I'm tasked with creating a Caesar cipher for a project I am working on. A Caesar cipher takes each letter in a string of text and replaces it with a letter a fixed number of places away from it (dictated by the cipher key). For instance if my text is "cat" and my cipher key is 3, my new word would be "fdw" (I'm assuming positive numbers move the letters to the right). I've been able to get my code to solve correctly for most strings of text, but I am finding that if my string includes > ? or # it will not work. Their ASCii codes are 62,63 and 64 if that helps. Any input is appreciated!
def caesar_cipher(str, num)
strArray = str.split('')
cipherArray = strArray.collect do |letter|
letter = letter.ord
if (65...90).include?(letter + num) || (97...122).include?(letter + num)
letter = letter + num
elsif (91...96).include?(letter + num) || (123...148).include?(letter + num)
letter = (letter - 26) + num
else
letter
end
end
cipherArray = cipherArray.collect {|x| x.chr}
cipher = cipherArray.join('')
end
caesar_cipher("Am I ill-prepared for this challenge?", 3)
#ord 62-64 DON'T work >, ?, #
You should create an alphabet variable, just think in if you use both ends then you will have 2 problems: negative numbers and an ASCii number that doesn't exist. You can handle this with module operator % or with a single subtraction.
alphabet = "abcde"
text_to_cipher= "aaee" => 0044 #number based in his position at aphabet var
key = 3
result will be 3377 => dd¡? or any other symbol since 7 is out of the string length "abcde" same happens with ASCii at its ends.
With module operator, you can restrict that.
size_of_your_alphabet = 5 # For this case
7%size_of_your_alphabet = 2
The Ruby builtin tr is ideal to implement substitution ciphers.
Step 1: assemble the characters you wish to transform.
chars = ["A".."Z", "a".."z", ">".."#"].flat_map(&:to_a)
Step 2: create a 1:1 array of the transformed characters
transformed = chars.map{|c| (c.ord + 3).chr}
Step 3: apply tr to transform the string.
str.tr(chars.join, transformed.join)
Full working example:
def caesar_cipher(str, num)
chars = ["A".."Z", "a".."z", ">".."#"].flat_map(&:to_a)
transformed = chars.map{|c| (c.ord + num).chr}
str.tr(chars.join, transformed.join)
end
Output:
> caesar_cipher("Am I ill-prepared for this challenge?", 3)
#=> "Dq L mpp-tvitevih jsv xlmw gleppirkiC"
Notes:
Most substitution ciphers actually rely on letter rotation, not ASCII values. My inital assumption was that you wanted a rotation, e.g. ("a".."z").to_a.rotate(num). See prior edit for a working example of that.
You can use ranges in tr() to create a really simple Caesar cipher: str.tr('A-Za-z','B-ZAb-za')
Edit: Also, because of the range specification feature, the \ is an escape character so that you can use literals like -. See this SO answer for details. I think the above exhibits a bug due to this, because it contains a \ which should be escaped by another \.
Related
I'm trying to take decimal number as an input and I need output of all numbers but without the point symbol in it.
Example input: 123.4
Wanted output 1234
The problem I have that when converting decimal number into string and trying to remove "." using :gsub('%.', '') its removing the point symbol but outputs 1234 1 .
I have tried :gsub('.', '') as well but it outputs 5.
I'm clueless where those numbers come from, here is the screenshot:
Use this syntax to get what you want and discard/ignore what you dont need...
local y = 123.4
-- Remove decimal point or comma here
local str, matches = tostring(y):gsub('[.,]', '')
-- str holds the first return value
-- The second return value goes to: matches
-- So output only the string...
print(str) -- Output: 1234
-- Or/And return it...
return str
There are two issues at play here:
string.gsub returns two values, the resulting string and the number of substitutions. When you pass the results of gsub to print, both will be printed. Solve this by either assigning only the first return value to a variable (more explicit) or surrounding gsub with parenthesis.
. is a pattern item that matches any character. Removing all characters will leave you with the empty string; the number of substitutions - 5 in your example - will be the number of characters. To match the literal dot, either escape it using the percent sign (%.) or enclose it within a character set ([.]), possibly adding further decimal separators ([.,] as in koyaanisqatsi's answer).
Fixed code:
local y = 123.4
local str = tostring(y):gsub("%.", "") -- discards the number of substitutions
print(str)
this is unreliable however since tostring guarantees no particular output format; it might as well emit numbers in scientific notation (which it does for very large or very small numbers), causing your code to break. A more elegant solution to the problem of shifting the number such that it becomes an integer would be to multiply the number by 10 until the fractional part becomes zero:
local y = 123.4
while y % 1 ~= 0 do y = y * 10 end
print(y) -- note: y is the number 1234 rather than the string "1234" here
how can I extract a few words separated by symbols in a string so that nothing is extracted if the symbols change?
for example I wrote this code:
function split(str)
result = {};
for match in string.gmatch(str, "[^%<%|:%,%FS:%>,%s]+" ) do
table.insert(result, match);
end
return result
end
--------------------------Example--------------------------------------------
str = "<busy|MPos:-750.222,900.853,1450.808|FS:2,10>"
my_status={}
status=split(str)
for key, value in pairs(status) do
table.insert(my_status,value)
end
print(my_status[1]) --
print(my_status[2]) --
print(my_status[3]) --
print(my_status[4]) --
print(my_status[5]) --
print(my_status[6]) --
print(my_status[7]) --
output :
busy
MPos
-750.222
900.853
1450.808
2
10
This code works fine, but if the characters and text in the str string change, the extraction is still done, which I do not want to be.
If the string change to
str = "Hello stack overFlow"
Output:
Hello
stack
over
low
nil
nil
nil
In other words, I only want to extract if the string is in this format: "<busy|MPos:-750.222,900.853,1450.808|FS:2,10>"
In lua patterns, you can use captures, which are perfect for things like this. I use something like the following:
--------------------------Example--------------------------------------------
str = "<busy|MPos:-750.222,900.853,1450.808|FS:2,10>"
local status, mpos1, mpos2, mpos3, fs1, fs2 = string.match(str, "%<(%w+)%|MPos:(%--%d+%.%d+),(%--%d+%.%d+),(%--%d+%.%d+)%|FS:(%d+),(%d+)%>")
print(status, mpos1, mpos2, mpos3, fs1, fs2)
I use string.match, not string.gmatch here, because we don't have an arbitrary number of entries (if that is the case, you have to have a different approach). Let's break down the pattern: All captures are surrounded by parantheses () and get returned, so there are as many return values as captures. The individual captures are:
the status flag (or whatever that is): busy is a simple word, so we can use the %w character class (alphanumeric characters, maybe %a, only letters would also do). Then apply the + operator (you already know that one). The + is within the capture
the three numbers for the MPos entry each get (%--%d+%.%d+), which looks weird at first. I use % in front of any non-alphanumeric character, since it turns all magic characters (such as + into normal ones). - is a magic character, so it is required here to match a literal -, but lua allows to put that in front of any non-alphanumerical character, which I do. So the minus is optional, so the capture starts with %-- which is one or zero repetitions (- operator) of a literal - (%-). Then I just match two integers separated by a dot (%d is a digit, %. matches a literal dot). We do this three times, separated by a comma (which I don't escape since I'm sure it is not a magical character).
the last entry (FS) works practically the same as the MPos entry
all entries are separated by |, which I simply match with %|
So putting it together:
start of string: %<
status field: (%w+)
separator: %|
MPos (three numbers): MPos:(%--%d+%.%d+),(%--%d+%.%d+),(%--%d+%.%d+)
separator: %|
FS entry (two integers): FS:(%d+),(%d+)
end of string: %>
With this approach you have the data in local variables with sensible names, which you can then put into a table (for example).
If the match failes (for instance, when you use "Hello stack overFlow"), nil` is returned, which can simply be checked for (you could check any of the local variables, but it is common to check the first one.
Probably it's an easy thing, but I'm a Lua beginner...
I'm creating a very simple QSC QSYS plugin to control a projection server using KVL API. Server API is based on hex strings.
For example this command asks the server to load a the playlist with 9bf5455689ed4c019731c6dd3c071f0e uuid:
Controls["LoadSPL"].EventHandler = function()
sock:Write(
"\x06\x0e\x2b\x34\x02\x05\x01\x0a\x0e\x10\x01\x01\x01\x03\x09\x00\x83\x00\x00\x14\x00\x00\x00\x01\x9b\xf5\x45\x56\x89\xed\x4c\x01\x97\x31\xc6\xdd\x3c\x07\x1f\x0e"
)
end
Now I need to be able to create a string with a variable UUID, according to the text indicated in a textbox (or a list of available UUIDs read from the server) in the user interface.
I will concatenate this string to the fixed part of the command.
How can I correctly make a string like
ad17fc696b49454db17d593db3e553e5 become
\xad\x17\xfc\x69\x6b\x49\x45\x4d\xb1\x7d\x59\x3d\xb3\xe5\x53\xe5?
Try this:
local input = "ad17fc696b49454db17d593db3e553e5"
local output = input:gsub("%w%w", function(s) return string.char(tonumber(s, 16)) end)
Explanation: this takes every pair of characters, interprets them as base 16 numeric string, and then takes the character with that number, and uses that to replace the original characters.
EDIT: To make it clear what's going on, and why the other answers are wrong, backslash escape sequences like \xad are a feature of the Lua source code, in memory it's represented by a byte with value 173, just like A is represented by a byte with value 65. Trying to concatenate a literal backslash character with hexadecimal characters does not create an escape code. So the way to do that is manually with string.char.
#! /usr/bin/env lua
str = 'ad17fc696b49454db17d593db3e553e5'
strx = ''
for i = 1, #str, 2 do -- loop through every-other position in your string
chars = str :sub( i, i+1 ) -- capture every 2 chars
strx = strx ..'\\x' ..chars
end -- append a literal backslash, the letter x, then those 2 chars
target = [[\xad\x17\xfc\x69\x6b\x49\x45\x4d\xb1\x7d\x59\x3d\xb3\xe5\x53\xe5]]
print( x, x == target ) -- print results, and test if it meets expected target
\xad\x17\xfc\x69\x6b\x49\x45\x4d\xb1\x7d\x59\x3d\xb3\xe5\x53\xe5 true
This can be code-golfed into a one-liner
x=''for i=1,#s,2 do x=x..'\\x'..s:sub(i,i+1)end
I want to extract the VALUE of lines containing key="VALUE", and I am trying to use a simple Lua pattern to solve this.
It works for lines except for those which contains a literal 1 in the VALUE. It seems the pattern parser is confusing my capture group for an escape sequence.
> return string.find('... key = "PHONE2" ...', 'key%s*=%s*(["\'])([^%1]-)%1')
5 18 " PHONE2
> return string.find('... key = "PHONE1" ...', 'key%s*=%s*(["\'])([^%1]-)%1')
nil
>
You do not need to use the [^%1] at all. Just use .- as it, by definition, matches the smallest possible string.
Also, you can use multiline string syntax, to not have to escape the quotes in your pattern:
> s=[[... key = "PHONE1" ...]]
> return s:find [[key%s*=%s*(["'])(.-)%1]]
5 18 " PHONE1
The pattern [^%1] actually means, do not search for characters % and 1 individually.
Is there anything better than string.scan(/(\w|-)+/).size (the - is so, e.g., "one-way street" counts as 2 words instead of 3)?
string.split.size
Edited to explain multiple spaces
From the Ruby String Documentation page
split(pattern=$;, [limit]) → anArray
Divides str into substrings based on a delimiter, returning an array
of these substrings.
If pattern is a String, then its contents are used as the delimiter
when splitting str. If pattern is a single space, str is split on
whitespace, with leading whitespace and runs of contiguous whitespace
characters ignored.
If pattern is a Regexp, str is divided where the pattern matches.
Whenever the pattern matches a zero-length string, str is split into
individual characters. If pattern contains groups, the respective
matches will be returned in the array as well.
If pattern is omitted, the value of $; is used. If $; is nil (which is
the default), str is split on whitespace as if ' ' were specified.
If the limit parameter is omitted, trailing null fields are
suppressed. If limit is a positive number, at most that number of
fields will be returned (if limit is 1, the entire string is returned
as the only entry in an array). If negative, there is no limit to the
number of fields returned, and trailing null fields are not
suppressed.
" now's the time".split #=> ["now's", "the", "time"]
While that is the current version of ruby as of this edit, I learned on 1.7 (IIRC), where that also worked. I just tested it on 1.8.3.
I know this is an old question, but this might be useful to someone else looking for something more sophisticated than string.split. I wrote the words_counted gem to solve this particular problem, since defining words is pretty tricky.
The gem lets you define your own custom criteria, or use the out of the box regexp, which is pretty handy for most use cases. You can pre-filter words with a variety of options, including a string, lambda, array, or another regexp.
counter = WordsCounted::Counter.new("Hello, Renée! 123")
counter.word_count #=> 2
counter.words #=> ["Hello", "Renée"]
# filter the word "hello"
counter = WordsCounted::Counter.new("Hello, Renée!", reject: "Hello")
counter.word_count #=> 1
counter.words #=> ["Renée"]
# Count numbers only
counter = WordsCounted::Counter.new("Hello, Renée! 123", rexexp: /[0-9]/)
counter.word_count #=> 1
counter.words #=> ["123"]
The gem provides a bunch more useful methods.
If the 'word' in this case can be described as an alphanumeric sequence which can include '-' then the following solution may be appropriate (assuming that everything that doesn't match the 'word' pattern is a separator):
>> 'one-way street'.split(/[^-a-zA-Z]/).size
=> 2
>> 'one-way street'.split(/[^-a-zA-Z]/).each { |m| puts m }
one-way
street
=> ["one-way", "street"]
However, there are some other symbols that can be included in the regex - for example, ' to support the words like "it's".
This is pretty simplistic but does the job if you are typing words with spaces in between. It ends up counting numbers as well but I'm sure you could edit the code to not count numbers.
puts "enter a sentence to find its word length: "
word = gets
word = word.chomp
splits = word.split(" ")
target = splits.length.to_s
puts "your sentence is " + target + " words long"
The best way to do is to use split method.
split divides a string into sub-strings based on a delimiter, returning an array of the sub-strings.
split takes two parameters, namely; pattern and limit.
pattern is the delimiter over which the string is to be split into an array.
limit specifies the number of elements in the resulting array.
For more details, refer to Ruby Documentation: Ruby String documentation
str = "This is a string"
str.split(' ').size
#output: 4
The above code splits the string wherever it finds a space and hence it give the number of words in the string which is indirectly the size of the array.
The above solution is wrong, consider the following:
"one-way street"
You will get
["one-way","", "street"]
Use
'one-way street'.gsub(/[^-a-zA-Z]/, ' ').split.size
This splits words only on ASCII whitespace chars:
p " some word\nother\tword|word".strip.split(/\s+/).size #=> 4