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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 →00011110
00011
10000

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.

Multiple choice1 mark

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.

Multiple choice1 mark

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.

  1. For ¬C ∧ B, use row C=0 and the columns in which B=1: AB=01 and AB=11.
  2. For A ∧ B, use column AB=11 in both C rows because C is absent from the term.
  3. The shared box still contains the single output 1.
  4. 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.

Fill in the blanks8 marks
Complete the map for (¬C ∧ B) ∨ (A ∧ B).
C ↓ / AB →00011110
0C0 AB00C0 AB01C0 AB11C0 AB10
1C1 AB00C1 AB01C1 AB11C1 AB10
Written answer5 marks

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 / AB00011110
00011
10011

All four 1s form one rectangle. Across it, A stays 1; B and C both change.

Multiple choice1 mark

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.

Fill in the blanks8 marks
Complete the map for A ∨ (¬B ∧ C).
C ↓ / AB →00011110
0entry 1entry 2entry 3entry 4
1entry 5entry 6entry 7entry 8
Multiple choice1 mark

After optimal grouping, which expression is read from the independent map?

  • AA ∧ ¬B ∧ C
  • BA ∨ (¬B ∧ C)
  • CA ∧ C
  • D¬A ∨ B
Multiple choice1 mark

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.

Fill in the blanks4 marks
A three-variable map has entry 1 cells. Its two-bit headings use entry 2 code order so neighbours differ in one bit. The first and last columns are also entry 3. Repeated placement leaves an occupied cell entry 4.

Review your work

Check that you have used the heading order 00, 01, 11, 10 and considered both edge adjacencies.