Solving 0/1 knapsack problem with dynamic programming

1|0Introduction

1|1Story of 0/1 knapsack problems

1|2Application of 0/1 knapsack problem

• On 0-1 Knapsack Problem Model for Security Investment Decision(economics)

  • A new security investment decision model is established

    • this model represents decision variables with each security investment project
    • represents weight with the number of investments in each security investment project,
    • represents value with the weight value of each security investment project.

  • LUO Jing-feng. On 0-1 Knapsack Problem Model for Security Investment Decision[J]. West Forum on Economy and Management, 2016, 27(1): 75-78.

• And all the problems that can be abstracted as this model.

2|0Content

2|1Why can we use dynamic programing(DP) to solve the problem.

• Principle of optimality
  • For this issue
    • The maximum value of the $i_{th}$ item with a total weight of $W$ can be derived from the maximum value of the first ${i-1}_{th}$ item with a weight of $W-w_i$
• Without aftereffect
  • For this issue
    • The maximum value at the current weight of $w$ is still the maximum value at w when the current weight is greater than $w$.

2|2Why should we use DP

• Faster than depth first search(DFS)
  • Time complexity achieved using DP is just $\mathrm{O}(nv)$
  • Despite extensive optimization, the time complexity of DFS remains high
• Higher accuracy than greedy algorithms
  • Obvious, an item cannot be separated.
  • as long as $w_i\mathrm{mod}{v}_i\ne 0$,Greedy algorithms cannot find maximum value.

2|3Our solution for the problem by DP

• Define the state
  • We define the state $F_{i,j}$ as the maximum value that a backpack can hold for the first $i$ items with a weight of $j$.
• Derive state transition equation
  • Facing the $i_{th}$ item, we only have two choice,then find which one can get maximum value.
    • Don't chose the $i_{th}$ item
      • Obviously, the $F_{i,j}$ can be transferred from $F_{i-1,j}$
    • Chose the $i_{th}$ item
      • Consider the weight of the current item
        • We can't chose the item without any weight consumption
        • We need $w_i$ weight to hold the item
        • then we should transfer from $F_{i-1,j-w_i}$
      • Consider the value of the current item
        • Obviously,when we chose the item, we get $v_i$ value。
        • $F_{i-1,j-w_i}+v_i$
    • $F_{i,j}=max(F_{i-1,j},F_{i-1,j-w_i}+v_i)$
• Optimize space consumption
  • Obviously,every $F_{i,j}$ is transferred from $F_{i-1,?}$
  • So we don't need to declare a two-dimensional array.
  • Instead of that, we use a one-dimensional array and do reverse enumeration so that we can use the $i-1$ state to update the $i$ state.
• Simple code implementation

for (int i = 1; i <= M; i++)
{
	for (int j = T; j >= v[i]; j--)
	{
		dp[j] = max(dp[j], dp[j - v[i]] + w[i]);
	}
}    

3|0Conclusion

In a word, use our DP solution to the 0/1 knapsack problem.

You can get:

• Better time complexity
  • save your time
• Better space complexity
  • reduce your space consumption
• Simple code implementation
  • Easy to maintain

__EOF__

本文作者：Kdlyh
本文链接：https://www.cnblogs.com/kdlyh/p/17844404.html
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