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# Mathematics under The Library of Babel

## 2022-11-24​

• To hold all occasions of a-z, a single character, requires 26 characters.
• ASCII: 128
• Considering any Unicode can be ASCII escaped, Unicode LoB will use the same equation but takes a much longer ASCII base string
• For two characters to hold all possible sequences, what should be the shortest sequence?
• Naively, it may take $2 \times 26 \times 26$ where $26 \times 26$ is all the combinations, and 2 is the length of a single combination.
• like aa ab ac ad ae ... without the space
• Naively adding, since any character can be shared bidirectionally, it seems possible to hold it inside $26 \times 26$
aaabacadaeafagahaiajakalamanaoapaqarasatauavawaxayazbabbbcbdbebfbgbhbibjbkblbmbnbobpbqbrbsbtbubvbwbxbybzcacbcccdcecfcgchcicjckclcmcncocpcqcrcsctcucvcwcxcyczdadbdcdddedfdgdhdidjdkdldmdndodpdqdrdsdtdudvdwdxdydzeaebecedeeefegeheiejekelemeneoepeqereseteuevewexeyezfafbfcfdfefffgfhfifjfkflfmfnfofpfqfrfsftfufvfwfxfyfzgagbgcgdgegfggghgigjgkglgmgngogpgqgrgsgtgugvgwgxgygzhahbhchdhehfhghhhihjhkhlhmhnhohphqhrhshthuhvhwhxhyhziaibicidieifigihiiijikiliminioipiqirisitiuiviwixiyizjajbjcjdjejfjgjhjijjjkjljmjnjojpjqjrjsjtjujvjwjxjyjzkakbkckdkekfkgkhkikjkkklkmknkokpkqkrksktkukvkwkxkykzlalblcldlelflglhliljlklllmlnlolplqlrlsltlulvlwlxlylzmambmcmdmemfmgmhmimjmkmlmmmnmompmqmrmsmtmumvmwmxmymznanbncndnenfngnhninjnknlnmnnnonpnqnrnsntnunvnwnxnynzoaobocodoeofogohoiojokolomonooopoqorosotouovowoxoyozpapbpcpdpepfpgphpipjpkplpmpnpopppqprpsptpupvpwpxpypzqaqbqcqdqeqfqgqhqiqjqkqlqmqnqoqpqqqrqsqtquqvqwqxqyqzrarbrcrdrerfrgrhrirjrkrlrmrnrorprqrrrsrtrurvrwrxryrzsasbscsdsesfsgshsisjskslsmsnsospsqsrssstsusvswsxsysztatbtctdtetftgthtitjtktltmtntotptqtrtstttutvtwtxtytzuaubucudueufuguhuiujukulumunuoupuqurusutuuuvuwuxuyuzvavbvcvdvevfvgvhvivjvkvlvmvnvovpvqvrvsvtvuvvvwvxvyvzwawbwcwdwewfwgwhwiwjwkwlwmwnwowpwqwrwswtwuwvwwwxwywzxaxbxcxdxexfxgxhxixjxkxlxmxnxoxpxqxrxsxtxuxvxwxxxyxzyaybycydyeyfygyhyiyjykylymynyoypyqyrysytyuyvywyxyyyzzazbzczdzezfzgzhzizjzkzlzmznzozpzqzrzsztzuzvzwzxzyzzlen: 1352
• aligns with initial calc $2 \times 26 \times 26$
• How can we compress it?
• Keep a map from the beginning; remove the substring if seen already
• may have a side effect
• early small sacrifices (redundancy) may save more significant future sacrifices. This does not consider such a case
• We can, for sure, trim three characters repeatedly char to 2
aabacadaeafagahaiajakalamanaoapaqarasatauavawaxayazbabbcbdbebfbgbhbibjbkblbmbnbobpbqbrbsbtbubvbwbxbybzcacbccdcecfcgchcicjckclcmcncocpcqcrcsctcucvcwcxcyczdadbdcddedfdgdhdidjdkdldmdndodpdqdrdsdtdudvdwdxdydzeaebecedeefegeheiejekelemeneoepeqereseteuevewexeyezfafbfcfdfeffgfhfifjfkflfmfnfofpfqfrfsftfufvfwfxfyfzgagbgcgdgegfgghgigjgkglgmgngogpgqgrgsgtgugvgwgxgygzhahbhchdhehfhghhihjhkhlhmhnhohphqhrhshthuhvhwhxhyhziaibicidieifigihiijikiliminioipiqirisitiuiviwixiyizjajbjcjdjejfjgjhjijjkjljmjnjojpjqjrjsjtjujvjwjxjyjzkakbkckdkekfkgkhkikjkklkmknkokpkqkrksktkukvkwkxkykzlalblcldlelflglhliljlkllmlnlolplqlrlsltlulvlwlxlylzmambmcmdmemfmgmhmimjmkmlmmnmompmqmrmsmtmumvmwmxmymznanbncndnenfngnhninjnknlnmnnonpnqnrnsntnunvnwnxnynzoaobocodoeofogohoiojokolomonoopoqorosotouovowoxoyozpapbpcpdpepfpgphpipjpkplpmpnpoppqprpsptpupvpwpxpypzqaqbqcqdqeqfqgqhqiqjqkqlqmqnqoqpqqrqsqtquqvqwqxqyqzrarbrcrdrerfrgrhrirjrkrlrmrnrorprqrrsrtrurvrwrxryrzsasbscsdsesfsgshsisjskslsmsnsospsqsrsstsusvswsxsysztatbtctdtetftgthtitjtktltmtntotptqtrtsttutvtwtxtytzuaubucudueufuguhuiujukulumunuoupuqurusutuuvuwuxuyuzvavbvcvdvevfvgvhvivjvkvlvmvnvovpvqvrvsvtvuvvwvxvyvzwawbwcwdwewfwgwhwiwjwkwlwmwnwowpwqwrwswtwuwvwwxwywzxaxbxcxdxexfxgxhxixjxkxlxmxnxoxpxqxrxsxtxuxvxwxxyxzyaybycydyeyfygyhyiyjykylymynyoypyqyrysytyuyvywyxyyzzazbzczdzezfzgzhzizjzkzlzmznzozpzqzrzsztzuzvzwzxzyzzlen: 1327
• Not a drastic difference
• Can we trim even further?
• I tried making a Python script to backtrack the combinations.
• Killed by the OS for using too much RAM
❯ python3 experiment3.pymaking list of length 52[1]    6919 killed     python3 experiment3.py
• Also made a C++ variant, but a meaningful difference is unlikely.

## 2022-11-25​

Continuing our discussion yesterday...

So given

aabacadaeafagahaiajakalamanaoapaqarasatauavawaxayazbabbcbdbebfbgbhbibjbkblbmbnbobpbqbrbsbtbubvbwbxbybzcacbccdcecfcgchcicjckclcmcncocpcqcrcsctcucvcwcxcyczdadbdcddedfdgdhdidjdkdldmdndodpdqdrdsdtdudvdwdxdydzeaebecedeefegeheiejekelemeneoepeqereseteuevewexeyezfafbfcfdfeffgfhfifjfkflfmfnfofpfqfrfsftfufvfwfxfyfzgagbgcgdgegfgghgigjgkglgmgngogpgqgrgsgtgugvgwgxgygzhahbhchdhehfhghhihjhkhlhmhnhohphqhrhshthuhvhwhxhyhziaibicidieifigihiijikiliminioipiqirisitiuiviwixiyizjajbjcjdjejfjgjhjijjkjljmjnjojpjqjrjsjtjujvjwjxjyjzkakbkckdkekfkgkhkikjkklkmknkokpkqkrksktkukvkwkxkykzlalblcldlelflglhliljlkllmlnlolplqlrlsltlulvlwlxlylzmambmcmdmemfmgmhmimjmkmlmmnmompmqmrmsmtmumvmwmxmymznanbncndnenfngnhninjnknlnmnnonpnqnrnsntnunvnwnxnynzoaobocodoeofogohoiojokolomonoopoqorosotouovowoxoyozpapbpcpdpepfpgphpipjpkplpmpnpoppqprpsptpupvpwpxpypzqaqbqcqdqeqfqgqhqiqjqkqlqmqnqoqpqqrqsqtquqvqwqxqyqzrarbrcrdrerfrgrhrirjrkrlrmrnrorprqrrsrtrurvrwrxryrzsasbscsdsesfsgshsisjskslsmsnsospsqsrsstsusvswsxsysztatbtctdtetftgthtitjtktltmtntotptqtrtsttutvtwtxtytzuaubucudueufuguhuiujukulumunuoupuqurusutuuvuwuxuyuzvavbvcvdvevfvgvhvivjvkvlvmvnvovpvqvrvsvtvuvvwvxvyvzwawbwcwdwewfwgwhwiwjwkwlwmwnwowpwqwrwswtwuwvwwxwywzxaxbxcxdxexfxgxhxixjxkxlxmxnxoxpxqxrxsxtxuxvxwxxyxzyaybycydyeyfygyhyiyjykylymynyoypyqyrysytyuyvywyxyyzzazbzczdzezfzgzhzizjzkzlzmznzozpzqzrzsztzuzvzwzxzyzzlen: 1327

This can be further trimmed because ba already appears at zero-based index $[2, 3]$. So we don't need it again after az. That is, we don't need azbabb..., we can jump to azbbc... This will create a triangular space complexity reduction, reducing the size to half. So

a: a → z (aabacadaea...)b: (start from b, skipping a)c: (start from c, skipping b and a)

Thus resulting in half in size. If this is true, this aligns with my naive conjecture at the beginning → to be lower-bounded at $26 \times 26$.

I made a script to confirm if this is right...

Yes, it is.

aabacadaeafagahaiajakalamanaoapaqarasatauavawaxayazbbcbdbebfbgbhbibjbkblbmbnbobpbqbrbsbtbubvbwbxbybzccdcecfcgchcicjckclcmcncocpcqcrcsctcucvcwcxcyczddedfdgdhdidjdkdldmdndodpdqdrdsdtdudvdwdxdydzeefegeheiejekelemeneoepeqereseteuevewexeyezffgfhfifjfkflfmfnfofpfqfrfsftfufvfwfxfyfzgghgigjgkglgmgngogpgqgrgsgtgugvgwgxgygzhhihjhkhlhmhnhohphqhrhshthuhvhwhxhyhziijikiliminioipiqirisitiuiviwixiyizjjkjljmjnjojpjqjrjsjtjujvjwjxjyjzkklkmknkokpkqkrksktkukvkwkxkykzllmlnlolplqlrlsltlulvlwlxlylzmmnmompmqmrmsmtmumvmwmxmymznnonpnqnrnsntnunvnwnxnynzoopoqorosotouovowoxoyozppqprpsptpupvpwpxpypzqqrqsqtquqvqwqxqyqzrrsrtrurvrwrxryrzsstsusvswsxsyszttutvtwtxtytzuuvuwuxuyuzvvwvxvyvzwwxwywzxxyxzyyzzalen: 677

Can we compress even further? The mathematic lower bound seems $26 \times 26$, given each character must at least pair with each other once (thus $26 \times 26$,) times the length of the search scope (thus $\times 2$,) and since each character can be reused by each other side, divide it in half (this $\div 2$)

Also, note that the mathematic lower bound could be attained if the above Babel String loops.

If this calculation is accurate, we can expand this to a string of size $3$. Similarly, the estimated lower bound would be $26 \times 26 \times 26 \times 3 \div 2 = 26364.$

## 2022-11-26​

I think the above is wrong.

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18578