How Can Boolean Algebra Simplify Complex Circuit Designs?

[Pariah]

Homework Statement

We have been given a task to develop a circuit which displays the square of a binary number on a 3 x 7 seq displays.

I have already gone through and done up the Karnaugh Maps for the task and have identified the minterms. However, I believe that these can still be simplified even more.

Homework Equations

Karnaugh Maps Output = A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'

The Attempt at a Solution

A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'
Factor out A' and C from minterms 1 and 3
a'c(b' + b) (d + d')
= a'c + a'bc'd + abcd + b'c'd' + ab'c'
Factor out B and D from minterms 2 and 3
bd(a' + a)(c' + c)
= a'c + bd + b'c'd' + ab'c'(Can I simplify this anymore?)
Can the minterm ab'c' absorb the minterm b'c'd' ?
Also by using the distributive law can I add B to the minterm a'c and then further simplify the equation?

Thank you for any assistance...

 

[Pariah]

I just had another look through and have come up with a different break down of the karnaugh maps

3. The Attempt at a Solution
A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'
Factor out A' and D from minterms 1 and 2
A'D(b' + b) (c + c')
= A'D + A'BC'D + ABCD + B'C'D' + AB'C'
Factor out B and D from minterms 2 and 3
bd(a' + a)(c' + c)
= A'D + BD + B'C'D' + AB'C'

This is where I am getting stuck. Is it possible to further simplify the equation or is this the final solution?

or

Can I do the following?

Factor out D from minterms 1 and 2
D(A' + B) + B'C'D' + AB'C'
A'D + B + B'C'D' + AB'C'

 

[piercas]

I would suggest using the Boolean algebra rules and laws to simplify the expression further. For example, you can use the distributive law to add B to the minterm a'c, but it is important to remember that Boolean algebra follows its own set of rules and laws, so it may not always align with traditional algebra.

You can also use the absorption law to simplify the expression. This law states that a term and its complement can be combined to form a simpler expression. In this case, if we have ab'c' and b'c'd', we can combine b'c' with its complement b'c'd' to get b'c'. This would then combine with the term bd to give us just bc.

It is always a good idea to check your simplification using the Karnaugh map or by plugging in different values to ensure that the simplified expression is equivalent to the original one. I hope this helps and good luck with your circuit development!