4 ms·
If it's 3 non-repeating unordered words, the minimum vocabulary will need to be 256*3 = 768. Realistically you'll need an order of magnitude larger. That's a pr
by pishpash 7y ago
If it's 3 non-repeating unordered words, the minimum vocabulary will need to be 256*3 = 768. Realistically you'll need an order of magnitude larger. That's a pretty large vocabulary.
- bottled_poe 7y agoThat’s not right. 256 * 255 * 254
- _0ffh 7y agoI suspect you didn't catch gp's meaning: To get to 24 bits of color info, you'd need at least 3 words of 8 bit info each. If you want these to be non-repeating and unordered that makes 256x3=768 unique words.
- samatman 7y agoThis isn't correct though. You need 256 base words, and two substitute words, one of which substitutes for the first repeat, and the other for the second, so 258 words total.
- IanCal 7y agoThat accounts for the repeating but not the ordering, right?
- samatman 7y agoOh, by 'unordered' you mean that the three words could be in any order and still mean the same thing? Okay but that's a strange requirement. A 'big burger house' isn't a 'big house burger', after all. I had read 'unordered' as 'the collation order of the words has no connection to the ordinal rank of the underlying byte', for some reason
- IanCal 7y agoI'm just jumping in where I saw a misunderstanding :) I think that's what they mean by unordered as that makes sense with the numbers (I think). To some degree I agree, but there are also a lot of cases where different orders aren't distinguishable easily. "Light bright red" and "bright light red" is the best I can come up with quickly but I'm sure you get the idea.
- suyjuris 7y agoIt is a minor point, but 467 words would suffice, as there are 16865705 possibilities to choose 3 elements out of 467. (This is the smallest possible.)
- _0ffh 7y agoStrictly speaking, yes, if you just view it as unordered sampling without replacement. Although that would squeeze out nearly all of the creative side of things.