Question 2:
(a)
Applying Master's second case method, for a=9 and b=3.
(b)
From above recursion tree,
(c)
Substituting, n=2m=>log n=m in above recursion.
T(2m)=T(2m/2)+1.
Now, substituting, T(2m)=S(m).
S(m)=S(m/2)+1.
Now, applying case 1 of Master's Method,(a=1, b=2) solution is:
Question 2 (8 marks) (From the DPU textbook, Exercise 2.5) Using the Master theorem or recursion...
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