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  MarioHok

  Wednesday, 01 October 2025 11:24

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  Richardpaf

  Wednesday, 01 October 2025 11:22

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  火绒安全下载

  Wednesday, 01 October 2025 11:03

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  anavar results

  Wednesday, 01 October 2025 11:01

  Anavar Results: How Long Until You See A Change?
  
  Title:
  
  Parallel Dynamic‑Programming for the 0/1 Knapsack: Complexity, Approximation, and CUDA Acceleration
  
  
  
  ---
  
  
  
  
  1. Introduction
  
  
  The 0/1 knapsack problem (KP) is a canonical NP‑hard combinatorial
  optimisation problem. Given a set of \(n\) items, each with an integer weight
  \(w_i>0\) and profit \(p_i>0\), together with a capacity constraint \(W>0\), the
  goal is to select a subset \(S\subseteq1,\dots,n\) such that
  
  
  
  [
  \sum_i\in S w_i \;\le\; W , \qquad
  \textmaximise \;\sum_i\in S p_i .
  ]
  
  
  
  The decision version—does there exist a subset achieving
  profit at least \(P\)?—is NP‑complete. Nonetheless, the classical
  dynamic programming (DP) algorithm solves the optimization problem in pseudo‑polynomial time:
  
  
  
  
  for i = 0..n
  for w = 0..W
  DPiw = max_subset of first i items with weight ≤ w total profit
  
  
  The recurrence is
  
  
  
  [
  \mathrmDPi,w =
  \begincases
  \max(\mathrmDPi-1,\,w, \; \mathrmDPi-1,\,w-w_i + p_i) & (w \ge w_i)\\
  \mathrmDPi-1,\,w & (w 0\), we can compute a solution with profit at least \((1-\epsilon)\)
  times optimum in time \(O(n^3 / \epsilon)\).
  The scheme is based on dynamic programming over scaled profits, similar to the classic FPTAS for 0/1‑KP.
  
  
  
  
  Practical algorithm.
  
  In practice, a simple greedy heuristic works well:
  sort items by decreasing value/weight ratio and fill the first knapsack until capacity; then continue with the second knapsack.
  This runs in \(O(n \log n)\). For more balanced loading, one can perform a local
  search that swaps items between the two knapsacks to improve total
  weight distribution.
  
  
  
  ---
  
  
  
  
  4. A Concrete Example
  
  
  Let us instantiate all three problems on a small dataset:
  
  
  
  
  Item Weight (kg) Value (USD)
  
  
  1 10 1000
  
  
  2 20 1200
  
  
  3 30 1500
  
  
  4 40 1800
  
  
  Assume we have two identical trucks, each with capacity \(C = 50\)
  kg.
  
  
  
  
  4.1 Minimum‑Load Problem
  
  
  We must assign items to trucks so that the total weight on each truck is at least \(30\) kg (i.e., at most \(20\) kg of empty space).
  Since we have only two trucks, each can carry at most one item because any pair
  would exceed capacity.
  
  
  
  Thus feasible assignments are:
  
  
  
  
  
  Truck 1: Item 4 (40 kg), Truck 2: Item 3 (30 kg).
  Empty spaces: 10 and 20 kg respectively. Total empty = 30 kg.
  
  
  
  Truck 1: Item 4, Truck 2: Item 2 (20 kg). Empty
  spaces: 10 and 30 kg. Total empty = 40 kg.
  
  
  
  We must choose the assignment with minimal total empty space:
  first option gives 30 kg empty, second gives 40 kg. So optimal is truck 1 holds item 4, truck 2 holds item
  3.
  
  If we had allowed loading more items, we might have filled both trucks
  more completely and reduced empty spaces further, but that would violate
  the rule of not loading other items beyond those two if
  they are the only ones loaded. Thus the optimal solution under the rule is as
  above.
  
  
  
  ---
  
  
  
  Conclusion
  
  
  
  The analysis shows that when you load at most one item per truck (or at
  most a limited number), the optimal strategy may involve leaving some trucks empty or
  partially filled to satisfy the constraint.
  This yields a different objective function than the standard multi-dimensional knapsack
  problem, where all available capacity is typically utilized as
  much as possible.
  
  
  
  In general, if you are allowed to load multiple items per truck
  but must restrict the total weight in each truck to be below a threshold \(W_\max\),
  then the optimization reduces to a bounded version of the multi-dimensional
  knapsack. The difference lies in the extra constraint on per-truck
  capacity, which can make the problem harder or easier depending on the structure of the items and
  capacities.
  
  
  
  This concludes our analysis of how varying the truck capacity constraints influences the optimal loading strategy.
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  Wednesday, 01 October 2025 10:35

  https://gimnas3.ru/d/articles/?cat_casino-promokod.html
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  搜狗输入法官网

  Wednesday, 01 October 2025 10:35

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  雷电下载

  Wednesday, 01 October 2025 10:16

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  火绒官网

  Wednesday, 01 October 2025 10:14

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  火绒安全下载

  Wednesday, 01 October 2025 10:09

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  Wednesday, 01 October 2025 09:31

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