(Codewars Kata) Memoised Log Cutting

Question

(I am not sure of the official CS name for the problem in question, so I just gave it the name of the kata).

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Problem

CLEAR CUTTER'S NEEDS YOUR HELP!

The logging company Clear Cutter's makes its money by optimizing the price-to-length of each log they cut before selling them. An example of one of their price tables is included:

# So a price table p
p = [ 0,  1,  5,  8,  9, 10]

# Can be imagined as:
# length i | 0  1  2  3  4  5 *in feet*
# price pi | 0  1  5  8  9 10 *in money* 

They hired an intern last summer to create a recursive function for them to easily calculate the most profitable price for a log of length n using price table p as follows:

def cut_log(p, n):
   if (n == 0):
      return 0
   q = -1
   for i in range(1, n+1)
      q = max(q, p[i] + cut_log(p, n-i))
   return q

An example of the working code:

# 5ft = $10, BUT 2ft + 3ft = 5ft -> $5 + $8 = $13 which is greater in value

However, their senior software engineer realized that the number of recursive calls in the function gives it an awful running time of 2^n (as this function iterates through ALL 2^n-1 possibilities for a log of length n).

Having discovered this problem just as you've arrived for your internship, he responsibly delegates the issue to you.

Using the power of Stack Overflow and Google, he wants you to create a solution that runs in Θ(n^2) time so that it doesn't take 5 hours to calculate the optimal price for a log of size 50. (He also suggests to research the problem using the keywords in the tags)

(Be aware that if your algorithm is not efficient, it will attempt to look at 2^49 = 562949953421312 nodes instead of 49^2 = 2401... The solution will automatically fail if it takes longer than 6 seconds... which occurs at right around Log 23)

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Solution

def cut_log(p, n):
    for i in range(2, len(p)):
        p[i] = max(p[i-k] + p[k] for k in range((i//2)+1))
    return p[n]

It passes all the tests. I'm putting it up for review because I'm pretty proud of it and experience as shown me that there's usually some improvement to be made.

Answer 1

I see three problems with this function:

• Why calculate the values up to len(p), when you are actually interested in p[n] (which may be larger or smaller than len(p))? In the case of n > len(p) your code actually raises an exception (and it seems not to be excluded by the rules, but also not to be checked by the online judge).
• This is not really using memoization, since you calculate everything again whenever the function is called.
• You mutate p, which might lead to unexpected results (in general), although here it happens to work out fine.

The second (and third) problem are not so easy to fix (probably only by memoizing all different p ever seen and for each of them the maximum n used). Also the online judge may or may not use the same p again.

The first one however is an easy fix:

def cut_log(p, n):
    for i in range(2, n + 1):
        p[i] = max(p[i-k] + p[k] for k in range((i//2)+1))
    return p[n]

Answer 2

I incorporated suggestions that @Graipher made in his answer. This is the improved function that resulted from it.

Solution

history = {}

def cut_log(p, n):
    global history
    ln = len(p)
    if id(p) not in history:
        p = p + [0]*max(n - ln, 0)
        for i in range(2, max(n+1, ln)):
            p[i] = max(p[i-k] + p[k] for k in range((i//2)+1))
        history[id(p)] = p
    else:
        p = history[id(p)]
        ln = len(p)
        if n+1 > ln:
            p = p + [0]*max(n - ln, 0)
            for i in range(ln, n+1):
                p[i] = max(p[i-k] + p[k] for k in range((i//2)+1))
    return p[n]

I wasn't memoising the function, and I added that for increased performance.