Euclidean Algorithm

Euclidean Algorithm

欧几里得算法也称辗转相除法，运用于求解两数的最大公因子

为了避免只会用库函数的尴尬境况，还是来复习一下它的原理

对于两个数a，b如何求解a，b的最大公因数呢？

欧几里得给出了如下算法

a=bq₁+r₁

b=r₁q₂+r₂

r₁=r₂q₃+r₃

···

r_n-1=r_nq_n+1

最后得到gcd(a,b)=r_n

but why?

首先算法必在有限步骤内结束，因为每一步的余数都在递减，有限步骤后为0

不妨设gcd(a,b)=d

=>d|a-bq₁ 即d|r₁

同样操作下去=>d|r_k直到r_k=0

于是我们可以得到d|r_n

现在我们从后往前由r_n|r_n-1=>r_n|r_n-2

=>r_n|a;r_n|b=>r_n|d

综上r_n=d

代码实现

def gcd(a,b):
    r=a%b
    if r:
        return gcd(b,r)
    else:
        return b

ps:

We also can think of (Z,+) to understand gcd

Let a,b ∈Z not both 0,then Z×a+Z×b=Z×d

a)d|a&d|b

b)if an integers e divides both a and b,it also divides d

c)∃r,s: d=ra+sb

pf:

1)∵a∈Z×d&b∈Z×d

=>d|a&d|b

2)e|a&e|b=>e|ra+sb

∵d∈Z×d=>d=k₁a+k₂b

=>e|d

3)by the conditions it's obvious

---

We know that d is the greatest common divisor of a and b

meanwhile we learn some property of d like c)

And then we prove the Euclidean algorithm use group structure

suppose we have a=qb+z

=>for any integers r,s: ra+sb=r(qb+z)+sb=(rq+s)b+rz

=>Z×a+Z×b⊆Z×b+Z×z

in a similar way

for any integers r,s: rb+sz=rb+s(a-qb)=sa+(r-sq)b

=>Z×b+Z×z⊆Z×a+Z×b

in conclusion Z×a+Z×b=Z×b+Z×z

on the basis of group theory wo know

Z×a+Z×b=Z×d₁

Z×b+Z×z=Z×d₂

=>d₁=d₂ means gcd(a,b)=gcd(b,z)(※)

(※) is the essence of Euclidean algorithm

Extended Euclidean Algorithm

前面我们用群论的知识证明了这样一个事实

对于不全为0的数a，b存在r，s使得

ra+sb=gcd(a,b)

欧几里得拓展算法就是描述的寻找这样的r与s的过程

首先我们有

r₁a+s₁b=gcd(a,b)

同样的方式可得

r₂b+s₂(a%b)=gcd(b,a%b)

由欧几里得算法=>r₁a+s₁b=r₂b+s₂(a%b)

a%b=a-[\(\frac{a}{b}\)]×b

=>r₁a+s₁b=r₂b+s₂(a-[\(\frac{a}{b}\)]×b)=s₂a+(r₂-s₂×[\(\frac{a}{b}\)])b

对比系数我们很容易得到一个寻找的方案

即r₁=s₂，s₁=r₂-s₂×[\(\frac{a}{b}\)]

由于欧几里得算法能在有限步骤内完成，可知扩展欧几里得算法也是在有限步骤内完成，完成时r=0;s=1

代码实现

def xgcd(a,b):
    if a%b==0:
        return 0,1
    else:
        r,s=xgcd(b,a%b)
        return s,r-s*(a//b)