First there are the answers to your questions:
Number of cells in every Karnaugh map matches the number of all possible input combinations. The way how the map's cells are indexed has to correspond to the truth table. In the following picture are examples of the different sizes of Karnaugh maps, where the visualisation of neighbouring cells is still pretty easy.
As you can see, the point is, that two neighbouring cells differ only in one variable's value, four neighbouring cells differ in two variables' values and so on. That is why you should look for groups that have the size of the 2^n. Map's indexing could look a bit mixed up, but that is for the purpose of displaying all the relationships between each and every row in the truth table.
If you index a K-map, but do not know, which lines are for which variables and in which order they should go in, then you can check it like this:
- index 0 = where not a single variable is true
- index 1 = where only the least significant bit is true (for a truth table ordered abcd that would be the d)
- index 2 = where only the second least significant bit is true (for the same truth table that would be the c)
- index 4 = where only the third least significant bit is true (for the same truth table that would be the b)
- index 8 = where only the fourth least significant bit is true (for the same truth table that would be the a)
As for an example: Here you can see a state diagram of a 01364 sequence generator implemented as a Moore machine. All edges of the machine are labelled by an input value of a reset button.
The machine's desired behaviour and the output values matching the states can be described by this transition table:
state || output (decimal) | reset || next state
-------------------------------------------------
S_0 || 0 | 0 || S_1
|| | 1 || S_0
-------------------------------------------------
S_1 || 1 | 0 || S_2
|| | 1 || S_0
-------------------------------------------------
S_2 || 3 | 0 || S_3
|| | 1 || S_0
-------------------------------------------------
S_3 || 6 | 0 || S_4
|| | 1 || S_0
-------------------------------------------------
S_4 || 4 | 0 || S_0
|| | 1 || S_0
After encoding the states' representation to match the decimal outputs in binary (q_2, q_1 and q_0; d_2, d_1 and d_0), the transition table looks like this:
state || q_2 | q_1 | q_0 | reset || d_2 | d_1 | d_0 || next state
-------------------------------------------------------------------
S_0 || 0 | 0 | 0 | 0 || 0 | 0 | 1 || S_1
|| | | | 1 || 0 | 0 | 0 || S_0
-------------------------------------------------------------------
S_1 || 0 | 0 | 1 | 0 || 0 | 1 | 1 || S_2
|| | | | 1 || 0 | 0 | 0 || S_0
-------------------------------------------------------------------
S_2 || 0 | 1 | 1 | 0 || 1 | 1 | 0 || S_3
|| | | | 1 || 0 | 0 | 0 || S_0
-------------------------------------------------------------------
S_3 || 1 | 1 | 0 | 0 || 1 | 0 | 0 || S_4
|| | | | 1 || 0 | 0 | 0 || S_0
-------------------------------------------------------------------
S_4 || 1 | 0 | 0 | 0 || 0 | 0 | 0 || S_0
|| | | | 1 || 0 | 0 | 0 || S_0
It is useful to study the transition table for every possible combination of inputs, because there are some 'do not care' (x) output values (for the states, that are not present in the sequence), which can be used for minimisation by Karnaugh maps.
index | state || q_2 | q_1 | q_0 | reset || d_2 | d_1 | d_0 || next state
---------------------------------------------------------------------------
0 | S_0 || 0 | 0 | 0 | 0 || 0 | 0 | 1 || S_1
1 | S_0 || 0 | 0 | 0 | 1 || 0 | 0 | 0 || S_0
---------------------------------------------------------------------------
2 | S_1 || 0 | 0 | 1 | 0 || 0 | 1 | 1 || S_2
3 | S_1 || 0 | 0 | 1 | 1 || 0 | 0 | 0 || S_0
---------------------------------------------------------------------------
4 | - || 0 | 1 | 0 | 0 || x | x | x || -
5 | - || 0 | 1 | 0 | 1 || 0 | 0 | 0 || S_0
---------------------------------------------------------------------------
6 | S_2 || 0 | 1 | 1 | 0 || 1 | 1 | 0 || S_3
7 | S_2 || 0 | 1 | 1 | 1 || 0 | 0 | 0 || S_0
---------------------------------------------------------------------------
8 | S_4 || 1 | 0 | 0 | 0 || 0 | 0 | 0 || S_0
9 | S_4 || 1 | 0 | 0 | 1 || 0 | 0 | 0 || S_0
---------------------------------------------------------------------------
10 | - || 1 | 0 | 1 | 0 || x | x | x || -
11 | - || 1 | 0 | 1 | 1 || 0 | 0 | 0 || S_0
---------------------------------------------------------------------------
12 | S_3 || 1 | 1 | 0 | 0 || 1 | 0 | 0 || S_4
13 | S_3 || 1 | 1 | 0 | 1 || 0 | 0 | 0 || S_0
---------------------------------------------------------------------------
14 | - || 1 | 1 | 1 | 0 || x | x | x || -
15 | - || 1 | 1 | 1 | 1 || 0 | 0 | 0 || S_0
Finally you can see, that the functions defining the d_2, d_1 and d_0 (that is the binary encoded state/output matching the number in the 01364 sequence) can be simply marked out in the following K-maps.
f(d_2) = q_1 ⋅ ¬(reset)
f(d_1) = q_0 ⋅ ¬(reset)
f(d_0) = ¬(q_2) ⋅ ¬(q_1) ⋅ ¬(reset)
(All images were generated using latex.)
