Dyalog APL competition phase 2: Add explanations for problems 1-4
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<a class="sourceLine" id="cb2-8" title="8"> :EndIf</a>
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<a class="sourceLine" id="cb2-9" title="9"> score←2(⍎⍕)dd×+/scores</a>
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<a class="sourceLine" id="cb2-10" title="10">∇</a></code></pre></div>
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<p>This is a very straightforward implementation of the algorithm describe in the problem description. I decided to switch explicitly on the size of the input vector because I feel it is more natural. For the cases with 5 or 7 judges, we use Drop (<code>↓</code>) to remove the lowest and highest scores.</p>
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<p>At the end, we sum up the scores with <code>+/</code> and multiply them by <code>dd</code>. The last operation, <code>2(⍎⍕)</code>, is a train using <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Format%20Dyadic.htm">Format (Dyadic)</a> to round to 2 decimal places, and <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Execute.htm">Execute</a> to get actual numbers and not strings.</p>
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<h2 id="problem-2-another-step-in-the-proper-direction">Problem 2 – Another Step in the Proper Direction</h2>
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<div class="sourceCode" id="cb3"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb3-1" title="1">∇ steps←{p}Steps fromTo;segments;width</a>
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<a class="sourceLine" id="cb3-2" title="2"> width←|-/fromTo</a>
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@ -103,147 +105,161 @@
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<a class="sourceLine" id="cb3-12" title="12"> ⍝ Take into account the start point and the direction.</a>
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<a class="sourceLine" id="cb3-13" title="13"> steps←fromTo{(⊃⍺)+(-×-/⍺)×⍵}segments</a>
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<a class="sourceLine" id="cb3-14" title="14">∇</a></code></pre></div>
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<p>This is an extension to <a href="./dyalog-apl-competition-2020-phase-1.html#stepping-in-the-proper-direction">Problem 5 of Phase I</a>. In each case, we compute the “segments”, i.e., the steps starting from 0. In a last step, common to all cases, we add the correct starting point and correct the direction if need be.</p>
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<p>To compute equally-sized steps, we first divide the segment <span class="math inline">\([0, 1]\)</span> in <code>p</code> equal segments with <code>(⍳p)÷p</code>. This subdivision can then be multiplied by the width to obtain the required segments.</p>
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<p>When <code>p</code> is the step size, we just divide the width by the step size (rounded to the next largest integer) to get the required number of segments. If the last segment is too large, we “crop” it to the width with Minimum (<code>⌊</code>).</p>
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<h2 id="problem-3-past-tasks-blast">Problem 3 – Past Tasks Blast</h2>
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<div class="sourceCode" id="cb4"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb4-1" title="1">∇ urls←PastTasks url;r;paths</a>
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<a class="sourceLine" id="cb4-2" title="2"> r←HttpCommand.Get url</a>
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<a class="sourceLine" id="cb4-3" title="3"> paths←('[a-zA-Z0-9_/]+\.pdf'⎕S'&')r.Data</a>
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<a class="sourceLine" id="cb4-4" title="4"> urls←('https://www.dyalog.com/'∘,)¨paths</a>
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<a class="sourceLine" id="cb4-5" title="5">∇</a></code></pre></div>
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<p>I decided to use <code>HttpCommand</code> for this task, since it is simply one <code>]load HttpCommand</code> away and should be platform-independent.</p>
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<p>Parsing XML is not something I consider “fun” in the best of cases, and I feel like APL is not the best language to do this kind of thing. Given how simple the task is, I just decided to find the relevant bits with a regular expression using <a href="https://help.dyalog.com/18.0/index.htm#Language/System%20Functions/r.htm">Replace and Search</a> (<code>⎕S</code>).</p>
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<p>After finding all the strings vaguely resembling a PDF file name (only alphanumeric characters and underscores, with a <code>.pdf</code> extension), I just concatenate them to the base URL of the Dyalog domain.</p>
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<h2 id="problem-4-bioinformatics">Problem 4 – Bioinformatics</h2>
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<p>The first task can be solved by decomposing it into several functions.</p>
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<div class="sourceCode" id="cb5"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb5-1" title="1">⍝ Test if a DNA string is a reverse palindrome.</a>
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<a class="sourceLine" id="cb5-2" title="2">isrevp←{⍵≡⌽'TAGC'['ATCG'⍳⍵]}</a>
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<a class="sourceLine" id="cb5-3" title="3"></a>
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<a class="sourceLine" id="cb5-4" title="4">⍝ Generate all subarrays (position, length) pairs, for</a>
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<a class="sourceLine" id="cb5-5" title="5">⍝ 4 ≤ length ≤ 12.</a>
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<a class="sourceLine" id="cb5-6" title="6">subarrays←{⊃,/(⍳⍵),¨¨3↓¨⍳¨12⌊1+⍵-⍳⍵}</a>
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<a class="sourceLine" id="cb5-7" title="7"></a>
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<a class="sourceLine" id="cb5-8" title="8">∇ r←revp dna;positions</a>
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<a class="sourceLine" id="cb5-9" title="9"> positions←subarrays⍴dna</a>
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<a class="sourceLine" id="cb5-10" title="10"> ⍝ Filter subarrays which are reverse palindromes.</a>
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<a class="sourceLine" id="cb5-11" title="11"> r←↑({isrevp dna[¯1+⍵[1]+⍳⍵[2]]}¨positions)/positions</a>
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<a class="sourceLine" id="cb5-12" title="12">∇</a></code></pre></div>
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<div class="sourceCode" id="cb6"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb6-1" title="1">sset←{((1E6|2∘×)⍣⍵)1}</a></code></pre></div>
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<a class="sourceLine" id="cb5-2" title="2">isrevp←{⍵≡⌽'TAGC'['ATCG'⍳⍵]}</a></code></pre></div>
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<p>First, we compute the complement of a DNA string (using simple indexing) and test if its Reverse (<code>⌽</code>) is equal to the original string.</p>
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<div class="sourceCode" id="cb6"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb6-1" title="1">⍝ Generate all subarrays (position, length) pairs, for 4 ≤ length ≤ 12.</a>
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<a class="sourceLine" id="cb6-2" title="2">subarrays←{⊃,/(⍳⍵),¨¨3↓¨⍳¨12⌊1+⍵-⍳⍵}</a></code></pre></div>
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<p>We first compute all the possible lengths for each starting point. For instance, the last element cannot have any (position, length) pair associated to it, because there is no three element following it. So we crop the possible lengths to <span class="math inline">\([3, 12]\)</span>. For instance for an array of size 10:</p>
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<div class="sourceCode" id="cb7"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb7-1" title="1"> {3↓¨⍳¨12⌊1+⍵-⍳⍵}10</a>
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<a class="sourceLine" id="cb7-2" title="2">┌──────────────┬───────────┬─────────┬───────┬─────┬───┬─┬┬┬┐</a>
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<a class="sourceLine" id="cb7-3" title="3">│4 5 6 7 8 9 10│4 5 6 7 8 9│4 5 6 7 8│4 5 6 7│4 5 6│4 5│4││││</a>
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<a class="sourceLine" id="cb7-4" title="4">└──────────────┴───────────┴─────────┴───────┴─────┴───┴─┴┴┴┘</a></code></pre></div>
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<p>Then, we just add the corresponding starting position to each length (1 for the first block, 2 for the second, and so on). Finally, we flatten everything.</p>
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<div class="sourceCode" id="cb8"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb8-1" title="1">∇ r←revp dna;positions</a>
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<a class="sourceLine" id="cb8-2" title="2"> positions←subarrays⍴dna</a>
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<a class="sourceLine" id="cb8-3" title="3"> ⍝ Filter subarrays which are reverse palindromes.</a>
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<a class="sourceLine" id="cb8-4" title="4"> r←↑({isrevp dna[¯1+⍵[1]+⍳⍵[2]]}¨positions)/positions</a>
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<a class="sourceLine" id="cb8-5" title="5">∇</a></code></pre></div>
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<p>For each possible (position, length) pair, we get the corresponding DNA substring with <code>dna[¯1+⍵[1]+⍳⍵[2]]</code> (adding <code>¯1</code> is necessary because <code>⎕IO←1</code>). We test if this substring is a reverse palindrome using <code>isrevp</code> above. <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Replicate.htm">Replicate</a> (<code>/</code>) then selects only the (position, length) pairs for which the substring is a reverse palindrome.</p>
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<p>The second task is just about counting the number of subsets modulo 1,000,000. So we just need to compute <span class="math inline">\(2^n \mod 1000000\)</span> for any positive integer <span class="math inline">\(n\leq1000\)</span>.</p>
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<div class="sourceCode" id="cb9"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb9-1" title="1">sset←{((1E6|2∘×)⍣⍵)1}</a></code></pre></div>
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<p>Since we cannot just compute <span class="math inline">\(2^n\)</span> directly and take the remainder, we use modular arithmetic to stay mod 1,000,000 during the whole computation. The dfn <code>(1E6|2∘×)</code> doubles its argument mod 1,000,000. So we just apply this function <span class="math inline">\(n\)</span> times using the <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Operators/Power%20Operator.htm">Power</a> operator (<code>⍣</code>), with an initial value of 1.</p>
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<h2 id="problem-5-future-and-present-value">Problem 5 – Future and Present Value</h2>
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<div class="sourceCode" id="cb7"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb7-1" title="1">⍝ First solution: ((1+⊢)⊥⊣) computes the total return</a>
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<a class="sourceLine" id="cb7-2" title="2">⍝ for a vector of amounts ⍺ and a vector of rates</a>
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<a class="sourceLine" id="cb7-3" title="3">⍝ ⍵. It is applied to every prefix subarray of amounts</a>
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<a class="sourceLine" id="cb7-4" title="4">⍝ and rates to get all intermediate values. However,</a>
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<a class="sourceLine" id="cb7-5" title="5">⍝ this has quadratic complexity.</a>
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<a class="sourceLine" id="cb7-6" title="6">⍝ rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)</a>
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<a class="sourceLine" id="cb7-7" title="7"></a>
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<a class="sourceLine" id="cb7-8" title="8">⍝ Second solution: We want to be able to use the</a>
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<a class="sourceLine" id="cb7-9" title="9">⍝ recurrence relation (recur) and scan through the</a>
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<a class="sourceLine" id="cb7-10" title="10">⍝ vectors of amounts and rates, accumulating the total</a>
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<a class="sourceLine" id="cb7-11" title="11">⍝ value at every time step. However, APL evaluation is</a>
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<a class="sourceLine" id="cb7-12" title="12">⍝ right-associative, so a simple Scan</a>
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<a class="sourceLine" id="cb7-13" title="13">⍝ (recur\amounts,¨values) would not give the correct</a>
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<a class="sourceLine" id="cb7-14" title="14">⍝ result, since recur is not associative and we need</a>
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<a class="sourceLine" id="cb7-15" title="15">⍝ to evaluate it left-to-right. (In any case, in this</a>
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<a class="sourceLine" id="cb7-16" title="16">⍝ case, Scan would have quadratic complexity, so would</a>
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<a class="sourceLine" id="cb7-17" title="17">⍝ not bring any benefit over the previous solution.)</a>
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<a class="sourceLine" id="cb7-18" title="18">⍝ What we need is something akin to Haskell's scanl</a>
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<a class="sourceLine" id="cb7-19" title="19">⍝ function, which would evaluate left to right in O(n)</a>
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<a class="sourceLine" id="cb7-20" title="20">⍝ time. This is what we do here, accumulating values</a>
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<a class="sourceLine" id="cb7-21" title="21">⍝ from left to right. (This is inspired from</a>
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<a class="sourceLine" id="cb7-22" title="22">⍝ dfns.ascan, although heavily simplified.)</a>
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<a class="sourceLine" id="cb7-23" title="23">rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}</a></code></pre></div>
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<div class="sourceCode" id="cb8"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb8-1" title="1">⍝ Simply apply the formula for cashflow calculations.</a>
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<a class="sourceLine" id="cb8-2" title="2">pv←{+/⍺÷×\1+⍵}</a></code></pre></div>
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<div class="sourceCode" id="cb10"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb10-1" title="1">⍝ First solution: ((1+⊢)⊥⊣) computes the total return</a>
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<a class="sourceLine" id="cb10-2" title="2">⍝ for a vector of amounts ⍺ and a vector of rates</a>
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<a class="sourceLine" id="cb10-3" title="3">⍝ ⍵. It is applied to every prefix subarray of amounts</a>
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<a class="sourceLine" id="cb10-4" title="4">⍝ and rates to get all intermediate values. However,</a>
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<a class="sourceLine" id="cb10-5" title="5">⍝ this has quadratic complexity.</a>
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<a class="sourceLine" id="cb10-6" title="6">⍝ rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)</a>
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<a class="sourceLine" id="cb10-7" title="7"></a>
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<a class="sourceLine" id="cb10-8" title="8">⍝ Second solution: We want to be able to use the</a>
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<a class="sourceLine" id="cb10-9" title="9">⍝ recurrence relation (recur) and scan through the</a>
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<a class="sourceLine" id="cb10-10" title="10">⍝ vectors of amounts and rates, accumulating the total</a>
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<a class="sourceLine" id="cb10-11" title="11">⍝ value at every time step. However, APL evaluation is</a>
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<a class="sourceLine" id="cb10-12" title="12">⍝ right-associative, so a simple Scan</a>
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<a class="sourceLine" id="cb10-13" title="13">⍝ (recur\amounts,¨values) would not give the correct</a>
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<a class="sourceLine" id="cb10-14" title="14">⍝ result, since recur is not associative and we need</a>
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<a class="sourceLine" id="cb10-15" title="15">⍝ to evaluate it left-to-right. (In any case, in this</a>
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<a class="sourceLine" id="cb10-16" title="16">⍝ case, Scan would have quadratic complexity, so would</a>
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<a class="sourceLine" id="cb10-17" title="17">⍝ not bring any benefit over the previous solution.)</a>
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<a class="sourceLine" id="cb10-18" title="18">⍝ What we need is something akin to Haskell's scanl</a>
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<a class="sourceLine" id="cb10-19" title="19">⍝ function, which would evaluate left to right in O(n)</a>
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<a class="sourceLine" id="cb10-20" title="20">⍝ time. This is what we do here, accumulating values</a>
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<a class="sourceLine" id="cb10-21" title="21">⍝ from left to right. (This is inspired from</a>
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<a class="sourceLine" id="cb10-22" title="22">⍝ dfns.ascan, although heavily simplified.)</a>
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<a class="sourceLine" id="cb10-23" title="23">rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}</a></code></pre></div>
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<div class="sourceCode" id="cb11"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb11-1" title="1">⍝ Simply apply the formula for cashflow calculations.</a>
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<a class="sourceLine" id="cb11-2" title="2">pv←{+/⍺÷×\1+⍵}</a></code></pre></div>
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<h2 id="problem-6-merge">Problem 6 – Merge</h2>
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<div class="sourceCode" id="cb9"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb9-1" title="1">∇ val←ns getval var</a>
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<a class="sourceLine" id="cb9-2" title="2"> :If ''≡var ⍝ literal '@'</a>
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<a class="sourceLine" id="cb9-3" title="3"> val←'@'</a>
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<a class="sourceLine" id="cb9-4" title="4"> :ElseIf (⊂var)∊ns.⎕NL ¯2</a>
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<a class="sourceLine" id="cb9-5" title="5"> val←⍕ns⍎var</a>
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<a class="sourceLine" id="cb9-6" title="6"> :Else</a>
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<a class="sourceLine" id="cb9-7" title="7"> val←'???'</a>
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<a class="sourceLine" id="cb9-8" title="8"> :EndIf</a>
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<a class="sourceLine" id="cb9-9" title="9">∇</a></code></pre></div>
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<div class="sourceCode" id="cb10"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb10-1" title="1">∇ text←templateFile Merge jsonFile;template;ns</a>
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<a class="sourceLine" id="cb10-2" title="2"> template←⊃⎕NGET templateFile 1</a>
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<a class="sourceLine" id="cb10-3" title="3"> ns←⎕JSON⊃⎕NGET jsonFile</a>
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<a class="sourceLine" id="cb10-4" title="4"> ⍝ We use a simple regex search and replace on the</a>
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<a class="sourceLine" id="cb10-5" title="5"> ⍝ template.</a>
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<a class="sourceLine" id="cb10-6" title="6"> text←↑('@[a-zA-Z]*@'⎕R{ns getval ¯1↓1↓⍵.Match})template</a>
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<a class="sourceLine" id="cb10-7" title="7">∇</a></code></pre></div>
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<div class="sourceCode" id="cb12"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb12-1" title="1">∇ val←ns getval var</a>
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<a class="sourceLine" id="cb12-2" title="2"> :If ''≡var ⍝ literal '@'</a>
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<a class="sourceLine" id="cb12-3" title="3"> val←'@'</a>
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<a class="sourceLine" id="cb12-4" title="4"> :ElseIf (⊂var)∊ns.⎕NL ¯2</a>
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<a class="sourceLine" id="cb12-5" title="5"> val←⍕ns⍎var</a>
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<a class="sourceLine" id="cb12-6" title="6"> :Else</a>
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<a class="sourceLine" id="cb12-7" title="7"> val←'???'</a>
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<a class="sourceLine" id="cb12-8" title="8"> :EndIf</a>
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<a class="sourceLine" id="cb12-9" title="9">∇</a></code></pre></div>
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<div class="sourceCode" id="cb13"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb13-1" title="1">∇ text←templateFile Merge jsonFile;template;ns</a>
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<a class="sourceLine" id="cb13-2" title="2"> template←⊃⎕NGET templateFile 1</a>
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<a class="sourceLine" id="cb13-3" title="3"> ns←⎕JSON⊃⎕NGET jsonFile</a>
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<a class="sourceLine" id="cb13-4" title="4"> ⍝ We use a simple regex search and replace on the</a>
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<a class="sourceLine" id="cb13-5" title="5"> ⍝ template.</a>
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<a class="sourceLine" id="cb13-6" title="6"> text←↑('@[a-zA-Z]*@'⎕R{ns getval ¯1↓1↓⍵.Match})template</a>
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<a class="sourceLine" id="cb13-7" title="7">∇</a></code></pre></div>
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<h2 id="problem-7-upc">Problem 7 – UPC</h2>
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<div class="sourceCode" id="cb11"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb11-1" title="1">CheckDigit←{10|-⍵+.×11⍴3 1}</a></code></pre></div>
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<div class="sourceCode" id="cb12"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb12-1" title="1">⍝ Left and right representations of digits. Decoding</a>
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<a class="sourceLine" id="cb12-2" title="2">⍝ the binary representation from decimal is more</a>
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<a class="sourceLine" id="cb12-3" title="3">⍝ compact than writing everything explicitly.</a>
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<a class="sourceLine" id="cb12-4" title="4">lrepr←⍉(7⍴2)⊤13 25 19 61 35 49 47 59 55 11</a>
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<a class="sourceLine" id="cb12-5" title="5">rrepr←~¨lrepr</a></code></pre></div>
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<div class="sourceCode" id="cb13"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb13-1" title="1">∇ bits←WriteUPC digits;left;right</a>
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<a class="sourceLine" id="cb13-2" title="2"> :If (11=≢digits)∧∧/digits∊0,⍳9</a>
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||||
<a class="sourceLine" id="cb13-3" title="3"> left←,lrepr[1+6↑digits;]</a>
|
||||
<a class="sourceLine" id="cb13-4" title="4"> right←,rrepr[1+6↓digits,CheckDigit digits;]</a>
|
||||
<a class="sourceLine" id="cb13-5" title="5"> bits←1 0 1,left,0 1 0 1 0,right,1 0 1</a>
|
||||
<a class="sourceLine" id="cb13-6" title="6"> :Else</a>
|
||||
<a class="sourceLine" id="cb13-7" title="7"> bits←¯1</a>
|
||||
<a class="sourceLine" id="cb13-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb13-9" title="9">∇</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb14"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb14-1" title="1">∇ digits←ReadUPC bits</a>
|
||||
<a class="sourceLine" id="cb14-2" title="2"> :If 95≠⍴bits ⍝ incorrect number of bits</a>
|
||||
<a class="sourceLine" id="cb14-3" title="3"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb14-4" title="4"> :Else</a>
|
||||
<a class="sourceLine" id="cb14-5" title="5"> ⍝ Test if the barcode was scanned right-to-left.</a>
|
||||
<a class="sourceLine" id="cb14-6" title="6"> :If 0=2|+/bits[3+⍳7]</a>
|
||||
<a class="sourceLine" id="cb14-7" title="7"> bits←⌽bits</a>
|
||||
<a class="sourceLine" id="cb14-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb14-9" title="9"> digits←({¯1+lrepr⍳⍵}¨(7/⍳6)⊆42↑3↓bits),{¯1+rrepr⍳⍵}¨(7/⍳6)⊆¯42↑¯3↓bits</a>
|
||||
<a class="sourceLine" id="cb14-10" title="10"> :If ~∧/digits∊0,⍳9 ⍝ incorrect parity</a>
|
||||
<a class="sourceLine" id="cb14-11" title="11"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb14-12" title="12"> :ElseIf (⊃⌽digits)≠CheckDigit ¯1↓digits ⍝ incorrect check digit</a>
|
||||
<a class="sourceLine" id="cb14-13" title="13"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb14-14" title="14"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb14-15" title="15"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb14-16" title="16">∇</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb14"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb14-1" title="1">CheckDigit←{10|-⍵+.×11⍴3 1}</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb15"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb15-1" title="1">⍝ Left and right representations of digits. Decoding</a>
|
||||
<a class="sourceLine" id="cb15-2" title="2">⍝ the binary representation from decimal is more</a>
|
||||
<a class="sourceLine" id="cb15-3" title="3">⍝ compact than writing everything explicitly.</a>
|
||||
<a class="sourceLine" id="cb15-4" title="4">lrepr←⍉(7⍴2)⊤13 25 19 61 35 49 47 59 55 11</a>
|
||||
<a class="sourceLine" id="cb15-5" title="5">rrepr←~¨lrepr</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb16"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb16-1" title="1">∇ bits←WriteUPC digits;left;right</a>
|
||||
<a class="sourceLine" id="cb16-2" title="2"> :If (11=≢digits)∧∧/digits∊0,⍳9</a>
|
||||
<a class="sourceLine" id="cb16-3" title="3"> left←,lrepr[1+6↑digits;]</a>
|
||||
<a class="sourceLine" id="cb16-4" title="4"> right←,rrepr[1+6↓digits,CheckDigit digits;]</a>
|
||||
<a class="sourceLine" id="cb16-5" title="5"> bits←1 0 1,left,0 1 0 1 0,right,1 0 1</a>
|
||||
<a class="sourceLine" id="cb16-6" title="6"> :Else</a>
|
||||
<a class="sourceLine" id="cb16-7" title="7"> bits←¯1</a>
|
||||
<a class="sourceLine" id="cb16-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb16-9" title="9">∇</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb17"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb17-1" title="1">∇ digits←ReadUPC bits</a>
|
||||
<a class="sourceLine" id="cb17-2" title="2"> :If 95≠⍴bits ⍝ incorrect number of bits</a>
|
||||
<a class="sourceLine" id="cb17-3" title="3"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb17-4" title="4"> :Else</a>
|
||||
<a class="sourceLine" id="cb17-5" title="5"> ⍝ Test if the barcode was scanned right-to-left.</a>
|
||||
<a class="sourceLine" id="cb17-6" title="6"> :If 0=2|+/bits[3+⍳7]</a>
|
||||
<a class="sourceLine" id="cb17-7" title="7"> bits←⌽bits</a>
|
||||
<a class="sourceLine" id="cb17-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb17-9" title="9"> digits←({¯1+lrepr⍳⍵}¨(7/⍳6)⊆42↑3↓bits),{¯1+rrepr⍳⍵}¨(7/⍳6)⊆¯42↑¯3↓bits</a>
|
||||
<a class="sourceLine" id="cb17-10" title="10"> :If ~∧/digits∊0,⍳9 ⍝ incorrect parity</a>
|
||||
<a class="sourceLine" id="cb17-11" title="11"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb17-12" title="12"> :ElseIf (⊃⌽digits)≠CheckDigit ¯1↓digits ⍝ incorrect check digit</a>
|
||||
<a class="sourceLine" id="cb17-13" title="13"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb17-14" title="14"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb17-15" title="15"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb17-16" title="16">∇</a></code></pre></div>
|
||||
<h2 id="problem-8-balancing-the-scales">Problem 8 – Balancing the Scales</h2>
|
||||
<div class="sourceCode" id="cb15"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb15-1" title="1">∇ parts←Balance nums;subsets;partitions</a>
|
||||
<a class="sourceLine" id="cb15-2" title="2"> ⍝ This is a brute force solution, running in</a>
|
||||
<a class="sourceLine" id="cb15-3" title="3"> ⍝ exponential time. We generate all the possible</a>
|
||||
<a class="sourceLine" id="cb15-4" title="4"> ⍝ partitions, filter out those which are not</a>
|
||||
<a class="sourceLine" id="cb15-5" title="5"> ⍝ balanced, and return the first matching one. There</a>
|
||||
<a class="sourceLine" id="cb15-6" title="6"> ⍝ are more advanced approach running in</a>
|
||||
<a class="sourceLine" id="cb15-7" title="7"> ⍝ pseudo-polynomial time (based on dynamic</a>
|
||||
<a class="sourceLine" id="cb15-8" title="8"> ⍝ programming, see the "Partition problem" Wikipedia</a>
|
||||
<a class="sourceLine" id="cb15-9" title="9"> ⍝ page), but they are not warranted here, as the</a>
|
||||
<a class="sourceLine" id="cb15-10" title="10"> ⍝ input size remains fairly small.</a>
|
||||
<a class="sourceLine" id="cb15-11" title="11"></a>
|
||||
<a class="sourceLine" id="cb15-12" title="12"> ⍝ Generate all partitions of a vector of a given</a>
|
||||
<a class="sourceLine" id="cb15-13" title="13"> ⍝ size, as binary mask vectors.</a>
|
||||
<a class="sourceLine" id="cb15-14" title="14"> subsets←{1↓2⊥⍣¯1⍳2*⍵}</a>
|
||||
<a class="sourceLine" id="cb15-15" title="15"> ⍝ Keep only the subsets whose sum is exactly</a>
|
||||
<a class="sourceLine" id="cb15-16" title="16"> ⍝ (+/nums)÷2.</a>
|
||||
<a class="sourceLine" id="cb15-17" title="17"> partitions←nums{((2÷⍨+/⍺)=⍺+.×⍵)/⍵}subsets⍴nums</a>
|
||||
<a class="sourceLine" id="cb15-18" title="18"> :If 0=≢,partitions</a>
|
||||
<a class="sourceLine" id="cb15-19" title="19"> ⍝ If no partition satisfy the above</a>
|
||||
<a class="sourceLine" id="cb15-20" title="20"> ⍝ criterion, we return ⍬.</a>
|
||||
<a class="sourceLine" id="cb15-21" title="21"> parts←⍬</a>
|
||||
<a class="sourceLine" id="cb15-22" title="22"> :Else</a>
|
||||
<a class="sourceLine" id="cb15-23" title="23"> ⍝ Otherwise, we return the first possible</a>
|
||||
<a class="sourceLine" id="cb15-24" title="24"> ⍝ partition.</a>
|
||||
<a class="sourceLine" id="cb15-25" title="25"> parts←nums{((⊂,(⊂~))⊃↓⍉⍵)/¨2⍴⊂⍺}partitions</a>
|
||||
<a class="sourceLine" id="cb15-26" title="26"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb15-27" title="27">∇</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb18"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb18-1" title="1">∇ parts←Balance nums;subsets;partitions</a>
|
||||
<a class="sourceLine" id="cb18-2" title="2"> ⍝ This is a brute force solution, running in</a>
|
||||
<a class="sourceLine" id="cb18-3" title="3"> ⍝ exponential time. We generate all the possible</a>
|
||||
<a class="sourceLine" id="cb18-4" title="4"> ⍝ partitions, filter out those which are not</a>
|
||||
<a class="sourceLine" id="cb18-5" title="5"> ⍝ balanced, and return the first matching one. There</a>
|
||||
<a class="sourceLine" id="cb18-6" title="6"> ⍝ are more advanced approach running in</a>
|
||||
<a class="sourceLine" id="cb18-7" title="7"> ⍝ pseudo-polynomial time (based on dynamic</a>
|
||||
<a class="sourceLine" id="cb18-8" title="8"> ⍝ programming, see the "Partition problem" Wikipedia</a>
|
||||
<a class="sourceLine" id="cb18-9" title="9"> ⍝ page), but they are not warranted here, as the</a>
|
||||
<a class="sourceLine" id="cb18-10" title="10"> ⍝ input size remains fairly small.</a>
|
||||
<a class="sourceLine" id="cb18-11" title="11"></a>
|
||||
<a class="sourceLine" id="cb18-12" title="12"> ⍝ Generate all partitions of a vector of a given</a>
|
||||
<a class="sourceLine" id="cb18-13" title="13"> ⍝ size, as binary mask vectors.</a>
|
||||
<a class="sourceLine" id="cb18-14" title="14"> subsets←{1↓2⊥⍣¯1⍳2*⍵}</a>
|
||||
<a class="sourceLine" id="cb18-15" title="15"> ⍝ Keep only the subsets whose sum is exactly</a>
|
||||
<a class="sourceLine" id="cb18-16" title="16"> ⍝ (+/nums)÷2.</a>
|
||||
<a class="sourceLine" id="cb18-17" title="17"> partitions←nums{((2÷⍨+/⍺)=⍺+.×⍵)/⍵}subsets⍴nums</a>
|
||||
<a class="sourceLine" id="cb18-18" title="18"> :If 0=≢,partitions</a>
|
||||
<a class="sourceLine" id="cb18-19" title="19"> ⍝ If no partition satisfy the above</a>
|
||||
<a class="sourceLine" id="cb18-20" title="20"> ⍝ criterion, we return ⍬.</a>
|
||||
<a class="sourceLine" id="cb18-21" title="21"> parts←⍬</a>
|
||||
<a class="sourceLine" id="cb18-22" title="22"> :Else</a>
|
||||
<a class="sourceLine" id="cb18-23" title="23"> ⍝ Otherwise, we return the first possible</a>
|
||||
<a class="sourceLine" id="cb18-24" title="24"> ⍝ partition.</a>
|
||||
<a class="sourceLine" id="cb18-25" title="25"> parts←nums{((⊂,(⊂~))⊃↓⍉⍵)/¨2⍴⊂⍺}partitions</a>
|
||||
<a class="sourceLine" id="cb18-26" title="26"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb18-27" title="27">∇</a></code></pre></div>
|
||||
<h2 id="problem-9-upwardly-mobile">Problem 9 – Upwardly Mobile</h2>
|
||||
<div class="sourceCode" id="cb16"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb16-1" title="1">∇ weights←Weights filename;mobile;branches;mat</a>
|
||||
<a class="sourceLine" id="cb16-2" title="2"> ⍝ Put your code and comments below here</a>
|
||||
<a class="sourceLine" id="cb16-3" title="3"></a>
|
||||
<a class="sourceLine" id="cb16-4" title="4"> ⍝ Parse the mobile input file.</a>
|
||||
<a class="sourceLine" id="cb16-5" title="5"> mobile←↑⊃⎕NGET filename 1</a>
|
||||
<a class="sourceLine" id="cb16-6" title="6"> branches←⍸mobile∊'┌┴┐'</a>
|
||||
<a class="sourceLine" id="cb16-7" title="7"> ⍝ TODO: Build the matrix of coefficients mat.</a>
|
||||
<a class="sourceLine" id="cb16-8" title="8"></a>
|
||||
<a class="sourceLine" id="cb16-9" title="9"> ⍝ Solve the system of equations (arbitrarily setting</a>
|
||||
<a class="sourceLine" id="cb16-10" title="10"> ⍝ the first variable at 1 because the system is</a>
|
||||
<a class="sourceLine" id="cb16-11" title="11"> ⍝ overdetermined), then multiply the coefficients by</a>
|
||||
<a class="sourceLine" id="cb16-12" title="12"> ⍝ their least common multiple to get the smallest</a>
|
||||
<a class="sourceLine" id="cb16-13" title="13"> ⍝ integer weights.</a>
|
||||
<a class="sourceLine" id="cb16-14" title="14"> weights←((1∘,)×(∧/÷))mat[;1]⌹1↓[2]mat</a>
|
||||
<a class="sourceLine" id="cb16-15" title="15">∇</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb17"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb17-1" title="1"> :EndNamespace</a>
|
||||
<a class="sourceLine" id="cb17-2" title="2">:EndNamespace</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb19"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb19-1" title="1">∇ weights←Weights filename;mobile;branches;mat</a>
|
||||
<a class="sourceLine" id="cb19-2" title="2"> ⍝ Put your code and comments below here</a>
|
||||
<a class="sourceLine" id="cb19-3" title="3"></a>
|
||||
<a class="sourceLine" id="cb19-4" title="4"> ⍝ Parse the mobile input file.</a>
|
||||
<a class="sourceLine" id="cb19-5" title="5"> mobile←↑⊃⎕NGET filename 1</a>
|
||||
<a class="sourceLine" id="cb19-6" title="6"> branches←⍸mobile∊'┌┴┐'</a>
|
||||
<a class="sourceLine" id="cb19-7" title="7"> ⍝ TODO: Build the matrix of coefficients mat.</a>
|
||||
<a class="sourceLine" id="cb19-8" title="8"></a>
|
||||
<a class="sourceLine" id="cb19-9" title="9"> ⍝ Solve the system of equations (arbitrarily setting</a>
|
||||
<a class="sourceLine" id="cb19-10" title="10"> ⍝ the first variable at 1 because the system is</a>
|
||||
<a class="sourceLine" id="cb19-11" title="11"> ⍝ overdetermined), then multiply the coefficients by</a>
|
||||
<a class="sourceLine" id="cb19-12" title="12"> ⍝ their least common multiple to get the smallest</a>
|
||||
<a class="sourceLine" id="cb19-13" title="13"> ⍝ integer weights.</a>
|
||||
<a class="sourceLine" id="cb19-14" title="14"> weights←((1∘,)×(∧/÷))mat[;1]⌹1↓[2]mat</a>
|
||||
<a class="sourceLine" id="cb19-15" title="15">∇</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb20"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb20-1" title="1"> :EndNamespace</a>
|
||||
<a class="sourceLine" id="cb20-2" title="2">:EndNamespace</a></code></pre></div>
|
||||
</section>
|
||||
</article>
|
||||
|
||||
|
|
|
|||
Loading…
Add table
Add a link
Reference in a new issue