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OCR H446 1.4.3 Three-variable Karnaugh maps
Part 6 of 11 · H446 1.4.3 · Boolean algebra
Eight input states, Gray-code column order and wraparound adjacency are what make three-variable maps harder than two, and OCR H446 1.4.3 expects all three. This worksheet decodes a cell before anything is placed in it, then builds towards overlapping and edge-wrapping groups.
Students will:
- combine a column heading and a row heading to recover a complete three-variable state
- explain why the headings run 00, 01, 11, 10 rather than in binary counting order
- place an expression on a map term by term, treating absent variables correctly
- form the largest legal groups, including a group that wraps across the edge
- justify why a single 1 may belong to more than one group
Inside: 8 explanation cells, 5 multiple-choice questions, 3 fill-in-the-blanks cells and 1 written answer. 30 marks, about 45 to 55 minutes.
Series: H446 1.4.3 · Boolean algebra, part 6 of 11.
Shared by Coding PathwayVerified teacher
- 17 cells
- About 45 minutes
- CC BY-SA 4.0
- Shared 31 Aug 2026
- Updated 15 Sept 2026
Preview
The whole resource, exactly as a class sees it. Answers and marking are held back.
Three-variable Karnaugh maps
A three-variable truth table has eight input combinations, so its Karnaugh map has eight boxes. Two variables label the columns and the third labels the rows.
You should already know: how to evaluate a Boolean expression and how a group removes variables that change.
Worked model: enter outputs directly into the grid
Use AB across the columns and C down the rows. The column headings follow 00, 01, 11, 10 so neighbouring boxes differ in one input.
For F = A ∧ ¬C, the map is:
| C ↓ / AB → | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 |
A is 1 in columns 11 and 10. ¬C is 1 in row 0. Only their intersections receive output 1. There is no need to write a full input code inside every box.
A truth-table row has A=0, B=1 and C=1. Where is its output placed?
- Acolumn AB=11, row C=0
- Bcolumn AB=01, row C=0
- Ccolumn AB=01, row C=1
- Dcolumn AB=10, row C=1
Why the headings use Gray-code order
Use 00, 01, 11, 10, not 00, 01, 10, 11. Each move to a neighbouring column changes only one variable. The first and last columns are neighbours too: 00 and 10 differ only in A.
That one-change rule is what allows a group to remove a variable. If two variables changed at once, the group would not show that one input is irrelevant.
Why is the column order 00, 01, 11, 10 used?
- AIt places all 1s on the right
- BIt is normal binary counting
- CIt prevents groups from overlapping
- DIt makes every adjacent pair differ in one variable
Worked placement: split at OR, then test each term
For (¬C ∧ B) ∨ (A ∧ B), treat the OR-separated product terms independently.
- For ¬C ∧ B, use row C=0 and the columns in which B=1: AB=01 and AB=11.
- For A ∧ B, use column AB=11 in both C rows because C is absent from the term.
- The shared box still contains the single output 1.
- Any box not made true by either term receives 0.
When a variable is absent from a product term, both of its possible values satisfy that term.
| C ↓ / AB → | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | C0 AB00 | C0 AB01 | C0 AB11 | C0 AB10 |
| 1 | C1 AB00 | C1 AB01 | C1 AB11 | C1 AB10 |
On paper, draw the largest groups for the completed map. Derive the simplified term from each group and give the final expression.
One horizontal pair and one vertical pair are needed; they overlap at one cell.
Students type their answer here.
Read a different map
| C / AB | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 |
All four 1s form one rectangle. Across it, A stays 1; B and C both change.
What is the simplest expression for this map?
- AA
- BB
- CA ∧ B
- DA ∨ C
Independent placement check
For A ∨ (¬B ∧ C), first fill every box in the columns where A=1. Then add the boxes where B=0 and C=1; A is absent from the second term.
| C ↓ / AB → | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | entry 1 | entry 2 | entry 3 | entry 4 |
| 1 | entry 5 | entry 6 | entry 7 | entry 8 |
After optimal grouping, which expression is read from the independent map?
- AA ∧ ¬B ∧ C
- BA ∨ (¬B ∧ C)
- CA ∧ C
- D¬A ∨ B
Why may one 1 appear in more than one Karnaugh group?
- AEvery 1 must be counted twice
- BGroups must cross diagonally
- COverlap is allowed when it enables larger legal groups and a simpler expression
- DIt changes the 1 into a don't-care value
Closed-book checkpoint
Retrieve the layout logic, not the completed example.
Review your work
Check that you have used the heading order 00, 01, 11, 10 and considered both edge adjacencies.