oppi.li
advent of code solutions
aoc.oppi.li
haskell
aoc
2025: add d9
Signed-off-by: oppiliappan <me@oppi.li>
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<h2 data-number="2.5" id="day-9">Day 9</h2>
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<div class="sourceCode" id="cb1"><pre
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class="sourceCode haskell"><code class="sourceCode haskell"><span id="cb1-1"><a href="#cb1-1" aria-hidden="true" tabindex="-1"></a><span class="kw">import</span> <span class="dt">Data.List.Split</span> (splitOn)</span>
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<span id="cb1-3"><a href="#cb1-3" aria-hidden="true" tabindex="-1"></a><span class="ot">parse ::</span> <span class="dt">String</span> <span class="ot">-></span> [[<span class="dt">Integer</span>]]</span>
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<span id="cb1-4"><a href="#cb1-4" aria-hidden="true" tabindex="-1"></a>parse <span class="ot">=</span> <span class="fu">map</span> (<span class="fu">map</span> <span class="fu">read</span> <span class="op">.</span> splitOn <span class="st">","</span>) <span class="op">.</span> <span class="fu">lines</span></span></code></pre></div>
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<p>Onto the problem itself.</p>
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<h3 data-number="2.5.1" id="part-1-6">Part 1</h3>
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<p>This part requires us to simply find a rectangle with the
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largest area among the points given to us, so first write a method
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of a point is <code>[x, y]</code>:</p>
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<div class="sourceCode" id="cb2"><pre
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class="sourceCode haskell"><code class="sourceCode haskell"><span id="cb2-1"><a href="#cb2-1" aria-hidden="true" tabindex="-1"></a><span class="ot">area ::</span> [<span class="dt">Integer</span>] <span class="ot">-></span> [<span class="dt">Integer</span>] <span class="ot">-></span> <span class="dt">Integer</span></span>
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<span id="cb2-2"><a href="#cb2-2" aria-hidden="true" tabindex="-1"></a>area [x, y] [x', y'] <span class="ot">=</span> </span>
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<span id="cb2-3"><a href="#cb2-3" aria-hidden="true" tabindex="-1"></a> (<span class="dv">1</span> <span class="op">+</span> <span class="fu">abs</span> (x <span class="op">-</span> x')) <span class="op">*</span> (<span class="dv">1</span> <span class="op">+</span> <span class="fu">abs</span> (y <span class="op">-</span> y'))</span></code></pre></div>
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<p>Thus, the solution to the first part is given by the following
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list comprehension:</p>
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<div class="code">
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<div class="sourceCode" id="cb3"><pre
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class="sourceCode haskell"><code class="sourceCode haskell"><span id="cb3-1"><a href="#cb3-1" aria-hidden="true" tabindex="-1"></a><span class="ot">p1 ::</span> [[<span class="dt">Integer</span>]] <span class="ot">-></span> <span class="dt">Integer</span></span>
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<span id="cb3-2"><a href="#cb3-2" aria-hidden="true" tabindex="-1"></a>p1 poly <span class="ot">=</span> <span class="fu">maximum</span> [area p p' <span class="op">|</span> p <span class="ot"><-</span> poly, p' <span class="ot"><-</span> poly]</span></code></pre></div>
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</div>
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</div>
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<h3 data-number="2.5.2" id="part-2-5">Part 2</h3>
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<div class="row">
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<p>This part is a whole lot trickier. We now require only
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rectangles that fall <em>within</em> the polygon. Rectangles that
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have areas outside the polygon should be ignored. The methodology
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here is to identify rectangles that do not intersect in any way
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with the polygon itself. Sharing an edge with the polygon is
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alright however.</p>
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<div class="code">
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<div class="row">
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<p>Luckily, the intersections are quite easily calculated, since
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all lines of the polygon are oriented along the axes, and so are
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the lines forming the rectangle. If a line of the polygon falls to
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the rightmost line, or above the topmost line, or below the
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bottommost line, then the line does not intersect the
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rectangle:</p>
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<div class="code">
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<div class="sourceCode" id="cb4"><pre
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class="sourceCode haskell"><code class="sourceCode haskell"><span id="cb4-1"><a href="#cb4-1" aria-hidden="true" tabindex="-1"></a><span class="ot">intersects ::</span> [<span class="dt">Integer</span>] <span class="ot">-></span> [<span class="dt">Integer</span>] <span class="ot">-></span> [[<span class="dt">Integer</span>]] <span class="ot">-></span> <span class="dt">Bool</span></span>
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<span id="cb4-2"><a href="#cb4-2" aria-hidden="true" tabindex="-1"></a>intersects [x, y] [x', y'] <span class="ot">=</span> <span class="fu">not</span> <span class="op">.</span> <span class="fu">all</span> away <span class="op">.</span> pairs</span>
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<span id="cb4-3"><a href="#cb4-3" aria-hidden="true" tabindex="-1"></a> <span class="kw">where</span></span>
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<span id="cb4-4"><a href="#cb4-4" aria-hidden="true" tabindex="-1"></a> pairs (p <span class="op">:</span> ps) <span class="ot">=</span> <span class="fu">zip</span> (p <span class="op">:</span> ps) (ps <span class="op">++</span> [p])</span>
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<span id="cb4-5"><a href="#cb4-5" aria-hidden="true" tabindex="-1"></a> away ([lx, ly], [lx', ly']) <span class="ot">=</span></span>
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<span id="cb4-6"><a href="#cb4-6" aria-hidden="true" tabindex="-1"></a> (<span class="fu">max</span> lx lx' <span class="op"><=</span> <span class="fu">min</span> x x')</span>
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<span id="cb4-7"><a href="#cb4-7" aria-hidden="true" tabindex="-1"></a> <span class="op">||</span> (<span class="fu">min</span> lx lx' <span class="op">>=</span> <span class="fu">max</span> x x')</span>
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<span id="cb4-8"><a href="#cb4-8" aria-hidden="true" tabindex="-1"></a> <span class="op">||</span> (<span class="fu">max</span> ly ly' <span class="op"><=</span> <span class="fu">min</span> y y')</span>
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<span id="cb4-9"><a href="#cb4-9" aria-hidden="true" tabindex="-1"></a> <span class="op">||</span> (<span class="fu">min</span> ly ly' <span class="op">>=</span> <span class="fu">max</span> y y')</span></code></pre></div>
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</div>
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</div>
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<div class="row">
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<p>To explain a bit further, we define <code>away</code> according
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the the rules above. Then the <code>intersects</code> method is
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given by simply ensuring that of all pairs of points in the
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polygon (we need pairs to compute line segments of the polygon),
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we have atleast one line segment that is not away from the
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rectangle.</p>
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<div class="code">
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</div>
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</div>
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<div class="row">
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<p>Thus, the solution to the second part is a similar list
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comprehension with an additional guard:</p>
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<div class="code">
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<div class="sourceCode" id="cb5"><pre
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class="sourceCode haskell"><code class="sourceCode haskell"><span id="cb5-1"><a href="#cb5-1" aria-hidden="true" tabindex="-1"></a><span class="ot">p2 ::</span> [[<span class="dt">Integer</span>]] <span class="ot">-></span> <span class="dt">Integer</span></span>
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<span id="cb5-2"><a href="#cb5-2" aria-hidden="true" tabindex="-1"></a>p2 poly <span class="ot">=</span></span>
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<span id="cb5-3"><a href="#cb5-3" aria-hidden="true" tabindex="-1"></a> <span class="fu">maximum</span></span>
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<span id="cb5-4"><a href="#cb5-4" aria-hidden="true" tabindex="-1"></a> [ area p p'</span>
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<span id="cb5-5"><a href="#cb5-5" aria-hidden="true" tabindex="-1"></a> <span class="op">|</span> p <span class="ot"><-</span> poly,</span>
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<span id="cb5-6"><a href="#cb5-6" aria-hidden="true" tabindex="-1"></a> p' <span class="ot"><-</span> poly,</span>
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<span id="cb5-7"><a href="#cb5-7" aria-hidden="true" tabindex="-1"></a> <span class="fu">not</span> (intersects p p' poly)</span>
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<span id="cb5-8"><a href="#cb5-8" aria-hidden="true" tabindex="-1"></a> ]</span></code></pre></div>
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</div>
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</div>
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<div class="row">
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<p>Finally, a main function to wrap it all up:</p>
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<div class="code">
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<div class="sourceCode" id="cb6"><pre
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class="sourceCode haskell"><code class="sourceCode haskell"><span id="cb6-1"><a href="#cb6-1" aria-hidden="true" tabindex="-1"></a>main <span class="ot">=</span> <span class="kw">do</span></span>
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<span id="cb6-2"><a href="#cb6-2" aria-hidden="true" tabindex="-1"></a> n <span class="ot"><-</span> parse <span class="op"><$></span> <span class="fu">getContents</span></span>
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<span id="cb6-3"><a href="#cb6-3" aria-hidden="true" tabindex="-1"></a> <span class="fu">print</span> <span class="op">$</span> p1 n</span>
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<span id="cb6-4"><a href="#cb6-4" aria-hidden="true" tabindex="-1"></a> <span class="fu">print</span> <span class="op">$</span> p2 n</span></code></pre></div>
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<title>The Book of Solves</title>
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<header id="title-block-header">
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<h1 class="title">The Book of Solves</h1>
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<p class="subtitle">Advent of Code solutions in Haskell</p>
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<p class="author">by <a
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href="https://tangled.org/oppi.li"><span class="citation"
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data-cites="oppi.li">@oppi.li</span></a></p>
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</header>
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<ul>
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<li><a href="1-section.html#section" id="toc-section">2016</a>
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<li><a href="2.2-day-4.html#day-4" id="toc-day-4">Day 4</a></li>
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<li><a href="2.3-day-5.html#day-5" id="toc-day-5">Day 5</a></li>
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<li><a href="2.4-day-6.html#day-6" id="toc-day-6">Day 6</a></li>
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<li><a href="2.5-day-9.html#day-9" id="toc-day-9">Day 9</a></li>
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{"section":{"id":"","level":"0","number":null,"path":"index.html","title":"The Book of Solves"},"subsections":[{"section":{"id":"section","level":"1","number":"1","path":"1-section.html#section","title":"2016"},"subsections":[{"section":{"id":"day-1","level":"2","number":"1.1","path":"1.1-day-1.html#day-1","title":"Day 1"},"subsections":[{"section":{"id":"part-1","level":"3","number":"1.1.1","path":"1.1-day-1.html#part-1","title":"Part 1"},"subsections":[]},{"section":{"id":"part-2","level":"3","number":"1.1.2","path":"1.1-day-1.html#part-2","title":"Part 2"},"subsections":[]}]},{"section":{"id":"day-2","level":"2","number":"1.2","path":"1.2-day-2.html#day-2","title":"Day 2"},"subsections":[{"section":{"id":"part-1-1","level":"3","number":"1.2.1","path":"1.2-day-2.html#part-1-1","title":"Part 1"},"subsections":[]}]}]},{"section":{"id":"section-1","level":"1","number":"2","path":"2-section-1.html#section-1","title":"2025"},"subsections":[{"section":{"id":"day-2-1","level":"2","number":"2.1","path":"2.1-day-2-1.html#day-2-1","title":"Day 2"},"subsections":[{"section":{"id":"part-1-2","level":"3","number":"2.1.1","path":"2.1-day-2-1.html#part-1-2","title":"Part 1"},"subsections":[]},{"section":{"id":"part-2-1","level":"3","number":"2.1.2","path":"2.1-day-2-1.html#part-2-1","title":"Part 2"},"subsections":[]}]},{"section":{"id":"day-4","level":"2","number":"2.2","path":"2.2-day-4.html#day-4","title":"Day 4"},"subsections":[{"section":{"id":"part-1-3","level":"3","number":"2.2.1","path":"2.2-day-4.html#part-1-3","title":"Part 1"},"subsections":[]},{"section":{"id":"part-2-2","level":"3","number":"2.2.2","path":"2.2-day-4.html#part-2-2","title":"Part 2"},"subsections":[]}]},{"section":{"id":"day-5","level":"2","number":"2.3","path":"2.3-day-5.html#day-5","title":"Day 5"},"subsections":[{"section":{"id":"part-1-4","level":"3","number":"2.3.1","path":"2.3-day-5.html#part-1-4","title":"Part 1"},"subsections":[]},{"section":{"id":"part-2-3","level":"3","number":"2.3.2","path":"2.3-day-5.html#part-2-3","title":"Part 2"},"subsections":[]}]},{"section":{"id":"day-6","level":"2","number":"2.4","path":"2.4-day-6.html#day-6","title":"Day 6"},"subsections":[{"section":{"id":"part-1-5","level":"3","number":"2.4.1","path":"2.4-day-6.html#part-1-5","title":"Part 1"},"subsections":[]},{"section":{"id":"part-2-4","level":"3","number":"2.4.2","path":"2.4-day-6.html#part-2-4","title":"Part 2"},"subsections":[]}]},{"section":{"id":"day-9","level":"2","number":"2.5","path":"2.5-day-9.html#day-9","title":"Day 9"},"subsections":[{"section":{"id":"part-1-6","level":"3","number":"2.5.1","path":"2.5-day-9.html#part-1-6","title":"Part 1"},"subsections":[]},{"section":{"id":"part-2-5","level":"3","number":"2.5.2","path":"2.5-day-9.html#part-2-5","title":"Part 2"},"subsections":[]}]}]}]}
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···
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## Day 9
2
+
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This day's problem is based on simple coordinate geometry.
4
+
5
+
Parse the input data first (pairs of numbers separated by commas, one pair per line):
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+
7
+
```haskell
8
+
import Data.List.Split (splitOn)
9
+
10
+
parse :: String -> [[Integer]]
11
+
parse = map (map read . splitOn ",") . lines
12
+
```
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+
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+
Onto the problem itself.
15
+
16
+
### Part 1
17
+
18
+
This part requires us to simply find a rectangle with the largest area among the points given to us, so first write a method to compute areas given a point. Remember that our representation of a point is `[x, y]`:
19
+
20
+
```haskell
21
+
area :: [Integer] -> [Integer] -> Integer
22
+
area [x, y] [x', y'] =
23
+
(1 + abs (x - x')) * (1 + abs (y - y'))
24
+
```
25
+
26
+
Thus, the solution to the first part is given by the following list comprehension:
27
+
28
+
```haskell
29
+
p1 :: [[Integer]] -> Integer
30
+
p1 poly = maximum [area p p' | p <- poly, p' <- poly]
31
+
```
32
+
33
+
### Part 2
34
+
35
+
This part is a whole lot trickier. We now require only rectangles that fall *within* the polygon. Rectangles that have areas outside the polygon should be ignored. The methodology here is to identify rectangles that do not intersect in any way with the polygon itself. Sharing an edge with the polygon is alright however.
36
+
37
+
Luckily, the intersections are quite easily calculated, since all lines of the polygon are oriented along the axes, and so are the lines forming the rectangle. If a line of the polygon falls to the left of the leftmost line of the rectangle, or to the right of the rightmost line, or above the topmost line, or below the bottommost line, then the line does not intersect the rectangle:
38
+
39
+
```haskell
40
+
intersects :: [Integer] -> [Integer] -> [[Integer]] -> Bool
41
+
intersects [x, y] [x', y'] = not . all away . pairs
42
+
where
43
+
pairs (p : ps) = zip (p : ps) (ps ++ [p])
44
+
away ([lx, ly], [lx', ly']) =
45
+
(max lx lx' <= min x x')
46
+
|| (min lx lx' >= max x x')
47
+
|| (max ly ly' <= min y y')
48
+
|| (min ly ly' >= max y y')
49
+
```
50
+
51
+
To explain a bit further, we define `away` according the the rules above. Then the `intersects` method is given by simply ensuring that of all pairs of points in the polygon (we need pairs to compute line segments of the polygon), we have atleast one line segment that is not away from the rectangle.
52
+
53
+
Thus, the solution to the second part is a similar list comprehension with an additional guard:
54
+
55
+
```haskell
56
+
p2 :: [[Integer]] -> Integer
57
+
p2 poly =
58
+
maximum
59
+
[ area p p'
60
+
| p <- poly,
61
+
p' <- poly,
62
+
not (intersects p p' poly)
63
+
]
64
+
```
65
+
66
+
Finally, a main function to wrap it all up:
67
+
68
+
```haskell
69
+
main = do
70
+
n <- parse <$> getContents
71
+
print $ p1 n
72
+
print $ p2 n
73
+
```