security代做 | 代做Algorithm | 代写aws | java代做 | 代做scheme | Python | assignment代做 | oracle – Written assignment


security代做 | 代做Algorithm | 代写aws | java代做 | 代做scheme | Python | assignment代做 | oracle – 本题是一个利用Python进行练习的代做, 对Python的流程进行训练解析, 是比较有代表性的security/Algorithm/aws/java/scheme/Python/oracle等代写方向, 这个项目是assignment代写的代写题目

数据库代写 代写oracle oracle代写

Total points: 80 (+6 bonus)

assignment 2

Written assignment

  1. There is an error in each of the following uses of cryptography that degrades its security. Identify the mistake and explain how it should be done instead. (a) [4 points] Alice wants to send Bob an encrypted e-mail. She generates two ECDH key pairs: one for encryption and one for signing. She uses the public encryption key to encrypt the e-mail, and:::::::::attaches::a::::::::::signature:::::::::created:::by::::::using:the private signing key to sign a SHA-256 hash of the e-mail. Thee-mailsare::::::e-mail:::is:sent through SMTP. She publicizes both public keys. (b) [4 points] Bob manages a website. To store Alices password securely, Bob first asks Alice to encrypt it using 128-bit AES with a secret key (the key is shared with Bob). Then, Bob stores a hash of the encrypted version of the password using SHA-512 and also stores the secret key. (c) [4 points] Alice and Bob use the Diffie-Hellman protocol to establish a shared 256- bit secret key. After doing so, Alice wants to send Bob her bank account number so Bob can transfer money to her. Alice encrypts her bank number using the shared 256-bit secret key under AES in counter mode. (d) [4 points] In a private end-to-end encrypted messaging app, Alice adds Bob as a friend, which involves downloading his public 512-bit RSA key from a trusted server manging the app. To ensure that the public key is correct, the server uses its private 256-bit ECC key to sign it, which is verified by the corresponding public key that comes with the app. Having Bobs public key allows Alice to start creating secure connections with him.
  1. [14 points] Cryptography relies on a series of assumptions regarding computational dif- ficulty and key use. In each of the following cases, a commonly held assumption is broken. Discuss the impact of breaking that assumption, focusing on current crypto- graphic tools, how people interact with them, and how we(including:::::::::::::all:::::::::relevant::::::::parties :::::such::as:::::::::::::cryptosuite::::::::::::developers):should respond to these discoveries. (a) [3 points] A quick polynomial Algorithm for integer factorization has been found. (b) [4 points] The private signing key used by Facebook to sign HTTPS certificates was stolen by an unknown group two months ago. (c) [3 points] An easy way to determine the input corresponding to a given SHA-2 hash has been found. (d) [4 points] Large, practical quantum computers have been constructed. (Hint: Start with Shors algorithm.)

Programming assignment

Breaking Cryptography
In this assignment, we will write programs to automatically break some weak ciphers.
Please make sure to read the submission instructions carefully.
Two-Time Pad [20 points]
Two files,ctext0andctext1, have been sent to you by e-mail. Those two files were

encrypted using the same one-time pad. They are exactly (^400) :::: (^600) :bytes each, and they both come from popular English Wikipedia articles::::::::::::downloaded::::on:::::27th::::::June:::::::::(around :^4 ::::::PM). Find the contents of both files using crib-dragging, and submit them asptext andptext1. You can flip around ptext0 and ptext1. You may assume the plaintext to consist only of ASCII characters with the following byte values, all ranges being inclusive of both ends:

  • Symbols: 32 to 41, 44 to 59, 63, 91, 93.
  • Capital letters: 65 to 90.
  • Small letters: 97 to 122.
The ctext file was derived by XORing the plaintext (as ASCII bytes) with the400-byte
:::::::::600-byte key. Each byte of the ctext file would be the XOR of the corresponding byte of
plaintext with the corresponding byte of the key (written as bit strings). If you cannot
find the full texts, submit as much of the text as you can find.
Padding  oracle Attack [30 points]
AES  the standard block cipher in use today  had a padding algorithm that intro-
duced vulnerabilities when combined with CBC (Ciphertext Block Chaining). In this
assignment, we will investigate why it was insecure. In fact, the attacker can arbitrar-
ily decrypt and encrypt in AES without knowledge of the key, and even without any
understanding of the operations of AES.
The following is an adaptation of Vaudenays  security Fl aws Induced by CBC Padding
 Applications to SSL, IPSEC, WTLS ... paper. However,the padding  scheme is
intentionally different from the one used in that paper, to discourage plagia-
rism. A solution derived from that paper will not work for this assignment.
AES encrypts plaintexts in blocks of 16 bytes at a time. If there are fewer than 16 bytes
of plaintext data, AES addspaddingbytes to the end of the plaintext until there are 16
bytes exactly. (During decryption, those padding bytes will be discarded.) If there are
more than 16 bytes of data, AES operates on each block one by one in order, and pads
the final block to 16 bytes. If there arenreal bytes, then the bytes to add is exactly
16 ncopies ofn. For example, suppose the plaintext we want to encrypt is:
x= (CA013AB4C561) 16

In the above, x is written in hexadecimal notation, and it has 6 bytes. We want to add 10 bytes to make 16 bytes, so we will add the byte ( 06 ) 16 ten times to makex, the padded version ofx:

x= (CA013AB4C56106060606060606060606) 16

Note that the minimum amount of padding is 1 byte: that is to say, if the original plaintext has a multiple of 16 bytes, then we will need to add 16 bytes of padding of ( 00 ) 16 (byte 0). There will be a whole block of padding at the end. The maximum value for padding is one byte of(0E) 16 = 15(:::::::::::0F) 16 = 15.

After paddingxtox, we can perform AES encryption (denote the operation asC) on xto get the ciphertextC(x).Cis dependent on the secret keyKand the initialization vectorIV; the attacker knowsIV because it is sent in the clear.

SupposexcontainsNblocks of data (in other words, the size ofxis 16Nbytes), denoted as (x 1 |x 2 |.. .|xN). | is the concatenation operation, meaning that the bytes ofx 1 are followed by that of x 2 , and then by x 3 , and so on. After encryption, the resulting ciphertext is (IV|y 1 |y 2 |.. .|yN). In CBC mode, we have:

y 1 =C(IV x 1 )
yi =C(yi 1 xi) fori= 2, 3 ,... , N

The inverse ofC, the AES block encryption function, is denoted asD, the block decryp- tion function. Note that bothCandDdo not perform any padding on their own; they both input and output 16 bytes of data. For any 16-byte blockz,D(C(z)) =z.

We will now break AES in CBC mode using a padding oracle. A padding oracle is some entity that tells the attacker if the padding of some ciphertext (IV|y 1 |.. .|yN) is correct after decryption. In other words, it decrypts (IV|y) using the correct key, gets the plaintextx, and checks ifxuses the correct padding scheme described above. The padding oracle has been shared with you. (See Notes on the Padding Oracle later for more details on how to run the padding oracle.)

Suppose we are deciphering some ciphertext (IV|y 1 |.. .|yN). There will be three steps. First, we will learn how to find the last byte ofxN(Decrypt byte). Then, we will find the wholexN(Decrypt block). Finally, we will find all of (x 1 |x 2 |.. .|xN) (Decrypt).

Decrypt byte

ExtractyNfrom the ciphertext by taking the last 16 bytes, andyN 1 as the last 32 to 16 bytes. Denote theith byte ofyNasyN,i. Here, we want to findxN, 16.

  1. First, generate a random blockr= (r 1 |r 2 |.. .|r 15 |i) with 15 random bytes, followed by a bytei. Initiallyi= 0.
  2. Ask the padding oracle if (r|yN) is valid. (r|yN) contains the 16 bytes ofr, followed by the 16 bytes ofy.
  1. If the padding oracle returns no, incrementiby 1, and then ask the padding oracle again. Keep incrementingiuntil the padding oracle returns yes.
  2. Replacer 1 with any other byte and ask the oracle if the new (r|yN) has valid padding. If the padding oracle returns yes, similarly replacer 2. Repeat until either we have finished replacingr 15 and the oracle always returned yes, or the oracle has returned no while we were replacing somerk.
  3. If the oracle always returned yes in Step 4, setD(yN) 16 =i15.
  4. If the oracle returned no when we replacedrkin Step 4, setD(yN) 16 =i(k1).
  5. The final byte ofxNisxN, 16 =D(yN) 16 yN 1 , 16.

Decrypt block

After finding xN, 16 , the attacker can proceed to find all other bytes ofxN, starting from the 15th bytexN, 15 , thenxN, 14 , and proceeding backwards toxN, 1. In this pro- cess, the attacker will also findD(yN) 16 , D(yN) 15 ,… , D(yN) 1 as above. The following describes how the attacker can find xN,k for anyk; the attacker has already found D(yN)k+1, D(yN)k+2,… , D(yN) 16.

  1. Setras (r 1 |r 2 |.. .|rk 1 |i|D(y)k+1(k1)|D(y)k+2(k1)|.. .|D(y) 16 (k1)). Initiallyi= 0.
  2. Ask the oracle ifr|yNis valid.
  3. If the padding oracle returns no, incrementiand ask the padding oracle again. Keep incrementingiuntil the padding oracle returns yes.
  4. When the padding oracle returns yes, setD(yN)k=i(k1)
  5. Thek-th byte ofxNisxN,k=D(yN)kyN 1 ,k.


The above shows how the attacker can decrypt the last blockyN to obtain XN. To decrypt thek-th blockyk, the attacker simply replaces all of the aboveyNwithykand yN 1 withyk 1.

Write a program,decrypt, which finds the plaintextxfor any ciphertextyand outputs it to standard output. It is run with:

./decrypt ciphertext

ciphertextis a file that contains an amount of data that is a multiple of 16 bytes, and at least 32 bytes. It is formatted asIV|y 1 |.. .|yN, where theIV is the first 16 bytes,y 1 are bytes 17 to 32, and so on. The plaintext is in ASCII.

After you get the plaintext, output it to standard output. Do not add a newline.

You should tackle the assignment step by step: do the Decrypt byte step, then the Decrypt block step, then the Decrypt step. In case you cannot finish the assignment, marks will be given for partially completing each step.

Hint: Suppose you are given the ciphertext (IV|y 1 |y 2 ). Write down the plaintext (x 1 |x 2 ) usingD,IV,y 1 , andy 2. (It is not simplyD(y 1 ) andD(y 2 ).)

Bonus (6 points)

Write a program,encrypt, which takes in some plaintextxand encryptsxusing the same encryption algorithm and key that is behind the padding oracle provided. It is run with:

./encrypt plaintext

plaintextcontains an amount of data that is a multiple of 16 bytes, and at least 16 bytes. It is formatted asx 1 |x 2 |.. .|xN. Output the ciphertext and the IV to standard output asIV|y 1 |.. .|yN.

(Hint:encryptshould calldecryptas a subroutine in order to guess the right ciphertext. You only need to calldecryptonce for each block. Note that you can choose your own IV.)

Notes on the Padding Oracle

The padding oracle should be run with: python3 ciphertext It will decrypt the ciphertext with the secret AES key, check the padding of the plaintext, and output 1 if the padding is correct and 0 if the padding is incorrect. The padding oracle was written in Python. It is not compiled, and it can be directly run on Unix-like systems such as Ubuntu and macOS. If you want to run it on Windows, you will have to install Python and then type:

python3 ciphertext You will also have to capture the output and feed it into your own code. The command to do so issystem() in C and C++,subprocess.checkoutput()in Python, andRuntime.getRuntime().exec()in Java. You may have to look up documentation for the relevant command. Since the oracle is not compiled, the key is hardcoded into the oracle code.Do not use this key in any way.When we test your code, we will set the oracle with a different key. Your code should work independent of what the actual key is. You are also provided with a ciphertext calledciphertextfor reference, with its genera- It was encrypted with the same key as the oracle, and you can see the IV and plaintext used to create it; see if you can decrypt it correctly.

Submission instructions

All submissions should be done through CourSys. Submit the following files:
  • a2.pdf, containing all your written answers. Make sure it is not the question file.
  • ptext0andptext1, for part (a) of the programming assignment.
  • decrypt.{cpp, py, java}, for part (b) of the programming assignment, as well as any other code necessary to run it. This may include a Makefile. Submit your code; do not submit any compiled files. You may also submitencrypt.{cpp, py, java} for the bonus marks. The bonus marks can only be applied to this assignment.
To run decrypt, for example, I will do the following:
C++: I will compile./g++ decrypt.cpp -o decryptand then run./decrypt ::::::::::::ciphertext.

Python3: I will callpython3 ciphertext::::::::::::.

Java: I will compilejavac decrypt.javaand then call java decrypt ::::::::::::ciphertext.
If you are using Python, pleaase make sure it is Python3 instead of Python2.
If there is a Makefile in your folder, the Makefile will override all of the above. I will call
maketo compile the code, and then I will callmake run.
Keep in mind that plagiarism is a serious academic offense; you may discuss the assign-
ment, but write your assignment alone and do not show anyone your answers and code.
The submission system will be closed exactly 48 hours after the due date of the assign-
ment. You will receive no marks if there is no submission within 48 hours after the due