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    <p><a href="http://en.wikipedia.org/wiki/Dynamic_programming" rel="nofollow noreferrer">Dynamic programming</a> is the solution you're looking for.</p> <p>Example with [4, 3, 10, 3, 2, 5]:</p> <pre> X-Axis: Reachable sum of group. max = sum(all numbers) / 2 (rounded up) Y-Axis: Count elements in group. max = count numbers / 2 (rounded up) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 | | | | 4| | | | | | | | | | | // 4 2 | | | | | | | | | | | | | | | 3 | | | | | | | | | | | | | | | </pre> <pre> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 | | | 3| 4| | | | | | | | | | | // 3 2 | | | | | | | 3| | | | | | | | 3 | | | | | | | | | | | | | | | </pre> <pre> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 | | | 3| 4| | | | | |10| | | | | // 10 2 | | | | | | | 3| | | | | |10|10| 3 | | | | | | | | | | | | | | | </pre> <pre> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 | | | 3| 4| | | | | |10| | | | | // 3 2 | | | | | | 3| 3| | | | | |10|10| 3 | | | | | | | | | | 3| | | | | </pre> <pre> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 | | 2| 3| 4| | | | | |10| | | | | // 2 2 | | | | | 2| 3| 3| | | | | 2|10|10| 3 | | | | | | | | 2| 2| 3| | | | | </pre> <pre> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 | | 2| 3| 4| 5| | | | |10| | | | | // 5 2 | | | | | 2| 3| 3| 5| 5| | | 2|10|10| 3 | | | | | | | | 2| 2| 3| 5| 5| | | ^ </pre> <p>12 is our lucky number! Backtracing to get the group:</p> <pre> 12 - 5 = 7 {5} 7 - 3 = 4 {5, 3} 4 - 4 = 0 {5, 3, 4} </pre> <p>The other set can then be calculated: {4,3,10,3,2,5} - {5,3,4} = {10,3,2}</p> <p>All fields with a number are possible solutions for one bag. Choose the one that is furthest in the bottom right corner.</p> <p>BTW: It's called the <a href="http://en.wikipedia.org/wiki/Knapsack_problem" rel="nofollow noreferrer">knapsack-problem</a>.</p> <blockquote> <p>If all weights (w1, ..., wn and W) are nonnegative integers, the knapsack problem can be solved in pseudo-polynomial time using dynamic programming.</p> </blockquote>
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    1. POAlgorithm to Divide a list of numbers into 2 equal sum lists
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    1. USGeorg Schölly
    1. USGeorg Schölly
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    1. COOkay. This makes sense to me...but then why are people claiming that this problem is NP-complete? Someone is wrong...and I can't figure out what (if anything) is wrong with this solution.
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    2. COYou would need a space of O(Sum(x[i]) in order to use the dynamic programming solution. In general, I believe that the problem is NP-Complete. (Think if each number is a floating point number, you can not easily use Dynamic programming)
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      1. USMatt W
    3. COThat's true, it works only for a limited subset of the problem.
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      1. USGeorg Schölly
 

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