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- /* * copyLSB - set all bits of result to least significant bit of x * Example: copyLSB(5) = 0xFFFFFFFF, copyLSB(6) = 0x00000000 * Legal ops: ! ~ & ^ | + << >> * Max ops: 5 * Rating: 2 */int copyLSB(int x) { return 2;}For example: Binary value of character ‘B’ from ASCII table is 01000010Dataword is 1000010 Even Parity Bit is 0 Codeword: 10000100 Even Parity – Complete the following table using Even Parity Character Binary Datawordfrom ASCII table(7-bit value) Parity BitValue TransmittedCodeword ReceivedCodewordat Receiver Check ifthecodewordisrejected t 1010100 10101101 h 1101000 11010000 i 1101001 11010010 s 1110011 11100110A string is called a palindrome if it reads the same when reversed. That is, the string x1 x2 ... xn is a palindrome if x1 x2 ... xn = xn ... x2 x1. The bit strings 11011 and 01011010 are palindromes.The bit strings 1011 and 01111 are not. How many bit-strings of length 7 are palindromes?
- You are hired to decrypt a message that has the following binary sequence: 0100 0101 0100 1100 0100 0011 0110 1111 0110 0100 0110 1001 0100 0111 0110 1111 0100 0101 0101 0011 0011 0011 0011 0111 0011 0101 0011 0001 0011 0110 0011 0010 The message must be displayed on a 7-segment display, it is only known that the encoding is in ASCII. 1.Approach, encoding of the message and truth table. 2.Functional electronic circuit to display the message.A binary code is a string of 0's and 1's. In particular, an n-bit binary code is a string of "n" 0's and 1's. For example, the code "01001" is a 5-bit binary code. How many 5-bit binary codes can be formed?Given that a computer uses 4-bit ones’s complement numbers, what value will be stored in the variable j after the following code segment terminates? j = 1 k = -2 while(k != 0) { j = j + 1; k = k + 1; }
- implement greatestBitPos function Compute a mask marking the most significant 1 bit. you are only allowed to use the following eight operators: ! ~ & ^ | + << >> “Max ops” field gives the maximum number of operators you are allowed to use to implement each function /* * greatestBitPos - return a mask that marks the position of the* most significant 1 bit. If x == 0, return 0* Example: greatestBitPos(96) = 0x40* Legal ops: ! ~ & ^ | + << >>* Max ops: 70* Rating: 4 */int greatestBitPos(int x) {return 2;}Given two numbers: n: number of consecutive letters beginning at a k: string length Output all possible k-length possible strings containing the first n letters of the alphabet. Letter repeats are allowed. Sample Run: # of letters: 4 length: 3 aaa aab aac aad aba abb abc abd aca acb acc acd ada adb adc add baa bab bac bad bba bbb bbc bbd bca bcb bcc bcd bda bdb bdc bdd caa cab cac cad cba cbb cbc cbd cca ccb ccc ccd cda cdb cdc cdd daa dab dac dad dba dbb dbc dbd dca dcb dcc dcd dda ddb ddc dddgreatestBitPos(x) Compute a mask marking the most significant 1 bit. program should return maskmarking not value of significant bit example: greatestBitPos(96) = 0x40 /* * greatestBitPos - return a mask that marks the position of the* most significant 1 bit. If x == 0, return 0* Example: greatestBitPos(96) = 0x40* Legal ops: ! ~ & ^ | + << >>* Max ops: 70* Rating: 4 */int greatestBitPos(int x) {return 2;}
- True/False In Floating Point Addition, the exponents of the two numbers being added need to be made equal before the mantissa fields can be added, as this ensures that the bit positional values are now alignedSuppose that we have the following 128-bit AES key, given in hexadecimalrepresentation:54 77 6F 20 4F 6E 65 20 4E 69 6E 65 20 54 77 6F Construct the round key for the first round.Given a single-precision IEEE 754 representation of a floating point number as: 0x40700000, do the following: (a) Express the value of the S field in binary without spaces (count carefully). Do not abbreviate your answer. (b) Express the value of the E field in binary without spaces (count carefully). Do not abbreviate your answer. (c) Express the value of the F field in binary without spaces (count carefully). Do not abbreviate your answer.