Thanks for answer
Surely Smallester Mouse
Surely Smallesterest Mouse
Since challenge 'Surely Smallesterest Mouse' apparently can only be solved with incredible luck, I would like to discuss potential solution ideas openly. Do you agree with that?
The most general solution which I have so far, returns max{cell[r ± 1], cell[s]}, where 0 < r < 9 and 0 ≤ s ≤ 9:
(r + 1)<(r - 1)<:r+< s< 11^3^::v p
With at most 19 instructions, is there a better approach than comparing three memory cells and returning their maximum? Are there perhaps memory cells where the largest number occurs more frequent?
The most general solution which I have so far, returns max{cell[r ± 1], cell[s]}, where 0 < r < 9 and 0 ≤ s ≤ 9:
(r + 1)<(r - 1)<:r+< s< 11^3^::v p
With at most 19 instructions, is there a better approach than comparing three memory cells and returning their maximum? Are there perhaps memory cells where the largest number occurs more frequent?
At least I am sending max of 3 cells algorithm each 1 hour. Average solution time is about 1.5 year. I am still unlucky.
Actually I use just
so your code is more general (I have 0,1 fixed instead of your's r+/-1) and I use ^ rather to v.
BTW: I have tested it on already solved mouses (where there is no pondering penalty) and number of tries till success was "random" as expected.
Actually I use just
Code: Select all
10<1<::<s<11^3^::^p
BTW: I have tested it on already solved mouses (where there is no pondering penalty) and number of tries till success was "random" as expected.
Surely Smallesterest Mouse
It would be nice to get some information from the five solvers, but I do not count on it. On the other hand, except for Enodo, all solvers seem to be still active. Perhaps it makes sense to send them a PM to draw their attention to the thread? I have to think about it.
Uniform distribution
I have inspected 5,000 test cases and here is how the maximum was distributed over the first 20 locations in memory:
This looks evenly distributed to me. In addition, your program would not have been successful and mine exactly once for the parameters r = 4, s = 6, r = 4, s = 9 and r = 7, s = 5.
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cell || 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
------++-----+-----+-----+-----+-----+-----+-----+-----+-----+---- ...
count || 260 | 259 | 242 | 253 | 251 | 240 | 253 | 237 | 254 | 238
10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19
... ----+-----+-----+-----+-----+-----+-----+-----+-----+----
224 | 244 | 268 | 241 | 258 | 243 | 228 | 269 | 269 | 269
I am clearly unlucky as my solution takes already more than expected value ... still expected value is 1.5 year from now solution.Hippo wrote:At least I am sending max of 3 cells algorithm each 1 hour. Average solution time is about 1.5 year. I am still unlucky.
Actually I use justso your code is more general (I have 0,1 fixed instead of your's r+/-1) and I use ^ rather to v.Code: Select all
10<1<::<s<11^3^::^p
BTW: I have tested it on already solved mouses (where there is no pondering penalty) and number of tries till success was "random" as expected.
BTW: a.goth ... I agree your solution can match more possible cases, but the success rate is same as trying memory 0,1,2 in each attempt. ... OK if the generator would not generate equally random cases, knowing it could favour you.
There is still harvestsnow's hint about 7/40 success probability per test (what was not confirmed nor rejected).
Surely Smallesterest Mouse
I have already been pondering a lot about harvestsnow's post. The only way I have found so far to achieve this probability of success is to use 4 of the 20 numbers and return their maximum in 21 of the 24 possible permutations. This would result in--correct me if I am wrong--a probability of success of exactly 4/20 * 21/24 = 7/40. Unfortunately, I have not found such a HVM program yet. Any ideas anyone?Hippo wrote:There is still harvestsnow's hint about 7/40 success probability per test (what was not confirmed nor rejected).