https://towardsdatascience.com/random-seeds-and-reproducibility-933da79446e3And how does this random.seed() work?
set some value and then Mersenne twister...
Mersenne twister 19937 bit (624·32 (2^32 = 4294967296) — 31)
for example, we take 3 random
random.seed(blablabla)
random.randrange(1,10)
random.randrange(1,10)
random.randrange(1,10)
we get for each of the 3 in order from the vortex of the first three?
1— 31
random.randrange(1,10) 1,4294967296 (624·2)
random.randrange(1,10) 1,4294967296 (624·3)
random.randrange(1,10) 1,4294967296 (624·4)
and if we take 624 random.randrange(1,10) period ends and a new one begins again
1— 31 next period
random.randrange(1,10) 1,4294967296 (625)624·2
random.randrange(1,10) 1,4294967296 (626)624·3
random.randrange(1,10) 1,4294967296 (627)624·4
and if we have a large sample of random.randrange(36893488147419103232,73786976294838206464) he spends 1 period for 1 sample or 624 and a new one (then why are they not repeated 2^32?)
or he these 19937 bit takes it all at once
in other words, to complete all puzzles with 1 seed() we need to iterate over this seed() to iterate over all variations of this 2^19937 bit ((2^32)^624)?
and seed() itself doesn’t matter (with brute force) you don’t need old computers to run it randomly in order to pick up the date and time, etc. to create the same bitcoin address.
random.seed(1, 2^19937) and all pz
random.randrange(36893488147419103232,73786976294838206464) 66
...
random.randrange(21267647932558653966460912964485513216,42535295865117307932921825928971026432) 125
...
random.randrange(730750818665451459101842416358141509827966271488,1461501637330902918203684832716283019655932542976) 160
and the puzzles themselves need to be multiplied 2^160×2^159×2^158...×2^66
and if it turns out that there may be collisions here , If 2^19937 the most options to open all the puzzles at once through 1 seed()
