K' Map Doubt!

kubeek

Joined Sep 20, 2005
5,661
Your K map is wrong. Only one bit can change between adjacent fields, so it needs to go 00 01 11 10, or 00 10 11 01.
 

hp1729

Joined Nov 23, 2015
2,304
Hey Guys I am studiyng full adders and I have a doubt.

In the youtube video tutorial the put the K' Map as follows:

00, 10, 11, 10. --> for this set up, anypne can be grouped.

And my K' Map is:

00, 01, 10, 11.

Any aclaration would be appreciated.

Thanx
Your map (IMG_4946) looks good to me. The incomplete one is concerned only with Carry Out.
A 0, B 0, Ci 0, then Co is a 0
A 1, B 0, Ci 0, then Co is a 0
A 0, B 1, Ci 0, then Co is a 0
A 0, B 0, Ci 1, then Co is a 0
A 1, B 1, Ci 0, then Co is a 1
A 1, B 0, Ci 1, then Co is a 1
A 0, B 1, Ci 1, then Co is a 1
A 1, B 1, Ci 1, then Co is a 1
 

hp1729

Joined Nov 23, 2015
2,304
Because in your map when you go from 01 to 10 you are changing two bits at once, which is not allowed and which is why the map then doesn´t work as it should.

We used to learn the map like this: http://www.spslevice.sk/ucebnice/SOC/pictures/Karnaughova mapa3.gif - where there is the bar the bit is 1, where the bar is missing the bit is not present.
It looks good to me. What is this with "only changing one bit at a time"? As long as all possible states are covered it is correct. What part of his map is incorrect?
 

kubeek

Joined Sep 20, 2005
5,661
If you don´t do the map like I showed, then when you try to make the largest minterms you cannot make rectangles like usual but you would have to divide one minterm into many parts when drawing it on the map.
 
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hp1729

Joined Nov 23, 2015
2,304
If you don´t do the map like I showed, then when you try to make the largest minterms you cannot make rectangles like usual but you would have to divide one minterm into many parts when drawing it on the map.
In his case, what rectangles?
 

kubeek

Joined Sep 20, 2005
5,661
In the correct K map you have four separate minterms. In the incorrect K map you get two rectangles, but you cannot cover either of them with a single minterm even though it at first looks like you could.

If it was the other way around, so you would get a rectangle in the correct K map, then you would not get a rectangle in the incorrect K map, which would lead to writing two minterms that could actually be minimized further into one.
 

hp1729

Joined Nov 23, 2015
2,304
In the correct K map you have four separate minterms. In the incorrect K map you get two rectangles, but you cannot cover either of them with a single minterm even though it at first looks like you could.

If it was the other way around, so you would get a rectangle in the correct K map, then you would not get a rectangle in the incorrect K map, which would lead to writing two minterms that could actually be minimized further into one.
I can see that Gray Code thing applying if there were four terms. You are technically correct. Does it make a difference in his case?
 

hp1729

Joined Nov 23, 2015
2,304
In the correct K map you have four separate minterms. In the incorrect K map you get two rectangles, but you cannot cover either of them with a single minterm even though it at first looks like you could.

If it was the other way around, so you would get a rectangle in the correct K map, then you would not get a rectangle in the incorrect K map, which would lead to writing two minterms that could actually be minimized further into one.
After looking at it a dozen times you are certainly correct. The single map is correct. The larger one is not.
Good stuff, thank you for the refresher course.
 
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