2つのパスが到達できるポイントの最大数

のリストが与えられたとしましょう $n$ ポイント、その $x$ そして $y$座標はすべて負ではありません。また、重複するポイントがないとします。ポイントからしか行けない$(x_i, y_i)$ ポイントへ $(x_j, y_j)$ もし $x_i \le x_j$ そして $y_i \le y_j$。問題は次のとおりです。$n$ポイント、上記のルールを使用してポイントを接続する2つのパスを描画することが許可されている場合に到達できるポイントの最大数はいくつですか？パスは原点から開始する必要があり、繰り返しポイントを含めることができます。$(0, 0)$ もちろん到達ポイントには含まれていません。

例：与えられた $(2, 0), (2, 1), (1, 2), (0, 3), (1, 3), (2, 3), (3, 3), (2, 4), (1, 5), (1, 6)$、答えは $8$ 私たちは取ることができるので $(0, 0) \rightarrow (2, 0) \rightarrow (2, 1) \rightarrow (2, 3) \rightarrow (2, 4)$ そして $(0, 0) \rightarrow (1, 2) \rightarrow (1, 3) \rightarrow (1, 5) \rightarrow (1, 6)$。

パスを1つだけ描画することが許可されている場合、実行される動的プログラミングによって問題を簡単に解決できます $O(n^2)$。最初にポイントを降順に並べ替えます$x_i+y_i$。しましょう$D[i]$ コインからピックアップできるコインの最大数 $1$ に $i$ソートされたリスト。その後$D[1] = 1$ そして $D[i] = \max\limits_{1\le j < i, x_j \le x_i, y_j \le y_i} D[j] + 1$。そのときの答えは$\max\limits_{1\le i \le n} D[i] + 1$。

しかし、私は2つの経路の繰り返し関係を思い付くことができません。このような再発関係について何かご存じの方がいらっしゃれば、ぜひお聞かせください。

問題、言い換えると一般化：有限集合が与えられた場合 $S$部分注文を装備 $\le$、チェーンを見つける $C_1, C_2 \subseteq S$ 最大化 $\lvert C_1 \cup C_2 \rvert$。問題は、$S \subseteq \mathbb R_+^2$ そして $(x, y) \le (z, w) \Longleftrightarrow x \le z \wedge y \le w$。

単純に、単一の最良のチェーンを見つけようとするかもしれません $S^2$、チェーンのコンポーネントが持つ明確な値の数によって最もよく測定されます。残念ながら、1つのコンポーネントは他のコンポーネントのステップをたどることができます。

$$\bigl((0,0),(0,0)\bigr) < \bigl((1,0),(0,0)\bigr) < \bigl((2,0),(0,0)\bigr) < \bigl((2,0),(1,0)\bigr),$$ したがって、この最良の概念には最適な部分構造はありません。

Instead, we look for chains in the set $T := \{(x, y) \mid (x, y) \in S^2 \wedge x \nless y \wedge y \nless x\}$. By requiring that the components be equal or incomparable, we prevent retracing but now need to argue that some best chain conforms to the new requirement.

Lemma 1 (no retracing). Let $C \subseteq T$ be a chain and define $C_1 := \{x \mid (x, y) \in C\}$ and $C_2 := \{y \mid (x, y) \in C\}$. For all $z \in S$, we have $z \in C_1 \cap C_2$ if and only if $(z, z) \in C$.

Proof. The if direction is trivial. In the only if direction, for all $z \in C_1 \cap C_2$, there exist $x, y \in S$ such that $(x, z), (z, y) \in C$. Since $C$ is a chain, $(x, z) \le (z, y) \vee (z, y) \le (x, z)$. Assume symmetrically that $(x, z) \le (z, y)$, which implies that $x \le z \le y$. We know by the definition of $T$ that $x \nless z \wedge z \nless y$, so $x = z = y$, and $(z, z) \in C$.

Lemma 2 (existence of restricted best chain). For all chains $C_1, C_2 \subseteq S$, there exists a chain $C \subseteq T$ such that $C_1 \subseteq \{x \mid (x, y) \in C\} \subseteq C_1 \cup C_2$ and $C_2 \subseteq \{y \mid (x, y) \in C\} \subseteq C_1 \cup C_2$.

Proof (revised). We give an algorithm to construct $C$. For convenience, define sentinels $\bot, \top$ such that $\bot < x < \top$ for all $x \in S$. Let $C_1' := C_1 \cup \{\top\}$ and $C_2' := C_2 \cup \{\top\}$.

1. Initialize $C := \varnothing$ and $x := \bot$ and $y := \bot$. An invariant is that $x \nless y \wedge y \nless x$.

2. Let $x'$ be the next element of $C_1$, that is, $x' := \inf \{z \mid z \in C_1' \wedge x < z\}$. Let $y'$ be the next element of $C_2$, that is, $y' := \inf \{w \mid w \in C_2' \wedge y < w\}$.

3. If $x' \nless y' \wedge y' \nless x'$, set $(x, y) := (x', y')$ and go to step 9.

4. If $y < x' < y'$, set $(x, y) := (x', x')$ and go to step 9.

5. If $y \nless x' < y'$, set $x := x'$ and go to step 9. Note that $x < x' \wedge x \nless y$ implies that $x' \nless y$.

6. If $x < y' < x'$, set $(x, y) := (y', y')$ and go to step 9.

7. If $x \nless y' < x'$, set $y := y'$ and go to step 9. Note that $y < y' \wedge y \nless x$ implies that $y' \nless x$.

8. This step is never reached, as the conditions for steps 3–7 are exhaustive.

9. If $x \ne \top$ (equivalently, $y \ne \top$), set $C := C \cup \{(x, y)\}$ and go to step 2.

Dynamic Program. For all $(x, y) \in T$, compute

$$D[x, y] := \sup\biggl(\Bigl\{D[z, w] + [x \ne z] + [y \ne w] - [x = y] \mathrel{\Bigl|\Bigr.} (z, w) \in T \wedge (z, w) < (x, y)\Bigr\} \cup \bigl\{2 - [x = y]\bigr\}\biggr),$$ where $[\textit{condition}] = 1$ if $\textit{condition}$ is true and $[\textit{condition}] = 0$ if $\textit{condition}$ is false. By Lemma 1, it follows that the bracket expressions correctly count the number of new elements. By Lemma 2, the optimal solution to the original problem is found.

Let $P = p_1\dots p_n$ to sorted list of points.

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Following your recurrence for one path, the first thing to note is that you have to keep track of which points have been visited by the paths; otherwise you can not count correctly. The second thing is that you have now four possibilities for every point: neither path may use it, one of them or both. So, we have to find maximising combinations for all three cases.

Formally, let $d : [0\dots n] \to (2^{[n]} \times 2^{[n]})^3$ with $d(i)$ the pair of (sets of) visited nodes of the two paths that maximise the number of visited points from the input set up to the $i$th one, with the first component the maximising pair of paths for which the first one uses $p_i$, the second component similar for the second path and the third component with both paths using $p_i$. $d$ is given by the recurrence

$\quad \displaystyle \begin{align} d(0) &= ((\emptyset, \emptyset),(\emptyset, \emptyset),(\emptyset, \emptyset)) \\ d(i) &= (\ \arg\max_{(L,R) \in (\mathcal{L}' \times \mathcal{R})_i} |L \cup R|, \\ &\phantom{= (\ } \arg\max_{(L,R) \in (\mathcal{L} \times \mathcal{R}')_i} |L \cup R|, \\ &\phantom{= (\ } \arg\max_{(L,R) \in (\mathcal{L}' \times \mathcal{R}')_i} |L \cup R| \ ) \\ \end{align}$

with

$\quad \displaystyle (\mathcal{L}' \times \mathcal{R})_i = \left\{ (L \cup \{i\}, R) \mid (L,R) \in \bigcup_{j=0}^{i-1} d(j), x_i \geq \max_{j \in L} x_j, y_i \geq \max_{j \in L} y_j \right\}$,

$(\mathcal{L}' \times \mathcal{R})_i$ similar with extending $R$ and $(\mathcal{L}' \times \mathcal{R}')_i$ similar with extending both $L$ and $R$.

Needless to say, this is not very nice. This is because the problem does not lend itself to dynamic programming very well: you can not combine many partial solutions because there is no nice total ordering on the points, and you can not discard intermediate results for the same reason.

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A nicer view on the problem is to model the set of points as weighted directed acyclic graph $G = (V, E, w)$ with

• $V = \{(0,0), p_1, \dots, p_n, (X,Y)\}$ with $X = \max x_i$, $Y = \max y_i$, and
• $E = \{ ((x_1, y_1),(x_2,y_2)) \in V^2 \mid x_i \leq x_j, y_i \leq y_j\}$ and
• $w(v_1,v_2) = \begin{cases} 0 &, v_2 = (X,Y) \\ 1 &, \text{ else}\end{cases}$.

Note that you can keep the graph smaller if you remove redundant edges, that is remove $(v_1,v_2)$ if there is a path $(v_1,\dots,v_2)$, because taking such "shortcuts" is never better advantageous.

For one path, the solution is clearly the length of the longest path $P^*$ from $(0,0)$ to $(X,Y)$. Now, if we change $w$ so that all edges leading to points on $P^*$ also have weight $0$ and compute the longest path in this modified graph, we get a path $P^+$ so that $P^*$ and $P^+$ together cover as many points as two paths can. This leaves us with a runtime in $O(|V| + |E|) \subseteq O(n^2)$ (see here).