Write a program that stores the maximum of three values. The values are stored in X19, X20 and X21. Store the result in X24. Ex: If the values of X19, X20, and X21 are initialized in the simulator as: the result is stored in X: Registers Data X19 5 X20 9 X21 8
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- For the DO WHILE AND GOTO loop below, generate an assembly language version of each DO WHILE loop int sumDoWhileLoop(int start, int end) { int sum = 0; int i = start; do { sum += i; i++; } while (i <= end); return sum;} GOTO loop int sumGotoLoop(int start, int end) { int sum = 0; int i = start; loop: sum += i; i++; if (i <= end) goto loop; return sum;} IS THE ASSEMBLY LANGUAGE VERSION OF EACH LOOP THE SAME OR DIFFERENT. COMPARE BASED ON No of registers used, jumps and total no of operationswrite in assebly language using lc3 simulator LC3 simulator-https://wchargin.com/lc3web/ Write a program that implements a for loop that adds 5 to R3 every iteration a. Increment a register by 1 in every cycle of the loop to keep track of the loop iterationsb. Store the maximum amount of iterations that the loop can have in a memory locationc. Your for loop should start at 0 iterations (i.e. i = 0) and stop looping once the maximum is reached Psuedo code: .ORIG x3000 ; Step 1: Clear R3 ; Step 2: Clear R0 ; Step 3: Load the value of MAX into R1 (maximum amount of iterations) ; Step 4: Take the two's compliment of R1 (2 instructions!) loop ; Step 5: Add #5 to R3 (result) ; Step 6: Increment R0 (loop iteration counter) ; Step 7: Add R1 and R0 and put the result in R2 (Since we did a 2's complement on R1, we are essentially subtracting R1 from R0) ; Step 8: Branch if negative back to loop (this means that R1 > R0 and we need to keep looping) ; Step 9: Halt CPU…Note: assembly language(8086 microprocessor) Write a program that takes a number N (2 digit decimal) as input. Then the user will enter that many numeric values ranging from 0 to 9 as input. Here the user may enter a negative number too. The program then finds a pair of values whose summation is closest to zero.Explanation: Here at first the user enters 10, which is a 2 digit decimal number. That means the user now has to enter 10 negative or positive digits ranging from 0 to 9. Then the program will calculate the sum of all possible pairs and find the pair which gives the minimum sum. Here from the given example, we see that the sum of -2 and -4 which is -6 is the least sum.Input:10 -2, 3, 7, 4, -4, 7, -8, 0, 9, 9 Output: Pair of values which has the smallest sum = -4, -2
- Using emu8086 write a full code that does the following - write your name in the middle of the screen - using the keyboard arrow keys, the user can move a blinking cursor on the screen - when the cursor is over a character, if the user hits "enter" then the character will move with the cursor - when "enter" is clicked again, the character is released to the new position -characters cannot be released on top of each other - when a character is moving with the cursor, if user click"R" the character is colored red, "B" for blue, and "G" for greenWrite a python code that does the following: - Asks student to enter his name - Ask them to enter their marks for 4 different modules. (Note: these marks should be out of 100 for each module) - Based on the average of the 4 different modules, the student should get the grade that they have achieved at that level. - The grades should follow the following trend - greater than or equal to 90 = A - from 80 to 89 inclusive = B - from 70 to 79 inclusive = C - from 60 to 69 inclusive = D - from 50 to 59 inclusive = E - below 50 = Fread the instruction carefully- This file should have multiple lines, each line has first name, last name followed by multiple integers "data.txt" file Samantha Johns 43, 32,99, 79, 85 Michael Smith 98, 84, 70, 100, 76, 81 John Michael 82,92, 92 Mary Hernandez 88, 91, 92, 79, 78 George Johnson 92,93,75 Sara Anderson 83,74. Susan John 78, 84,65, 71,81 Mark Smith 69, 67,59,62
- Implement a subprogram which takes 4 numbers in the argument registers $a0...$a3, andreturns the largest value and the average in $v0 and $v1 to the calling program. Theprogram must be structured as follows:Subprogram largestAndAverage($a1, $a2, $a3, $a4){ int var0 = $a0, var1 = $a1, var2 = $a2, var3 = $a3; $s0 = getLarger($a1, $a2); $s0 = getLarger($s0, $a3); $v0 = getLarager(s0, $a4); // Largest is in $v0 $v1 = (var0 + var1 + var2 + var3)/ 4; // Aversge is in $v1 return;}Subprogram getLarger($a0, $a1) { $v0 = $a0 if ($a1 > $a0) $v0 = $a1 return;}Note the use of the variables var0...var3. Because the values of $a0 and $a1 (at least) arechanged on the call to getLarger, they will not be available when they are needed to calculatethe average, and must be stored on the stack. To do this problem correctly, you mustcalculate the maximum value using the getLarger subprogram shown here, and it must becalled before the average is calculated. This implies that at a minimum $a0 and $a1 must…Implement a subprogram which takes 4 numbers in the argument registers $a0...$a3, andreturns the largest value and the average in $v0 and $v1 to the calling program. Theprogram must be structured as follows:Subprogram largestAndAverage($a1, $a2, $a3, $a4){ int var0 = $a0, var1 = $a1, var2 = $a2, var3 = $a3; $s0 = getLarger($a1, $a2); $s0 = getLarger($s0, $a3); $v0 = getLarager(s0, $a4); // Largest is in $v0 $v1 = (var0 + var1 + var2 + var3)/ 4; // Aversge is in $v1 return;}Subprogram getLarger($a0, $a1) { $v0 = $a0 if ($a1 > $a0) $v0 = $a1 return;}Take note of how var0...var3 are used. The values of $a0 and $a1 (at least) must be placed on the stack because they are not immediately available when needed to calculate the average because they are modified during the call to getLarger. You must use the getLarger subprogram displayed above to calculate the greatest value for this problem, and it must be called before the average calculation. This indicates that at the very least $a0…Implement the counter increment, and branch back to the start of the loop. When you havecompleted these steps, the basic structure of the counter control loop has been implemented,and your code should look similar to the following:.text li $s0, 0 lw $s1, n start_loop: sle $t1, $s0, $s1 beqz $t1, end_loop # code block addi $s0, $s0, 1 b start_loop end_loop:.data n: .word 5
- use the c# programming language While and For loops (Solve using all three loops), you may code all three loops in the same project. //do-while loop int counter = 0, stop = 5; //initializer do { Console.WriteLine(counter); //code statement counter++; //updater }while(counter < stop); //checker //while loop int counter = 0, stop = 5; //initializer while(counter < stop) //checker { Console.WriteLine(counter); //code statement counter++; //updater } //for loop // initializer checker updater for(int counter = 0, stop = 5;counter < stop; counter++) { Console.WriteLine(counter); //code statement } Write a program to print the numbers 2 to 10 in increments of two on the same line. The output of your program should be 2 4 6 8 10 Write a Program to compute the sum of eight numbers. You will prompt the user…Write in assembly language LC3 LC3 simulator to use: https://wchargin.com/lc3web/ Write a program that implements a while loopa. Loop while(R0 > -10)b. For every iteration of the loop subtract 2 from R0c. R0 should start at 0 Use LC3 simulator : https://wchargin.com/lc3web/ Psuedo code: .ORIG x3000 ; NOTE: it says to use R0 for the result; instead, use R3 so the result does not get overwritten when HALT command executes. ; Initialize R3 with #0 ; Set R1 to #10 LOOP ; Decrease R3 by #2 ; Add R1 and R3 and store into R2 ; Branch if positive back to top of loop (we have reached our end condition) DONE ; HALT CPU .ENDWrite a new program using your LFSR function from the first problem. Start from state 0x00000001, then run a loop counting all the possible states in the LFSR before it cycles back to 0x00000001. This is shown by the following pseudocode: count = 0; initial = state = 0x00000001; do { state = LFSR(state); count++; while (state != initial); print count; Verify that the number of states is (2^15) - 1 = 32767 . Your program should complete within 30 seconds. If it takes longer then you likely have entered an infinite loop.