# 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.

• Top left is the naive solution that implements the function blindly
• Top right is when you move the logical NOT gates backwards and join them
• Bottom is when you factorize $$\\bar{A}\bar{D}+\bar{A}B+\bar{A}C\$$ into $$\\bar{A}(\bar{D}+B+C)\$$

All 3 of them give the same output as the karnaugh map. All 3 of them are viable. If you have many NAND or NOR gates laying around then you can apply De Morgan's law and then you have a solution that uses NAND and NOR heavily and you can use up some NAND and NOR gates.

The correct logic circuit for you boils down to what you are allowed to use or some other requirements/constraints on your project. Keep in mind that NAND and NOR gates usually requires fewer transistors to implement and are therefor usually the gates you want to use.

• Great answer. My 2 cents: logisim is a free tool that might help a lot creating these circuits. Since this seems like a homework question you probably need to show your answers and not just use a tool, but this might help visualize the differences. Dec 15, 2020 at 22:48