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2012-10-31T13:30:51.593
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<p>My answer is in Java, I am just getting into Python so I will update this answer when I get around to it.</p> <pre class="lang-java prettyprint-override"><code>boolean nand(boolean[] a) { boolean b = true if (a.length > 1) { b = a[0] && a[1]; if (a.length > 2) for (int i = 2; i < a.length; i++) b &= a[i]; } return !b; } </code></pre> <p>Usage (Should return true):</p> <pre><code>nand(training_set[i][0]) == training_set[i][1] </code></pre> <h1>Edit</h1> <p>Over two years later, I have come back to this question to add to my answer...</p> <p>I have created an iterative and recursive solution in python. I have also attempted to code golf the recursive. I got it down to 106 bytes.</p> <h2>Implementations</h2> <h3>Iterative</h3> <pre class="lang-py prettyprint-override"><code>def nand(*a): if len(a) > 1: b = a[0] and a[1] if len(a) > 2: for i in range(2, len(a)): b = b and a[i] return not b return None </code></pre> <h3>Recursive</h3> <pre class="lang-py prettyprint-override"><code>def nand(*a): if len(a) < 2: return None elif len(a) == 2: return not (a[0] and a[1]) else: return nand(*([a[0] and a[1]] + list(a[2:]))) </code></pre> <h3>Recursive (lambda)</h3> <pre class="lang-py prettyprint-override"><code>nand=lambda*a:None if len(a)<2 else not(a[0]and a[1])if len(a)==2 else nand(*([a[0]and a[1]]+list(a[2:]))) </code></pre> <h2>Results</h2> <pre class="lang-py prettyprint-override"><code>print nand(True, True, True) # ¬(1 ∧ 1 ∧ 1) == 0 print nand(True, True, False) # ¬(1 ∧ 1 ∧ 0) == 1 print nand(True, False, True) # ¬(1 ∧ 0 ∧ 1) == 1 print nand(True, False, False) # ¬(1 ∧ 0 ∧ 0) == 1 print nand(False, True, True) # ¬(0 ∧ 1 ∧ 1) == 1 print nand(False, True, False) # ¬(0 ∧ 1 ∧ 0) == 1 print nand(False, False, True) # ¬(0 ∧ 0 ∧ 1) == 1 print nand(False, False, False) # ¬(0 ∧ 0 ∧ 0) == 1 </code></pre>
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1. POA simple perceptron in Python
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1. USMr. Polywhirl
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