Module4 - has been ended 1,2,3,4?

Signed-off-by: Marcin Woźniak <y0rune@aol.com>

module.rb

@@ -565,7 +565,16 @@ def algorytmDeSzyfrowania(x,y,u)
   return m
 end
 
-# Module 4
+########################### Module 4 ############################################
+# Podstawowe operacje na Galois Field GF(2^8)
+# Źródła:
+# * https://people.scs.carleton.ca/~maheshwa/courses/4109/Seminar11/The_Advanced_Encryption_Standard_AES_.pdf
+# * https://cs465.internet.byu.edu/static/lectures/w19/AES.pdf
+# * https://en.wikipedia.org/wiki/Finite_field_arithmetic
+# * https://swarm.cs.pub.ro/~mbarbulescu/cripto/Understanding%20Cryptography%20by%20Christof%20Paar%20.pdf
+# * http://www.cs.man.ac.uk/~banach/COMP61411.Info/CourseSlides/Wk2.2.FinField.pdf
+#################################################################################
+
 #################################################################################
 # Zadanie 1
 # Funkcja suma(a,b) wykorzystujac liczby hex
@@ -573,13 +582,8 @@ end
 def suma(a,b)
   binA = a.to_i(16).to_s(2)
   binB = b.to_i(16).to_s(2)
-
-  if (a =~ /[^01]/).nil?  && (b =~ /[^01]/).nil?
-    return (a.to_i(2) ^ b.to_i(2)).to_s(2)
-  else
   return (binA.to_i(2) ^ binB.to_i(2)).to_s(16)
 end
-end
 
 #################################################################################
 # Zadanie 2
@@ -588,11 +592,13 @@ end
 def xtime(a)
   binA = a.to_i(16).to_s(2)
   const = "1B".to_i(16).to_s(2)
+
   dl = binA.length
   while dl != 8
     binA = "0" + binA
     dl = dl + 1
   end
+
   if binA[0].to_i == 1
     binA[0] = ''
     return ((binA.to_i(2) << 1)  ^ const.to_i(2)).to_s(16)
@@ -604,16 +610,12 @@ end
 #################################################################################
 # Zadanie 3
 # Funkcja iloczyn(a,b) wykorzystujac liczby hex
+# {53} • {CA} = {01}
+# {53} • {13} = {fe}
 #################################################################################
 def iloczyn(a,b)
   solve = "0"
-
   binA = a.to_i(16).to_s(2)
-  dl = binA.length
-  while dl != 8
-    binA = "0" + binA
-    dl = dl + 1
-  end
 
   len = binA.length - 1
   binA.split('').each { |a|
@@ -628,6 +630,47 @@ def iloczyn(a,b)
     end
     len = len - 1
   }
-
   return solve
 end
+
+#################################################################################
+# Zadanie 4
+# Funkcja odwrotnosc(a,b) wykorzystujac liczby hex
+#################################################################################
+def extended_euklidesHex(a, b)
+  aDec = a.to_i(16)
+  bDec = b.to_i(16)
+
+  if bDec == 0
+    return a
+  else
+    return extended_euklidesHex(bDec.to_s(16), (aDec % bDec).to_s(16))
+  end
+end
+
+
+def bpd(ax,fx)
+  a = ax.to_i(16).to_s(2)
+  f = fx.to_i(16).to_s(2)
+  r = 1
+  q = 1
+
+  lenA = a.length
+  lenF = f.length
+
+  puts "#{a} #{f}"
+  puts "#{lenF} #{lenA}"
+  while (lenF >= lenA)
+    puts "#{lenF} #{lenA}"
+    a = (a.to_i(2) << (lenF - lenA)).to_s(2)
+    r = suma(a,fx)
+    lenR = r.to_i(16).to_s(2).length
+    if (lenR >= lenA)
+      q = (q << (lenF - lenR)) + 1
+    else
+      q = (q << (lenF - lenA))
+    end
+    f = r
+    return r,q
+  end
+end