adding solution to problem 3

README.markdown

@@ -7,4 +7,4 @@ Problem 1. Finished but unable to find my original solution.
 
 Problem 2. Finished. 9.9.11
 
-Problem 3. In progress. 9.11.11
+Problem 3. Finished. 9.13.11

problem3/problem3.rb

@@ -9,23 +9,38 @@
 # http://stackoverflow.com/questions/241691/sieve-of-eratosthenes-in-ruby
 # http://stackoverflow.com/questions/719049/fixnum-and-prime-numbers-in-ruby
 # http://on-ruby.blogspot.com/2006/07/rubyinline-making-making-things-faster.html
+# http://en.wikibooks.org/wiki/Algorithm_Implementation/Mathematics/Prime_number_generation
+# http://snippets.dzone.com/posts/show/4636
+
+#  AJ suggests:
+#- find all primes n/2
+#- start from greatest prime divide into n, first integer solves the problem.
+#- with the solution being the prime
+#- actualy, don't have to goto n/2 since n is odd
+#- start by dividing n by the integers greater than 0... testing each integer for primeness... the first prime is the solution
 
 class Factorization
 
 
   def initialize(args)
+    @prime_factors = []
     @factors = []
     self.trial_division(args)
+    self.find_prime_factors
   end
 
-  attr_reader :factors
+  attr_reader :factors, :prime_factors
 
   def trial_division(num)
     primes = []
     if num == 1
       primes.push(1)
     else
+      #primes = primeSieve(num/2)
+      # not quite sure on the math, but this works much faster num/2 or num
+      # (num**0.5)+1 >> num/2 >>>>> num
       primes = primeSieve((num**0.5)+1)
+      #primes = sieve_of_eratosthenes(num/2)
     end
     for p in primes
       if p*p > num
@@ -42,6 +57,15 @@ def trial_division(num)
     end
   end
 
+  def find_prime_factors
+    for f in @factors
+      if f.is_prime?
+        @prime_factors.push(f)
+      end
+    end
+  end
+
+  # faster than sieve below
   def primeSieve(n)
     s = (0..n).to_a
       s[0] = s[1] = nil
@@ -53,17 +77,96 @@ def primeSieve(n)
       s.compact
   end
 
+  def sieve_of_eratosthenes(max)
+    arr=(2..max).to_a
+    (2..Math::sqrt(max)).each do |i|
+       arr.delete_if {|a|a % i == 0 && a!=i}
+     end
+     arr
+  end
 
 end
 
-number = 600851475143
+# http://snippets.dzone.com/posts/show/4636
+# all this below is to test for primeness
+class Integer
+def bitarray(n) 
+    b=Array::new  
+    i=0       
+    v=n
+    while v > 0
+      b[i] = (0x1 & v)
+      v = v/2
+      i = i + 1
+    end
+    return b
+end
+def miller_rabin(n,s)
+    b=bitarray(n-1)
+    i=b.size 
+    j =1
+    while j <= s
+      a = 1 + (rand(n-2).to_i)
+      if witness(a,n,i,b) == true
+	return false
+      end
+      j+=1
+    end
+    return true
+end
+def witness(a,n,i,b)
+    d=1
+    x=0
+    while i > 0 
+      x = d 
+      d = (d**2) % n
+      if ( (d == 1) && (x != 1) && (x != (n-1)) )
+	return true
+      end
+      if ( b[i-1] == 1 )
+	d = (d * a ) % n
+      end
+      i -= 1
+    end
+    if ( d != 1) 
+      return true
+    end
+    return false
+end
+
+  #def prime?
+  def is_prime?
+    a = self
+    return miller_rabin(a,30)    
+  end
 
-factor = Factorization.new(number/2)
+  def bytesize
+    n = self
+    i=0
+    while n > 0
+      n = n >> 8
+      i += 1
+    end
+    return i
+  end
+
+  def bitsize
+    n = self
+    i=0
+    while n > 0
+      n = n >> 1
+      i += 1
+    end
+    return i
+  end
+end
+
+number = 600851475143
+#number = 13195
+factor = Factorization.new(number)
+puts "potential factors"
 puts factor.factors
+puts "---"
+puts "potential prime factors"
+puts factor.prime_factors
 
-#  AJ suggests
-#- find all primes n/2
-#- start from greatest prime divide into n, first integer solves the problem.
-#- with the solution being the prime
-#- actualy, don't have to goto n/2 since n is odd
-#- start by dividing n by the integers greater than 0... testing each integer for primeness... the first prime is the solution