8. We have the following code word: 01111010101. This string has 7 data bits, plus 4 check bits. We are using an error-correcting code that can correct single bit errors. In fact, there is a single bit error in this code. Assuming the use of the Hamming algorithm, locate the bit that has been altered.
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8. We have the following code word: 01111010101. This string has 7 data bits, plus 4
check bits. We are using an error-correcting code that can correct single bit errors.
In fact, there is a single bit error in this code. Assuming the use of the Hamming
algorithm, locate the bit that has been altered.
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- Consider the following code which includes a parity check for the third digit: C=(000,011,101,110). You get the third digit by adding the first two and then using the remainder on division by 2. Can this code detect and count single errors? how can this be implemented?You have two 32-bit numbers, N and M, as well as two bit positions, I and j. Create a method for inserting Minto N in such a way that M begins at bit j and ends at bit i. You can assume that bits j through I have enough space to accommodate all of M. That is, if M = 10011, you can presume at least 5 bits exist between j and i. You wouldn't have j = 3 and I = 2, for example, because M couldn't completely fit between bit 3 and bit 2.Input: N=10000000000, M=10011,i=2,j=6Output: N=10001001100Transcribed Image Text You are given a sequence of 16N positive integers a1, a2, . .. , a16N - You may shuffle this sequence in any way you choose, i.e. change it to any one of its permutations. Then, let (a1 Đ a2) ® (az e a4) ...® (asN-1 asN), (a8N+1 O a8N+2) ® (asN+3 O a8N+4) ®...® (a16N-1 O A16N), where O and O denote bitwise AND and XOR respectively. Find the maximum possible value of x – y
- Create two n-bit codes C1 and C2 (where n is the minimum number of bits) sufficient to uniquely encode each letter (in upper and lower case) of your full name (exclude spaces between first/second/third names). Each letter, even if it repeats in the name (as S repeats in COMSATS), should get only one unique code. No two codes in C1 and C2 should be the same. For example, to encode upper and lower case letters of COMSATS, the binary codes C1 and C2 are given in the Table 1. Sum all the bits of code C2 and produce a sum (S) and a carry (C) bit for each row of code C2 as shown in the table 1. Upper case letters of name Code C1 Lower case letters of name Code C2 SUMing bits of C2 Carry bit (C) Sum bit (S) C 0 0 0 0 c 0 1 1 0 1 0 O 0 0 0 1 o 0 1 1 1 1 1 M 0 0 1 0 m 1 0 0 0 0 1 S 0 0 1 1 s 1 0 0 1 1 0 A 0 1 0 0 a 1 0 1 0 1 0 T 0 1 0 1 t 1 0You are given two bit locations, I and j, together with the 32-bit values N and M. To insert Minto N, create a method that starts M at bit j and finishes it at bit i. It is safe to presume that all of M can fit in the bits j through i. Therefore, you may infer that there are at least 5 bits between j and I if M = 10011. M could not completely fit between bit 3 and bit 2, thus you would not, for instance, have j = 3 and I = 2.EXAMPLE Key in: NN 10000000000, M 10001001100 as outputimplement anyEvenBit(x) Return 1 if any even bit in x is set to 1 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 /* * anyEvenBit - return 1 if any even-numbered bit in word set to 1* Examples anyEvenBit(0xA) = 0, anyEvenBit(0xE) = 1* Legal ops: ! ~ & ^ | + << >>* Max ops: 12*/int anyEvenBit(int x) {return 2;}
- You are given two bit positions, I and j, along with the 32-bit numbers N and M. To insert Minto N, create a method that begins M at bit j and ends it at bit i. It is safe to presume that all of M can fit in the bits j through i. Therefore, you can assume that there are at least 5 bits between j and I if M = 10011. M could not completely fit between bit 3 and bit 2, so you would not, for instance, have j = 3 and I = 2.Input: N=10000000000, M=10011,i=2,j=6Output: N=10001001100Write a program that swaps 5th~11th bits in data_a with 25th~31th bits in data_bYour program must work for any data given, not just the example belowIn this question, we assume that the positions of bits count from right to left.That is, the first bit is the least significant bit.data_a DCD 0x77FFD1D1data_b DCD 0x12345678Let L={ w in(0 + 1)* 1 | w has even number of 1s}, i.e. L is the set of all bit strings with even number of 1s. Which one of the regular expression below represents L? (A) (0* 10* 1)* (B) 0* (10* 10* )* (C) 0* (10* 1* )* 0* (D) 0* 1(10* 1)* 10*
- 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;}What is the Huffman code for a string whose characters are all from a twocharacter alphabet? Give an example showing the maximum number of bits that couldbe used in a Huffman code for an N-character string whose characters are all from atwo-character alphabet.In this assignment you will practice using bitwise operators as well as binary, decimal, and hexadecimal representations. Given the following numbers: A = 0110 1100 B = 1100 0011 C = 1111 0000 D = 0000 1111 1.1 Questions Write the result of the following expressions in a plain text le. Write the answer in binary, hexadecimal, and decimal (unsigned integer). 1. A & B 2. A | B 3. A ^ B 4. A & C 5. A | C 6. A ^ C 7. A & D 8. A | D 9. A ^ D 10. B & C 11. B | C 12. B ^ C 13. C & D 14. C | D 15. C ^ D