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Larger 4 variable Karnaugh maps
Knowing how to generate Gray code should allow us to build larger maps. Actually, all we need to do is look at the left to right sequence across the top of the 3- variable map, and copy it down the left side of the 4 - variable map. See below. The following four variable Karnaugh maps illustrate reduction of Boolean expressions too tedious for Boolean algebra. Reductions could be done with Boolean algebra. However, the Karnaugh map is faster and easier, especially if there are many logic reductions to do.
The above Boolean expression has seven product terms. They are mapped top to bottom and left to right on the K-map above. For example, the first P- term A'B'CD is first row 3rd cell, corresponding to map location A=0, B=0, C=1, D=1. The other product terms are placed in a similar manner. Encircling the largest groups possible, two groups of four are shown above. The dashed horizontal group corresponds the the simplified product term AB. The vertical group corresponds to Boolean CD. Since there are two groups, there will be two product terms in the Sum-Of-Products result of Out=AB+CD. Fold up the corners of the map below like it is a napkin to make the four cells physically adjacent. The four cells above are a group of four because they all have the Boolean variables B' and D' in common. In other words, B=0 for the four cells, and D=0 for the four cells. The other variables (A, B) are 0 in some cases, 1 in other cases with
The six product terms of four Boolean variables map in the usual manner above as single cells. The three Boolean variable terms (three each) map as cell pairs, which is shown above. Note that we are mapping p-terms into the K-map, not pulling them out at this point. For the simplification, we form two groups of eight. Cells in the corners are shared with both groups. This is fine. In fact, this leads to a better solution than forming a group of eight and a group of four without sharing any cells. Final Solution is Out=B'+D' Below we map the unsimplified Boolean expression to the Karnaugh map.
Above, three of the cells form into a groups of two cells. A fourth cell cannot be combined with anything, which often happens in "real world" problems. In this case, the Boolean p-term ABCD is unchanged in the simplification process. Result: Out= B'C'D'+A'B'D'+ABCD Often times there is more than one minimum cost solution to a simplification problem. Such is the case illustrated below.
On a cautionary note, do not attempt to form groups of three. Groupings must be powers of 2, that is, 1, 2, 4, 8 ... Below we have another example of two possible minimal cost solutions. Start by forming a couple of groups of four after mapping the cells. The two solutions depend on whether the single remaining cell is grouped with the first or the second group of four as a group of two cells. That cell either comes out as either ABC' or ABD , your choice. Either way, this cell is covered by either Boolean product term. Final results are shown above. Below we have an example of a simplification using the Karnaugh map at left or Boolean algebra at right. Plot C' on the map as the area of all cells covered by address C=0 , the 8-cells on the left of the map. Then, plot the single ABCD cell.
That single cell forms a group of 2-cell as shown, which simplifies to P-term ABD , for an end result of Out = C' + ABD. This (above) is a rare example of a four variable problem that can be reduced with Boolean algebra without a lot of work, assuming that you remember the theorems.
Source: http://www.allaboutcircuits.com/vol_4/chpt_8/7.html
