"a string".xor("another string")

I want to do an XOR of two strings:

“a string”.xor(“another string”)

It’s not that hard to implement, but it’s not fast either,
since it walks through the data string, byte-by-byte.

Any ideas? For example: “It’s memory-hungry!”. Any solutions?

gegroet,
Erik V. - http://www.erikveen.dds.nl/


class String
def xor(other)
if other.empty?
self
else
a1 = self.unpack(“c*”)
a2 = other.unpack(“c*”)

   a2 *= 2   while a2.length < a1.length

   a1.zip(a2).collect{|c1,c2| c1^c2}.pack("c*")
 end

end
end

On 28.01.2007 21:28, Erik V. wrote:

Erik V. - http://www.erikveen.dds.nl/

   a2 *= 2   while a2.length < a1.length

   a1.zip(a2).collect{|c1,c2| c1^c2}.pack("c*")
 end

end
end


irb(main):008:0> a=“a string”
=> “a string”
irb(main):009:0> b=“another string”
=> “another string”
irb(main):014:0> s="";[a.size, b.size].max.times {|i| s << ((a[i]||0) ^
(b[i]||0))}
=> 14
irb(main):015:0> s
=> “\000N\034\000\032\f\034Gstring”

Hm…

robert

You’re padding the shortest string with low values. When using
XOR to encrypt a string (my goal…), uh, you’re not encrypting
most of the data:

“very long secret data”.xor(“secret”) # ==> “\005\000\021\vE\030ong
secret data”

The secret key should be repeated several times, so every byte
in the data gets XOR’ed.

In a more general situation, your solution (padding low values)
should do. And String#xor should be general, so I’m wrong…

Does “s << a” generate a new string? If so, we have a lot of
intermediate long strings.

gegroet,
Erik V. - http://www.erikveen.dds.nl/


class String
def xor(other, encrypting=false)
if other.empty?
self
else
a1 = self.unpack(“c*”)
a2 = other.unpack(“c*”)

   if encrypting
     a2 *= 2 while a2.length < a1.length
   else
     l       = [a1.length, a2.length].max

     a1.concat([0x00]*(l - a1.length))
     a2.concat([0x00]*(l - a2.length))
   end

   a1.zip(a2).collect{|c1,c2| c1^c2}.pack("c*")
 end

end
end

p “very long secret data”.xor(“secret”) # ==>
“\005\000\021\vE\030ong secret data”

p “long string”.xor(“short”) # ==> “\037\a
\001\025Tstring”
p “short”.xor(“long string”) # ==> “\037\a
\001\025Tstring”

message = “This is a secret message.”
secret = “secret”

p message.xor(secret, false) # ==> “’\r\n
\001E\035s a secret message.”
p message.xor(secret, true) # ==> “’\r\n
\001E\035\000E\002R\026\021\020\027\006\006E
\031\026\026\020\023\002\021]”

p message.xor(secret, false).xor(secret, false) # ==> “This
is a secret message.”
p message.xor(secret, true).xor(secret, true) # ==> “This
is a secret message.”

Erik V. wrote:

You’re padding the shortest string with low values. When using
XOR to encrypt a string (my goal…), uh, you’re not encrypting
most of the data:

“very long secret data”.xor(“secret”) # ==> “\005\000\021\vE\030ong
secret data”

The secret key should be repeated several times, so every byte
in the data gets XOR’ed.
Robert’s solution can be modified:

irb(main):008:0> a=“a string”
=> “a string”
irb(main):009:0> b=“another string”
=> “another string”
irb(main):014:0> s="";[a.size, b.size].max.times {|i| s << ((a[i]||0) ^ (b[i]||0))}
=> 14
This becomes:

s=""
[a.size, b.size].max.times {|i|
s << ((a[i%a.size]||0)^(b[i%b.size]||0))
}

That takes care of the looping. << doesn’t create a new string (at
least, not to my knowledge…) but it will (I presume) cause
reallocation. This should get around that:

s = " " * [a.size,b.size].max
[a.size, b.size].max.times {|i|
s[i] = ((a[i%a.size]||0)^(b[i%b.size]||0))
}

No idea if it’s faster or slower, and it’s past my bedtime so I’m not
going to benchmark it right now :slight_smile:

String#crypt is a hash function, not an encrypt function,
although it’s called “encrypt”… ;]

gegroet,
Erik V. - http://www.erikveen.dds.nl/

Erik V. wrote:

You’re padding the shortest string with low values. When using
XOR to encrypt a string (my goal…), uh, you’re not encrypting
most of the data:

If you want to encrypt the data, maybe you should use actual
cryptographic functions?
“a string”.crypt(“another string”)
That’s probably more secure than a homegrown xor function

Daniel

On Mon, 29 Jan 2007 05:28:59 +0900, Erik V. wrote:

Erik V. - http://www.erikveen.dds.nl/

   a2 *= 2   while a2.length < a1.length

   a1.zip(a2).collect{|c1,c2| c1^c2}.pack("c*")
 end

end
end


I tried changing your code to work on machine words (on my x86) by using
“I*” instead of “c*”. The result wasn’t any faster. I also tried

class String
def xor(other)
ret=dup
ret.length.times{|n| ret[n]^=other[n]}
ret
end
end

once again, the result wasn’t any faster. If you want fast, your best
bet
is probably to write it in C.

THANK YOU!

I added a bit of code to detect the presence of this narray. It
falls back to the Ruby implementation if narray isn’t installed.

gegroet,
Erik V. - http://www.erikveen.dds.nl/


class String
def xor_slow(other)
if other.empty?
self
else
a1 = self.unpack(“c*”)
a2 = other.unpack(“c*”)

   a2 *= a1.length/a2.length + 1

   a1.zip(a2).collect{|c1,c2| c1^c2}.pack("c*")
 end

end

def xor_fast(other)
if other.empty?
self
else
if other.length < self.length
div, mod = self.length.divmod(other.length)
other = other * div + other[0, mod]
end

   a1        = NArray.to_na(self, "byte")
   a2        = NArray.to_na(other, "byte")

   (a1 ^ a2).to_s
 end

end

begin
require “narray”

 alias :xor :xor_fast

rescue LoadError
if $VERBOSE
$stderr.puts “I strongly advise to install the library
‘narray’.”
end

 alias :xor :xor_slow

end
end

On Tue, 30 Jan 2007, Erik V. wrote:

THANK YOU!

I added a bit of code to detect the presence of this narray. It
falls back to the Ruby implementation if narray isn’t installed.

gegroet,
Erik V. - http://www.erikveen.dds.nl/

sure would be great to have narray in the core eh? sigh.

-a

I’ve found a little bug in your version:

long = “this is a very long string”
short = “a short one”

p long.xor(short) # ==> OK
p short.xor(long) # ==> Array size mismatch: 11 != 26

See this fix.

gegroet,
Erik V. - http://www.erikveen.dds.nl/


require “narray”

class String
def xor(other)
if other.empty?
self
else
s1 = self
s2 = other

   if s1.length < s2.length
     n, r    = s2.length.divmod(s1.length)
     s1      = s1 * n + s1[0, r]
   elsif s2.length < s1.length
     n, r    = s1.length.divmod(s2.length)
     s2      = s2 * n + s2[0, r]
   end

   a1        = NArray.to_na(s1, "byte")
   a2        = NArray.to_na(s2, "byte")

   (a1 ^ a2).to_s
 end

end
end

On Mon, 29 Jan 2007 15:08:08 +0000, Ken B. wrote:

once again, the result wasn’t any faster. If you want fast, your best bet
is probably to write it in C.

Or use an existing C extension :slight_smile:
Using NArray:

require ‘narray’
require ‘benchmark’

class String
def xor1(other)
if other.empty?
self
else
a1 = self.unpack(“c*”)
a2 = other.unpack(“c*”)

   a2 *= 2   while a2.length < a1.length

   a1.zip(a2).collect{|c1,c2| c1^c2}.pack("c*")
 end

end

def xor2(other)
if other.empty?
self
else
if other.length < self.length
div, mod = self.length.divmod(other.length)
other = other * div + other[0, mod]
end

     a1 = NArray.to_na(self, "byte")
     a2 = NArray.to_na(other, "byte")

     (a1 ^ a2).to_s
 end

end
end

$a1 = “abcdefg” * 1000
$a2 = “hijkl” * 1000
Benchmark.bm do |x|
x.report(“xor1:”) {$a1.xor1($a2)}
x.report(“xor2:”) {$a1.xor2($a2)}
end
user system total real
xor1: 0.030000 0.000000 0.030000 ( 0.029777)
xor2: 0.000000 0.000000 0.000000 ( 0.000427)

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