Question

Given the number N, write a program that computes the numbers E1, E2, …En with the following properties:

1) N = E1 + E2 + … + En;
2) E1 * E2 * … En is maximum.
3) E1..En, are integers. No negative values 🙂

How would you do that ? I have a solution based on divide et impera but i want to check if is optimal.

Example: N=10

5,5        S=10,P=25
3,2,3,2    S=10,P=36

Answer 1

No need for an algorithm, mathematic intuition can do it on its own:

Step 1: prove that a result set with numbers higher than 3 is at most as good as a result set with only 3’s and 2’s

Given any number x in your result set, one might consider whether it would be better to divide it into two numbers.

The sum should still be x.

• When x is even, The maximum for t (x – t) is reached when t = x/2 , and except for the special case x = 2, then it is greater than x, and for the special case x = 4, equal to x (see note 1).
• When x is odd, The maximum for t (x – t) is reached when t = (x ± 1)/2.

What does this show? Only that you should only have 3’s and 2’s in your final set, because otherwise it is suboptimal (or equivalent to an optimal set).

Step 2: you should have as many 3’s as possible

Now, as 3² > 2³, you should have as many 3’s as possible as long as the remainder is not 1.

Conclusion: for every N >= 3:

• If N = 0 mod 3, then the result set is only 3’s
• If N = 1 mod 3, then the result set has one pair of 2’s (or a 4) and the rest is 3’s
• If N = 2 mod 3, then the result set has one 2 and the rest is 3’s

Please correct this post. The times when I was writing well-structured mathematical proofs is far away…

Note 1: (2,4) is the only pair of distinct integers such that x^y = y^x. You can prove that with:

x^y = y^x
y ln(x) = x ln(y)
ln(x)/x = ln(y) / y

and the function ln(t)/t is strictly decreasing after its global maximum, reached between 2 and 3, so if you want two distinct integers such that ln(x)/x = ln(y)/y, one of them must be lower or equal to 2. From that you can infer that only (2,4) works