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UCSD CSE 127 - Homework #2

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CSE 127: Computer Security Spring 2011Homework #2Due: Thursday, June 2nd, 2011, 11:00 am.Note: You may take no more than five late days on this homework.Problem 1 The Internet is, slowly, transitioning from the version of the TCP/IP protocolsuite currently in use — IPv4 — to a new version, IPv6.Unlike IPv4 IP addresses, which are 32 bits long (e.g., 192.168.10.1), IPv6 IP addressesare 128 bits long (e.g., 2001:1890:1112:0001:0000:0000:0000:0020).(a) Consider random-scanning Internet worms. These worms spread by choosinga random IP address, connecting to any host answering to that address, andattempting to infect it.Is the random-scanning strategy feasible if the Internet switches from IPv4 toIPv6? Why or why not?Hint : Consider the density of Internet-connected machines in the IP addressspace.(b) On the IPv6 Internet, what are some specific ways that a worm, executing on acompromised computer, can discover IP addresses of other hosts to try to infect?Hint : Consider sources of IP address on the infected computer itself and on thelocal-area network to which it is connected.(c) Suppose the worm targets Web servers running some application (say, bulletinboard software written in PHP). Can Google searches help the worm find potentialtargets? How?Problem 2 The most commonly used keyed lock is a pin tumbler lock, shown in Figure 1.This type of lock features k pin stacks (typically 4 to 7, with the lock in the figurehaving 6), each cut at one of n possible depths (typically 4 to 10), which are held inthe plug by springs.A key has bitings at a certain depth for each pin stack. When the key is inserted intothe keyway, its bitings push each pin stack up by a certain amounts. The correct keywill cause each to be pushed so that its cut aligns with the shear line that separatesthe plug from the shell, allowing the plug to rotate, as shown in Figure 2.In many institutional settings a master-keying system is used, where each of severallocks (say, one for each room in a dorm) has a different change key, but all share asingle master key. This is most commonly implemented by including two cuts in eachFigure 1: On the left, the lock face. On the right, a cutaway view showing six pin stacksand (horizontal) shear line. The pin stacks keep the plug from rotating within the shell andopening the door.Figure 2: On the left, a cutaway view showing the correct key inserted in the lock. Note thatthe cuts in the pin stacks are aligned with the shear line. This allows the plug to rotate, asshown on the right.Figure 3: On the left, a cutaway view of a master-keyed lock. Note that each pin stack hastwo cuts. On the right, the lock with the change (non-master) key inserted. One set of cutsis at the shear line; the other cut for each stack is either above or below.pin stack, one for the change key, the other for the master key, as shown in Figure 3.The locks will each have different change cuts, but all share the same master cuts, sothe same master key will open all of them. The change key and the master key nevershare a cut: in each pin stack there is one cut for the change key, and another at adifferent depth (either above or below) for the master key.a. Suppose that a lock is keyed with the change key at 11111 (five pins, all at depth1), and the master key at 44444 (five pins again, all at depth 4 this time). Will akey cut at 11411 open the lock? Explain.b. Suppose you have access to a master-keyed lock, a change key that opens it, butnot the master key. In addition, you have a supply of blanks that can be fileddown to specific bitings of your choice. Explain how to use the observation frompart a. to recover the master key.As a function of k and n, how many blanks will you need? How many tries in thelock?c. Suppose you have access to the change keys for many locks all keyed for the samemaster key (but, again, no access to the master key). For example, suppose thatyou measure the biting on the keys given to all the residents in a dorm. Explainhow you can (with high probability) recover the master key without any blanks —and even without access to any of the


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UCSD CSE 127 - Homework #2

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