# [Crypto] Valentine's Day
{%hackmd M1bgOPoiQbmM0JRHWaYA1g %}

> -👤堇姬
## Step.1
我們知這個題目是Affine cipher

Solve this code before opening the box for me~♡
cipher text: hphxhHYK{fkknsj_h1um3w_hfs_xqtaj_gd_jdj5} Dtz wjhjnaji hmthtqfyjx mfsirfij gd dtzw lnwqkwnjsi

## Step.2
根據Affine cipher解密的方法

**$P \equiv keya^{-1}(C-keyb)\pmod {26}$**

因為這題有給keya，那只要暴力破解keyb(1~25)即可。

**Solve Script :**

def multiplicative_inverse(a, m):
    for x in range(1, m):
        if (a * x) % m == 1:
            return x

def affine_decrypt(cipher_text, a, b):
    decrypted_text = ""
    a_inverse = multiplicative_inverse(a, 26)
    
    for char in cipher_text:
        if char.isalpha():
            if char.isupper():
                decrypted_char = chr(((a_inverse * (ord(char) - ord('A') - b)) % 26) + ord('A'))
            else:
                decrypted_char = chr(((a_inverse * (ord(char) - ord('a') - b)) % 26) + ord('a'))
            decrypted_text += decrypted_char
        else:
            decrypted_text += char
    return decrypted_text

a = 646458645
cipher_text="hphxhHYK{fkknsj_h1um3w_hfs_xqtaj_gd_jdj5} Dtz wjhjnaji hmthtqfyjx mfsirfij gd dtzw lnwqkwnjsi"
for b in range(1,26):
    decrypted_text = affine_decrypt(cipher_text, a, b)
    print("keyB={},明文".format(b), decrypted_text)

## Another way
其實你會發現keya=1，所以其實他是一題凱薩密碼，直接推偏移量或是暴力破解即可。

~~至於為啥他會變凱薩密碼，是因為出題者隨便打了一串keya然後沒有驗www~~