[QUIZ] One-Liners Mashup (#177 again)

ruby-talk
41

def per(n, p = 16)
  (1..(p/2)).detect { |x| ("%.#{p}f" %
(1.0/n)).split(".").last.scan(%r{.{#{x}}}).uniq.size == 1 } || 0
end

This produces correct output for the numbers you specified. The optional 'p'
param sets the precision level for converting the number to a string.

···

2008/9/20 Holger Mack <hmack.gm@googlemail.com>

Periodicity of 1/n -> length of the recurring sequence of the decimal
fraction of 1/n
- 1/3 = 0.3333333.... -> "3" -> 1
- 1/7 = 0.1428571428571.... -> "142957" -> 6
- 1/11 = 0.09090909.... -> "09" -> 2

a bonus point: other than decimal fraction

42

...and I got n and s reversed :frowning:

Regards,
Pit

···

2008/9/20 Pit Capitain <pit.capitain@gmail.com>:

I don't think that your method rolls the dice s times.

43

James Coglan wrote:

def per(n, p = 16)
(1..(p/2)).detect { |x| ("%.#{p}f" %
(1.0/n)).split(".").last.scan(%r{.{#{x}}}).uniq.size == 1 } || 0
end

This produces correct output for the numbers you specified.

It produces 0 for 6 though (because printf rounds instead of truncating so you
get a 7 at the end).
Here's one that should always be accurate and it's even in 1 line and handles
different bases:
def per(n, b = 10)
  i=1;x=b;h={};loop {x=x%n*b;break 0 if x==0;h?(break i-h):h=i; i+=1}
end

···

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Jabber: sepp2k@jabber.org
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44

Here's a ruby 1.9 version that doesn't use semicolons or mutable data
structures.

def roll n, s
  ((1..s).to_a+n.times.map{|x| rand(s)+1}).group_by{|x| x}.map{|k,v| "#{k}|"+"#"*(v.size-1)}.join("\n")
end

--Ken

···

In article <a245658d0809201035q6a76601ao140ffe56ff219334@mail.gmail.com> you wrote:

2008/9/20 James Coglan <jcoglan@googlemail.com>:

def roll(n,s)
(1..n).map { |x| "#{x}|#{'#' * (1 + rand(s))}" } * "\n"
end

puts roll(20,4)

I don't think that your method rolls the dice s times.

def roll(n, s)
a=(1..n).map{|x|"#{x}|"};s.times{a[rand(n)]<<?#};puts a
end

--
Chanoch (Ken) Bloom. PhD candidate. Linguistic Cognition Laboratory.
Department of Computer Science. Illinois Institute of Technology.
http://www.iit.edu/~kbloom1/

45

Would be interested to see if you could produce a version with no
assignments or semicolons. With Ruby's functional and OO capabilities, it's
always more satisfying if you can solve a problem with a single statement,
and I think these little one-liner problems are great for getting you
thinking in functional style.

···

2008/9/21 Sebastian Hungerecker <sepp2k@googlemail.com>

James Coglan wrote:
> def per(n, p = 16)
> (1..(p/2)).detect { |x| ("%.#{p}f" %
> (1.0/n)).split(".").last.scan(%r{.{#{x}}}).uniq.size == 1 } || 0
> end
>
> This produces correct output for the numbers you specified.

It produces 0 for 6 though (because printf rounds instead of truncating so
you
get a 7 at the end).
Here's one that should always be accurate and it's even in 1 line and
handles
different bases:
def per(n, b = 10)
i=1;x=b;h={};loop {x=x%n*b;break 0 if x==0;h?(break i-h):h=i;
i+=1}
end

46

James Coglan wrote:

> def per(n, b = 10)
> i=1;x=b;h={};loop {x=x%n*b;break 0 if x==0;h?(break i-h):h=i;
> i+=1}
> end

Would be interested to see if you could produce a version with no
assignments or semicolons.

def per(n, b=10)
  (1..1.0/0).inject([b,{}]) { |(x,h),i| if x==0 then break 0 elsif h[x%n*b]
then break i-h[x%n*b] else [x%n*b, h.merge(x%n*b=>i)] end}
end
But frankly, I like my first solution better. And this one clearly does not
fit in one line.

···

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