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Program to find expected value of maximum occurred frequency values of expression results in Python

PythonServer Side ProgrammingProgramming

Suppose we have M different expressions, and the answers of these expressions are in range 1 to N (both inclusive) So consider x = max(f(i)) for each i in range 1 through N, we have to find the expected value of x.

So, if the input is like M = 3, N = 3, then the output will be 2.2, because

SequenceMaximum frequency
1113
1122
1132
1222
1231
1331
2223
2232
2332
3333

$$E(x) = \sum P(x) * x = P(1) + 2P(2) + 3P(3) = \frac{1}{10} + 2 * \frac{6}{10} + 3 * \frac{3}{10} = \frac{22}{10}$$

To solve this, we will follow these steps −

  • combination := a new map
  • Define a function nCr() . This will take n, k_in
  • k := minimum of k_in and (n - k_in)
  • if n < k or k < 0, then
    • return 0
  • otherwise when (n, k) is in combination, then
    • return combination[n, k]
  • otherwise when k is same as 0, then
    • return 1
  • otherwise when n is same as k, then
    • return 1
  • otherwise,
    • a := 1
    • for cnt in range 0 to k - 1, do
      • a := a * (n - cnt)
      • a := floor of a/(cnt + 1)
      • combination[n, cnt + 1] := a
    • return a
  • From the main method, do the following:
  • arr := a new list
  • for k in range 2 to M + 1, do
    • a := 1
    • s := 0
    • for i in range 0 to floor of M/k + 2, do
      • if M < i * k, then
        • come out from loop
      • s := s + a * nCr(N, i) * nCr(N-1+M-i*k, M-i*k)
      • a := -a
    • insert s at the end of arr
  • total := last element of arr
  • diff := an array where insert arr[0] at beginning then add a list where (arr[cnt + 1] - arr[cnt]) for each cnt in range 0 to M - 2
  • output := sum of all elements present in (diff[cnt] *(cnt + 1) / total for cnt in range 0 to M-1)
  • return output

Example

Let us see the following implementation to get better understanding −

combination = {}
def nCr(n, k_in):
   k = min(k_in, n - k_in)
   if n < k or k < 0:
      return 0
   elif (n, k) in combination:
      return combination[(n, k)]
   elif k == 0:
      return 1
   elif n == k:
      return 1
   else:
      a = 1
      for cnt in range(k):
         a *= (n - cnt)
         a //= (cnt + 1)
         combination[(n, cnt + 1)] = a
      return a

def solve(M, N):
   arr = []
   for k in range(2, M + 2):
      a = 1
      s = 0
      for i in range(M // k + 2):
         if (M < i * k):
            break
         s += a * nCr(N, i) * nCr(N - 1 + M - i * k, M - i * k)
         a *= -1
      arr.append(s)
   total = arr[-1]
   diff = [arr[0]] + [arr[cnt + 1] - arr[cnt] for cnt in range(M - 1)]
   output = sum(diff[cnt] * (cnt + 1) / total for cnt in range(M))
   return output

M = 3
N = 3
print(solve(M, N))

Input

3, 3

Output

1
raja
Arnab Chakraborty
Updated on 11-Oct-2021 09:20:44

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