4

我怎样才能纠正他们的工作？我正在尝试根据之前的建议优化我的筛子，但在这两种情况下，代码都会中断：

递增j = j + ( i * 2)会破坏代码。

显然，我和其他人一样缺少一些关于优化的概念。但总的来说，您只需将素数的所有倍数标记为非素数即可。优化是下一步。

// prime-3
// sieve implementation
function prime3(n) {
  const sieve = (new Array(n)).fill(true);
  // Only iterate up to the square root n for marking
  for (let i = 2; i * i <= n; i += 1) {
    if (sieve[i]) {
      // recommended optimization breaks the code
      // j = j + ( i * 2 )
      for (let j = i * i; j <= n; j = j + i) {
        sieve[j] = false;
      }
    }
  }
  return makePrimes(sieve, n);
};

function makePrimes(sieve, n) {
  let primes = [];
  for (let i = 2; i < n; i++) {
    if (sieve[i]) {
      primes.push(i);
    }
  }
  return primes;
}
console.log(prime3(100));

4

3 回答 3

2

你的优化需要先去掉偶数后应用（除了2）。因为当 时i==2，您通过递增有效地跳过了所有偶数i*2。

这是一个工作代码：

// prime-3
// sieve implementation
function prime3(n) {
  let sieve = (new Array(n)).fill(true);
  for (let i = 4; i < n; i+=2) {
    sieve[i] = false;
  }
  
  // Only iterate up to the square root n for marking
  for (let i = 2; i * i <= n; i += 1) {
    if (sieve[i]) {
      // now it works
      // j = j + ( i * 2 )
      for (let j = i * i; j <= n; j = j + i*2) {
        sieve[j] = false;
      }
    }
  }
  return makePrimes(sieve, n);
};

function makePrimes(sieve, n) {
  let primes = [];
  for (let i = 2; i < n; i++) {
    if (sieve[i]) {
      primes.push(i);
    }
  }
  return primes;
}
console.log(prime3(100));

编辑

刮这个。进一步的测试确实表明 prime3 比简单的筛子快 3 倍。

该代码虽然有效，但似乎有太多技巧要做，并引入了额外的计算和混乱。在我的比较中，一个简单的筛选代码（如下所示）执行答案中的代码。同样，KISS 是原则。

原始答案中的脚本用了 317 毫秒来筛选 1M 个数字，而简单的筛选只用了 241 毫秒。

function simpleSieve(n) {
  let a = new Array(n)
  let answer = []
  let p
  for (let i = 2; i < n; i ++) {
    a[i] = i
  }
  for (let i = 2; i < n; i++) {
    if (a[i]) {
      answer.push(a[i])
      p = a[i]
      for(let j = p; j < n;  j += p) {
        a[j] = 0
      }
    }
  }
  return answer
}

编辑 2

用 cpp 和 prime3 重新测试确实比简单的筛子快 3 倍：

p3:
n = 100000000, t = 866717 microseconds.
n = 200000000, t = 2354425 microseconds.
n = 300000000, t = 3689165 microseconds.
n = 400000000, t = 4950224 microseconds.
n = 500000000, t = 6119779 microseconds.
n = 600000000, t = 7375925 microseconds.
n = 700000000, t = 8647293 microseconds.
n = 800000000, t = 10477116 microseconds.
n = 900000000, t = 11589894 microseconds.
n = 1000000000, t = 12806997 microseconds.
simple:
n = 100000000, t = 2316019 microseconds.
n = 200000000, t = 6063749 microseconds.
n = 300000000, t = 9783295 microseconds.
n = 400000000, t = 13315450 microseconds.
n = 500000000, t = 16640474 microseconds.
n = 600000000, t = 20282461 microseconds.
n = 700000000, t = 24219469 microseconds.
n = 800000000, t = 29203786 microseconds.
n = 900000000, t = 32965856 microseconds.
n = 1000000000, t = 37694084 microseconds.

为了完整性，这里的代码：

void simpleSieve(int n) {
  bool *a = (bool *)calloc(n, sizeof(bool));
  int p;
  memset(a, true, sizeof(bool) * n);
  for (int i = 2; i < n; i++) {
    if (a[i]) {
      p = i;
      for (int j = p; j < n; j += p) {
        a[j] = 0;
      }
    }
  }
  free(a);
}

void prime3(int n) {
  bool *sieve = (bool*)calloc(n, sizeof(bool));
  sieve[2] = true;
  for (int i = 3; i < n; i+=2) {
    sieve[i] = true;
  }
  int step;
  for (int i = 2; i * i <= n; i += 1) {
    if (sieve[i]) {
      step = i*2;
      for (int j = i * i; j <= n; j = j + step) {
        sieve[j] = false;
      }
    }
  }
  free(sieve);
}
于 2020-03-11T14:38:34.803 回答
1

你犯了一个简单的错误，没有准备筛子。您应该消除所有 2 的倍数以开始：

function makeSieve(n){
  const sieve = new Array(n).fill(true);
  for(let i = 2; i < sieve.length; i += 2){
    sieve[i] = false;
  }
}

现在，当您标记非素数时，您可以递增i * 2

例如

3, 6, 9, 12, 15, 18, 21

会变成

3, 9, 15, 21
于 2020-03-15T13:54:41.333 回答
0

实际上优化对我有用：

 // prime-3
    // sieve implementation
    function prime3 (n) {
      const sieve = (new Array(n)).fill(true);
    
      // Only iterate up to the square root n for marking
      for (let i = 2; i * i <= n; i += 1) {
        if (sieve[i]) {
    
          // recomended optimizations break the code
          // j = i * i 
          // j = j + ( i * 2 )
          for (let j = i * i; j <= n; j = j+i) {
            sieve[j] = false;
          }
        }
      }
      return makePrimes(sieve, n);
    };
    
    function makePrimes(sieve, n){
      let primes = [];
      for (let i = 2; i < n; i++) {
        if(sieve[i]) {
          primes.push(i);
        }
      }
      return primes;
    }
    
    console.log(prime3(100));

但是您j = j + ( i * 2 )的“优化”实际上破坏了算法。结果将包含非素数。这里有一个要检查的片段：

function prime3 (n) {
      const sieve = (new Array(n)).fill(true);
    
      // Only iterate up to the square root n for marking
      for (let i = 2; i * i <= n; i += 1) {
        if (sieve[i]) {
    
          // recomended optimizations break the code
          // j = i * i 
          // j = j + ( i * 2 )
          for (let j = i * i; j <= n; j = j+(i*2)) {
            sieve[j] = false;
          }
        }
      }
      return makePrimes(sieve, n);
    };
    
    function makePrimes(sieve, n){
      let primes = [];
      for (let i = 2; i < n; i++) {
        if(sieve[i]) {
          primes.push(i);
        }
      }
      return primes;
    }
    
    console.log(prime3(100));

于 2020-03-11T14:15:11.217 回答