# Logical Puzzles

Logical Puzzles, It includes everything you learned. Permutation and combination, algebra etc. It is not really hard, but you need time to solve the problems.

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It can be any type of problems, and almost each one includes many topic you've learned. For example, Richard Hovasse's bridge and torch problem, which is the problem below, and you can find more and more! You can even create one.

**HISTORY**
The logic puzzle was first produced by Charles Lutwidge Dodgson, who is better known under his pen name Lewis Carroll, the author of Alice's Adventures in Wonderland. In his book The Game of Logic he introduced a game to solve problems such as confirming the conclusion "Some greyhounds are not fat" from the statements "No fat creatures run well" and "Some greyhounds run well". Puzzles like this, where we are given a list of premises and asked what can be deduced from them, are known as syllogisms. Dodgson goes on to construct much more complex puzzles consisting of up to 8 premises.

In the second half of the 20th century mathematician Raymond M. Smullyan has continued and expanded the branch of logic puzzles with books such as The Lady or the Tiger?, To Mock a Mockingbird and Alice in Puzzle-Land. He popularized the "knights and knaves" puzzles, which involve knights, who always tell the truth, and knaves, who always lie.

There are also logic puzzles that are completely non-verbal in nature. Some popular forms include Sudoku, which involves using deduction to correctly place numbers in a grid; the nonogram, also called "Paint by Numbers", which involves using deduction to correctly fill in a grid with black-and-white squares to produce a picture; and logic mazes, which involve using deduction to figure out the rules of a maze.

Four people come to a river in the night. There is a narrow bridge, but it can only hold two people at a time. They have one torch and, because it's night, the torch has to be used when crossing the bridge. Person A can cross the bridge in one minute, B in two minutes, C in five minutes, and D in eight minutes. When two people cross the bridge together, they must move at the slower person's pace. The question is, can they all get across the bridge in 15 minutes or less?

Assume that a solution minimizes the total number of crosses. This gives a total of five crosses - three pair crosses and two solo-crosses. Also, assume we always choose the fastest for the solo-cross. First, we show that if the two slowest persons (C and D) cross separately, they accumulate a total crossing time of 15. This is done by taking persons A, C, & D: C+A+B+A = 8+1+5+1=15. (Here we use A because we know that using A to cross both C and D separately is the most efficient.) But, the time has elapsed and person A and B are still on the starting side of the bridge and must cross. So it is not possible for the two slowest (C & D) to cross separately. Second, we show that in order for C and D to cross together that they need to cross on the second pair-cross: i.e. not C or D, so A and B, must cross together first. Remember our assumption at the beginning states that we should minimize crosses and so we have five crosses - 3 pair-crossings and 2 single crossings. Assume that C and D cross first. But then C or D must cross back to bring the torch to the other side, and so whoever solo-crossed must cross again. Hence, they will cross separately. Also, it is impossible for them to cross together last, since this implies that one of them must have crossed previously, otherwise there would be three persons total on the start side. So, since there are only three choices for the pair-crossings and C and D cannot cross first or last, they must cross together on the second, or middle, pair-crossing. Putting all this together, A and B must cross first, since we know C and D cannot and we are minimizing crossings. Then, A must cross next, since we assume we should choose the fastest to make the solo-cross. Then we are at the second, or middle, pair-crossing so C and D must go. Then we choose to send the fastest back, which is B. A and B are now on the start side and must cross for the last pair-crossing. This gives us, B+A+D+B+B = 2+1+8+2+2 = 15.

And our final answer is 15 mins.