continue crypto article

content/from-0-to-crypto-by-projects/episode-2-proof-demonstration.md

@@ -25,11 +25,9 @@ The question is to proove that there exists a function $ rev(hashed) $ so that $
 
 We intuitevely see points to split the issue into easier pieces:
 
-- the algorithm threat data block by blocks with blocks of two opcodes (numbers between 0 and 256) knowns bigram.
-
-The algorith treat bigrams as following:
-
+The algorith treat bigrams (blocks of two opcodes) as following:
 - the two opcodes are both xored to the hardcoded password.
+- 
 
 
 

static/from-0-to-crypto-by-projects/episode-2-proof-demonstration/latex-reverse-type7.tex

@@ -103,17 +103,47 @@
 \end{multline}
 $$
 
-## I/ exclusive or
+## I/ substraction to reverse the addition
 
-According to the [Karnaught table](https://fr.wikipedia.org/wiki/Table_de_v%C3%A9rit%C3%A9#Disjonction_exclusive), $ \forall x [(x \xor x) \implies (x = 0)] $.
+$\forall x [(x = y + z) \implies (y = e \minus z)]$ then it follow that as the previous part of the function contains: $ h = x + 0x30 $, then $ h - 0x30 = x $ so $ \exists rev(h)[rev(H(p)) =p - 0x30] $ 
 
-Then as $ xlat \xor xlat = 0 $, and as $ p \xor 0 = p $, we know that the original password $p = xlat \xor h $.
+## II/ exclusive or
+
+According to the Karnaught table at: https://fr.wikipedia.org/wiki/Table_de_v%C3%A9rit%C3%A9#Disjonction_exclusive, $ \forall x [(x \oplus x) \implies (x = 0)] $.
+$$
+\\
+$$
+Then as $ xlat \oplus xlat = 0 $, and as $ p \oplus 0 = p $, we know that the original password $p = xlat \oplus h $.
+$$
+\\
+\\
+$$
+
+## III/ rotating 4 first to 4 last bits
+
+$ \forall x [(x \ggg y) \implies (x \lll y = x)] $.
+$$
+\\
+$$
+Then as $z = (x \ggg y) = (x \lll y) $, we know that the original password $ p = H(p) \lll 4 $.
+$$
+\\
+\\
+$$
+
+## IV/ unmasking different signatures (recurrent marks) in the hash
+
+In the previous chapter one `I/ substraction to reverse the addition`, we told we can reverse the previous addition. We still need to guess which addition/substraction has been done previously.
+
+As both addition values are made depending of:
+
+```
+if (password[left] & 0xfffffff0 < 0xa0) -> password[left] + 0x30 or else password[left] + 0x37
+if (password[right] & 0x0000000f < 0x0a) -> password[right] & 0x0f + 0x30 or else password[right] & 0x0f + 0x37
+```
 
-## II/ substraction to reverse the addition
 
-$\forall x [(x = y + z) \implies (y = e \minus z)]$
 
-## III/ truncating 4 first and 4 last bits
 
 
 Then we have proven that: