東京大学 情報理工学系研究科 創造情報学専攻 2021年8月実施 筆記試験 第1問

标签：

• Tokyo-University
• Greedy-Algorithm
• Knapsack-Problem

Author

itsuitsuki, 祭音Myyura

Description

Let $n$ and $r$ be positive integers. For $i=1,2,\dots,n$, let $f_i$ be a univariate real-valued function defined in the integer domain and let $f_i(x_i)$ be $-\infty$ for negative integer $x_i$. Any non-negative integer solution $(x_1,\dots,x_n)$ that satisfies $\sum_{i=1}^n x_i=r$ is called a feasible solution. In addition, a feasible solution that maximizes the objective function $\sum_{i=1}^n f_i(x_i)$ is called an optimal solution and the objective function value at the solution is called the optimal value. This problem is expressed as follows.

$$(P) \ \left| \ \begin{aligned} &\text{Maximize} && \sum_{i=1}^n f_i(x_i) \\ &\text{subject to} && \sum_{i=1}^n x_i=r \\ &&& x_i \text{ is a non-negative integer, } i=1,\dots,n \end{aligned} \right.$$

(1)

For $i=1,2,\dots,n$ and non-negative integer $\alpha$, define the function $d_i(\alpha):=f_i(\alpha)-f_i(\alpha-1)$ and assume that $d_i(\alpha)$ is non-increasing in terms of $\alpha$. Apply the following greedy algorithm $\mathcal{A}_G$ to $(P)$.

Step 0: For $i=1,2,\dots,n$, set $x_i\leftarrow 0$.

Step 1: Repeat the following procedure for $r$ times: Let $\gamma$ be any index $i$ that maximizes $d_i(x_i+1)$ among $i=1,2,\dots,n$, and set $x_{\gamma}\leftarrow x_{\gamma}+1$.

Answer the following questions.

(1-1) Let $r=5,n=3$, and let $f_1,f_2,f_3$ take the following values. Notice that $d_1,d_2,d_3$ are non-increasing. Answer the solution obtained by the greedy algorithm $\mathcal{A}_G$.

$\alpha$	0	1	2	3	$\dots$
$f_1(\alpha)$	0	0	-8	-24	$\dots$
$f_2(\alpha)$	-2	1	-14	-40	$\dots$
$f_3(\alpha)$	0	-3	-12	-22	$\dots$

(1-2) Let $(x_1^*,x_2^*,\dots,x_n^*)$ be a feasible solution. Show that it is an optimal solution of $(P)$ if and only if the following condition holds.

$$\max_{i=1,2,\dots,n}d_i(x_i^*+1)\le\min_{i=1,2,\dots,n}d_i(x_i^*)$$

(1-3) Show that the greedy algorithm $\mathcal{A}_G$ outputs an optimal solution of $(P)$.

(2)

Unless the non-increasing assumption of (1) holds, the greedy algorithm $\mathcal{A}_G$ does not always output an optimal solution of $(P)$. To apply dynamic programming, we consider the following problem $(P_N^R)$ in which $n$ and $r$ in $(P)$ are replaced with $N\in\{1,2,\dots,n\}$ and $R\in\{0,1,\dots,r\}$, respectively.

$$(P_N^R) \ \left| \ \begin{aligned} &\text{Maximize} && \sum_{i=1}^N f_i(x_i) \\ &\text{subject to} && \sum_{i=1}^N x_i = R \\ &&& x_i \text{ is a non-negative integer, } i = 1, \dots, N \end{aligned} \right.$$

The optimal value of the problem is denoted by $g_N(R)$. Answer the following questions.

(2-1) Express $g_N(R)$ only with $g_{N-1}(c)$ and $f_N(c)$ for any non-negative integer $c$ in the case of $N\ge 2$.

(2-2) Write a pseudo-code of a dynamic programming algorithm within 15 lines to output the optimal value $g_n(r)$ of $(P)$. Hereafter, this algorithm is called $\mathcal{A}_D$.

(2-3) Show that the optimal value of $(P)$ is obtained by the dynamic programming algorithm $\mathcal{A}_D$.

(2-4) Answer the computational complexity of the dynamic programming algorithm $\mathcal{A}_D$ and the computational complexity of the greedy algorithm $\mathcal{A}_G$. Ignore the computational cost of calculating $f_1,\dots,f_n$.

题目描述

给定正整数 $n,r$。对 $i=1,\ldots,n$，$f_i$ 是定义在整数域上的一元实值函数，并约定负整数 $x_i$ 时 $f_i(x_i)=-\infty$。满足 $x_i$ 为非负整数且 $\sum_{i=1}^nx_i=r$ 的解为可行解；使 $\sum_i f_i(x_i)$ 最大者为最优解，其目标值为最优值：

$$(P)\quad \begin{array}{ll} \text{最大化}&\displaystyle\sum_{i=1}^nf_i(x_i)\\ \text{约束}&\displaystyle\sum_{i=1}^nx_i=r,\quad x_i\in\mathbb Z_{\ge0}. \end{array}$$

1. 边际收益非增时的贪心算法

定义

$$d_i(\alpha)=f_i(\alpha)-f_i(\alpha-1)$$

并假设对非负整数 $\alpha$，每个 $d_i(\alpha)$ 随 $\alpha$ 非增。算法 $\mathcal A_G$：

• 初始令所有 $x_i=0$。
• 重复 $r$ 次：任选使 $d_i(x_i+1)$ 最大的下标 $\gamma$，令 $x_\gamma\leftarrow x_\gamma+1$。

1. 当 $r=5,n=3$ 且函数值如下时，求 $\mathcal A_G$ 输出的解：

   $\alpha$	0	1	2	3	$\dots$
   $f_1(\alpha)$	0	0	-8	-24	$\dots$
   $f_2(\alpha)$	-2	1	-14	-40	$\dots$
   $f_3(\alpha)$	0	-3	-12	-22	$\dots$

2. 对任一可行解 $(x_1^*,\ldots,x_n^*)$，证明它最优当且仅当

   $$\max_i d_i(x_i^*+1)\le\min_i d_i(x_i^*).$$

3. 证明贪心算法 $\mathcal A_G$ 必输出最优解。

2. 一般情形的动态规划

若边际收益非增假设不成立，贪心不一定最优。把 $P$ 中 $n,r$ 分别替换为 $N\in\{1,\ldots,n\}$、$R\in\{0,\ldots,r\}$，定义子问题

$$(P_N^R)\quad \begin{array}{ll} \text{最大化}&\displaystyle\sum_{i=1}^Nf_i(x_i)\\ \text{约束}&\displaystyle\sum_{i=1}^Nx_i=R,\quad x_i\in\mathbb Z_{\ge0}, \end{array}$$

其最优值记为 $g_N(R)$。

1. 当 $N\ge2$ 时，只使用非负整数 $c$ 对应的 $g_{N-1}(c)$ 与 $f_N(c)$ 写出 $g_N(R)$ 的递推式。
2. 用不超过 15 行伪代码写动态规划算法 $\mathcal A_D$，输出 $g_n(r)$。
3. 证明 $\mathcal A_D$ 得到 $P$ 的最优值。
4. 分别给出 $\mathcal A_D$ 与 $\mathcal A_G$ 的计算复杂度，忽略计算各 $f_i$ 的代价。

Kai

(1)

(1-1)

The selected indices are $2,1,3,1,3$. Therefore

$$\boxed{(x_1,x_2,x_3)=(2,1,2)}.$$

(1-2)

Set $d_i(0)=+\infty$, as follows from $f_i(-1)=-\infty$.

If the condition fails, choose $p,q$ such that $d_p(x_p^*+1)>d_q(x_q^*)$. Necessarily $x_q^*>0$. Moving one unit from $q$ to $p$ changes the objective by

$$d_p(x_p^*+1)-d_q(x_q^*)>0,$$

so $x^*$ is not optimal.

Conversely, compare $x^*$ with any feasible $x$. Every marginal added where $x_i>x_i^*$ is at most $\max_i d_i(x_i^*+1)$, while every marginal removed where $x_i<x_i^*$ is at least $\min_i d_i(x_i^*)$. The numbers added and removed are equal. Under the stated inequality, the objective at $x$ cannot exceed that at $x^*$; hence $x^*$ is optimal.

(1-3)

Fix $j$ with final $x_j>0$ and consider the last iteration that increments $x_j$. Greedy selection gives, for every $i$, the selected marginal $d_j(x_j)$ at least the then-available marginal of $i$. Since each $d_i$ is non-increasing, the final value satisfies

$$d_i(x_i+1)\le d_j(x_j).$$

For $x_j=0$, $d_j(0)=+\infty$. Thus the condition in (1-2) holds, so the greedy output is optimal.

(2)

(2-1)

Assigning $c$ units to the last variable leaves $R-c$ units for the first $N-1$ variables:

$$\boxed{g_N(R)=\max_{0\le c\le R}\{g_{N-1}(R-c)+f_N(c)\}}, \qquad g_1(R)=f_1(R).$$

(2-2)

for R = 0 to r:
    g[1,R] = f_1(R)
for N = 2 to n:
    for R = 0 to r:
        g[N,R] = -infinity
        for c = 0 to R:
            g[N,R] = max(g[N,R],
                         g[N-1,R-c] + f_N(c))
return g[n,r]

(2-3)

The base row is the exact optimum for one variable. Assuming row $N-1$ is optimal, every feasible solution for $(P_N^R)$ has one value $x_N=c$ and an optimal value at most $g_{N-1}(R-c)+f_N(c)$. Conversely, combining the maximizing $c$ with an optimizer of that subproblem is feasible. Induction proves that the algorithm returns $g_n(r)$.

(2-4)

The dynamic program takes $\Theta(nr^2)$ time and $\Theta(nr)$ space, reducible to $\Theta(r)$ space with two rows. A direct implementation of the stated greedy algorithm scans $n$ marginals in each of $r$ iterations, taking $\Theta(nr)$ time and $\Theta(n)$ space.