1.1 The Fifth Trial pages 18 and 25

Transcription

1 HW03 Deadline: Tuesday, October , before 9:00 am Mailto: Subject line: COMP110 HW03 your name Textbook: J. G. Brookshear. Computer Science an Overview, 11 th ed., 2012 Book: R. Smullyan. The Lady or the Tiger?, 1982 Reference: Gates.pdf 1. Read the book The Lady or the Tiger? (pages and 23-26). Represent the following problem statements (not their solutions) by logic gates. Then solve the problems using the constructed circuits: 1.1 The Fifth Trial pages 18 and 25 Before looking at the messages, let us implement the main rules: Rule I If a lady is in Room I, then the sign on the door is true, but if a tiger is in it the sign is false (page 17); Rule II In the Room II the situation is the opposite: a lady in the room means the sign on the door is false, and a tiger in the room means the sign is true. The easiest way is to fill in the truth tables: Room I Sign I Rule I Room II Sign II Rule II 0 (tiger) 0 (false) 1 (satisfied) 0 (tiger) 0 (false) 0 (not satisfied) 0 (tiger) 1 (true) 0 (not satisfied) 0 (tiger) 1 (true) 1 (satisfied) 1 (lady) 0 (false) 0 (not satisfied) 1 (lady) 0 (false) 1 (satisfied) 1 (lady) 1 (true) 1 (satisfied) 1 (lady) 1 (true) 0 (not satisfied) The first gate is NOT XOR XNOR, the second XOR. Both rules must be satisfied. Therefore, they are connected by AND: Now the signs: I AT LEAST ONE ROOM CONTAINS A LADY II THE OTHER ROOM CONTAINS A LADY

2 Combining with the circuit above, we get It is easy to show that the constructed circuit implements the following truth table, which corresponds to the correct solution the lady is in the Room I: Room I Room II Solution The Seventh Trial pages 18 and 25-26; The signs: I IT DOES MAKE A DIFFERENCE WHICH ROOM YOU PICK II YOU ARE BETTER OFF CHOOSING THE OTHER ROOM This trial is similar to the previous one the rules and the Room II sign are the same, and instead of the OR gate, here the Room I sign is implemented by an XOR gate: It has similar solution the lady is in the Room I.

3 1.3 The Eighth Trial pages 19 and 26; It is not known which rooms the following signs belong to: THIS ROOM CONTAINS BOTH ROOMS CONTAIN A TIGER TIGERS First, let s assume the left sign is for the Room I and the right one for the Room II. The inputs of the top XNOR gate will always be of opposite values. Therefore, its output will be 0, which is an input for the concluding AND gate. So, the overall result will remain 0 the problem does not have solutions. The circuit of the above wrong assumption is shaded, and the correct one is shown below: Note that now the inputs of the bottom XOR gate will always be of opposite values. Therefore, its output will be 1, which is an input for the concluding AND gate. So, the overall result will depend only on the top part, which reduces the circuit to the following:

4 1.4 The Tenth Trial pages and 26. The rules change compared to the previous trials. It is stated that there are at least one true and one false signs. Let s implement this statement starting from the truth table: Sign I Sign II Sign III Rule 0 (false) 0 (false) 0 (false) 0 (not satisfied) 0 (false) 0 (false) 1 (true) 1 (satisfied) 0 (false) 1 (true) 0 (false) 1 (satisfied) 0 (false) 1 (true) 1 (true) 1 (satisfied) 1 (true) 0 (false) 0 (false) 1 (satisfied) 1 (true) 0 (false) 1 (true) 1 (satisfied) 1 (true) 1 (true) 0 (false) 1 (satisfied) 1 (true) 1 (true) 1 (true) 0 (not satisfied) The rule is symmetric relative to the signs. Therefore, if there is a gate connecting, for example, Sign I and Sign II, then a similar gate should connect Sign II and Sign III. And the output must not change whether the former or the latter are taken. The signs are: I II III A TIGER IS IN ROOM II A TIGER IS IN THIS ROOM A TIGER IS IN ROOM I Combining the signs with the above circuit, we get: Note that the bottom XOR gets similar inputs and, therefore, results in 0. So, the overall result is determined by the upper XOR, and the circuit can be obviously simplified:

5 2. Using logic gates solve the simplified version of the problem 1197 Lonesome Knight ( assuming a knight is placed in the bottom left quadrant of the chessboard with the rows and columns enumerated from 0 to 3, count the number of its available moves. Solve the problem in the following steps: 2.1 Create a binary circuit that maps 3 to 2 and leaves 0, 1, and 2 unchanged. More specifically, implement the following truth table: Input Output Create a binary circuit that adds two 2-bit numbers and outputs the sum as a 3-bit result. Exclude 11 from the possible inputs. More specifically, implement the following truth table: Input1 Input2 Output

6 2.3 Create a binary circuit that maps 3 to 4, 4 to 6 and leaves 0, 1, and 2 unchanged. More specifically, implement the following truth table: Input Output

7 2.4 Combine the circuits from the Tasks 2.1, 2.2 and Read the Chapter 1 Data Storage, part of Section 1.1 Bits and their Storage, pages and Chapter 2 Data Manipulation, part of Section 2.4 Arithmetic / Logic Instructions, pages Fill in the following tables that represent the logical operations AND, OR and XOR applied to hexadecimal inputs: OR AND 8 9 A B C D E F A A A 8 8 A A B B 8 9 A B C C C C C D D C D E E E F F XOR 8 9 A B C D E F A B C D E F B A D C F E 2 A B 8 9 E F C D 3 B A 9 8 F E D C 4 C D E F 8 9 A B 5 D C F E 9 8 B A 6 E F C D A B F E D C B A 9 8