General Computer Science 320201 GenCS I & II Lecture ... - Kwarc

c○: Michael Kohlhase 249
The analysis for the cost is much more complex, we also have to solve a recursive equation, but a
more difficult one. Instead of just guessing the correct closed form, we will use the opportunity to
show a more general technique: using Master’s theorem for recursive equations. There are many
similar theorems which can be used in situations like these, going into them or proving Master’s
theorem would be beyond the scope of the course.
The Cost of CSA
C(CSA n ) = 3C(CSA n/2 ) + C(MUX n/2+1 ).
Problem: How to solve this recursive equation?
Solution: Guess a closed formula, prove by induction. (if we are lucky)
Solution2: Use a general tool for solving recursive equations.
Theorem 424 (Master’s Theorem for Recursive Equations) Given the recursively
defined function f : N → R, such that f(1) = c ∈ R and f(b k ) = af(b k−1 ) + g(b k ) for some
a ∈ R, 1 ≤ a, k ∈ N, and g : N → R, then f(b k ) = ca k + k−1
i=0 ai g(b k−i )
We have C(CSA n ) = 3C(CSA n/2 ) + C(MUX n/2+1 ) = 3C(CSA n/2 ) + 3(n/2 + 1) + 1 = 3C(CSA n/2 ) + 3
2n + 4
So, C(CSA n ) is a function that can be handled via Master’s theorem with a = 3, b = 2,
n = b k , g(n) = 3/2n + 4, and c = C(f 1 CSA ) = C(FA1 ) = 5
thus C(CSA n ) = 5 · 3log2 (n) + log2 (n)−1
i=0 3i · 3
2n · 2−i + 4
Note: a log 2 (n) = 2 log 2 (a)log 2 (n) = 2 log 2 (a)·log 2 (n) = 2 log 2 (n)log 2 (a) = n log 2 (a)
C(CSA n ) = 5 · 3 log2 (n) log2 (n)−1
+
i=0
= 5n log2 (3) log2 (n)
+
i=1
3 i · 3
2 n · 2−i + 4
n 3 i
n + 4
2
= 5n log2 (3) log2 (n) i 3
+ n · + 4log2 (n)
2
i=1
= 5n log2 (3) + 2n · 3 log2 (n)+1
− 1 + 4log
2
2(n)
= 5n log 2 (3) + 3n · n log 2 ( 3 2 ) − 2n + 4log 2 (n)
= 8n log 2 (3) − 2n + 4log 2 (n) ∈ O(n log 2 (3) )
Theorem 425 The cost and the depth of the conditional sum adder are in the following
complexity classes:
C(CSA n ) ∈ O(n log 2 (3) ) dp(CSA n ) ∈ O(log 2(n))
Compare with: C(CCA n ) ∈ O(n) dp(CCA n ) ∈ O(n)
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