128 lines
2.8 KiB
Markdown
128 lines
2.8 KiB
Markdown
# CSE442T Introduction to Cryptography (Lecture 1)
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## Chapter 1: Introduction
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### Alice sending information to Bob
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Assuming _Eve_ can always listen
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Rule 1. Message, Encryption to Code and Decryption to original Message.
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### Kerckhoffs' principle
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It states that the security of a cryptographic system shouldn't rely on the secrecy of the algorithm (Assuming Eve knows how everything works.)
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**Security is due to the security of the key.**
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### Private key encryption scheme
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Let $M$ be the set of message that Alice will send to Bob. (The message space) "plaintext"
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Let $K$ be the set of key that will ever be used. (The key space)
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$Gen$ be the key generation algorithm.
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$k\gets Gen(K)$
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$c\gets Enc_k(m)$ denotes cipher encryption.
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$m'\gets Dec_k(c')$ $m'$ might be null for incorrect $c'$.
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$P[k\gets K:Dec_k(Enc_k(M))=m]=1$ The probability of decryption of encrypted message is original message is 1.
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*_in some cases we can allow the probability not be 1_
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### Some examples of crypto system
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Let $M=\text{all five letter strings}$.
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And $K=[1,10^{10}]$
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Example:
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$P[k=k']=\frac{1}{10^{10}}$
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$Enc_{1234567890}("brion")="brion1234567890"$
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$Dec_{1234567890}(brion1234567890)="brion"$
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Seems not very secure but valid crypto system.
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### Early attempts for crypto system
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#### Caesar cipher
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$M=\text{finite string of texts}$
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$K=[1,26]$
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$Enc_k=[(i+K)\% 26\ for\ i \in m]=c$
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$Dec_k=[(i+26-K)\% 26\ for\ i \in c]$
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```python
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def caesar_cipher_enc(s: str, k:int):
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return ''.join([chr((ord(i)-ord('a')+k)%26+ord('a')) for i in s])
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def caesar_cipher_dec(s: str, k:int):
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return ''.join([chr((ord(i)-ord('a')+26-k)%26+ord('a')) for i in s])
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```
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#### Substitution cipher
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$M=\text{finite string of texts}$
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$K=\text{set of all bijective linear transformations (for English alphabet},|K|=26!\text{)}$
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$Enc_k=[iK\ for\ i \in m]=c$
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$Dec_k=[iK^{-1}\ for\ i \in c]$
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Fails to frequency analysis
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#### Vigenere Cipher
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$M=\text{finite string of texts with length }m$
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$K=\text{[0,26]}^n$ (assuming English alphabet)
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```python
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def viginere_cipher_enc(s: str, k: List[int]):
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res=''
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n,m=len(s),len(k)
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j=0
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for i in s:
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res+=caesar_cipher_enc(i,k[j])
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j=(j+1)%m
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return res
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def viginere_cipher_dec(s: str, k: List[int]):
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res=''
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n,m=len(s),len(k)
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j=0
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for i in s:
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res+=caesar_cipher_dec(i,k[j])
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j=(j+1)%m
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return res
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```
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#### One time pad
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Completely random string, sufficiently long.
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$M=\text{finite string of texts with length }n$
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$K=\text{[0,26]}^n$ (assuming English alphabet)$
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$Enc_k=m\oplus k$
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$Dec_k=c\oplus k$
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```python
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def one_time_pad_enc(s: str, k: List[int]):
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return ''.join([chr((ord(i)-ord('a')+k[j])%26+ord('a')) for j,i in enumerate(s)])
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def one_time_pad_dec(s: str, k: List[int]):
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return ''.join([chr((ord(i)-ord('a')+26-k[j])%26+ord('a')) for j,i in enumerate(s)])
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```
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