Students sometimes compare a K-map answer with a key and find a different expression. That difference does not automatically mean one answer is wrong. A Boolean function can have more than one minimum cover: separate sets of prime implicants may cover the same on-set using the same number of terms and literals. K-map geometry exposes those choices as alternative ways to cover cells.
This guide works through a three-variable function with two distinct minimum SOP expressions. It checks each group, proves that fewer terms are impossible, and shows how to compare answers under a stated cost model. It also covers a subtle point: two expressions can be logically equivalent yet have different literal counts, or have equal term counts but map differently to a gate library. “Both are minimal” is meaningful only after the objective is clear.
The function and its map
Use the input order A, B, C, with A as the most significant bit. Consider F(A,B,C)=Σm(0,2,3,4,5). In binary, the required 1s are 000, 010, 011, 100, and 101. The zero cells are 001, 110, and 111. There are no don't-cares. On a three-variable map with BC arranged in Gray order 00, 01, 11, 10, the five 1s form four possible adjacent pairs. No valid group of four exists because any four-cell rectangle would include at least one of the specified zeros.
The map is compact, but it already contains the reason for non-uniqueness. Minterm 0 can pair with 2 in one direction or with 4 in another. Both choices are equally large and valid. Those alternatives correspond to different fixed literals: the first pair holds A and C at zero, while the second holds B and C at zero. The other two pairs, {2,3} and {4,5}, cover the two cells that lie next to their partners. Each pair is prime; none can be expanded without swallowing a zero.
| Pair | Cells in binary | Fixed values | SOP term |
|---|---|---|---|
| {2,3} | 010, 011 | A=0, B=1 | A′B |
| {0,2} | 000, 010 | A=0, C=0 | A′C′ |
| {4,5} | 100, 101 | A=1, B=0 | AB′ |
| {0,4} | 000, 100 | B=0, C=0 | B′C′ |
Identify the prime implicants
Begin with the cells that are hard to cover. Minterm 3 (011) is adjacent to minterm 2 (010) among the on-set; its other one-bit neighbors are zeros. Therefore every valid implicant covering 3 must use pair {2,3}, which gives A′B. Minterm 5 (101) has only one adjacent on-set neighbor, 4 (100), so every cover must include {4,5}, which gives AB′. These two terms cover 2, 3, 4, and 5, but not minterm 0.
There are now two ways to cover 0 without adding a zero. Pair it with 2, producing A′C′, or pair it with 4, producing B′C′. Either way, the extra pair overlaps a minterm already covered by a forced term. That overlap is legitimate and useful: the purpose is to cover the whole on-set with implicants, not to assign each cell to exactly one group. The two possible additional terms are equally large and each adds two literals.
Two minimum covers
Select {2,3}, {4,5}, and {0,2}. The resulting expression is:
Its coverage is direct. A′B covers assignments 010 and 011, or minterms 2 and 3. AB′ covers 100 and 101, or minterms 4 and 5. A′C′ covers 000 and 010, or minterms 0 and 2. The union is {0,2,3,4,5}; none of the three terms is true for 1, 6, or 7. So F₁ matches the specified function.
Instead select {2,3}, {4,5}, and {0,4}. This gives:
The first two terms are unchanged. B′C′ covers 000 and 100, or minterms 0 and 4. Together, the terms again cover exactly {0,2,3,4,5}. The two formulas are not textually identical, but their truth tables agree for all eight assignments. The choice is between two alternative ways to extend coverage to minterm 0.
Both expressions have three product terms and six literal occurrences. Under the common lexicographic SOP objective—first minimize the term count, then minimize literals—the answers tie. They also have the same maximum product width, two literals. A solver that returns one cover may use a different deterministic tie-breaker than a textbook, another solver, or a human map reader.
Prove the term count is minimum
Showing that two formulas work is not enough to prove they are minimum. Use the forced cells. Minterm 3 requires the implicant {2,3}; minterm 5 requires {4,5}. These are two different groups, so any cover needs at least two terms. But those groups do not contain minterm 0. A third implicant is required to cover 0. Therefore no two-term cover is possible, and a cover with three terms is minimum by term count.
The literal count is also easy to compare. Each of the three required prime implicants is a pair on a three-variable map, so each term has two literals. Any three-term prime cover thus has six literal occurrences. If a valid larger group existed, it could change that comparison, but the map’s zeros prevent every size-four group. As a result, the two displayed covers are also minimum by literal count among minimum-term SOP covers.
This kind of proof scales conceptually. Identify cells that force particular prime implicants; then check what remains after those forced rows are selected. If alternatives remain, compare their costs. For larger charts, a prime-implicant table makes these relationships explicit. The article on essential versus nonessential implicants explains the chart terminology, and Petrick’s method can enumerate exact covers when the residual choices are more complex.
Check equivalence independently
When an answer differs from a reference, do not decide by visual resemblance. First expand each product term into its covered assignments. For F₁, the sets are {2,3}, {4,5}, and {0,2}. Their union is the on-set and their union contains no zero. For F₂, the sets are {2,3}, {4,5}, and {0,4}; the same two checks pass. This set-based audit works even if your algebraic notation uses overbars, primes, or a programming language’s complement operator.
A truth-table check is another independent method. Evaluate both formulas for A,B,C from 000 through 111. A′B is true at 010 and 011; AB′ at 100 and 101; A′C′ at 000 and 010; B′C′ at 000 and 100. Taking the OR yields 1 at 000, 010, 011, 100, and 101, and 0 at the other inputs. Compare that column to the original truth table, not merely to another simplified expression.
For larger inputs, a symbolic equivalence check can compare the exclusive-OR of the two functions: if F₁ XOR F₂ is 0 for every assignment, they are equivalent. A solver can also enumerate all input rows. That verification is especially useful when translating between minterm index conventions or when a complement symbol is visually ambiguous. The page on truth tables and minimal expressions shows the full row-by-row workflow.
Minimum depends on cost
Term count and literal count are classroom-friendly measures, but neither is a universal measure of hardware quality. A library may implement A′B + AB′ as an XOR gate plus an inverter arrangement, or it may have a compact complex cell that favors a factored form. Output polarity, available gate fan-in, delay targets, and fanout can all change the best implementation. A minimum SOP is not automatically a minimum-area, minimum-delay, or lowest-power circuit.
For this example, the algebraic identity A′B + AB′ = A XOR B lets both formulas be rewritten as (A XOR B) + A′C′ or (A XOR B) + B′C′. That observation may help a library with XOR cells, but it does not make one cover universally better. The two alternatives differ in which signal appears in the final correction term. If one input arrives late or has limited drive, that practical detail may matter more than the equal literal count.
Hazard behavior is another independent criterion. If an output must remain stable during a particular input transition, a minimum cover can have a static hazard because neighboring 1-cells are covered by separate implicants. Adding a redundant consensus implicant can bridge that transition. The result then has more terms than the Boolean minimum but may satisfy a timing-related design requirement. See the static-hazard and consensus-term example before treating redundancy as an error.
How to present alternate answers
On an assignment, write down the convention and show the selected groups. Label the minterm set or the 1-cells, draw every group including overlaps, and derive each term from the variables that remain constant. Then verify coverage. If your answer differs from the key, this evidence lets a grader see that you minimized the same function rather than made a transcription error.
If asked for “the simplified expression,” a single valid minimum answer is usually enough unless the problem requests all minimal forms. When a question asks for a unique answer, it may be using a specific tie-break rule from class; state your cost assumptions and ask what convention is expected. A reference answer is often one representative cover, not a claim that every other cover is incorrect. Course notes sometimes explicitly mention multiple equally minimal solutions when a chart contains alternative prime implicants.
For a solver comparison, record more than the displayed expression. Compare the truth table, term count, literal count, group coverage, and any tie-breaking rule. Two different strings can encode the same Boolean function and share the same minimum cost. Conversely, similar-looking strings can disagree on one assignment. Equivalence and optimality are separate checks: verify the function first, then verify the cost.
Practice and answers
Practice 1: H(A,B,C)=Σm(0,1,2,3,5). The group of four covering 0,1,2,3 gives A′. Minterm 5 can pair with 1, producing B′C. So H=A′+B′C. The first group is forced if you want a one-term coverage of four cells; the remaining 5 is not already covered. Check the other possible implicants before claiming there is another equally minimum answer.
Practice 2: J(A,B,C)=Σm(0,1,4,5,6,7). The pair {0,1} gives A′B′, while {4,5,6,7} forms a group of four and gives A. Thus J=A+A′B′, which reduces by absorption to A+B′. This illustrates a different issue: two valid covers can have different apparent term counts before algebraic reduction. A K-map prime cover should use maximal groups, and Boolean identities can reveal an even simpler notation.
Practice 3: If a reference answer has two product terms but your expression has three, check whether your groups form a complete cover and whether a larger wraparound group was missed. If both answers have the same term count but different literals, compare all required and zero assignments and state the measure used. Never reject an answer solely because its symbols differ from the key.
Multiple minima are a normal property of Boolean minimization, not a loophole in the rules. Draw accurate groups, prove coverage, and make the cost criterion explicit. Once those steps are documented, different but equally minimum answers are easy to recognize and explain.
Tie-breaking in a tool is a policy
When two covers have the same number of terms and literal count, an implementation or solver still has to return one result. It may prefer a particular implicant ordering, variable order, stable text format, or gate pattern. Those tie-breaks make output repeatable; they do not prove the other cover is invalid. A different tool version may choose a different representative while preserving the same minimum cost.
When reproducibility matters, record the truth table, cost criterion, and tool settings alongside the expression. For a human explanation, show one valid minimum cover and note that alternatives exist if the chart has cyclic choices. This is more informative than forcing a canonical-looking string that the Boolean function does not uniquely determine.