Add final annotations
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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="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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<p>First solution: <code>((1+⊢)⊥⊣)</code> computes the total return for a vector of amounts <code>⍺</code> and a vector of rates <code>⍵</code>. It is applied to every prefix subarray of amounts and rates to get all intermediate values. However, this has quadratic complexity.</p>
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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">rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)</a></code></pre></div>
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<p>Second solution: We want to be able to use the recurrence relation (<code>recur</code>) and scan through the vectors of amounts and rates, accumulating the total value at every time step. However, APL evaluation is right-associative, so a simple <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Operators/Scan.htm">Scan</a> (<code>recur\amounts,¨values</code>) would not give the correct result, since <code>recur</code> is not associative and we need to evaluate it left-to-right. (In any case, in this case, Scan would have quadratic complexity, so would not bring any benefit over the previous solution.) What we need is something akin to Haskell’s <code>scanl</code> function, which would evaluate left to right in <span class="math inline">\(O(n)\)</span> time<span><label for="sn-1" class="margin-toggle sidenote-number"></label><input type="checkbox" id="sn-1" class="margin-toggle" /><span class="sidenote">There is an interesting <a href="https://stackoverflow.com/a/25100675/8864368">StackOverflow answer</a> explaining the behaviour of Scan, and compares it to Haskell’s <code>scanl</code> function.<br />
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<br />
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</span></span>. This is what we do here, accumulating values from left to right. (This is inspired from <a href="https://dfns.dyalog.com/c_ascan.htm"><code>dfns.ascan</code></a>, although heavily simplified.)</p>
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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">rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}</a></code></pre></div>
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<p>For the second task, there is an explicit formula for cashflow calculations, so we can just apply it.</p>
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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">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="cb12"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb12-1" title="1">∇ text←templateFile Merge jsonFile;template;ns</a>
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<a class="sourceLine" id="cb12-2" title="2"> template←⊃⎕NGET templateFile 1</a>
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<a class="sourceLine" id="cb12-3" title="3"> ns←⎕JSON⊃⎕NGET jsonFile</a>
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<a class="sourceLine" id="cb12-4" title="4"> ⍝ We use a simple regex search and replace on the</a>
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<a class="sourceLine" id="cb12-5" title="5"> ⍝ template.</a>
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<a class="sourceLine" id="cb12-6" title="6"> text←↑('@[a-zA-Z]*@'⎕R{ns getval ¯1↓1↓⍵.Match})template</a>
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<a class="sourceLine" id="cb12-7" title="7">∇</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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<p>We first read the template and the JSON values from their files. The <a href="https://help.dyalog.com/18.0/index.htm#Language/System%20Functions/nget.htm"><code>⎕NGET</code></a> function read simple text files, and <a href="https://help.dyalog.com/18.0/index.htm#Language/System%20Functions/json.htm"><code>⎕JSON</code></a> extracts the key-value pairs as a namespace.</p>
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<p>Assuming all variable names contain only letters, we match the regex <code>@[a-zA-Z]*@</code> to match variable names enclosed between <code>@</code> symbols. The function <code>getval</code> then returns the appropriate value, and we can replace the variable name in the template.</p>
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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">∇ val←ns getval var</a>
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<a class="sourceLine" id="cb13-2" title="2"> :If ''≡var ⍝ literal '@'</a>
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<a class="sourceLine" id="cb13-3" title="3"> val←'@'</a>
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<a class="sourceLine" id="cb13-4" title="4"> :ElseIf (⊂var)∊ns.⎕NL ¯2</a>
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<a class="sourceLine" id="cb13-5" title="5"> val←⍕ns⍎var</a>
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<a class="sourceLine" id="cb13-6" title="6"> :Else</a>
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<a class="sourceLine" id="cb13-7" title="7"> val←'???'</a>
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<a class="sourceLine" id="cb13-8" title="8"> :EndIf</a>
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<a class="sourceLine" id="cb13-9" title="9">∇</a></code></pre></div>
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<div class="sourceCode" id="cb14"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb14-1" title="1">∇ val←ns getval var</a>
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<a class="sourceLine" id="cb14-2" title="2"> :If ''≡var ⍝ literal '@'</a>
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<a class="sourceLine" id="cb14-3" title="3"> val←'@'</a>
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<a class="sourceLine" id="cb14-4" title="4"> :ElseIf (⊂var)∊ns.⎕NL ¯2</a>
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<a class="sourceLine" id="cb14-5" title="5"> val←⍕ns⍎var</a>
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<a class="sourceLine" id="cb14-6" title="6"> :Else</a>
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<a class="sourceLine" id="cb14-7" title="7"> val←'???'</a>
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<a class="sourceLine" id="cb14-8" title="8"> :EndIf</a>
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<a class="sourceLine" id="cb14-9" title="9">∇</a></code></pre></div>
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<p>This function takes the namespace matching the variable names to their respective values, and the name of the variable.</p>
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<ul>
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<li>If the variable name is empty, we matched the string <code>@@</code>, which corresponds to a literal <code>@</code>.</li>
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@ -154,42 +137,42 @@
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<li>Otherwise, we have an unknown variable, so we replace it with <code>???</code>.</li>
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</ul>
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<h2 id="problem-7-upc">Problem 7 – UPC</h2>
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<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>
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<div class="sourceCode" id="cb15"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb15-1" title="1">CheckDigit←{10|-⍵+.×11⍴3 1}</a></code></pre></div>
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<p>The check digit satisfies the equation <span class="math display">\[ 3 x_{1}+x_{2}+3 x_{3}+x_{4}+3 x_{5}+x_{6}+3 x_{7}+x_{8}+3 x_{9}+x_{10}+3 x_{11}+x_{12} \equiv 0 \bmod 10, \]</span> therefore, <span class="math display">\[ x_{12} \equiv -(3 x_{1}+x_{2}+3 x_{3}+x_{4}+3 x_{5}+x_{6}+3 x_{7}+x_{8}+3 x_{9}+x_{10}+3 x_{11}) \bmod 10. \]</span></p>
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<p>Translated to APL, we just take the dot product between the first 11 digits of the barcode with <code>11⍴3 1</code>, negate it, and take the remainder by 10.</p>
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<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>
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<a class="sourceLine" id="cb15-2" title="2">⍝ the binary representation from decimal is more</a>
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<a class="sourceLine" id="cb15-3" title="3">⍝ compact than writing everything explicitly.</a>
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<a class="sourceLine" id="cb15-4" title="4">lrepr←⍉(7⍴2)⊤13 25 19 61 35 49 47 59 55 11</a>
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<a class="sourceLine" id="cb15-5" title="5">rrepr←~¨lrepr</a></code></pre></div>
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<div class="sourceCode" id="cb16"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb16-1" title="1">⍝ Left and right representations of digits. Decoding</a>
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<a class="sourceLine" id="cb16-2" title="2">⍝ the binary representation from decimal is more</a>
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<a class="sourceLine" id="cb16-3" title="3">⍝ compact than writing everything explicitly.</a>
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<a class="sourceLine" id="cb16-4" title="4">lrepr←⍉(7⍴2)⊤13 25 19 61 35 49 47 59 55 11</a>
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<a class="sourceLine" id="cb16-5" title="5">rrepr←~¨lrepr</a></code></pre></div>
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<p>For the second task, the first thing we need to do is save the representation of digits. To save space, I did not encode the binary representation explicitly, instead using a decimal representation that I then decode in base 2. The right representation is just the bitwise negation.</p>
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<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>
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<a class="sourceLine" id="cb16-2" title="2"> :If (11=≢digits)∧∧/digits∊0,⍳9</a>
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<a class="sourceLine" id="cb16-3" title="3"> left←,lrepr[1+6↑digits;]</a>
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<a class="sourceLine" id="cb16-4" title="4"> right←,rrepr[1+6↓digits,CheckDigit digits;]</a>
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<a class="sourceLine" id="cb16-5" title="5"> bits←1 0 1,left,0 1 0 1 0,right,1 0 1</a>
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<a class="sourceLine" id="cb16-6" title="6"> :Else</a>
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<a class="sourceLine" id="cb16-7" title="7"> bits←¯1</a>
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<a class="sourceLine" id="cb16-8" title="8"> :EndIf</a>
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<a class="sourceLine" id="cb16-9" title="9">∇</a></code></pre></div>
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<div class="sourceCode" id="cb17"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb17-1" title="1">∇ bits←WriteUPC digits;left;right</a>
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<a class="sourceLine" id="cb17-2" title="2"> :If (11=≢digits)∧∧/digits∊0,⍳9</a>
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<a class="sourceLine" id="cb17-3" title="3"> left←,lrepr[1+6↑digits;]</a>
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<a class="sourceLine" id="cb17-4" title="4"> right←,rrepr[1+6↓digits,CheckDigit digits;]</a>
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<a class="sourceLine" id="cb17-5" title="5"> bits←1 0 1,left,0 1 0 1 0,right,1 0 1</a>
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<a class="sourceLine" id="cb17-6" title="6"> :Else</a>
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<a class="sourceLine" id="cb17-7" title="7"> bits←¯1</a>
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<a class="sourceLine" id="cb17-8" title="8"> :EndIf</a>
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<a class="sourceLine" id="cb17-9" title="9">∇</a></code></pre></div>
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<p>First of all, if the vector <code>digits</code> does not have exactly 11 elements, all between 0 and 9, it is an error and we return <code>¯1</code>.</p>
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<p>Then, we take the first 6 digits and encode them with <code>lrepr</code>, and the last 5 digits plus the check digit encoded with <code>rrepr</code>. In each case, adding 1 is necessary because <code>⎕IO←1</code>. We return the final bit array with the required beginning, middle, and end guard patterns.</p>
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<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>
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<a class="sourceLine" id="cb17-2" title="2"> :If 95≠⍴bits ⍝ incorrect number of bits</a>
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<a class="sourceLine" id="cb17-3" title="3"> digits←¯1</a>
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<a class="sourceLine" id="cb17-4" title="4"> :Else</a>
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<a class="sourceLine" id="cb17-5" title="5"> ⍝ Test if the barcode was scanned right-to-left.</a>
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<a class="sourceLine" id="cb17-6" title="6"> :If 0=2|+/bits[3+⍳7]</a>
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<a class="sourceLine" id="cb17-7" title="7"> bits←⌽bits</a>
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<a class="sourceLine" id="cb17-8" title="8"> :EndIf</a>
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<a class="sourceLine" id="cb17-9" title="9"> digits←({¯1+lrepr⍳⍵}¨(7/⍳6)⊆42↑3↓bits),{¯1+rrepr⍳⍵}¨(7/⍳6)⊆¯42↑¯3↓bits</a>
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<a class="sourceLine" id="cb17-10" title="10"> :If ~∧/digits∊0,⍳9 ⍝ incorrect parity</a>
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<a class="sourceLine" id="cb17-11" title="11"> digits←¯1</a>
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<a class="sourceLine" id="cb17-12" title="12"> :ElseIf (⊃⌽digits)≠CheckDigit ¯1↓digits ⍝ incorrect check digit</a>
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<a class="sourceLine" id="cb17-13" title="13"> digits←¯1</a>
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<a class="sourceLine" id="cb17-14" title="14"> :EndIf</a>
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<a class="sourceLine" id="cb17-15" title="15"> :EndIf</a>
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<a class="sourceLine" id="cb17-16" title="16">∇</a></code></pre></div>
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<div class="sourceCode" id="cb18"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb18-1" title="1">∇ digits←ReadUPC bits</a>
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<a class="sourceLine" id="cb18-2" title="2"> :If 95≠⍴bits ⍝ incorrect number of bits</a>
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<a class="sourceLine" id="cb18-3" title="3"> digits←¯1</a>
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<a class="sourceLine" id="cb18-4" title="4"> :Else</a>
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<a class="sourceLine" id="cb18-5" title="5"> ⍝ Test if the barcode was scanned right-to-left.</a>
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<a class="sourceLine" id="cb18-6" title="6"> :If 0=2|+/bits[3+⍳7]</a>
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<a class="sourceLine" id="cb18-7" title="7"> bits←⌽bits</a>
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<a class="sourceLine" id="cb18-8" title="8"> :EndIf</a>
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<a class="sourceLine" id="cb18-9" title="9"> digits←({¯1+lrepr⍳⍵}¨(7/⍳6)⊆42↑3↓bits),{¯1+rrepr⍳⍵}¨(7/⍳6)⊆¯42↑¯3↓bits</a>
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<a class="sourceLine" id="cb18-10" title="10"> :If ~∧/digits∊0,⍳9 ⍝ incorrect parity</a>
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<a class="sourceLine" id="cb18-11" title="11"> digits←¯1</a>
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<a class="sourceLine" id="cb18-12" title="12"> :ElseIf (⊃⌽digits)≠CheckDigit ¯1↓digits ⍝ incorrect check digit</a>
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<a class="sourceLine" id="cb18-13" title="13"> digits←¯1</a>
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<a class="sourceLine" id="cb18-14" title="14"> :EndIf</a>
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<a class="sourceLine" id="cb18-15" title="15"> :EndIf</a>
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<a class="sourceLine" id="cb18-16" title="16">∇</a></code></pre></div>
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<ul>
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<li>If we don’t have the correct number of bits, we return <code>¯1</code>.</li>
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<li>We test the first digit for its parity, to determine if its actually a left representation. If it’s not, we reverse the bit array.</li>
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@ -197,51 +180,53 @@
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<li>Final checks for the range of the digits (i.e., if the representations could not be found in the <code>lrepr</code> and <code>rrepr</code> vectors), and for the check digit.</li>
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</ul>
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<h2 id="problem-8-balancing-the-scales">Problem 8 – Balancing the Scales</h2>
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<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>
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<a class="sourceLine" id="cb18-2" title="2"> ⍝ This is a brute force solution, running in</a>
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<a class="sourceLine" id="cb18-3" title="3"> ⍝ exponential time. We generate all the possible</a>
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<a class="sourceLine" id="cb18-4" title="4"> ⍝ partitions, filter out those which are not</a>
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<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>
|
||||
<div class="sourceCode" id="cb19"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb19-1" title="1">∇ parts←Balance nums;subsets;partitions</a>
|
||||
<a class="sourceLine" id="cb19-2" title="2"> ⍝ This is a brute force solution, running in</a>
|
||||
<a class="sourceLine" id="cb19-3" title="3"> ⍝ exponential time. We generate all the possible</a>
|
||||
<a class="sourceLine" id="cb19-4" title="4"> ⍝ partitions, filter out those which are not</a>
|
||||
<a class="sourceLine" id="cb19-5" title="5"> ⍝ balanced, and return the first matching one. There</a>
|
||||
<a class="sourceLine" id="cb19-6" title="6"> ⍝ are more advanced approach running in</a>
|
||||
<a class="sourceLine" id="cb19-7" title="7"> ⍝ pseudo-polynomial time (based on dynamic</a>
|
||||
<a class="sourceLine" id="cb19-8" title="8"> ⍝ programming, see the "Partition problem" Wikipedia</a>
|
||||
<a class="sourceLine" id="cb19-9" title="9"> ⍝ page), but they are not warranted here, as the</a>
|
||||
<a class="sourceLine" id="cb19-10" title="10"> ⍝ input size remains fairly small.</a>
|
||||
<a class="sourceLine" id="cb19-11" title="11"></a>
|
||||
<a class="sourceLine" id="cb19-12" title="12"> ⍝ Generate all partitions of a vector of a given</a>
|
||||
<a class="sourceLine" id="cb19-13" title="13"> ⍝ size, as binary mask vectors.</a>
|
||||
<a class="sourceLine" id="cb19-14" title="14"> subsets←{1↓2⊥⍣¯1⍳2*⍵}</a>
|
||||
<a class="sourceLine" id="cb19-15" title="15"> ⍝ Keep only the subsets whose sum is exactly</a>
|
||||
<a class="sourceLine" id="cb19-16" title="16"> ⍝ (+/nums)÷2.</a>
|
||||
<a class="sourceLine" id="cb19-17" title="17"> partitions←nums{((2÷⍨+/⍺)=⍺+.×⍵)/⍵}subsets⍴nums</a>
|
||||
<a class="sourceLine" id="cb19-18" title="18"> :If 0=≢,partitions</a>
|
||||
<a class="sourceLine" id="cb19-19" title="19"> ⍝ If no partition satisfy the above</a>
|
||||
<a class="sourceLine" id="cb19-20" title="20"> ⍝ criterion, we return ⍬.</a>
|
||||
<a class="sourceLine" id="cb19-21" title="21"> parts←⍬</a>
|
||||
<a class="sourceLine" id="cb19-22" title="22"> :Else</a>
|
||||
<a class="sourceLine" id="cb19-23" title="23"> ⍝ Otherwise, we return the first possible</a>
|
||||
<a class="sourceLine" id="cb19-24" title="24"> ⍝ partition.</a>
|
||||
<a class="sourceLine" id="cb19-25" title="25"> parts←nums{((⊂,(⊂~))⊃↓⍉⍵)/¨2⍴⊂⍺}partitions</a>
|
||||
<a class="sourceLine" id="cb19-26" title="26"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb19-27" title="27">∇</a></code></pre></div>
|
||||
<h2 id="problem-9-upwardly-mobile">Problem 9 – Upwardly Mobile</h2>
|
||||
<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>
|
||||
<p>This is the only problem that I didn’t complete. It required parsing the files containing the graphical representations of the trees, which was needlessly complex and, quite frankly, hard and boring with a language like APL.</p>
|
||||
<p>However, the next part is interesting: once we have a matrix of coefficients representing the relationships between the weights, we can solve the system of equations. <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Matrix%20Divide.htm">Matrix Divide</a> (<code>⌹</code>) will find one solution to the system. Since the system is overdetermined, we fix <code>A=1</code> to find one possible solution. Since we want integer weights, the solution we find is smaller than the one we want, and may contain fractional weights. So we multiply everything by the <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/And%20Lowest%20Common%20Multiple.htm">Lowest Common Multiple</a> (<code>∧</code>) to get the smallest integer weights.</p>
|
||||
<div class="sourceCode" id="cb20"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb20-1" title="1">∇ weights←Weights filename;mobile;branches;mat</a>
|
||||
<a class="sourceLine" id="cb20-2" title="2"> ⍝ Put your code and comments below here</a>
|
||||
<a class="sourceLine" id="cb20-3" title="3"></a>
|
||||
<a class="sourceLine" id="cb20-4" title="4"> ⍝ Parse the mobile input file.</a>
|
||||
<a class="sourceLine" id="cb20-5" title="5"> mobile←↑⊃⎕NGET filename 1</a>
|
||||
<a class="sourceLine" id="cb20-6" title="6"> branches←⍸mobile∊'┌┴┐'</a>
|
||||
<a class="sourceLine" id="cb20-7" title="7"> ⍝ TODO: Build the matrix of coefficients mat.</a>
|
||||
<a class="sourceLine" id="cb20-8" title="8"></a>
|
||||
<a class="sourceLine" id="cb20-9" title="9"> ⍝ Solve the system of equations (arbitrarily setting</a>
|
||||
<a class="sourceLine" id="cb20-10" title="10"> ⍝ the first variable at 1 because the system is</a>
|
||||
<a class="sourceLine" id="cb20-11" title="11"> ⍝ overdetermined), then multiply the coefficients by</a>
|
||||
<a class="sourceLine" id="cb20-12" title="12"> ⍝ their least common multiple to get the smallest</a>
|
||||
<a class="sourceLine" id="cb20-13" title="13"> ⍝ integer weights.</a>
|
||||
<a class="sourceLine" id="cb20-14" title="14"> weights←((1∘,)×(∧/÷))mat[;1]⌹1↓[2]mat</a>
|
||||
<a class="sourceLine" id="cb20-15" title="15">∇</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb21"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb21-1" title="1"> :EndNamespace</a>
|
||||
<a class="sourceLine" id="cb21-2" title="2">:EndNamespace</a></code></pre></div>
|
||||
</section>
|
||||
</article>
|
||||
]]></summary>
|
||||
|
|
|
|||
|
|
@ -137,50 +137,33 @@
|
|||
<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>
|
||||
<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>
|
||||
<h2 id="problem-5-future-and-present-value">Problem 5 – Future and Present Value</h2>
|
||||
<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>
|
||||
<a class="sourceLine" id="cb10-2" title="2">⍝ for a vector of amounts ⍺ and a vector of rates</a>
|
||||
<a class="sourceLine" id="cb10-3" title="3">⍝ ⍵. It is applied to every prefix subarray of amounts</a>
|
||||
<a class="sourceLine" id="cb10-4" title="4">⍝ and rates to get all intermediate values. However,</a>
|
||||
<a class="sourceLine" id="cb10-5" title="5">⍝ this has quadratic complexity.</a>
|
||||
<a class="sourceLine" id="cb10-6" title="6">⍝ rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)</a>
|
||||
<a class="sourceLine" id="cb10-7" title="7"></a>
|
||||
<a class="sourceLine" id="cb10-8" title="8">⍝ Second solution: We want to be able to use the</a>
|
||||
<a class="sourceLine" id="cb10-9" title="9">⍝ recurrence relation (recur) and scan through the</a>
|
||||
<a class="sourceLine" id="cb10-10" title="10">⍝ vectors of amounts and rates, accumulating the total</a>
|
||||
<a class="sourceLine" id="cb10-11" title="11">⍝ value at every time step. However, APL evaluation is</a>
|
||||
<a class="sourceLine" id="cb10-12" title="12">⍝ right-associative, so a simple Scan</a>
|
||||
<a class="sourceLine" id="cb10-13" title="13">⍝ (recur\amounts,¨values) would not give the correct</a>
|
||||
<a class="sourceLine" id="cb10-14" title="14">⍝ result, since recur is not associative and we need</a>
|
||||
<a class="sourceLine" id="cb10-15" title="15">⍝ to evaluate it left-to-right. (In any case, in this</a>
|
||||
<a class="sourceLine" id="cb10-16" title="16">⍝ case, Scan would have quadratic complexity, so would</a>
|
||||
<a class="sourceLine" id="cb10-17" title="17">⍝ not bring any benefit over the previous solution.)</a>
|
||||
<a class="sourceLine" id="cb10-18" title="18">⍝ What we need is something akin to Haskell's scanl</a>
|
||||
<a class="sourceLine" id="cb10-19" title="19">⍝ function, which would evaluate left to right in O(n)</a>
|
||||
<a class="sourceLine" id="cb10-20" title="20">⍝ time. This is what we do here, accumulating values</a>
|
||||
<a class="sourceLine" id="cb10-21" title="21">⍝ from left to right. (This is inspired from</a>
|
||||
<a class="sourceLine" id="cb10-22" title="22">⍝ dfns.ascan, although heavily simplified.)</a>
|
||||
<a class="sourceLine" id="cb10-23" title="23">rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}</a></code></pre></div>
|
||||
<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>
|
||||
<a class="sourceLine" id="cb11-2" title="2">pv←{+/⍺÷×\1+⍵}</a></code></pre></div>
|
||||
<p>First solution: <code>((1+⊢)⊥⊣)</code> computes the total return for a vector of amounts <code>⍺</code> and a vector of rates <code>⍵</code>. It is applied to every prefix subarray of amounts and rates to get all intermediate values. However, this has quadratic complexity.</p>
|
||||
<div class="sourceCode" id="cb10"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb10-1" title="1">rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)</a></code></pre></div>
|
||||
<p>Second solution: We want to be able to use the recurrence relation (<code>recur</code>) and scan through the vectors of amounts and rates, accumulating the total value at every time step. However, APL evaluation is right-associative, so a simple <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Operators/Scan.htm">Scan</a> (<code>recur\amounts,¨values</code>) would not give the correct result, since <code>recur</code> is not associative and we need to evaluate it left-to-right. (In any case, in this case, Scan would have quadratic complexity, so would not bring any benefit over the previous solution.) What we need is something akin to Haskell’s <code>scanl</code> function, which would evaluate left to right in <span class="math inline">\(O(n)\)</span> time<span><label for="sn-1" class="margin-toggle sidenote-number"></label><input type="checkbox" id="sn-1" class="margin-toggle" /><span class="sidenote">There is an interesting <a href="https://stackoverflow.com/a/25100675/8864368">StackOverflow answer</a> explaining the behaviour of Scan, and compares it to Haskell’s <code>scanl</code> function.<br />
|
||||
<br />
|
||||
</span></span>. This is what we do here, accumulating values from left to right. (This is inspired from <a href="https://dfns.dyalog.com/c_ascan.htm"><code>dfns.ascan</code></a>, although heavily simplified.)</p>
|
||||
<div class="sourceCode" id="cb11"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb11-1" title="1">rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}</a></code></pre></div>
|
||||
<p>For the second task, there is an explicit formula for cashflow calculations, so we can just apply it.</p>
|
||||
<div class="sourceCode" id="cb12"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb12-1" title="1">pv←{+/⍺÷×\1+⍵}</a></code></pre></div>
|
||||
<h2 id="problem-6-merge">Problem 6 – Merge</h2>
|
||||
<div class="sourceCode" id="cb12"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb12-1" title="1">∇ text←templateFile Merge jsonFile;template;ns</a>
|
||||
<a class="sourceLine" id="cb12-2" title="2"> template←⊃⎕NGET templateFile 1</a>
|
||||
<a class="sourceLine" id="cb12-3" title="3"> ns←⎕JSON⊃⎕NGET jsonFile</a>
|
||||
<a class="sourceLine" id="cb12-4" title="4"> ⍝ We use a simple regex search and replace on the</a>
|
||||
<a class="sourceLine" id="cb12-5" title="5"> ⍝ template.</a>
|
||||
<a class="sourceLine" id="cb12-6" title="6"> text←↑('@[a-zA-Z]*@'⎕R{ns getval ¯1↓1↓⍵.Match})template</a>
|
||||
<a class="sourceLine" id="cb12-7" title="7">∇</a></code></pre></div>
|
||||
<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>
|
||||
<a class="sourceLine" id="cb13-2" title="2"> template←⊃⎕NGET templateFile 1</a>
|
||||
<a class="sourceLine" id="cb13-3" title="3"> ns←⎕JSON⊃⎕NGET jsonFile</a>
|
||||
<a class="sourceLine" id="cb13-4" title="4"> ⍝ We use a simple regex search and replace on the</a>
|
||||
<a class="sourceLine" id="cb13-5" title="5"> ⍝ template.</a>
|
||||
<a class="sourceLine" id="cb13-6" title="6"> text←↑('@[a-zA-Z]*@'⎕R{ns getval ¯1↓1↓⍵.Match})template</a>
|
||||
<a class="sourceLine" id="cb13-7" title="7">∇</a></code></pre></div>
|
||||
<p>We first read the template and the JSON values from their files. The <a href="https://help.dyalog.com/18.0/index.htm#Language/System%20Functions/nget.htm"><code>⎕NGET</code></a> function read simple text files, and <a href="https://help.dyalog.com/18.0/index.htm#Language/System%20Functions/json.htm"><code>⎕JSON</code></a> extracts the key-value pairs as a namespace.</p>
|
||||
<p>Assuming all variable names contain only letters, we match the regex <code>@[a-zA-Z]*@</code> to match variable names enclosed between <code>@</code> symbols. The function <code>getval</code> then returns the appropriate value, and we can replace the variable name in the template.</p>
|
||||
<div class="sourceCode" id="cb13"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb13-1" title="1">∇ val←ns getval var</a>
|
||||
<a class="sourceLine" id="cb13-2" title="2"> :If ''≡var ⍝ literal '@'</a>
|
||||
<a class="sourceLine" id="cb13-3" title="3"> val←'@'</a>
|
||||
<a class="sourceLine" id="cb13-4" title="4"> :ElseIf (⊂var)∊ns.⎕NL ¯2</a>
|
||||
<a class="sourceLine" id="cb13-5" title="5"> val←⍕ns⍎var</a>
|
||||
<a class="sourceLine" id="cb13-6" title="6"> :Else</a>
|
||||
<a class="sourceLine" id="cb13-7" title="7"> val←'???'</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">∇ val←ns getval var</a>
|
||||
<a class="sourceLine" id="cb14-2" title="2"> :If ''≡var ⍝ literal '@'</a>
|
||||
<a class="sourceLine" id="cb14-3" title="3"> val←'@'</a>
|
||||
<a class="sourceLine" id="cb14-4" title="4"> :ElseIf (⊂var)∊ns.⎕NL ¯2</a>
|
||||
<a class="sourceLine" id="cb14-5" title="5"> val←⍕ns⍎var</a>
|
||||
<a class="sourceLine" id="cb14-6" title="6"> :Else</a>
|
||||
<a class="sourceLine" id="cb14-7" title="7"> val←'???'</a>
|
||||
<a class="sourceLine" id="cb14-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb14-9" title="9">∇</a></code></pre></div>
|
||||
<p>This function takes the namespace matching the variable names to their respective values, and the name of the variable.</p>
|
||||
<ul>
|
||||
<li>If the variable name is empty, we matched the string <code>@@</code>, which corresponds to a literal <code>@</code>.</li>
|
||||
|
|
@ -188,42 +171,42 @@
|
|||
<li>Otherwise, we have an unknown variable, so we replace it with <code>???</code>.</li>
|
||||
</ul>
|
||||
<h2 id="problem-7-upc">Problem 7 – UPC</h2>
|
||||
<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">CheckDigit←{10|-⍵+.×11⍴3 1}</a></code></pre></div>
|
||||
<p>The check digit satisfies the equation <span class="math display">\[ 3 x_{1}+x_{2}+3 x_{3}+x_{4}+3 x_{5}+x_{6}+3 x_{7}+x_{8}+3 x_{9}+x_{10}+3 x_{11}+x_{12} \equiv 0 \bmod 10, \]</span> therefore, <span class="math display">\[ x_{12} \equiv -(3 x_{1}+x_{2}+3 x_{3}+x_{4}+3 x_{5}+x_{6}+3 x_{7}+x_{8}+3 x_{9}+x_{10}+3 x_{11}) \bmod 10. \]</span></p>
|
||||
<p>Translated to APL, we just take the dot product between the first 11 digits of the barcode with <code>11⍴3 1</code>, negate it, and take the remainder by 10.</p>
|
||||
<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">⍝ Left and right representations of digits. Decoding</a>
|
||||
<a class="sourceLine" id="cb16-2" title="2">⍝ the binary representation from decimal is more</a>
|
||||
<a class="sourceLine" id="cb16-3" title="3">⍝ compact than writing everything explicitly.</a>
|
||||
<a class="sourceLine" id="cb16-4" title="4">lrepr←⍉(7⍴2)⊤13 25 19 61 35 49 47 59 55 11</a>
|
||||
<a class="sourceLine" id="cb16-5" title="5">rrepr←~¨lrepr</a></code></pre></div>
|
||||
<p>For the second task, the first thing we need to do is save the representation of digits. To save space, I did not encode the binary representation explicitly, instead using a decimal representation that I then decode in base 2. The right representation is just the bitwise negation.</p>
|
||||
<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">∇ bits←WriteUPC digits;left;right</a>
|
||||
<a class="sourceLine" id="cb17-2" title="2"> :If (11=≢digits)∧∧/digits∊0,⍳9</a>
|
||||
<a class="sourceLine" id="cb17-3" title="3"> left←,lrepr[1+6↑digits;]</a>
|
||||
<a class="sourceLine" id="cb17-4" title="4"> right←,rrepr[1+6↓digits,CheckDigit digits;]</a>
|
||||
<a class="sourceLine" id="cb17-5" title="5"> bits←1 0 1,left,0 1 0 1 0,right,1 0 1</a>
|
||||
<a class="sourceLine" id="cb17-6" title="6"> :Else</a>
|
||||
<a class="sourceLine" id="cb17-7" title="7"> bits←¯1</a>
|
||||
<a class="sourceLine" id="cb17-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb17-9" title="9">∇</a></code></pre></div>
|
||||
<p>First of all, if the vector <code>digits</code> does not have exactly 11 elements, all between 0 and 9, it is an error and we return <code>¯1</code>.</p>
|
||||
<p>Then, we take the first 6 digits and encode them with <code>lrepr</code>, and the last 5 digits plus the check digit encoded with <code>rrepr</code>. In each case, adding 1 is necessary because <code>⎕IO←1</code>. We return the final bit array with the required beginning, middle, and end guard patterns.</p>
|
||||
<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>
|
||||
<div class="sourceCode" id="cb18"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb18-1" title="1">∇ digits←ReadUPC bits</a>
|
||||
<a class="sourceLine" id="cb18-2" title="2"> :If 95≠⍴bits ⍝ incorrect number of bits</a>
|
||||
<a class="sourceLine" id="cb18-3" title="3"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb18-4" title="4"> :Else</a>
|
||||
<a class="sourceLine" id="cb18-5" title="5"> ⍝ Test if the barcode was scanned right-to-left.</a>
|
||||
<a class="sourceLine" id="cb18-6" title="6"> :If 0=2|+/bits[3+⍳7]</a>
|
||||
<a class="sourceLine" id="cb18-7" title="7"> bits←⌽bits</a>
|
||||
<a class="sourceLine" id="cb18-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb18-9" title="9"> digits←({¯1+lrepr⍳⍵}¨(7/⍳6)⊆42↑3↓bits),{¯1+rrepr⍳⍵}¨(7/⍳6)⊆¯42↑¯3↓bits</a>
|
||||
<a class="sourceLine" id="cb18-10" title="10"> :If ~∧/digits∊0,⍳9 ⍝ incorrect parity</a>
|
||||
<a class="sourceLine" id="cb18-11" title="11"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb18-12" title="12"> :ElseIf (⊃⌽digits)≠CheckDigit ¯1↓digits ⍝ incorrect check digit</a>
|
||||
<a class="sourceLine" id="cb18-13" title="13"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb18-14" title="14"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb18-15" title="15"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb18-16" title="16">∇</a></code></pre></div>
|
||||
<ul>
|
||||
<li>If we don’t have the correct number of bits, we return <code>¯1</code>.</li>
|
||||
<li>We test the first digit for its parity, to determine if its actually a left representation. If it’s not, we reverse the bit array.</li>
|
||||
|
|
@ -231,51 +214,53 @@
|
|||
<li>Final checks for the range of the digits (i.e., if the representations could not be found in the <code>lrepr</code> and <code>rrepr</code> vectors), and for the check digit.</li>
|
||||
</ul>
|
||||
<h2 id="problem-8-balancing-the-scales">Problem 8 – Balancing the Scales</h2>
|
||||
<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>
|
||||
<div class="sourceCode" id="cb19"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb19-1" title="1">∇ parts←Balance nums;subsets;partitions</a>
|
||||
<a class="sourceLine" id="cb19-2" title="2"> ⍝ This is a brute force solution, running in</a>
|
||||
<a class="sourceLine" id="cb19-3" title="3"> ⍝ exponential time. We generate all the possible</a>
|
||||
<a class="sourceLine" id="cb19-4" title="4"> ⍝ partitions, filter out those which are not</a>
|
||||
<a class="sourceLine" id="cb19-5" title="5"> ⍝ balanced, and return the first matching one. There</a>
|
||||
<a class="sourceLine" id="cb19-6" title="6"> ⍝ are more advanced approach running in</a>
|
||||
<a class="sourceLine" id="cb19-7" title="7"> ⍝ pseudo-polynomial time (based on dynamic</a>
|
||||
<a class="sourceLine" id="cb19-8" title="8"> ⍝ programming, see the "Partition problem" Wikipedia</a>
|
||||
<a class="sourceLine" id="cb19-9" title="9"> ⍝ page), but they are not warranted here, as the</a>
|
||||
<a class="sourceLine" id="cb19-10" title="10"> ⍝ input size remains fairly small.</a>
|
||||
<a class="sourceLine" id="cb19-11" title="11"></a>
|
||||
<a class="sourceLine" id="cb19-12" title="12"> ⍝ Generate all partitions of a vector of a given</a>
|
||||
<a class="sourceLine" id="cb19-13" title="13"> ⍝ size, as binary mask vectors.</a>
|
||||
<a class="sourceLine" id="cb19-14" title="14"> subsets←{1↓2⊥⍣¯1⍳2*⍵}</a>
|
||||
<a class="sourceLine" id="cb19-15" title="15"> ⍝ Keep only the subsets whose sum is exactly</a>
|
||||
<a class="sourceLine" id="cb19-16" title="16"> ⍝ (+/nums)÷2.</a>
|
||||
<a class="sourceLine" id="cb19-17" title="17"> partitions←nums{((2÷⍨+/⍺)=⍺+.×⍵)/⍵}subsets⍴nums</a>
|
||||
<a class="sourceLine" id="cb19-18" title="18"> :If 0=≢,partitions</a>
|
||||
<a class="sourceLine" id="cb19-19" title="19"> ⍝ If no partition satisfy the above</a>
|
||||
<a class="sourceLine" id="cb19-20" title="20"> ⍝ criterion, we return ⍬.</a>
|
||||
<a class="sourceLine" id="cb19-21" title="21"> parts←⍬</a>
|
||||
<a class="sourceLine" id="cb19-22" title="22"> :Else</a>
|
||||
<a class="sourceLine" id="cb19-23" title="23"> ⍝ Otherwise, we return the first possible</a>
|
||||
<a class="sourceLine" id="cb19-24" title="24"> ⍝ partition.</a>
|
||||
<a class="sourceLine" id="cb19-25" title="25"> parts←nums{((⊂,(⊂~))⊃↓⍉⍵)/¨2⍴⊂⍺}partitions</a>
|
||||
<a class="sourceLine" id="cb19-26" title="26"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb19-27" title="27">∇</a></code></pre></div>
|
||||
<h2 id="problem-9-upwardly-mobile">Problem 9 – Upwardly Mobile</h2>
|
||||
<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>
|
||||
<p>This is the only problem that I didn’t complete. It required parsing the files containing the graphical representations of the trees, which was needlessly complex and, quite frankly, hard and boring with a language like APL.</p>
|
||||
<p>However, the next part is interesting: once we have a matrix of coefficients representing the relationships between the weights, we can solve the system of equations. <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Matrix%20Divide.htm">Matrix Divide</a> (<code>⌹</code>) will find one solution to the system. Since the system is overdetermined, we fix <code>A=1</code> to find one possible solution. Since we want integer weights, the solution we find is smaller than the one we want, and may contain fractional weights. So we multiply everything by the <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/And%20Lowest%20Common%20Multiple.htm">Lowest Common Multiple</a> (<code>∧</code>) to get the smallest integer weights.</p>
|
||||
<div class="sourceCode" id="cb20"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb20-1" title="1">∇ weights←Weights filename;mobile;branches;mat</a>
|
||||
<a class="sourceLine" id="cb20-2" title="2"> ⍝ Put your code and comments below here</a>
|
||||
<a class="sourceLine" id="cb20-3" title="3"></a>
|
||||
<a class="sourceLine" id="cb20-4" title="4"> ⍝ Parse the mobile input file.</a>
|
||||
<a class="sourceLine" id="cb20-5" title="5"> mobile←↑⊃⎕NGET filename 1</a>
|
||||
<a class="sourceLine" id="cb20-6" title="6"> branches←⍸mobile∊'┌┴┐'</a>
|
||||
<a class="sourceLine" id="cb20-7" title="7"> ⍝ TODO: Build the matrix of coefficients mat.</a>
|
||||
<a class="sourceLine" id="cb20-8" title="8"></a>
|
||||
<a class="sourceLine" id="cb20-9" title="9"> ⍝ Solve the system of equations (arbitrarily setting</a>
|
||||
<a class="sourceLine" id="cb20-10" title="10"> ⍝ the first variable at 1 because the system is</a>
|
||||
<a class="sourceLine" id="cb20-11" title="11"> ⍝ overdetermined), then multiply the coefficients by</a>
|
||||
<a class="sourceLine" id="cb20-12" title="12"> ⍝ their least common multiple to get the smallest</a>
|
||||
<a class="sourceLine" id="cb20-13" title="13"> ⍝ integer weights.</a>
|
||||
<a class="sourceLine" id="cb20-14" title="14"> weights←((1∘,)×(∧/÷))mat[;1]⌹1↓[2]mat</a>
|
||||
<a class="sourceLine" id="cb20-15" title="15">∇</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb21"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb21-1" title="1"> :EndNamespace</a>
|
||||
<a class="sourceLine" id="cb21-2" title="2">:EndNamespace</a></code></pre></div>
|
||||
</section>
|
||||
</article>
|
||||
|
||||
|
|
|
|||
217
_site/rss.xml
217
_site/rss.xml
|
|
@ -99,50 +99,33 @@
|
|||
<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>
|
||||
<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>
|
||||
<h2 id="problem-5-future-and-present-value">Problem 5 – Future and Present Value</h2>
|
||||
<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>
|
||||
<a class="sourceLine" id="cb10-2" title="2">⍝ for a vector of amounts ⍺ and a vector of rates</a>
|
||||
<a class="sourceLine" id="cb10-3" title="3">⍝ ⍵. It is applied to every prefix subarray of amounts</a>
|
||||
<a class="sourceLine" id="cb10-4" title="4">⍝ and rates to get all intermediate values. However,</a>
|
||||
<a class="sourceLine" id="cb10-5" title="5">⍝ this has quadratic complexity.</a>
|
||||
<a class="sourceLine" id="cb10-6" title="6">⍝ rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)</a>
|
||||
<a class="sourceLine" id="cb10-7" title="7"></a>
|
||||
<a class="sourceLine" id="cb10-8" title="8">⍝ Second solution: We want to be able to use the</a>
|
||||
<a class="sourceLine" id="cb10-9" title="9">⍝ recurrence relation (recur) and scan through the</a>
|
||||
<a class="sourceLine" id="cb10-10" title="10">⍝ vectors of amounts and rates, accumulating the total</a>
|
||||
<a class="sourceLine" id="cb10-11" title="11">⍝ value at every time step. However, APL evaluation is</a>
|
||||
<a class="sourceLine" id="cb10-12" title="12">⍝ right-associative, so a simple Scan</a>
|
||||
<a class="sourceLine" id="cb10-13" title="13">⍝ (recur\amounts,¨values) would not give the correct</a>
|
||||
<a class="sourceLine" id="cb10-14" title="14">⍝ result, since recur is not associative and we need</a>
|
||||
<a class="sourceLine" id="cb10-15" title="15">⍝ to evaluate it left-to-right. (In any case, in this</a>
|
||||
<a class="sourceLine" id="cb10-16" title="16">⍝ case, Scan would have quadratic complexity, so would</a>
|
||||
<a class="sourceLine" id="cb10-17" title="17">⍝ not bring any benefit over the previous solution.)</a>
|
||||
<a class="sourceLine" id="cb10-18" title="18">⍝ What we need is something akin to Haskell's scanl</a>
|
||||
<a class="sourceLine" id="cb10-19" title="19">⍝ function, which would evaluate left to right in O(n)</a>
|
||||
<a class="sourceLine" id="cb10-20" title="20">⍝ time. This is what we do here, accumulating values</a>
|
||||
<a class="sourceLine" id="cb10-21" title="21">⍝ from left to right. (This is inspired from</a>
|
||||
<a class="sourceLine" id="cb10-22" title="22">⍝ dfns.ascan, although heavily simplified.)</a>
|
||||
<a class="sourceLine" id="cb10-23" title="23">rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}</a></code></pre></div>
|
||||
<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>
|
||||
<a class="sourceLine" id="cb11-2" title="2">pv←{+/⍺÷×\1+⍵}</a></code></pre></div>
|
||||
<p>First solution: <code>((1+⊢)⊥⊣)</code> computes the total return for a vector of amounts <code>⍺</code> and a vector of rates <code>⍵</code>. It is applied to every prefix subarray of amounts and rates to get all intermediate values. However, this has quadratic complexity.</p>
|
||||
<div class="sourceCode" id="cb10"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb10-1" title="1">rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)</a></code></pre></div>
|
||||
<p>Second solution: We want to be able to use the recurrence relation (<code>recur</code>) and scan through the vectors of amounts and rates, accumulating the total value at every time step. However, APL evaluation is right-associative, so a simple <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Operators/Scan.htm">Scan</a> (<code>recur\amounts,¨values</code>) would not give the correct result, since <code>recur</code> is not associative and we need to evaluate it left-to-right. (In any case, in this case, Scan would have quadratic complexity, so would not bring any benefit over the previous solution.) What we need is something akin to Haskell’s <code>scanl</code> function, which would evaluate left to right in <span class="math inline">\(O(n)\)</span> time<span><label for="sn-1" class="margin-toggle sidenote-number"></label><input type="checkbox" id="sn-1" class="margin-toggle" /><span class="sidenote">There is an interesting <a href="https://stackoverflow.com/a/25100675/8864368">StackOverflow answer</a> explaining the behaviour of Scan, and compares it to Haskell’s <code>scanl</code> function.<br />
|
||||
<br />
|
||||
</span></span>. This is what we do here, accumulating values from left to right. (This is inspired from <a href="https://dfns.dyalog.com/c_ascan.htm"><code>dfns.ascan</code></a>, although heavily simplified.)</p>
|
||||
<div class="sourceCode" id="cb11"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb11-1" title="1">rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}</a></code></pre></div>
|
||||
<p>For the second task, there is an explicit formula for cashflow calculations, so we can just apply it.</p>
|
||||
<div class="sourceCode" id="cb12"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb12-1" title="1">pv←{+/⍺÷×\1+⍵}</a></code></pre></div>
|
||||
<h2 id="problem-6-merge">Problem 6 – Merge</h2>
|
||||
<div class="sourceCode" id="cb12"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb12-1" title="1">∇ text←templateFile Merge jsonFile;template;ns</a>
|
||||
<a class="sourceLine" id="cb12-2" title="2"> template←⊃⎕NGET templateFile 1</a>
|
||||
<a class="sourceLine" id="cb12-3" title="3"> ns←⎕JSON⊃⎕NGET jsonFile</a>
|
||||
<a class="sourceLine" id="cb12-4" title="4"> ⍝ We use a simple regex search and replace on the</a>
|
||||
<a class="sourceLine" id="cb12-5" title="5"> ⍝ template.</a>
|
||||
<a class="sourceLine" id="cb12-6" title="6"> text←↑('@[a-zA-Z]*@'⎕R{ns getval ¯1↓1↓⍵.Match})template</a>
|
||||
<a class="sourceLine" id="cb12-7" title="7">∇</a></code></pre></div>
|
||||
<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>
|
||||
<a class="sourceLine" id="cb13-2" title="2"> template←⊃⎕NGET templateFile 1</a>
|
||||
<a class="sourceLine" id="cb13-3" title="3"> ns←⎕JSON⊃⎕NGET jsonFile</a>
|
||||
<a class="sourceLine" id="cb13-4" title="4"> ⍝ We use a simple regex search and replace on the</a>
|
||||
<a class="sourceLine" id="cb13-5" title="5"> ⍝ template.</a>
|
||||
<a class="sourceLine" id="cb13-6" title="6"> text←↑('@[a-zA-Z]*@'⎕R{ns getval ¯1↓1↓⍵.Match})template</a>
|
||||
<a class="sourceLine" id="cb13-7" title="7">∇</a></code></pre></div>
|
||||
<p>We first read the template and the JSON values from their files. The <a href="https://help.dyalog.com/18.0/index.htm#Language/System%20Functions/nget.htm"><code>⎕NGET</code></a> function read simple text files, and <a href="https://help.dyalog.com/18.0/index.htm#Language/System%20Functions/json.htm"><code>⎕JSON</code></a> extracts the key-value pairs as a namespace.</p>
|
||||
<p>Assuming all variable names contain only letters, we match the regex <code>@[a-zA-Z]*@</code> to match variable names enclosed between <code>@</code> symbols. The function <code>getval</code> then returns the appropriate value, and we can replace the variable name in the template.</p>
|
||||
<div class="sourceCode" id="cb13"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb13-1" title="1">∇ val←ns getval var</a>
|
||||
<a class="sourceLine" id="cb13-2" title="2"> :If ''≡var ⍝ literal '@'</a>
|
||||
<a class="sourceLine" id="cb13-3" title="3"> val←'@'</a>
|
||||
<a class="sourceLine" id="cb13-4" title="4"> :ElseIf (⊂var)∊ns.⎕NL ¯2</a>
|
||||
<a class="sourceLine" id="cb13-5" title="5"> val←⍕ns⍎var</a>
|
||||
<a class="sourceLine" id="cb13-6" title="6"> :Else</a>
|
||||
<a class="sourceLine" id="cb13-7" title="7"> val←'???'</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">∇ val←ns getval var</a>
|
||||
<a class="sourceLine" id="cb14-2" title="2"> :If ''≡var ⍝ literal '@'</a>
|
||||
<a class="sourceLine" id="cb14-3" title="3"> val←'@'</a>
|
||||
<a class="sourceLine" id="cb14-4" title="4"> :ElseIf (⊂var)∊ns.⎕NL ¯2</a>
|
||||
<a class="sourceLine" id="cb14-5" title="5"> val←⍕ns⍎var</a>
|
||||
<a class="sourceLine" id="cb14-6" title="6"> :Else</a>
|
||||
<a class="sourceLine" id="cb14-7" title="7"> val←'???'</a>
|
||||
<a class="sourceLine" id="cb14-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb14-9" title="9">∇</a></code></pre></div>
|
||||
<p>This function takes the namespace matching the variable names to their respective values, and the name of the variable.</p>
|
||||
<ul>
|
||||
<li>If the variable name is empty, we matched the string <code>@@</code>, which corresponds to a literal <code>@</code>.</li>
|
||||
|
|
@ -150,42 +133,42 @@
|
|||
<li>Otherwise, we have an unknown variable, so we replace it with <code>???</code>.</li>
|
||||
</ul>
|
||||
<h2 id="problem-7-upc">Problem 7 – UPC</h2>
|
||||
<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">CheckDigit←{10|-⍵+.×11⍴3 1}</a></code></pre></div>
|
||||
<p>The check digit satisfies the equation <span class="math display">\[ 3 x_{1}+x_{2}+3 x_{3}+x_{4}+3 x_{5}+x_{6}+3 x_{7}+x_{8}+3 x_{9}+x_{10}+3 x_{11}+x_{12} \equiv 0 \bmod 10, \]</span> therefore, <span class="math display">\[ x_{12} \equiv -(3 x_{1}+x_{2}+3 x_{3}+x_{4}+3 x_{5}+x_{6}+3 x_{7}+x_{8}+3 x_{9}+x_{10}+3 x_{11}) \bmod 10. \]</span></p>
|
||||
<p>Translated to APL, we just take the dot product between the first 11 digits of the barcode with <code>11⍴3 1</code>, negate it, and take the remainder by 10.</p>
|
||||
<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">⍝ Left and right representations of digits. Decoding</a>
|
||||
<a class="sourceLine" id="cb16-2" title="2">⍝ the binary representation from decimal is more</a>
|
||||
<a class="sourceLine" id="cb16-3" title="3">⍝ compact than writing everything explicitly.</a>
|
||||
<a class="sourceLine" id="cb16-4" title="4">lrepr←⍉(7⍴2)⊤13 25 19 61 35 49 47 59 55 11</a>
|
||||
<a class="sourceLine" id="cb16-5" title="5">rrepr←~¨lrepr</a></code></pre></div>
|
||||
<p>For the second task, the first thing we need to do is save the representation of digits. To save space, I did not encode the binary representation explicitly, instead using a decimal representation that I then decode in base 2. The right representation is just the bitwise negation.</p>
|
||||
<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">∇ bits←WriteUPC digits;left;right</a>
|
||||
<a class="sourceLine" id="cb17-2" title="2"> :If (11=≢digits)∧∧/digits∊0,⍳9</a>
|
||||
<a class="sourceLine" id="cb17-3" title="3"> left←,lrepr[1+6↑digits;]</a>
|
||||
<a class="sourceLine" id="cb17-4" title="4"> right←,rrepr[1+6↓digits,CheckDigit digits;]</a>
|
||||
<a class="sourceLine" id="cb17-5" title="5"> bits←1 0 1,left,0 1 0 1 0,right,1 0 1</a>
|
||||
<a class="sourceLine" id="cb17-6" title="6"> :Else</a>
|
||||
<a class="sourceLine" id="cb17-7" title="7"> bits←¯1</a>
|
||||
<a class="sourceLine" id="cb17-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb17-9" title="9">∇</a></code></pre></div>
|
||||
<p>First of all, if the vector <code>digits</code> does not have exactly 11 elements, all between 0 and 9, it is an error and we return <code>¯1</code>.</p>
|
||||
<p>Then, we take the first 6 digits and encode them with <code>lrepr</code>, and the last 5 digits plus the check digit encoded with <code>rrepr</code>. In each case, adding 1 is necessary because <code>⎕IO←1</code>. We return the final bit array with the required beginning, middle, and end guard patterns.</p>
|
||||
<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>
|
||||
<div class="sourceCode" id="cb18"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb18-1" title="1">∇ digits←ReadUPC bits</a>
|
||||
<a class="sourceLine" id="cb18-2" title="2"> :If 95≠⍴bits ⍝ incorrect number of bits</a>
|
||||
<a class="sourceLine" id="cb18-3" title="3"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb18-4" title="4"> :Else</a>
|
||||
<a class="sourceLine" id="cb18-5" title="5"> ⍝ Test if the barcode was scanned right-to-left.</a>
|
||||
<a class="sourceLine" id="cb18-6" title="6"> :If 0=2|+/bits[3+⍳7]</a>
|
||||
<a class="sourceLine" id="cb18-7" title="7"> bits←⌽bits</a>
|
||||
<a class="sourceLine" id="cb18-8" title="8"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb18-9" title="9"> digits←({¯1+lrepr⍳⍵}¨(7/⍳6)⊆42↑3↓bits),{¯1+rrepr⍳⍵}¨(7/⍳6)⊆¯42↑¯3↓bits</a>
|
||||
<a class="sourceLine" id="cb18-10" title="10"> :If ~∧/digits∊0,⍳9 ⍝ incorrect parity</a>
|
||||
<a class="sourceLine" id="cb18-11" title="11"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb18-12" title="12"> :ElseIf (⊃⌽digits)≠CheckDigit ¯1↓digits ⍝ incorrect check digit</a>
|
||||
<a class="sourceLine" id="cb18-13" title="13"> digits←¯1</a>
|
||||
<a class="sourceLine" id="cb18-14" title="14"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb18-15" title="15"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb18-16" title="16">∇</a></code></pre></div>
|
||||
<ul>
|
||||
<li>If we don’t have the correct number of bits, we return <code>¯1</code>.</li>
|
||||
<li>We test the first digit for its parity, to determine if its actually a left representation. If it’s not, we reverse the bit array.</li>
|
||||
|
|
@ -193,51 +176,53 @@
|
|||
<li>Final checks for the range of the digits (i.e., if the representations could not be found in the <code>lrepr</code> and <code>rrepr</code> vectors), and for the check digit.</li>
|
||||
</ul>
|
||||
<h2 id="problem-8-balancing-the-scales">Problem 8 – Balancing the Scales</h2>
|
||||
<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>
|
||||
<div class="sourceCode" id="cb19"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb19-1" title="1">∇ parts←Balance nums;subsets;partitions</a>
|
||||
<a class="sourceLine" id="cb19-2" title="2"> ⍝ This is a brute force solution, running in</a>
|
||||
<a class="sourceLine" id="cb19-3" title="3"> ⍝ exponential time. We generate all the possible</a>
|
||||
<a class="sourceLine" id="cb19-4" title="4"> ⍝ partitions, filter out those which are not</a>
|
||||
<a class="sourceLine" id="cb19-5" title="5"> ⍝ balanced, and return the first matching one. There</a>
|
||||
<a class="sourceLine" id="cb19-6" title="6"> ⍝ are more advanced approach running in</a>
|
||||
<a class="sourceLine" id="cb19-7" title="7"> ⍝ pseudo-polynomial time (based on dynamic</a>
|
||||
<a class="sourceLine" id="cb19-8" title="8"> ⍝ programming, see the "Partition problem" Wikipedia</a>
|
||||
<a class="sourceLine" id="cb19-9" title="9"> ⍝ page), but they are not warranted here, as the</a>
|
||||
<a class="sourceLine" id="cb19-10" title="10"> ⍝ input size remains fairly small.</a>
|
||||
<a class="sourceLine" id="cb19-11" title="11"></a>
|
||||
<a class="sourceLine" id="cb19-12" title="12"> ⍝ Generate all partitions of a vector of a given</a>
|
||||
<a class="sourceLine" id="cb19-13" title="13"> ⍝ size, as binary mask vectors.</a>
|
||||
<a class="sourceLine" id="cb19-14" title="14"> subsets←{1↓2⊥⍣¯1⍳2*⍵}</a>
|
||||
<a class="sourceLine" id="cb19-15" title="15"> ⍝ Keep only the subsets whose sum is exactly</a>
|
||||
<a class="sourceLine" id="cb19-16" title="16"> ⍝ (+/nums)÷2.</a>
|
||||
<a class="sourceLine" id="cb19-17" title="17"> partitions←nums{((2÷⍨+/⍺)=⍺+.×⍵)/⍵}subsets⍴nums</a>
|
||||
<a class="sourceLine" id="cb19-18" title="18"> :If 0=≢,partitions</a>
|
||||
<a class="sourceLine" id="cb19-19" title="19"> ⍝ If no partition satisfy the above</a>
|
||||
<a class="sourceLine" id="cb19-20" title="20"> ⍝ criterion, we return ⍬.</a>
|
||||
<a class="sourceLine" id="cb19-21" title="21"> parts←⍬</a>
|
||||
<a class="sourceLine" id="cb19-22" title="22"> :Else</a>
|
||||
<a class="sourceLine" id="cb19-23" title="23"> ⍝ Otherwise, we return the first possible</a>
|
||||
<a class="sourceLine" id="cb19-24" title="24"> ⍝ partition.</a>
|
||||
<a class="sourceLine" id="cb19-25" title="25"> parts←nums{((⊂,(⊂~))⊃↓⍉⍵)/¨2⍴⊂⍺}partitions</a>
|
||||
<a class="sourceLine" id="cb19-26" title="26"> :EndIf</a>
|
||||
<a class="sourceLine" id="cb19-27" title="27">∇</a></code></pre></div>
|
||||
<h2 id="problem-9-upwardly-mobile">Problem 9 – Upwardly Mobile</h2>
|
||||
<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>
|
||||
<p>This is the only problem that I didn’t complete. It required parsing the files containing the graphical representations of the trees, which was needlessly complex and, quite frankly, hard and boring with a language like APL.</p>
|
||||
<p>However, the next part is interesting: once we have a matrix of coefficients representing the relationships between the weights, we can solve the system of equations. <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Matrix%20Divide.htm">Matrix Divide</a> (<code>⌹</code>) will find one solution to the system. Since the system is overdetermined, we fix <code>A=1</code> to find one possible solution. Since we want integer weights, the solution we find is smaller than the one we want, and may contain fractional weights. So we multiply everything by the <a href="https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/And%20Lowest%20Common%20Multiple.htm">Lowest Common Multiple</a> (<code>∧</code>) to get the smallest integer weights.</p>
|
||||
<div class="sourceCode" id="cb20"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb20-1" title="1">∇ weights←Weights filename;mobile;branches;mat</a>
|
||||
<a class="sourceLine" id="cb20-2" title="2"> ⍝ Put your code and comments below here</a>
|
||||
<a class="sourceLine" id="cb20-3" title="3"></a>
|
||||
<a class="sourceLine" id="cb20-4" title="4"> ⍝ Parse the mobile input file.</a>
|
||||
<a class="sourceLine" id="cb20-5" title="5"> mobile←↑⊃⎕NGET filename 1</a>
|
||||
<a class="sourceLine" id="cb20-6" title="6"> branches←⍸mobile∊'┌┴┐'</a>
|
||||
<a class="sourceLine" id="cb20-7" title="7"> ⍝ TODO: Build the matrix of coefficients mat.</a>
|
||||
<a class="sourceLine" id="cb20-8" title="8"></a>
|
||||
<a class="sourceLine" id="cb20-9" title="9"> ⍝ Solve the system of equations (arbitrarily setting</a>
|
||||
<a class="sourceLine" id="cb20-10" title="10"> ⍝ the first variable at 1 because the system is</a>
|
||||
<a class="sourceLine" id="cb20-11" title="11"> ⍝ overdetermined), then multiply the coefficients by</a>
|
||||
<a class="sourceLine" id="cb20-12" title="12"> ⍝ their least common multiple to get the smallest</a>
|
||||
<a class="sourceLine" id="cb20-13" title="13"> ⍝ integer weights.</a>
|
||||
<a class="sourceLine" id="cb20-14" title="14"> weights←((1∘,)×(∧/÷))mat[;1]⌹1↓[2]mat</a>
|
||||
<a class="sourceLine" id="cb20-15" title="15">∇</a></code></pre></div>
|
||||
<div class="sourceCode" id="cb21"><pre class="sourceCode default"><code class="sourceCode default"><a class="sourceLine" id="cb21-1" title="1"> :EndNamespace</a>
|
||||
<a class="sourceLine" id="cb21-2" title="2">:EndNamespace</a></code></pre></div>
|
||||
</section>
|
||||
</article>
|
||||
]]></description>
|
||||
|
|
|
|||
|
|
@ -173,34 +173,41 @@ operator (~⍣~), with an initial value of 1.
|
|||
|
||||
* Problem 5 -- Future and Present Value
|
||||
|
||||
#+begin_src default
|
||||
⍝ First solution: ((1+⊢)⊥⊣) computes the total return
|
||||
⍝ for a vector of amounts ⍺ and a vector of rates
|
||||
⍝ ⍵. It is applied to every prefix subarray of amounts
|
||||
⍝ and rates to get all intermediate values. However,
|
||||
⍝ this has quadratic complexity.
|
||||
⍝ rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)
|
||||
First solution: ~((1+⊢)⊥⊣)~ computes the total return for a vector of
|
||||
amounts ~⍺~ and a vector of rates ~⍵~. It is applied to every prefix
|
||||
subarray of amounts and rates to get all intermediate values. However,
|
||||
this has quadratic complexity.
|
||||
|
||||
⍝ Second solution: We want to be able to use the
|
||||
⍝ recurrence relation (recur) and scan through the
|
||||
⍝ vectors of amounts and rates, accumulating the total
|
||||
⍝ value at every time step. However, APL evaluation is
|
||||
⍝ right-associative, so a simple Scan
|
||||
⍝ (recur\amounts,¨values) would not give the correct
|
||||
⍝ result, since recur is not associative and we need
|
||||
⍝ to evaluate it left-to-right. (In any case, in this
|
||||
⍝ case, Scan would have quadratic complexity, so would
|
||||
⍝ not bring any benefit over the previous solution.)
|
||||
⍝ What we need is something akin to Haskell's scanl
|
||||
⍝ function, which would evaluate left to right in O(n)
|
||||
⍝ time. This is what we do here, accumulating values
|
||||
⍝ from left to right. (This is inspired from
|
||||
⍝ dfns.ascan, although heavily simplified.)
|
||||
#+begin_src default
|
||||
rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)
|
||||
#+end_src
|
||||
|
||||
Second solution: We want to be able to use the recurrence relation
|
||||
(~recur~) and scan through the vectors of amounts and rates,
|
||||
accumulating the total value at every time step. However, APL
|
||||
evaluation is right-associative, so a simple [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Operators/Scan.htm][Scan]]
|
||||
(~recur\amounts,¨values~) would not give the correct result, since
|
||||
~recur~ is not associative and we need to evaluate it
|
||||
left-to-right. (In any case, in this case, Scan would have quadratic
|
||||
complexity, so would not bring any benefit over the previous
|
||||
solution.) What we need is something akin to Haskell's ~scanl~
|
||||
function, which would evaluate left to right in $O(n)$
|
||||
time[fn:apl-scan]. This is what we do here, accumulating values from
|
||||
left to right. (This is inspired from [[https://dfns.dyalog.com/c_ascan.htm][~dfns.ascan~]], although heavily
|
||||
simplified.)
|
||||
|
||||
[fn:apl-scan] There is an interesting [[https://stackoverflow.com/a/25100675/8864368][StackOverflow answer]] explaining
|
||||
the behaviour of Scan, and compares it to Haskell's ~scanl~ function.
|
||||
|
||||
|
||||
#+begin_src default
|
||||
rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}
|
||||
#+end_src
|
||||
|
||||
For the second task, there is an explicit formula for cashflow
|
||||
calculations, so we can just apply it.
|
||||
|
||||
#+begin_src default
|
||||
⍝ Simply apply the formula for cashflow calculations.
|
||||
pv←{+/⍺÷×\1+⍵}
|
||||
#+end_src
|
||||
|
||||
|
|
@ -358,6 +365,20 @@ with the required beginning, middle, and end guard patterns.
|
|||
|
||||
* Problem 9 -- Upwardly Mobile
|
||||
|
||||
This is the only problem that I didn't complete. It required parsing
|
||||
the files containing the graphical representations of the trees, which
|
||||
was needlessly complex and, quite frankly, hard and boring with a
|
||||
language like APL.
|
||||
|
||||
However, the next part is interesting: once we have a matrix of
|
||||
coefficients representing the relationships between the weights, we
|
||||
can solve the system of equations. [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Matrix%20Divide.htm][Matrix Divide]] (~⌹~) will find one
|
||||
solution to the system. Since the system is overdetermined, we fix
|
||||
~A=1~ to find one possible solution. Since we want integer weights,
|
||||
the solution we find is smaller than the one we want, and may contain
|
||||
fractional weights. So we multiply everything by the [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/And%20Lowest%20Common%20Multiple.htm][Lowest Common
|
||||
Multiple]] (~∧~) to get the smallest integer weights.
|
||||
|
||||
#+begin_src default
|
||||
∇ weights←Weights filename;mobile;branches;mat
|
||||
⍝ Put your code and comments below here
|
||||
|
|
|
|||
Loading…
Add table
Add a link
Reference in a new issue