如何隨機生成一個數字，其條件與其他兩個隨機生成的數字沒有共同因素

Question

p1 = 1097
p2 = 1279
p = p1*p2
phi = (p1-1)*(p1-1)

我想隨機生成一個數字>phi，它在“p”和“phi”之間不共享公因數

Answer 1

從已知素數構造數字以使它們不共享相同的素數（作為因子）應該比創建隨機數、分解它們並檢查它們的非重疊因子更快。

算法：

• 計算/提供足夠的素數（從你的下限開始）
• 然后在素數列表中創建 2 個隨機索引列表
  • 第一個列表畫出你想要多少個數字作為因素
  • 從第二個列表中刪除已經在第一個列表中的任何索引，直到兩個列表都確認您想要的因素數量
• 然后將素數相乘

可以這樣做：

# get or calculate some primes that are big enough
# starting with 
primes = [1097,1103,1109,1117,1123,1129,1151,
  1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,
  1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,
  1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,
  1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,
  1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,
  1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,
  1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,
  1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,
  1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,
  1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,
  2063,2069,2081,2083,2087,2089,2099,2111,2113,2129,2131,2137,
  2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,
  2251,2267,2269,2273,2281,2287]

然后在其中選擇索引並將它們相乘以獲得結果：

import random
from operator import mul
from functools import reduce

random.seed(42)
choices = random.choices

def m(numbers):
    return reduce(mul, numbers, 1)

# how many numbers to use to factorize the number you want    
size = 5

# draw f.e. 5 factors from then
p1 = [primes[i] for i in choices(range(len(primes)), k=size)]

# draw 5 others that are not in p1
p2 = []
while len(p2) != size:
  p2 = [primes[i] for i in choices(range(len(primes)), k=3 * size) if i not in p1][:size]

# calculate them 
print( m(p1), m(p2)) # 8058608274530453 8642866553052643

您選擇的要使用的素數應該大於您想要用來乘以結果的數字的兩倍，否則您可能會陷入長時間運行的循環中。 提供大約 5 到 10 倍的要用作因子的素數應該足夠了。

Answer 2

不確定“不要在 p 和 phi 之間共享公因數”是什么意思，因為 p 和 phi 不會有公因數

One of the "bruteforce" ways to do it would be to run a Poisson random variable to get an integer n, then to compute the decomposition of each integer bigger than psi until you get the n-th integer bigger than psi satisfying your condition.

我會看到的另一種方法是采用大於 psi 的隨機 integer （例如泊松隨機變量）並檢查它們是否滿足您的條件。

取決於你的意思和你的數字的大小。

編輯：我建議泊松變量，因為它們允許您避免為隨機數設置任意硬編碼上限

Answer 3

您可以使用 python 中的random模塊生成隨機數。

要發現p和phi沒有公因數，您可以使用p*phi檢查最大公約數 (GCD)。

from math import gcd as calcGcd
from random import randint

def getRandom(lower, upper, *noFacWithThese):
    for _ in range(upper - lower): # to avoid infinite loop
        randNum = randint(lower, upper)
        if noCommonFactors(randNum, noFacWithThese):
            return randNum
    return None

def noCommonFactors(base, withThese):
    withThis = 1
    for n in withThese: # find mutiplication of numbers inside 'withThese' list
        withThis *= n
    return calcGcd(base, withThis) == 1

p1 = 1097
p2 = 1279
p = p1*p2
phi = (p1-1)*(p1-1)

while True:
    print( getRandom(1, 10000, p, phi) )
    input()
    

確保為randint function 使用必要的下限和上限。