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Topic: Tunnels of Callicrates (Read 6270 times) 

rmsgrey
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Re: Tunnels of Callicrates
« Reply #25 on: Jun 19^{th}, 2007, 7:18am » 
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on Jun 19^{th}, 2007, 3:10am, Sir_Rogers wrote:3) 12456 Now the longest way only uses 5 of the circles 
 What options do you have at 6 when you arrive from 5?


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Hippo
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Re: Tunnels of Callicrates
« Reply #26 on: Jun 19^{th}, 2007, 10:52am » 
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I can read this problem several ways, it seems is not well formulated. The variants depend on answers to following questions: 1) When you arive to "L" (no crossing) place from right ... a) you continue up (with prob=1) b) you return (with prob=1) c) your path ends here (with prob 1) 2) When you arrive to T from left a) you go down with prob=1/2 you continue right with prob=1/2 b) you go down with prob 1/3, left with 1/3 and right with 1/3 3) When you arrive to T from buttom (1a) case a) go left with prob=1/2, go right with prob=1/2 b) go left with prob=1/3, go right with prob=1/3 and go down with prob=1/3 4) When you arrive to "+ with missing left" from right a) go down with prob=1 b) go down with prob 1/2 and right with prob 1/2 5) When you arrive to "+ with missing left" from bottom (1a) case) a) go right with prob 1 b) go down with prob 1/2, go right with prob 1/2 6) When you arrive to "+ with missing left" from top a) go right with prob 1/2 and down with prob 1/2. (Suppose rightleft symmetry) I suppose 1a) 2a) 3a) 4a) 5a) 6a) case is the basic variant and 1a) 2b) 3b) 4b) 5b) 6a) is the backtrack variant. The basic variant is easier ... Number all crossings left to right from top to bottom starting with 0 to extend Sir_Rogers notation. Denote w_{x}^{top} winning probability when arriving to vertex x from top, similarly for bot,rt,lft. We can start with w_{14}^{lft}=1, w_{13}^{rt}=0, w_{12}^{rt}=0, w_{7}^{rt}=0. Continue with easiest equations: w_{14}^{top}=1/2+1/2 w_{10}^{bot}= 1/2+1/2 w_{11}^{lft}= 1/2+1/2w_{14}^{top}. Therefore w_{14}^{top}=1 =w_{10}^{bot}=w_{11}^{lft}. Now w_{10}^{top}= 1/2(w_{11}^{lft}+w_{14}^{lft})=1 and similarly w_{11}^{top}= 1/2(w_{10}^{rt}+w_{14}^{top})= 1/2(w_{14}^{lft}+w_{14}^{top})=1. Follow with more complicated equations: w_{13}^{top}= 0+1/2w_{5}^{bot}= 1/4w_{4}^{rt}+1/4w_{6}^{lft}= 1/4w_{9}^{top}+1/4w_{10}^{top}= 1/8w_{12}^{rt}+1/8w_{13}^{top}+1/4= 0+1/8w_{13}^{top}+1/4. Therefore 7/8 w_{13}^{top}=1/4 and w_{13}^{top}=2/7, w_{4}^{rt}= w_{9}^{top}=1/7, Now w_{12}^{top}= 0+1/2w_{9}^{lft}= 1/2w_{9}^{top}=1/14 and w_{8}^{lft}= w_{12}^{top}=1/14 and w_{7}^{top}= 0+1/2w_{8}^{lft}=1/28. Similarly w_{8}^{top}= 0+1/2w_{12}^{top}= 1/2w_{8}^{lft}=1/28. Now w_{0}^{rt}= 1/2(w_{7}^{top}+w_{8}^{top})=1/28. Now w_{6}^{top}= 1/2w_{5}^{rt}+1/2= 1/4w_{4}^{rt}+0+1/2=15/28 and w_{4}^{top}= 1/2w_{9}^{top}+1/2w_{5}^{lft}= 1/14+1/4=18/56. Finaly w_{3}^{lft}= 1/2(w_{6}^{top}+1)= 43/56 and w_{2}^{lft}= 1/2w_{4}^{top}+1/2w_{3}^{lft}= 61/112 and w_{1}^{top}= 1/2w_{0}^{rt}+1/2w_{2}^{lft}= 69/224.

« Last Edit: Jun 22^{nd}, 2007, 11:38am by Hippo » 
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Sir_Rogers
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Re: Tunnels of Callicrates
« Reply #27 on: Jun 19^{th}, 2007, 11:21am » 
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Erm ... your assumptions are flawed, because it says you can't go up, and there's two different solutions, one for backtracking and one for none. Backtracking means you can go back to where you came from ... I'm just not sure about the dead end one. Any ideas? That would throw my results all over ..


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ThudnBlunder
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Re: Tunnels of Callicrates
« Reply #28 on: Jun 19^{th}, 2007, 12:49pm » 
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I first saw the problem here.


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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



Grimbal
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Re: Tunnels of Callicrates
« Reply #29 on: Jun 20^{th}, 2007, 1:39am » 
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on Jun 19^{th}, 2007, 3:10am, Sir_Rogers wrote:1) 123 2) 1236 3) 12456 
 In 3) 12456 you have no choice at 6, so the probability is 1/16. And the problem of the last junction right needs clarification before giving an answer. I'd say if the marble gets stuck, it should count as a loss.

« Last Edit: Jun 20^{th}, 2007, 1:40am by Grimbal » 
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Hippo
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Re: Tunnels of Callicrates
« Reply #30 on: Jun 20^{th}, 2007, 4:25am » 
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on Jun 19^{th}, 2007, 11:21am, Sir_Rogers wrote:Erm ... your assumptions are flawed, because it says you can't go up, and there's two different solutions, one for backtracking and one for none. Backtracking means you can go back to where you came from ... 
 OK, so what are the options for which you are solving the problem? No backtracking is inconsistent in 1b) case, too (at least if I understand backtracking well). The 1c) case does not correspond to "marble" interpretation. If 1a) is not correct choice, there is only one consistent variant ... 1b) 2b) 3b) 4b) 5b) 6a). Do we agree? For the 1b) 2b) 3b) 4b) 5b) 6a) case the situation is much more easier. We neednot distinguish the direction from which we arrived. w_{14}=1, w_{13}=0, w_{12}=0 The bar from 10 to 11 does not play any role ... (an easy equation) w_{11}=1/2w_{10}+1/2w_{14}= 1/4w_{11}+3/4w_{14}=1/4w_{11}+3/4 and we get w_{11}=1=w_{10}. Furthermore w_{9}=1/2w_{12}+1/2w_{13}=0, w_{8}=w_{12}=0, and w_{7}=0+1/2w_{8}=0. Now the interesting part begins w_{5}=1/3w_{4}+1/3w_{13}+1/3w_{6}=1/6w_{9}+1/6w_{5}+0+1/6w_{5}+1/6w_{10 }= 0+1/3w_{5}+1/6 and therefore w_{5}=3/2*1/6=1/4, w_{6}=5/8, and w_{4}=1/8. Finaly w_{2}=1/3w_{0}+1/3w_{4}+1/3w_{3}= 1/9w_{7}+1/9w_{8}+1/9w_{2}+ 1/24+1/9w_{2}+1/9w_{6}+1/9w_{11}= 0+0+2/9w_{2}+1/24+5/72+1/9 and w_{2}=9/7*(1/24+5/72+1/9)=2/7, w_{0}=1/3w_{2}=2/21, and w_{3}=1/3w_{2}+1/3w_{6}+1/3w_{11}=2/21+5/24+1/3=107/168. w_{1}=1/2w_{0}+1/2w_{2}=1/21+107/336=123/336 oops ... I've used w_{3} instead of w_{2}. Correct answer is 1/21+1/7=4/21.

« Last Edit: Jun 22^{nd}, 2007, 11:37am by Hippo » 
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Wardub
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Re: Tunnels of Callicrates
« Reply #31 on: Apr 8^{th}, 2008, 12:03am » 
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What about the probability if back tracking isn't allowed but you can move upward. Also assume you start moving down from the top. So at the first intersection you can only move left or right. This lets solutions like left, down, right, up, right, up, right, right, down, down, right, down. I believe I have the answer. .102996826 or 3375/32768. I found 48 unique paths. I also found it interesting that the probability lowered when you allowed moving up. I think this is because at the last intersection right before the end. There is always a 50% chance you take the wrong path and end up stopped. Could someone check my answer, there is a high chance I did something wrong. Also could someone clarify something about backtracking allowed. When backtracking is allowed. Can you go up? If so wouldn't you eventually end at the win place?

« Last Edit: Apr 8^{th}, 2008, 12:26am by Wardub » 
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