I don't understand a step in my book proving the fractional knapsack problem:

Let value of items $v_1\ge v_2\ge \dots\ge v_n$, and assume $X=\langle x_1, \dots,x_n\rangle$ are the solution by greedy, where $0\le x_i\le 1$ is the fraction packed into the knapsack.

Assume $j$ is the first index s.t. $x_j<1$. Let $Y=\langle y_1,\dots, y_n\rangle$ be any solution not $X$.

Consider $$\dfrac{v_i}{w_i}(x_i-y_i)\ge\dfrac{v_j}{w_j}(x_i-y_i),\tag{????}$$

So $\displaystyle\sum_{i=1}^n v_i(x_i-y_i)=\sum\color{blue}v_i\dfrac{w_i}{\color{blue}w_i}(x_i-y_i)\ge\color{blue}{\dfrac{v_j}{w_j}}\sum w_i(x_i-y_i)\ge0.$

I can understand the blue part I highlighted, but can anyone help me understand the (????) part? Why it must hold?


The following condition is implicitly included in the question.

$$\dfrac{v_1}{w_1}, \dfrac{v_2}{w_2}, \cdots, \dfrac{v_n}{w_n} \text{ is in the descending order.}$$

Let $W$ be the total weight to be filled. The greedy algorithm for fractional knapsack problem is the following procedure.

Let $k$ loop through $1, 2, ..., n$ in that order.

  1. Set $$x_k = \dfrac{W- \sum_{1\le i\lt k} x_i}{w_i}$$
  2. If $x_k<1$, set $x_l=0$ for all $l>k$. Break the loop.


There are three cases for $i$.

  • The case when i < j, i.e., $\dfrac{v_i}{w_i}\ge\dfrac{v_j}{w_j}$. Since $j$ is the first index s.t. $x_j<1$, $x_i=1$, i.e., $x_i-y_i\ge0$.
  • The case when i = j. Both sides are 0.

  • The case when i > j, i.e., $\dfrac{v_i}{w_i}\le\dfrac{v_j}{w_j}$. Since $j$ is the first index s.t. $x_j<1$, according to the definition of the greedy algorithm, $x_i=0$. That means $x_i-y_i\le0$.

  • $\begingroup$ The condition "$v_1\ge v_2\ge \dots\ge v_n$" should be $\dfrac{v_1}{w_1}\ge\dfrac{v_2}{w_2}\ge \cdots\ge\dfrac{v_n}{w_n}$ instead. $\endgroup$ – Apass.Jack Jan 1 at 6:02
  • $\begingroup$ So it seems like the condition $\dfrac{v_1}{w_1}\ge\dfrac{v_2}{w_2}\ge \cdots\ge\dfrac{v_n}{w_n}$ is actually not needed? Only need to know $x_j$ is the last one greedy choose. $\endgroup$ – Bit_hcAlgorithm Jan 1 at 6:07
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    $\begingroup$ Yes, if we change "Assume $j$ is the first index s.t. $x_j<1$" to "Assume $j$ is the only index s.t. $x_j<1$". $\endgroup$ – Apass.Jack Jan 1 at 6:08
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    $\begingroup$ It should be "Assume $j$ is the only index s.t. $0<x_j<1$". But it might happen all $x_k$ are either 1 or 0. So it looks like there is some problem with your book or your formulation of the question. (It seems my answer is valid.) Anyway, I would believe you have understood what is happening here. $\endgroup$ – Apass.Jack Jan 1 at 6:18

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