程序代做CS代考 scheme dns DHCP cache algorithm Popa & Wagner CS 161 – cscodehelp代写

Popa & Wagner CS 161
Midterm 2
Spring 2020
Computer Security
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You have 80 minutes. There are 10 questions of varying credit (200 points total).
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We will not be answering any clarifications about the exam. If there are any glaring problems with wording, we will consider dropping the question from the exam after solutions/grades are released.
Q1 MANDATORY – Honor Code (3 points) On your Gradescope answer sheet, read the honor code and type your name. Failure to do so will result in a grade of 0 for this exam.
This is the end of Q1. Proceed to Q2 on your Gradescope answer sheet.
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Q2 True/False (48 points) Each true/false is worth 2 points unless otherwise specified.
Q2.1
Q2.2
Q2.3
TrueorFalse:IfBobusestheauthenticate-then-encryptparadigm,theintegrityofhisplaintext is guaranteed.
True False
True or False: A hash function must be collision-resistant to be considered safe for password hashing.
True False
True or False: Alice wants to use authenticated encryption to send a message to Bob. She should use Enc(M), MAC(H(Enc(M))) over Enc(M), MAC(H(M)).
True False
Suppose we increase the entropy of the DNS ID field to 128 bits. It is infeasible for an on-path adversary to spoof a DNS answer.
True False
True or False: By default, in a TLS connection, both the server and client are authenticated to each other.
Solution: True. Authenticate-then-encrypt guarantees integrity for the plaintext, just not the ciphertext.
Solution: False. Only the one-way property is needed. Collisions are okay as long as one cannot find a preimage for a given function value.
Solution: True. The former scheme provides confidentiality, whereas the second scheme may not, depending on what MAC algorithm is used. For instance, MAC(H(M)) may be a deterministic function of M, so an eavesdropper might be able to identify when the same message is sent twice. Also the second scheme may be vulnerable to padding oracle attacks if it is not implemented very carefully, though this is beyond the scope of this class.
Q2.4
Q2.5
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Solution: False. The adversary is on-path so they can see the QID in plaintext.
True
False
Solution: False. TLS only authenticates the server by default.

Q2.6
Q2.7
Q2.8
Q2.9
Q2.10
Q2.11
True or False: If weak passwords are salted and hashed before being stored, then the attacker cannot easily learn the plaintext values of the weak passwords.
True False
True or False: A DNS lookup for en.wikipedia.org will always force the recursive resolver to send at least 3 DNS queries.
True False
True or False: If the server’s random number 𝑎 in Diffie- LS is the same in every handshake, Diffie- LS no longer has forward secrecy. Assume the value 𝑎 is stored on the server along with its secret key.
True False
TrueorFalse:IfBobisanon-pathattackerwhocanguaranteethathisspoofedresponsearrives before the legitimate response, Bob only needs the victim to make one request for a nonexistent domain in order to successfully execute a Kaminsky attack with 100% probability.
True False
True or False: Randomizing the client port helps defend TCP against on-path attackers. True False
True or False: TLS provides end-to-end security, so it is secure even if the server has a buffer overflow vulnerability.
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Solution: False. The salted password is still susceptible to a dictionary attack.
Solution: False. Answers could be cached.
Solution: True. An attacker who steals 𝑎 will be able to reconstruct the PS and decrypt past recorded messages by computing (𝑔𝑏)𝑎 mod 𝑝.
Solution: True. Bob doesn’t need to guess any random fields or try several times to race the legitimate response.
Solution: False. The on-path attacker can see the port values.
True
False
Solution: False. If an attacker exploits the buffer overflow vulnerability to gain control of the server, TLS doesn’t stop you from talking to the compromised server.

Q2.12
True or False: Suppose we modified TCP so that the sequence number increases by 2 for every byte sent, but the initial sequence numbers are still randomly chosen. This modified protocol has the same security guarantees as standard TCP.
True False
True or False: If IP spoofing is eliminated from the Internet (all attackers must send messages from their real IP), then an on-path attacker is no longer more powerful than an off-path attacker.
True False
True or False: Consider a modified version of DHCP, where in the server offer step, the server signs its message and sends its public key along with the signed message. This version of DHCP is secure against the DHCP spoofing attack.
True False
TrueorFalse:TCPissecureagainstaDoSattackbyaman-in-the-middle(MITM)becauseTCP guarantees delivery and will re-send messages until they are delivered.
True False
True or False: RSA-TLS is still secure if we use publically known lottery numbers as the value of the premaster secret (PS).
True False
True or False: Under the SOP, it is possible for two webpages with different origins to commu- nicate through narrowly defined APIs.
Solution: True. Incrementing the sequence number differently doesn’t make it any easier for an off-path attacker to guess it, and if you’re on-path or MITM, you can see everything anyway.
Q2.13
Q2.14
Q2.15
Q2.16
Q2.17
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Solution: False. On-path attackers can still read messages, while off-path attackers can’t.
Solution: False. The client has no way to verify the public key. An attacker could easily send their own malicious public key and use that to sign a spoofed response.
Solution: False. The MITM could just keep dropping packets so that messages never arrive. Also, the MITM can inject a RST packet, which ends the connection.
Solution: False. An on-path or MITM attacker would know 𝑅𝑏 and 𝑅𝑠 (sent in plaintext) as well as PS, which would allow them to generate the symmetric keys and decrypt everything.
True
False
Solution: True. This is the postMessage API.

Q2.18
Q2.19
Q2.20
Q2.21
Q2.22
True or False: Under the SOP, the webpage at https://example.com/randompic.html cannot fetch the image at https://cute-cats.com/cutest.jpg because they have different origins.
True False
True or False: Suppose the webpage at https://example.com/index.html contains a child frame that loads https://another-example.com/index.html. Under the SOP, the parent frame can read and modify the properties of the child frame.
True False
(2 points) True or False: Suppose the webpage at https://example.com/index.html con- tains a child frame that loads https://example.com/views.html. Under the SOP, the child frame can read and modify the properties of the parent frame.
True False
True or False: Suppose the webpage at https://example.com/index.html loads and runs an external script from https://sample.com/script.js. Under the SOP, the script runs with the same origin as https://sample.com/script.js.
True False
TrueorFalse:MalloryconvincesAlicetoconnecttoherprivateWi-Finetwork.Webpagesthat Alice visits while on this network may no longer be subject to the SOP.
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Solution: False. A page can fetch images and content regardless of origin; the SOP only prevents it from determining detailed properties of cross-origin content.
Solution: False. Because example.com and another-example.com have different domains, they are considered different origins and are therefore unable to access each other directly.
Solution: True. Pages are allowed to directly access child frames from the same origin, and vice versa.
Solution: False. External scripts run with the origin of the page that fetched them (in this case, https://example.com).
True
False
Solution: False. The SOP is unrelated to the network.

Q2.23 True or False: Mallory convinces Alice to try out her custom browser, FireFaux. Webpages Alice visits using this browser may no longer be subject to the SOP.
True False
Q2.24 True or False: Consider a modified version of Diffie- LS where the server does not include the signature when sending 𝑔𝑎 mod 𝑝. This version of TLS does not provide confidentiality against a MITM.
True False
Q2.25 (0 points) True or False: EvanBot is a bot.
True False
This is the end of Q2. Proceed to Q3 on your Gradescope answer sheet.
Solution: True. The SOP is enforced by the browser; if the browser is compromised, there is no guarantee that webpages will play by the rules.
Solution: Withoutthesignature,weloseprotectionagainstaMan-in-the-Middlesinceweare now effectively using standard Diffe-Hellman key exchange. An adversary can now pretend to be the client to the server and pretend to be the server to the client, and therefore we lose all security guarantees because we now trust the Man-in-the-Middle.
Solution: True. How dare you doubt our trusty AI.
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Q3 (18 points)
For each public-key infrastructure (PKI) scheme, mark whether it provides the same trust guarantees as the standard PKI from lecture for all certificates, some certificates, or no certificates at all. Assume that everyone has the root certificate hardcoded into their machines.
Q3.1
(3points)Eachservercanonlysignthepublickeysofitsgrandchildren(twodescendantsbelowthe current level). For example, the root server can sign the public key of berkeley.edu but not .edu, and the .edu server can sign the public key of eecs.berkeley.edu but not berkeley.edu.
(A) All certificates (C) No certificates (E) (B) Some certificates (D) (F)
(3 points) As in the previous part, each server can only sign the public keys of its grandchildren. However, the root is additionally allowed to sign the public key of its direct children. For example, the root server can sign the public key of .edu and berkeley.edu. The .edu server can sign the public key of eecs.berkeley.edu but not berkeley.edu.
(G) All certificates (I) No certificates (K) (H) Some certificates (J) (L)
(3 points) Same setup as the previous part, but an attacker has compromised a server one level below the root (e.g. .edu).
Solution: This works for any certificates located an even number of levels below the root. However, there is no way to create a path of trust from the root to a certificate located an odd number of levels below the root, such as eecs.berkeley.edu.
Q3.2
Solution: Skipping two levels at a time, all certificates must have a path of trust that ends at either root or a server one level below root. Since the root is allowed to sign public keys of servers one level below it, this scheme now works for all certificates.
Q3.3
(A) All certificates (B) Some certificates
(C) No certificates (E) (D) (F)
Solution: Any certificate whose path to the root doesn’t go through the compromised server (e.g. google.com) is unaffected, but a certificate whose path goes through the compromised server (e.g. berkeley.edu) cannot be trusted.
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Q3.4 (3points)Theroothandlesallrequestsandsendstherequestedpublickeyandacertificatedirectly through a TLS connection.
(G) All certificates (I) No certificates (K) (H) Some certificates (J) (L)
Q3.5 (3 points) Instead of signing, use a cryptographic hash to create a certificate. For example, the root server signs the public key of .edu by hashing it.
(A) All certificates (C) No certificates (E) (B) Some certificates (D) (F)
Q3.6 (3points)Insteadofsigning,useHMACtocreateacertificate.Forexample,therootserversignsthe public key of berkeley.edu by applying HMAC(𝐾 , berkeley.edu), where 𝐾 is the root’s private signing key.
(G) All certificates (I) No certificates (K) (H) Some certificates (J) (L)
This is the end of Q3. Proceed to Q4 on your Gradescope answer sheet.
Solution: TLS provides end-to-end integrity.
Solution: Hashes don’t provide integrity. An attacker can create a valid signature on their malicious public key just by hashing it.
Solution: HMACs use symmetric keys, so there is no way for the signatures to be verified without knowing the server’s secret key.
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Q4 (29 points)
Alice is using a DNS resolver to perform a DNS lookup for www.google.com. A single, valid nameserver is authoritative for each of the following zones:
Zone
.
.com
google.com
Nameserver
a.root-servers.net
a.gtld-servers.net
ns1.google.com
Assume no other legitimate clients will query the resolver (but the adversary can query it if they wish), the resolver’s cache is initially empty, and the resolver uses iterative querying.
Assume that in DNSSEC, no one will accept a record unless it has a valid signature.
The attacker is on-path between the resolver and ns1.google.com, but off-path to the other name servers. The attacker also knows when Alice makes a request. Assume DNS uses a static source port known to the attacker.
For each part, select all of the records that the attacker can poison. Q4.1 (4 points) Standard DNS is used.
(A) Alice’s cached A record for www.google.com (B) Resolver’s cached NS record for .com
(C) Resolver’s cached NS record for google.com (D) Resolver’s cached NS record for .
(E) (F)
Solution: The adversary can spoof the DNS response from the resolver to the client even though they are off-path since we are using vanilla DNS and the source port isn’t randomized (poisoning Alice’s A record). Furthermore, the adversary can mount a Kaminsky attack against the resolver by querying the resolver directly; since the attacker can predict the source port, this will be successful. So, they can poison the cache for all of the NS records except the root since this is hardcoded. The adversary can also use this method to poison the A record for www.google.com cached on the resolver, and when Alice queries it, the resolver will respond to the client with the poisoned cache entry.
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Q4.2 (3 points) Standard DNS is used. Also, the resolver has a hardcoded NS record that maps the google.com zone to ns1.google.com, and a hardcoded A record with the IP address of ns1.google.com.
(G) Alice’s cached A record for www.google.com (H) Resolver’s cached NS record for .com
(I) Resolver’s cached NS record for google.com (J)
(K) (L)
Q4.3 (3 points) The resolver and nameservers use DNSSEC, and Alice uses standard DNS. (A) Alice’s cached A record for www.google.com
(B) Resolver’s cached NS record for .com
(C) Resolver’s cached NS record for google.com (D)
(E)
(F)
Q4.4 (3 points) The resolver and nameservers use DNSSEC, and Alice uses standard DNS. The adversary compromises a.gtld-servers.net.
(G) Alice’s cached A record for www.google.com (H) Resolver’s cached NS record for .com
(I) Resolver’s cached NS record for google.com (J)
(K) (L)
Solution: Similartoabove,theadversarycanpoisoneverythingintheresolver’scache,except the hardcoded records.
Solution: Same reasoning as above for the first option. The adversary can’t poison the cache for any of the NS records since that would require forging a signature.
Solution: Same reasoning as above for the first option. Controlling the .com domain allows the attacker to poison the resolver’s cached NS record for google.com and .com since they have the private signing key and both of these domains are in-bailiwick. However, the fact that .com is in-bailiwick was considered outside the scope of the course and so this option wasn’t graded.
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Q4.5 (3 points) The resolver and nameservers use DNSSEC, and Alice uses standard DNS. The adversary compromises ns1.google.com.
(A) Alice’s cached A record for www.google.com (B) Resolver’s cached NS record for .com
(C) Resolver’s cached NS record for google.com (D)
(E) (F)
Q4.6 (3 points) All parties use standard DNS, but the resolver and Alice encrypt their DNS messages with TLS.
(G) Alice’s cached A record for www.google.com (H) Resolver’s cached NS record for .com
(I) Resolver’s cached NS record for google.com (J)
(K) (L)
Q4.7 (3points)AllpartiesusestandardDNS,butAlice,theresolver,andns1.google.comencrypttheir DNS messages with TLS.
(A) Alice’s cached A record for www.google.com (B) Resolver’s cached NS record for .com
(C) Resolver’s cached NS record for google.com (D)
(E) (F)
Solution: Controlling the google.com nameserver allows the attacker to poison the final result which they were already able to do. Same as the previous question, the attacker can also poison the cache for google.com but this option wasn’t graded. Bailiwick rules stop them from poisoning the cache for higher zones.
Solution: The attacker can perform the Kaminsky attack to poison the .com + google.com NS records and on-path spoofing for the final DNS result.
Solution: The attacker can use the Kaminsky attack to poison the NS records for the .com zone since the root nameserver and .com nameservers don’t use TLS, and then can subsequently poison the google.com NS record and the final result.
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Q4.8 (3 points) All parties use standard DNS, but everyone encrypts their DNS messages with TLS. (G) Alice’s cached A record for www.google.com
(H) Resolver’s cached NS record for .com
(I) Resolver’s cached NS record for google.com (J)
(K)
(L)
Q4.9 (4 points) Alice and the resolver use standard DNS, but encrypt their DNS messages with TLS. The resolver and nameservers use DNSSEC.
(A) Alice’s cached A record for www.google.com (B) Alice’s cached NS record for google.com
(C) Resolver’s cached NS record for .com
(D) Resolver’s cached NS record for google.com (E)
(F)
This is the end of Q4. Proceed to Q5 on your Gradescope answer sheet.
Solution: Theattackerhasn’tcompromisedanyofthenameserverssotheycan’tdoanything here. TLS prevents an on-path adversary from tampering with the messages between the parties.
Solution: The attacker can’t poison any caches here due to DNSSEC, and can’t compromise the client’s connection with the resolver due to TLS.
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Q5 (37 points)
FastCash is a fast banking service which requires users to log in before making a transfer, and uses TLS with ephemeral Diffie Hellman and RSA certificates to secure all their connections. They implemented a TLS extension called 0-Round Trip (0-RTT) to speed up the connection process. 0-RTT changes the initial handshake as follows:
• Users authenticate themselves during the second round of the handshake
• If the user authenticates correctly, the server stores a session_token for that user
(Recall that in TLS, PS, RS, and RC generate a master key set MK which contains all the symmetric keys. EncPKS denotesRSAencryptionusingtheserver’spublic RSA key.)
A user only needs to perform the modified TLS handshake once. To send an HTTP request after the initial connection ends, a user encrypts it using the keys derived in the initial handshake and attaches the session_token. The server verifies that the entry session_token ∶ (uname, MK) exists and, if so, decrypts and executes the request as the user uname using the keys derived from MK.
Simplified diagram of modified initial TLS handshake
Assume that any on-path TCP injection attacks are impossible, and that once a user makes the initial modified TLS handshake, they will use the 0-RTT extension for future requests to the server.
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Q5.1 (6 points) An on-path attacker observes an initial TLS handshake between a user and server, as well as a subsequent 0-RTT packet which contains an encrypted HTTP request. What can they do?
(A) Read the user’s future communications
(B) Break forward secrecy for that user’s communications (C) Pretend to be the server to the user
(D) Pretend to be the user to the server in a new handshake (E) Replay the encrypted HTTP request to the server
(F) Learn the master key set
Solution: The adversary can’t learn any of the keys and so can’t decrypt anything or fake being the server. While normally TLS doesn’t authenticate the client, the 0-RTT extension involves authentication so without knowledge of the username/password the adversary can’t pretend to be the user either. Including 𝑅𝑆 in the encryption stops the ciphertext from being replayed in a different session.
The adversary knows the session_token, so they can use the 0-RTT extension to replay an encrypted HTTP request they observed.
We gave credit for all answers for ’Break forward secrecy..’ due to two separate interpretations of forward secrecy.
Q5.2 (6 points) Suppose we removed 𝑅𝑆 from the user’s KeyExchange in the third step of the handshake. After observing an initial handshake between a user and the server, what can an on-path adversary do?
(G) Read the user’s future communications
(H) Break forward secrecy for that user’s communications (I) Pretend to be the server to the user
(J) Pretend to be the user to the server in a new handshake (K) Learn the premaster secret
(L) Learn the master key set
Solution: The adversary can’t derive the premaster secret due to DH, and thus can’t learn the master key set, violate forward secrecy, or learn future communications. Furthermore, they can’t pretend to be the server to the user: the server’s KeyExchange message, so the attacker can’t modify or forge it, and consequently the adversary cannot predict the result of the Diffie-Hellman key exchange, the premaster secret, or the master key set.
However, the adversary can initiate a new handshake and replay the EncPKS (uname; pwd) ciphertext observed in the first handshake to log in as the user.
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Q5.3 (6 points) Due to a bug, an on-path adversary is able to choose the server’s 𝑅𝑆 . After observing an initial handshake between a user and the server, what can they do?
(A) Read the user’s future communications
(B) Break forward secrecy for that user’s communications (C) Pretend to be the server to the user
(D) Pretend to be the user to the server in a new handshake (E) Learn the premaster secret
(F) Learn the master key set
Q5.4 (6 points) An on-path adversary observes a user and the server communicating using 0-RTT for some time (without observing the initial handshake). At some point in the future, the adversary manages to learn all of the server’s session_token ∶ (uname, MK) entries. What can they do?
(G) Read the user’s future communications
(H) Break forward secrecy for that user’s communications (I) Pretend to be the server to the user
(J) Pretend to be the user to the server in a new handshake (K) Learn the premaster secret
(L) Learn the master key set
Solution: Same reasoning as above. The only thing that’s different is the adversary has to force the server’s RS to be the same as used in the initial handshake to get the replay to work.
Solution: The adversary can learn the master key set (but not the premaster secret). This allows them to decrypt all future communications. Note that since we are essentially using a long-term private key (it is reused for all subsequent 0-RTT handshakes from the same user), we nullify the forward secrecy of using the Diffie-Hellman key exchange once the adversary has the master key set.
The adversary has no way to learn a valid ciphertext for the user’s password so they can’t pretend to be them. Since in future connections the user doesn’t check the server’s certificate, the adversary can pretend to be the server.
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Q5.5 (10 points) Consider a MITM adversary during the initial handshake between a user and the server. Describe how this adversary can send a malicious HTTP request that appears to come from the legitimate user (Be specific with what is sent). Disregard any bugs from previous parts.
Solution: The adversary should do a DH MITM. When the server sends 𝐑𝐒, the adversary should relay that same value on to the client. When the client sends the encrypted password, the adversary forces the client’s connection to end by sending a RST packet. Next, the adversary replays this encrypted password to the server. This will be accepted by the server since RS will be the same as the server was expecting, but replaces the client’s 𝑔𝑏 value with the adversary’s own 𝑔𝑏′ . The adversary can then compute the result of the DH key exchange with the server, derive the premaster secret and master key set, and uses the derived MAC key to finish the handshake with the server. The adversary can now log in as the user using the session token returned by the server.
Normally TLS does not authenticate the client, so a MITM can always take over a connec- tion initiated by the client. The key difference here is that the 0-RTT extension effectively authenticates the user. So the MITM can not only take over the connection, but also does so authenticated as the client; the server thinks messages are coming from the user, when actually they are coming from the adversary.
Q5.6 (3 points) Because of the vulnerability from the previous part, the company decides that it’s too dangerous to allow all web pages to be accessible via 0-RTT. suppose they support the following three HTTP requests:
1. GET request for bank’s homepage
2. GET request for bank’s transfer page 3. POST request to execute a transfer
Below are different possible combinations of pages which will be made accessible via 0-RTT. Select the combination with no vulnerability or None if they are all vulnerable.
(G) 2, 3 (I) 1, 3 (K) (H) 1,2 (J) None (L)
This is the end of Q5. Proceed to Q6 on your Gradescope answer sheet.
Solution: We can never allow the POST request since this changes state. The question didn’t specify how the webpages worked. If the webpages are static, and don’t leak private information, then allowing the GET requests is fine. Otherwise, we can’t allow those either. So 1,2 and None were accepted answers.
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Q6 (8 points)
Q6.1 (2 points) True or False: A NIDS always provides the most insight about ongoing network traffic.
(A) True (B) False (C) (D) (E) (F)
Q6.2 (3 points) An edgy hacker, xXOskiTheHackerXx, downloads a ransomware tool on GitHub and, without modifying it, tries to target the CDC. Which is the best detection strategy to detect this type of hacker?
Solution: False, a NIDS can’t be used to monitor TLS traffic.
(G) Signature based (H) Behavior based (I) Anomaly based
(J) Specification based
(K) (L)
Solution: Signature based. The tools are public (on GitHub) and xXOskiTheHackerXx won’t be able to modify the program to avoid signature detection.
Q6.3 (3 points) Andrew needs to decide between two burglar alarm systems – system A and system B. System A has a false positive rate of 0.05% and a false negative rate of 1%. System B has a false positive rate of 1% and a false negative rate of 0.05%.
The cost of a false positive is $100, because his parents fine him for causing a ruckus, and the cost of a false negative is $10000, because the burglar steals all his stuff. Which system should Andrew pick?
(A) System A (D) (B) System B (E) (C) Not enough information (F)
This is the end of Q6. Proceed to Q7 on your Gradescope answer sheet.
Solution: Not enough information — we don’t know how often attacks happen.
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Q7 (15 points)
Q7.1 (3 points) Alice clears all her network settings and broadcasts a DHCP discover message. What information should she expect to receive in the DHCP offer in response?
(A) DNS server (B) Source port (C) Lease time
(D) Premaster secret (E) Gateway router (F) IP address
Solution: The DHCP offer will include IP address, DNS server, gateway router, and how long the client can have these (“lease time”). The source port is determined by the user’s machine. DHCP does not involve any premaster secret.
Q7.2 (6points)AfterreceivingtheDHCPoffer,Alicetriesconnectingtowww.cutecats.com,butinstead of pictures of cats, the site she gets is filled with dog photos.
How did the attacker compromise DHCP to accomplish this?
Which of the following could the attacker have replaced?
Solution: Since the DHCP discover message is broadcasted, any local attacker can hear the host’s request. The attacker then spoofed the DHCP response by racing the actual server to send the DHCP offer to the client.
(G) DNS server (H) Source port (I) Lease time
(J) Premaster secret (K) Gateway router (L) IP address
Solution: Replacing the DNS server allows the attacker to redirect address lookups to a machine of the attacker’s choosing. Replacing the gateway router allows the attacker to intercept all of the host’s off-subnet traffic.
Q7.3 (3 points) Alice clears all her network settings and starts a new connection to www.cutecats.com with TCP. Now an off-path attacker wants to send a packet to the server to interfere with Alice’s connection. What information do they need to know?
(A) Server sequence number (B) Source port
(C) Client sequence number
(D) Destination IP address (E) Destination port
(F) Source IP address
Solution: An off-path attacker needs the IP addresses, ports, and sequence numbers to inject a packet. However, the server sequence number is not necessary because the server won’t reject a packet with an incorrect ACK number.
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Q7.4 (3 points) At some point, Alice’s connection with www.cutecats.com is suddenly terminated. Assuming some information was leaked and the attacker correctly guessed the fields from the previous part, how was the attacker able to execute this attack?
(G) (H) (I) ( J) (K) (L)
This is the end of Q7. Proceed to Q8 on your Gradescope answer sheet.
Solution: TheattackerinjectedaRSTpacketwiththecorrectlyguessedfields.Thisterminates the TCP connection.
Midterm 2 Page 19 of 26 CS 161 – Spring 2020

Q8 (18 points)
Q8.1 (4 points) Write a stateful firewall rule that would allow all TLS traffic from an external host 161.20.2.0 into your network 16.120.20.0/24.
(A) (B) (C) (D) (E) (F)
Q8.2 (4 points) Recall that an attacker can spoof source IPs to hide themselves while executing a DoS attack. Assume the attacker securely randomly generates these IPv4 addresses. Describe a pattern in the packets that a network operator could observe to best discern whether or not their network is a victim of a DoS attack.
(G) (H) (I) ( J) (K) (L)
Solution: allow tcp 161.20.2.0:* -> 16.120.20.0/24:*
Common mistakes were not including the ports, including an incorrect port, forgetting to include the CIDR notation for 16.120.20.0/24, specifying TLS as the protocol when a firewall would not have application layer context, etc.
Solution: Look at the distribution of the source IP addresses of the incoming packets. If they are roughly uniformly distributed across the IP address space, this is likely to be the result of a DoS attack (see backscatter analysis).
Another viable option is to see that some source IP addresses are routed to private or non- routable IP addresses. Other accepted solutions mentioned the logic for maximum or minium sized packets.
Q8.3 (6 points) What intrusion detection method would be best fit to perform the previous analysis? Justify your answer.
(A) HIDS (B) NIDS
(C) Logging (E) (D) (F)
Solution: NIDS allows for real-time analysis, and by looking at the IP address source fields on the IP packets, there is no need for any visibility or context from the host. A NIDS is cheap to deploy.
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Q8.4 (4 points) Describe a major drawback or exploit to the intrusion detection method you described above.
(G) (H) (I) ( J) (K) (L)
Solution: The NIDS could itself be overwhelmed by the volume of traffic. Also, if the bottle- neck network link is upstream, the DoS attack might overwhelm that bottleneck link, causing many packets to be dropped before they reach the NIDS, making it harder for the NIDS to have full visibility of the attack.
Also accepted due to question ambiguity: a drawback of the intrusion detection method that is irrelevant in the context of DoS detection (e.g., traffic being encrypted).
This is the end of Q8. Proceed to Q9 on your Gradescope answer sheet.
Midterm 2 Page 21 of 26 CS 161 – Spring 2020

Q9 (12 points)
EvanBot has decided to switch career paths and pursue creating new cryptographic hash functions. EvanBot proposes two new hash functions, 𝐸 and 𝐵:
𝐸(𝑥) = 𝐻(𝑥1𝑥2 …𝑥𝑀−1) 𝐵(𝑥) = 𝐻(𝑥1𝑥2 …𝑥𝑀||0)
where 𝐻 is a preimage-resistant and collision-resistant hash function, 𝑥 = 𝑥1𝑥2 … 𝑥𝑀 , 𝑥𝑖 ∈ {0, 1} and || denotes concatenation.
In other words, 𝐸(𝑥) calls 𝐻 with the last bit of 𝑥 removed, and 𝐵(𝑥) calls 𝐻 with a 0 bit appended to 𝑥.
Q9.1
Q9.2
(3 points) Is 𝐸(𝑥) preimage-resistant? Provide a counter-example if it is not. (A) Yes (C) (E)
(B) No (D) (F) Counterexample:
(3 points) Is 𝐸(𝑥) collision-resistant? Provide a counter-example if it is not. (G) Yes (I) (K)
(H) No (J) (L) Counterexample:
Solution: 𝐸(𝑥) is preimage-resistant. Suppose not, i.e., given 𝐸(𝑥) we could find an 𝑥′ such
that 𝐸(𝑥) = 𝐸(𝑥′). We will argue this means that 𝐻 is not preimage-resistant, either. Suppose
we are given 𝐻(𝑦). Let 𝑥 = 𝑦0, so that 𝐸(𝑥) = 𝐻(𝑦). By assumption, we can find 𝑥′ such that
𝐸(𝑥) = 𝐸(𝑥′). Let 𝑦′ = 𝑥′ ⋯𝑥′ . Then it follows that 𝐻(𝑦) = 𝐸(𝑥) = 𝐸(𝑥′) = 𝐻(𝑦′), so given 1 𝑀−1
𝐻(𝑦) we can find 𝑦′ such that 𝐻(𝑦) = 𝐻(𝑦′). This implies that 𝐻 is not preimage resistant. That is a contradiction, so our assumption that 𝐸 was not preimage-resistant must have been wrong.
𝐸(𝑥) is not collision-resistant. Counter example: 𝐸(1 ⋯ 010) = 𝐸(1 ⋯ 011),
Q9.3
(3 points) Is 𝐵(𝑥) preimage-resistant? Provide a counter-example if it is not.
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(A) Yes (B) No
Counterexample:
(C) (E) (D) (F)

Q9.4 (3 points) Is 𝐵(𝑥) collision-resistant? Provide a counter-example if it is not. (G) Yes (I) (K)
(H) No (J) (L) Counterexample:
Solution:
𝐵(𝑥) is preimage resistant, using the same reasoning as 𝐸(𝑥). (If there is an attack 𝐵’s preimage- resistance, then we can construct an attack against 𝐻 ’s preimage-resistance that succeeds half as often, which is often enough to show that 𝐻 is not preimage-resistant — but we were promised that 𝐻 is preimage-resistant, so it follows that 𝐵 must be preimage-resistant, too.)
𝐵(𝑥) is collision-resistant. If 𝐵(𝑥) was not collision resistant, then we can find 𝑥 ≠ 𝑦 such that 𝐵(𝑥) = 𝐵(𝑦). This can be rewritten as 𝐻(𝑥||0) = 𝐻(𝑦||0). Letting 𝑥′ = 𝑥′||0 and 𝑦′ = 𝑦′||0, this means we found 𝑥′ ≠ 𝑦′ such that 𝐻(𝑥′) = 𝐻(𝑦′), which proves that 𝐻(⋅) is not collision- resistant, which is a contradiction. Thus 𝐵(𝑥) must be collision-resistant.
This is the end of Q9. Proceed to Q10 on your Gradescope answer sheet.
Midterm 2 Page 23 of 26 CS 161 – Spring 2020

Q10 (12 points)
Alice comes up with a couple of schemes to securely send messages to Bob. Assume that Bob and Alice have known RSA public keys.
For this question, 𝐸𝑛𝑐 denotes AES-CBC encryption, 𝐻 denotes a collision-resistant hash function, || denotes concatenation, and ⨁ denotes bitwise XOR.
Consider each scheme below independently and select whether each one guarantees confidentiality, integrity, and authenticity in the face of a MITM.
Q10.1
(3 points) Alice and Bob share two symmetric keys 𝑘1 and 𝑘2. Alice sends over the pair [𝐸𝑛𝑐(𝑘1, 𝐸𝑛𝑐(𝑘2, 𝑚)), 𝐸𝑛𝑐(𝑘2, 𝑚)].
(A) Confidentiality (C) Authenticity (E) (B) Integrity (D) (F)
(3 points) Alice and Bob share a symmetric key 𝑘, have agreed on a PRNG, and implement a stream cipher as follows: they use the key 𝑘 to seed the PRNG and use the PRNG to generate message-length codes as a one-time pad every time they send/receive a message. Alice sends the pair [𝑚⨁code,𝐻𝑀𝐴𝐶(𝑘,𝑚⨁code)].
(G) Confidentiality (I) Authenticity (K) (H) Integrity (J) (L)
(3 points) Alice and Bob share a symmetric key 𝑘. Alice sends over the pair [𝐸𝑛𝑐(𝑘, 𝑚), 𝐻 (𝐸𝑛𝑐(𝑘, 𝑚))].
Q10.2
Solution: Note that 𝐸𝑛𝑐 denotes AES-CBC, not AES-EMAC, so we can only provide confi- dentiality. An attacker can forge a pair [𝐸𝑛𝑐(𝑘1, 𝑐1), 𝑐1] given [𝐸𝑛𝑐(𝑘1, 𝑐1||𝑐2), 𝑐1||𝑐2].
Solution: This stream cipher scheme has confidentiality since the attacker has no way of coming up with the pseudorandomly generated one-time pads. 𝐻𝑀𝐴𝐶 provides the integrity and authentication.
Q10.3
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(A) Confidentiality (B) Integrity
(C) Authenticity (E) (D) (F)
Solution: Public hash functions alone do not provide integrity or authentication. Anyone can forge a pair 𝑐, 𝐻 (𝑐), which will pass the integrity check and can be decrypted.

Q10.4 (3 points) Alice and Bob share a symmetric key 𝑘. Alice sends over the pair [𝐸𝑛𝑐(𝑘, 𝑚), 𝐻 (𝑘||𝐸𝑛𝑐(𝑘, 𝑚))].
(G) Confidentiality (I) Authenticity (K) (H) Integrity (J) (L)
This is the end of Q10. You have reached the end of the exam.
Solution: 𝐻(𝑘||𝐸𝑛𝑐(𝑘,𝑚)) is not a valid substitute for 𝐻𝑀𝐴𝐶 because it is vulnerable to a length extension attack.
Midterm 2 Page 25 of 26 CS 161 – Spring 2020

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