Home > Applied Algorithms, Dynamic Programming > Forming Cricket Teams (Balanced Partition problem)

Forming Cricket Teams (Balanced Partition problem)

Problem Statement: You have been given N players, with a corresponding strength value S for each player (Minimum strength being 0 and maximum being K). You are supposed to divide them into two teams such that the difference in the total strengths of both teams is minimized.

This is an everyday problem that we tackle while forming equally matched teams. You might try a greedy approach here but in many cases it will fail. It is actually known as a Balanced Partition problem and utilizes the method we used while solving the integer knapsack (1/0) problem. Great! So you know half of it! But let us give it a quick look.

We defined a term P(i,j) which specified that I would find a subset of elements in the first ‘i’ elements which sum up to ‘j’. Now we can define the term recursively, as,

P(i,j) = maximum {P(i-1,j), P(i-1, j-Si)} 

While you can refer to the link given to the earlier post, what this formula specifies is that we either have a subset that sums up to ‘j’ in the first ‘i-1’ elements, or that we have a subset in the first ‘i-1’ elements whose sum when added to the ith value Si gives us ‘j’.

Now that we are done refreshing our memory, we’ll move on to the new problem.

We have now generated a 2D array giving us the possibility of a sum (j) being formed using some (<i) elements at each step. Now we need to divide this into 2 sets Team 1 and Team 2 such that the different of their totals is minimized.

For this, we’ll first find the mean value of the list of N elements. If a subset’s sum hits this particular value, we have found a solution with difference 0. Otherwise, we keep trying to find a solution where the sum of strengths of teams is closest to the mean. For this, we can once again check for all ‘i'<=’Mean’,

Minimum {Mean-i : P(n,i)=1}

where 1 indicates we have such a subset. What we are doing is that we are checking for every value less than the Mean it can be formed from any subset of the N elements. Checking this one by one, we land up with the required value closest to the Mean.

There it is. You have your solution! Divide you teams and game on! 🙂

(Retrieving the values for the subsets is as simple as maintaining back-pointers while creating the 2D array and filling it up.)

Doubts? Ask away!

PS: Need to be spoonfed (euphemism for “Me.Want.Code.”) ? Write it yourself! 😛 If you just can’t, drop me an email and I’ll mail it to you!

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  1. Rookieatalogo
    July 4, 2012 at 8:50 am

    This is a good explanation.. can you provide a example based on this algorithm??. As a beginner like me finds it a bit hard to translate P(i-1,j), P(i-1, j-Si) into an example and then into code

  2. Asalle
    February 9, 2014 at 6:32 pm

    Yes, the explanation is brilliant, but I still can’t understand without the poorest example.

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