1. Part a)
2. Part b)
   1. Explanation
3. Part c)
4. Part d) i)
   1. Explanation/Working:
5. Part d) ii)
6. Part d) iii)
7. Part d) iv)
8. Links to Notes
   1. Attributions to media used in this post
9. Footnotes

Part a)

AND Gate:

  A | B | A AND B
  ---------------
  0 | 0 |    0
  0 | 1 |    0
  1 | 0 |    0
  1 | 1 |    1

Or Gate:

  A | B | A OR B
  ---------------
  0 | 0 |    0
  0 | 1 |    1
  1 | 0 |    1
  1 | 1 |    1

Not Gate:

  A | NOT A
  ---------
  0 |   1
  1 |   0

See this note¹

Part b)

Suggested response:

XOR truth table:

Output is 1 in XOR for (NOT A AND B) OR (A AND NOT B)

Circuit:

Explanation

Assume we want to use an AND gate as a building block for our resulting circuit. Wherever we have an output of 1  it means that the inputs to that and gate should also be 1 .

If we imagine an AND gate being the last gate before the output, wherever there is a 1 we need to understand that the inputs to that gate would have to be 1.

-For each output that produces a 1, we modify the variable input to match  the modded input listed in the specific row of the table:

We see that either row ii) OR row iii)  produces a 1, i.e

(NOT A AND B) OR (A AND NOT B)

From this expression, we draw the circuit as shown in the suggested response above.

Part c)

Data Lines (Input) are denoted as x₁…x₄, select line are denoted as s₁ and s₂, Output is denoted as f

Alternative Diagram in ASCII :

        _______________________
       |        4-to-1         |
D0 ----|      Multiplexer      |
       |                       |
D1 ----|                       |---> Output
       |                       |
D2 ----|                       |
       |                       |
D3 ----|_______________________|
              |         |
              |         |
              |         |
              |         |
              S1        S0
Data Lines (Input) are denoted as D0...D3, select line are denoted as S0 and S1.

See Note²

Part d) i)

16+8+0+2+1 =27

Explanation/Working:

(0×2⁷)+(0×2⁶)+(0×2⁵)+(1×2⁴)+(1×2³)+(0×2²)+(1×2¹)+(1×2⁰)

0+0+0+16+8+0+2+1=27

Part d) ii)

  0111 +
  1110
———-
10101 ← this result cannot be stored in 4 bits.

Part d) iii)

Largest number = 0111 =7 (Show conversion)
Smallest number= 1111= -ve 7 (Show conversion)

Part d) iv)

Suggestion solution 1:

Algorithm, copy all numbers from LSB to MSB up to and including the first 1, then flip remaining bits in that order, i.e
5= 0101

Applying algorithm : 1011

Suggestion solution 2:

+5= 0101
Ones = 0101+1 = 1010
Twos = 1010+1 = 1011

Links to Notes

Attributions to media used in this post

Inductiveload, Public domain, via Wikimedia Commons

Footnotes

1. https://islandclass.org/2021/10/20/logic-gates-formal-introduction/ ↩︎
2. https://islandclass.org/2021/10/21/multiplexers/ ↩︎

© 2023  Vedesh Kungebeharry. All rights reserved. 

1. Part a) i)
2. Part a) ii)
3. Part a) iii)
4. Part b) i)
5. Part b) ii)
6. Part b) iii)
7. Part c)
8. Part d) i)
9. Part d) ii)
10. Part d) iii)

Part a) i)

        _______________________
       |        4-to-1         |
D0 ----|      Multiplexer      |
       |                       |
D1 ----|                       |---> Output
       |                       |
D2 ----|                       |
       |                       |
D3 ----|_______________________|
              |         |
              |         |
              |         |
              |         |
              S1        S0

Part a) ii)

Set the select inputlines to 01 to select the 2^nd line, and when input from the 4^th line is needed after 1 second (i.e i3), set the select input lines to 11

Part a) iii)

– It always exists in 1 of 2 stable states.
-It can retain its state to achieve storage (in the case the set goes from 1 to 0)

Can be used in RAM Arrays (to implement 1 bit storage) or to Implement a Ripple Counter

Part b) i)

1101

(First bit is 1 for to represent a negative value,

working for 5₁₀ to binary :

       2 | 5  

       2 |2  remainder 1

       2 |1  remainder 0 

            0  remainder 1

i.e 5₁₀  = 101₂

combining the result we get 1101₂

)

Part b) ii)

1010 (flipped bits of 0101)

Part b) iii)

1010 +1 = 1011

Part c)

(Show binary-decimal conversions)

For an unbiased exponent and unnormalized mantissa:

sign is -ve, Exponent = 2, Mantissa= 0101

i.e -0.0101 x 2_­² = -1.01

-1.01₂   =  (-1 x 2⁰)+ (0 x 2^-1)+ (1 x 2^-2)

          = -1 + 0+ ¼

          = -1.25

Part d) i)

Not (Unary operator)

Part d) ii)

AND

Part d) iii)

OR

© 2023  Vedesh Kungebeharry. All rights reserved. 

Part A

i) Debugging is the process of detecting removing and/or correcting errors in code.

ii) Coded is indented every time a control structure is entered. i.e.  thus isolating blocks of logic which is easier to read and understand.

Comments are used to explain what the code is intended to accomplish, thus making it easier understand and modify

Descriptive variable names explains the purpose of each variable making code easier to understand and maintain.

Part B

//this is a complete program, but starting from main() would be sufficient.
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

int main()
{
    //Start:declare and initialize all variables
    double redTime     =   -999.99;
    double greenTime   =   -999.99;
    double orangeTime  =   -999.99;
    bool valid=false;//used to signify that no
                    //valid inpuha been enterd
                    //at the start of execution

    //End:declare and initialize all variables

    //while input is not valid
    while(valid==false)
    {
        //ASSUMPTION: the usere only enters positive values
        //get input from user
        puts("Please enter a value for Redtime");
        scanf("%lf",&redTime);
        puts("Please enter a value for greenTime");
        scanf("%lf",&greenTime);
        puts("Please enter a value for orangeTime");
        scanf("%lf",&orangeTime);
        //evaluate each condition, and check that they are all true.
        valid = (redTime>=greenTime+orangeTime)&&
                (orangeTime>=5)&&
                (greenTime>4*orangeTime);

        //if input is invalid
        if(valid==false)
        {
            //prompt user
            puts("Invalid timings: Please re-enter\n");
        }
        //continue looping if valid == false
    }
    //if we have exited the loop, then timings are valid
    puts("Timings are valid\n");
    return 0;
}

Part C(i)

    /*Code presented in question
    int i j;
    for (i=1;<= 3;i++)
        for (j=1;j<i;j=j+1)
            printf("%d" j);
        printf(\n);
    */

//Corrected code
    int i, j;  //comma was missing between i and j
    for (i=1;i<= 3;i++)//no left operand for comparison, "i" dentifier missing
        for (j=1;j<i;j=j+1)
            printf("%d", j);//comma missing between format string and parameter
        printf("\n");  //String was not in quotes

/*Note that this code seems to contain a logical error , 
as implied by the indentation of the code.  
It seems that the 2nd for loop and the printf statement are 
2 lines of code that are meant to be executed within the first for loop, 
and should be enclosed in curly braces. 
This question asks us to correct the syntax errors only. 
One may feel the need to fix the logical error also,  
but it is not required.  */

Part C(ii)

Answer:

112

Note, if you assumed a logical error and fixed it, the output would be

1

12

Based on the code:

#include <stdio.h>

int main() {
        //Corrected code
    int i, j;  //comma was missing between i and j
    for (i=1;i<= 3;i++)//no left operand for comparison, "i" dentifier missing
    {    
        for (j=1;j<i;j=j+1)
            printf("%d", j);//comma missing between format string and parameter
        printf("\n");  //String was not in quotes
    }
    
    return 0;
}

This fix however, would render your response incorrect.

© 2021  Vedesh Kungebeharry. All rights reserved. 

Part A

Identifying and evaluating possible solutions involves generating different approaches to solving the problem.  Solutions will include different methods of solving the problem, e.g contrasting data structures :  one solution may use arrays only, another will use a mixture of arrays and structs.  Each solution is compared against each other to determine individual advantages and disadvantages when compared to each other in terms of efficiency and how well it solves the problem.[1]

[1] https://islandclass.org/2020/02/17/stages-in-problem-solving/

Part B

Strategy – keep track of all variables in a trace table, draw output as you go along.

Table – write all variables in the order of its logical use within main control structures, use 1 variable per column:

Answer:

----------* 
---------**
--------*-*
-------*--*
------*---*
-----*----*
----*-----*
---*------*
--*-------*
-*--------*
***********

Part C

START
Sum<--0
For i<--14 to 126 DO
	If i%7 ==  0  Then
		Sum<--sum +i
	Endif
Endfor
PRINT Sum
END

Coded solution:

#include <stdio.h>
#include <stdlib.h>

int main()
{
   
    int sum = 0;
    for (int i= 14; i<=126; i++)
        if (i%7==0)
            sum+=i;
    printf("The sum is : %d", sum);

    return 0;
}

© 2021  Vedesh Kungebeharry. All rights reserved. 

Part a

An algorithm is a well-defined, precise, finite set  of steps for solving a problem.[1]

Part b

Sequence (procedure), Selection (if-then-else) , Iteration (Looping)

Part c

Note:  j goes from 2,4,6,… 96,98 during looping and is output to the screen. Odd numbers are not output. See the attached flowgorithm file here:

https://drive.google.com/file/d/1H21gL4E0YGtgzXgwUD-1cFuEHsmC8IY0/view?usp=sharing

Image:

https://drive.google.com/file/d/1smeXp9ArthIPi7ZY1NfnJqJrYKIJx4rK/view?usp=sharing

Part C(i)

J = 100, C=50

Part c(ii)

2. J=1
3. C=1

Note, for this part we arrive at a solution that solves the problem exactly as stated (only numbers from odd numbers from 1 to 99 and the count.  See flowgorithm file for the corrected algorithm:

Flowgorithm file

Image

Part c(iii)

c=50

Part  d

Use “Start” instead of main.  See Files Below:

Flowgorithm

Image

[1] VK. Blogged at https://islandclass.org/2020/02/10/what-is-an-algorithm-definition/

© 2021  Vedesh Kungebeharry. All rights reserved.