Extended Euclidean Algorithm
Author: Benjamin Qi
Prerequisites
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Euclidean Algorithm
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The original Euclidean Algorithm computes and looks like this:
C++
1ll euclid(ll a, ll b) {2 for (;b;swap(a,b)) {3 ll k = a/b;4 a -= k*b;5 }6 return a; // gcd(a,b)7}
Extended Euclidean Algorithm
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The extended Euclidean algorithm computes integers and such that
We can slightly modify the version of the Euclidean algorithm given above to return more information!
C++
1array<ll,3> extendEuclid(ll a, ll b) {2 array<ll,3> x = {1,0,a}, y = {0,1,b};3 // we know that 1*a+0*b=a and 0*a+1*b=b4 for (;y[2];swap(x,y)) { // run extended Euclidean algo5 ll k = x[2]/y[2];6 F0R(i,3) x[i] -= k*y[i];7 // keep subtracting multiple of one equation from the other8 }9 return x; // x[0]*a+x[1]*b=x[2], x[2]=gcd(a,b)10}
Recursive Version
C++
1ll euclid(ll a, ll b) {2 if (!b) return a;3 return euclid(b,a%b);4}
becomes
C++
1pl extendEuclid(ll a, ll b) { // returns {x,y}2 if (!b) return {1,0};3 pl p = extendEuclid(b,a%b); return {p.s,p.f-a/b*p.s};4}
The pair will equal to the first two returned elements of the array in the iterative version. Looking at this version, we can prove by induction that when and are distinct positive integers, the returned pair will satisfy and . Furthermore, there can only exist one pair that satisfies these conditions!
Note that this works when are quite large (say, ) and we won't wind up with overflow issues.
Application: Modular Inverse
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It seems that when multiplication / division is involved in this problem, .
C++
1ll invGeneral(ll a, ll b) {2 array<ll,3> x = extendEuclid(a,b);3 assert(x[2] == 1); // gcd must be 14 return x[0]+(x[0]<0)*b;5}67int main() {8 FOR(b,1,101) F0R(a,101) if (__gcd(a,b) == 1) {9 ll x = invGeneral(a,b);10 assert((a*x-1)%b == 0);
Application: Chinese Remainder Theorem
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Module Progress:
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