Fix the remainder group size in binary grouping solution of the multiple knapsack problem #1387
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…ple knapsack problem
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I think you're right! Good catch! |
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The existing groups are:$2^0, 2^1, \dots, 2^{\lfloor \log_2(k+1) \rfloor - 1}$ and they sum up to $2^{\lfloor \log_2(k+1) \rfloor} - 1$ . So, the remaining groups should be $k_i - sum$ .