python求素數(shù)算法

There are various methods through which we can calculate prime numbers upto n.

我們可以通過多種方法來計算最大為n的素數(shù) 。

1) General Method

1)一般方法

In this method, we usually run two for loops in which the First one is used to increase the number and the second one is used to check whether the number is prime or not. Second loop runs from 2 to ( n / 2 + 1 ) ( for better performance).

在這種方法中，我們通常運行兩個for循環(huán)，其中第一個循環(huán)用于增加數(shù)字，第二個循環(huán)用于檢查數(shù)字是否為質(zhì)數(shù)。 第二個循環(huán)從2到(n / 2 +1)運行(以獲得更好的性能)。

Note: This is the least efficient method (one should not use this if efficiency is required.)

注意：這是效率最低的方法(如果需要效率，則不應(yīng)使用此方法。)

2) Square-Root Method

2)平方根法

In this method, two loops run first one is to increase the number and the second one is to check whether the number is prime or not. The second loop runs from 2 to square root (number) (a number which is to be check), that’s why the run length of second for loop is relatively small, that’s why it’s efficient than the na?ve approach.

在此方法中，運行兩個循環(huán)，第一個循環(huán)是增加數(shù)字，第二個循環(huán)是檢查數(shù)字是否為質(zhì)數(shù)。 第二個循環(huán)從2到平方根(數(shù)字)(要檢查的數(shù)字)，這就是為什么第二個for循環(huán)的運行長度相對較小的原因，這就是為什么它比幼稚的方法有效的原因。

3) Sieve of Eratosthenes

3)Eratosthenes篩

This is the best and most efficient method to calculate the prime numbers upto n.

這是計算最高達(dá)n的素數(shù)的最佳和最有效的方法。

Algorithm for Sieve of Eratosthenes:

Eratosthenes篩分算法：

Let A be an array from 2 to n.

設(shè)A為2到n的數(shù)組。

Set all the values to

將所有值設(shè)置為

True (we are considering every number to be Prime)

正確 (我們認(rèn)為每個數(shù)字都是素數(shù))

For loop from p == 2 (smallest prime number)

從p == 2開始的循環(huán)(最小質(zhì)數(shù))

For loop from p² to n

從p ²到n的循環(huán)

Mark all the multiples of

標(biāo)記所有的倍數(shù)

p as False and increase the value of p to the next prime number

數(shù)p作為假和增加p值到下一個素數(shù)

End of second FOR loop

第二個FOR循環(huán)結(jié)束

End of first FOR loop

第一個FOR循環(huán)結(jié)束

At the end of both the for loops, all the values that are marked as TRUE are primes and all the composite numbers are marked as FALSE in step 3.

在兩個for循環(huán)的末尾，在步驟3中，所有標(biāo)記為TRUE的值均為質(zhì)數(shù)，所有復(fù)合數(shù)字均標(biāo)記為FALSE。

Time complexity : O(n*log(log(n)))

時間復(fù)雜度：O(n * log(log(n)))

Note: Performance of General Method and SquareRoot Method can be increased a little bit if we check only ODD numbers because instead of 2 no even number is prime.

注意：如果只檢查ODD數(shù)，則通用方法和SquareRoot方法的性能可以提高一點，因為不是2的偶數(shù)不是素數(shù)。

Example:

例：

from time import time from math import sqrtdef general_approach(n):'''Generates all the prime numbers from 2 to n - 1.n - 1 is the largest potential prime considered.'''start = time()count = 0for i in range(2, n):flag = 0x = i // 2 + 1for j in range(2, x):if i % j == 0:flag = 1breakif flag == 0:count += 1stop = time()print("Count =", count, "Elapsed time:", stop - start, "seconds")def count_primes_by_sqrt_method(n):'''Generates all the prime numbers from 2 to n - 1.n - 1 is the largest potential prime considered.'''start = time()count = 0for val in range(2, n):root = round(sqrt(val)) + 1for trial_factor in range(2, root):if val % trial_factor == 0:breakelse:count += 1stop = time()print("Count =", count, "Elapsed time:", stop - start, "seconds")def seive(n):'''Generates all the prime numbers from 2 to n - 1.n - 1 is the largest potential prime considered.Algorithm originally developed by Eratosthenes.'''start = time()# Each position in the Boolean list indicates# if the number of that position is not prime:# false means "prime," and true means "composite."# Initially all numbers are prime until proven otherwisenonprimes = n * [False]count = 0nonprimes[0] = nonprimes[1] = Truefor i in range(2, n):if not nonprimes[i]:count += 1for j in range(2*i, n, i):nonprimes[j] = Truestop = time()print("Count =", count, "Elapsed time:", stop - start, "seconds")# Time complexity : O(n*log(log(n)))def main():print("For N == 200000\n")print('Sieve of Eratosthenes Method')seive(200000)print('\nSquare Root Method')count_primes_by_sqrt_method(200000)print('\nGeneral Approach')general_approach(200000)main()

Output

輸出量

For N == 200000Sieve of Eratosthenes Method Count = 17984 Elapsed time: 0.050385475158691406 secondsSquare Root Method Count = 17984 Elapsed time: 0.9392056465148926 secondsGeneral Approach Count = 17984 Elapsed time: 101.83296346664429 seconds

翻譯自: https://www.includehelp.com/python/calculate-prime-numbers-using-different-algorithms-upto-n-terms.aspx

python求素數(shù)算法

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