Can someone confirm this? (About boolean algebra and reduced cost)

[tamtam402]

I was asked to make the simplest possible circuit to solve this problem:

for an entry of 0000 to 1001 (0 to 9), detect the numbers that are divisible by 2 or 3. I decided to write the truth table with an output of 1 for functions that are NOT divisible by either 2 or 3, since that seemed like less trouble to me. I plotted the karnaugh map and solved as usual. However, I put an inverter on the output of the last gate since the question was asking to find the numbers that ARE divisible by 2 or 3. Is my solution still in the simplest possible form? If not, why not?

 

[berkeman]

I was asked to make the simplest possible circuit to solve this problem:

for an entry of 0000 to 1001 (0 to 9), detect the numbers that are divisible by 2 or 3. I decided to write the truth table with an output of 1 for functions that are NOT divisible by either 2 or 3, since that seemed like less trouble to me. I plotted the karnaugh map and solved as usual. However, I put an inverter on the output of the last gate since the question was asking to find the numbers that ARE divisible by 2 or 3. Is my solution still in the simplest possible form? If not, why not?

Can you show us your K-map and your final circuit? And can you show us what the circuit would look like if you did it the other way instead (implemented the positive output version)?

 

[tamtam402]

Can you show us your K-map and your final circuit? And can you show us what the circuit would look like if you did it the other way instead (implemented the positive output version)?

Sorry I don't have these anymore, I answered the question on an exam. I was simply wondering if finding F' (by switching 0's and 1's in the truth table), then finding a sum of product and finally inverting the last output (since the signal was F' and I wanted F) gave me an equation as good (cost wise) as if I had simply found F directly.

 

[Joffan]

So you detected 0001, 0101 and 0111 and inverted? Since numbers >10 are irrelevant you could use that space on the K-map to come to something like BD + A'C'D - is that what you had before inversion?

 

[tamtam402]

Yes that's what I had if I recall correctly. I drew the schematics for the 2 AND gates as well as the OR gate, then I inverted it's output.

edit: I added "don't cares" (that's how we call these, I'm not sure if the name is universal) to numbers from 10 to 15 as you suggested, since these entries are impossible in BCD.

 

[Joffan]

You (should) have a total of three NOTs - using a bit of logic on ((B+A'C')D)' you can get that down to two.