Let's crack Sloot algorithm - infinite "compression"

[lancaIV]

The best heads works for solution    : https://en.wikipedia.org/wiki/Beale_ciphers


https://en.wikipedia.org/wiki/The_Code_Book


https://en.wikipedia.org/wiki/Quantum_cryptography




https://www.degruyter.com/view/journals/nanoph/ahead-of-print/article-10.1515-nanoph-2020-0110/article-10.1515-nanoph-2020-0110.xml?language=en


https://en.wikipedia.org/wiki/Quantum_superposition  Experiments and applications


The code-key has ever to be "perfect identical" = finite stage  ! Decoding input = Encoding output


Pi and https://www.youtube.com/watch?v=io11cbrTTNg are "endless" numbers, infinite !


1      = I  Unum                 /1                          /= per or pro = in-version
10    = X/D ecem              / 10
100  = C    entum             / 100
1000=M     ile                  /  1000
........


0 = nullus or zero

[nix85]

i used to think calculating is the key too, that was my first idea, to represent big numbers as short formulas, of course that's a bad idea

its not a problem to generate long strings of numbers, so many ways to do that. to do it in a predictable way for arbitrary values is different story

[nix85]

when i first got into it i also gave a lot of thought to prime numbers, like indexing first say 1000 primes and breaking down superlong numbers using these and similar. blind alley.

there must be a simple workaround that allows a repatable compression, that is the only way that makes sense how he achieved 1:2 million ratio


                              4
                             34
                            234
                           1234
                          01234

[WhatIsIt]

http://endlesscompression.com

You probably saw this page...

[nix85]

http://endlesscompression.com

You probably saw this page...

yes, he talks a lot but does not have a solution.

to elaborate more on one of first ideas i had, how to turn 2 digit number into 1

let's say you want to compress 45

random numbers are also 2 digit and at that moment number is 56

each random number could have it's own complex mathematical formula

but let's keep it simple and say we just divide, we calculate

56/45=1.24444444444

let's say we are writing down 3rd digit after decimal point so 4

we could as well calculate 56-45=11 and write down numerical value 2 or something else, and infinity of other formulas

so on decoder side, it sees answer is 4, random number 56, it knows which operation has been performed for that random number...

so it calculates or looks in the table of precalculated values

if 45 is the only of 100 numbers that has 4 at that decimal point for that operation, then we got a score

what are the chances

let's see


56/99=0.56565656565
56/98=0.57142857142
56/97=0.57731958762
56/96=0.58333333333
56/95=0.58947368421
56/94=0.59574468085
56/93=0.60215053763
56/92=0.60869565217
56/91=0.61538461538
56/90=0.62222222222
56/89=0.62921348314
56/88=0.63636363636
56/87=0.64367816092
56/86=0.65116279069
56/85=0.65882352941
56/84=0.66666666666
56/83=0.67469879518 and ups another one, so no good