# How do I convert a Karnaugh map into a logic gate circuit?

I currently have some questions that I need to work out in uni. I've been given 2 lengthy boolean expressions and I need to simplify them.

Question a I feel like I've done right but question b I'm stuck on as its so long and I'm fairly new to logic gates still. I'll attach a picture of the question and what I've come up with so far for the Karnaugh map.

I feel like I understand how the terms work and how to get them with say ABC = 2 terms, AB = 3 A = 4 etc. I know that the + symbols are OR gates.

Added another image below, I knew about being able to overlap when circling the inputs but didnt know and still have no idea what they mean or how to replicate those circled inputs into a logic gate circuit. I'll have to do some research on Karnaugh maps. Also apologies for the mistype earlier, I put AND instead of OR.

I assume the 4 shown below would replicate the 4x1, 4x1, 4x1, and 2x1 inputs, but I assume iI have to change quite a few things.

My attempt at creating the circuit:

• You can simply the formula more, for example the second and third part (between the plus signs): A.notC.notD + A.notC.D = A.notC(notD + D) and notD + D = true (1), so it is a.notC ... you can simplify more after this. Also check deMorgan's law. Nov 6, 2019 at 23:23
• + is not AND. You might need to revisit your material if this is your understanding of boolean math (and not just a typo) Nov 6, 2019 at 23:24
• You should look for larger groupings on the K-map. The groups can overlap. Nov 6, 2019 at 23:24
• You haven't used your Karnaugh map correctly. Your bubbles can be optimized further. See the area as the surface on a donut, the edges are connected (left and right, top and bottom) and repeating forever. - The upper left bubble 2x1 can be put in a 4x4 bubble. The center 1x1 bubble can be placed in a 1x4 bubble. The bottom right 1x1 bubble can be placed in a 2x1 bubble. - You will get same amount of bubbles but the larger the bubbles are the less circuitry they require. Nov 6, 2019 at 23:38
• @HarrySvensson its something that I did know about to an extent but didnt actually think about doing that to create bigger bubbles of inputs, Ive uploaded a new picture of the new karnaugh map, ill just be looking over de morgans theorem to see how I can convert those 4 bubbles into a simplified boolean expression so i can create a circuit
– Meck
Nov 7, 2019 at 0:03

After an update to your question you came up with

Here's the same equation you wrote but with "Y=" added to it for clarity:
$$\Y = \bar{A}\bar{D}+\bar{A}B+\bar{A}C+\bar{B}C\bar{D}\$$

The equation is correct.

How do I convert a Karnaugh map into a Logic gate circuit?

Now, look at what the equation actually say,
it says that $$\Y\$$ is equal to $$\\bar{A}\$$ AND $$\\bar{D}\$$ OR $$\\bar{A}\$$ AND $$\B\$$ OR $$\\bar{A}\$$ AND $$\C\$$ OR $$\\bar{B}\$$ AND $$\C\$$ AND $$\\bar{D}\$$

So let's just make that logic circuit three times in three different ways.

• Bottom is when you factorize $$\\bar{A}\bar{D}+\bar{A}B+\bar{A}C\$$ into $$\\bar{A}(\bar{D}+B+C)\$$