Compress specific text for storage.
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Nothing wrong with that. In fact, having your reply so clear and having my reply which elaborates on it probably will help the OP to understand both approaches more clearly. Just thought I'd add some additional input to this little brainstorm session. :)
I can only hope he is not studying DCT algorithms and the like. :laugh: I don't think he'll bring it well below 13 bytes, unless there is a lot he hasn't told. And that is the main reason I offered him Shannon, so he can calculate the optimum himself. :)
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Luc Pattyn wrote:
if corners can only have 4 or fewer values that saves 1 bit each
With 4 corners and each corner allowing only 3 possible values, all 4 corners can be stored using 7 bits (because 3^4 = 81 possible combinations, which is less than the 128 combinations 7 bits can store). So, rather than use 2 bits per corner (for a total of 8 bits), only 7 bits are required. It's only a bit, but that might just help to make it down to the next lower byte if the OP can come up with other compression/encoding techniques to save space. :)
And if all the cells on the edge can only have 5 different values, you can combine each corner cell with an edge cell, giving 4 * 15 possibilities (6 bits), and then you have 5^12 possibilities left for all other edge cells together which fits in 28 bits, giving 85 bits in total.
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An interesting concept, however it mainly makes apparent a 4-byte seed isn't sufficient to generate all conceivable board set-ups: rather than some 2^108 it will produce no more than 2^32 of them, a mere drop in the ocean. BTW: the OP mentioned trillions, not sure he meant that in the long or short scale of ways (see here[^]) :)
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True, but 2^32 ought to be enough for anybody :)
Proud to have finally moved to the A-Ark. Which one are you in?
Author of the Guardians Saga (Sci-Fi/Fantasy novels) -
And if all the cells on the edge can only have 5 different values, you can combine each corner cell with an edge cell, giving 4 * 15 possibilities (6 bits), and then you have 5^12 possibilities left for all other edge cells together which fits in 28 bits, giving 85 bits in total.
harold aptroot wrote:
And if all the cells on the edge can only have 5 different values
I don't really follow what you said (e.g., what do you mean by "combine each corner cell with an edge cell"?), but keep this in mind:
some specific cells (Corner cells) can have only 3 of those values
We already know each corner cell can only use 3 (rather than 8) values. I didn't just pick an arbitrary value, if that's what you are implying (I honestly don't know what you mean).
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True, but 2^32 ought to be enough for anybody :)
Proud to have finally moved to the A-Ark. Which one are you in?
Author of the Guardians Saga (Sci-Fi/Fantasy novels)like 640KB? :-D
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True, but 2^32 ought to be enough for anybody :)
Proud to have finally moved to the A-Ark. Which one are you in?
Author of the Guardians Saga (Sci-Fi/Fantasy novels)I'd say nobody will ever need more than 640KB of RAM. :rolleyes:
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like 640KB? :-D
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LOL, you just beat me.
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LOL, you just beat me.
It wasn't a very obscure joke, for a site like this :)
Proud to have finally moved to the A-Ark. Which one are you in?
Author of the Guardians Saga (Sci-Fi/Fantasy novels) -
harold aptroot wrote:
And if all the cells on the edge can only have 5 different values
I don't really follow what you said (e.g., what do you mean by "combine each corner cell with an edge cell"?), but keep this in mind:
some specific cells (Corner cells) can have only 3 of those values
We already know each corner cell can only use 3 (rather than 8) values. I didn't just pick an arbitrary value, if that's what you are implying (I honestly don't know what you mean).
If the corner cells can have 3 values and the middle cells 8, then I see the pattern "they can hold as many values as there are adjacent cells". The OP did not say it, but he's been a bit slow in providing information anyway. It's just a guess, and that's why there is an if near the beginning of my post. The combining is .. just combining. Combine a corner cell (3 states) with an edge cell (might have 5 states) to get 15 in total, which is very close to a power of 2.
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It wasn't a very obscure joke, for a site like this :)
Proud to have finally moved to the A-Ark. Which one are you in?
Author of the Guardians Saga (Sci-Fi/Fantasy novels)Haha, yeah, I saw it posted the other day. I wondered if perhaps it was you who posted it. :)
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Haha, yeah, I saw it posted the other day. I wondered if perhaps it was you who posted it. :)
Don't think it was me... But hey, it's an old joke :)
Proud to have finally moved to the A-Ark. Which one are you in?
Author of the Guardians Saga (Sci-Fi/Fantasy novels) -
I'd say nobody will ever need more than 640KB of RAM. :rolleyes:
for 6*6 cells with no more than 8 states each, not really compression, is it? :)
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modified on Thursday, August 26, 2010 9:21 PM