Logical reasoning problem solving, Logical reasoning questions: If: 1 = 1,2 = 2,3 = 3,4 = 5,5 = 8,6 = 10,7 = 13,8 = 16,9 = 23,10 = 32 So: 11 =? Why? I'm very sorry, because my title was not written well, it was deleted by the system, so I had to write it again! Please say why!

Logical reasoning problem solving, Logical reasoning questions: If: 1 = 1,2 = 2,3 = 3,4 = 5,5 = 8,6 = 10,7 = 13,8 = 16,9 = 23,10 = 32 So: 11 =? Why? I'm very sorry, because my title was not written well, it was deleted by the system, so I had to write it again! Please say why!

I think it's 64

In the family circuit, the most reasonable connection method of switch, lamp and socket is ()
A. Switch, lamp and socket in series B. switch and socket in series and then in parallel C. switch, lamp and socket in parallel D. switch and lamp in series and then in parallel with socket

Switch to control the lamp, should be in series with the lamp, socket is to facilitate access to electrical appliances, should be in parallel with the switch controlled lamp

Solving a logical reasoning problem
A professor of logic has three students, and they are all very smart! One day, the professor gave them a question. The professor pasted a note on the door of each person's brain and told them that each person's note had a positive integer, and the sum of some two numbers was equal to the third! (everyone can see the other two numbers, The professor asked the first student: can you guess your own number? Answer: No, ask the second, no, the third, no, ask the first, no, the second, no, the third: I guess, it's 144! The professor was very satisfied with the smile. Can you guess the number of the other two?

The answer is: 36 and 108. The idea is as follows: first of all, the person who says this number should be the sum of the two numbers, because the other two addends should get equal information. Under the same conditions, if one can't deduce, the other should not. (of course, I just say that this is more likely, because after all, there is a sequence of answers, In addition, only when the third person sees that the number of the other two people is the same, can he immediately say his own number. The above two points are known conditions that can be deduced according to the meaning of the question. If the third person says 144 after only one round of asking, then according to reasoning, it is easy to get that the other two are 48 and 96, How to make the teacher ask for two rounds to get the answer? This requires further consideration: A: 36 (36 / 152) B: 108 (108 / 180) C: 144 (144 / 72) in brackets is the number that the student may guess his head after seeing the other two numbers, When B knows 36 and 72, he will infer like this: "my number should be 36 or 108, but if it is 36, C should be able to say his own number immediately, but C didn't say it, so it should be 108!" however, in the next round, B still doesn't know, so C can judge whether his hypothesis is false, and his number can only be 144! - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Why does the professor who taught you say 169? You want QM to spit blood! -- there is a special kind of problem in logical reasoning -- "thinking nesting", that is, how B thinks about a in C's mind. This kind of problem is usually very abstract, considering many situations, and the thinking process is extremely complex, A professor of logic had three students a, B and C who were good at reasoning and mental arithmetic. One day, the professor gave them a question: the professor pasted a note on each person's forehead and told them that each person's note had an integer greater than 0, and the sum of some two numbers was equal to the third, Every student can see the integers pasted on the heads of the other two students, but they can't see their own numbers. The professor asked a, B and C in turn whether they could guess their own numbers. After several questions, when the professor asked someone again, he suddenly showed a proud smile, Our question is: prove whether someone can guess the number on his head. If someone can guess it, calculate the first time someone guessed the number on his head in the first question. Let's analyze a simple example and observe how everyone reasoned. Suppose a, B and C, the number on their head is L, 2 and 3. L. ask a first, then a can see that the number on B and C is 2 and 3 respectively. A will find that his head can only be 3 + 2 = 5, or 3-2 = 1. But a can't judge whether it is l or 5, so a can only answer "no". 2. Ask B again, B will find that his head can only be 3 + 1 = 4, or 3-1 = 2, B can only analyze from a's answer: (the following is the analysis in B's brain) if there is 2 on his head, then a can see that the numbers on B and C are 2 and 3 respectively, and a will find that there is only 3 + 2 = 5, or 3 - 2 = 1 on his head. A can't judge whether it is l or 5, but can only answer "no". This is the same as a's actual answer, So B can't rule out this situation. If it's 4 on his head, a can see that the numbers on B and C are 4 and 3 respectively. A will find that it can only be 4 + 3 = 7 or 4-3 = 1 on his head. Whether it's l or 7, a can't judge, but can only answer "no". This is the same as a's actual answer, so B can't rule out this situation. B can't judge, but can only answer "no". 3. Ask C again C will find that his head can only be 2 + 1 = 3, or 2-1 = 1. But whether it is l or 3. C can only start from the answer of a or B: (the following is the analysis in C's brain) if his head is 1. A will find that his head can only be 2 + 1 = 3, or 2-1 = 1. But whether it is l or 3 is impossible to judge, and can only answer "no". This is the same as the actual answer of A, B will find that his head can only be 1 + 1 = 2 (because the head of B is an integer greater than 0, so the head of B can not be 1-L = 0). B should answer "can". But this is in contradiction with B's actual answer. C can rule out the head is 1. Continue to analyze the head of C is 3, you will find that there is no contradiction (consistent with the actual situation). C will accurately judge the number on the head is 3, So answer "yes". So in the third question, someone guessed the number on the head. From everyone's point of view, we analyzed the situation that the number on the head is l, 2 and 3. This method is also a common method we use to solve simple logical reasoning problems. But if we enlarge the scale of the problem, we will find that the complexity of the problem will rise sharply, almost one more reasoning, The complexity of the problem will be doubled. It is not easy to solve the problem by such tedious reasoning. The reason is that there are a lot of "thinking nesting". That is, in C's mind, we should consider how B thinks a's idea, Let's use the first, the second and the third students to represent a, B and C respectively. It is inferred that no matter how the three numbers change, no matter who starts the question, the person with the largest number on his head must be the first to guess the number on his head, k) The recursive formula of F (A1, A1, A2, A2, A3, A3, K) can be defined as the following: when k = 1, when k = 1, when A2 = A3, f (A1, A2, A2, a3,3,1) = 1, when A2 > A3, f (A1, A1, A2, A2, A2, A3, A2, A2, A2, A3, a3,3,3, K (k = 1, when k = 1, when A2 = 1, when A2 = A3 = A3, f (A1, A1, A2, A2, A2, A2, A2, A2, A3, a3,3,3,3,3,3,3,1) when k = 2, when A1 = 2, when A1 = A3 = A3, when A1 = A3, when A1 = A3, when A1 = A3, as as as as, when A2 = A3, when A1 = A3, as as, as as as as, as as as as as as as as as as as as a3 > A3 > A3, as as as as as as as as as as as as as as as 3,1) + 1 when A2 < A3, f (A1, A2, a3,2) = f (A1, a3-a1, a3,3) + 2 when k = 3, when A1 = A2, f (A1, A2, A3,3) = 3 when A1 > A2, f (al, A2, a3,3) = f (A1, A2, a1-a2,1) + 2 when Al < A2, f (A1, A2, a3,3) = f (A1, A2, a2-a1,2) + 1. Since we only consider (A1, A2, A3, k) ∈ = S3, K can be directly determined by A1, A2, A3, so f (A1, A2, A3, K) can be simplified to f (A1, A2, A3), Therefore, it avoids the problem scale increasing exponentially with the number of questions, and effectively solves the problem, The solution is based on the in-depth analysis of the problem. Now let's summarize the main line of thinking in solving the problem: refining important preconditions → considering which situation is the "end situation" The whole process starts from the analysis of the essence of the problem, rather than from everyone's thought, and deduces the conclusion of universal significance, One day, the professor gave them a question: the professor pasted a note on everyone's head and told them, Everyone wrote an integer greater than 0 on the note, and a certain number was equal to the sum of the other N-1 numbers. Therefore, each student could see the integer pasted on the head of another n-1 student, but could not see his own number. The professor asked the students in turn whether he could guess the number on his head, Our question is: prove whether someone can guess the number on his head. If someone can guess it, calculate the first time when someone first guesses the number on his head. Analyze the whole reasoning process and draw a conclusion, No matter who starts to ask questions, the person with the largest number on his head must be the first to guess the number on his head We can define f ((A1, A2 When 2w-m ≤ 0, f ((A1, A2 When 2w-m > O, let AI '= AI, where I ≠ K, AK' = 2w-m, when V < K, f (A1, A2 ,an,k)=f(a1’,a2’… When v > k, f (A1, A2 ,an,k)=f(a1’,a2’… Because we only consider (A1, A2 So K can be determined directly by N numbers, so f (A1, A2 , an, K) can be simplified as f (A1, A2 Using the above formula, we can solve the problem by computer programming. So far, the first generalized case is solved. We can find the proof of the case when n = 3, which provides a good contrast for solving the general case, so that we can solve the problem more easily, In fact, this is also based on the analysis of the situation when n = 3. Third, the second kind of promotion. A logic professor has n (n ≥ 3) students who are very good at reasoning and mental arithmetic. One day, the professor gave them a question: the professor pasted a note on each person's brain and told them that each person's note had an integer greater than 0, They are divided into two groups (one group has m students, (m ≥ n / 2), and students do not know how to group), and the sum of the numbers on the heads of the two groups of students is equal. Therefore, each student can see the integer pasted on the heads of the other n-1 students, but can not see his own number. The teaching wheel asks the students whether they can guess the number on their heads, When the professor asked someone again, the person suddenly showed a proud smile and reported the number on his head accurately. Our question is: to prove whether someone can guess the number on his head. If someone can guess the number, calculate the earliest time when someone first guessed the number on his head. Because when n = 3, M can only be 2, which is the original form of the problem, For M = n-1, that is, the first generalized case, we only discuss the case when n > 3 and m < n-1