Detecting XOR in Karnaugh Maps - xor

Question

I got the following Karnaugh Maps but I am still having problems working out the expression for XOR from each table.
Table 1
-------
WZ
00 01 11 10
-----------------------
00 | | | | 1 |
-----------------------
01 | 1 | | | |
-----------------------
XY 11 | | | | 1 |
-----------------------
10 | 1 | | | |
-----------------------
Table 2
-------
WZ
00 01 11 10
-----------------------
00 | | 1 | | |
-----------------------
01 | | | 1 | |
-----------------------
XY 11 | | 1 | | |
-----------------------
10 | | | 1 | |
-----------------------
It is XORs, but how can I easily deduce the XOR expressions?

Answer 1

I would not dismiss the variable z from the expression, because I think, the expression ¬z·(¬x·y·¬w + ¬x·w·¬y + ¬y·¬w·x + w·y·x) is not equal to (¬x·y·¬w + ¬x·w·¬y + ¬y·¬w·x + w·y·x). That would mean, that the K-map contains four doubles of ones, but there is only four singles.
I would rather find the expression in the K-map and then use the laws of Boolean algebra.
For the first table:
¬x·¬y·w·¬z + ¬x·y·¬w·¬z + x·y·w·¬z + x·¬y·¬w·¬z
¬z·((¬x + ¬y + w)·(¬x + y + ¬w)·(x + y + w)·(x + ¬y + ¬w)) //distributivity
¬z· (¬x + ¬y + w)·(¬x + y + ¬w)·(x + y + w)·(x + ¬y + ¬w) //relaxed syntax
¬z· (¬x·¬x + ¬x·y + ¬x·¬w + ¬y·¬x + ¬y·y + ¬y·¬w + w·¬x + w·y + w·¬w)·
(x·x + x·¬y + x·¬w + y·x + y·¬y + y·¬w + w·x + w·¬y + w·¬w) //distributivity
Because of the laws of
idempotence (e.g.: ¬x·¬x=¬x),
absorption (e.g.:¬x + ¬x·y=¬x)
and complementation (e.g.: ¬x·x=0)
the expression is equivalent to:
¬z· (¬x + 0 + ¬y·¬w + w·y + 0)·
( x + + 0 + y·¬w + + w·¬y + 0 )
¬z· (¬x + ¬y·¬w + w·y)·(x + y·¬w + w·¬y) //just formatted
¬z· (¬x·x + ¬x·y·¬w + ¬x·w·¬y
+ ¬y·¬w·x + ¬y·¬w·y·¬w + ¬y·¬w·w·¬y
+ w·y·x + w·y·y·¬w + w·y·w·¬y) //distributivity
¬z· ( 0 + ¬x·y·¬w + ¬x·w·¬y
+ ¬y·¬w·x + 0 + 0
+ w·y·x + 0 + 0 ) //using the three laws↑ again
¬z· (¬x·y·¬w + ¬x·w·¬y + ¬y·¬w·x + w·y·x) //how the 3-input XOR is defined
¬z· (x xor y xor w)
For the second table:
¬x·¬y·¬w·z + ¬x·y·w·z + x·y·¬w·z + x·¬y·w·z
z·((¬x + ¬y + ¬w)·(¬x + y + w)·(x + y + ¬w)·(x + ¬y + w)) //distributivity
z· (¬x + ¬y + ¬w)·(¬x + y + w)·(x + y + ¬w)·(x + ¬y + w) //relaxed syntax
z· (¬x·¬x + ¬x·y + ¬x·w + ¬y·¬x + ¬y·y + ¬y·w + ¬w·¬x + ¬w·y + ¬w·w)·
(x·x + x·¬y + x·w + y·x + y·¬y + y·w + ¬w·x + ¬w·¬y + ¬w·w) //distributivity
z· ( ¬x + + 0 + ¬y·w + + ¬w·y + 0 )·
( x + + 0 + y·w + + ¬w·¬y + 0 )
z· (¬x + ¬y·w + ¬w·y)·(x + y·w + ¬w·¬y) //just formatted
z· (¬x·x + ¬x·y·w + ¬x·¬w·¬y
+ ¬y·w·x + ¬y·w·y·w + ¬y·w·¬w·¬y
+ ¬w·y·x + ¬w·y·y·w + ¬w·y·¬w·¬y) //distributivity
z· ( 0 + ¬x·y·w + ¬x·¬w·¬y
+ ¬y·w·x + 0 + 0
+ ¬w·y·x + 0 + 0) //using the three laws↑ again
z· (¬x·y·w + ¬x·¬w·¬y + ¬y·w·x + ¬w·y·x) //how the 3-input XNOR is defined
z· (x xnor y xnor w)

Answer 2

The first table contains an Xor expression :
`First table`
w
\ wz ___________
xy \-----------------------+
| | | | 1 |
+-----+-----+-----+-----+
| 1 | | | | |
+-----+-----+-----+-----+ | y
| | | | | 1 | |
x | +-----+-----+-----+-----+
| | 1 | | | |
+-----------------------+
___________
z
as you could see the middle of the table (Z area) is fake.
that is, the table function is :
F(Table1) = w'x'yz' + wx'y'z' + w'xy'z' + wxyz'
in binary form you could see a zero column :
F(Table) = 0010 eliminating Z F(xor)= 001
0100 ---------------\ 010
1110 ---------------/ 111
1000 100
^--> fake
and the final table must be something like this :
`simplified xor table`
w
\ w 0 __1__
xy \-----------+
00 | | 1 |
+-----+-----+
01 | 1 | | |
+-----+-----+ | y And " F = wy' + w'y " is an Xor only
|10 | 1 | | | between 2 variables, right?
x | +-----+-----+
|11 | | 1 |
+-----------+
The second table just contains an Xnor expression of the first one :
`Second Table`
F(Table2) = w'xyz + wxy'z + w'x'y'z + wx'yz
w
\ wz ___________
xy \-----------------------+ negation of table 2 is table 1 and vise versa
| | 1 | | | F(Table2) = 1101 F(Table2)'= F(Table1) = 0010
+-----+-----+-----+-----+ 1011 0100
| | | 1 | | | 0001 1110
+-----+-----+-----+-----+ | y 0111 1000
| | | 1 | | | | ^--> fake ^
x | +-----+-----+-----+-----+
| | | | 1 | |
+-----------------------+
^ ___________ ^
^ z ^
^ ^
^--------z'-------^
the final table is:
w
\ w 0 __1__
xy \-----------+
00 | 1 | |
+-----+-----+
01 | | 1 | |
+-----+-----+ | y And " F = w'y' + wy " is an Xnor
|10 | | 1 | |
x | +-----+-----+
|11 | 1 | |
+-----------+
Always remember the tables that contain the zigzag pattern
are either an Xor or Xnor expression.

Answer 3

Just put a copy of this map on the right hand side of it (or left, no difference) and then choose two tilted cubes.
Now, we write the simplified function for both of them:
(A = 1) (AND) (B=0 when C=1 and B=1 when C=0)
(OR)
(A = 0) (AND) (B=0 when C=0 and B=1 when C=1)
that finaly gives this:
(A AND (B XOR C)) OR (¬A AND (B XNOR C))

Answer 4

Basic rule for xor is that it gives 1 when odd number of input are 1.
So in KMAP just see if 1 is present in all the odd number of 1's.
Like WXYZ ( 0010, 1110 etc) if all gives 1 than there is a XOR in kmap.