Which of the following is not the resultant sequence after first run of quicksort? (A)11,5,19,1,15,26,59,61,48,37 (B)2,3,1,4,9,5,6,7,10 (C)15,8,32,4,831,52,64,41 (D)1,4,3,2,5
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Which of the following is not the resultant sequence after first run of quicksort? (A)11,5,19,1,15,26,59,61,48,37
(B)2,3,1,4,9,5,6,7,10
(C)15,8,32,4,831,52,64,41
(D)1,4,3,2,5
Step by step
Solved in 2 steps
- 63, 52, 10, 42, 32, 17, 60, 45, 47, 39, 71, 55, 41, 95, 70, 48, 42, 32, 13 as so on the following been series.'list_1': [15, 20, 26, 432, 23, 34, 4, 3, 2, 15, 10, 28, 39, 56, 10, 10, 10, 432, 3, 3] 'list_2': [250, 101, 210, 150, 150 ,201, 205, 250, 143, 158, 250, 11, 101, 205] 'list_3': [350, 1000, 12, 450, 960, 966, 852, 350, 1000, 960, 852, 1200, 1000] There are 3 lists in above. These lists will be created with the specified names. You must remove the elements that is repeated in each listafter creating them. At last, calculate the averages of each list, and append the average values to each list with append function. Print eachlist. Example outputs: [1,2,3,4, 2.5] [3,6,9,12, 7.5] [3,4,8,5, 4]Key Permutation 57, 10. 58. 42, 60. 3. 46, 34, 52, 38, 20, 26, 44, 30, 12, 49. 41, 2, 33, 51, 39, 53, 25, 43, 31, 45, 17, 35, 23, 37, 9, 27, 15, 29, 1, 11, 7, 50. 59, 47, 6, 18, 36, 22, 19, 63, 14, 55, 62, 13, 54. 61, 21, 5, 28. 4 If the following plaintext is processed using the above Key Permutation table, the last two bits in the output will be Plaintext: 1100100101010111000111000000110101110001111000101100011011000101 O a. 11 O b. 10 O c. 00 O d. 01
- Choose: 16,3,8,5,12,13,17,15Page545 #36 (a) Page545 #36 (c) E(5,6)+d(2,7,9,13,14,15)=bdtbs E(3,4,5,10)+d(2,11,13,15)= d. d. 1 b b d d. a a d. d Page545 #36 (d) E(5,6,12,15)+d(0,4,10,14)= Page545 #36 (f) Z(0,2,3,4)+d(8,9,10,11,13,14,15)= b. b a a d[1, 3, 4, 6, 7, 7, 7, 9, 11, 12, 13, 14, 14, 15, 16, 18, 19, 22, 22, 24, 25, 27, 29, 30, 31, 34, 40, 43, 45, 46, 47, 48, 49, 49, 50, 50, 61, 64, 70, 71, 78, 96, 133, 135, 152 ] please fill these blanks using the above data my sample data within the limits of the first standard deviation is _(pick one: normally distributed, or not normally distributed)_at this measurement. my sample data within the limits of the second standard deviation is _(pick one: normally distributed, or not normally distributed)_ at this measurement. my sample data within the limits of the third standard deviation is _(pick one: normally distributed, or not normally distributed)_ at this measurement.
- Construct a BST with the following numbers and answer the questions 19-20. 26 - 4 - 15 - 12 - 76 - 43 - 23 - 55 - 8 - 27 - 63 - 22 - 1 - 45 - 2 - 88 - 98 Which number is at the leaves?8. Consider the LEGV8 code below. Assume that X1 is initialized to 11 and X2 is initialized to 22. ADDI X1, X2, #5 ADD X3, X1, X2 ADDI X4, X1, #15 ADD X5, X1, X1 (a) Suppose you executed the code on a version of the pipeline from Section 4.5 that does not handle data hazards (i.e., the programmer is responsible for addressing data hazards by inserting NOP instructions where necessary). Show the pipeline timing diagram below when the code is executed. ADDI X1, X2, #5 IF ID EX МЕМ WB ADD X3, X1, X2 ADDI X4, X1, #15 ADD X5, X1, X1 (b) What would the final values of registers X3 and X4 be? (c) What would the final values of register X5 be? Assume the register file is written at the beginning of the cycle and read at the end of a cycle. Therefore, an ID stage will return the results of a WB state occurring during the same cycle. See Section 4.7 and Figure 4.51 for details. (d) Suppose you executed the code below on a pipeline from Section 4.5 that uses data forwarding for handling data…For the following data: 11| {0,0,0,1,3,5, 5,5, 5,5,7,8,8,9} 12| {1, 2, 4,5,5,5,6,7,8,9,9,9,9,9,9} 13| {0,0,1, 2, 4, 6, 8, 9} 17| {1, 2,2, 2, 2,2, 2, 2,2, 2,2} 20| {1} 241 {4,7} The total number of data equals Select one: O NONE O 27 O 6 O 51
- 2. Use the defined GPR of TMP (it is a DATA) and RST (It is an ADDRESS), present your ASM codes that can perform the followings: TMP EQU 0X30 RST EQU OX36 (a) TMP – OXDF Result in W (b) 0X8A - TMP → Result in RSTQuestion 3: On a byte addressable machine a string COMSATS UNIVERSITY ISLAMABAD PAKISTAN, starting from logical address 0 (zero) to 36 is to be loaded into memory, in noncontiguous fashion, using 'Paging' technique. The string is divided mechanically into equal sized pages of 4 bytes; similarly the memory is also divided into equal sized frames of 4 bytes. The ASCII code of each Character occupies one byte. The free frame list is given as follows: 8,3,4,2,6,7 ,5,1,9,21,24, 11 Answer the following questions a. What is the page map table? b. What are the linear logical addresses of characters 'U' and_'B'? c. What are the linear physical addresses of characters 'U' and_'B'? d. What is the difference between Internal and External fragmentation? e. What is the internal fragmentation in loading given string?The complement of the function F(A,B ,C, D) = E (2, 4, 7. 10, 12, 14) is F =7 (0,1.3.5,6,8,9,11,13.15) F =E (0,1,3,5,6,8,9.11.13.15) O Other