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Topic: Tunnels of Callicrates (Read 6289 times) |
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rmsgrey
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Re: Tunnels of Callicrates
« Reply #25 on: Jun 19th, 2007, 7:18am » |
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on Jun 19th, 2007, 3:10am, Sir_Rogers wrote:3) 1-2-4-5-6
Now the longest way only uses 5 of the circles |
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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 19th, 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 right-left 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 wxtop winning probability when arriving to vertex x from top, similarly
for bot,rt,lft. We can start with w14lft=1, w13rt=0,
w12rt=0, w7rt=0.
Continue with easiest equations: w14top=1/2+1/2 w10bot=
1/2+1/2 w11lft=
1/2+1/2w14top.
Therefore w14top=1
=w10bot=w11lft.
Now w10top=
1/2(w11lft+w14lft)=1 and similarly
w11top=
1/2(w10rt+w14top)=
1/2(w14lft+w14top)=1.
Follow with more complicated equations:
w13top=
0+1/2w5bot=
1/4w4rt+1/4w6lft=
1/4w9top+1/4w10top=
1/8w12rt+1/8w13top+1/4=
0+1/8w13top+1/4.
Therefore 7/8 w13top=1/4 and w13top=2/7,
w4rt=
w9top=1/7,
Now w12top=
0+1/2w9lft=
1/2w9top=1/14 and
w8lft=
w12top=1/14 and
w7top=
0+1/2w8lft=1/28.
Similarly w8top=
0+1/2w12top=
1/2w8lft=1/28.
Now w0rt=
1/2(w7top+w8top)=1/28.
Now w6top=
1/2w5rt+1/2=
1/4w4rt+0+1/2=15/28 and
w4top=
1/2w9top+1/2w5lft=
1/14+1/4=18/56.
Finaly w3lft=
1/2(w6top+1)=
43/56 and w2lft=
1/2w4top+1/2w3lft=
61/112 and w1top=
1/2w0rt+1/2w2lft=
69/224.
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| « Last Edit: Jun 22nd, 2007, 11:38am by Hippo » |
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Sir_Rogers
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Re: Tunnels of Callicrates
« Reply #27 on: Jun 19th, 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 19th, 2007, 12:49pm » |
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I first saw the problem here.
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Grimbal
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Re: Tunnels of Callicrates
« Reply #29 on: Jun 20th, 2007, 1:39am » |
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on Jun 19th, 2007, 3:10am, Sir_Rogers wrote:1) 1-2-3
2) 1-2-3-6
3) 1-2-4-5-6
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In
3) 1-2-4-5-6
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.
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| « Last Edit: Jun 20th, 2007, 1:40am by Grimbal » |
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Hippo
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Re: Tunnels of Callicrates
« Reply #30 on: Jun 20th, 2007, 4:25am » |
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on Jun 19th, 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 ...
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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.
w14=1, w13=0, w12=0
The bar from 10 to 11 does not play any role ... (an easy equation) w11=1/2w10+1/2w14=
1/4w11+3/4w14=1/4w11+3/4 and we get
w11=1=w10.
Furthermore w9=1/2w12+1/2w13=0, w8=w12=0,
and w7=0+1/2w8=0.
Now the interesting part begins
w5=1/3w4+1/3w13+1/3w6=1/6w9+1/6w5+0+1/6w5+1/6w10 =
0+1/3w5+1/6 and therefore w5=3/2*1/6=1/4, w6=5/8, and w4=1/8.
Finaly w2=1/3w0+1/3w4+1/3w3=
1/9w7+1/9w8+1/9w2+
1/24+1/9w2+1/9w6+1/9w11=
0+0+2/9w2+1/24+5/72+1/9 and w2=9/7*(1/24+5/72+1/9)=2/7, w0=1/3w2=2/21,
and w3=1/3w2+1/3w6+1/3w11=2/21+5/24+1/3=107/168.
w1=1/2w0+1/2w2=1/21+107/336=123/336
oops ... I've used w3 instead of w2. Correct answer is 1/21+1/7=4/21.
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| « Last Edit: Jun 22nd, 2007, 11:37am by Hippo » |
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Wardub
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Re: Tunnels of Callicrates
« Reply #31 on: Apr 8th, 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?
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| « Last Edit: Apr 8th, 2008, 12:26am by Wardub » |
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