algorithm - 将给定数字表示为四个平方和

Question

我正在寻找一种算法，将给定数字表示为(最多)四个平方的总和。

例子

120 = 8² + 6² + 4² + 2²
6 = 0² + 1² + 1² + 2²
20 = 4² + 2² + 0²+ 0²

我的方法

取平方根并对余数重复此操作:

while (count != 4) {
    root = (int) Math.sqrt(N)
    N -= root * root
    count++
} 

但是当 N 为 23 时，即使有解决方案，这也会失败:

3² + 3²+ 2² + 1²

问题

1. 还有其他算法可以做到这一点吗？

2. 总是可行吗？

Answer 1

###总是可能的？

是的，Lagrange's four square theorem指出:

every natural number can be represented as the sum of four integer squares.

已经从几个方面证明了这一点。

###算法

有一些更智能的算法，但我建议使用以下算法:

将数字分解为质因数。它们不一定是质数，但它们越小越好:所以质数是最好的。然后解决以下每个因素的任务，并将任何生成的 4 个方 block 与先前找到的 4 个方 block 与 Euler's four-square identity 组合起来。 .

(a² + b² + c² + d²)(A² + B² + C² + D²) =
(aA + bB + cC + dD)² +
(aB − bA + cD − dC)² +
(aC − bD − cA + dB)² +
(aD + bC − cB − dA)²

1. 给定一个数n(上述因素之一)，求出不大于n的最大平方，看是否n 减去这个平方可以写成使用 Legendre's three-square theorem 的三个平方和: 这是可能的，当且仅当此数字不是以下形式时:

4^a(8b+7)

If this square is not found suitable, try the next smaller one, ... until you find one. It guaranteed there will be one, and most are found within a few retries.

1. 尝试以与步骤 1 相同的方式找到实际的第二个平方项，但现在使用 Fermat's theorem on sums of two squares 测试其可行性这在扩展意味着:

   if all the prime factors of n congruent to 3 modulo 4 occur to an even exponent, then n is expressible as a sum of two squares. The converse also holds.

如果找不到合适的正方形，请尝试下一个较小的正方形，...直到找到一个。保证会有一个。

1. 现在我们在减去两个正方形后得到余数。尝试减去第三个方 block ，直到产生另一个方 block ，这意味着我们有一个解决方案。可以通过首先分解出最大的平方除数来改进此步骤。然后，当确定两个平方项时，每个平方项都可以再次乘以该平方除数的平方根。

大致思路是这样的。为了找到主要因素，有 several solutions .下面我将只使用 Sieve of Eratosthenes .

这是 JavaScript 代码，因此您可以立即运行它——它会生成一个随机数作为输入并将其显示为四个平方和:

function divisor(n, factor) {
    var divisor = 1;
    while (n % factor == 0) {
        n = n / factor;
        divisor = divisor * factor;
    }
    return divisor;
}

function getPrimesUntil(n) {
    // Prime sieve algorithm
    var range = Math.floor(Math.sqrt(n)) + 1;
    var isPrime = Array(n).fill(1);
    var primes = [2];
    for (var m = 3; m < range; m += 2) {
        if (isPrime[m]) {
            primes.push(m);
            for (var k = m * m; k <= n; k += m) {
                isPrime[k] = 0;
            }
        }
    }
    for (var m = range + 1 - (range % 2); m <= n; m += 2) {
        if (isPrime[m]) primes.push(m);
    }
    return {
        primes: primes,
        factorize: function (n) {
            var p, count, primeFactors;
            // Trial division algorithm
            if (n < 2) return [];
            primeFactors = [];
            for (p of this.primes) {
                count = 0;
                while (n % p == 0) {
                    count++;
                    n /= p;
                }
                if (count) primeFactors.push({value: p, count: count});
            }
            if (n > 1) {
                primeFactors.push({value: n, count: 1});
            }
            return primeFactors;
        }
    }
}

function squareTerms4(n) {
    var n1, n2, n3, n4, sq, sq1, sq2, sq3, sq4, primes, factors, f, f3, factors3, ok,
        res1, res2, res3, res4;
    primes = getPrimesUntil(n);
    factors = primes.factorize(n);
    res1 = n > 0 ? 1 : 0;
    res2 = res3 = res4 = 0;
    for (f of factors) { // For each of the factors:
        n1 = f.value;
        // 1. Find a suitable first square
        for (sq1 = Math.floor(Math.sqrt(n1)); sq1>0; sq1--) {
            n2 = n1 - sq1*sq1;
            // A number can be written as a sum of three squares
            // <==> it is NOT of the form 4^a(8b+7)
            if ( (n2 / divisor(n2, 4)) % 8 !== 7 ) break; // found a possibility
        }
        // 2. Find a suitable second square
        for (sq2 = Math.floor(Math.sqrt(n2)); sq2>0; sq2--) {
            n3 = n2 - sq2*sq2;
            // A number can be written as a sum of two squares
            // <==> all its prime factors of the form 4a+3 have an even exponent
            factors3 = primes.factorize(n3);
            ok = true;
            for (f3 of factors3) {
                ok = (f3.value % 4 != 3) || (f3.count % 2 == 0);
                if (!ok) break;
            }
            if (ok) break;
        }
        // To save time: extract the largest square divisor from the previous factorisation:
        sq = 1;
        for (f3 of factors3) {
            sq *= Math.pow(f3.value, (f3.count - f3.count % 2) / 2); 
            f3.count = f3.count % 2;
        }
        n3 /= sq*sq;
        // 3. Find a suitable third square
        sq4 = 0;
        // b. Find square for the remaining value:
        for (sq3 = Math.floor(Math.sqrt(n3)); sq3>0; sq3--) {
            n4 = n3 - sq3*sq3;
            // See if this yields a sum of two squares:
            sq4 = Math.floor(Math.sqrt(n4));
            if (n4 == sq4*sq4) break; // YES!
        }
        // Incorporate the square divisor back into the step-3 result:
        sq3 *= sq;
        sq4 *= sq;
        // 4. Merge this quadruple of squares with any previous 
        // quadruple we had, using the Euler square identity:
        while (f.count--) {
            [res1, res2, res3, res4] = [
                Math.abs(res1*sq1 + res2*sq2 + res3*sq3 + res4*sq4),
                Math.abs(res1*sq2 - res2*sq1 + res3*sq4 - res4*sq3),
                Math.abs(res1*sq3 - res2*sq4 - res3*sq1 + res4*sq2),
                Math.abs(res1*sq4 + res2*sq3 - res3*sq2 - res4*sq1)
            ];
        }
    }
    // Return the 4 squares in descending order (for convenience):
    return [res1, res2, res3, res4].sort( (a,b) => b-a );
}

// Produce the result for some random input number
var n = Math.floor(Math.random() * 1000000);
var solution = squareTerms4(n);
// Perform the sum of squares to see it is correct:
var check = solution.reduce( (a,b) => a+b*b, 0 );
if (check !== n) throw "FAILURE: difference " + n + " - " + check;
// Print the result
console.log(n + ' = ' + solution.map( x => x+'²' ).join(' + '));

The article by by Michael Barr on the subject可能代表了一种更省时的方法，但文本更像是一种证明而不是一种算法。但是，如果您需要更高的时间效率，您可以考虑将其与更高效的因式分解算法一起考虑。