From e92615f88621f3d7e3396d3c32a2dd2399e2ef74 Mon Sep 17 00:00:00 2001 From: virinci Date: Thu, 31 Oct 2024 22:30:30 +0530 Subject: [PATCH] Fix the remainder group size in binary grouping solution of the multiple knapsack problem --- src/dynamic_programming/knapsack.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/dynamic_programming/knapsack.md b/src/dynamic_programming/knapsack.md index c6d14debd..9f73d173c 100644 --- a/src/dynamic_programming/knapsack.md +++ b/src/dynamic_programming/knapsack.md @@ -116,7 +116,7 @@ Let $A_{i, j}$ denote the $j^{th}$ item split from the $i^{th}$ item. In the tri The grouping is made more efficient by using binary grouping. -Specifically, $A_{i, j}$ holds $2^j$ individual items ($j\in[0,\lfloor \log_2(k_i+1)\rfloor-1]$).If $k_i + 1$ is not an integer power of $2$, another bundle of size $k_i-2^{\lfloor \log_2(k_i+1)\rfloor-1}$ is used to make up for it. +Specifically, $A_{i, j}$ holds $2^j$ individual items ($j\in[0,\lfloor \log_2(k_i+1)\rfloor-1]$).If $k_i + 1$ is not an integer power of $2$, another bundle of size $k_i-(2^{\lfloor \log_2(k_i+1)\rfloor}-1)$ is used to make up for it. Through the above splitting method, it is possible to obtain any sum of $\leq k_i$ items by selecting a few $A_{i, j}$'s. After splitting each item in the described way, it is sufficient to use 0-1 knapsack method to solve the new formulation of the problem. pFad - Phonifier reborn

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