Logical Puzzles

We can find many puzzles on mathematical logic principles for learning and fun purposes. Raymond Smullyan (1999—2017) is an American mathematician who is the author of many books on logical puzzles. Let’s look at a few exciting puzzles by Smullyan.

Puzzle 1

There are two kinds of inhabitants of an island: knaves and knights. The knaves always lie, and the knights always tell the truth. We meet two people on this island, $X$ and $Y$. The person $X$ says, “I am a knave, or $Y$ is a knight.” And $Y$ does not say anything. We want to determine whether $X$ and $Y$ are knave or knight.

Solution

There are four possibilities here as each one of $X$ and $Y$ can be knave or knight. We will examine all four cases and discard those leading to a contradiction.

Case 1: If both $X$ and $Y$ are knights.

In this case, the statement of $X$ should be true as they are a knight. I am a knave, or $Y$ is a knight, is a true statement because the disjunction of false and true is true.

Case 2: If both $X$ and $Y$ are knaves.

In this case, $X$ must lie. But we observe that $X$ gives a true statement that I am a knave, which is a contradiction.

Case 3: If $X$ is a knave and $Y$ is a knight.

In this case, $X$ must lie as they are a knave. But $X$ says, “I am a knave,” which is true. Therefore, this can not be the case as it leads to a contradiction.

Case 4: If $X$ is a knight and $Y$ is a knave.

In this case, $X$ must tell the truth as a knight. But $X$ says that I am a knave or that $Y$ is a knight, which is not true. Therefore, this case also leads to a contradiction.

Hence, we can conclude the only possibility that both $X$ and $Y$ are knights.

Puzzle 2

There are two kinds of inhabitants of an island: knaves and knights. The knaves always lie, and the knights always tell the truth. We meet two people on this island, $X$ and $Y$. The person $X$ says, “We both are knights.” And $Y$ says that “$X$ is a knave.” We want to determine whether $X$ and $Y$ are knave or knight.

Solution

There are four possibilities here as each one of $X$ and $Y$ can be knave or knight. We will examine all four cases and discard those leading to a contradiction.

Case 1: If both $X$ and $Y$ are knights.

In this case, the statement of $X$ should be true as they are a knight. $X$ says that we both are knights, which is true. But if $Y$ is a knight, then the statement that $X$ is a knave must be true, which is not the case. Hence, this case leads to a contradiction.

Case 2: If both $X$ and $Y$ are knaves.

In this case, $X$ must lie. The statement of $X$ that we both are knights is a lie. But the words of $Y$ that $X$ is a knave are not a lie. Hence, this case leads to a contradiction.

Case 3: If $X$ is a knave and $Y$ is a knight.

In this case, $X$ must lie as they are a knave. The statement of $X$, “We both are knights,” is a false statement. As $Y$ is a knight, the statement of $Y$ must be true. $Y$ says, “$X$ is a knave, which is a truth. Hence, there is no contradiction.

Case 4: If $X$ is a knight and $Y$ is a knave.

In this case, $X$ must tell the truth as a knight. But $X$ says that we both are knights, which is not the case. Hence, this case also leads to a contradiction.

Hence, we can conclude the only possibility that $X$ is a knave and $Y$ is a knight.

Puzzle 3

Assume three kinds of inhabitants of an island: knaves, knights, and knatives. The knaves always lie, and the knights always tell the truth. The knatives can speak the truth, or they can lie. Assume we meet three people on this island: $W$, $X$, and $Y$. One of them is a knight, one is a knave, and one is a knative. We want to determine who is the knight, who is the knave, and who is the knative using the following information:

• $W$ says: “I am the knight.”
• $X$ says: “I am the knave.”
• $Y$ says: “$X$ is the knight.”

Solution

There are six possibilities, as shown in the following table:

Possibility	$W$	$X$	$Y$
1	Knight	Knave	Knative
2	Knight	Knative	Knave
3	Knave	Knight	Knative
4	Knave	Knative	Knight
5	Knative	Knight	Knave
6	Knative	Knave	Knight

We’ll look at each case and discard those which lead to a contradiction.

Cases 1 and 6

If $X$ is the knave, he can not say I am the knave; that is a contradiction.

Cases 3 and 5

If $X$ is the knight, he can not say I am the knave; that is a contradiction.

Case 4

If $Y$ is the knight, and he says $X$ is a knight, then $X$ should be a knight. But, $X$ can not be the knight because it is contradictory, as shown in cases three and five. Therefore, $Y$ is not telling the truth, which is contradictory.

Case 2

If $W$ is a knight, he says I am the knight, which is consistent. If $X$ is knative and says that I am the knave, that means $X$ is telling a lie, which is consistent. If $Y$ is the knave, and he says $X$ is the knight, he is telling a lie, which is again consistent.

So, after analyzing each case, we can conclude that $W$ is the knight, $X$ is knative, and $Y$ is the knave.

Puzzle 4

There are two kinds of inhabitants of an island: knaves and knights. The knaves always lie, and the knights always tell the truth. Assume we meet two people on this island, $X$ and $Y$. The person $X$ says, “I am a knave if and only if $Y$ is a knave.” And $Y$ says, “Our kind is different.” We want to determine whether $X$ and $Y$ are knave or knight.

Solution

There are four possibilities here as each one of $X$ and $Y$ can be knave or knight. We will examine all four cases and discard those leading to a contradiction.

Case 1: If both $X$ and $Y$ are knights.

In this case, the statement of $X$ should be true as he is a knight. I am a knave, if and only if $Y$ is a knave is a true statement because $F\Leftrightarrow F\equiv T$. The statement of $Y$ should also be true as $Y$ is a knight. $Y$ says our kind is different, which is not the case, giving a contradiction.

Case 2: If both $X$ and $Y$ are knaves.

In this case, $X$ must lie. But we observe that $X$ gives a true statement that I am a knave if and only if $Y$ is a knave, which is a contradiction.

Case 3: If $X$ is a knave and $Y$ is a knight.

In this case, $X$ must lie as he is a knave. As $X$ says, “I am a knave if and only if $Y$ is a knave,” which is false; therefore, it is consistent. The statement given by $Y$ should be true as $Y$ is a knight. $Y$ says our kind is different, which is true. Consequently, this case is consistent and does not lead to a contradiction.

Case 4: If $X$ is a knight and $Y$ is a knave.

In this case, $X$ must tell the truth as a knight. But $X$ says that I am a knave if and only if $Y$ is a knave, which is not true. Therefore, this case also leads to a contradiction.

So, we can conclude the only possibility is that $X$ is a knave and $Y$ is a knight.