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First day of exam!!!!
I have 2 exams today. One is of programming and other one is literature.
I think I am prepared. Let's rock thissss.
🎉🍀
ALRIGHT!
Buckle up everybody. Mid sems are coming up and this one's FUCKING important. We got this. Lezzgoooooo!
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tch im tired bas ek exam deke
Midsem - 2016
==== Part A ==== This part is worth 12 marks and consists of 6 two mark questions. Answer all questions.
=== Question 0 === (2 Marks) Which of the below best describes the concept of a NONCE?
.[A] A key which is only guessable by brute force .[B] A number which is not used again .[C] A random number or phrase .[D] A salt value for a hash .[E] An idiot
A nonce is a pseudo-randomized number chosen for an authentication protocol such that old messages cannot be used in a replay attack - B.
=== Question 1 === (2 Marks) Buffer overflows can affect the control flow of a program by? .[A] Storing a negative number in an integer .[B] Changing a loop counter .[C] Overwriting a return access in the stack frame .[D] Changing data in the heap .[E] All of the above
All of the above can happen from buffer overflow attacks, but buffer overflows can affect the control flow of a program by overwriting the return access in the stack frame - C.
=== Question 2 === (2 Marks) A MAC is designed to provide .[A] Neither Confidentiality nor Authentication .[B] Authentication but not Confidentiality .[C] Confidentiality but not Authentication .[D] Confidentiality and Authentication .[E] A small hamburger
A message authentication code (MAC) is a form of symmetric key encryption technique used to verify that a message came from the intended source and is unaltered. A sender and receiver shares a secret key. When the sender sends a message, the secret key is appended to the message, then hashed (the MAC). The message and the MAC are both sent to the receiver. Upon receipt, the receiver can verify that the message came from the source and that the message is unaltered from its original form by hashing the message with the secret key and comparing with the MAC. Hence, integrity and authentication are provided. Confidentiality is not guaranteed as any party could still access the message in transit - B.
=== Question 3 === (2 Marks) Given a cryptographic hash function, which is most likely easier: .[A] Finding two different messages with identical hash values .[B] Given the hash of a message, finding the original message .[C] Given a hash value, finding a message which hashes to that value .[D] Given the hash of a long message, and the contents of the second half of the message, finding the first half of the message. .[E] It's not this one
Lets consider the choices:
A - property of collision resistance. Recall the birthday paradox - if we are looking for at least a 50% chance that a particular person shares the same birthday with someone else in a room, we need at least 253 people; but if we are looking for just any one pair of people sharing the same birthday, we only need 23. A birthday attack exploits the exponential decrease of the latter scenario by brute forcing a hash function to try find any random pair of inputs that output the same hash (i.e. a collision). Brute forcing to find any two messages with the same hash would utilise the exponential decrease of work as the number of bits increases.
B - Property of first preimage resistance which means that it is easy to get from input to hash, but very hard to get from hash to input. Finding the original message will be a plain brute force.
C - Property of second preimage resistance. Refer to the reasoning for A, where this would apply as the former scenario.
D - We may have some of the contents, but this still won’t help us at all in finding the first half of the message, because even a small change in the text will lead to a completely different hash like an avalanche effect. This would still be just as good as a normal brute force.
E - Don’t be dumb
The solution is likely A.
=== Question 4 === (2 Marks) If all of the 100 agents in a system need to be able to communicate with each other confidentially using an asymmetric cipher, how many distinct private keys would be needed? Note: only enter DIGITS in your answer (or else the automarker will mark it wrong.)
Every agent needs their own private key to encrypt and decrypt message. If a message wants to be sent, then the private key must first be shared securely with the receiver. Hence, 100 distinct keys will be needed.
=== Question 5 ===
(2 Marks) You are going to use a system based on Merkle Puzzles to allow other people to communicate with you securely even if you have never met or shared a secret before. You encode 1,000,000 messages, each consisting of a different 200 character English phrase and a different 80 bit Merkle key, each message encoded using a different 40 bit CDMF key. If someone has a system which can decode an CDMF ciphertext into a plaintext using a given 40 bit CDMF key in 1 billionth of a second, roughly how many seconds will it take them to discover a Merkle key to use to privately communicate with you on average? Give your answer in digits only, and rounded to the nearest power of 2. So if you thought the answer was 70 seconds you would enter 64. Note: only enter DIGITS in your answer (or else the auto marker will mark it wrong.) except you may use scientific notation for very big numbers if needed. In order to find a key to communicate privately, only one of the 1,000,000 messages needs to be decoded. The key that encrypts a message is one of 2^40. The average time required to find the key for the message is therefore 2^39.
To discover a key to use, they need to decode at least one message which is encoded in a 40 bit CDMF key (that is, 2^40 amount of work). The person can decode one guess in a billionth of a second, or 1 billion guesses per second. On average, the CDMF key will be decoded in 2^40/2 = 2^39 amount of work.
2^39 work / (1,000,000,000 guesses/second) = 2^39 / 2^30 = 2^9 = 512 seconds
==== Part B ==== This part is worth 12 marks and consists of 1 question. Working is not marked, only your final answer. Note: only enter DIGITS in your answer (or else the automarker will mark it wrong.) === Question 6 === (12 Marks) A bank has invented a new way of turning a message into a number - called a CDC code, and uses this as part of a procress to verify messages with so called VERIFY codes. The bank authenticates and protects the integrity of its messages by computing a VERIFY CODE which is the CDC code of {password,message} I.E. The CDC code of a secret password with the message appended to then end of it. The Bank sends the resultant VERIFY CODE along with the message (but does not send the password of course!) All messages each day use the same password. Below is how a CDC code is computed: Start with the first character of the message, convert it into a code using the following table (only some values shown) {{{ A->23 B->47 I->397 L->507 O->581 P->635 R->687 U->763 Y->901 0->405 1->73 }}} Then multiply this by 521, mod it by 10,000, add 450, mod it by 967, then add in the code for the next character, and repeat... For example the CDC code for the message "AU" is 216. Working: {{{ 'A' 23x521 = 11983 mod 10,000 = 1983 +450 = 2433 mod 967 = 499 +763 'U' = 1262 x521 = 657502 mod 10,000 = 7502 +450 = 7952 mod 967 = 216 }}} You have intercepted the following messages and VERIFY CODES: {{{ "PAYBOB100" has VERIFY CODE of 481 "BILLBOB1000" has VERIFY CODE of 573 "PAYROB1000" has VERIFY CODE of 301 }}} What is the VERIFY code for "PAYBOB1000" ? (enter digits only in your answer) Notice this question is asking for the VERIFY code not the CDC code.
Can perform a length extension attack by using “PAYBOB100″ and appending “0″ and calculating the new verification code. We get 122
481 + 405 = 886 886 * 521 = 461606 461606 mod 10000 = 1606 1606 + 450 = 2056 2056 mod 967 = 122
Midsems - 2017
=== Question 0 === (3 Marks)
What is the most important property for a cryptographic hash function to have to make it as resistant as possible to birthday attacks?
.[A] Preimage resistance .[B] Collision resistance .[C] Confidentiality .[D] Symmetry .[E] Second-preimage resistance .[F] Non-determinism
A birthday attack is a cryptographic attack that has its roots in the following statistical scenario: If we are looking for at least a 50% chance that a particular person shares the same birthday with someone else in a room, we need at least 253 people; but if we are looking for just any one pair of people sharing the same birthday, we only need 23. A birthday attack exploits the exponential growth of the latter scenario by brute forcing a hash function to try find any random pair of inputs that output the same hash (i.e. a collision). Hence, a hash function will need strong collision resistance - B.
=== Question 1 === (3 Marks)
You are a developer of opensource fake-news software called "The trumpet shall sound", or "Trumpet" for short. You wish to use a cryptographic algorithm to generate a 64 character "digital fingerprint" of the "Trumpet" binary, which you will display on your web site so that people can check that they have downloaded a legitimate and unaltered copy of "Trumpet".
What is the most important property for this cryptographic fingerprint algorithm to have?
.[A] Preimage resistance .[B] Collision resistance .[C] Confidentiality .[D] Symmetry .[E] Second-preimage resistance .[F] Non-determinism
Lets quickly revise the properties of a solid cryptographic function:
First pre-image resistance: it is easy to give an input to turn into a hash, but hard to reverse a hash to find the original input
Second pre-image resistance: given an input and its hash, it is hard to find another input that gives the same hash
Collision resistance: it is tough to find any pair of inputs that give the same hash
In this case, the idea of a digital fingerprint is to make sure you have an authentic version of product. That purpose would be defeated if a hacker could have an unauthentic product passing off as real by giving the same digital fingerprint as the real thing. Thus, the second pre-image resistance property would be the most important in this case - E.
=== Question 2 === (3 Marks)
A MAC is designed to provide .[A] Neither Integrity nor Authentication .[B] Authentication but not Integrity .[C] Integrity but not Authentication .[D] Integrity and Authentication .[E] A Very Fashionable Device
A message authentication code (MAC) is a form of symmetric key encryption technique used to verify that a message came from the intended source and is unaltered. A sender and receiver shares a secret key. When the sender sends a message, the secret key is appended to the message, then hashed (the MAC). The message and the MAC are both sent to the receiver. Upon receipt, the receiver can verify that the message came from the source and that the message is unaltered from its original form by hashing the message with the secret key and comparing with the MAC. Hence, integrity and authentication are provided - D.
=== Question 3 ===
(3 Marks)
You have a password file of 1,000 password hashes. Assuming that 50% of the passwords used were weak passwords which can be found in a dictionary of 1,000,000 common passwords, and that hashing a password takes 12 bits of work on average, roughly how many bits of work would it take to discover all these 500 or so weak passwords?
Note: only enter DIGITS in your answer (or else the automarker will mark it wrong.)
To find the 500 or so weak passwords, we can hash the 1,000,000 common passwords in the dictionary, then compare it with the 1,000 password hashes in the file. Each password hash takes 12 bits of work on average, that is, 2^12 guesses on average. So:
2^12 amount of work * 1,000,000 passwords =~ 2^12 * 2^20 = 2^32. Therefore 32 bits of work. The work associated with the actual comparing of the hashes to the file is negligible.
=== Question 4 === (3 Marks)
As for Question 3 but now assume that each password is appended to a different 32 bit number (a salt) before the hash was computed, and that the salt number for each password hash is stored, unencrypted, next to the hash in the password file.
Roughly how many bits of work would it take to discover all the 500 or so weak passwords in this case?
Note: only enter DIGITS in your answer (or else the automarker will mark it wrong.)
For this question, we can’t use the previous method of just computing a rainbow table and comparing it with our password file hashes because the file hashes have been salted.
On average to discover a weak password, we will need 1,000,000/2 = 500,000 hashing attempts. There are 500 weak passwords. So:
2^12 work * 500,000 attempts * 500 passwords =~ 2^12 * 2^19 * 2^9 = 2^40 work
On average when encountering a strong password, we will need to exhuasatively try all 1,000,000 dictionary words before determining it is not one of the weak passwords in the file. There are 500 strong passwords. So:
2^12 work * 1,000,000 attempts * 500 passwords =~ 2^12 * 2^20 * 2^9 = 2^41 work.
2^40 + 2^41 =~ 2^41, therefore 41 bits of work.
=== Question 5 === (3 Marks)
Below are six lists of expenses submitted by six politicians. Sadly 3 of the politicians are dishonest and submitted made-up fake expense claims, not costs that they really incurred. :(
{{{ Albus Bert Carl Dave Egon Fred
269 6,296 543 1,137 503 206 1,428 3,217 220 638 4,195 74 3,840 520 3,080 696 5,780 257 2,163 456 5,070 543 7,774 350 165 111 108 113 986 50 2,701 2,176 8,078 145 884 426 339 476 7,963 313 6,530 3,226 64 2,607 68 236 50 108 942 824 2,457 701 9,801 94 17 10 7,952 422 5,717 17 423 112 5,332 5,723 29 158 37 13 5,705 964 3,933 609 723 9,048 8,851 750 6,122 5,769 172 295 99 7,905 91 19 498 165 6,424 9,953 89 275 5,583 3,008 48 491 19 898 56 1,396 8,339 2,675 758 506 158 19 1,945 8,212 427 3,467 1,244 238 916 897 8,503 9,217 2,445 51 3,618 392 3,617 26 866 921 4,219 941 338 1,349
}}}
Correctly identify the status of each politician.
==== 5.1 Albus ==== .[A] Honest .[B] Dishonest
==== 5.2 Bert ==== .[A] Honest .[B] Dishonest
==== 5.3 Carl ==== .[A] Honest .[B] Dishonest
==== 5.4 Dave ==== .[A] Honest .[B] Dishonest
==== 5.5 Egon ==== .[A] Honest .[B] Dishonest
==== 5.6 Fred ==== .[A] Honest .[B] Dishonest
One approach to this question is understanding that generally fake-up expenses should be inflated because human-nature of greed. Summing up all the expenses of individual politicians, the hugely inflated expenses will likely be the dishonest ones - C, D, E.
==== Part B ==== This part is worth 10 marks and consists of 1 question. Spend about 10 minutes on it.
=== Question 6 === (10 Marks)
What is the 5th word in the plaintext for the ciphertext below?
{{{ HHPET IAESR DTEHN SRAKA UASMI UEOYR CNWOK FNERA TRONE TNYLO SOENO OHDAS RYUSW DONUR BNSGI STLAU HTTOO DSAHE EWSYO SRTEC IFHWO EHNOC TIHGM AENUB EAERW DEYRV EYHTA OUGNS WSODE ROEVN HLTLA ESSEE SRTEC ONSDA YVREE SNIEO ENHTI IAKRD WOENN AYROA EOHTN OIYFR SAERU TNABU IIGNH RAAPN IFROK NSATN TEUBC OODUY NOKTN AWHTO CAOLT AECDK TIENB RSUBI IEFDI ETFYF NTEBE YAHTE LUBRO TNEKA YHNET EURAO ENHTI EAKRD HETNV YUHGO EURAO COATN LULAT HITNY KDRAE }}}
The cipher questions we will receive in the exam will most likely be either substitution or transposition, anything else will probably be way too hard in the time pressures of a 1 hour exam. It is good to first scan the cipher for any possible patterns, and consider various methods such as letters across different blocks, reverse order, forward order, rearrangement of letters in blocks, etc. A quick inspection of the cipher, I notice that I see the group of letters T,H,E occuring together several times. Since a tranposition cipher only rearranges the leters, it is highly likely that this group of letter could refer to the common word THE. The pattern turns out to be a rearrangement of letters in the blocks with the original order of the text formed by reading in the order {5, 2, 4, 3, 1} -> {1, 2, 3, 4, 5}. We get: THEPHRASEINTHEDARKASIAM... So the 5th word is DARK.
(==== Part C ==== This part is worth 20 Marks and consists of 1 question. Spend about 20 minutes on it.
=== Question 7 === (20 Marks)
Suppose we convert letters and other characters into numbers in the range 1..54 using the following table and then separately encipher each of these numbers using RSA with a public encryption key of 23 and a modulus of 55.
{{{ 1 < 21 L 41 4 2 > 22 M 42 5 3 ? 23 N 43 6 4 / 24 O 44 7 5 : 25 P 45 8 6 ; 26 Q 46 9 7 “ 27 R 47 ( 8 ' 28 S 48 ) 9 @ 29 T 49 * 10 A 30 U 50 + 11 B 31 V 51 , 12 C 32 W 52 - 13 D 33 X 53 . 14 E 34 Y 54 / 15 F 35 Z 16 G 36 SPACE 17 H 37 0 18 I 38 1 19 J 39 2 20 K 40 3 }}}
For example the plaintext ":)" would be converted into 5,48, and then each of these numbers would be individually encrypted using RSA to produce the cipertext 15,42.
The remainder of this question has been encrypted using RSA with an encryption key of 23 and a modulus of 55.
{{{ 24,18,49,16,48,49,33,10, 2,12, 52,49,48,16,19,20,16,24,18, 2, 7,16,31,50,49, 7,24, 2,19,12, 16,18,10, 7,16,11,49,49,12,16, 49,12,23,48,34, 5,24,49,52,16, 50, 7, 2,12,26,16,48, 7,10,16, 43, 2,24,18,16,10,12,16,49,12, 23,48,34, 5,24, 2,19,12,16,25, 49,34,16,19,20,16,29,35,16,10, 12,52,16,10,16,33,19,52,50,21, 50, 7,16,19,20,16, 3, 3,47,16, 16,24,18,49,16,10,12, 7,43,49, 48,16,23,19,52,49,43,19,48,52, 16, 2, 7,16,53,53,53,53,53,53, 53,53. }}}
What is the answer codeword?
You’ll need to understand how RSA cryptography works to solve this question. RSA works by having any user maintain a public key for encrypting messages (e,N) and private key (d,N) for decrypting messages. As the name implies, anyone can access the user’s public key to encrypt messages, but only the user can decrypt the messages with their private key.
To decode the codeword in the example, we need to brute force and find the private key. In this case, the user has chosen the public key (23,55). Lets consider how we can form our private key.
The traditional steps to generating a set of public and private key are as follows:
Choose any 2 prime numbers P and Q.
Calculate N = PQ.
Calculate Euler’s totient φ(N), which refers to how many numbers less than N are also co-prime (does not share common factors) with N. An easy formula to quickly calculate this is φ(N) = (P - 1)(Q - 1)
Choose an encryption key e by choosing a number that satisfies the properties 1 < e < φ(N), and co-prime with φ(N) and N
Choose a decryption key d by choosing a number that satisfies e*d mod φ(N) = 1
We know that e = 23, N = 55. We can easily work out that the only possible combination of P and Q that gives N = 55 is P = 5 and Q = 11. φ(N) = (5 - 1)(11 - 1) = 40. So we can find d by solving for 23*d mod 40 = 1. Probably in this exam d will be the first instance that satisfies the condition (otherwise we could go on forever brute forcing). So lets try d = 7 to give the decryption key (7,55).
For a text, we can encode with RSA by using (t^d) mod 55 = c. Similarly, to decode a cipher text c, we need to simply apply (c^d) mod 55 = t. Working through the text applying t = (c^7) mod 55, we get our text “THE REMAINDER...” (You can decode the rest yourself :).
There is a calm in this library, a calm that is not polluted by the noise of stark silence nor by the kind of lighting that craves attention.
