Generate binary sequences of length n?

I’m rather new to Ruby. I feel this should be very simple, but I’m
having trouble:

I’d like to write a script to work through all possible binary sequences
of length n.

The script would work as follows:

mylength = 4
resultz = binary_seq_generator(mylength)
puts “#{resultz}”

#resultz, not necessarily in this order:
[“0000”,“0001”,“0010”,“0011”,“0100”,“0101”,“0110”,“0111”,“1000”,“1001”,“1010”,“1011”,“1100”,“1101”,“1110”,“1111”]

Important to my use: This binary_seq_generator method must be written
in such a way so that each subsequent member of the resultz array would
be completely generated before the next member starts. This is where
I’m stuck. I have trying manipulating the graycode algorithm here:
http://yagni.com/graycode/

This graycode algorithm seems to produce an array where each member is
only finalized on the final recurse. I may be wrong…but by this
method, I can’t use resultz[i] for something before the algorithm starts
to build resultz[i+1], and I need to use each member of the resultz
array before moving on to the next binary string in the sequence.

Thanks much for helping a newbie!

Hi –

On Mon, 10 Aug 2009, Tom Best wrote:

mylength = 4
resultz = binary_seq_generator(mylength)
puts “#{resultz}”

#resultz, not necessarily in this order:
[“0000”,“0001”,“0010”,“0011”,“0100”,“0101”,“0110”,“0111”,“1000”,“1001”,“1010”,“1011”,“1100”,“1101”,“1110”,“1111”]

Try this:

def binary_seq_generator(n)
(0…(1 << n)).map {|e| “%0#{n}d” % e.to_s(2) }
end

Doesn’t necessarily roll off the fingers as readily as some Ruby
idioms do :slight_smile: But I think all or most of what you need is there, and
there are some interesting bits to it.

David

On Sun, Aug 9, 2009 at 8:32 PM, David A. Black[email protected]
wrote:

Try this:

def binary_seq_generator(n)
 (0…(1 << n)).map {|e| “%0#{n}d” % e.to_s(2) }
end

Doesn’t necessarily roll off the fingers as readily as some Ruby
idioms do :slight_smile: But I think all or most of what you need is there, and
there are some interesting bits to it.
Well 1.9 has to offer some elegance here

(1<<n).times.map{ | d | “%0{n}b” % d }

HTH
Robert

(1<<n).times.map{ | d | “%0{n}b” % d }
Oh I just forgot, maybe you need the “combinatoric” method :wink:

n.times.inject( [ “” ] ){ |s,| s.map{ |e| [ e + “0”, e + “1” ] }.flatten
}

Robert D. wrote:

On Sun, Aug 9, 2009 at 8:32 PM, David A. Black[email protected]
wrote:

Try this:

def binary_seq_generator(n)
 (0…(1 << n)).map {|e| “%0#{n}d” % e.to_s(2) }
end

Doesn’t necessarily roll off the fingers as readily as some Ruby
idioms do :slight_smile: But I think all or most of what you need is there, and
there are some interesting bits to it.
Well 1.9 has to offer some elegance here

(1<<n).times.map{ | d | “%0{n}b” % d }

HTH
Robert

They worked! Thank you both very much - having a comprehensive
understanding of these operators will move my coding to the next level!
Very much appreciated, Tom

Robert D. wrote:

(1<<n).times.map{ | d | “%0{n}b” % d }

Perhaps safer to avoid the interpolation in the format string, using ‘*’
to give the number of digits.

“%0*b” % [8,123]
=> “01111011”

On Mon, Aug 10, 2009 at 10:43 AM, Brian C.[email protected]
wrote:

Robert D. wrote:

(1<<n).times.map{ | d | “%0{n}b” % d }

Perhaps safer to avoid the interpolation in the format string, using ‘*’
to give the number of digits.

“%0*b” % [8,123]
=> “01111011”
Well safer, you mean regarding to my typo, well spotted ;).
This is a fascinating idiom I was not aware of. I too prefer it, thx
for sharing.

Cheers
Robert

Hi –

On Mon, 10 Aug 2009, Robert D. wrote:

[“0000”,“0001”,“0010”,“0011”,“0100”,“0101”,“0110”,“0111”,“1000”,“1001”,“1010”,“1011”,“1100”,“1101”,“1110”,“1111”]
Well 1.9 has to offer some elegance here

(1<<n).times.map{ | d | “%0{n}b” % d }

Good idea – I was definitely doing an extra round trip to binary :slight_smile:

David

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