put pdf

.github/workflows/gh-pages.yml

@@ -10,15 +10,15 @@ jobs:
   deploy:
     runs-on: ubuntu-20.04
     steps:
-      - uses: actions/checkout@v2
+      - uses: actions/checkout@v4
         with:
           submodules: true  # Fetch Hugo themes (true OR recursive)
           fetch-depth: 0    # Fetch all history for .GitInfo and .Lastmod
 
       - name: Setup Hugo
-        uses: peaceiris/actions-hugo@v2
+        uses: peaceiris/actions-hugo@v3
         with:
-          hugo-version: 'latest'
+          hugo-version: '0.135.0'
           extended: true
 
       - name: Clean

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

@@ -9,7 +9,7 @@ draft: false
 
 According to the documentation and as it is mentionned that type 7 is an hashing algorithm.
 
-A secure hash algorithm is an hash algorithm so that for any function hash that transform the original (plaintext) value $ hased = h(plain) $ there does not exist a function $ rev() $ so that $ rev(hashed) = plain $. 
+A secure hash algorithm is an hash algorithm so that for any function hash that transform the original (plaintext) value $ hased = H(plain) $ there does not exist a function $ rev(hashed) $ so that $ rev(hashed) = plain $. 
 
 ## 1 - Analysis under mathematical thinking
 
@@ -20,4 +20,22 @@ We could then guess that the researchers thanks then that:
 
 ![image](/gogo-s-blog-cpe/from-0-to-crypto-by-projects/episode-2-proof-demonstration/theory-behind-type7-hash.png)
 
-We now have the mathematical proof (demonstration) that the algorithm is vulnerable.
+The question is to proove that there exists a function $ rev(hashed) $ so that $ \forall plain [rev(H(plain)) = plain] $ then $ \forall x [x = H(plain)]$
+
+
+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 two opcodes are both xored to the hardcoded password.
+
+
+
+
+
+
+We now have the mathematical proof (demonstration) that the algorithm is vulnerable:
+
+[Download the pdf]( /gogo-s-blog-cpe/from-0-to-crypto-by-projects/episode-2-proof-demonstration/latex-reverse-type7.pdf )

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

@@ -0,0 +1,154 @@
+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{mathtools, nccmath}
+\usepackage{amssymb, amsthm, mathrsfs}
+\begin{document}
+
+According to the decompilation of the Ciso Vigenere hash algorithm, when the password length is less than 16 the idea behind Ciso Vigenere hash algorithm is: \\
+Let p be the password that the user types. \\
+Let hp be the hardcoded password in the code of Packet Tracer. \\
+Let lp be the length of the user input password. \\
+Let h be the hash value obtained from the custom algorithm. \\
+So that:
+$$
+\begin{flushleft}
+\begin{multline}
+\[
+
+\forall h \forall lp \forall hp [(hp = (d, s, f, d, ;, k, f, o, A, ,, ., i, y, e, w, r, k, l, d, J, K, D, H, S, U, B, s, g, v, c, a, 6, 9, 8, 3, 4, n, c, x , v), \\
+0 \textless lp \textless 16, \\
+h_{0} = 0, \\
+h_{1} = 8, \\
+h = \Sigma_{i=2}^{lp}
+\begin{cases}
+    ((p_i \oplus hp_{8 + i}) \ggg 4) + 0x30,                                   & \text{if } (p_{i} \oplus hp_{i+8} \land 0xfffffff0 < 0xa0)        \text{ and if } i \equiv 0 \pmod 2 \\
+    ((p_i \oplus hp_{8 + i}) \ggg 4) + 0x37,                                   & \text{if } (p_{i} \oplus hp_{i+8} \land 0xfffffff0 \geq 0xa0)     \text{ and if } i \equiv 0 \pmod 2 \\
+    ((p_i \oplus hp_{8 + i}) \land 0xf) + 0x30,                               & \text{if } (p_{i} \oplus hp_{i+8} \land 0xf < 0x0a)               \text{ and if } i \equiv 1 \pmod 2 \\
+    ((p_i \oplus hp_{8 + i}) \land 0xf) + 0x37,                               & \text{if } (p_{i} \oplus hp_{i+8} \land 0xf \geq 0x0a)            \text{ and if } i \equiv 1 \pmod 2
+\end{cases} \\
+) \implies \nexists p[p = \mathbf{rev}(h)] \\
+
+\]
+\end{flushleft}
+\end{multline}
+$$
+
+So let's split each sub steps of the algorithm. In this wayt, we could start prooving that if $ P \implies Q $ and if $ Q \implies R $ then $ P \implies R $
+
+So for any P so that:
+
+$$
+\begin{flushleft}
+\begin{multline}
+\[
+
+h = \Sigma_{i=2}^{lp}
+\begin{cases}
+    (p_{i} \oplus hp_{i+8} \land 0xfffffff0 < 0xa0)        \text{ if } i \equiv 0 \pmod 2 \\
+    (p_{i} \oplus hp_{i+8} \land 0xfffffff0 \geq 0xa0)     \text{ if } i \equiv 0 \pmod 2 \\
+    (p_{i} \oplus hp_{i+8} \land 0xf < 0x0a)               \text{ if } i \equiv 1 \pmod 2 \\
+    (p_{i} \oplus hp_{i+8} \land 0xf \geq 0x0a)            \text{ if } i \equiv 1 \pmod 2
+\end{cases} \\
+) \implies \nexists p[p = \mathbf{rev}(h)] \\
+
+\]
+\end{flushleft}
+\end{multline}
+$$
+
+
+So for any Q so that:
+
+$$
+\begin{flushleft}
+\begin{multline}
+\[
+
+h = \Sigma_{i=2}^{lp}
+\begin{cases}
+    (p_{i} \oplus hp_{i+8} \land 0xfffffff0 < 0xa0),                 \text{ if } i \equiv 0 \pmod 2 \\
+    (p_{i} \oplus hp_{i+8} \land 0xfffffff0 \geq 0xa0)               \text{ if } i \equiv 0 \pmod 2 \\
+    (p_{i} \oplus hp_{i+8} \land 0xf < 0x0a),                        \text{ if } i \equiv 1 \pmod 2 \\
+    (p_{i} \oplus hp_{i+8} \land 0xf \geq 0x0a),                     \text{ if } i \equiv 1 \pmod 2
+\end{cases} \\
+) \implies \forall p[p = \mathbf{rev}(h)] \\
+\]
+\end{flushleft}
+\end{multline}\\
+$$
+
+
+Let's start by prooving 
+
+$$
+\begin{flushleft}
+\begin{multline}
+\[
+
+\forall h \forall lp \forall hp [(hp = (d, s, f, d, ;, k, f, o, A, ,, ., i, y, e, w, r, k, l, d, J, K, D, H, S, U, B, s, g, v, c, a, 6, 9, 8, 3, 4, n, c, x , v), \\
+0 \textless lp \textless 16, \\
+h_{0} = 0, \\
+h_{1} = 8, \\
+h = \Sigma_{i=2}^{lp}
+\begin{cases}
+    ((p_i \oplus hp_{8 + i}) \ggg 4) + 0x30,                                   & \text{if } (p_{i} \oplus hp_{i+8} \land 0xfffffff0 < 0xa0)        \text{ and if } i \equiv 0 \pmod 2 \\
+    ((p_i \oplus hp_{8 + i}) \ggg 4) + 0x37,                                   & \text{if } (p_{i} \oplus hp_{i+8} \land 0xfffffff0 \geq 0xa0)     \text{ and if } i \equiv 0 \pmod 2 \\
+    ((p_i \oplus hp_{8 + i}) \land 0xf) + 0x30,                               & \text{if } (p_{i} \oplus hp_{i+8} \land 0xf < 0x0a)               \text{ and if } i \equiv 1 \pmod 2 \\
+    ((p_i \oplus hp_{8 + i}) \land 0xf) + 0x37,                               & \text{if } (p_{i} \oplus hp_{i+8} \land 0xf \geq 0x0a)            \text{ and if } i \equiv 1 \pmod 2
+\end{cases} \\
+) \implies \nexists p[p = \mathbf{rev}(h)] \\
+
+\]
+\end{flushleft}
+\end{multline}
+$$
+
+## I/ exclusive or
+
+According to the [Karnaught table](https://fr.wikipedia.org/wiki/Table_de_v%C3%A9rit%C3%A9#Disjonction_exclusive), $ \forall x [(x \xor x) \implies (x = 0)] $.
+
+Then as $ xlat \xor xlat = 0 $, and as $ p \xor 0 = p $, we know that the original password $p = xlat \xor h $.
+
+## II/ substraction to reverse the addition
+
+$\forall x [(x = y + z) \implies (y = e \minus z)]$
+
+## III/ truncating 4 first and 4 last bits
+
+
+Then we have proven that:
+
+$hp = (d, s, f, d, ;, k, f, o, A, ,, ., i, y, e, w, r, k, l, d, J, K, D, H, S, U, B, s, g, v, c, a, 6, 9, 8, 3, 4, n, c, x , v) \implies (\forall x \in hp[0 \geq x 0 \geq 256 \implies x \in hp]) $
+
+then:
+
+$$
+\begin{flushleft}
+\begin{multline}
+
+Let p be the password that the user types. \\
+Let hp be the hardcoded password in the code of Packet Tracer. \\
+Let lp be the length of the user input password. \\
+Let h be the hash value obtained from the custom algorithm. \\
+So that:
+
+\[
+
+\forall h \forall lp \forall hp [(hp \in N \land 0 \geq hp, \\
+0 \textless lp \textless 16, \\
+h_{0} = 0, \\
+h_{1} = 8, \\
+h = \Sigma_{i=2}^{lp}
+\begin{cases}
+    (((p_{i} \oplus hp_{i+8}) \lll 4) - 0x30),                               & \text{if } p_i < 0xa0             \text{ and if } i \equiv 0 \pmod 2 \\
+    (((p_{i} \oplus hp_{i+8}) \lll 4) - 0x37),                               & \text{if } p_i \geq 0x0a0         \text{ and if } i \equiv 0 \pmod 2 \\
+    (((p_{i} \oplus hp_{i+8}) \land 0xffffffff0) - 0x30),                    & \text{if } p_i < 0x0a             \text{ and if } i \equiv 1 \pmod 2 \\
+    (((p_{i} \oplus hp_{i+8}) \land 0xffffffff0) - 0x37),                    & \text{if } p_i \geq 0x0a          \text{ and if } i \equiv 1 \pmod 2
+\end{cases} \\
+) \implies \forall p[p = \mathbf{rev}(h)] \\
+
+\]
+\end{flushleft}
+\end{multline}\\
+$$
+\end{document}