Key takeaways
- Let A stand for 'A is a knight'. If A says S, the puzzle gives you the biconditional A ↔ S: A is a knight exactly when S is true.
- Solve by cases: assume a speaker is a knight, follow the consequences, and discard any assumption that leads to a contradiction.
- For two or three people, a truth table with one row per assignment lists every possibility; keep the rows where every speaker's biconditional holds.
- Nobody can say 'I am a knave': a knight would be lying and a knave would be telling the truth.
- Always check the final answer against every statement, and check that it is the only consistent assignment.
On a certain island, every inhabitant is either a knight, who always tells the truth, or a knave, who always lies. You meet some of them, they make statements, and you must work out who is which. Puzzles of this kind turn propositional logic into a game, and textbooks use them as exercises: Levin's Discrete Mathematics: An Open Introduction dresses them up as trolls guarding a bridge, each either a knight or a knave, who will not let you pass until you identify them.
The translation that solves everything
Use a letter for each person, standing for the statement "this person is a knight". So A means "A is a knight" and ¬A means "A is a knave".
Now the key observation. If A says S, then:
- if A is a knight, S is true;
- if A is a knave, S is false.
Both together say exactly A ↔ S: A is a knight if and only if S is true. Every statement in every puzzle becomes a biconditional, and the puzzle becomes a question about which truth assignments make all of them true at once.
The impossible sentence
What if A says "I am a knave"? The translation is A ↔ ¬A. A knight would be lying; a knave would be telling the truth. A ↔ ¬A is false in every row of its truth table, a contradiction in the sense of tautology, contradiction or contingency? So nobody on the island can say it, and a puzzle in which someone does has no solution.
Puzzle 1: "At least one of us is a knave"
A says: "At least one of us is a knave," referring to A and B. B says nothing.
Translation: A ↔ (¬A ∨ ¬B).
- Suppose A is a knave. Then the statement must be false, so neither is a knave, so A is a knight. Contradiction.
- So A is a knight. Then the statement is true: at least one of A, B is a knave. A is not, so B is.
Answer: A is a knight, B is a knave. Check: A's statement, "at least one of us is a knave", is true, and A is a knight. ✓
Puzzle 2: two statements that refer to each other
A says: "B is a knave." B says: "A and I are the same type."
Translation: A ↔ ¬B, and B ↔ (A ↔ B).
- Suppose B is a knight. Then B's statement is true, so A is also a knight. But then A's statement "B is a knave" is true, while B is a knight. Contradiction.
- So B is a knave. Then B's statement is false, so A and B are different types, which makes A a knight. A's statement "B is a knave" is then true, consistent with A being a knight. ✓
Answer: A is a knight, B is a knave.
Puzzle 3: three people, by truth table
A says: "B is a knight." B says: "A and C are different types." C says: "B is a knave."
Translation: A ↔ B; B ↔ (A ⊕ C), where ⊕ means exactly one is true; C ↔ ¬B.
With three people there are 2³ = 8 possible assignments. Build a truth table with a column for each biconditional and keep the rows where all three hold. OpenStax's Contemporary Mathematics uses the same idea of listing every combination of truth values for compound statements.
| A | B | C | A ↔ B | B ↔ (A ⊕ C) | C ↔ ¬B | All hold? |
|---|---|---|---|---|---|---|
| T | T | T | T | F | F | No |
| T | T | F | T | T | T | Yes |
| T | F | T | F | T | T | No |
| T | F | F | F | F | F | No |
| F | T | T | F | T | F | No |
| F | T | F | F | F | T | No |
| F | F | T | T | F | T | No |
| F | F | F | T | T | F | No |
Exactly one row survives: A and B are knights, C is a knave. Check in words: A says B is a knight, true. B says A and C differ, true, since A is a knight and C a knave. C says B is a knave, false, as a knave's statements must be. ✓
The table also proves the solution is unique, which reasoning by cases can miss if you stop at the first consistent answer.
The two guards
A classic variation: two guards stand at a fork, one road leading to the village. One guard is a knight and one a knave, and you do not know which. You may ask one question of one guard.
Ask either guard: "If I asked you whether the left road leads to the village, would you say yes?"
- A knight answers the inner question truthfully and reports that answer truthfully.
- A knave would lie to the inner question, and then lie about what he would say, so the two lies cancel.
Either way, "yes" means the left road is correct. The question works because negating twice gives back the original, the logic law ¬¬p ≡ p.
A method for any puzzle
- Give each person a variable meaning "is a knight".
- Translate each statement S by person X into X ↔ S, rewriting "same type" as ↔ and "different types" as ⊕.
- For one or two people, reason by cases, using proof by contradiction to eliminate assumptions.
- For three or more, build the truth table or eliminate rows systematically.
- Check the answer against every statement, and confirm it is the only one.
If you enjoy these, logic grid puzzles train the same systematic elimination, and our guide to deductive reasoning collects everyday examples. IQ Test & Brain Training Games has logic games covering conditional reasoning and deduction, and Discrete Math: Study & Quiz covers the propositional logic behind these puzzles in its first unit, with a Truth Table Generator for checking tables like the one above. Both are among our education apps; the discrete math hub has the rest of the logic series.
Good to know
Frequently asked questions
How do you solve a knights and knaves puzzle?
Why can't anyone say 'I am a knave'?
What question finds the right road if one guard lies and one tells the truth?
Can a knights and knaves puzzle have more than one answer?
Sources
- Mathematical Statements (Discrete Mathematics: An Open Introduction, 3rd edition, Oscar Levin) (opens in a new tab)Open Math Books — accessed
- Propositional Logic (Discrete Mathematics: An Open Introduction, 3rd edition, Oscar Levin) (opens in a new tab)Open Math Books — accessed
- 2.3 Constructing Truth Tables (Contemporary Mathematics) (opens in a new tab)OpenStax — accessed
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