Cyber Security Summer Camp 2024 - Crypto writeup

講師好帥

Sun Jul 28 2024
2769 words · 22 minutes
Daily writeup

小提醒

講師Curious的get-flag function:

print(f'FLAG{{{hashlib.sha256(東東.encode()).hexdigest()}}}')

1 - Classical Cipher

1 - Crypto Playground - ASCII & 進位轉換

把以下三段數字轉成 ASCII 編碼後的字串


70 76 65 71 123 74 117 115 116 95 //這是十進位> FLAG{Just_


1110011 1101111 1101101 1100101 1011111 1100011 1101111 //這是二進位> some_co


6e 76 65 72 73 69 6f 6e 7e 7e 7e 7d //這是十六進位> nversion~~~}

FLAG{Just_some_conversion~~~}

2 - ASCII & 進位轉換

寫 Python 來把以下的數字轉成 ASCII 編碼後的字串


46 4c 41 47 7b 4a 75 73 74 5f 61 5f 73 69 6d 70 6c 65 5f 50 79 74 68 6f 6e 5f 73 63 72 69 70 74 21 7d
slove.py
def hex_to_ascii(hex_string):
    hex_list = hex_string.split()
    ascii_string = ''.join([chr(int(hex_num, 16)) for hex_num in hex_list])
    return ascii_string
hex_string = "46 4c 41 47 7b 4a 75 73 74 5f 61 5f 73 69 6d 70 6c 65 5f 50 79 74 68 6f 6e 5f 73 63 72 69 70 74 21 7d"
ans = hex_to_ascii(hex_string)
print(ans)

FLAG{Just_a_simple_Python_script!}

3 - Crypto Playground - Hex String

把以下的 hex string 轉成一般被 ASCII 編碼後的字串


464c41477b495f6b6e6f775f7468655f616e737765725f69735f6062797465732e66726f6d686578602e2e2e7d
slove.py
def hex_to_ascii(hex_string):
    byte_data = bytes.fromhex(hex_string)
    ascii_string = byte_data.decode('ascii')
    return ascii_string


hex_string = "464c41477b495f6b6e6f775f7468655f616e737765725f69735f6062797465732e66726f6d686578602e2e2e7d"
ans = hex_to_ascii(hex_string)
print(ans)

FLAG{Iknow_the_answer_isbytes.fromhex…}

4 - Chinese To Hex String

把以下的中文用 Python 轉換成電腦內儲存原始數據的 hex string，再把這個 hex string 放到 FLAG{} 的大括號之間（hex string 的英文全部都小寫）


蹦蹦炸彈
slove.py
ans = '蹦蹦炸彈'.encode()
ans = ans.hex()
print(f'FLAG{{{ans}}}')

FLAG{e8b9a6e8b9a6e782b8e5bd88}

5 - Crypto Playground - Base64

把以下的字串拿去做 base64 解碼


RkxBR3tCYXNlNjRfaXNfbW9yZV9vZnRlbl91c2VkX2luX3JlYWxpdHl9
slove.py
import base64
ans = base64.b64decode("RkxBR3tCYXNlNjRfaXNfbW9yZV9vZnRlbl91c2VkX2luX3JlYWxpdHl9").decode('utf-8')
print(ans)

FLAG{Base64_is_more_often_used_in_reality}

6 - Base64 & Hex String

把以下的字串用 Python 先 base64 解碼然後再從 hex string 轉成一般經過 ASCII 編碼的字串


NDY0YzQxNDc3YjYwNjI3OTc0NjU3MzYwNWYyNjVmNjA3Mzc0NzI2MDVmNjE3MjY1NWY2NDY5NjY2NjY1NzI2NTZlNzQ1ZjY5NmU1ZjUwNzk3NDY4NmY2ZTdk
slove.py
import base64
ans = base64.b64decode("NDY0YzQxNDc3YjYwNjI3OTc0NjU3MzYwNWYyNjVmNjA3Mzc0NzI2MDVmNjE3MjY1NWY2NDY5NjY2NjY1NzI2NTZlNzQ1ZjY5NmU1ZjUwNzk3NDY4NmY2ZTdk").decode('utf-8')
ans = bytes.fromhex(ans).decode('ascii')
print(ans)

FLAG{`bytes`_&_`str`_are_different_in_Python}

7 - Caesar Cipher With

cipher.txt
cipher.txt 裡面是一段被凱薩加密的數據，把這段數據解密後（不需要分析這段數據解密後在講什麼，理論上解密後會拿到一段是全大寫的數據）直接丟到 Crypto Playground 的 Get Flag 裡拿 flag


ZNKIGKYGXIOVNKXOYGXKGRREURJIOVNKXCNOINOYXKGRRECKGQOSTUZYAXKNUCURJHKIGAYKOSZUURGFEZURUUQGZZNKCOQOVGMKGZZNKSUSKTZHAZOLOMAXKOZYMUZZUHKGZRKGYZROQKLOLZEEKGXYURJUXCNGZKBKXBGPJADLIVBAYKZNUYKRGYZZKTINGXGIZKXYGYZNKYURAZOUT

8 - Caesar Cipher With Python

cipher.txt
cipher.txt 裡面是一段被凱薩加密的數據（和上一題一模一樣的數據），這題的目標是用 Python 把這段數據解密，解密後的操作和上一題一樣


ZNKIGKYGXIOVNKXOYGXKGRREURJIOVNKXCNOINOYXKGRRECKGQOSTUZYAXKNUCURJHKIGAYKOSZUURGFEZURUUQGZZNKCOQOVGMKGZZNKSUSKTZHAZOLOMAXKOZYMUZZUHKGZRKGYZROQKLOLZEEKGXYURJUXCNGZKBKXBGPJADLIVBAYKZNUYKRGYZZKTINGXGIZKXYGYZNKYURAZOUT
slove.py
import hashlib
def caesar_decrypt(ciphertext, shift):
    plaintext = ""
    for char in ciphertext:
        if 'A' <= char <= 'Z':
            decrypted_char = chr((ord(char) - ord('A') - shift) % 26 + ord('A'))
        else:
            decrypted_char = char
        plaintext += decrypted_char
    return plaintext


cipher_text='ZNKIGKYGXIOVNKXOYGXKGRREURJIOVNKXCNOINOYXKGRRECKGQOSTUZYAXKNUCURJHKIGAYKOSZUURGFEZURUUQGZZNKCOQOVGMKGZZNKSUSKTZHAZOLOMAXKOZYMUZZUHKGZRKGYZROQKLOLZEEKGXYURJUXCNGZKBKXBGPJADLIVBAYKZNUYKRGYZZKTINGXGIZKXYGYZNKYURAZOUT'


for shift in range(26):
    decrypted_text = caesar_decrypt(cipher_text, shift)
    if decrypted_text.isupper():
        print(f"{shift}:")
        print(decrypted_text)


# 發現6的時候最像，帶入get-flag
plain = 'THECAESARCIPHERISAREALLYOLDCIPHERWHICHISREALLYWEAKIMNOTSUREHOWOLDBECAUSEIMTOOLAZYTOLOOKATTHEWIKIPAGEATTHEMOMENTBUTIFIGUREITSGOTTOBEATLEASTLIKEFIFTYYEARSOLDORWHATEVERVAJDUXFCPVUSETHOSELASTTENCHARACTERSASTHESOLUTION'
print(f'FLAG{{{hashlib.sha256(plain.encode()).hexdigest()}}}')

FLAG{e241bb8575436ca05489e433b3f0782bf30c5ec286459cccd3a0830e718b6ded}

9 - Substitution Cipher

cipher.txt
cipher.txt 裡面是一段被替換式加密後的數據，把這段數據解密後（不需要分析這段數據解密後在講什麼，理論上解密後會拿到一段是全大寫的數據）直接丟到 Crypto Playground 的 Get Flag 裡拿 flag


EANOEHHNFJLFXDGFANOYDJINTNOEDJXFOSBFESDOLGDWWSNUEDJNQCSJNUFEFFOUIDTTCOSIFESDOLNJKSINLEDNOXSONNJEZNSJMJDUCIELEDXCFJFOENNGFANOYDJINTNOEFIINLLEDFGGUFEFFYENJGNOXEZHUNWFENFOUKSXDJDCLMJNUSIESDOLDYNOYDJINTNOEIZFOONGLXDSOXUFJREZNLNFEENTMELEDJNXCGFENEZNNTNJXSOXSOENJONEANJNFWFOUDONUSOEZNSOENJKNOSOXHNFJLSOODKFESDODOEZNSOENJONEYGDCJSLZNUFOUGFANOYDJINTNOEFXNOISNLYDCOUONAFOUTDJNNYYNIESKNTNFOLDYFIINLLSOXKFLEGHGFJXNJQCFOESESNLDYUFEFEDUFHANFJNFXFSOZNFJSOXIFGGLYDJJNXCGFESDOEDTFOUFENEZNMJDKSLSDODYNPINMESDOFGFIINLLTNIZFOSLTLSOEZSLJNMDJEFXJDCMDYIDTMCENJLISNOESLELFOULNICJSEHNPMNJELTFOHDYAZDTMFJESISMFENUSOFLECUHDYEZNLNLFTNEDMSILZFLIDOKNONUEDNPMGDJNEZNGSRNGHNYYNIELDYSTMDLSOXNPEJFDJUSOFJHFIINLLTFOUFENLANZFKNYDCOUEZFEEZNUFTFXNEZFEIDCGUWNIFCLNUWHGFANOYDJINTNOENPINMESDOFGFIINLLJNQCSJNTNOELADCGUWNNKNOXJNFENJEDUFHEZFOSEADCGUZFKNWNNOHNFJLFXDSOEZNAFRNDYEZNXJDASOXNIDODTSIFOULDISFGIDLEDYEZNYCOUFTNOEFGSOLNICJSEHDYEDUFHLSOENJONENOKSJDOTNOEFOHMJDMDLFGLEZFEFGENJEZNLNICJSEHUHOFTSILDOGSONLZDCGUWNFMMJDFIZNUASEZIFCESDONPINMESDOFGFIINLLADCGUYDJINSOENJONELHLENTUNKNGDMNJLEDJNKNJLNYDJAFJULNIJNIHUNLSXOMJFIESINLEZFELNNREDTSOSTSBNEZNSTMFIEDOCLNJMJSKFIHAZNOLHLENTLFJNWJNFIZNUEZNIDTMGNPSEHDYEDUFHLSOENJONENOKSJDOTNOEASEZTSGGSDOLDYFMMLFOUXGDWFGGHIDOONIENULNJKSINLTNFOLEZFEONAGFANOYDJINTNOEJNQCSJNTNOELFJNGSRNGHEDSOEJDUCINCOFOESISMFENUZFJUEDUNENIELNICJSEHYGFALWNHDOUEZNLNFOUDEZNJENIZOSIFGKCGONJFWSGSESNLEZNMJDLMNIEDYXGDWFGGHUNMGDHNUNPINMESDOFGFIINLLLHLENTLJFSLNLUSYYSICGEMJDWGNTLFWDCEZDALCIZFONOKSJDOTNOEADCGUWNXDKNJONUFOUZDAEDNOLCJNEZFELCIZLHLENTLADCGUJNLMNIEZCTFOJSXZELFOUEZNJCGNDYGFA
slove.py
from collections import Counter
import string
import hashlib


cipher_text = 'TWENTYYEARSAGOLAWENFORCEMENTORGANIZATIONSLOBBIEDTOREQUIREDATAANDCOMMUNICATIONSERVICESTOENGINEERTHEIRPRODUCTSTOGUARANTEELAWENFORCEMENTACCESSTOALLDATAAFTERLENGTHYDEBATEANDVIGOROUSPREDICTIONSOFENFORCEMENTCHANNELSGOINGDARKTHESEATTEMPTSTOREGULATETHEEMERGINGINTERNETWEREABANDONEDINTHEINTERVENINGYEARSINNOVATIONONTHEINTERNETFLOURISHEDANDLAWENFORCEMENTAGENCIESFOUNDNEWANDMOREEFFECTIVEMEANSOFACCESSINGVASTLYLARGERQUANTITIESOFDATATODAYWEAREAGAINHEARINGCALLSFORREGULATIONTOMANpubDateTHEPROVISIONOFEXCEPTIONALACCESSMECHANISMSINTHISREPORTAGROUPOFCOMPUTERSCIENTISTSANDSECURITYEXPERTSMANYOFWHOMPARTICIPATEDINASTUDYOFTHESESAMETOPICSHASCONVENEDTOEXPLORETHELIKELYEFFECTSOFIMPOSINGEXTRAORDINARYACCESSMANpubDateSWEHAVEFOUNDTHATTHEDAMAGETHATCOULDBECAUSEDBYLAWENFORCEMENTEXCEPTIONALACCESSREQUIREMENTSWOULDBEEVENGREATERTODAYTHANITWOULDHAVEBEENYEARSAGOINTHEWAKEOFTHEGROWINGECONOMICANDSOCIALCOSTOFTHEFUNDAMENTALINSECURITYOFTODAYSINTERNETENVIRONMENTANYPROPOSALSTHATALTERTHESECURITYDYNAMICSONLINESHOULDBEAPPROACHEDWITHCAUTIONEXCEPTIONALACCESSWOULDFORCEINTERNETSYSTEMDEVELOPERSTOREVERSEFORWARDSECRECYDESIGNPRACTICESTHATSEEKTOMINIMIZETHEIMPACTONUSERPRIVACYWHENSYSTEMSAREBREACHEDTHECOMPLEXITYOFTODAYSINTERNETENVIRONMENTWITHMILLIONSOFAPPSANDGLOBALLYCONNECTEDSERVICESMEANSTHATNEWLAWENFORCEMENTREQUIREMENTSARELIKELYTOINTRODUCEUNANTICIPATEDHARDTODETECTSECURITYFLAWSBEYONDTHESEANDOTHERTECHNICALVULNERABILITIESTHEPROSPECTOFGLOBALLYDEPLOYEDEXCEPTIONALACCESSSYSTEMSRAISESDIFFICULTPROBLEMSABOUTHOWSUCHANENVIRONMENTWOULDBEGOVERNEDANDHOWTOENSURETHATSUCHSYSTEMSWOULDRESPECTHUMANRIGHTSANDTHERULEOFLAW'
def calculate_frequencies(text):
    text = text.upper()
    total = len(text)
    frequencies = {ch: text.count(ch) / total for ch in string.ascii_uppercase}
    return frequencies


#  https://en.wikipedia.org/wiki/Letter_frequency
english_frequencies = {
    'A': 0.0817, 'B': 0.0150, 'C': 0.0278, 'D': 0.0425, 'E': 0.1270,
    'F': 0.0223, 'G': 0.0202, 'H': 0.0609, 'I': 0.0697, 'J': 0.0015,
    'K': 0.0077, 'L': 0.0403, 'M': 0.0241, 'N': 0.0675, 'O': 0.0751,
    'P': 0.0193, 'Q': 0.0010, 'R': 0.0599, 'S': 0.0633, 'T': 0.0906,
    'U': 0.0276, 'V': 0.0098, 'W': 0.0236, 'X': 0.0015, 'Y': 0.0197,
    'Z': 0.0007
}


def caesar_decrypt(text, shift):
    decrypted_text = ""
    for char in text:
        if 'A' <= char <= 'Z':
            decrypted_char = chr((ord(char) - ord('A') - shift) % 26 + ord('A'))
        else:
            decrypted_char = char
        decrypted_text += decrypted_char
    return decrypted_text


cipher_frequencies = calculate_frequencies(cipher_text)


most_frequent_cipher_letter = max(cipher_frequencies, key=cipher_frequencies.get)
shift = (ord(most_frequent_cipher_letter) - ord('E')) % 26


decrypted_text = caesar_decrypt(cipher_text, shift)


print(f"Decrypted text with shift {shift}:")
print(decrypted_text)
print(f'FLAG{{{hashlib.sha256(decrypted_text.encode()).hexdigest()}}}')

10 - Frequency Analysis On Caesar Cipher

ciphet.txt
cipher.txt 裡面是一段被凱薩加密的數據（和前面凱薩加密題目一樣的數據），這題的目標是用 Python 把這段數據用頻率分析的方法解密，解密後的操作和上一題一樣


ZNKIGKYGXIOVNKXOYGXKGRREURJIOVNKXCNOINOYXKGRRECKGQOSTUZYAXKNUCURJHKIGAYKOSZUURGFEZURUUQGZZNKCOQOVGMKGZZNKSUSKTZHAZOLOMAXKOZYMUZZUHKGZRKGYZROQKLOLZEEKGXYURJUXCNGZKBKXBGPJADLIVBAYKZNUYKRGYZZKTINGXGIZKXYGYZNKYURAZOUT
slove.py
from collections import Counter
import string
import hashlib


cipher_text = 'ZNKIGKYGXIOVNKXOYGXKGRREURJIOVNKXCNOINOYXKGRRECKGQOSTUZYAXKNUCURJHKIGAYKOSZUURGFEZURUUQGZZNKCOQOVGMKGZZNKSUSKTZHAZOLOMAXKOZYMUZZUHKGZRKGYZROQKLOLZEEKGXYURJUXCNGZKBKXBGPJADLIVBAYKZNUYKRGYZZKTINGXGIZKXYGYZNKYURAZOUT'
def calculate_frequencies(text):
    text = text.upper()
    total = len(text)
    frequencies = {ch: text.count(ch) / total for ch in string.ascii_uppercase}
    return frequencies


#  https://en.wikipedia.org/wiki/Letter_frequency
english_frequencies = {
    'A': 0.0817, 'B': 0.0150, 'C': 0.0278, 'D': 0.0425, 'E': 0.1270,
    'F': 0.0223, 'G': 0.0202, 'H': 0.0609, 'I': 0.0697, 'J': 0.0015,
    'K': 0.0077, 'L': 0.0403, 'M': 0.0241, 'N': 0.0675, 'O': 0.0751,
    'P': 0.0193, 'Q': 0.0010, 'R': 0.0599, 'S': 0.0633, 'T': 0.0906,
    'U': 0.0276, 'V': 0.0098, 'W': 0.0236, 'X': 0.0015, 'Y': 0.0197,
    'Z': 0.0007
}


def caesar_decrypt(text, shift):
    decrypted_text = ""
    for char in text:
        if 'A' <= char <= 'Z':
            decrypted_char = chr((ord(char) - ord('A') - shift) % 26 + ord('A'))
        else:
            decrypted_char = char
        decrypted_text += decrypted_char
    return decrypted_text


cipher_frequencies = calculate_frequencies(cipher_text)


most_frequent_cipher_letter = max(cipher_frequencies, key=cipher_frequencies.get)
shift = (ord(most_frequent_cipher_letter) - ord('E')) % 26


decrypted_text = caesar_decrypt(cipher_text, shift)


print(f"Decrypted text with shift {shift}:")
print(decrypted_text)
print(f'FLAG{{{hashlib.sha256(decrypted_text.encode()).hexdigest()}}}')

FLAG{e241bb8575436ca05489e433b3f0782bf30c5ec286459cccd3a0830e718b6ded}

11 - Affine Cipher

cipher.txt 裡面是一段被仿射密碼加密後的數據，把這段數據用 Python 解密後（不需要分析這段數據解密後在講什麼，理論上解密後會拿到一段是全大寫的數據）直接丟到 Crypto Playground 的 Get Flag 裡拿 flag


注意仿射密碼的 charset 是 ABCDEFGHIJKLMNOPQRSTUVWXYZ ,.


//cipher.txt
BOHHIKBI,OZ,REI,WZRIKZIR,EX.,BOHI,RO,KISU,XSHO.R,ICBSG.WYISU,OZ, WZXZBWXS,WZ.RWRGRWOZ.,.IKYWZP,X.RKG.RIT,REWKT,DXKRWI.,RO,DKOBI..,ISIBRKOZWB,DXUHIZR.F,NEWSI,REI,.U.RIH,NOKA.,NISS,IZOGPE, OKHO.R,RKXZ.XBRWOZ.Q,WR,.RWSS,.G  IK., KOH,REI,WZEIKIZR,NIXAZI..I.,O ,REI,RKG.R,MX.IT,HOTISF,BOHDSIRISU,ZOZKIYIK.WMSI,RKXZ.XBRWOZ.,XKI,ZOR,KIXSSU,DO..WMSIQ,.WZBI, WZXZBWXS,WZ.RWRGRWOZ.,BXZZORXYOWT,HITWXRWZP,TW.DGRI.F,REI,BO.R,O ,HITWXRWOZ,WZBKIX.I.,RKXZ.XBRWOZ,BO.R.Q,SWHWRWZP,REIHWZWHGH,DKXBRWBXS,RKXZ.XBRWOZ,.WJI,XZT,BGRRWZP,O  ,REI,DO..WMWSWRU, OK,.HXSS,BX.GXS,RKXZ.XBRWOZ.QXZT,REIKI,W.,X,MKOXTIK,BO.R,WZ,REI,SO..,O ,XMWSWRU,RO,HXAI,ZOZKIYIK.WMSI,DXUHIZR., OK,ZOZKIYIK.WMSI.IKYWBI.F,NWRE,REI,DO..WMWSWRU,O ,KIYIK.XSQ,REI,ZIIT, OK,RKG.R,.DKIXT.F,HIKBEXZR.,HG.RMI,NXKU,O ,REIWK,BG.ROHIK.Q,EX..SWZP,REIH, OK,HOKI,WZ OKHXRWOZ,REXZ,REIU,NOGST,OREIKNW.I,ZIITF,X,BIKRXWZ,DIKBIZRXPI,O , KXGT,W.,XBBIDRIT,X.,GZXYOWTXMSIF,REI.I,BO.R.,XZT,DXUHIZR,GZBIKRXWZRWI.BXZ,MI,XYOWTIT,WZ,DIK.OZ,MU,G.WZP,DEU.WBXS,BGKKIZBUQ,MGR,ZO,HIBEXZW.H,ICW.R.,RO,HXAI,DXUHIZR.OYIK,X,BOHHGZWBXRWOZ.,BEXZZIS,NWREOGR,X,RKG.RIT,DXKRUF

先用 https://www.guballa.de/substitution-solver↗ 我是人工刪除標點符號 要把標點符號考量在內

slove.py
import hashlib
ans = 'COMMERCE ON THE INTERNET HAS COME TO RELY ALMOST EXCLUSIVELY ON FINANCIAL INSTITUTIONS SERVING ASTRUSTED THIRD PARTIES TO PROCESS ELECTRONIC PAYMENTS. WHILE THE SYSTEM WORKS WELL ENOUGH FORMOST TRANSACTIONS, IT STILL SUFFERS FROM THE INHERENT WEAKNESSES OF THE TRUST BASED MODEL. COMPLETELY NONREVERSIBLE TRANSACTIONS ARE NOT REALLY POSSIBLE, SINCE FINANCIAL INSTITUTIONS CANNOTAVOID MEDIATING DISPUTES. THE COST OF MEDIATION INCREASES TRANSACTION COSTS, LIMITING THEMINIMUM PRACTICAL TRANSACTION SIZE AND CUTTING OFF THE POSSIBILITY FOR SMALL CASUAL TRANSACTIONS,AND THERE IS A BROADER COST IN THE LOSS OF ABILITY TO MAKE NONREVERSIBLE PAYMENTS FOR NONREVERSIBLESERVICES. WITH THE POSSIBILITY OF REVERSAL, THE NEED FOR TRUST SPREADS. MERCHANTS MUSTBE WARY OF THEIR CUSTOMERS, HASSLING THEM FOR MORE INFORMATION THAN THEY WOULD OTHERWISE NEED. A CERTAIN PERCENTAGE OF FRAUD IS ACCEPTED AS UNAVOIDABLE. THESE COSTS AND PAYMENT UNCERTAINTIESCAN BE AVOIDED IN PERSON BY USING PHYSICAL CURRENCY, BUT NO MECHANISM EXISTS TO MAKE PAYMENTSOVER A COMMUNICATIONS CHANNEL WITHOUT A TRUSTED PARTY.'
print(f'FLAG{{{hashlib.sha256(ans.encode()).hexdigest()}}}')

FLAG{510fc8a1bf88260c1a1f24491927d5419fadafafc3764aaba07900cf9f4b397a}

2 - Symmetric Cipher

1 - Crypto Playground - AES

cipher 是一段被 AES 加密後的密文，加密的金鑰是 key，解密 cipher 並嘗試不同的 key、cipher 對解密有什麼影響（同樣也可以試試看不同的 key 和 plain 對加密有什麼影響）


cipher : beb500cb0298d5a427d3c6532fa0d88e
key : e4cb2aa8ff51a7209a923d59932b5841
slove.py
from Crypto.Cipher import AES
from binascii import *


key = unhexlify("e4cb2aa8ff51a7209a923d59932b5841")
cipher = unhexlify("beb500cb0298d5a427d3c6532fa0d88e")
# 一定要16N，而且要對齊


cipher_aes = AES.new(key, AES.MODE_ECB)
ans = cipher_aes.decrypt(cipher).decode("utf-8")
print(ans)

FLAG{A—E—S}

2 - AES

cipher 是一段被 AES 加密後的密文，加密的金鑰是 key，用 Python 來解密 cipher


cipher : 75924759583675f4a56c2c322c72be86
key : 77e9ed71d1d4ea27816aa0538c091f14
slove.py
from Crypto.Cipher import AES
from binascii import *


key = unhexlify("77e9ed71d1d4ea27816aa0538c091f14")
cipher = unhexlify("75924759583675f4a56c2c322c72be86")
# 一定要16N，而且要對齊


cipher_aes = AES.new(key, AES.MODE_ECB)
ans = cipher_aes.decrypt(cipher).decode("utf-8")
print(ans)

FLAG{PyCrypt…}

3 - Crypto Playground - AES ECB Mode

cipher 是一段被 AES ECB Mode 加密後的密文，加密的金鑰是 key，嘗試把 cipher 解密


cipher : 8d8f543098319a54c0c44c265a50bc2f3ff33f67cc44c6c4f992bdc0b0adcbc86aece1f2f0b59772a59c5060d9669d0b
key : 872b456c9b4e3b320555e62b6ad9ee32
slove.py
from Crypto.Cipher import AES
from binascii import *


key = unhexlify("872b456c9b4e3b320555e62b6ad9ee32")
cipher1 = unhexlify("8d8f543098319a54c0c44c265a50bc2f")
cipher2 = unhexlify("3ff33f67cc44c6c4f992bdc0b0adcbc8")
cipher3 = unhexlify("6aece1f2f0b59772a59c5060d9669d0b")
# 一定要16N，而且要對齊


ans=''
cipher_aes = AES.new(key, AES.MODE_ECB)
ans += cipher_aes.decrypt(cipher1).decode("utf-8")
ans += cipher_aes.decrypt(cipher2).decode("utf-8")
ans += cipher_aes.decrypt(cipher3).decode("utf-8")
print(ans)

FLAG{Plain’s_length_needs_to_be_multiple_of_16}

4 - AES ECB Mode

cipher 是一段被 AES ECB Mode 加密後的密文，加密的金鑰是 key，用 Python 把 cipher 解密


cipher : 6b2f26db534ee26ab1c987cea73a2fecd1a3f8335c50412ffa785d3ce5b90ad310a064dfa4bf824bc7968d876c9e5d6917e1157fc27a29c2e042a82f08ed6c96
key : 85f1c9f80a5b9e440380cf9b02b31ff46de4f864cd7c2ba32f7ec2881b07d71c
slove.py
from Crypto.Cipher import AES
from binascii import *


key = unhexlify("85f1c9f80a5b9e440380cf9b02b31ff46de4f864cd7c2ba32f7ec2881b07d71c")
cipher1 = unhexlify("6b2f26db534ee26ab1c987cea73a2fecd1a3f8335c50412ffa785d3ce5b90ad3")
cipher2 = unhexlify("10a064dfa4bf824bc7968d876c9e5d6917e1157fc27a29c2e042a82f08ed6c96")
# 一定要16N，而且要對齊


ans=''
cipher_aes = AES.new(key, AES.MODE_ECB)
ans += cipher_aes.decrypt(cipher1).decode("utf-8")
ans += cipher_aes.decrypt(cipher2).decode("utf-8")
print(ans)

FLAG{Check_plain’s_length_before_AES_ECB_Encrypt~}

5 - Cut & Paste - 0

理解 Cut & Paste 後端的所有邏輯

image 一個買旗子的網頁 要觀察他的cookie token image 觀察甚麼被加密了

FLAG{Now_you_know_Flask!!!}

6 - Cut & Paste - 1

嘗試更改 money 來拿到 flag

利用token拼接的方式去將100元替換掉

username=aaaaaaa
1234567891234567
cccccc&money=100

用這樣的結構去看以及對比token image

token =
26c90abbbe466b6536a86d0153096a3e  >username=aaaaaaa
f27e4b014aaa9dd6620ab19c810729c6  >1234567891234567
4ffeda74080d40ea1384c26ee231aab5  >cccccc&money=100
146af8b1750e6b69fcb3fc0d5e9a938a  >空值，AES的特性

我們只要轉成

token =
26c90abbbe466b6536a86d0153096a3e  >username=aaaaaaa
4ffeda74080d40ea1384c26ee231aab5  >cccccc&money=100
f27e4b014aaa9dd6620ab19c810729c6  >1234567891234567
146af8b1750e6b69fcb3fc0d5e9a938a  >空值，AES的特性

這樣金額就被遞補的數字銜接上了

完整token =
26c90abbbe466b6536a86d0153096a3e4ffeda74080d40ea1384c26ee231aab5f27e4b014aaa9dd6620ab19c810729c6146af8b1750e6b69fcb3fc0d5e9a938a

image

FLAG{why_do_you_have_so_much_money???}

7 - Crypto Playground - XOR

已知


A ⊕ FLAG = a3eb43a265fb02a62f6db2e7bf749b6aab4350fb004031c920d1
FLAG ⊕ B = e322463f629716f9c07bd310a1ffcd84c5bcf2baf736b7b749b8
A ⊕ B ⊕ FLAG = 068544da7c287b00967914a875e53999319fe507df44af1e5614
求 FLAG

用XOR的特性 (A ⊕ FLAG) ⊕ (FLAG ⊕ B) = (A ⊕ B) 接著 (A ⊕ B) ⊕ (A ⊕ B ⊕ FLAG) = FLAG

slove.py
from pwn import *
from binascii import *


A_XOR_FLAG = unhexlify("a3eb43a265fb02a62f6db2e7bf749b6aab4350fb004031c920d1")
FLAG_XOR_B = unhexlify("e322463f629716f9c07bd310a1ffcd84c5bcf2baf736b7b749b8")
A_XOR_B_XOR_FLAG = unhexlify("068544da7c287b00967914a875e53999319fe507df44af1e5614")


A_XOR_B = xor(A_XOR_FLAG, FLAG_XOR_B)
FLAG = xor(A_XOR_B, A_XOR_B_XOR_FLAG)
FLAG = FLAG.decode('ascii')
print(FLAG)

FLAG{Doyou_knowGF(2)?}

8 - Crypto Playground - AES CBC Mode

cipher 是一段被 AES CBC Mode 加密後的密文，加密的金鑰是 key，嘗試把 cipher 解密


cipher : 829ad9df660fd928274eb552bc6bc88dd00a62c48d221f37b427a0db7b6d9e5dbcf0f8a52d156fa0fa881dc46eb5555b
key : 62bdc4885ca8a8be1d1e8826a2187528
iv : 7ae66a4b7e87a6d374ce66e73a112bf0
slove.py
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
import binascii


def hex_to_ascii(hex_string):
    byte_data = bytes.fromhex(hex_string)
    ascii_string = byte_data.decode('ascii')
    return ascii_string


cipher_hex = '829ad9df660fd928274eb552bc6bc88dd00a62c48d221f37b427a0db7b6d9e5dbcf0f8a52d156fa0fa881dc46eb5555b'
key_hex = '62bdc4885ca8a8be1d1e8826a2187528'
iv_hex = '7ae66a4b7e87a6d374ce66e73a112bf0'


cipher = binascii.unhexlify(cipher_hex)
key = binascii.unhexlify(key_hex)
iv = binascii.unhexlify(iv_hex)


cipher_obj = AES.new(key, AES.MODE_CBC, iv)


try:
    decrypted_padded = cipher_obj.decrypt(cipher)
    ans = binascii.hexlify(decrypted_padded).decode('utf-8')
    ans = hex_to_ascii(ans)
    print(ans)
except ValueError as e:
    print("Error:", e)

FLAG{AES_CBC_Mode_needs_padding_too!}

9 - AES CBC Mode

cipher 是一段被 AES CBC Mode 加密後的密文，加密的金鑰是 key，嘗試用 Python 把 cipher 解密


cipher : af6ab08a9380c038328bec3c47e70b915bbf4faf0e4b72db5f56d274bd6be5458552b6023e64f10a492bb88e839e51516b50903363bc637a1fe2550317392564
key : d1e138810051ac8deaef9fd33c969e080003d3f87d264dd431190f0e7795528b
iv : 25741419227aece3cb4f715f1c2b3878
slove.py
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
import binascii


def hex_to_ascii(hex_string):
    byte_data = bytes.fromhex(hex_string)
    ascii_string = byte_data.decode('ascii')
    return ascii_string


cipher_hex = 'af6ab08a9380c038328bec3c47e70b915bbf4faf0e4b72db5f56d274bd6be5458552b6023e64f10a492bb88e839e51516b50903363bc637a1fe2550317392564'
key_hex = 'd1e138810051ac8deaef9fd33c969e080003d3f87d264dd431190f0e7795528b'
iv_hex = '25741419227aece3cb4f715f1c2b3878'


cipher = binascii.unhexlify(cipher_hex)
key = binascii.unhexlify(key_hex)
iv = binascii.unhexlify(iv_hex)


cipher_obj = AES.new(key, AES.MODE_CBC, iv)


try:
    decrypted_padded = cipher_obj.decrypt(cipher)
    ans = binascii.hexlify(decrypted_padded).decode('utf-8')
    ans = hex_to_ascii(ans)
    print(ans)
except ValueError as e:
    print("Error:", e)

FLAG{It’s_better_to_know_to_use_Python_to_encrypt/decrypt}

3 - Asymmetric Cipher

1 - Crypto Playground - Calculate RSA

首先嘗試照著 Crypto Playground 的 RSA 步驟一步一步（每一步都要讓右邊的檢查通過才能繼續下一步）把字元 w 加密後解密


接著嘗試把要加密的東西從 w 變成 www 後再操作一次


操作完成後就可以繳交這題的 flag : FLAG{w->www}

2 - Calculate RSA

如果有以下的參數，想辦法還原 m（也就是 c 對應的明文）


p : 9835210563207632773
q : 10100395802203591453
e : 65537
c : 38901730400489769669866798139902101966
slove.py
import sympy


p = 9835210563207632773
q = 10100395802203591453
e = 65537
c = 38901730400489769669866798139902101966
n = p * q
phi_n = (p - 1) * (q - 1)
d = sympy.mod_inverse(e, phi_n)
m = pow(c, d, n)
m_bytes = m.to_bytes((m.bit_length() + 7) // 8)
ans = m_bytes.decode('utf-8')


print(ans)

FLAG{^_^}

3 - Baby RSA

https://factordb.com/↗

嘗試利用以下參數破解 c


n : 13910994564373802167
e : 65537
c : 1407190981767527409
slove.py
import sympy


p = 3538692379
q = 3931111573
e = 65537
c = 1407190981767527409
# n = 3538692379 * 3931111573


n = 13910994564373802167
phi_n = (p - 1) * (q - 1)
d = sympy.mod_inverse(e, phi_n)
m = pow(c, d, n)
m_bytes = m.to_bytes((m.bit_length() + 7) // 8)
ans = m_bytes.decode('utf-8')


print(ans)

FLAG{><}

4 - Fermat Factorization

chal.txt
n, e = (19587702115512625820726258933115749290382334844612502424053794718163357592187340905802291862587895104495960381932789753299005813251948996260611184897562326247682666901501448353047521698592249729834516283480674732105780439116269761997349396715933581609849899129728708922840663620676291576240621853120362443923808843760440882014006768364922012481526403094238006375367123013507968974573352190073292257633888791365308369702512213666185905143646188820887176281342149612272108589506611923214505057652995032329959141214241834219457937303114204179131653711364915208078857401621642027638259900918528159843126369443751432654581, 65537)
c = 14809508634249394893558452640498090344470474977233370068642995529577239063125903654339193829089342286599958099423774430315364606689897708158792085795087913537413061325614538809362728179967067658551451631961128472298086158947566207472019480325726689080054858772427589292753172987340834339682656397647927344725017560020482638665085591254911640209651814182071721929923296769089646566223761778492179802372108005014115841061248387428023122508966754739541458191646360889128169311625988454734758608425890600356875991590804236732489527856297719688715806423429351972578716069190339992852705790784452278002937884704968039421527
chal.py
from Crypto.Util.number import getPrime, bytes_to_long, isPrime


from secret import FLAG


base = getPrime(1024) + 1


p = base + getPrime(32)
while not isPrime(p):
    p = base + getPrime(32)


q = base + getPrime(32)
while not isPrime(q):
    q = base + getPrime(32)


n = p * q
e = 65537


c = pow(bytes_to_long(FLAG), e, n)


print(f'{n, e = }')
print(f'{c = }')
slove.py
from sympy import isprime, factorint
from Crypto.Util.number import long_to_bytes
n = 19587702115512625820726258933115749290382334844612502424053794718163357592187340905802291862587895104495960381932789753299005813251948996260611184897562326247682666901501448353047521698592249729834516283480674732105780439116269761997349396715933581609849899129728708922840663620676291576240621853120362443923808843760440882014006768364922012481526403094238006375367123013507968974573352190073292257633888791365308369702512213666185905143646188820887176281342149612272108589506611923214505057652995032329959141214241834219457937303114204179131653711364915208078857401621642027638259900918528159843126369443751432654581
e = 65537
c = 14809508634249394893558452640498090344470474977233370068642995529577239063125903654339193829089342286599958099423774430315364606689897708158792085795087913537413061325614538809362728179967067658551451631961128472298086158947566207472019480325726689080054858772427589292753172987340834339682656397647927344725017560020482638665085591254911640209651814182071721929923296769089646566223761778492179802372108005014115841061248387428023122508966754739541458191646360889128169311625988454734758608425890600356875991590804236732489527856297719688715806423429351972578716069190339992852705790784452278002937884704968039421527
factors = factorint(n)
p, q = list(factors.keys())
phi_n = (p - 1) * (q - 1)
d = pow(e, -1, phi_n)
m = pow(c, d, n)
ans = long_to_bytes(m).decode('utf-8')


print(ans)

5 - Crypto Playground - RSA Signature

首先嘗試照著 Crypto Playground 的 RSA Signature 步驟一步一步（每一步都要讓右邊的檢查通過才能繼續下一步）把字元 w 簽張後驗證簽章


接著嘗試把要簽張的東西從 w 變成 www 後再操作一次


操作完成後就可以繳交這題的 flag : FLAG{w->www_AGAIN!!!}

6 - Easy RSA 0

嘗試利用以下參數破解 c


n : 2639362258825246588480470347939064671401603842072135672799
e : 65537
c : 1746618984754679181914163060955061486968283106432500087220

image

slove.py
from sympy import *


n = 2639362258825246588480470347939064671401603842072135672799
e = 65537
c = 1746618984754679181914163060955061486968283106432500087220


p = 11469368108335263743
q = 12783666266952877523
r = 18001308980066920891


phi_n = (p - 1) * (q - 1) * (r - 1)


d = mod_inverse(e, phi_n)
m = pow(c, d, n)
ans = bytes.fromhex(hex(m)[2:]).decode('utf-8')


print(ans)

FLAG{3_Primes?}

7 - Easy RSA 1

嘗試利用以下參數破解 c


n : 558098643612442982869412022469246232193159102034581572625352272974227508724473952162960296832221266344918324209865053779262023996763482052514433441095525096832124022399320433830134804734321051616877085252534068802372830522359759248304816981536234329416042954494512044857893369730186700141160304115005148441761425180202201733261896417588308517257278921240631686899602960449771608786644313798116795435002176570933880511306119378180015212643414069934184998512031405143417
e : 65537
c : 516159413447458442301604958221763534692609641988597871082605764283823956573795619128264278802865129137415684054075645301940202655838462944593710475306765753445494124622433361044097665307358761301082095390555467396319971408496508662769321921809765119753151071758945708372163145352828897339448061094186267610588750722986326168345483999990218154775909728920834392462419411374276336477276222786342486282130486976724901696575277929058329049260487708653226358970107854753057

image 要注意 有平方 為^2 第二個要+1

slove.py
from sympy import mod_inverse
n = 558098643612442982869412022469246232193159102034581572625352272974227508724473952162960296832221266344918324209865053779262023996763482052514433441095525096832124022399320433830134804734321051616877085252534068802372830522359759248304816981536234329416042954494512044857893369730186700141160304115005148441761425180202201733261896417588308517257278921240631686899602960449771608786644313798116795435002176570933880511306119378180015212643414069934184998512031405143417
e = 65537
c = 516159413447458442301604958221763534692609641988597871082605764283823956573795619128264278802865129137415684054075645301940202655838462944593710475306765753445494124622433361044097665307358761301082095390555467396319971408496508662769321921809765119753151071758945708372163145352828897339448061094186267610588750722986326168345483999990218154775909728920834392462419411374276336477276222786342486282130486976724901696575277929058329049260487708653226358970107854753057
phin = (33493-1)*(33637-1)*(33647-1)*(34313-1)*(34781-1)*(35153-1)*(35447-1)*(35573-1)*(35801-1)*(35899-1)*(35993-1)*(36277-1)*(36353-1)*(36653-1)*(37021-1)*(37139-1)*(37273-1)*(37619-1)*(37897-1)*(38287-1)*(38723-1)*(39251-1)*(39419-1)*(39727-1)*(39863-1)*(40609-1)*(40639-1)*(41269-1)*(41579-1)*(41621-1)*(42179-1)*(42181-1)*(42571-1)*(42899-1)*(43189-1)*(43271-1)*(43607-1)*(43711-1)*(44053-1)*(44621-1)*(44641-1)*(44657-1)*(44711-1)*(45307-1)*(45863-1)*(45971)* (45971-1)* (45917-1)*(46723-1)*(47123-1)*(47513-1)*(47743-1)*(47857-1)*(48023-1)*(48491-1)*(49069-1)*(49261-1)*(49871-1)*(49937-1)*(49999-1)*(51059-1)*(51109-1)*(51203-1)*(51817-1)*(51869-1)*(52223-1)*(52999-1)*(53593-1)*(54293-1)*(54323-1)*(54449-1)*(55163-1)*(55667-1)*(55843-1)*(56431-1)*(56509-1)*(56941-1)*(56951-1)*(57037-1)*(57269-1)*(57809-1)*(58679-1)*(61091-1)*(61141-1)*(61297-1)*(61543-1)*(61651-1)*(62401-1)*(62683-1)*(63031-1)*(63131-1)*(63197-1)*(63601-1)*(63617-1)*(63781-1)*(63977-1)*(64037-1)*(64067-1)*(64567-1)*(65119-1)*(65203-1)
d = mod_inverse(e, phin)
m = pow(c, d, n)
ans = bytes.fromhex(hex(m)[2:]).decode('utf-8')
print(ans)

FLAG{WWWait???_Too_many_primes!!!}

Thanks for reading!

Cyber Security Summer Camp 2024 - Crypto writeup

Sun Jul 28 2024
2769 words · 22 minutes
Daily writeup

© I-Hung Wu | CC BY-NC-SA 4.0

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