Logic reasoning

Logic reasoning is the skill of systematically analyzing a problem to arrive at a solution. It is a basic yet powerful skill in problem-solving.

River crossing: Four people, A, B, C, and D need to get across a river. The only way to cross the river is by an old bridge, which holds at most 2 people at a time. Because it is dark, they can't cross the bridge without a torch, of which they only have one. Each pair can only walk at the speed of the slower person. They need to get all of them across to the other side as quickly as possible. A is the slowest and takes 10 minutes to cross; B takes 5 minutes; C takes 2 minutes; D takes 1 minute.

What is the minimum time to get all of them across to the other side?[15]

Solution: The key point is to realize that the 10-minute person should go with the 5-minute person, and this should not happen in the first crossing. Otherwise, one of them has to go back. Therefore, C and D should go across first (2 min); then send D back (1min); A and B go across (10 min); send C back (2min); C and D go across again (2 min).

It takes 17 minutes in total. Alternatively, we can send C back first and then D back in the second round, which takes 17 minutes as well.

A ship leaves Le Havre for New York at noon every day, and a ship leaves New York for Le Havre at noon every day. If the trip takes exactly 7 days, how many other ships will a ship leaving Le Havre pass on its way to New York?

Solution: Legend has it that this was a problem that French Mathematician Édouard Lucas challenged his friends with. It can be solved with a little logical reasoning. When a ship (let’s call it S) leaves Le Havre, the ship that leaves New York 7 days ago (day -7) arrives at Le Havre. It is the first ship that S passes. When S arrives at New York, the ship that leaves New York 7 days later (day +7) leaves New York. It is the last ship that S passes. Therefore, S passes ships ‑7, -6, …, 6, and 7, and there are 15 ships in total.

Birthday problem: You and your colleagues know that your boss A’s birthday is one of the following 10 dates:

Mar 4, Mar 5, Mar 8

Jun 4, Jun 7

Sep 1, Sep 5

Dec 1, Dec 2, Dec 8

A told you only the month of his birthday and told your colleague C only the day. After that, you first said: “I don’t know A’s birthday; C doesn’t know it either.” After hearing what you said, C replied: “I didn’t know A’s birthday, but now I know it.” You smiled and said: “Now I know it, too.” After looking at the 10 dates and hearing your comments, your administrative assistant wrote down A’s birthday without asking any questions. What did the assistant write?

Solution: Don’t let the “he said, she said” part confuse you. Just interpret the logic behind each individual’s comments and try your best to derive useful information from these comments.

Let D be the day of the month of A’s birthday; we have . If the birthday is on a unique day, C will know the A’s birthday immediately. Among possible Ds, 2 and 7 are unique days. Considering that you are sure that C does not know A’s birthday, you must infer that the day the C was told of is not 2 or 7. Conclusion: the month is not June or December. (If the month had been June, the day C was told of might have been 2; if the month had been December, the day C was told of might have been 7.)

Now C knows that the month must be either March or September. He immediately figures out A’s birthday, which means the day must be unique in the March and September list. It means A’s birthday cannot be Mar 5, or Sep 5. Conclusion: the birthday must be Mar 4, Mar 8 or Sep 1.

Among these three possibilities left, Mar 4 and Mar 8 have the same month. If the month you have is March, you still cannot figure out A’s birthday. Since you can figure out A’s birthday, A’s birthday must be Sep 1. Therefore, the assistant must have written Sep 1.

Card game: A casino offers a card game using a normal deck of 52 cards. The rule is that you turn over two cards each time. For each pair, if both are black, they go to the dealer’s pile; if both are red, they go to your pile; if one black and one red, they are discarded. The process is repeated until you two go through all 52 cards. If you have more cards in your pile, you win $100; otherwise (including ties), you get nothing. The casino allows you to negotiate the price you want to pay for the game. How much would you be willing to pay to play this game?[16]

Solution: This surely is an insidious casino. No matter how the cards are arranged, you and the dealer will always have the same number of cards in your piles. Why? Because each pair of discarded cards have one black card and one red card, so an equal number of red and black cards are discarded. As a result, the number of red cards left for you and the number of black cards left for the dealer are always the same. The dealer always wins! Therefore, you should not pay anything to play the game.

«« You are in a room with two doors. One door leads to your freedom, and the other leads to a jail cell. In front of either door is a guard. One guard always tells lies, and the other always tells the truth. You can only ask one guard one yes/no question. What question will you ask to guarantee your freedom?

Solution: This is a classic logic puzzle. One popular answer is to ask one guard: “Would the other guard say that you are guarding the door to freedom?” If he answers yes, choose the other door; if he answers no, choose the door that this guard is standing in front of.

There are two possible scenarios:

1. Truth-teller guards the door to freedom; Liar guards the door to jail.
2. Truth-teller guards the door to jail; Liar guards the door to freedom.

If we ask a guard a direct question such as, “Are you guarding the door to the freedom?” For scenario 1, both guards will answer yes; for scenario 2, both guards will answer no. Thus, a direct question does not help us solve the problem. The key is to involve both guards in the question as the popular answer does. For scenario 1, if we happen to choose the truth-teller, he will answer no since the liar will say no; if we happen to choose the liar guard, he will answer yes since the truth-teller will say no. For scenario 2, if we happen to choose the truth-teller, he will answer yes since the liar will say yes; if we happen to choose the liar guard, he will answer no since the truth-teller will say yes. Therefore, for both scenarios, if the answer is no, we choose that door; if the answer is yes, we choose the other door.

Burning ropes: You have two ropes, each of which takes 1 hour to burn. But either rope has different densities at different points, so there's no guarantee of consistency in the time it takes different sections within the rope to burn. How do you use these two ropes to measure 45 minutes?

Solution: For a rope that takes x minutes to burn, if you light both ends of the rope simultaneously, it takes minutes to burn. We should light both ends of the first rope and light one end of the second rope. 30 minutes later, the first rope will get completely burned, while that second rope now becomes a 30-min rope. At that moment, we can light the second rope at the other end (with the first end still burning), and when it is burnt out, the total time is exactly 45 minutes.

Horse race: There are 25 horses, each of which runs at a constant speed that is different from the other horses’. Since the track only has 5 lanes, each race can have at most 5 horses. Let’s assume that you have no stopwatch or similar timing device. If you need to find the 3 fastest horses, what is the minimum number of races needed to identify them?

Solution: This problem tests your basic analytical skills. To find the 3 fastest horses, surely all horses need to be tested. A natural first step is to divide the horses into 5 groups (with horses 1-5, 6-10, 11-15, 16-20, 21-25 in each group). After 5 races, we will have the order within each group, let’s assume the order follows the order of numbers (e.g., 6 is the fastest and 10 is the slowest in the 6-10 group)[17]. That means 1, 6, 11, 16, and 21 are the fastest within each group.

Surely the last two horses within each group are eliminated. What else can we infer? We know that within each group, if the fastest horse ranks the 5th or the 4th among 25 horses, then all horses in that group cannot be in the top 3. If it ranks the 3rd, no other horse in that group can be within the top 3. If it ranks the 2nd, then one other horse in that group may be in the top 3. If it ranks the first, then two other horses in that group may be in the top 3.

Thus, let’s race horses 1, 6, 11, 16, and 21. Again without loss of generality, let’s assume the order is 1, 6, 11, 16, and 21. Then we immediately know that horses 4-5, 8-10, 12-15, 16-20, and 21-25 are eliminated. Since 1 is fastest among all the horses, 1 is in. We need to determine which two among horses 2, 3, 6, 7, and 11 are in the top 3, which only takes one extra race.

Therefore, we need 7 races (in 3 rounds) to identify the 3 fastest horses.

««« Glass ball: You are holding two glass balls in a 100-story building. If a ball is thrown out of the window, it will not break if the floor number is less than X, and it will always break if the floor number is equal to or greater than X. You would like to determine X. What is the strategy that will minimize the number of drops for the worst-case scenario? [18]

Solution: Suppose that we have a strategy with a maximum of N throws. For the first throw of ball one, we can try the N-th floor. If the ball breaks, we can start to try the second ball from the first floor and increase the floor number by one until the second ball breaks. At most, there are floors to test. Thus, a maximum of N throws is enough to cover all possibilities. If the first ball thrown out of N-th floor does not break, we have throws left. This time we can only increase the floor number by for the first ball since the second ball can only cover floors if the first ball breaks. If the second ball thrown out of (2N-1)-th floor does not break, we have throws left. Thus, we can only increase the floor number by for the first ball since the second ball can only cover floors if the first ball breaks…

Using such logic, we can see that the number of floors that these two balls can cover with a maximum of throws is . To cover 100 stories, we need to have . Taking the smallest integer, we have .

Basically, we start the first ball on the 14th floor. If the ball breaks, we can use the second ball to try floors with a maximum number of throws of 14 (when the 13th or the 14th floor is X). If the first ball does not break, we will try the first ball on the floor. If it breaks, we can use the second ball to cover floors with a total maximum number of throws of 14 as well...