Timeline for Maze Solving Robot
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 17, 2015 at 12:39 | comment | added | kasperd | @ypercube Yes, that's the idea. | |
| Jul 17, 2015 at 9:21 | comment | added | ypercubeᵀᴹ | @kasperd I have no idea what you are talking about. ("to perform what calculation")? Ah, ok, I think I understand what your point is. Create a (long enough) random list of directions, then check it against each configuration. If it solves all of them good. If it doesn't solve some them, extend the list and check again (until you have a long enough list that solves all.) Right? | |
| Jul 16, 2015 at 19:28 | comment | added | kasperd | @ypercube You are making the same erroneous assumption as Taemyr did. I proposed to look at the actual random values and use that to perform the calculation. Each time that algorithm is used it will produce a finite sequence of steps, which is long enough. If you execute that algorithm again it will produce a different sequence which most likely will be of a different length, but each sequence in itself is long enough. | |
| Jul 16, 2015 at 16:57 | comment | added | ypercubeᵀᴹ |
@kasperd you can't compute how many random steps will be needed. For all we know the first trillion trillion trillion moves can be North, South, North, South, ... which will not get you from many mazes ;)
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| Jul 16, 2015 at 11:19 | comment | added | ypercubeᵀᴹ | @Taemyr True. And a non-zero probability of reaching the end. With a long enough sequence, the probability will be high. | |
| Jul 16, 2015 at 11:02 | comment | added | kasperd | @Taemyr Only if you assume the random choices are made after the length of the sequence has been decided. But that is not the case with the method I sketched, because it computes the length of the sequence based on the random choices being made. | |
| Jul 16, 2015 at 10:54 | comment | added | Taemyr | @kasperd For any initial condition any finite length of random steps will have a non-zero probability of not reaching the end. | |
| Jul 16, 2015 at 10:23 | comment | added | kasperd | Since there is only a finite set of initial conditions, you can compute how many of the random steps are needed to solve each initial condition. Taking the maximum of all these numbers will tell you how long the list needs to be. But I am thinking this will be less efficient than xnor's accepted solution. | |
| Jul 16, 2015 at 9:55 | history | edited | ypercubeᵀᴹ | CC BY-SA 3.0 |
added 118 characters in body
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| Jul 16, 2015 at 9:50 | review | First posts | |||
| Jul 16, 2015 at 10:03 | |||||
| Jul 16, 2015 at 9:49 | history | answered | ypercubeᵀᴹ | CC BY-SA 3.0 |