Moritz Schauer

 The Nineteenth Byte

The Nineteenth Byte: General discussion for codegolf.stackexc...
Nov 7, 2017 20:47
hi, does anybody know the link to a first prototype of I believe J in something like 15 lines of C which was mentioned in an article available online I cannot find anymore? it really bothers me to have lost that
Nov 1, 2017 14:55
i posted a script generating the answer. i am a bit more lucky than you that julia has a isidentifier function
Nov 1, 2017 14:24
julia can have 126427 snippets
Nov 1, 2017 14:02
thanks
Nov 1, 2017 14:00
Sad.
Nov 1, 2017 13:59
Q: How do you deal with tied winners usually?
Nov 29, 2016 21:50
it agrees quite well with your graph
Nov 29, 2016 21:49
the first formula is valid if only numbers with digit 7 are unspeakable, the second is closer when also numbers divisible by 7 are unspeakable
Nov 29, 2016 21:48
@NiklasB. f(k0) above is an estimate of i given k. a slightly improved version is
k=k0=big(7)^77;i=0;while k > 1;i+=1; k = k*(1 - (9/10)^(log(10,k)-2)*0.73);end;
Nov 29, 2016 21:46
f(k0) = begin k=k0;i=0;while k > 1;i+=1; k = k*(1 - k^(log(10,9/10)));end; println(log(i,k0)); i; end
Nov 29, 2016 21:42
i have a probabilistic way to compute it
Nov 29, 2016 21:42
thank you
Nov 29, 2016 21:38
to be specific, when i is the number of ones and k the length, I thought i = log(alpha,k) and you showed i = k^alpha
Nov 29, 2016 21:36
and you showed that it is infact polynomial in the length of the sequence
Nov 29, 2016 21:36
no, so initially i and some others thought that the number of new levels in the sequence is of logarithmic order
Nov 29, 2016 21:34
i was curious why your algorithm breaks down when the number of ones grows logarithmically instead of polynomially (or did I not get this right?)
Nov 29, 2016 21:33
ah, hi
Nov 29, 2016 20:36
see previous message
Nov 29, 2016 20:35
@GabrielBenamy