Tonelli–Shanks Algorithm 二次剩余系解法 (Ural 1132. Square Root)

wikipedia上的解释和证明：http://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm

The Tonelli–Shanks algorithm (referred to by Shanks as the RESSOL algorithm) is used within modular arithmetic to solve a congruence of the form

where n is a quadratic residue (mod p), and p is an odd prime.

Tonelli–Shanks cannot be used for composite moduli; finding square roots modulo composite numbers is a computational problem equivalent to integer factorization.^[1]

An equivalent, but slightly more redundant version of this algorithm was developed by Alberto Tonelli in 1891. The version discussed here was developed independently by Daniel Shanksin 1973, who explained:

"My tardiness in learning of these historical references was because I had lent Volume 1 of Dickson's History to a friend and it was never returned."^[2]

(Note: All  are taken to mean , unless indicated otherwise).[edit]The algorithm

Inputs: p, an odd prime. n, an integer which is a quadratic residue (mod p), meaning that the Legendre symbol .

Outputs: R, an integer satisfying .

1. Factor out powers of 2 from p − 1, defining Q and S as:  with Q odd. Note that if , i.e. , then solutions are given directly by .
2. Select a z such that the Legendre symbol  (that is, z should be a quadratic non-residue modulo p), and set .
3. Let 
4. Loop:
   1. If , return R.
   2. Otherwise, find the lowest i, , such that ; e.g. via repeated squaring.
   3. Let , and set  and .

Once you have solved the congruence with R the second solution is p − R.

Example

Solving the congruence . It is clear that  is odd, and since , 10 is a quadratic residue (by Euler's criterion).

• Step 1: Observe  so , .

• Step 2: Take  as the quadratic nonresidue (2 is a quadratic nonresidue since  (again, Euler's criterion)). Set 

• Step 3: 

• Step 4: Now we start the loop:  so ; i.e. 
  • Let , so .
  • Set . Set , and 
  • We restart the loop, and since  we are done, returning 

Indeed, observe that  and naturally also . So the algorithm yields two solutions to our congruence.

Proof

First write . Now write  and , observing that . This latter congruence will be true after every iteration of the algorithm's main loop. If at any point,  then  and the algorithm terminates with .

If , then consider , a quadratic non-residue of . Let . Then  and , which shows that the order of  is .

Similarly we have , so the order of  divides . Suppose the order of  is . Since  is a square modulo ,  is also a square, and hence .

Now we set  and with this ,  and . As before, holds; however with this construction both  and  have order . This implies that  has order  with .

If  then , and the algorithm stops, returning . Else, we restart the loop with analogous definitions of , , and  until we arrive at an  that equals 0. Since the sequence of S is strictly decreasing the algorithm terminates.

 

 

算法的大体过程已经说的很清楚，然后模拟那个过程就可以了，不过对于Ural上的这个题，要对n = 2特判一下。详见代码：

 

 

//#pragma comment(linker,"/STACK:327680000,327680000")
#include 
#include 
#include 
#include 
#include 
#include 
#include <string>
#include <set>
#include 
#include 
#include 
#include 
#include 

 

 

 POJ 1808 勒让德符号(Legendre symbol)判定

 

View Code

//#pragma comment(linker,"/STACK:327680000,327680000")
#include 
#include 
#include 
#include 
#include 
#include 
#include <string>
#include <set>
#include 
#include 
#include 
#include 
#include