Week 3 at a glance

We will be learning and practicing to:

TODO:

Review quiz based on Week 2 class material (due Monday)

Schedule your Test 1 and Test 2 times at PrairieTest (http://us.prairietest.com)

Review quiz based on Week 3 class material (due Monday next week)

Week 3 Monday: Fixed-width Addition and Circuits

Example digital circuit:

Last time we considered this circuit:

image

Design a digital circuit that, given a sign-magnitude width \(4\) representation of an integer, outputs a sign-magnitude width \(4\) represention of \(-1\) times that integer. Are there any corner cases where this circuit doesn’t work?

Design a digital circuit that, given a 2s complement width \(4\) representation of an integer, outputs a 2s complement width \(4\) representation of an \(-1\) times that integer. Are there any corner cases where this circuit doesn’t work?

Fixed-width addition: adding one bit at time, using the usual column-by-column and carry arithmetic, and dropping the carry from the leftmost column so the result is the same width as the summands. In many cases, this gives representation of the correct value for the sum when we interpret the summands in fixed-width binary or in 2s complement.

For single column:

2

Input Output
\(x_0\) \(y_0\) \(c_0\) \(s_0\)
\(1\) \(1\)
\(1\) \(0\)
\(0\) \(1\)
\(0\) \(0\)

image

Draw a logic circuit that implements binary addition of two numbers that are each represented in fixed-width binary:

First approach: half-adder for each column, then combine carry from right column with sum of left column

Write expressions for the circuit output values in terms of input values:

\(z_0 = \underline{\phantom{x_0 \oplus y_0\hspace{3in}}}\)

\(z_1 = \underline{\phantom{(x_1 \oplus y_1) \oplus c_0}\hspace{2.5in}}\)

\(z_2 = \underline{\phantom{(c_0 \land (x_1 \oplus y_1)) \oplus c_1}\hspace{2in}}\)
image

There are other approaches, for example: for middle column, first add carry from right column to \(x_1\), then add result to \(y_1\)

Week 3 Wednesday: Propositional Logic

Logical operators aka propositional connectives

Conjunction AND \(\land\) \land 2 inputs Evaluates to \(T\) exactly when both inputs are \(T\)
Exclusive or XOR \(\oplus\) \oplus 2 inputs Evaluates to \(T\) exactly when exactly one of inputs is \(T\)
Disjunction OR \(\lor\) \lor 2 inputs Evaluates to \(T\) exactly when at least one of inputs is \(T\)
Negation NOT \(\lnot\) \lnot 1 input Evaluates to \(T\) exactly when its input is \(F\)

Truth tables: Input-output tables where we use \(T\) for \(1\) and \(F\) for \(0\).

Input Output
Conjunction Exclusive or Disjunction
\(p\) \(q\) \(p \land q\) \(p \oplus q\) \(p \lor q\)
\(T\) \(T\) \(T\) \(F\) \(T\)
\(T\) \(F\) \(F\) \(T\) \(T\)
\(F\) \(T\) \(F\) \(T\) \(T\)
\(F\) \(F\) \(F\) \(F\) \(F\)
image image image
Input Output
Negation
\(p\) \(\lnot p\)
\(T\) \(F\)
\(F\) \(T\)
image
Input Output
\(p\) \(q\) \(r\) \((p \land q) \oplus (~ ( p \oplus q) \land r~)\) \((p \land q) \vee (~ ( p \oplus q) \land r~)\)
\(T\) \(T\) \(T\)
\(T\) \(T\) \(F\)
\(T\) \(F\) \(T\)
\(T\) \(F\) \(F\)
\(F\) \(T\) \(T\)
\(F\) \(T\) \(F\)
\(F\) \(F\) \(T\)
\(F\) \(F\) \(F\)

Given a truth table, how do we find an expression using the input variables and logical operators that has the output values specified in this table?

Application: design a circuit given a desired input-output relationship.

Input Output
\(p\) \(q\) \(mystery_1\) \(mystery_2\)
\(T\) \(T\) \(T\) \(F\)
\(T\) \(F\) \(T\) \(F\)
\(F\) \(T\) \(F\) \(F\)
\(F\) \(F\) \(T\) \(T\)

Expressions that have output \(mystery_1\) are

Expressions that have output \(mystery_2\) are

Idea: To develop an algorithm for translating truth tables to expressions, define a convenient normal form for expressions.

Definition An expression built of variables and logical operators is in disjunctive normal form (DNF) means that it is an OR of ANDs of variables and their negations.

Definition An expression built of variables and logical operators is in conjunctive normal form (CNF) means that it is an AND of ORs of variables and their negations.

Week 3 Friday: Logical Equivalence

Proposition: Declarative sentence that is true or false (not both).

Propositional variable: Variable that represents a proposition.

Compound proposition: New proposition formed from existing propositions (potentially) using logical operators. Note: A propositional variable is one example of a compound proposition.

Truth table: Table with one row for each of the possible combinations of truth values of the input and an additional column that shows the truth value of the result of the operation corresponding to a particular row.

Logical equivalence : Two compound propositions are logically equivalent means that they have the same truth values for all settings of truth values to their propositional variables.

Tautology: A compound proposition that evaluates to true for all settings of truth values to its propositional variables; it is abbreviated \(T\).

Contradiction: A compound proposition that evaluates to false for all settings of truth values to its propositional variables; it is abbreviated \(F\).

Contingency: A compound proposition that is neither a tautology nor a contradiction.

Label each of the following as a tautology, contradiction, or contingency.

\(p \land p\)

\(p \oplus p\)

\(p \lor p\)

\(p \lor \lnot p\)

\(p \land \lnot p\)

Extra Example: Which of the compound propositions in the table below are logically equivalent?

Input Output
\(p\) \(q\) \(\lnot (p \land \lnot q)\) \(\lnot (\lnot p \lor \lnot q)\) \((\lnot p \lor q)\) \((\lnot q \lor \lnot p)\) \((p \land q)\)
\(T\) \(T\)
\(T\) \(F\)
\(F\) \(T\)
\(F\) \(F\)

(Some) logical equivalences

Can replace \(p\) and \(q\) with any compound proposition

\(\lnot ( \lnot p) \equiv p\) Double negation
\(p \lor q \equiv q \lor p\) \(p \land q \equiv q \land p\) Commutativity Ordering of terms
\((p \lor q) \lor r \equiv p \lor (q \lor r)\) \((p \land q) \land r \equiv p \land (q \land r)\) Associativity Grouping of terms
\(p \land F \equiv F\) \(p \lor T \equiv T\) \(p \land T \equiv p\) \(p \lor F \equiv p\) Domination aka short circuit evaluation
\(\lnot (p \land q) \equiv \lnot p \lor \lnot q\) \(\lnot (p \lor q) \equiv \lnot p \land\lnot q\) DeMorgan’s Laws