递归迭代方法是否比纯迭代方法更好地确定一个数是否为素数? [英] Is a Recursive-Iterative Method Better than a Purely Iterative Method to find out if a number is prime?

问题描述

我用 C 编写了这个程序，用于测试数字是否为质数.我还不熟悉算法复杂性和所有大 O 的东西，所以我不确定我的方法(迭代和递归的组合)实际上是否比使用 >纯迭代方法.

I made this program in C that tests if a number is prime. I'm as yet unfamiliar with Algorithm complexity and all that Big O stuff, so I'm unsure if my approach, which is a combination of iteration and recursion, is actually more efficient than using a purely iterative method.

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

typedef struct primenode{
    long int key;
    struct primenode * next;
}primenode;

typedef struct{
    primenode * head;
    primenode * tail;
    primenode * curr;
    unsigned long int size;
}primelist;

int isPrime(long int number, primelist * list ,long int * calls, long int * searchcalls);
primenode * primelist_insert(long int prime, primelist * list);
int primelist_search(long int searchval, primenode * searchat, long int * calls);
void primelist_destroy(primenode * destroyat);

int main(){
    long int n;
    long int callstoisprime = 0;
    long int callstosearch = 0;
    int result = 0;
    primelist primes;

    //Initialize primelist
    primes.head = NULL;
    primes.tail = NULL;
    primes.size = 0;

    //Insert 2 as a default prime (optional step)
    primelist_insert(2, &primes);

    printf("

Please enter a number: ");
    scanf("%d",&n);
    printf("Please wait while I crunch the numbers...");
    result = isPrime(n, &primes, &callstoisprime, &callstosearch);
    switch(result){
        case 1: printf("
%ld is a prime.",n); break;
        case -1: printf("
%ld is a special case. It's neither prime nor composite.",n); break;
        default: printf("
%ld is composite.",n); break;
    }
    printf("

%d calls made to function: isPrime()",callstoisprime);
    printf("
%d calls made to function: primelist_search()",callstosearch);

    //Print all prime numbers in the linked list
    printf("

Here are all the prime numbers in the linked list:

");
    primes.curr = primes.head;
    while(primes.curr != NULL){
        printf("%ld ", primes.curr->key);
        primes.curr = primes.curr->next;
    }
    printf("

Note: Only primes up to the square root of your number are listed.
"
                "If your number is negative, only the smallest prime will be listed.
"
                "If your number is a prime, it will itself be listed.

");

    //Free up linked list before exiting
    primelist_destroy(primes.head);

    return 0;
}

int isPrime(long int number, primelist * list ,long int * calls, long int *searchcalls){
//Returns 1 if prime
//          0 if composite
//          -1 if special case
    *calls += 1;
    long int i = 2;
    if(number==0||number==1){
        return -1;
    }
    if(number<0){
        return 0;
    }
    //Search for it in the linked list of previously found primes
    if(primelist_search(number, list->head, searchcalls) == 1){
        return 1;
    }
    //Go through all possible prime factors up to its square root
    for(i = 2; i <= sqrt(number); i++){ 
        if(isPrime(i, list,calls,searchcalls)){
            if(number%i==0) return 0; //It's not a prime
        }
    }
    primelist_insert(number, list); /*Insert into linked list so it doesn't have to keep checking
                                                if this number is prime every time*/
    return 1;
}

primenode * primelist_insert(long int prime, primelist * list){
    list->curr = malloc(sizeof(primenode));
    list->curr->next = NULL;

    if(list->head == NULL){
        list->head = list->curr;
    }
    else{
        list->tail->next = list->curr;
    }
    list->tail = list->curr;
    list->curr->key = prime;
    list->size += 1;

    return list->curr;
}

int primelist_search(long int searchval, primenode * searchat, long int * calls){
    *calls += 1;
    if(searchat == NULL) return 0;
    if(searchat->key == searchval) return 1;
    return primelist_search(searchval, searchat->next, calls);
}

void primelist_destroy(primenode * destroyat){
    if(destroyat == NULL) return;
    primelist_destroy(destroyat->next);
    free(destroyat);
    return;
}

基本上，我看到的很多简单的原始测试所做的是:0. 2 是素数.1. 循环所有整数，从 2 到被测数的一半或平方根.2.如果数字可以被任何东西整除，则break并返回false；它是复合的.3. 否则，在最后一次迭代后返回true；这是主要的.

Basically, a lot of what I've seen simple primalty tests do is: 0. 2 is a prime. 1. Cycle through all integers from 2 to half or the square root of the number being tested. 2. If the number is divisible by anything, break and return false; it's composite. 3. Otherwise, return true after the last iteration; it's prime.

我认为您不必针对从 2 到平方根的每个数字进行测试，只需针对每个 质数 数字进行测试，因为所有其他数字都是质数的倍数.因此，该函数调用自身以在使用模数之前确定一个数字是否为素数.这有效，但我认为一遍又一遍地测试所有这些素数有点乏味.因此，我也使用了一个链表来存储在其中找到的每个素数，以便程序在测试素数之前先搜索列表.

I figured that you don't have to test against every number from 2 to the square root, just every prime number, because all other numbers are multiples of primes. So, the function calls itself to find out if a number is prime before using the modulus on it. This works, but I thought it a bit tedious to keep testing all those primes over and over again. So, I used a linked list to store every prime found in it as well, so that before testing primalty, the program searches the list first.

它真的更快或更高效，还是我只是浪费了很多时间?我确实在我的计算机上对其进行了测试，对于较大的素数，它确实看起来更快，但我不确定.我也不知道它是否使用了更多的内存，因为无论我做什么，任务管理器都会保持恒定的 0.7 MB.

Is it really faster, or more efficient, or did I just waste a lot of time? I did test it on my computer, and for the larger primes it did seem faster, but I'm not sure. I also don't know if it uses significantly more memory since Task Manager just stays a constant 0.7 MB whatever I do.

感谢您的回答！

推荐答案

由于您的程序只测试一个数字，因此您在尝试避免通过组合进行测试是在浪费时间.您执行了大量计算以节省一个微不足道的模运算.

Since your program tests just one number, you're wasting time trying to avoid testing by composites. You perform a lot of computations to save one meager modulo operation.

如果您连续测试多个数字的素数，那么预先计算最多为该范围上限的 sqrt 的素数，并在测试候选者时遍历这些素数是有意义的.

If you were testing more than a few numbers in a row for primality, then it would make sense to pre-compute the primes up to a sqrt of the top limit of that range, and go through those primes when testing the candidates.

更好的是进行偏移 Eratosthenes 筛分(此处的 C 代码)，找到给定范围内的素数.Eratosthenes筛法寻找2到N的素数的时间复杂度为O(N log log N)；和由素数直到 sqrt 的试除，O(N^1.5/(log N)^2)(这更糟；例如 与 100k 相比，高达 100k 的筛子的运行时间比是 10.7 倍，而试验划分是 22 倍；2mln 对 100 万是 2.04 倍筛子，2.7x 用于试验分裂).

Better yet will be to perform an offset sieve of Eratosthenes (C code here), to find the primes in a given range. The time complexity of the sieve of Eratosthenes to find primes from 2 to N is O(N log log N); and trial division by primes up to the sqrt, O(N^1.5 / (log N)^2) (which is worse; e.g. the ratio of run times for the sieve up to 1mln compared to 100k is 10.7x, vs 22x for trial division; 2mln vs 1 mln is 2.04x for the sieve, 2.7x for the trial division).

埃拉托色尼偏移筛的伪代码:

Pseudocode for the offset sieve of Eratosthenes:

Input: two Integers n >= m > 1

Let k = Floor(Sqrt(n)),
Let A be an array of Boolean values, indexed by Integers 2 to k, and
    B an array of Booleans indexed by Integers from m to n,
    initially all set to True.

for i = 2, 3, 4, ..., not exceeding k:
  if A[i] is True:
    for j = i^2, i^2+i, i^2+2i, ..., not greater than k:
      A[j] := False
    for j = i^2, i^2+i, i^2+2i, ..., between m and n, inclusive:
      B[j] := False

Output: all `i`s such that B[i] is True, are all the primes 
                                     between m and n, inclusive.

一个常见的优化是只使用赔率，i = 3,5,7,...，从一开始就避免任何偶数(2 无论如何都是素数，并且任何偶数都是合数).然后2i的步骤，而不仅仅是i，可以在两个内循环中使用.因此，偶数索引被完全排除在处理之外(通常使用压缩寻址方案，val = start + 2*i).

A common optimization is to work with odds only, i = 3,5,7,..., avoiding any even numbers from the outset (2 is known to be a prime anyway, and any even number is a composite). Then the step of 2i, and not just i, can be used in both inner loops. Thus the even indices are excluded from processing altogether (usually with condensed addressing schemes, val = start + 2*i).

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