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Answer the following using a truth table and a Karnaugh map.
A superhero is allowed access to a secure Avengers facility if he / she meets any of the following…
A superhero is allowed access to a secure Avengers facility if he / she meets any of the following criteria:
• The superhero has Avengers' membership and possesses a high-security clearance badge
OR
• The superhero does not have Avengers membership but holds a special permit issued by S.H.I.E.L.D. along with a high-security clearance badge
OR
• The superhero is not a recognised ally but holds a special permit issued by S.H.I.E.L.D. along with a high-security clearance badge
The inputs are:
(In all the above cases, 1 indicates YES and 0 indicates NO)
Output: X – Denotes allowed access [1 indicates YES and 0 indicates NO in all cases]
| INPUTS | |
|---|---|
| A | Superhero has Avengers membership. |
| S | Superhero holds a special permit issued by S.H.I.E.L.D. |
| C | Superhero possesses a high-security clearance badge |
| L | Superhero is a recognised ally |
(i)[5.0]
Draw the truth table for the inputs and outputs given above. Write the POS expression for X (A, S, C, L).
(ii)(a)[2.5]
Reduce the above expression X (A, S, C, L) by using 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs).
(ii)(b)[2.5]
Draw the logic gate diagram using NOR gates only for the reduced expression. Assume that the variables and their complements are available as inputs.
Draw: logic gate diagram using NOR gates only for the reduced expression
Answer
Answer
AI3(i): X = A.C + A'.S.C + L'.S.C
(access allowed when: has membership AND clearance; OR lacks membership but has a SHIELD permit AND clearance; OR is not a recognised ally but has a SHIELD permit AND clearance)
Truth Table:
POS expression (product of maxterms where X=0, in order A,S,C,L):
$X(A,S,C,L) = (A+S+C+L)(A+S+C+L')(A+S+C'+L)(A+S+C'+L')(A+S'+C+L)(A+S'+C+L')(A'+S+C+L)(A'+S+C+L')(A'+S'+C+L)(A'+S'+C+L')$
3(ii): Reducing X(A,S,C,L) using a 4-variable K-map, grouping the zeros (0s) since the circuit is to be built with NOR gates only, which naturally realise a POS (Product of Sums) expression:
- Octet (8 cells): all cells where C=0 -> gives the POS factor $C$
- Quad (4 cells): the cells where A=0 and S=0 -> gives the POS factor $(A+S)$
Reduced (minimal) POS expression: $X = (A+S) \cdot C$
With NOR gates only: a NOR gate gives $(A+S)'$, and a second NOR gate with inputs $(A+S)'$ and $C'$ gives $[(A+S)' + C']' = (A+S) \cdot C$ by De Morgan's law.
| A | S | C | L | X |
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
From ISC 2025 Computer Science Paper 1, question 3.
Check your working with the Boolean algebra solver: the steps law by law, the K-map and the logic circuit for any expression.