HOMEWORK # 2 SOLUTIO Problem 1 (2 points) a. There are 313 characters in the Tamil language. If every character is to be encoded into a unique bit pattern, what is the minimum number of bits required to do this? 8 bits can used to encode 2 8 = 256 characters and 9 bits can be used to encode 2 9 = 512 characters. So, we would need 9 bits. b. How many more characters can be accommodated in the language without requiring additional bits for each character? 512 313 = 199 Problem 2 (4 points) Convert the following 2's complement binary numbers to decimal numbers. a. 1010 First bit is 1. So it is a ve number. 2 s complement of 1010 = 0101 + 1 = 0110. So the answer is -6. b. 0010 This is a +ve number since it starts with 0 Answer is 2. c. 111111 This is a ve number since it starts with 1. Its 2 s complement is 000000 + 1 = 000001. So the answer is -1 d. 011111 This is a +ve number since it starts with 0. The answer is 31. Problem 3 (4 points) a. What is the largest positive number one can represent in a 16-bit 2's complement code? Write your result in binary and decimal. 0111 1111 1111 1111 binary and 2 15-1 = 32767 decimal
b. What is the greatest magnitude negative number one can represent in a 16-bit 2's complement code? Write your result in binary and decimal. 1000 0000 0000 0000 binary and -2 15 = -32768 decimal c. What is the largest positive number one can represent in a 16-bit signed magnitude code? Write your result in binary and decimal. 0111 1111 1111 1111 binary and 2 15-1 = 32767 decimal d. What is the greatest magnitude negative number one can represent in a 16-bit signed magnitude code? Write your result in binary and decimal. 1111 1111 1111 1111 binary and (2 15 1) = -32767 decimal Problem 4 (2 points) What are the 8-bit patterns used to represent each of the characters in the string "This Is Easy!"? (Only represent the characters between the quotation marks.) Character Hex (from ASCII table) Binary equivalent T 54 0101 0100 h 68 0110 1000 i 69 0110 1001 s 73 0111 0011 Space 20 0010 0000 I 49 0100 1001 s 73 0111 0011 Space 20 0010 0000 E 45 0100 0101 a 61 0110 0001 s 73 0111 0011 y 79 0111 1001! 21 0010 0001 Problem 5 (4 points) Convert the following decimal numbers to 8-bit 2's complement binary numbers. If there is problem while doing this, describe it. a. 102 0110 0110
b. 64 c. 128 0100 0000 Does not fit in an 8-bit signed number d. -128 1000 0000 Problem 6 (4 points) The following binary numbers are 4-bit 2's complement binary numbers. Which of the following operations generate overflow? Justify your answers by translating the operands and results into decimal. a. 0111 + 1101 No overflow. 0111 1101 10100 Answer is 0100 binary = 4 decimal [7 + (-3)] b. 1001 + 1110 Overflow. 1001 1110 10111 Answer is 0111 binary = 7 decimal. But actual answer is -9 [(-7) + (-2)]
c. 1111 + 1001 No overflow. 1111 1001 11000 Answer is 1000 binary = -8 decimal [(-1) + (-7)] d. 0011 + 0101 Overflow. 0011 0101 1000 Answer is 1000 binary = -8 decimal. But actual answer is 8 [3 + 5] Problem 7 (2 points) A computer programmer wrote a program that adds two numbers. The programmer ran the program and observed that when 5 is added to 8, the result is the character m. Explain why this program is behaving erroneously. The error that is occurring here is that 5 and 8 are being interpreted as characters 5 and 8 respectively. As a result, the addition that is taking place is not 5 + 8; rather, it is 5 + 8. If we look up values in the ASCII table, 5 is 0x35 and 8 is 0x38. 0x35 + 0x38 = 0x6d, which is the ASCII value for m.
Problem 8 (2 points) Compute the following: a. OT(1011) OR (1011) NOT(1011) = 0100 Answer = (0100) OR (1011) = 1111 b. OT(1001 A D (0100 OR 0110)) 0100 OR 0110 = 0110 1001 AND 0110 = 0000 Answer = NOT(0000) = 1111 Problem 9 (4 points) Write the decimal equivalents for these IEEE floating point numbers. a. 0 01111111 11000000000000000000000 Sign bit is 0 (+ve). Exponent = 127. Fraction = 1*2-1 + 1*2-2 = 0.75 Answer = (+) 1.fraction * 2exponent 127 = 1.75 * 2 0 = 1.75 b. 1 01111101 10000000000000000000000 Sign bit is 1 (-ve). Exponent = 125. Fraction = 1*2-1 = 0.5 Answer = (-) 1.fraction * 2exponent 127 = - 1.5 * 2-2 = - 0.375
Problem 10 (2 points) Given a black box which takes n bits as input and produces one bit for output, what is the maximum number of unique functions that the black box can implement? (Hint: Try to visualize a truth table for a single function of n bits. Determine how many rows such a truth table has. Then determine how many combinations are possible with the number of rows that you just found) Consider a single function that this black box implements. If there are n binary inputs, the truth table contains 2 n rows. Now, each of these rows in the truth table can be filled with 0 or 1. The number of ways in which we can fill in these rows (using 0 and 1) gives us the number of unique functions. Since each of the rows can be filled in using 2 possible values and since the number of rows is 2 n, the number of ways = 2 power (2 n ).