# Forum: Ruby Exponential calcs with very large exponents

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on 2006-05-09 22:14
```Greetings all.

Does anyone have a clue how to use Ruby to do modular exponentiation
using the binary left-to-right method?  I looked through the
documentation and searched the forums and found the String.each_byte
method.  However I had no luck finding anything showing how one might
manipulate bits of bytes.

Below is an example of what I am talking about.

The calculation a = b^e mod n (or in Ruby: a = (b ** e).modulo(n) )is
known as modular exponentiation.  One efficient method to carry this out
on a computer is the binary left-to-right method. To solve y = x^e mod n
let e be represented in base 2 as

e = e(k-1)e(k-2)...e(1)e(0)

where e(k-1) is the most significant non-zero bit and bit e(0) the
least.
set y = x
for bit j = k - 2 downto 0
begin
y = y * y mod n   /* square */
if e(j) == 1 then
y = y * x mod n  /* multiply */
end
return y

Thanks for looking,

Doug```
on 2006-05-10 02:17
```#
## Use Fixnum#[] - Bit Reference
#
def modexp x, e, n
return 1%n if e.zero?
# k - most significant bit posistion
ee, k = e, 0
# linear search
(ee>>=1;k+=1) while ee>0
y = x
(k-2).downto(0) do |j|
y=y*y%n  # square
(y=y*x%n) if e[j] == 1 # multiply
end
y
end
__END__

Btw: do you know how to find most significant bit faster?
Sergey Volkov

doug meharry wrote:
> Greetings all.
>
> Does anyone have a clue how to use Ruby to do modular exponentiation
> using the binary left-to-right method?  I looked through the
> documentation and searched the forums and found the String.each_byte
> method.  However I had no luck finding anything showing how one might
> manipulate bits of bytes.
>
> Below is an example of what I am talking about.
>
> The calculation a = b^e mod n (or in Ruby: a = (b ** e).modulo(n) )is
> known as modular exponentiation.  One efficient method to carry this out
> on a computer is the binary left-to-right method. To solve y = x^e mod n
> let e be represented in base 2 as
>
> e = e(k-1)e(k-2)...e(1)e(0)
>
> where e(k-1) is the most significant non-zero bit and bit e(0) the
> least.
> set y = x
> for bit j = k - 2 downto 0
> begin
>   y = y * y mod n   /* square */
>   if e(j) == 1 then
>     y = y * x mod n  /* multiply */
> end
> return y
>
>
> Thanks for looking,
>
> Doug```
on 2006-05-10 02:59
```On Wed, May 10, 2006 at 05:15:06AM +0900, doug meharry wrote:
> for bit j = k - 2 downto 0
> begin
>   y = y * y mod n   /* square */
>   if e(j) == 1 then
if e[j] == 1
>     y = y * x mod n  /* multiply */
> end
> return y
>
>
> Thanks for looking,
>
> Doug

Integers in ruby has a [] method that returns the bit at the specified
offset (0 for lsb)```
on 2006-05-10 07:36
```> > Does anyone have a clue how to use Ruby to do modular exponentiation
> > using the binary left-to-right method?  I looked through the
> > documentation and searched the forums and found the String.each_byte
> > method.  However I had no luck finding anything showing how one might
> > manipulate bits of bytes.

No, but in case its useful (a long shot), I have some source that does
it using the "square and multiply" method. Used it to implement RSA in
pure ruby.

In the hopes it will be useful, here it is.

--------------------------------------------------------------------------------
# \$Id: modn.rb,v 1.3 2004/12/04 20:39:41 sam Exp \$

class String
# Convert String to a string of binary digits, similar to
Integer.to_s(2).
def to_bin
n = self.to_str

s = ''

n.each_byte do |b|
s << b.to_s(2)
end

s
end

# Do I need a Integer#to_integer and a String.to_integer? Strings
should be
# allowed as inputs to a lot of the crypto APIs, but they will be
treated as
# integers, internally! How to deal with this?
def to_integer
Integer.from_unsigned_bytes(self)
end

def to_bytes
self
end
end

class Integer
# +bytes+ is a sequence of octets in network byte order (most
significant
# byte first) that comprises an unsigned integer.
def self.from_unsigned_bytes(bytes)
bytes = bytes.to_str
n = 0
bytes.each_byte do |b|
n <<= 8
n |= b
end
n
end

# Return self as a String of bytes in network byte order.
def to_bytes
a = []
n = self.to_int

while n != 0
a.unshift( (n & 0xff).chr )
n >>= 8
end
a.join
end

# Return self.
#
# Purpose is to allow a set of classes to declare themselves to be
"duck-typed"
# to Integer.  This set of classes includes String, see
String#to_integer.
def to_integer
self
end

# Why isn't this a standard ruby method?
def []=(position, value)
bit = 2 ** position
i = self.to_int
if value
i |= bit
else
i &= ~bit
end
i
end

# Determine size of +self+ in bits.
def bit_size
i = self.to_int

hibit = i.size * 8 - 1

while( i[hibit] == 0 ) do
hibit = hibit - 1

break if hibit < 0
end

hibit + 1
end
end

class Integer
# Calculate the inverse of an Integer modulo +n+. The modular inverse
of +a mod n+,
# +a^-1 mod n+, is a number +a^-1+ such that:
#
#   a^-1 * a = 1 mod n
#
# There may not be such a number, in which case a RangeError is
raised.
#
# Uses the 'Extended Euclidean Algorithm' implementation
# from Figure 4.1, +Cryptography Theory and Practice+, Stinson.
def modular_inverse(n)
n = n.to_int
b = self.to_int
n0 = n
b0 = b
t0 = 0
t = 1
q = (n0/b0).floor
r = n0 - q * b0
while r > 0 do
temp = t0 - q * t
if temp > 0 then temp = temp.modulo(n); end
if temp < 0 then temp = n - ((-temp).modulo(n)); end
t0 = t
t = temp
n0 = b0
b0 = r
q = (n0/b0).floor
r = n0 - q * b0
end

if b0 != 1
raise RangeError, "#{b} has no inverse modulo #{n}"
else
t.modulo(n)
end
end

# Calculate +self+ ** +exp+ modulo +n+.
#
# This method uses the "square and multiply" approach. Why should be
fairly
# obvious from the code, see +Cryptography Theory and Practice+,
Stinson,
# Chapter 4.4 for a description of the method.
def modular_exp(exp, n)
# x ** b mod n
x = self.to_int
b = exp.to_int
n = n.to_int

z = 1

(n.bit_size - 1).downto(0) do |i|
z = z ** 2 % n

if b[i] == 1 then
z = z * x % n
end
end

z
end

# Return whether +self+ is even, that is, evenly divisible by 2.
def even?
self[0] == 0
end

# True if +self+ is probably prime, false otherwise. Probabalistic
primality
# test is the Miller-Rabin algorithm, aka "strong pseudo-prime test".
#
# +accuracy+ is the number of times to try the test, and error
probablity
# will be aproximately 1 time out of 4**+accuracy+. Default is 10,
wich gives
# an error rate of 1 in 1,048,076.
def prime?(accuracy = 10)
miller_rabin_prime?(accuracy)
end

# Determines if an odd number is prime, with an error probability of
1/4, at
# most.  Implementation from Stinson, Figure 4.9.
def miller_rabin_prime?(accuracy)
# Two is prime
return true if self == 2
# Not prime if its even!
return false if self.even?

n = self.to_int

# Find k, m such that n - 1 = (2 ** k) * m, where m is odd

m = n - 1
k = 0

# Since n is odd, n-1 is even, and this will loop at least once
while m.even?
m >>= 1
k += 1
end

# Answers of 'composite' are always correct - n is not prime.
# 'prime' are not necessarily true, so we try again. If we answered
'prime'
# accuracy number of times, then maybe it is prime.
accuracy.times do

catch(:isprime) do
# Choose a, 1 <= a <= n - 1
a = Kernel.rand(n - 1)  # 0..(n-2)
a = a + 1               # 1..n-1

# Compute b = a ** m mod n
b = a.modular_exp(m, n)

#       puts "n #{n} m #{m} k #{k} a #{a} b #{b}"

# If b == 1 mod n, n is prime
if( b == 1 )
throw :isprime
end

# For i = 0 to k - 1 do
k.times do
# if b == -1 (mod n), n is prime
if( b == (n - 1) )
throw :isprime
else
b = b.modular_exp(2, n)
end
end

# It's composite.
return false
end

end

return true
end

end```
on 2006-05-10 09:02
```2006/5/10, Sergey Volkov <gm.vlkv@gmail.com>:

> Btw: do you know how to find most significant bit faster?

Dunno whether any of these is faster

>> require 'mathn'
=> true
>> x=1<<40
=> 1099511627776
>> k=("%b" % x).length - 1
=> 40
>> x[k]
=> 1
>> k=Math.log(x)/Math.log(2)
=> 40.0
>> x[k]
=> 1

Kind regards

robert```
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