[Solved] time complexity of recursion function

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We have:

1 : if statement

3 : * operator

3 : function statement

Totally: 7

So we have: 

T(n)=t(n-1) + t(n-2) + t(n-3) + 7
T(99)=1
T(98)=1
...
T(1)=1

Then to reduce T(n), we have T(n-3)<T(n-2)<T(n-1) therefore:

T(n) <= T(n-1)+T(n-1)+T(n-1) + 7
T(n) <= 3T(n-1) + 7

So we can solve T(n) = 3T(n-1) + 7 (in big numbers T(n-1) is almost equal T(n-2) and … )

Then we can calculate T(n) by following approach:

T(n) = 3.T(n-1) + 7
     =3.(3.T(n-2)+7) + 7     = 3^2.T(n-2) + 7.(3^1 + 3^0)
     =3^2.(3.T(n-3)+7) + ... = 3^3.T(n-3) + 7.(3^2 + 3^1 + 3^0)
                             = 3^4.T(n-4) + 7.(3^4 + 3^2 + 3^1 + 3^0)
     ...
     =3^(n-99).T(n-(n-99)) + 7.(3^(n-98) + 3^(n-97) + ...+ 3^1 + 3^0)
     =3^(n-99) + 7.(3^(n-98) + 3^(n-97) + ...+ 3^1 + 3^0)

So consider that 1 + 3 + 3^2 ...+3^(n-1) = 3^n - 1/2

Therefore we reached:

T(n) = 3^(n-99) + 7.(3^(n-99) + 1/2) = 8.3^(n-99) - 7/2
     = 8/(3^99) . 3^n -7/2

Finally: Big O is O(3^n)

If you just have T(1)=1, the answer is the same.

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solved time complexity of recursion function