Truth-tellers, liars and binary logic in CAT DILR

In a truth-teller and liar set, each person is one of two types, and every statement they make tells you something about the types of the others. With five people there are 2 × 2 × 2 × 2 × 2 = 32 possible type patterns. The skill is to avoid testing all 32, by spotting which speakers are linked and testing only the cases that matter.

CAT uses these as "binary logic" sets, sometimes with a third type who alternates between truth and lies. The logic is strict, which is good news: a correct deduction is certain, and a contradiction is final. This guide shows how to link speakers, how to handle statements about counts, and how to deal with alternators, using three original sets, each checked by listing every possible pattern.

The rules of the game

Two translation rules do most of the work.

• "X is a liar", said by Y. Y and X are of opposite types. If Y is truthful, X lies; if Y lies, X is truthful.
• "X is a truth-teller", said by Y. Y and X are of the same type.

A statement with "and" is false if either part is false. A statement with "or" is true if either part is true.

The case method

1. Link the speakers using the rules above.
2. Pick the pivot: the person whose type decides the most other types, often the one who makes a statement about a count ("exactly two of us lie").
3. Assume the pivot is truthful. Follow every consequence.
4. At a contradiction, the assumption is false. Switch it.
5. Check the survivor against every statement, not only the ones you used.

Set 1: three speakers

Each of A, B and C is either a truth-teller or a liar.

• A says: "B is a liar."
• B says: "C is a liar."
• C says: "A and B are both liars."

Links. A and B are opposite (A's statement). B and C are opposite (B's statement). So A and C are the same type, and B is the other.

Test C. If C is truthful, A and B are both liars. But A and C are the same type, so A would be truthful. Contradiction. C is a liar, so A is a liar and B is truthful.

Check. A: "B is a liar" is false, and A lies. B: "C is a liar" is true, and B is truthful. C: "A and B are both liars" is false because B is truthful, and C lies. All consistent.

Set 2: five speakers with a counting statement

Each of V, W, X, Y and Z is either a truth-teller or a liar.

• V says: "Exactly two of us five are liars."
• W says: "V is a liar."
• X says: "W and Y are of the same type."
• Y says: "Z is a truth-teller."
• Z says: "X is a liar, or V is a truth-teller."

Links. V and W are opposite. Y and Z are the same type. V is the pivot, because V's statement is about everyone.

Case 1: V is truthful. Then exactly two are liars, and W (opposite of V) is one of them. So exactly one of X, Y, Z lies. Y and Z are the same type, so they cannot contain exactly one liar between them; both must be truthful, and X is the liar.

• X's statement: W is a liar, Y is truthful, so "same type" is false. X lies. Consistent.
• Z's statement: "V is a truth-teller" is true, so the "or" is true. Z is truthful. Consistent.

Case 2: V is a liar. Then W is truthful.

• X's statement "W and Y are of the same type" is true exactly when Y is truthful. So X has the same type as Y, and therefore as Z.
• Z's statement "X is a liar, or V is a truth-teller": V is a liar, so it is true exactly when X is a liar. If Z is truthful, X is a liar; but X has Z's type. If Z lies, X must be truthful; again X has Z's type. Contradiction either way.

Case 2 fails. The unique solution:

Q1. How many liars are there? 2 (W and X), which is exactly what V said.

Q2. Which two people are certain to be of opposite types, from their statements alone, before any case is tested? V and W.

Set 3: a truth-teller, a liar and an alternator

Among P, Q and R, one is a truth-teller, one is a liar and one is an alternator. Each makes two statements. The alternator makes one true and one false statement.

• P: "I am not the liar." "R is the alternator."
• Q: "P is the truth-teller." "I am the alternator."
• R: "Q is the liar." "P is the alternator."

Start with Q's second statement. If Q were the truth-teller, "I am the alternator" would be false. So Q is not the truth-teller, and the truth-teller is P or R.

Case 1: P is the truth-teller. Then "R is the alternator" is true, so Q is the liar. But Q's first statement, "P is the truth-teller", would then be true, and a liar cannot say something true. Fails.

Case 2: R is the truth-teller. Both of R's statements are true: Q is the liar and P is the alternator.

• Q (liar): "P is the truth-teller" is false; "I am the alternator" is false. Both false. Consistent.
• P (alternator): "I am not the liar" is true; "R is the alternator" is false. One of each. Consistent.

Answer: P is the alternator, Q the liar, R the truth-teller. A check of all six role assignments confirms this is the only one that works.

Common mistakes

• Testing people in alphabetical order. Start with the pivot, not with A.
• Treating "or" as "exactly one". "X is a liar, or V is truthful" is true when both parts are true.
• Forgetting to check the survivor. A case can avoid contradiction in the statements you used and still break one you did not.
• Assuming the alternator starts with the truth. Unless the set says so, the true statement can come first or second.
• Stopping at the first consistent case in a set that does not promise a unique answer. If the questions assume one answer, make sure the other cases fail.

Practice set

1. E, F, G and H are each a truth-teller or a liar. E says: "Exactly three of us are liars." F says: "E is a truth-teller." G says: "F is a liar." H says: "G is a truth-teller." Who are the liars?
2. E, F, G and H are each a truth-teller or a liar. E says: "Exactly one of us is a liar." F says: "E is a liar." G says: "E and F are both truth-tellers." H says: "G is a liar." Who are the truth-tellers?
3. In set 2, suppose V's statement were changed to "Exactly three of us are liars." Does Case 1 still work?

Answers:

1. E and F. F has E's type, G has the opposite type, and H has G's type. So E and F are one type and G and H the other: exactly two liars, whatever happens. E's statement ("exactly three") is therefore false, so E lies, F lies, and G and H are truthful.
2. F and H. E and F are opposite, so they cannot both be truthful, and G's statement is false: G lies, so H (who says G lies) is truthful. The liars are G and one of E or F, which makes two. E's "exactly one" is false, so E lies and F is truthful.
3. No. If V is truthful, there are three liars: W plus two of X, Y and Z. Y and Z share a type, so the two must be Y and Z, leaving X truthful. X's statement holds (W and Y are both liars) and Y's statement is false as it should be. But Z's statement, "X is a liar, or V is a truth-teller", is true because V is truthful, and Z is supposed to be a liar. Contradiction, so Case 1 fails. (Case 2 fails too, as in the original set, so the changed set has no solution at all: a reminder that a set must be checked, not assumed, to have an answer.)

What to do next

• On every binary set, draw the same-type and opposite-type links before assuming anything.
• Identify the pivot and write it down; test it first.
• Solve one alternator set this week and write both statements of each person as true or false in a small table.
• See DILR mistakes for the checking habits that catch the "or" and "and" traps.