Foundations of Computation

Transcription

1 The Australian National University Semester 2, 2018 Research School of Computer Science Tutorial 1 Dirk Pattinson Foundations of Computation The tutorial contains a number of exercises designed for the students to practice the course content. During the lab, the tutor will work through some of these exercises while the students will be responsible for completing the remaining exercises in their own time. There is no expectation that all the exercises will be covered in lab time. In this tutorial, you will use the Logisim tool ( along with pen and paper, to analyse and create simple combinatorial circuits. It is not installed on the CSIT lab computers, so to use it, you will have to install it on your own computer or on your student account. Download the Logisim v JAR file from and save it in a convenient location (e.g., your home directory). You will probably be able to start the program by simply (double-)clicking on the JAR file. If that doesn?t work, on a GNU/Linux system you can start a commandline ( terminal ) and type java -jar logisim-generic jar (assuming you are in the directory where the file is). Note that Logisim requires Java version 5 or later. Exercise 1 Understanding Circuits First, do the following simple exercise to familiarise yourself a bit with the Logisim environment. 1. Download the file at and then open the file in Logisim (select Open... under the File menu). If everything worked, you should now see a simple combinatorial circuit. 2. What does this circuit do? It computes some function Output = F (A, B, C) where all three inputs and the output are binary (0 or 1). But what is the function F? To find out, you can: a. Figure it out by looking at the circuit. b. Select the poke tool ( ), click on the inputs A, B and C to change their value, and observe how the output changes. c. From the Project menu, select Analyze Circuit. This opens a window with information about the circuit. Under Table you can find the complete truth table for the circuit, and under Expression you can see the equivalent Boolean expression. (Note that Logisim does not have a symbol for logical AND in expressions: x AND y is written just x y. The symbol for logical OR is + and for NOT is.) Make sure you try the first two methods before the last. 3. Is there a simpler/smaller circuit that computes the same function? Using the truth table for the circuit, attempt to find a smaller expression that generates the same table. Build the corresponding circuit next to the one given (using the same inputs, but a new output) and test it to see that it gives the same result. (Again, Logisim offers you a way to cheat: The Minimized tab of the circuit analysis window will give you a minimal expression, and can even rebuild the circuit automatically for you. Use this to check what you came up with by hand.) Solution. The circuit (just) computes the logical AND of B and C, as a formula: B C. A simpler circuit is the following:

2 Exercise 2 Consider the formula φ = ( (x (y x))) (y z). From Formulae to Gates 1. Construct a truth table for φ that gives the truth value of φ, for all assignments of truth values for the variables x, y, and z. Can the formula be simplified, i.e. is there a formula with the same truth table that is simpler than the formula given? Solution. We write φ for the formula and obtain the following truth table: x y z x y x x (y x) (x (y x)) y z φ F F F T T F T F T F F T T T F T F T F T F T T F T F T F T T T T F T T T T F F F F F T F T T F T F F F T F T T T F F T T F F F T T T F T T F T T From the truth table, we see that there s only one row where the formula evaluates to F : precisely when x = y = T and z = F. This means that we can simplify the formula to (x y z). The reasoning is the following: the formula x y z will evaluate to T precisely if x = y = T and z = F. For that reason, the negation of this formula, i.e. (x y z) evaluates to F if and only if x = y = T and z = F. 2. Draw a circuit with three inputs (labelled x, y and z) that computes the same boolean function. You can do this either by hand or use the Logism tool. Solution. Given that the formula simplifies to (x y z), we can use either a three-way AND-gate or a combination of two AND-gates, as follows: Exercise 3 Consider the following circuits with inputs x, y, and z: From Gates to Formulae This circuit uses a three-way OR-gate. Given inputs x, y and z, the three-way OR-gate delivers 1 if at least one of the inputs x, y and z are 1.

3 1. Convert the circuit to a formula (using the variables x, y, and z) Solution. We obtain, for example ( x) (( x y) ( z)). 2. Is there a simpler circuit or formula that represents the same boolean function? If so, give an example of a simpler formula or circuit that represents the same function. How can you justify that both circuits (or formulae) are really representing the same function? Exercise 4 Solution. Looking at the formula, we see that ( x y) x can be simplified to just x (think about it... or draw a truth table. We will see this more systematically later in the course). Therefore, a simpler formula is just x z. 1-Bit Switch We construct a one-bit switch, that is, a circuit with three inputs x, y and s and one output z so that if s = 0, then the output has the value of x if s = 1, then the ouput has the value of y. 1. Construct a truth table for the boolean function f(x, y, s) that behaves as described above. Solution. We obtain the following truth table: x y s f(x, y, s) F F F F F F T F F T F F F T T T T F F T T F T F T T F T T T T T To make the functionality clearer, we could use variables in the truth table that represent any boolean value and denote the truth table as follows: x y s f(x, y, s) a b F a a b T b 2. Give a logical formula in variables x, y and s that behaves according to the truth table. Solution. The idea here is that either s = T or s = F. If s = F, we want the truth value of x, and in case s = F, the formula s x always evaluates to the truth value of x. Similarly, if s = T, we are after the truth value of y, and the formula s y evaluates to the truth value of y in case s = T. Putting things together, we can therefore write for a formula that represents the above function. 3. construct a circuit that implements the boolean function. Solution. We obtain the following circuit: ( s x) (s y)

4 Exercise 5 4-Bit Switch The aim of the exercise is to use the circuit that we have built previously and construct a 4-bit switch. That is, we build a circuit with two four-bit inputs, x 1,... x 4 and y 1,... y 4, one input s (for switch ), and four one-bit outputs, z 1,..., z 4 so that if s = 0, the bits x 1,..., x 4 are being copied to the outputs z 1,..., z 4 if s = 1, the bits y 1,..., y 4 are being copied to the outputs z 1,..., z 4. In other words, if s = 0, then the circuit switches the inputs x 1,..., x 4 to the outputs, and if s = 1, the circuit switches y 1,..., y 4 to the outputs. The idea is to construct a 4-bit switch by combining four 1-bit switches. 1. In Logism, use Project Add Circuit to create a new circuit. Call this new circuit 1-bit switch and draw (or import) the circuit of the previous exercise. You can edit the circuit s appearance by clicking Project Edit circuit appearance. 2. Now double-click on main to construct the 4-bit switch using several copies of the 1-bit switch. You can use the 1-bit switch just like any other gate: single-click on it and drag it onto the canvas. 3. Simulate your circuit to make sure that you have implemented the switch correctly. Solution. We obtain the following: Exercise 6 Consider the boolean function given by the following truth table: Construct a Formula x y z f(x, y, z) F F F T F F T F F T F F F T T T T F F F T F T F T T F T T T T T 1. Give a formula (in variables x, y and z) that represents the boolean function given above. Briefly argue why the formula indeed represents the boolean function. Solution. There are two ways to go about this. The first way is following the algorithm given in the lectures. We look at those rows of the truth table where the function evaluates to T and obtain: ( x y z) ( x y z) (x y z) (x y z). This formula represents the boolean function, because:

5 every row in the truth table for which the formula evaluates to T corresponds precisely to a (conjunctive) term in the formula that evaluates to T under the assignment in the corresponding row. taking the disjunction (or) covers precisely all possibilities where the function evaluates to T. For the last two terms, we see that they are independent of z, so the above simplifies to ( x y z) ( x y z) (x y). 2. Convert the boolean function to a circuit, and give the circuit diagram. Can you use gates other than AND, OR and NOT to obtain a simpler circuit? If you use the Logisim tool, you can use Project Analyse Circuit to verify that your circuit indeed corresponds to the given truth table. Solution. It is routine to convert the above formula into a circuit. For the first two rows where the function evaluates to TRUE, we see that this happens if and only if x = F and y = z. We can use an XOR gate, together with a negation to represent this. This gives the following circuit: Exercise 7 The parity function Consider three variables x, y, and z. The (ternary) parity function p(x, y, z) is a boolean function with value F, if an even number of the inputs x, y, and z have truth value T T, if an odd number of inputs have truth value T. This exercise is about designing a circuit, and a formula, for the ternary parity function. 1. Draw a truth table for the ternary parity function. The truth table should have the following form: x y z p(x, y, z) F F F F... where we have just given the first row, and list all combinations of truth values for x, y, and z, as well as the result of the parity function. Solution. The full truth table is as follows: x y z p(x, y, z) F F F F F F T T F T F T F T T F T F F T T F T F T T F F T T T T 2. Convert the formula to a circuit, and draw the ensuing circuit diagram. You may use circuits other than AND, OR and NOT. If you use the Logisim tool, you can use Project Analyze Circuit to verify your solution. Solution. The easiest way is to just use XOR gates as follows:

6 3. Based on the truth table, the circuit, or otherwise, give a logical formula with variables x, y, and z that represents the parity function. Briefly argue why your formula indeed represents the parity function. Solution. We can encode the XOR of x and y using the formula ( x y) (x y), but the ensuing formula is rather big and difficult to understand. We go for the simpler option and write ( x y z) ( x y z) (x y z) (x y z) where each conjunctive term corresponds to one line of the truth table for which the formula evaluates to true. Appendix: How to build circuits in Logisim To view and edit the circuit that implements one of the defined elements (1-bit swap, etc), double-click on it in the left-hand-side menu. You can also double-click main to get back to the main circuit. To add a gate to the circuit, select the gate you want from the menu on the left. Then just click on the circuit board to add the gate. When you have selected the edit tool ( to remove a component or wire. ) you can move components and add wires. Right-click and select Delete To connect components, add or extend wires. Components have a small marker where they can connect (you will see a small green circle when the pointer is on the mark). You can also add a branch off an existing wire, at any point.