Vidul Nikolaev Petrov

import re
 
greeting = "Merry Christmas"
 
greeting_list = map(chr, [ord(i) - len(greeting) - 1 for i in greeting])
greeting_list = set(filter(lambda x: re.match('[a-z]|[A-Z]', x), greeting_list))
 
# greeting_list is set(['c', 'b', 'd', 'i', 'Q', 'U', 'Y', 'X'])
# the 1st 3 elements are not clear to me: c,b,d
# 'i' means 'I', 'Q' means 'question', 'U' means 'you', 'Y' means 'why', 'X' means 'Christmas'
# which makes the compound: 'I question you - why Christmas?'

arr = (1..10).to_a
arr.select{ |e| e % 2 != 0 }

arr = range(1,11)
[e for e in arr if e % 2]

@arr = (1..10);
grep($_ % 2, @arr);

#!/usr/bin/python
 
import re
 
hex_encode = lambda x: ("&#%i;" * len(x)) % (tuple(map(ord, re.findall('.', x))))
 
# hex_encode("root@example.com") =>;
# root@vidul.com

a = [1, 2, 3, 4, 5, 6]
a[0..-1] # [1, 2, 3, 4, 5, 6]
a[(0..-1)] # tha same

a = [1, 2, 3, 4, 5, 6]
a[:-1] # [1, 2, 3, 4, 5]
a[:] # [1, 2, 3, 4, 5, 6]
a[0:len(a):2] # [2, 4, 6]  Nice, isn't it?

"a".upto("z"){ |i| puts i }

a = String.new "a"
0.upto(0){ p a }
1.upto(25){ p a.next! }

("a".."z").each{ |e| puts e }

($a, $b) = (1, 2);
@arr    = (1, 2);

 a, b = 1, 2
arr = 1, 
arr = 1,2

 a, b = 1, 2
arr  = 1, 2
arr  = 1,2, *[11,22] # the same as
arr  = [1,2, [11,22]].flatten