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<!--replace-end-7--><!--replace-end-4--><!--replace-end-1--></head><body><div class="ui fluid container universe"><!--replace-start-2--><!--replace-start-3--><!--replace-start-6--><div class="ui text container" id="zettel-container" style="position: relative"><div class="zettel-view"><article class="ui raised attached segment zettel-content"><div class="pandoc"><h1 id="title-h1">Boolean function synthesis</h1><p>When we looked at <span class="zettel-link-container cf"><span class="zettel-link" title="Zettel: Boolean functions"><a href="Boolean_functions.html">boolean functions</a></span></span> we were working in a particular direction: from a function to a truth table. When we do Boolean function synthesis we work in the opposite direction: from a truth table to a function.</p><p>This is an important skill that we will use when constructing <span class="zettel-link-container errors"><span class="zettel-link" title="Wiki-link does not refer to any existing zettel"><a>Digital_circuits</a></span></span>. We will go from truth conditions (i.e. what we want the circuit to do and when we want it to do it) to a function expression which is then reduced to its simplest form and implemented with <span class="zettel-link-container cf"><span class="zettel-link" title="Zettel: Logic gates"><a href="Logic_gates.html">logic gates</a></span></span>. Specifically, NAND gates.</p><p>We will show here that a complex logical expression can be reduced to an equivalent expression that uses only the NAND operator.</p><h2 id="the-process">The process</h2><p>The process proceeds as follows:</p><ol><li>Work out the truth conditions for the circuit we want to construct</li><li>Identify the rows where the output is equal to 1</li><li>For each of these rows construct a Boolean expression that evaluates to that output</li><li>Join each expression with OR</li><li>Reduce these expressions to a single expression in its simplest form</li></ol><h2 id="example">Example</h2><p>Let’s say we have the following truth table:</p><table class="ui table"><thead><tr><th>Line</th><th><span class="math inline">\(x\)</span></th><th><span class="math inline">\(y\)</span></th><th><span class="math inline">\(z\)</span></th><th><span class="math inline">\(f\)</span></th></tr></thead><tbody><tr><td>1</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>2</td><td>0</td><td>0</td><td>1</td><td>0</td></tr><tr><td>3</td><td>0</td><td>1</td><td>0</td><td>1</td></tr><tr><td>4</td><td>0</td><td>1</td><td>1</td><td>0</td></tr><tr><td>5</td><td>1</td><td>0</td><td>0</td><td>1</td></tr><tr><td>6</td><td>1</td><td>0</td><td>1</td><td>0</td></tr><tr><td>7</td><td>1</td><td>1</td><td>0</td><td>0</td></tr><tr><td>8</td><td>1</td><td>1</td><td>1</td><td>0</td></tr></tbody></table><p>We only need to focus on lines 1, 3, and 5 since they have the output 1:</p><table class="ui table"><thead><tr><th>Line</th><th><span class="math inline">\(x\)</span></th><th><span class="math inline">\(y\)</span></th><th><span class="math inline">\(z\)</span></th><th><span class="math inline">\(f\)</span></th></tr></thead><tbody><tr><td>1</td><td>0</td><td>0</td><td>0</td><td>1</td></tr><tr><td>3</td><td>0</td><td>1</td><td>0</td><td>1</td></tr><tr><td>5</td><td>1</td><td>0</td><td>0</td><td>1</td></tr></tbody></table><p>For each line we construct a Boolean expression that would result in the value in the <span class="math inline">\(f\)</span> column. In other words we construct the function:</p><table class="ui table"><thead><tr><th>Line</th><th><span class="math inline">\(x\)</span></th><th><span class="math inline">\(y\)</span></th><th><span class="math inline">\(z\)</span></th><th><span class="math inline">\(f\)</span></th></tr></thead><tbody><tr><td>1</td><td>0</td><td>0</td><td>0</td><td><span class="math inline">\(\lnot(x) \land \lnot (y) \land \lnot(z)\)</span></td></tr><tr><td>3</td><td>0</td><td>1</td><td>0</td><td><span class="math inline">\(\lnot(x) \land y \land \lnot(z)\)</span></td></tr><tr><td>5</td><td>1</td><td>0</td><td>0</td><td><span class="math inline">\(x \land \lnot(y) \land \lnot(z)\)</span></td></tr></tbody></table><p>We can now join each expression to create a complex expression that covers the entire truth table using OR:</p><p><span class="math display">$$
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(\lnot(x) \land \lnot (y) \land \lnot(z)) \\ \lor \\ (\lnot(x) \land y \land \lnot(z)) \\ \lor \\ (x \land \lnot(y) \land \lnot(z))
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$$</span></p><p>It’s clear that we have transcribed the truth conditions accurately but that we are doing so in a rather verbose way. We can simplify by just looking at the position of the 1s in the truth table. Notice:</p><ul><li><span class="math inline">\(z\)</span> is always 0</li><li><span class="math inline">\(x\)</span> and <span class="math inline">\(y\)</span> are either 0 or 1 but never both 1 in the same row</li></ul><p>So we simplify:</p><p><span class="math display">$$
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(\lnot(x) \land \lnot(z)) \lor (\lnot(y) \land \lnot(z))
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$$</span></p><p>Notice that <span class="math inline">\(\lnot(z)\)</span> is repeated so we can remove the repetition through <a href="Boolean_algebra.md#idempotent-law">idempotence</a>:</p><p><span class="math display">$$
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\lnot z \land (\lnot(x) \lor \lnot(y))
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$$</span></p><p>The upshot is that we now have a simpler expression that uses only NOT, OR and AND. These are the fundamental logic gates thus we are able to construct a circuit that embodies the logic of the expression.</p><blockquote><p>This is important and is an instance of the general theorem that <em>any Boolean function</em> can be represented using an expression containing AND, OR and NOT operations</p></blockquote><p>But even this is too complex. We could get rid of the OR and just use AND and NOT:</p><p>We can prove this theorem by showing that an expression with AND, NOT, and OR can be formulated as an equivalent expression using just NOT and AND:</p><p><span class="math display">$$
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x \lor y = \lnot(\lnot(x) \land \lnot(y))
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$$</span></p><table class="ui table"><thead><tr><th><span class="math inline">\(x\)</span></th><th><span class="math inline">\(y\)</span></th><th><span class="math inline">\(x \lor y\)</span></th><th><span class="math inline">\(\lnot(\lnot(x) \land \lnot(y)\)</span></th></tr></thead><tbody><tr><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>0</td><td>1</td><td>1</td><td>1</td></tr><tr><td>1</td><td>0</td><td>1</td><td>1</td></tr><tr><td>1</td><td>1</td><td>1</td><td>1</td></tr></tbody></table><p>Finally, we can simplify even further by doing away with AND and NOT and using a single <a href="Logic_gates.md#nand-gate">NAND gate</a> which embodies the logic of both, being true in all instances where AND would be false: <span class="math inline">\(\lnot (x \land y)\)</span>.</p><p>Let’s prove the theorem that every logical expression can be formulated as a NAND function. To do this we need to show that both NOT and AND can be converted to NAND.</p><p>NOT:</p><p><span class="math display">$$
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\lnot(x) = x \lnot\land x
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$$</span></p><p>AND:</p><p><span class="math display">$$
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x \land y = \lnot(x \lnot\land y)
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$$</span></p></div></article><nav class="ui attached segment deemphasized bottomPane" id="neuron-tags-pane"><div><span class="ui basic label zettel-tag" title="Tag">logic</span><span class="ui basic label zettel-tag" title="Tag">nand-to-tetris</span><span class="ui basic label zettel-tag" title="Tag">propositional-logic</span></div></nav><nav class="ui bottom attached icon compact inverted menu blue" id="neuron-nav-bar"><!--replace-start-9--><!--replace-end-9--><a class="right item" href="impulse.html" title="Open Impulse"><i class="wave square icon"></i></a></nav></div></div><!--replace-end-6--><!--replace-end-3--><!--replace-end-2--><div class="ui center aligned container footer-version"><div class="ui tiny image"><a href="https://neuron.zettel.page"><img alt="logo" src="https://raw.githubusercontent.com/srid/neuron/master/assets/neuron.svg" title="Generated by Neuron 1.9.35.3" /></a></div></div></div></body></html> |