Logika: Programming Logics
3. Circuits that Compute on Numbers: Algebra

# Circuits that Compute on Numbers: Algebra¶

Circuits are “programs” that are constructed from AND, OR, NOT operations and compute on the value set, { t, f }. We use equations to name the wires within the circuit, and we saw how to deduce properties of the circuit even when we did not know the exact values of the circuit’s inputs. This is critical for circuit design and analysis.

We can also make “circuits” from the arithmetic operations, +, -, *, /, that compute on numbers – numbers travel along the wires. We apply the principles of algebra to deduce properties of the circuits we write.

In this chapter, we review principles of algebra and apply them to equationally defined systems (circuits).

## Algebra¶

An algebra is a set of (equational) laws for a set of operators. For example, here are laws for the binary operators, + and *:

(a + b) + c = a + (b + c)           law assoc+
a + b = b + a                       law comm+
a + 0 = a                           law unit+

(a * b) * c = a * (b * c)           law assoc*
a * b = b * a                       law comm*
a * 1 = a                           law unit*

a * (b + c) = (a * b) + (a * c)     law dist


(By the way, this set of laws defines a ring for + and *.)

Say we use the numbers, 0, 1, 2,..., as the arguments to the operators; then we add some equational laws to define how the numbers are related by +1, namely:

1 + 1 = 2
2 + 1 = 3
3 + 1 = 4
and so on, and also
0 * 0 = 0


We use the operators and arguments to write expressions, like (3 * 2) + 1. We use the equational laws to calculate the expression’s answer, its output, its meaning. The short story is that (3 * 2) + 1 = 7; the long story is that the above equations calculate, deduce, that the expression’s meaning is 7.

Here’s why 3 * 2 = 6:

3 * 2 =  3 * (1 + 1)          because  1 + 1 = 2
=  (3 * 1) + (3 * 1)    because  dist
=  3 + 3                because  unit* (twice)
=  3 + (2 + 1)          because  2 + 1 = 3
=  3 + (1 + 2)          because  comm+
=  (3 + 1) + 2          because  assoc+
=  4 + 2                because  3 + 1 = 4
= ... =  6              (you do the remaining steps)


So, 3 * 2 = 6 and then (3 * 2) + 1 = 6 + 1 = 7. A CPU does arithmetic similarly, implementing the laws by shifting and flipping bits in its gate layout.

## Arithmetic Circuits¶

Let’s write an expression that uses variable names like x and y. One example is (2 * (x + y)) - (y + 1). Like the logic expressions in the previous chapter, this expression uses x and y as its “inputs.” So, let’s pretend we have hardware gates for the operations, +, -, *. Now, we have circuits where numbers travel along the wires. The circuit corresponding to (2 * (x + y)) - (y + 1) looks like this:

In the 1970’s, this form of circuit, called a data-flow program, was used with computers with multiple processors for parallel computation of subexpressions. They are now again being investigated for use with multi-core processors. Here is the equational system for the above circuit, where each internal wire is labelled with its own name:

a = x + y
b = 2 * a
c = y + 1
d = b - c


This equation set uses one operation in each equation. It is called three-address code and is produced by a compiler when the compiler translates arithmetic expressions into machine language. Let’s repeat the exercises at the end of the previous chapter for this equational system. First, say that x = 3 and y = 2; we can apply the algebra laws to calculate the knowledge that holds true after each equation is “executed” within the program:

            premises: x = 3,  y = 2
a = x + y
therefore,  x = 3,  y = 2,  a = x + y

therefore,  x = 3,  y = 2,  a = 5
(by application of the previously viewed algebra laws)
b = 2 * a
therefore,  x = 3,  y = 2,  a = 5,  b = 10
c = y + 1
therefore,  x = 3,  y = 2,  a = 5,  b = 10,  c = 3
d = b - c
therefore,  x = 3,  y = 2,  a = 5,  b = 10,  c = 3,  d = 7


Algebra calculates the numerical values for all of a, b, c , d. A computer does the same, but it saves the values of the variables in storage cells. The above is a test case of the program. Next, say that we know that the circuit will be plugged into an assembly that supplies input 1 to x. What can we deduce about the circuit’s output (in terms of y)? This is a bread-and-butter use of algebra:

           premise:  x = 1
a = x + y
therefore,  a = x + y,  x = 1
a = 1 + y                                by substitution
b = 2 * a
therefore,  b = 2 * a,  a = 1 + y
b = 2 * (1 + y)                          by substitution
b = (2 * y) + 2                          by algebra
c = y + 1
therefore,  c = y + 1,  b = (2 * y) + 2
d = b - c
therefore, d = b - c,  c = y + 1,  b = (2 * y) + 2
d = (2 * y) + 2 - (y + 1)                 by substitution
d = y + 1                                 by algebra


This reasoning is similar to what we use when we write assignment commands in a programming language.

### Validating Code for an Embedded System¶

Let’s do one more exercise. Say we have this equation set/circuit/program:

a = x - y
b = 2 * a
c = b + 1


Pretend that the algebra equations define a real-time controller for the rudder of an airplane, and inputs x and y are supplied from the plane’s gyroscope, refreshed every few milliseconds. Assume that the input value supplied by the x sensor will always be larger than the value supplied by the y sensor, that is, x > y must hold true. Assume also that x > 0.

Say that we want to validate, to prove, that the value of the output, c, is greater than one, that is, c > 1. If so, then the code can be burned into the hardware controller and installed in the plane.

We construct the proof with algebra. Along the way, we will prove that a and b are positive:

           premises:  x > y,  x > 0
a = x - y
since  x > y, then  x - y > y - y  then  x - y > 0  by algebra
since  a = x - y,  then  a > 0  by substitution
b = 2 * a
since a > 0,  then  2 * a > 0  by algebra
then  b > 0   by substitution
c = b + 1
since b > 0  then  b + 1 > 0 + 1  then  b + 1 > 1  by algebra
then  c > 1  by substitution


So, we have proved that when the inputs satisfy the precondition that x > y ^ x > 0, then the equations satisfy the postcondition that c > 1. This style of reasoning is critical to developing correct code for embedded, real-time systems.

Like an electronics engineer, who analyzes a circuit for voltage and resistance levels, we have analyzed the program and calculated in advance of execution its behavioral properties.

## Single-Assignment Programs¶

The equation sets we have just studied are well behaved because there is exactly one assignment equation for each variable in the program. A program with this feature is called a single-assignment program. Further, there is no self reference, that is, no equation, x = e, where e references x directly or indirectly. If we violate either property, our analysis breaks down. Consider this “equation set”:

x = 0
y = x
x = x + 1


A naive application of algebra lets us deduce a falsehood:

x = 0
therefore,  x == 0
y = x
therefore,  y == x  and  x == 0
x = x + 1
therefore,  y == x  (not any more!)
x == 0  (not any more!)
and also    x == x + 1  (impossible!)
and also    0 == 0 + 1  (impossible!)


The assignment command we use in programming is different from an algebraic equality. We must distinguish between the “old” value of x prior to the assignment, x = x + 1, and the “new” value of x that results from the assignment’s “output.” This leads us to programming logic in the next chapter.

## Some history¶

By the way, the first “high level” programming language, Fortran, was designed in the 1950s by John Backus for IBM’s engineers, who wanted to write and solve equation sets. Backus knew that computer hardware used storage cells for variables and that the cells’ values could be updated. So, he introduced the equational version of assignment to Fortran, which wrecks the correspondence between algebra equations and program code. Since then, there are many modern languages, the functional programming languages, which prohibit Fortran-style assignment and restore the connection between algebra and program code. Backus himself, in his Turing Award Lecture of 1977, disavowed assignment as a fundamental programming notion.

This note was adapted from David Schmidt's CIS 301, 2008, Chapter 1 course note.