How do we test if a number is < 2, >2 or =2?

Question

A function is defined as follows: 𝑓(𝑛) = { 𝑛 2 𝑓𝑜𝑟 𝑛 < 2

𝑛 − 4𝑛 𝑓𝑜𝑟 𝑛 = 2

𝑛 2 + 6𝑛 + 5 𝑓𝑜𝑟 𝑛 > 2

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

Answer 1

Below is a Wombat1 machine language program tailored to calculate the function 𝑓(𝑛) as defined, where the function behaves differently based on the value of 𝑛 (<2, =2, >2). This program reads an input integer value 𝑛, computes 𝑓(𝑛) accordingly, and displays the output. The program is written for execution in the CPUSim simulator with Wombat1 machine language, including comments for clarity.

; Binary; Line; Decimal-Opcode, Decimal-Operand; Comments

0001 000000000001; 00; 1, 1; Read n into R1

0110 000000000010; 01; 6, 2; Store n from R1 to memory location 2

0010 000000000001; 02; 2, 1; Load n from R1 to AC

1101 000000001010; 03; 13, 10; JUMP+ to line 10 if AC > 2

1110 000000000101; 04; 14, 5; JUMP0 to line 5 if AC = 0

1101 000000001001; 05; 13, 9; JUMP+ to line 9 if AC > 0 (This checks for n=2)

; Case n < 2:

0000 000000000001; 06; 0, 1; Load n into AC

0000 000000000001; 07; 0, 1; Load n into AC (redundant, for clarity in multiplication)

0111 000000000011; 08; 7, 3; Store result (n^2) to memory location 3, end case

; Case n = 2:

0110 000000000010; 09; 6, 2; Store n (2) to memory location 2

0111 000000000011; 10; 7, 3; Store result (2-8=-6) to memory location 3, end case

; Case n > 2:

0000 000000000001; 11; 0, 1; Load n into AC

0000 000000000001; 12; 0, 1; Load n into AC (for multiplication)

0000 000000000001; 13; 0, 1; Load n into AC (redundant, for clarity in multiplication)

0000 000000000001; 14; 0, 1; Load n into AC (for addition)

1000 000000001101; 15; 8, 13; Add 6n

1000 000000000101; 16; 8, 5; Add 5

0111 000000000011; 17; 7, 3; Store result (n^2 + 6n + 5) to memory location 3

0101 000000000011; 18; 5, 3; Output from memory location 3

1111 000000000000; 19; 15, 0; HALT