A six-variable Karnaugh map has 64 input combinations, so drawing one flat map with six labeled axes is not practical. The standard layout splits the function into four 4×4 tiles. Each tile fixes two variables; the remaining four variables define its rows and columns. Cells at corresponding positions in neighboring tiles can be adjacent, just as cells next to each other inside a tile are adjacent.
This walkthrough solves one complete six-variable function. It shows how to place minterms, form a group that spans all four tiles, add a second overlapping group, and derive a two-term expression. The example is designed to make tile adjacency visible without listing all 64 cell values.
Set up four tiles
Use variable order A, B, C, D, E, F for the minterm indices. In this example, A and B select the tile, C and D label the rows, and E and F label the columns. The row and column labels use Gray order 00, 01, 11, 10. The four tiles are arranged in Gray order for AB as well: 00, 01, 11, 10.
The Gray order applies to tile placement because tiles also represent adjacent input assignments. The tiles AB=00 and AB=01 are neighbors; AB=01 and AB=11 are neighbors; AB=11 and AB=10 are neighbors; and AB=10 wraps around to AB=00. The diagonal pairs 00/11 and 01/10 differ in both A and B, so they are not adjacent.
Within each tile, rows CD and columns EF follow the same Gray sequence. For example, row CD=00 is adjacent to row CD=01 and to row CD=10, but row CD=01 is not directly adjacent to row CD=10. The first and last Gray entries wrap because 00 and 10 differ by one bit. Treat the entire layout as a three-dimensional arrangement: two variables choose the tile, two choose the row, and two choose the column.
Place the minterms
Consider the function:
Split the minterm list into two patterns. The first sixteen indices are 0, 1, 4, 5, 16, 17, 20, 21, 32, 33, 36, 37, 48, 49, 52, and 53. They form the same four-cell block in every tile. The remaining two indices, 8 and 9, are in the AB=00 tile. The table makes the repeated block explicit:
| Tile AB | Minterms in the repeated block | Values that stay fixed |
|---|---|---|
| 00 | 0, 1, 4, 5 | C=0, E=0 |
| 01 | 16, 17, 20, 21 | C=0, E=0 |
| 11 | 48, 49, 52, 53 | C=0, E=0 |
| 10 | 32, 33, 36, 37 | C=0, E=0 |
Within one tile, the repeated block uses rows CD=00 and CD=01, and columns EF=00 and EF=01. Those are adjacent row and column pairs. The same four coordinates are occupied in each of the other three tiles, so the group can span all four tiles.
Now place minterms 8 and 9. They are in tile AB=00, row CD=10, and columns EF=00 and EF=01. They are adjacent horizontally because only F changes between the two indices. They also sit in a row adjacent to CD=00 through wraparound. That means a group can overlap minterms 0 and 1 to include all four cells {0,1,8,9}.
Find the group spanning all tiles
The first group covers sixteen cells: four cells in each of the four tiles. Across those sixteen minterms, A and B vary because every tile is included. D and F vary within the block. C remains 0, and E remains 0. Therefore the product term for the group of ones is C′E′.
This group is valid because its size is 16, a power of two, and it is a rectangle in the full tile layout. It is not enough to look only at a single 4×4 drawing: the four pieces must occupy corresponding coordinates in all four tiles. If one tile were missing a required 1 at that location, the sixteen-cell group would be invalid.
A group of sixteen on a six-variable map removes four variables from its product term, leaving two literals. The group does not assert that A, B, D, or F are always zero or always one. Those variables take both values across the group and therefore disappear. Reading the fixed bits is safer than guessing the term from the apparent shape.
Cover the remaining cells
The first group covers minterms 0, 1, 4, 5, 16, 17, 20, 21, 32, 33, 36, 37, 48, 49, 52, and 53. It does not cover 8 or 9 because those cells have C=1. The pair 8 and 9 alone would produce a five-literal term, A′B′CD′E′, since A=0, B=0, C=1, D=0, and E=0 remain fixed while F changes.
But that pair is not the best candidate. Minterms 0 and 1 are adjacent to 8 and 9 by row wraparound in the AB=00 tile. Including them creates a group of four: {0,1,8,9}. In that group, A=0, B=0, D=0, and E=0 remain fixed; C and F change. The resulting term is A′B′D′E′, which has four literals instead of five.
Using minterms 0 and 1 twice is allowed. They belong to the sixteen-cell group and to the four-cell group. Overlap does not count a minterm twice in the function: Boolean addition is idempotent, so X + X = X. A cell may participate in several groups when that helps cover all required ones with a lower-cost expression.
Derive and verify the expression
Combining the two product terms gives:
The first term covers the sixteen minterms where C=0 and E=0. The second term adds the two remaining minterms, 8 and 9, and overlaps the first term at 0 and 1. Together they cover exactly the eighteen listed minterms. There are no extra cells: every assignment that satisfies either product term appears in the supplied minterm set.
It is useful to check this algebraically. C′E′ is true for any values of A, B, D, and F when C=0 and E=0, which produces the sixteen entries listed in the table. A′B′D′E′ is true for either value of C and F when A=0, B=0, D=0, and E=0. The C=0 half of that four-cell cube is already covered by the first term; the C=1 half contributes exactly 8 and 9.
The expression has two terms. One term cannot cover all eighteen cells as a valid K-map implicant: the sixteen-cell region holds C=0 fixed, while the extra cells require C=1, and the other combinations that would complete a larger cube are not in the function. The two-term cover is therefore necessary. Among two-term covers, using the wrapped group of four avoids the longer isolated pair term.
Check that the cover is complete
For any multi-tile map, check both directions. First, every listed minterm must lie in at least one selected group. Second, every selected group must contain only listed minterms or explicitly declared don't-cares. The first check catches omissions; the second catches accidental inclusion of a zero cell.
Then verify the term literals. Pick one minterm from each group, write it in six-bit binary using the declared order A through F, and confirm that it satisfies the term. Also test a nearby zero assignment that differs in a supposedly fixed variable. If the term still evaluates to 1 for that zero, the group is too large or one literal is missing.
Do not infer a unique answer from a map. Several prime implicant covers may have the same number of terms, and a cover with the same term count may differ in literal count. The exact result depends on the cost rule. For the solver's selection process and the prime-implicant chart, see the minimum-cover section of the main guide.
Avoid common six-variable mistakes
- Using binary order for tiles: the order 00, 01, 11, 10 is needed for one-bit adjacency.
- Joining diagonal tiles: tiles differing in both selector bits are not adjacent, even if a drawing puts them near each other.
- Forgetting within-tile wraparound: first and last rows, and first and last columns, still touch.
- Assuming equal coordinates are always enough: a cross-tile group must contain the corresponding cells in every tile it spans.
- Rejecting overlap: selected groups may share minterms; the goal is a valid cover, not a partition.
- Calling the visually largest group the final answer: identify candidate implicants, then choose a cover for all required cells.
A six-variable map can feel large because there are 64 assignments, but each tile still uses familiar four-variable geometry. The key new task is tracking which tiles differ in one selector bit. Start by sorting the minterms by tile, row, and column; mark each repeated coordinate; then search for groups that combine local adjacency with cross-tile adjacency.
For a smaller introduction to tile adjacency, see the five- and six-variable map guide. If you want to compare this result with an exact cover rather than solve only by eye, enter the same minterm list in the six-variable solver and inspect the selected terms.
Tile-by-tile audit
A useful audit is to record each cube as fixed variables plus free variables, then derive its covered minterms from that description. For the sixteen-cell group, C=0 and E=0 are fixed while A, B, D, and F are free. Enumerating the four AB tiles and the four within-tile combinations of D and F gives four minterms per tile, matching the table. For the second group, A=0, B=0, D=0, and E=0 are fixed while C and F vary; those four combinations are exactly 0, 1, 8, and 9.
This check is especially helpful when a diagram compresses four maps into a single picture. A drawing can make cells appear aligned even when their tile labels differ in two bits. The fixed/free-variable description is independent of page layout and makes it possible to compare a hand map with a solver output. If the expanded cube includes even one minterm absent from the function, the group is invalid unless that row was explicitly declared a don't-care.