# Minimizing using a Karnaugh map when given as subscripts F4,2655

I have to minimize the expression using minterms and a Karnaugh map:

$F_{4,2655}$

How might I get this expression I am given into a form much like a typical boolean algebra minification question? I do not understand the given notation.

From a classmate they commented that the notation means the following:

2655 would be the sum of all the minterm values of function F, expressed in decimal (which when converted to binary gives the Boolean representation that corresponds to truth table).

So...

$2655_{10} = 0000 1010 0101 1111_2$

Then I need to do something with this from here, but I am not sure what. I included an extra 4 bits as I believe the $2^4$ in the question represents the amount of bits. Now having this somewhat converted I am not sure where to go from here.

• @HenningMakholm en.wikipedia.org/wiki/Karnaugh_map k-map is just the shortform – ComputerLocus Oct 12 '15 at 15:30
• But what does $F_{4,2655}$ mean? – hmakholm left over Monica Oct 12 '15 at 15:34
• @HenningMakholm That is exactly what I am here for. I have no idea what it means yet have to solve that. – ComputerLocus Oct 12 '15 at 15:35
• Then I think you should reformulate your question such that that is what you actually ask -- the stuff about Karnaugh maps then seems to be more context for the real question ("the notation $F_{4,2655}$ seems to denote a particular truth function, but which?") than an actual pat of the question. – hmakholm left over Monica Oct 12 '15 at 15:36
• @HenningMakholm I would be more than happy to clarify that in the question, however where exactly needs altering? My last sentence states what I am looking for. – ComputerLocus Oct 12 '15 at 15:37

I assume that F4,2655 is a shorthand notation for the Boolean expression with four inputs which has a truthtable with the binary equivalent of 2655 as output column.

The truthtable looks as follows:

A B C D F
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1
0 1 0 1 0
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 0
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1


Translated into a Karnaugh map

             AB
00  01  11  10
+---+---+---+---+
00 | 0 | 1 | 1 | 0 |
+---+---+---+---+
01 | 0 | 0 | 1 | 1 |
CD    +---+---+---+---+
11 | 0 | 0 | 1 | 1 |
+---+---+---+---+
10 | 0 | 1 | 1 | 0 |
+---+---+---+---+


Resulting minimized expression:

A & D  +  B & !D


Reversing the order of bits in the output column results in:

Truthtable:

A B C D  F
0 0 0 0  1
0 0 0 1  1
0 0 1 0  1
0 0 1 1  1
0 1 0 0  1
0 1 0 1  0
0 1 1 0  1
0 1 1 1  0
1 0 0 0  0
1 0 0 1  1
1 0 1 0  0
1 0 1 1  1
1 1 0 0  0
1 1 0 1  0
1 1 1 0  0
1 1 1 1  0


Karnaugh map:

             AB
00  01  11  10
+---+---+---+---+
00 | 1 | 1 | 0 | 0 |
+---+---+---+---+
01 | 1 | 0 | 0 | 1 |
CD    +---+---+---+---+
11 | 1 | 0 | 0 | 1 |
+---+---+---+---+
10 | 1 | 1 | 0 | 0 |
+---+---+---+---+


Minimized expression:

!B & D + !A & !D


So, both cases yield a somewhat similar expression

• I think your answer is incorrect. I believe the extra zero's from the 16 bit number should end up on the bottom and you have it the other way around. – ComputerLocus Oct 13 '15 at 1:05
• Yes, that might be. It is a matter of convention, if you are looking on the output column from the left or from the right. The general approach should be ok. – Axel Kemper Oct 13 '15 at 7:25