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Riddles, brain puzzles and mathematical problems - Printable Version +- Frictional Games Forum (read-only) (https://www.frictionalgames.com/forum) +-- Forum: Frictional Games (https://www.frictionalgames.com/forum/forum-3.html) +--- Forum: Off-Topic (https://www.frictionalgames.com/forum/forum-16.html) +--- Thread: Riddles, brain puzzles and mathematical problems (/thread-19232.html) |
RE: Riddles, brain puzzles and mathematical problems - Romulator - 08-24-2014 I thought I had the answer, but if he can only make one cut, then I don't think I am correct. Spoiler below!
RE: Riddles, brain puzzles and mathematical problems - BAndrew - 08-24-2014 A link can be both horizontal and vertical. The chain in this image has 7 links (not counting the two at the edge of the picture). @Romulator You are almost there! Peter can receive change as long as the inn keeper has enough links to give him. That doesn't count as paying in advance. In fact you solved it. But he doesn't need to make three cuts to achieve this strategy. Figured out why? PS: By one cut I mean one removal of a link. RE: Riddles, brain puzzles and mathematical problems - BAndrew - 08-24-2014 By cutting/removing the third link then you have three parts A,B,C: A: 1 link (the one you removed) B: 2 links (on the left) C: 4 links (on the right) RE: Riddles, brain puzzles and mathematical problems - BAndrew - 04-17-2015 I know it is a long time, but I've got two new problems if anyone is interested. Problem 1: Consider this a game. Starting from the square on the top left corner (marked as green) you must end up in the bottom right square (marked as red). The rules you must follow are the following:
![]() Can you find a solution? What is the best solution? Or equivalently what is the route with the minimum number of steps? If you feel like it, write an algorithm or a computer program that finds the best possible route (or routes). Problem 2 (This actually is hard) What is the average of the number of ways you can express a positive integer as the sum of the squares of two integers. In other words: Let s>0 be an integer. In average, how many times will it satisfy: n² + m² = s (where n,m are integers) For example, if s = 2 we have: 1² + 1² = 2 (-1)² + (-1)² = 2 1² + (-1)² = 2 (-1)² + 1² = 2 So there are 4 ways for number 2. RE: Riddles, brain puzzles and mathematical problems - Romulator - 04-17-2015 (04-17-2015, 01:49 AM)BAndrew Wrote: Problem 1: Probably a faster way, but in five minutes, I deduced; Spoiler below!
First way I found, in prepping for Sociology class. RE: Riddles, brain puzzles and mathematical problems - BAndrew - 04-17-2015 That's right, good job. There is a "better" way though. EDIT: Just noticed on your first solution, you go "outside" the table and circle back in (on step #4). I forgot to say that this is not allowed. If that were the case you would only need 6 steps: 1 up and 1 right (land in 4) 4 right (land in red square) Your second solution is solid. RE: Riddles, brain puzzles and mathematical problems - FlawlessHappiness - 04-17-2015 Spoiler below!
RE: Riddles, brain puzzles and mathematical problems - BAndrew - 04-17-2015 Congratulations @Flawless! That is a possible solution with the least number of steps! RE: Riddles, brain puzzles and mathematical problems - Froge - 04-17-2015 ![]() RE: Riddles, brain puzzles and mathematical problems - BAndrew - 04-17-2015 (04-17-2015, 11:29 AM)Rainfroge Wrote: Let x be the number of rooks on the board. If x = 0, then the claim is true trivially If x > 0, then we are going to show that the claim is also true. Let S be the largest possible set of mutually non-attacking rooks. or S = {R1, R2, ..., Rk} where Ri (1≤i≤k) is a rook on the board. The rooks selected are mutually non-attacking. The order the rooks are selected is random. Also, R1 ≠ R2 ≠ ... ≠ Rk. |S| = k and this is the largest number of mutually non-attacking rooks. We choose the rows/columns where these rooks ∈ S are located. By this choice we are going to show that all the rooks on the board are contained. Assume to the contrary that there exists a rook (let it be r) on the board that isn't contained with the choice we made. Then this rook is by definition mutually non-attacking with every other rook ∈ S. That implies that S isn't the largest possible set of mutually attacking rooks, because by adding r to S , we get a new set U = {R1, R2, ..., Rk, r} where |U| = k+1 > k. Contradiction! Therefore, all the rooks are contained by this choice. That completes the proof. |