Is time complexity O(K^3*Log(N))?

#1

F(i) = T^(n-1) * F(i-1)

How is T^(n-1) a O(LogN) here ?

Because,

Matrix exponentiation of T^(n-1) involves matrix multiplication also, inside recursive function “multiply”.

Fast exponentiation is O(LogN).

And, matrix multiplication is O(K^3).

Thus, calculating T^(n-1) should be greater than O(K^3*Log(N)).

#2

@anuj_it,
Yes, complexity of matrix exponentiation is O(k^3. log(n)), what is the doubt…??

#3

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#4

@abhijeet.srivastava6499 We are calling functions recursively, multiple times, so it needs to be greater isn’t it ?

#5

@anuj_it,
We are calling the recursive function O(log(n)) times, and in each call we are doing O(k^3) operations, thus total complexity = O(k^3.log(n))

#6

@abhijeet.srivastava6499
Okkk… So it means, call to recursive fn() from inside same recursive fn() doesn’t gets multiplied again for computing time complexity ?

Like we do in nested for loop, we multiply the complexity…

#7

@anuj_it,
It’s something like,

for(int i=1; I<=log(n); I==){
for(int j=1;j<=k^3;j++){
//O(const.) operation here
}
}