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A school intends to select candidates for an Inter-School Essay Competition as per the criteria…
A school intends to select candidates for an Inter-School Essay Competition as per the criteria given below:
• The student has participated in an earlier competition and is very creative.
OR
• The student is very creative and has excellent general awareness, but has not participated in any competition earlier.
OR
• The student has excellent general awareness and has won prize in an inter-house competition.
The inputs are:
(In all the above cases 1 indicates yes and 0 indicates no).
Output : X [1 indicates yes, 0 indicates no for all cases]
Draw the truth table for the inputs and outputs given above and write the POS expression for $X(A, B, C, D)$.
| INPUTS | |
| A | participated in a competition earlier |
| B | is very creative |
| C | won prize in an inter-house competition |
| D | has excellent general awareness |
Answer
Answer
AIFrom the criteria: X = 1 when (A.B) OR (A'.B.D) OR (C.D), i.e. $X = AB + A'BD + CD$.
Truth table (verified with the boolean tool):
POS expression (product of the maxterms where X = 0, i.e. rows 0, 1, 2, 4, 6, 8, 9, 10):
$X(A,B,C,D) = \pi(0,1,2,4,6,8,9,10) = (A+B+C+D)(A+B+C+D')(A+B+C'+D)(A+B'+C+D)(A+B'+C'+D)(A'+B+C+D)(A'+B+C+D')(A'+B+C'+D)$
In reduced form this is $X = (A + D)(B + C)(B + D)$.
| A | B | C | D | X |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
From ISC 2017 Computer Science Paper 1, question 5(a).