如何找到这个递归函数的递归关系？

Question

Here I got a function like this and I want to find the recursive relation of that and after that calculate time complexity of that recursive relation.

 public static void f(int n) {
    if (n == 1) {
        //" Do sth "
    } else {
        for (int i = 1; i < n; i++) {
            f(i - 1);
            //" Do sth "
        }
    }
}

actually I tried a lot for that and I got T(n) = n * f( n-1) for this function as the relation but I am not sure about that . could you help me find the correct relation and solve it ?

Answer 1

Assuming T(1) = "Do sth" is constant work ie it doesn't depend on the input size n , you could write the recursive time function as :

T(n) =  T(1) + T(2) + ... + T(n-1)
     =  { T(1) } +  { T(1) } + { T(1) + T(2) } + { T(1) + T(2) + T(3) } + { T(1) + T(2) + T(3) + T(4) } +....

     [let T(1) = x]

     =  x + x + {x + x} + {x + x + (x + x)} + {x + x + (x + x) + x + x + (x + x)} +....

     = x + x + 2x + 4x + 8x + ...

     ~ x.2^(n-2)

     ~ O(2^n)

Here is a python program to demonstrate the sequence of coefficients for the summation:

t = [0 for i in range(10)]
for i in range(1,10):
  if i == 1:
    t[i] = 1
  else:
    val = 0
    for j in range(1,i):
      val += t[j]
    t[i] = val
print(t[1:])

prints : [1, 1, 2, 4, 8, 16, 32, 64, 128]

You can see that 2 ^(n-2) for n >= 2 holds good at each 'n' and complexity is O(2 ⁿ )