# Program To Simulate Bit Adder Using Logic Gate Simulation In Scheme Assignment Solution

July 01, 2024
Prof. Benjamin
🇬🇧 United Kingdom
Computer Science
Professor Harper, with a Master's degree in Electrical Engineering from a renowned university in the United Kingdom, brings a wealth of knowledge and expertise to our team. With over 800 completed assignments, he excels in Multisim Simulator, particularly in signal processing and power electronics. Prof. Harper's dedication to excellence ensures that our clients receive impeccable solutions that meet the highest academic standards.
Key Topics
• Instructions
• Objective
• Requirements and Specifications
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## Instructions

### Objective

Write a Scheme assignment program to simulate the bit adder implementation using the knowledge of logic gates simulation.

## Requirements and Specifications

Logic gates and adder are basic components of building a computer and any digital devices. All high-level language programs will be translated into instructions that can be executed on these basic components. The correctness and performance of these components are extremely important.

A computer system consists of hardware and software. Before we implement a hardware component, the design is normally simulated by a program first, so that we can verify its correctness and evaluate its performance. In this project, you will write Scheme programs to simulate an n-bit Adder.

Write three Scheme procedures to simulate these three gates (functions): AND, OR, and XOR, shown in the diagram in Figure 1. Test your procedures using all possible input combinations.

Screenshots of output

Source Code

```;Your Name: ;Course: CSE 240 ; Use the template for your programming assignment #lang scheme ; added to test the program in DrRacket ;Q1 (6 points) (define OR-GATE (lambda (a b) (if (or (= a 1) (= b 1)) ; returns 1 if either a or b are 1, return 0 otherwise 1 0 ))) (define AND-GATE (lambda (a b) (if (and (= a 1) (= b 1)) ; returns 1 only if a and b are 1, return 0 otherwise 1 0 ))) (define XOR-GATE (lambda (a b) (if (= a b) ; returns 0 if a and b are equal, return 1 otherwise 0 1 ))) ;Test cases (AND-GATE 0 0) (AND-GATE 0 1) (AND-GATE 1 0) (AND-GATE 1 1) ;0 ;0 ;0 ;1 (newline) (OR-GATE 0 0) (OR-GATE 0 1) (OR-GATE 1 0) (OR-GATE 1 1) ;0 ;1 ;1 ;1 (newline) (XOR-GATE 0 0) (XOR-GATE 0 1) (XOR-GATE 1 0) (XOR-GATE 1 1) ;0 ;1 ;1 ;0 ;Q2 (define bitAdder (lambda (x a b) ;(4 points each) (cons (sum-bits x a b) (carry-out x a b)))) ; make list with sum and carry (define sum-bits (lambda (x a b) ;(5 points each) (XOR-GATE (XOR-GATE a b) x))) (define carry-out (lambda (x a b) ;(5 points each) (OR-GATE (AND-GATE (XOR-GATE a b) x) (AND-GATE a b)))) ;Test cases (newline) (bitAdder 0 0 0) (bitAdder 0 0 1) (bitAdder 0 1 0) (bitAdder 0 1 1) (bitAdder 1 0 0) (bitAdder 1 0 1) (bitAdder 1 1 0) (bitAdder 1 1 1) ;(0 . 0) ;(1 . 0) ;(1 . 0) ;(0 . 1) ;(1 . 0) ;(0 . 1) ;(0 . 1) ;(1 . 1) ;Q3.1 (4 points) (define tail (lambda (lst) (if (null? (cdr lst)) ; if we have empty tail we are at the last element (car lst) ; return the last element ; else list remainder is not empty, recurse using list remainder (tail (cdr lst))))) ;Q3.2 (4 points) (define rmtail (lambda (lst) (if (null? (cdr lst)) ; if we have empty tail we are at the last element '() ; return empty list to remove the last element ; else list remainder is not empty, recurse using list remainder, ; keep the current element in the list (cons (car lst) (rmtail (cdr lst)))))) ;Q3.3 (4 points) ;(1) size-n is the number of bits to sum, the size-n problem will be finding the ; recursive procedure n-bit-adder1 that takes 2 n-bit binary numbers ; (lists of n elements), the n and the previous result (resulti) and returns ; the sum of n-bits and a carry-out ;(2) Stopping condition is n = 0, we added all bits, we return the sum and the ; carry-out of all previous bits which is passed in resulti ;(3) The size-(n-1) problem removes the last bit in each list and decrements n: ; (n-bit-adder (rmtail L1) (rmtail L2) (- n 1)) ;(4) To build the size-n problem from size-(n-1), first we test the stopping ; condition, if test is true then we return the previous resulti ; if false, we recurse to the recursive size-(n-1) problem passing as resulti ; the sum of the current bits appended to the previous sum bit list and ; the previous carry. ;Q3.4 (14 points) (define X1 '(0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0) ) (define X2 '(1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1) ) (define X3 '(0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1) ) (define X4 '(1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0) ) (define X5 '(0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1) ) (define X6 '(1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1) ) (define n-bit-adder1 (lambda (L1 L2 n) ; use helper to calculate sum, start with a carry-in of 0 and an empty list (n-bit-adderhelper1 L1 L2 n '((0))) )) (define n-bit-adderhelper1 (lambda (L1 L2 n resulti) (if (= n 0) ; if stopping condition resulti ; return previous result ; calculate the sum of the current bits and the previous carry, ; label result as sum, (car sum) will be the sum and (cdr sum) will be the carry (let ((sum (bitAdder (tail L1) (tail L2) (caar resulti)))) ; recurse removing the tail of the lists and decrementing n (n-bit-adderhelper1 (rmtail L1) (rmtail L2) (- n 1) ; pass the current result as a list of a list containing only the carry ; and a list of the previous sums appending the current result at the start (cons (list (cdr sum)) (cons (car sum) (cdr resulti)))))))) ; this definition was not used but it calculates the carry for the n-bit sum (define n-bit-adder-carry-out (lambda (L1 L2 n resulti) (if (= n 0) ; if stopping condition resulti ; return previous result ; else, recurse removing the tail of the lists and decrementing n (n-bit-adder-carry-out (rmtail L1) (rmtail L2) (- n 1) ; calculate the carry between the current bits and the previous result, ; pass the carry result on the recursive call as the resulti (carry-out resulti (tail L1) (tail L2)))))) ;Test cases (newline) (n-bit-adder1 X1 X2 32) (n-bit-adder1 X3 X4 32) (n-bit-adder1 X5 X6 32) (n-bit-adder1 X2 X3 32) (n-bit-adder1 X4 X5 32) (n-bit-adder1 X1 X6 32) (newline) (n-bit-adder1 X1 X3 32) (n-bit-adder1 X1 X4 32) (n-bit-adder1 X1 X5 32) (n-bit-adder1 X2 X4 32) (n-bit-adder1 X2 X5 32) (n-bit-adder1 X2 X6 32) ;((0) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1) ;((0) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1) ;((1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0) ;((1) 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0) ;((1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1) ;((0) 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1) ;((0) 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1) ;((0) 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0) ;((0) 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1) ;((1) 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1) ;((1) 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0) ;((1) 1 0 1 0 1 ```

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