Algorithms JONATHAN SORENSON Department of Mathematics and Computer Science Butler University Journal of algorthms 報告者：張圻毓

2. 2 Outline The right-shift Binary The right-shift k-ary GCD The left-shift Binary The left-shift k-ary GCD Experiment conclusion

3. 3 The right-shift Binary 1.g=1 while u and v are both even do 2. g=2g,u=u/2,v=v/2 3. while u≠0 and v ≠0 do 4. if u is even,u=u/2 5. else if v is even,v=v/2 6. else max{u,v}=|u-v|/2 return(g(u+v)) 濃縮版 1. if u and v are both even, gcd(u,v)=2gcd(u/2,v/2) 2. if u is even and v is odd, gcd(u,v)=gcd(u/2,v) 3. if u and v are both odd, gcd(u,v)=gcd(|u-v|/2,v) 4. gcd(u,0)=u

4. 4 The right-shift k-ary GCD 1.while u≠0 and v≠0 do: 2. if gcd(u,k)>1,u=u/gcd(u,k) 3. else if gcd(v,k)>1,v=v/gcd(v,k) 4. else 5. find nonzero integers a,b satisfying au+bv=0 (mod k) 6. max{u,v}=|au+bv|/k

5. 5 Example u=561 v=627 取掉 K=7 以下的 GCD 為 3 u=209 v=187 k=7 209 mod k=6,187 mod k=5 a=5,b=-6 使得滿足 5. |5u-6v|/7=11 u=11 v=187 k=7 11 mod k=4, 187 mod k=5 a=5, b=-4 |5u-4v|/7=99

6. 6 Example u=11 v=99 k=7 11 mod k=4,99 mod k=1 a=-1,b=4 |-1u+4v|/7=55 u=11 v=55 11 mod k=4, 55 mod k=6 a=-6, b=4 |-6u+4v|/7=22

7. 7 Example u=11 v=22 k=7 11 mod k=4,22 mod k=1 a=-1,b=4 |-1u+4v|/7=11 u=11 v=11 11 mod k=4, 11 mod k=4 a=1, b=-1 |u-v|/7=0 u=11 v=0 11*3=33 GCD=33

8. 8 The left-shift Binary 1.if u<v then swap(u,v) 2.while v≠0 do 3. compute t=2 e v such that t ≦ u<2t 4. u=min{u-t,2t-u} 5. if u<v then swap(u,v) 6.return(u); 濃縮版 1.gcd(u,v)=gcd(|u+bv|,v),while b is any interger 2.gcd(u,0)=u

9. 9 Example u=627 v=561 e=0 t=2 e v=561 t ≦ u 561 ≦ 627<2*561 u=min{u-t,2t-u}=66 u=561 v=66 e=3 t=2 3 *66 528 ≦ 561<16*66 u=min{u-t,2t-u}=33 u=66 v=33 e=1 t=2*33=66 66 ≦ 66<2*66 u=min{u-t,2t-u}=0 u=33 v=0

10. 10 The left-shift k-ary GCD 1.if u<v then swap(u,v) 2.while v≠0 do 3. compute t=k e v such that t ≦ u ﹤ kt 4. find a,b ≦ k such that t/u=b/a u=|at-bu| 5. if u<v then swap(u,v)

11. 11 Example u=561 v=627 取掉 K=7 以下的 GCD 為 3 u=209 v=187 e=0 t=187 187 ≦ 209<7*187 演算法 3. t/u=187/209 ≒ 0.89 … ≒ 1/1=b/a u=|1*t-1*u|=22 u=187 v=22 e=1 t=7*22=154 154 ≦ 187<7*154 t/u=154/187 ≒ 0.82 … ≒ 1/1=b/a u=|1*t-1*u|=33

12. 12 Example u=33 v=22 e=0 t=22 22 ≦ 33<7*22 t/u=22/33 ≒ 0.66 … ≒ 3/4=b/a u=|4*t-3*u|=11 u=22 v=11 e=0 t=11 11 ≦ 22<7*11 t/u=11/22=0.5=1/2=b/a u=|2*t-1*u|=0 u=11 v=0 11*3=33 GCD=33

13. 13 Experiment conclusion