139 lines
4.5 KiB
Markdown
139 lines
4.5 KiB
Markdown
---
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tags: [propositional-logic, nand-to-tetris]
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---
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# Boolean function synthesis
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When we looked at
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[boolean functions](Boolean_functions.md) we were
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working in a particular direction: from a function to a truth table. When we do
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Boolean function synthesis we work in the opposite direction: from a truth table
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to a function.
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This is an important skill that we will use when constructing
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[logic circuits](Digital_circuits.md).
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We will go from truth conditions (i.e. what we want the circuit to do and when
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we want it to do it) to a function expression which is then reduced to its
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simplest form and implemented with
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[logic gates](Logic_gates.md).
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Specifically, NAND gates.
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We will show here that a complex logical expression can be reduced to an
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equivalent expression that uses only the NAND operator.
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## The process
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The process proceeds as follows:
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1. Work out the truth conditions for the circuit we want to construct
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2. Identify the rows where the output is equal to 1
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3. For each of these rows construct a Boolean expression that evaluates to that
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output
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4. Join each expression with OR
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5. Reduce these expressions to a single expression in its simplest form
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## Example
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Let's say we have the following truth table:
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| Line | $x$ | $y$ | $z$ | $f$ |
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| ---- | --- | --- | --- | --- |
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| 1 | 0 | 0 | 0 | 0 |
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| 2 | 0 | 0 | 1 | 0 |
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| 3 | 0 | 1 | 0 | 1 |
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| 4 | 0 | 1 | 1 | 0 |
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| 5 | 1 | 0 | 0 | 1 |
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| 6 | 1 | 0 | 1 | 0 |
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| 7 | 1 | 1 | 0 | 0 |
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| 8 | 1 | 1 | 1 | 0 |
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We only need to focus on lines 1, 3, and 5 since they have the output 1:
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| Line | $x$ | $y$ | $z$ | $f$ |
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| ---- | --- | --- | --- | --- |
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| 1 | 0 | 0 | 0 | 1 |
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| 3 | 0 | 1 | 0 | 1 |
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| 5 | 1 | 0 | 0 | 1 |
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For each line we construct a Boolean expression that would result in the value
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in the $f$ column. In other words we construct the function:
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| Line | $x$ | $y$ | $z$ | $f$ |
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| ---- | --- | --- | --- | ----------------------------------------- |
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| 1 | 0 | 0 | 0 | $\lnot(x) \land \lnot (y) \land \lnot(z)$ |
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| 3 | 0 | 1 | 0 | $\lnot(x) \land y \land \lnot(z)$ |
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| 5 | 1 | 0 | 0 | $x \land \lnot(y) \land \lnot(z)$ |
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We can now join each expression to create a complex expression that covers the
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entire truth table using OR:
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$$
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(\lnot(x) \land \lnot (y) \land \lnot(z)) \\ \lor \\ (\lnot(x) \land y \land \lnot(z)) \\ \lor \\ (x \land \lnot(y) \land \lnot(z))
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$$
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It's clear that we have transcribed the truth conditions accurately but that we
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are doing so in a rather verbose way. We can simplify by just looking at the
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position of the 1s in the truth table. Notice:
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- $z$ is always 0
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- $x$ and $y$ are either 0 or 1 but never both 1 in the same row
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So we simplify:
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$$
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(\lnot(x) \land \lnot(z)) \lor (\lnot(y) \land \lnot(z))
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$$
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Notice that $\lnot(z)$ is repeated so we can remove the repetition through
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[idempotence](Boolean_algebra.md#idempotent-law):
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$$
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\lnot z \land (\lnot(x) \lor \lnot(y))
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$$
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The upshot is that we now have a simpler expression that uses only NOT, OR and
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AND. These are the fundamental logic gates thus we are able to construct a
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circuit that embodies the logic of the expression.
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> This is important and is an instance of the general theorem that _any Boolean
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> function_ can be represented using an expression containing AND, OR and NOT
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> operations
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But even this is too complex. We could get rid of the OR and just use AND and
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NOT:
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We can prove this theorem by showing that an expression with AND, NOT, and OR
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can be formulated as an equivalent expression using just NOT and AND:
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$$
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x \lor y = \lnot(\lnot(x) \land \lnot(y))
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$$
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| $x$ | $y$ | $x \lor y$ | $\lnot(\lnot(x) \land \lnot(y)$ |
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| --- | --- | ---------- | ------------------------------- |
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| 0 | 0 | 0 | 0 |
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| 0 | 1 | 1 | 1 |
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| 1 | 0 | 1 | 1 |
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| 1 | 1 | 1 | 1 |
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Finally, we can simplify even further by doing away with AND and NOT and using a
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single
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[NAND gate](Logic_gates.md#nand-gate)
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which embodies the logic of both, being true in all instances where AND would be
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false: $\lnot (x \land y)$.
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Let's prove the theorem that every logical expression can be formulated as a
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NAND function. To do this we need to show that both NOT and AND can be converted
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to NAND.
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NOT:
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$$
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\lnot(x) = x \lnot\land x
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$$
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AND:
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$$
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x \land y = \lnot(x \lnot\land y)
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$$
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