How do you create integers 0..9 and math operators + - * / in to binary strings. For example:
0 = 0000,
1 = 0001,
...
9 = 1001
Is there a way to do this with Ruby 1.8.6 without using a library?
How do you create integers 0..9 and math operators + - * / in to binary strings. For example:
0 = 0000,
1 = 0001,
...
9 = 1001
Is there a way to do this with Ruby 1.8.6 without using a library?
You have Integer#to_s(base)
and String#to_i(base)
available to you.
Integer#to_s(base)
converts a decimal number to a string representing the number in the base specified:
9.to_s(2) #=> "1001"
while the reverse is obtained with String#to_i(base)
:
"1001".to_i(2) #=> 9
I asked a similar question. Based on @sawa's answer, the most succinct way to represent an integer in a string in binary format is to use the string formatter:
"%b" % 245
=> "11110101"
You can also choose how long the string representation to be, which might be useful if you want to compare fixed-width binary numbers:
1.upto(10).each { |n| puts "%04b" % n }
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
Picking up on bta's lookup table idea, you can create the lookup table with a block. Values get generated when they are first accessed and stored for later:
>> lookup_table = Hash.new { |h, i| h[i] = i.to_s(2) }
=> {}
>> lookup_table[1]
=> "1"
>> lookup_table[2]
=> "10"
>> lookup_table[20]
=> "10100"
>> lookup_table[200]
=> "11001000"
>> lookup_table
=> {1=>"1", 200=>"11001000", 2=>"10", 20=>"10100"}
You would naturally use Integer#to_s(2)
, String#to_i(2)
or "%b"
in a real program, but, if you're interested in how the translation works, this method calculates the binary representation of a given integer using basic operators:
def int_to_binary(x)
p = 0
two_p = 0
output = ""
while two_p * 2 <= x do
two_p = 2 ** p
output << ((two_p & x == two_p) ? "1" : "0")
p += 1
end
#Reverse output to match the endianness of %b
output.reverse
end
To check it works:
1.upto(1000) do |n|
built_in, custom = ("%b" % n), int_to_binary(n)
if built_in != custom
puts "I expected #{built_in} but got #{custom}!"
exit 1
end
puts custom
end
If you're only working with the single digits 0-9, it's likely faster to build a lookup table so you don't have to call the conversion functions every time.
lookup_table = Hash.new
(0..9).each {|x|
lookup_table[x] = x.to_s(2)
lookup_table[x.to_s] = x.to_s(2)
}
lookup_table[5]
=> "101"
lookup_table["8"]
=> "1000"
Indexing into this hash table using either the integer or string representation of a number will yield its binary representation as a string.
If you require the binary strings to be a certain number of digits long (keep leading zeroes), then change x.to_s(2)
to sprintf "%04b", x
(where 4
is the minimum number of digits to use).
If you are looking for a Ruby class/method I used this, and I have also included the tests:
class Binary
def self.binary_to_decimal(binary)
binary_array = binary.to_s.chars.map(&:to_i)
total = 0
binary_array.each_with_index do |n, i|
total += 2 ** (binary_array.length-i-1) * n
end
total
end
end
class BinaryTest < Test::Unit::TestCase
def test_1
test1 = Binary.binary_to_decimal(0001)
assert_equal 1, test1
end
def test_8
test8 = Binary.binary_to_decimal(1000)
assert_equal 8, test8
end
def test_15
test15 = Binary.binary_to_decimal(1111)
assert_equal 15, test15
end
def test_12341
test12341 = Binary.binary_to_decimal(11000000110101)
assert_equal 12341, test12341
end
end