---
# System prepended metadata

title: wco4_13/10/2025

---

Xin chào mọi người, mình là Mai Đăng Khoa. Đây là writeup cho những bài tập cryptography mà mình đã làm

---
# Cryptopals Challenges
## 1. Convert hex to base64
### Đề bài
![img](https://hackmd.io/_uploads/SJqdr4qalg.png)
### Ý tưởng
Đề cho cho ta một string hex và yêu cầu ta phải chuyển thành base64.
 
Trước tiên mình sẽ chuyển hex thành bytes, rồi sau đó chuyển từ bytes sang base64. Và decode nó để về string mà đề yêu cầu.
### Code
```python=
from base64 import b64encode
a = "49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d"
b = b64encode(bytes.fromhex(a)).decode()
print(b)
```
## 2. Fixed XOR
### Đề bài
![img](https://hackmd.io/_uploads/S1qktV5Txx.png)
### Ý tưởng
Bài này yêu cầu ta xor 2 strings có độ dài cùng nhau.

Theo mình hiểu thì ta sẽ tách 2 strings này thành từng kí tự và xor lần lượt chúng lại với nhau, sau đó gộp lại thành một string mới.

Và mình sẽ thực hiện bằng 2 cách: 
    - 1 là xor truyền thống
    - 2 là sử dụng thư viện để giải quyết.
### Code
```python=
a_hex = "1c0111001f010100061a024b53535009181c"
b_hex = '686974207468652062756c6c277320657965'

a = bytes.fromhex(a_hex)
b = bytes.fromhex(b_hex)

#1
c = bytes([i ^ j for i,j in zip(a,b)])
print(c.hex())

#2
from pwn import xor
d = xor(a,b)
print(d.hex())
```
## 3. Single-byte XOR cipher
### Đề bài
![img](https://hackmd.io/_uploads/BycmcEqage.png)
### Ý tưởng
Đề yêu cầu ta phải tìm một string bị xor bởi string đề cho và một kí tự đơn. 

Mà ta đã biết thì trong hệ ascii, một kí tự đơn có giá trị cao nhất là 255. Nên đầu tiên ta sẽ bruteforce nó từ 1-255.

Và bài toán đồng thời yêu cầu ta phải đếm tần suất của những kí tự đó để xem nó có giống đoạn mã thật không. Và mình dùng link [này](https://pi.math.cornell.edu/~mec/2003-2004/cryptography/subs/frequencies.html) để làm cơ sở cho score, tuy nhiên nó lại thiếu một kí tự quan trọng nhất là ' ' nên mình đã tự cho nó số điểm 13 (tức số điểm cao nhất). 
### Code
```python=
from pwn import xor

eng_fre_char = {
    ' ' : 13.0,
    'e' : 12.02,
    't' : 9.10,
    'a' : 8.12,
    'o' : 7.68,
    'i' : 7.31,
    'n' : 6.95,
    's' : 6.28,
    'r' : 6.02,
    'h' : 5.92,
    'd' : 4.32,
    'l' : 3.98,
    'u' : 2.88,
    'c' : 2.71,
    'm' : 2.61,
    'f' : 2.30,
    'y' : 2.11,
    'w' : 2.09,
    'g' : 2.03,
    'p' : 1.82,
    'b' : 1.49,
    'v' : 1.11,
    'k' : 0.69,
    'x' : 0.17,
    'q' : 0.11,
    'j' : 0.10,
    'z' : 0.07
}

def solve(text: bytes) -> bytes:
    score = 0
    index = 0
    for i in range(256):
        b=xor(text,i)
        cur_score = 0
        for j in b:
            char = chr(j).lower()
            if char in eng_fre_char:
                cur_score += eng_fre_char[char]
        if cur_score>score:
            score = cur_score
            index = i
    return xor(text,index)

a_hex = "1b37373331363f78151b7f2b783431333d78397828372d363c78373e783a393b3736"
a=bytes.fromhex(a_hex)
print(solve(a).decode())
```
> Cooking MC's like a pound of bacon
## 5. Implement repeating-key XOR
### Đề bài
![img](https://hackmd.io/_uploads/SJrGYr96ge.png)
### Ý tưởng
Đề yêu cầu ta xor một string và một key để ra string mới.

Tuy nhiên key này lại không cùng độ dài với string, vì thế ta phải nhân bản nó lên ("ICEICEICE...").

Bài này mình dùng 2 cách: 
    - 1 là sử dụng cycle trong thư viện itertools để nhân bản, hoặc cách truyền thống hơn là ta sẽ dùng tính chất module để giải (ví dụ như I thì VỊ TRÍ của những ký tự ở string gốc module 3 = 0 thì lúc này ta sẽ xor nó với I). 
    - 2 là sử dụng thư viện như bài 2 để giải =))))).

### Code
```python=
a = """Burning 'em, if you ain't quick and nimble
I go crazy when I hear a cymbal"""
key = "ICE"

a_bytes = a.encode()
key_bytes = key.encode()

#1
from itertools import cycle
key_cycle = cycle(key_bytes)
c = bytes([i ^ j for i, j in zip(a_bytes,key_cycle)])
print(c.hex())

#2 
from pwn import xor
d = xor(a_bytes,key_bytes)
print(d.hex())
```
## 6. Break repeating-key XOR
### Đề bài
![img](https://hackmd.io/_uploads/ryVvlYcpex.png)
:::spoiler
```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```
:::
### Ý tưởng
Yêu cầu của đề bài chính là từ một đoạn base64 và giải mã nó thành một đoạn text. Và mính sẽ vừa trình bày code cũng như ý tưởng theo tiến trình giải của đề.

1-2:
```python=
def hamming_distance(a: bytes, b: bytes) -> int:
    dist = 0
    for x, y in zip(a, b):
        diff = x ^ y 
        dist += bin(diff).count("1")  
    return dist
```
Đề yêu cầu ta phải tính hamming distance, giải thích đơn giản thì ví dụ với số `36` thì theo hệ 2 nó sẽ là `100100`, và hamming distance là:
$$
1+0+0+1+0+0=2
$$
Thì đối với bài này ta sẽ tính tổng đó sau khi ta xor xong.

3-4:
```python=
def find_best_keysize(data: bytes):
    best_keysize = 0
    smallest_normalized = 9999

    for keysize in range(2, 41):
        blocks = [data[i:i+keysize] for i in range(0, keysize * 4, keysize)]

        distances = []
        for a, b in combinations(blocks, 2):
            d = hamming_distance(a, b)
            distances.append(d)

        avg_distance = sum(distances) / len(distances)
        normalized = avg_distance / keysize

        if normalized < smallest_normalized:
            smallest_normalized = normalized
            best_keysize = keysize
    return best_keysize
```
Những gì 3, 4 yêu cầu ta là phải tìm được best keysize. Mà best keysize lại được xác định bằng normalized, nomralized được tính bằng cách lấy tổng disctance chia cho số lượng (khúc này bạn nên đọc code sẽ dễ hiểu hơn). Từ đó ta sẽ tìm ra được best keysize bằng cách tìm ra smallest normalized.

Note: dòng 3. chính là trường hợp cơ bản để có thể qua dòng 4. nên mình chọn rút gọn luôn dòng 3. mà chỉ giải thích dòng 4. và đoạn code của mình.

5-6:
```python=
best_keysize = find_best_keysize(data)
blocks = [data[i:i+best_keysize] for i in range(0, len(data), best_keysize)]
transposed_blocks = [data[i::best_keysize] for i in range(best_keysize)]
```
2 dòng này mình không giải thích, mọi người có thể đọc code và script của đề để hiểu.

7:
```python=
eng_fre_char = {
    ' ' : 13.0,
    'e' : 12.02,
    't' : 9.10,
    'a' : 8.12,
    'o' : 7.68,
    'i' : 7.31,
    'n' : 6.95,
    's' : 6.28,
    'r' : 6.02,
    'h' : 5.92,
    'd' : 4.32,
    'l' : 3.98,
    'u' : 2.88,
    'c' : 2.71,
    'm' : 2.61,
    'f' : 2.30,
    'y' : 2.11,
    'w' : 2.09,
    'g' : 2.03,
    'p' : 1.82,
    'b' : 1.49,
    'v' : 1.11,
    'k' : 0.69,
    'x' : 0.17,
    'q' : 0.11,
    'j' : 0.10,
    'z' : 0.07
}

def solve(text: bytes):
    score = 0
    index = 0
    for i in range(256):
        b=xor(text,i)
        cur_score = 0
        for j in b:
            char = chr(j).lower()
            if char in eng_fre_char:
                cur_score += eng_fre_char[char]
        if cur_score>score:
            score = cur_score
            index = i
    return index

key_bytes = []
for block in transposed_blocks:
    k = solve(block)
    key_bytes.append(k)
key = bytes(key_bytes)
```
Nó chính xác là trường hợp phức tạp hơn của bài 3. mình đã giải, và mình cũng tận dụng code từ bài đó nên có thể hàm solve sẽ không ăn nhập với tổng thể bài này lắm.

8:
```python=
plaintext = xor(data, key)
print(plaintext.decode())
```
no cmt =))
### Code
```python=
from pwn import xor
import base64
from itertools import combinations

with open("key.txt", "r", encoding="utf-8") as f:
    b64 = "".join(f.read().split())
data = base64.b64decode(b64)

def hamming_distance(a: bytes, b: bytes) -> int:
    dist = 0
    for x, y in zip(a, b):
        diff = x ^ y 
        dist += bin(diff).count("1")  
    return dist

def find_best_keysize(data: bytes):
    best_keysize = 0
    smallest_normalized = 9999

    for keysize in range(2, 41):
        blocks = [data[i:i+keysize] for i in range(0, keysize * 4, keysize)]

        distances = []
        for a, b in combinations(blocks, 2):
            d = hamming_distance(a, b)
            distances.append(d)

        avg_distance = sum(distances) / len(distances)
        normalized = avg_distance / keysize

        if normalized < smallest_normalized:
            smallest_normalized = normalized
            best_keysize = keysize
    return best_keysize

best_keysize = find_best_keysize(data)
blocks = [data[i:i+best_keysize] for i in range(0, len(data), best_keysize)]
transposed_blocks = [data[i::best_keysize] for i in range(best_keysize)]

eng_fre_char = {
    ' ' : 13.0,
    'e' : 12.02,
    't' : 9.10,
    'a' : 8.12,
    'o' : 7.68,
    'i' : 7.31,
    'n' : 6.95,
    's' : 6.28,
    'r' : 6.02,
    'h' : 5.92,
    'd' : 4.32,
    'l' : 3.98,
    'u' : 2.88,
    'c' : 2.71,
    'm' : 2.61,
    'f' : 2.30,
    'y' : 2.11,
    'w' : 2.09,
    'g' : 2.03,
    'p' : 1.82,
    'b' : 1.49,
    'v' : 1.11,
    'k' : 0.69,
    'x' : 0.17,
    'q' : 0.11,
    'j' : 0.10,
    'z' : 0.07
}

def solve(text: bytes):
    score = 0
    index = 0
    for i in range(256):
        b=xor(text,i)
        cur_score = 0
        for j in b:
            char = chr(j).lower()
            if char in eng_fre_char:
                cur_score += eng_fre_char[char]
        if cur_score>score:
            score = cur_score
            index = i
    return index

key_bytes = []
for block in transposed_blocks:
    k = solve(block)
    key_bytes.append(k)
key = bytes(key_bytes)

plaintext = xor(data, key)
print(plaintext.decode())
```
## 7. AES in ECB mode
### Đề bài
:::spoiler
```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npwtoOLsZPNbPw6K2jpko9NeZAx7PYFmamR4D50KtzgELQcaEsi5aCztMg7f
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RA3PIfmvGfMUGFVWlyS7+O73l5oIJHxuaJrR4EenzAu4Avpa5d+VuiYbM10a
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```
:::
### Ý tưởng
Bài này cho ta một đoạn mã base64 và có key là "YELLOW SUBMARINE"

Trước tiên ta cần hiểu rằng cách để mã hoá theo ECB mode: ta sẽ tách đoạn mã mình cần phải encode thành các blocks với mỗi blocks là 16 byte. Với trường hợp không đủ byte thì ta sẽ phải chủ động đệm vào. Sau đó ta encrypt các block đó với key, quá trình encrypt bao gồm (SubBytes (Thay thế Byte), ShiftRows (Dịch hàng), MixColumns (Trộn cột), AddRoundKey (Kết hợp với khóa vòng)) và ta sẽ gộp những khối đã bị mã hoá lại với nhau.

Và bài này thì giải mã theo ECB mode, tuy nhiên ta đã có thư viện xử lý nên không cần quan tâm làm gì.

### Code
```python=
import base64
from Crypto.Cipher import AES

with open('text.txt', 'r') as fh:
    ciphertext = base64.b64decode(fh.read())

key = b"YELLOW SUBMARINE"
tool = AES.new(key, AES.MODE_ECB)
plaintext = tool.decrypt(ciphertext)
print(plaintext)
```
## 8. Detect AES in ECB mode
### Đề bài
:::spoiler
```
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4c0a7faa755a8b877860d60f62e4a85e7319b39628d509211a00026d4d278dd7746c6feb6bf232d38298a321e48f8172eadb7782e188e1342bf088814329608ae756700d7786ab99602e83ab084407e05c349a0d72ee7662005784694d98fdf1d2e08efce71d14d940e7c4105d3fa5095454fe478401ba38f98a4eebd38477c613972c86f08e69f9e82e1ac09e67d81238271bb492da526bb1897bd330fe7b75
dd6a92b0659968f01d7b638960d747f7f0a0b20460de239b8f16d5a95936d1a4d9f4d3c77f5b14942d91304ce572dab54c8e4c01cab0df8d7653f0ef9fd486191a0ead2f1cfa71b30d7653322fde828b83f4ffafd2060727acf2c0d4062ed24fc9608bae7ab851cce4fde56d4ad593ed775ce856d7299e17d5f88325dddf7e268534710d3510ed24093d217f199afdb650581ac7962d0e281469e040beae01e1
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3bd94f9b4f7ec90219acb46c1ac6a8b597db73f8945f3f3d88a5b99c72f7d112e39bbe961feb0b635d4e5709c6270ba848a0233cf216ab252dbf129efc1b0af8c8c3cf54bebaa72d5252848155b84d73486bbe6f49d4393a6c22ccf3c103666ebd738f24dc573a69f6f58672d100c5aa57a893eabfb34e4270027dbd9df9b7dd94bbe5ada3f5045776133b9d0cf826dcd1b458ab635cb005493167560cc23a4d
c12d6deacc0cb3256e8bb49888cfe3580bb7c397270ff4dec562b6c2a81468c831c303982112e4fab409b3fb51af69e4ef6abbbd1e52f12de44693fcead0b601da48dfa4781839f7c2b43fa11f86f727def889835e10ce34afbc985cc978dbe3eef04d3533b49d2fcbb4d9ad61311a4859cd802f5e6502ec87fe987f13ea54082bb6e5f6d6abadf273be65ac58221d9da10e8c928c8a90627858fb222a672161
ac204457694bdbddc21a634ab13743a37f67360df1adb7bb090b5649ea63136dfc5408c3feecfd3b556a428bdafb58cfccc58ab4c96b409708a58e5a8edf944f95a710bcbebf004ef37654d125314b4f447d34ede6936bb1ec51643aa668456d7a440e947deb1eed3355d23678c5feebe0c9bd24514d07d03a9212c33f5aa0c8d8b75e6388204caf4d7264ddda2e062f2ef24dc441fbf9e928559daad5922057
6303b8bae1b14c332df030685bc0a0adf32bd8b6f1431fca361cc555ba91b013cd3d07c525bc110760ab36cb47897e777c5c83c6896a3b70d0a4ca8f405ba22554079ef658f976351e334ece622acfa29c03842761c7dad92b03da3e80be471388d955f6cc73261e889086be607567a1ddf9bf4c6813e194157cb928bc597ffcbad2e851eb11d0bda4e8b3806d37d7e5eca24546e2ccf68542d9dad23b9335e2
2b05d320464f3fd68e9c8461689a47dad22ca4877d10539062ffb1c22b599586daa9233e66fe12c3bf08527e1ca5ed250a6544c492bee68284559aa29516e83e6849e0dfc569301e77d3beee392ce5a134367cc78ea5283be538e88152be0230de9e7dc3ce05737a7645788cda9202362c5b4c5d792fc20f24f5a232fec877f659c558e2d4639c7630064f8ab91d585170387740910f96fc92a2a188d98d31f6
e8ed25e6f2b5cd72277cff0831b34d9980566aa183ec0e171ee3539bcdd70c831154e2c415c0a4dbba166cedb8ff10e677546845f85c081aed78a280fde896eab49ee37f88afb3e85676bcaf4cdb8acdf528ee703dd5ec05d7290ec200bbda77df57251e31ea70c6785c72c9a4b19439fa6f45f1713ba7890d983dedd45e2cf05f07606db3000dd60b69c3363efa71f9a6ee1dd8fdbb15b15257ef8b8754e349
e59604ffc914374ae1ef25c09e3eae08bcec726f68b58b1d8b514e26a0f32e091329dd591da4c203d6cdb2b47ed82e608965ccee9f4ae34374bcf6a6f6d982004382287087c322ba96e1a2e178f142a302725a434d03a6e9ae4f9241bc252ee92e7752be3ce483f5b89ce8e455958ed4d1f60eb23724136c04f54f03434925636d6fcf2ca884cb0b91031cf905b9386b3bb623e4b25dc59770b7bff8bc772068
c4f4ea4fdd7233ec7d425387cfb2ed5ff71cf31624bac99b18e80d1976968ac7eb62c1fbe008951b781f4ace195573b3ecc4cdcd8f4cb7241d963cab67c030d609cb23f00c71e8f7c885ce8e948d61cba17657e9875e15f80c06168f798646b1e618801123929851a4c7838f97efd8b2418a894d3e7438c3331cc63cd87d1057e6b3f29f40c7a8da7a5a2501718e991c30f4f91fe32a19e6e313603424a56f89
f1798850ee36bc07a1c9a95352c7147b9a083c24a0b3c655e419e02358f17a08235d4478815b065565d3c366044d19629d3420074961ee72583908d8d0de84949ee3a7889cbbbee77df74273b97ba5d96710df88816a3a6e180610d2c7c43305939cfcb36486cb994f843fc4acaa467ea34eae07d3170f0d40eb1e5a86ff9d17650844b33086f5e36b70c1383ea59d0a442b6b7bcf7b52f2a2fa91a2a2b0e6b6
4cbd2002032ec927bdf62aaef9c2f8490484db405c68f6b65ce06f27fe1253bfec1e66a9aa34e0ff290ced66e0efd026f69082c91e78cb8b5006972d77ea501cc2e969f52069bb97b8fd3ad9cb70dd832c39904837de32195ee6642ac274117eab545e58caacd00b31f5d4603fda91fc249eb5d2268d95e1a323da630be718e5cd7c82722ea347322befcd8cd7a85e180a34487319ae14c105e8db5754c347b3
acceaf70e88969fcc1ec9286388b5fb2d086d0d22617186b509b0675193070b773eb057772cd1d573fc29dfa756372c2b3c0e7610f479bf871fbc8f7fbeca8bd31c480abcf18b94ba2c9b2af81687330bda4aa170341c41c6282f58bb420120a6d3ac54e7eefa12cdba25724753f840f4a886817a566ca62e7d195a5766b6ecfc2df80f0401af48020da44c0eb7b387a502f075946707492f125452aab1849b0
7a415b762daa914e1e9e29e466d4e780cc67fc0f3e662b2652c0864bb83eb7cbdac4ac6e56c249b2e38228790ae1628d95f16a04832d70941d44a3d39409a942a64e66dc1995605dff569a140724a6ac7faa5247f27d66e026accf9dc6fdc2db8040d2519c6838601a3cb8b10afb4db257dfd9743f53ca09d940b18b9989cd64f174dd8f74a19006ac446b12c03c1df38446b08d05a701f19e1c4169698df731
4c251a5a914cab979d07f19c3da38e3ad678c87d17822b1109cb718fc909fb9ce285ffcbf05facd6eb1f348ccfbcc5b3c634946a49213564aaa34408e8d8f1b26dc531d3f56deec0ec5223589964b296a62b0c7eb3785c559447b73a098859d8dfb6a762c808d190f1439048a9993aef5cd637eadf28499239abbe7d9dec6e2399581c427ffe2321e8288ae147174bde7cd811bb3c78751dab36fc0f87092a7e
27a9a955ad770f38912dade0a36ff8530fded4bf57aceb950a10afe66bc88b2c5efc7a563ef269d00d90d5b169bf2c6791eff0cf9507160b653d0e19eb050ef90b04c8e21597d44705a6267f05977cd3a0cf2b7d6ed8130e7e80bdf668bb8ed75df11c4ea7def35cfbbedb043dd449c900f3dd213eb5d52ebeece6071589016a56ef0d4bfd8fe621b58fe1dbfafe0658a96cecce18eea088e4606346dce5e7c2
95b3a1adbcf8d2987c2f2cba58d5f0a4ef6e0301e186b8d62a59acb6eb4be54867136f319f97470fedae743acd6bb0231ab7d60450e9856989e92c06ec014341af3e69f5e54da348868528df895b56458661d8fdebb3eb4d9aba2b854e897b7ae43cbc1c0f53c959e9ee3d2c3c0e87bed291053da3aaf181a40cc6f25c067be5bb329fdc22068ef11566a08c7918125c6eaffee31147b8004fcba000ceb9802e
a6cadd53a2621482b7d66ecc82dc4ea6431bc0191c3801ac6b705df38c7fffe469043e5002096aca4aaca5ef033cc2c5f884d5c339d9a648ffa9400098c7851b9f5990b3771fcf55b196b7c2723085ad11268daadd411967a6a545986c93b86b7f72387bf68dc6ae8dcfd21c7f57ba70b15f3f5517c5585f345ff751c7bf21dc1a33d396ba180a2cb22998fc05ca47b5525a2c150effb15ffd681006c479f0c5
06df04188832b10afff94209d2aa1c8a123702de28234dcd3e0a7d36c1aa8449e6fa55e3e1e3d77d8424e87a45e38697755f84c49a99473797268113eb69098888947526035b246d00a630f6201ecc4075d8aa6604de73e2119e264e4c96751f2a67a2e46cf467a0df8f0520bcf4762b2715aba266d9b3f5e8fa67d12f9caac928b07ac3be99f41120655aa77f6433fc264673a92929e792187f87b5fda50cf2
```
:::
### Ý tưởng
Đề cho ta một list các đoạn mã đã bị mã hoá theo aes, tuy nhiên chỉ có một đoạn mã là bị mã hoá theo ECB mode. Và ta phải tìm ra nó

Trước hết, tính chất của ECB mode là với với cùng một đoạn mã gốc, khi nó bị biến đổi thì mọi đoạn mã bị biến đổi đều giống nhau. Và mục đích của bài này là xem xét đoạn mã nào có nhiều blocks bị trùng lặp nhất thì khả năng cao nó bị mã hoá theo ECB mode nhất.
### Code
```python=
def score_ecb_mode(cipher: bytes) -> float:
    blocks = []
    for i in range(0, len(cipher), 16):
        blocks.append(cipher[i:i+16])
    return len(set(blocks)) / len(blocks)

with open('text.txt', 'r') as fh:
    lines = fh.readlines()
    ciphertext_list = list(map(bytes.fromhex, lines))

min = 9999
index_of_ecb_mode = 0

for idx, ciphertext in enumerate(ciphertext_list):
    count = score_ecb_mode(ciphertext)
    if count < min:
        min = count
        index_of_ecb_mode = idx

print(index_of_ecb_mode)
```
## 28. Implement a SHA-1 keyed MAC 
### Đề bài
![img](https://hackmd.io/_uploads/H1mKrwjTlx.png)
### Ý tưởng
Đề chỉ đơn giản là yêu cầu ta viết lại hàm sha1 để tạo ra MAC, tức là nối 2 string key và message lại với nhau sau đó hash. Và ta không được sử dụng thư viện mà buộc phải code bằng tay.

Sau khi tham khảo từ nhiều nguồn (cụ thể là trên [wiki](https://en.wikipedia.org/wiki/SHA-1#SHA-1_pseudocode) và gemini pro) thì mình xin phép tóm tắt gồm các bước như sau:
1. Khởi tạo
```
h0 = 0x67452301
h1 = 0xEFCDAB89
h2 = 0x98BADCFE
h3 = 0x10325476
h4 = 0xC3D2E1F0
```
2. Tiền xử lý: như trên wiki đề cập thì nó sẽ bao gồm thêm bit 1,
thêm các bit 0 sao cho độ dài của nó cách bội số của 521 đúng 64 bit, và cuối cùng ta sẽ thêm chuỗi gốc với dạng là 1 số nguyên 64 bit.

3. Xử lý thông điệp theo từng khối: khối ở trên bước tiền xử lý sẽ được tách ra thành các khối 512 bit, và từng khối đó sẽ được chuyển thành 16 chữ có 32 bit và ta sẽ dùng công thức:`w[i] = (w[i-3] xor w[i-8] xor w[i-14] xor w[i-16]) leftrotate 1` này để thêm 64 bit còn lại. Sau đó, ta khởi tạo a, b, c, d, e có giá trị từ thằng 1. và bắt đầu vòng lặp chính (khúc này mình không hiểu lắm nên xin phép bỏ qua). Sau khi hoàn tất vòng lặp, các giá trị cuối cùng của a, b, c, d, e được cộng vào các h0, h1, h2, h3, h4

4. Sau khi hoàn thành, 5 giá trị 32 bit cuối của h0, h1, h2, h3, h4 được nối lại với nhau. Kết quả cuối cùng là một số 160 bit và thường được biểu diễn dưới dạng hex.
### Code
```python=
import struct

def left_rotate(n, b):
    return ((n << b) | (n >> (32 - b))) & 0xffffffff

def sha1(message_bytes):
    h0 = 0x67452301
    h1 = 0xEFCDAB89
    h2 = 0x98BADCFE
    h3 = 0x10325476
    h4 = 0xC3D2E1F0
    
    original_len_bits = len(message_bytes) * 8
    message_bytes += b'\x80'
    while len(message_bytes) % 64 != 56:
        message_bytes += b'\x00'
    message_bytes += struct.pack('>Q', original_len_bits)
    
    for i in range(0, len(message_bytes), 64):
        chunk = message_bytes[i:i+64]

        w = [0] * 80
        for j in range(16):
            w[j] = struct.unpack('>I', chunk[j*4:j*4+4])[0]

        for j in range(16, 80):
            w[j] = left_rotate(w[j-3] ^ w[j-8] ^ w[j-14] ^ w[j-16], 1)

        a, b, c, d, e = h0, h1, h2, h3, h4

        for j in range(80):
            if 0 <= j <= 19:
                f = (b & c) | ((~b) & d)
                k = 0x5A827999
            elif 20 <= j <= 39:
                f = b ^ c ^ d
                k = 0x6ED9EBA1
            elif 40 <= j <= 59:
                f = (b & c) | (b & d) | (c & d)
                k = 0x8F1BBCDC
            elif 60 <= j <= 79:
                f = b ^ c ^ d
                k = 0xCA62C1D6

            temp = (left_rotate(a, 5) + f + e + k + w[j]) & 0xffffffff
            e = d
            d = c
            c = left_rotate(b, 30)
            b = a
            a = temp

        h0 = (h0 + a) & 0xffffffff
        h1 = (h1 + b) & 0xffffffff
        h2 = (h2 + c) & 0xffffffff
        h3 = (h3 + d) & 0xffffffff
        h4 = (h4 + e) & 0xffffffff

    digest = '%08x%08x%08x%08x%08x' % (h0, h1, h2, h3, h4)
    return digest

def generate_secret_prefix_mac(key: str, message: str) -> str:
    key_bytes = key.encode()
    message_bytes = message.encode()
    
    data_to_hash = key_bytes + message_bytes
    
    mac = sha1(data_to_hash)
    
    return mac

key = "Sky1nNorth"
message = "Sky1nNorth is the goat <(\")"
print(generate_secret_prefix_mac(key, message))
```
## 39. Implement RSA
### Đề bài
![img](https://hackmd.io/_uploads/H1bTQOsalx.png)
### Ý tưởng
Bài này yêu cầu ta viết lại một hàm để hiểu rõ về cách hoạt động của rsa.

Tóm tắt thì, để mã hoá:
$$
c = m^e \ (mod \ n)
$$
Và để giải mã:
$$
m = c^d \ (mod \ n)
$$
Ta tìm d bằng cách:
$$
d = e^{-1} \ (mod \ et)
$$
Và:
$$
et = (p-1)(q-1)
$$
Tuy nhiên, như đề đã đề cập thì ta có 2 thứ rắc rối ở đây, 1 là tạo ra p, q; 2 là tìm d. Và may mắn là python hoàn toàn hỗ trợ việc tìm p,q cũng như d rất dễ nên nó đã khiến cho bài này dễ hơn rất nhiều.
### Code
```python=
from Crypto.Util.number import getPrime, long_to_bytes, bytes_to_long
p = getPrime(64)
q = getPrime(64)
n = p * q
et = (p-1) * (q-1)
e = 65537
d = pow(e, -1, et)

def encrypt(num: int) -> int:
    return pow(num, e, n)

def decrypt(num: int) -> int:
    return pow(num,d, n)


print(f"public key: {e}, {n}" )
print(f"private key: {d}, {n}")

a = b"Sky1nNorth"

c = encrypt(bytes_to_long(a))
print(c)

m = decrypt(c)
print(long_to_bytes(m))
```
# picoCTF
## EVEN RSA CAN BE BROKEN???
### Đề bài
![img](https://hackmd.io/_uploads/BJPHr95Txe.png)
### Ý tưởng
Đề bài này cho chúng ta ciphertext, N và e. Yêu cầu ta phải tìm lại plaintext.

Để giải được bài thì ta phải tìm được p, q. Nhận thấy rằng trong mọi trường hợp N luôn là một số nguyên chia hết cho 2. Vậy từ đó ta có thể suy ra rằng:
\begin{align}
p &= 2 \\
q &= \frac{N}{2}
\end{align}

Lúc này ta chỉ cần giải như một bài rsa thông thường.
### Code
```python=
from Crypto.Util.number import long_to_bytes
N = 23744210594394204962412591705837075848697163881031243327283152733386510119952476418908078253776365173453135514776170874842457430213531843947588015453628194
e = 65537
cyphertext = 16970058499961397133320520675431394618441246049997866708768391309017252831164238091868192934180038374070948783515091578062815885121609688128761114275568909

p = 2
q = N//2
phi = q-1
d = pow(e, -1, phi)
c = pow(cyphertext, d, N)
plaintext = long_to_bytes(c)
print(plaintext.decode())
```
> picoCTF{tw0_1$_pr!m375129bb1}
## 13
### Đề bài
```
Cryptography can be easy, do you know what ROT13 is? 
cvpbPGS{abg_gbb_onq_bs_n_ceboyrz}
```
### Ý tưởng
Đề yêu cầu giải theo rot 13, nên ta chỉ việc theo đó mà làm =)))
### Code
```python=
a = "cvpbPGS{abg_gbb_onq_bs_n_ceboyrz}"

#1
alphabet = "abcdefghijklmnopqrstuvwxyz"
result = ""
for j in a:
    if j.isalpha():
        idx = alphabet.index(j.lower())
        newidx = (idx+13) % 26
        new_char = alphabet[newidx]
        if j.isupper():
            new_char = new_char.upper()
        result += new_char
    else:
        result += j
print(result)

#2
from codecs import decode
print(decode(a,"rot 13"))
```
> picoCTF{not_too_bad_of_a_problem}
## The Numbers
### Đề bài
![the_numbers](https://hackmd.io/_uploads/ByIZp5cTll.png)
### Ý tưởng
Đề cho ta một bức ảnh và yêu cầu ta giải mã, nhận thấy flag format của pico là picoCTF, và chữ C, T, F ứng đúng với vị trí 3, 20, 6 trong bảng chữ cái. Ta chỉ cần chuyển đổi lại từ số sang chữ là được.
> PICOCTF{THENUMBERSMASON}
## interencdec
### Đề bài
`YidkM0JxZGtwQlRYdHFhR3g2YUhsZmF6TnFlVGwzWVROclgyeG9OakJzTURCcGZRPT0nCg==`
### Ý tưởng
Đề yêu cầu ta giải mã đoạn đó. Nhận thấy rằng, đoạn mã đó là base64 nên mình đã decode nó. Sau khi decode nó tiếp tục là base64 nên mình lại decode nó thêm một lần nữa. Và lúc này mình đoán nó là rot 13 nhưng lại sai nên mình đã bruteforce nó.
### Code
```python=
from base64 import b64decode

a = "YidkM0JxZGtwQlRYdHFhR3g2YUhsZmF6TnFlVGwzWVROclgyeG9OakJzTURCcGZRPT0nCg=="
print(b64decode(a))

b = "d3BqdkpBTXtqaGx6aHlfazNqeTl3YTNrX2xoNjBsMDBpfQ=="
print(b64decode(b))

c = "wpjvJAM{jhlzhy_k3jy9wa3k_lh60l00i}"
alphabet = "abcdefghijklmnopqrstuvwxyz"
for i in range(26):
    result = ""
    for j in c:
        if j.isalpha():
            idx = alphabet.index(j.lower())
            newidx = (idx + i) % 26
            new_char = alphabet[newidx]
            if j.isupper():
                new_char = new_char.upper()
            result += new_char
        else:
            result += j
    if "picoCTF" in result:
        print(result)
        break
```
> picoCTF{caesar_d3cr9pt3d_ea60e00b}
## Mod 26
### Đề bài
`cvpbPGS{arkg_gvzr_V'yy_gel_2_ebhaqf_bs_ebg13_MAZyqFQj}`
### Ý tưởng
Bài này mình sẽ rot 13 bruteforce. Lý do là vì mình đã thử rot 13 nhưng không ra nên chắc chắn ta phải bruteforce =))))))
### Code
```python==
a = "cvpbPGS{arkg_gvzr_V'yy_gel_2_ebhaqf_bs_ebg13_MAZyqFQj}"
alphabet = "abcdefghijklmnopqrstuvwxyz"
for i in range(26):
    result = ""
    for j in a:
        if j.isalpha():
            idx = alphabet.index(j.lower())
            newidx = (idx + i) % 26
            new_char = alphabet[newidx]
            if j.isupper():
                new_char = new_char.upper()
            result += new_char
        else:
            result += j
    if "picoCTF" in result:
        print(result)
        break
```
> picoCTF{next_time_I'll_try_2_rounds_of_rot13_ZNMldSDw}
## hashcrack
![img](https://hackmd.io/_uploads/B1lx7ic6ex.png)
Đây là đề cũng như lời giải của bài này.

Mình lên web [này](https://md5hashing.net/hash) để decode. 

Và theo mình biết thì hash là cách encode một chiều nên bản chất bài này sử dụng những password đơn giản và các password này đều nằm trên rockyou.txt chứ ta không có cách để decode lại từ hash.
> picoCTF{UseStr0nG_h@shEs_&PaSswDs!_93e052d7}

