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"Simple" Maths Problems - CLOSED!
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Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Dominique solves Q43.... Q43 Answer: 43 Congratulations!!! Anybody that wants to have a go at solving Q41 & Q42, have a look at Bill's last three posts, there could be some useful information to get everybody started in there.... Problem Solver Stats: 1. William Rothamel - 9 1/2 Points 2. Mr. Kevvy - 7 Points 3. Guido.Man - 6 Points 4. Sarge - 5 1/2 Points 5. Dominique - 4 1/2 Points 6. WinterKnight - 5 Points 7. Fred W - 4 1/2 Points 8. John McLeod VII - 1 Point 9. Labbie - 1 Point 10. Scary Capitalist - 1/2 Point 11. TBD... With Dominique in 5th, WinterKnight, the original problem solving champion, has fallen to 6th.... Two Questions Remain: (To solve Q42, you need to read Q41)
Best Regards, Luke. - Luke. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Re-iterating the current questions left: Q41: An ant and a blind spider are on opposite corners of a cube. The ant is stationary and the spider moves at random from one corner to another along the edges only. What is the expected number of turns before the spider reaches the ant? (1 Point) Q42: Also solve the above question but with different shapes: Square? Octahedron? Dodecahedron? Icosahedron? (1 Point for all shapes combined) Best Regards, Luke. - Luke. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Hint--it's not too hard at each possible configuration of ant and spider, the spider has three options. Whether he gets the ant or moves to the same face or moves in a penultimate position and then gets the ant is also broken into different numbers of 1/3 probability. Let's go --you can do it Bunky !! |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Hint--it's not too hard Very true... Q41 is not that hard... Thanks for everyone's participation!!! Dedications today go-to: Sarge! Bill! Fred! Best Regards, Luke. Here again are the questions! Q41: An ant and a blind spider are on opposite corners of a cube. The ant is stationary and the spider moves at random from one corner to another along the edges only. What is the expected number of turns before the spider reaches the ant? (1 Point) Q42: Also solve the above question but with different shapes: Square? Octahedron? Dodecahedron? Icosahedron? (1 Point for all shapes combined) - Luke. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
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William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
One last chance, or this thread is going to the fishes.... Warning : if youse guys don't solve this real soon Daddio will have to come out of retirement and stumble around to the answer. I think it would involve solving a few simultaneous equations is all. Where are the other stalwarts ?? Regards, Bill AKA DADDIO |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
One last chance, or this thread is going to the fishes.... Sorry but it's over 40 years since my brain was in this sort of gear. Given the equations - no problemo; it's deriving the relevant equations from the question that's giving me headaches ;( F. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
I also had math about 40-50 years ago. Doing these poblems has refreshed some of these notions and added new ones. Remember your Math teacher's advice . "ALWAYS draw a picture of the problem--once you understand the problem completely you are half way there to the solution ) I have re-learned this myself from playing with Luke's problems. Let X = to the number of trials to catch the ant from the initial position, Let Y = to the number of trails to catch the ant from the diagonal corner on the same face as the ant. Let Z =to the number of trials to catch the ant from one corner away from the ant. Choose the starting position ( e. g. Assume that you are diagonally across the cube on another face of the cube from the ant) At this point and at every other point in the random walk there is a 1/3 , 2/3 probability that the spider will move to condition X, Y or Z. Write equations for how these probabilities compare. How X Y and Z compare on average, There would be three equations. Express the relationships for these three conditions and solve the equations. You could probably find a matrix tool on the Internet or maybe one to solve the equations directly so you wont have to wade through all the substitution. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
I also had math about 40-50 years ago. Doing these poblems has refreshed some of these notions and added new ones. Nope, I'm still not getting it. And I have to be on-site with a client for the next 2 days so I think you'll have to enlighten us all or the master might pull the plug :( F. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Phred et al One more hint which I think should lead to the solution. It isn't obvious until you draw the picture--draw the cube (put the ant and spider diagonally opposed across the cube. Now you can see that after one move the spider will be on a common face with the ant and diagonally across from it. This is state Y in my earlier post. Important to realize that the expected value of the number of trials is the AVERAGE number of trials. So starting from state X will always be exactly one more move than the solution than if you started from state Y (same face diagonally) A(X) =1 + A(Y) where A is average or expected number of trials Now you want to get to an adjacent node to the ant (State Z) there is a 2/3 chance that you will do this from the same face and a 1/3 chance that you will go back to state X. So now you have: A(Y)=(2/3)*(1+A(Z))+(1/3)*(1+A(X). When you are in state Z there is a 1/3 chance you will get the ant and a 2/3 chance you will get back to state Y. So you have: A(Z)=(1/3)*1+(2/3)*(1+A(Y)). Solve for A(X) , A(Y), A(Z) It's too late here now for messy algebra so I will leave this for someone else to finish. You should now be able to solve the puzzle for the other figures as well. The difficulty here initially is not having been told and having to realize that there are three different answers for the three different starting points that are mixed up in the words ( "OPPOSITE" side of the cube) |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Yikes. I just realized that most of the other figures are solids so here is another hint: Octahedron: is really |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
For the square: The first move will put the spider in an adjacent corner. Then there is a 1/2 probability that he will get the spider but 1/2 probability that he will go back to the original starting point so on average: 1/2*1/2=1/4 or 4 moves on average |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
For the square: But since the question asks for the number of "turns", by which I take it to means "corners encountered", then 4 moves would involve only 3 "turns"? F. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Well Done Bill! 1 Point for your Q41 answer... Q41 Answer: 4 moves - Also... 1/2 a point to Fred for his reasoning.... More Questions later... can anyone answer Q42? Q42: (continuation of 41) Also solve the above question but with different shapes: Square? Octahedron? Dodecahedron? Icosahedron? (1 Point for all shapes combined) Have I left the thread to die, not yet.... ;) Best Regards, Luke. - Luke. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
Well Done Bill! 1 Point for your Q41 answer... Ermmmm - Didn't Q41 ask for the cube and Bill answer for a square?? F. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Well Done Bill! 1 Point for your Q41 answer... The answer "4" is the same for a square and a cube, I should have changed the question (which I took from a maths website).... The thing is that he answered the original question correctly - I'll see what Bill makes of this in his next post.... Best Regards, Luke. - Luke. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
Well Done Bill! 1 Point for your Q41 answer... Got to run but that defies logic. For a cube there should be one more corner to turn than for a square? F. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Lets see.... here is the link to the question.... http://mathproblems.info/prob4a.htm I've mixed up the answers.... damn..... Scrap the points for Q41/42.... geez.... I hate doing that..... Sorry Bill & Fred.... I'll award you each with 1 point anyway.... Answers: Cube - 10 Square - 4 Octahedron - 6 Dodecahedron - 35 Icosahedon - 15 Sorry 'bout that guys.... I'm exhausted... maybe some easier questions tomorrow.... Best Regards, Luke. - Luke. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
Lets see.... here is the link to the question.... http://mathproblems.info/prob4a.htm Oopssss ;) Don't worry about it, Luke; happens to the best on occasion. I still contest the definition of "turn" though it is not your fault (I had a look at the web-site). A "turn" seems to be used to indicate a movement from one corner to the next (perhaps as in "It's my turn to ...") e.g. in the official solution: "After one turn the spider is on the same face as the ant" but at this point the spider has not "turned" any corners, just traversed an edge. I have long maintained that examiners in science subjects should be more careful in the use of language. Keep on Boinc'ing F. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Follow-ups: A "turn" in this context is a trial or an occurance of a random event as in "take your turn" or "it's my turn". Colloquial English can be confusing to different dialect speakers. I had a Scottish friend in Teheran --he had a nasty burr in his speech. I could understand him; but it make my head hurt from the extra brainpower to parse and decipher his speech. For the cube: 10 moves (trials or turns) from the diagonal across the cube and nine from the diagonal on the same face as the ant and 7 from an adjacent corner on the same face as the ant. In retrospect with the added solids --this is a problem only for the most professional of mathematicians. I probably spent a few hours on it for 4 days or so.. I am expecting to be working full time in a few weeks so I imagine that I like most people won't have a lot of time to struggle along with ver y difficult problems --even if it is instructive in the end. Perhaps Luke --you should strive for a wider audience --we could still have fun over some problems that required some insight. Best regards to the gang, Daddio |
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