Solve computational problems using assembly and machine language.

Question

Wombat 1.1

A function is defined as follows:

𝑓(𝑛) = { 𝑛 + 1 𝑓𝑜𝑟 𝑛 < 1

3𝑛 + 5 𝑓𝑜𝑟 𝑛 = 1

𝑛^3 − 3𝑛 + 5 𝑓𝑜𝑟 𝑛 > 1

Write Wombat1 machine language program that reads an input integer value n, and computes and displays the output of the above function f(n).

In the binary version of your programs that can be executed on the CPUSim simulator, you are also required to write as comments the line number/address, equivalent decimal version of the binary and appropriate comments to explain each line in your program. The decimal version of the program represents each binary line as two decimal values (first value is the opcode – 4 bits; second value is the operand/memory address – 12 bits). Therefore, each line of your program should be formatted as follows:

Binary-number, Semi-colon, Line-number/address, Decimal-opcode, Decimal-operand/address, Comments

As an example, suppose you are writing a program to read and display a number. This program could be written as follows:

0011000000000000 ; 0 : 3 0 read into the accumulator

0010000000001010 ; 2 : 2 10 store n at location 10

0001000000001010 ; 4 : 1 10 load n from location 10

0100000000000000 ; 6 : 4 0 output n

0000000000000000 ; 8 : 0 0 stop

0000000000000000 ; 10 : 0 0 storage location for n

write the code in notepad++

Answer 1

𝑓(𝑛) = { 𝑛 + 1 𝑓𝑜𝑟 𝑛 < 1

3𝑛 + 5 𝑓𝑜𝑟 𝑛 = 1

𝑛^3 − 3𝑛 + 5 𝑓𝑜𝑟 𝑛 > 1

0011000000000000; 0: 3 0 This is f(n), read n into the accumulator

0010000000101100; 2: 2 44 n <- accumulator

0110000000101110; 4: 6 46 accumulator <- n - 1

1010000000100100; 6: 10 36 Jump to handle n < 1 case

1001000000011010; 8: 9 26 Jump to handle n = 1 case

0001000000101100; a: 1 44 accumulator <- n

0111000000101100; c: 7 44 accumulator <- n * n = n^2

0111000000101100; e: 7 44 accumulator <- n * n = n^3

0110000000101100; 10: 6 44 accumulator <- n^3 - n

0110000000101100; 12: 6 44 accumulator <- n^3 - 2n

0110000000101100; 14: 6 44 accumulator <- n^3 - 3n

0101000000110000; 16: 5 48 accumulator <- n^3 - 3n + 5

1101000000101000; 18: 11 40 go finish up

0001000000101100; 1a: 1 44 accumulator <- n

0101000000101100; 1c: 5 44 accumulator <- 2n

0101000000101100; 1e: 5 44 accumulator <- 3n

0101000000110000; 20: 5 48 accumulator <- 3n + 5

1101000000101000; 22: 11 40 go finish up

0001000000101100; 24: 1 44 accumulator <- n

0101000000101110; 26: 5 46 accumulator <- n + 1

0100000000000000; 28: 4 0 display the result

0000000000000000; 2a: 0 0 stop

0000000000000000; 2c: 0 A place to hold n

0000000000000001; 2e: 1 A place to hold 1

0000000000000101; 30: 5 A place to hold 5