Time complexity for count-change procedure in SICP
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16-10-2019 - |
Pergunta
In famous Structure and Interretation of Computer Programs, there is an exercise (1.14), that asks for the time complexity of the following algorithm - in Scheme - for counting change (the problem statement suggests drawing the tree for (cc 11 5)
- which looks like this):
; count change
(define (count-change amount)
(define (cc amount kinds-of-coins)
(cond ((= amount 0) 1)
((or (< amount 0) (= kinds-of-coins 0)) 0)
(else (+ (cc (- amount
(first-denomination kinds-of-coins))
kinds-of-coins)
(cc amount
(- kinds-of-coins 1))))))
(define (first-denomination kinds-of-coins)
(cond ((= kinds-of-coins 1) 1)
((= kinds-of-coins 2) 5)
((= kinds-of-coins 3) 10)
((= kinds-of-coins 4) 25)
((= kinds-of-coins 5) 50)))
(cc amount 5))
Now... there are sites with solutions to the SICP problems, but I couldn't find any easy to understand proof for the time complexity of the algorithm - there is a mention somewhere that it's polynomial O(n^5)
Solução 2
Probably this was not the right place for this question, but anyway, I found the answer in the meantime, in the form of a mostly "digestible" proof at http://wqzhang.wordpress.com/2009/06/09/sicp-exercise-1-14/.
Outras dicas
Order of growth of number of steps: $\theta (n^5)$
We can prove that, in general, the order of growth of number of steps is $\theta (n^m)$, where $m$ is the number of types of coin available. Here is my (very) crude reasoning using induction:
When there is only one type of coin, the number of steps is obviously proportional to n.
Suppose it would take
(cc n m)
steps to change an amount of $n$ with $m$ types of coin. Now let's consider(cc n m+1)
: ($A$ is the denomination of the $m$th kind of coin.)
(cc $n$ $m+1$)
= (cc $n$ $m$) + (cc $n-A$ $m+1$)
= (cc $n$ $m$) + (cc $n-A$ $m$) + (cc $n-2A$ $m+1$)
= (cc $n$ $m$) + (cc $n-A$ $m$) + (cc $n-2A$ $m$) + (cc $n-3A$ $m+1$)
= ......
It would eventually computes to
(cc n m) + (cc n-A m) + ... + (cc <something-negative> m+1)
There are approximately $n/A$ items. So the total number of steps would be proportional to $n/A*n^m$, which is proportional to $n^{m+1}$. Thus, the order of growth for number of steps of (cc n m)
is $\theta (n^m)$.
Let $m$ be 5, and the order of growth of number of steps is $\theta (n^5)$.