table of Contents
Euclidean algorithm (Euclidean algorithm)
Euclidean algorithm is also called the tossing and dividing method, which is mainly used to calculate the greatest common divisor of two integers a and b.
Simply talk about the principle of the algorithm: the greatest common divisor of two integers is equal to the greatest common divisor of the smaller one and the larger divided by the small remainder.
That is: gcd(a,b)=gcd(b,a mod b).
Give a simple example: For
example, find the greatest common divisor a = gcd(10, 24) of 10 and 24:
- Finding the greatest common divisor of 10 and 24 is equal to finding the greatest common divisor of 10 and 4:
a = gcd(10, 24) = gcd(10, 4) - Finding the greatest common divisor of 10 and 4 is equal to finding the greatest common divisor of 4 and 2, which is 2:
a = gcd(10, 24) = gcd(10, 4) = gcd(4, 2) = 2
# python
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
print(gcd(10,24)) # 2
Extended Euclidean algorithm
Algorithm principle: If a and b are positive integers, there are integers x, y such that gcd(a,b)=ax+by; in
layman's terms, gcd(a,b) can be expressed as a linear combination of integers of a and b.
Give a simple example:
gcd(10, 24) = 2
2 = 10*(-7) + 24*3
The main applications are as follows:
-
Solve the indeterminate equation;
example: find a set of integer solutions of 435x + 783y = 87:First get through Euclid's algorithm:
783 = 1× 435 + 348 435 = 348×1 + 87 348 = 87 × 4 + 0 ∴ 87 = 435 – 348 87 = 435 – (783 – 435) 87 = (–1)(783) + 2(435) ∴ x = 2, y = −1是此不定方程的一组整数解。 -
Solve the inverse element of the modulus (multiplicative inverse element), refer to the previous article on congruence equations, Euler functions, multiplicative inverse elements, and matrix inversion defined on Zm ;
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Solve modular linear equations (linear congruence equations);
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Solve the congruence equation ax ≡ b (mod m), x =?
Take an extremely representative example: 15x = 1 mod 26 This
problem is transformed into 15x-26y = 1 can be used as 1 to solve the indefinite equation, or as 2 Find the multiplicative inverse解法如下： 26 = 1× 15 + 11 15 = 11×1 + 4 11 = 4 × 2 + 3 4 = 1 × 3 + 1 3 = 1 × 3 ∴ 1 = 4 – 3 = 4 – (11 – 4×2) = 4×3 – 11 = (15-11) ×3 - 11 = 15×3 - 11×4 = (26-11)×3 - 11 ×4 = 26×3 - (26 - 15)×7 =26×(-4) + 15×7 ∴ x = 7, y = −4 为此不定方程的一组整数解，15关于模26的乘法逆元为7 -
Solve the system of congruence equations and continue to look at the Chinese remainder theorem
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Chinese remainder theorem
There is a question in "Sun Tzu Suan Jing": "Today there are things that do not know the number, two of the three or three are left (divide by 3 and remain 2), and
three of the five and five are left (divide by 5 and remain 3), seven or seven. Two of the remaining numbers (divide by 7 remain 2), what is the matter?"
Song Dynasty mathematician Qin Jiushao made a complete and systematic answer to the problem of "things do not know the number" in Volume 1 and 2 of "Nine Chapters of Mathematics" in 1247. Ming Dynasty mathematician Cheng Dawei compiled the solution into the easy-to-spond "Sun Tzu Ge Jue":
三人同行七十稀，
五树梅花廿一支，
七子团圆正半月，
除百零五便得知。
It means that as long as there is a 1 remaining after dividing by 3, a 70 is added;
as long as a 1 is remaining after dividing by 5, a 21 is added;
as long as a 1 is remaining after dividing by 7, a 15 is added. Then add up.
Finally, calculate the remainder of this sum divided by 105.
That is (2×70 + 3×21 + 15×2) mod 105 = 23
The solution is as follows:
First find out the smaller numbers 15, 21, 70 that are divided by 7, 5, and 3 from the common multiples of 3 and 5, 3 and 7, 5 and 7, respectively (this step is also called "modulo inverse" "Operation, refer to the multiplication inverse element solution). That is:
15÷7=2……Remaining 1,
21÷5=4……Remaining 1,
70÷3=23……Remaining 1.
Then use the three smaller numbers found to multiply the required number by 7. The product of the remainder obtained by dividing, 5, and 3 is continuously added,
15×2+21×3+70×2=233.
Finally, divide 233 by the least common multiple of the three divisors of 3, 5, and 7.
233÷105=2 ...... The remainder is 23,
this remainder 23 is the smallest number that meets the conditions.
Extend to the general situation:
assuming that the integers m1, m2,…, mn are mutually prime, then for any integer: a1, a2,…, an equation system:
there are integer solutions, and if X, Y satisfy the equation system , There must be X ≡ Y (mod N) where: The
formula is as follows:
I really don’t want to look at the formula symbols in the textbook. Let’s take the homework and give two examples.
Assignment 1:
Solve the system of congruence equations:
x ≡ 12 (mod 25)
x ≡ 9 (mod 26)
x ≡ 23 (mod 27)
The above equations are equivalent to x = 25a + 12 = 26b + 9 = 27c + 23.
Shift the terms to get:
①: 25a-27c = 23-12 = 11
②: 26b-25a = 12-9 = 3
Firstly, use Euclidean to expand Euclidean:
27 = 25×1 + 2
25 = 2×7 + 11
则：
11 = 25 - 2×7
= 25 - (27-25) ×7
= 25×8 - 27×7
所以a=8, c=7
代入x = 25a + 12 = 27c + 23 得：
x = 212
Get the combined equation x = 212 + 25 × 27t, that is: x ≡ 212 (mod 675)
and then merge with x ≡ 9 (mod 26)
x = 212 + 675t = 26b + 9
26b - 675t = 203
675 = 26×25+25
25 = 25×1
所以：
203 = (26-25)×203
= (26 - (675-26*25))×203
= 26×5278 - 675×203
b=5278 , t=203
代入得x = 137237
The combined equation x = 137237 + 25 × 27×26t is obtained: x ≡ 137237 (mod 17550), x=14387
x = 14387 + 17550n (n∈Z)
Assignment 2:
Solve the following congruence equations:
13x ≡ 4 (mod 99)
15x ≡ 56 (mod 101)
This kind of congruence equations with coefficients makes people daunting, but it does not prevent the use of extended Euclidean,
first remove the coefficients:
x ≡ 46 (mod 99)
x ≡ 98 (mod 101)
求解方法很多，这里列举利用二元一次不定方程方法：
13x ≡ 4 (mod 99) 转化为 13x-99y = 4
然后用拓展欧几里德：
13×46-99×6 = 4
x=46, y=6
所以不定方程13x-99y = 4 的所有解为
x=46 + 99t
y=6+13t
所以原同余方程解为：x ≡ 46 (mod 99)
Eliminate x to get: 99a-101b = 52
Expand Euclidean to walk you: x = 7471 (mod 9999)
x = 9999 n + 7471 (n ∈ Z)