recursion - 行列式递归算法的复杂度计算

Question

我写了一个算法来计算矩阵的n x n行列式，基于拉普拉斯展开：

我得到了下面的递归关系： T(n) = n(n² + T(n-1))

我在维基百科中读到这应该会产生结果T(n) = O(n!)，但我不知道如何证明这一点（尽管它很直观）。

Answer 1

公式是正确的，但您的递归关系不正确。您不需要n², 因为您不必保存或生成子矩阵。

Mij是(n-1) x (n-1)子矩阵的行列式。所以你必须计算不同矩阵n的行列式。n所以正确的递推关系是T(n) = n⋅T(n-1) + 2n-1。这简化为

T(n) = n ⋅ T(n-1) + 2n-1
     = n ⋅ (n-1) ⋅ T(n-2) + n ⋅ (n-1)
     = n ⋅ ( (n-1) ⋅ ( (n-2) ⋅ (...) + n-3 ) + n-2 ) + n-1
     = 2n-1 + n ⋅ (2(n-1)-1) + n ⋅ (n-1) ⋅ (2(n-2)-1) + ... + n!
     < 2 ⋅ n + 2 ⋅ n ⋅ (n-1) + 2 ⋅ n ⋅ (n-1) ⋅ (n-2) + ... + 2 ⋅ n! + n!
     = 2 ⋅ (n + n ⋅ (n-1) + ... + n!/2) + 3 ⋅ n!
     < 2 ⋅ (n!/(n-1)! + n!/(n-2)! + ... + n!/2!) + 3 ⋅ n!

由于2⋅n!/k! ≤ n!/(k-1)!所有的k ≥ 2，我们得到

  n!/(n-1)! + n!/(n-2)! + n!/(n-3)! + ... + n!/2!
≤ n!/(n-2)! + n!/(n-2)! + n!/(n-3)! + ... + n!/2!
≤ n!/(n-3)! + n!/(n-3)! + ... + n!/2!
≤ n!/(n-4)! + ... + n!/2!
≤ ...
≤ n!/2! + n!/2!
≤ n!

所以

T(n) = n ⋅ T(n-1) + 2n-1
     < 2 ⋅ (n!/(n-1)! + n!/(n-2)! + ... + n!/2!) + 3 ⋅ n!
     ≤ 2 ⋅ n! + 3 ⋅ n!
     = 5 ⋅ n!
     = O(n!)

所以你的算法运行在O(n!)