Timestamps: 0:00 Introduction 0:26 Construction of 4-variable K-map 4:31 Mapping the 4 variable Boolean Function in K-map 6:00 Minimization of 4-variable Boolean Function using K-map 10:25 Example 1 11:28 Example 2 13:11 Example 3 14:41 Example 4 16:08 Example 5 18:15 Example 6
There are some K-map rules. One of them is the group size should be in the power of 2. I have made a separate video for K-map grouping rules. Here is the link: th-cam.com/video/A_LFVBWYZME/w-d-xo.htmlsi=0CyRX9ZYx4HjzW1e
Here the goal is to cover all the 1s in the k-map by making the minimum groups. Of course, you can make the group of 5 and 13, but If you make the group of 5 and 13, then to cover 9, you need to make another group with 13. So, group (5, 13) is redundant over here, More technically if I say, then these three groups (shown in the video) are essential prime implicants. The essential prime implicants will always be the part of the final minimized expression. The (5, 13) is prime implicant. For more info, you can check this video : th-cam.com/video/fmAwCosFSRs/w-d-xo.htmlsi=3XL4fZlaaXm40y4l
On the top if you see, then in the four columns both C and D are changing. But for the same group, A remains 0 while B remains 1. Therefore, the given group of four 1s represents A'B. I hope, it will clear your doubt.
You can if you want. But here the idea is to cover all the minterms (with '1') by making minimum number of groups. Here, just by making three groups {(4,5,6,7) (13,9) and (14,15)}, we are able to cover all the 1s in the K-map. If you make a group of 4 1s with (5,7,13, and 15) then to cover remaining 1s, you need to make another three groups. So, that is why, here (5,7,13 and 15) are not grouped, and instead 4,5,6 and 7 is grouped. I hope, it will clear your doubt.
After making two groups of 4 1s, only two mimterms m13 and m9 are remaining. And it can be covered by making the group of 2 1s. If there is one more mimterms m11, then we would have made a group of 4 1s. I hope, it will clear your doubt.
You mean during grouping ? During grouping, you just need to see in the group which variables remains the same (not changing) and what is the value of those variables. For example, in a group, if the variable A and B is not changing and if their values (just see the 00, 01, 10, 11 written horizontally and vertically in K-map) are 1 and 0. It means that in that group A = 1 and B = 0. And hence, the group represents AB'. I hope, it will clear your doubt.
No, all the minterms for which output is 1 needs to be covered. If that 1 ( minterm) cannot be grouped with any other minterm then you just need to make a group of only that 1.
Which group ? I can only see 1 group of 8. That is C'. The other groups are groups of 4. The group should be in the power of 2. And it has to be symmetrical. It can be either square or rectangle. As per that, there are only three groups.
8 and 9 can be grouped but to cover the minterm 1, one needs to make the group of 1 and 9. Similarly, to cover 10, one needs to make group of 1 and 10. So, yes we can make the group of 8 and 9, but during the minimization or for finding the minimal expression, that group will be redundant. Because just by making these two groups, we are able to cover all the 1s in the K-map. In the next video, I am going to cover it in detail. (The concept of Prime implicant and the essential prime implicant) After that, I think your all doubts related to K-map will get clear.
you can make the group of 2. For more info, you can check this video, where couple of questions related to K-map have been solved. th-cam.com/video/cTRDJrq5ySM/w-d-xo.html
On the vertical side of the map, we have two variables A and B. And on the horizontal side we have C and D. Now, in this group if you see, on the vertical side the variable B is not changing. ( 01 and 11, the second variable B value is 1). Likewise, on the horizontal side, the variable D is not changing. ( 01 and 11, the second digit is representing variable D, and its value is 1). Therefore, we can say that this group of 4 represents B.D. I hope, it will clear your doubt.
No, you cannot group them. You can make the group of four 1s in the corner, but you cannot group the two 1s which are in the opposite corner. You can combine these groups of 1s. (0,2) (0,8) (8,10) (2,10). But you cannot make the group of (0,10) and (2,8)
You can make those groups. But those are redundant groups. Because to cover 9 and 14, you need to make another groups. Just by making these three groups, we are able to cover all the 1s in the K-map. (basically the two groups you were talking are the non-essential prime implicants) For more info, please check this video. Your doubts will get clear. th-cam.com/video/fmAwCosFSRs/w-d-xo.html
As far as I can see, there seems no mistake in example 4. But still, if you feel there is a mistake, please mention the timestamp where you feel, there is a mistake.
You should thank him for his efforts rather than this. There is an Arabic proverb that says : "They did not find any flaw in the gold, and they said its sparkle blinds the eyes."
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Timestamps:
0:00 Introduction
0:26 Construction of 4-variable K-map
4:31 Mapping the 4 variable Boolean Function in K-map
6:00 Minimization of 4-variable Boolean Function using K-map
10:25 Example 1
11:28 Example 2
13:11 Example 3
14:41 Example 4
16:08 Example 5
18:15 Example 6
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Sir, your K-map explanation is very clear. Thanks
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True
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At 15:36 , why don't you group the 3 1's together and stop with 2? If overlapping is allowed...yoi can go all the way right?
There are some K-map rules. One of them is the group size should be in the power of 2. I have made a separate video for K-map grouping rules.
Here is the link: th-cam.com/video/A_LFVBWYZME/w-d-xo.htmlsi=0CyRX9ZYx4HjzW1e
Thank you! amazing
Thanks a lot for this
Excellent 👍
12:27 why didn't you made pair of 13 and 5
Here the goal is to cover all the 1s in the k-map by making the minimum groups. Of course, you can make the group of 5 and 13, but If you make the group of 5 and 13, then to cover 9, you need to make another group with 13. So, group (5, 13) is redundant over here, More technically if I say, then these three groups (shown in the video) are essential prime implicants. The essential prime implicants will always be the part of the final minimized expression. The (5, 13) is prime implicant.
For more info, you can check this video : th-cam.com/video/fmAwCosFSRs/w-d-xo.htmlsi=3XL4fZlaaXm40y4l
Thanks a lot
Does that mean you have to group
the ones without anyone remaining
No reply yet😢
Yes, all the 1s needs to be covered in the k-map.
Thanks again
God richly bless u
Please l need the laws of boolean algebra and l also want to know if a boolean expression must be simplified to the lowest level
Hi Sir, at 14:02 is the term we got from the four 1's not supposed to be A'BD' since D is not changing?
On the top if you see, then in the four columns both C and D are changing. But for the same group, A remains 0 while B remains 1. Therefore, the given group of four 1s represents A'B. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS Thank you l now understand
great work
Hats off
why in the example 3 you haven't taken the group of 4 with5,7,13,15 ?
You can if you want. But here the idea is to cover all the minterms (with '1') by making minimum number of groups. Here, just by making three groups {(4,5,6,7) (13,9) and (14,15)}, we are able to cover all the 1s in the K-map. If you make a group of 4 1s with (5,7,13, and 15) then to cover remaining 1s, you need to make another three groups. So, that is why, here (5,7,13 and 15) are not grouped, and instead 4,5,6 and 7 is grouped. I hope, it will clear your doubt.
it's like ur reading from a textbook?
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Beautifuly explained thank you.
My tomorrow's exam is about digital electronics 😲👍
thank you sir
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Do we need to minimize it further ?
good video
Waiting for prime implicant and essential prime implicant 🙋🙋🙋
Tomorrow am getting everything.......
all the best for tomorrow's exam 👍👍
I want all the difficult examples
i think at 14:33 another 4 1's can be done instead of 2 1's.
After making two groups of 4 1s, only two mimterms m13 and m9 are remaining. And it can be covered by making the group of 2 1s. If there is one more mimterms m11, then we would have made a group of 4 1s.
I hope, it will clear your doubt.
goodjob
Hi,
I don’t understand how it comes to ABC’ or A’B…stuff like that
You mean during grouping ? During grouping, you just need to see in the group which variables remains the same (not changing) and what is the value of those variables. For example, in a group, if the variable A and B is not changing and if their values (just see the 00, 01, 10, 11 written horizontally and vertically in K-map) are 1 and 0. It means that in that group A = 1 and B = 0. And hence, the group represents AB'. I hope, it will clear your doubt.
(Example:6) , I can't stop myself to like this video, 🥲
What if there is only one variable group formed do we ignore it???????
No, all the minterms for which output is 1 needs to be covered. If that 1 ( minterm) cannot be grouped with any other minterm then you just need to make a group of only that 1.
17:05 why you have not made second group as group of 8 ?
Which group ? I can only see 1 group of 8. That is C'. The other groups are groups of 4. The group should be in the power of 2. And it has to be symmetrical. It can be either square or rectangle. As per that, there are only three groups.
Bro at 8:18 cant 8 9 be included in Grouping
8 and 9 can be grouped but to cover the minterm 1, one needs to make the group of 1 and 9. Similarly, to cover 10, one needs to make group of 1 and 10. So, yes we can make the group of 8 and 9, but during the minimization or for finding the minimal expression, that group will be redundant. Because just by making these two groups, we are able to cover all the 1s in the K-map.
In the next video, I am going to cover it in detail. (The concept of Prime implicant and the essential prime implicant)
After that, I think your all doubts related to K-map will get clear.
What if they are in column and can't be paired 8... Do we seperate it to 2:2
you can make the group of 2. For more info, you can check this video, where couple of questions related to K-map have been solved.
th-cam.com/video/cTRDJrq5ySM/w-d-xo.html
Are you from kolhapur ?
6:40 I didn’t understood like how do you that it will come BD ?
On the vertical side of the map, we have two variables A and B. And on the horizontal side we have C and D. Now, in this group if you see, on the vertical side the variable B is not changing. ( 01 and 11, the second variable B value is 1). Likewise, on the horizontal side, the variable D is not changing. ( 01 and 11, the second digit is representing variable D, and its value is 1). Therefore, we can say that this group of 4 represents B.D. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS ok yes it cleared my doubts thank you and fabulous explanation 👍🏻👍🏻👍🏻
Today is digital lab exam 😌
can i make a pair of two with elements in block 10 and 0
No, you cannot group them. You can make the group of four 1s in the corner, but you cannot group the two 1s which are in the opposite corner. You can combine these groups of 1s. (0,2) (0,8) (8,10) (2,10). But you cannot make the group of (0,10) and (2,8)
Thank you@@ALLABOUTELECTRONICS
Tommrrow is my exams 😉
Electronics students are here 😢
can you teach how to make state diagram, state table and kmap when given 8bit input?
I will cover the state diagram very soon.
my question is that at11:59 can we not make the group of two 1s that are in box no 5 and 13 and in 6 and 4
You can make those groups. But those are redundant groups. Because to cover 9 and 14, you need to make another groups. Just by making these three groups, we are able to cover all the 1s in the K-map. (basically the two groups you were talking are the non-essential prime implicants)
For more info, please check this video. Your doubts will get clear.
th-cam.com/video/fmAwCosFSRs/w-d-xo.html
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You made mistake in example 4
As far as I can see, there seems no mistake in example 4. But still, if you feel there is a mistake, please mention the timestamp where you feel, there is a mistake.
kadavuley all about electronics
Bro don't care condition example you didn't include 🤐
It will be covered seperately. ( In the seperate video)
@@ALLABOUTELECTRONICS okay 😉❤️
Whyareyoutalkinglike - thiisss
Lol😂
Like what
You should thank him for his efforts rather than this.
There is an Arabic proverb that says : "They did not find any flaw in the gold, and they said its sparkle blinds the eyes."
Is your keyboard doesn't have spacecap
Whyareyoustupidlike - thiiss
vik
k map
Well explained, thanks to you man!🎉