K-Map minimization question - why it not optimal?!

[Kruz89]

I have a K-Map and need to solve it.
My minimization is:

but my teacher tells me that this is not a minimal form, and requires me to find a more minimal form. I can't understand why this minimization is not most minimal. Anybody can explain me what I am doing wrong?

Thank you for help!!!

 

[Ratchit]

Kruz89,

but my teacher tells me that this is not a minimal form, and requires me to find a more minimal form. I can't understand why this minimization is not most minimal. Anybody can explain me what I am doing wrong?

Your teacher is correct. You have not covered all the terms as efficiently as you could. The answer is x2' x4' x5' + x1 x2' x5' +x2 x6 +x1 x4' x5 + x2' x3' x4'

Ratch

 

[Kruz89]

x2' x4' is the same for !x2&!x4 ? Your record is not absolutely clear

I will check up your results, but I would like to learn, how you have made it?
Can you explain the algorithm to me?

 

[Ratchit]

Kruz89,

x2' x4' is the same for !x2&!x4 ? Your record is not absolutely clear

Using an apostrophe is a standard method to designate an inverted variable. Using nothing between variables is a standard method of designating an AND operation. Using a plus sign is a standard method of designating an OR operation. I am surprised you have not come across this method, which looks cleaner than ! and &.

I will check up your results, but I would like to learn, how you have made it?
Can you explain the algorithm to me?

I used no algorithm. I just looked at the K-map and used all the "don't cares" I could. You can see if it is correct by mapping each of the 5 terms I gave onto the K-map, and see if they cover all the 1s. That will also tell you how you missed some of the "don't cares" you could have included.

Ratch

 

[Kruz89]

Thank you for explain about apos after name for inverted variable. Usually we use the '~' or '!' before var, and apos is special sign for function or variable. Plus, vertical bar, ampersand or mult sign are known for me, thank you, I'm not stupid

 

[Kruz89]

Thanks for help!!
I found the solution: y1 = x2 x6 | !x2 !x5 !x6 | x1 !x4 x5 | !x1 !x2 !x3 x5;
My general miss is that the standalone covers can be merged

 

[Ratchit]

Kruz89,

The last term, x1' x2' x3' x5, can be changed to x2' x3' x4' for an even greater optimization.

Ratch

 

[Kruz89]

You are right, thank you