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Security in Computing

by: Jenifer Moen

Security in Computing CIS 3360

Jenifer Moen
University of Central Florida
GPA 3.73


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This 6 page Class Notes was uploaded by Jenifer Moen on Thursday October 22, 2015. The Class Notes belongs to CIS 3360 at University of Central Florida taught by Staff in Fall. Since its upload, it has received 33 views. For similar materials see /class/227531/cis-3360-university-of-central-florida in Computer Information Systems at University of Central Florida.

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Date Created: 10/22/15
Notes about mod Arup Guha First we ll de ne divisibility We say that a l b if and only if there is some integer c such that b ac In English a l b would be read as b is divisible by a For example 6 l 18 1971 0 and 341 34 Now let s de ne mod a E b mod n if and only if n l a E b This just means there exists some integer c such that a E b nc In essence this is true if n divides evenly into the difference of a and b Alternatively we can think of it as follows when a and b are divided by n they leave the same remainder In our class typically we will make some mathematical calculation and then we d like to know what letter a particular number corresponds to What we really want is give some integer a we want to nd a value b such that 0 S b lt 26 and a E b mod 26 For example if we get 194 after some calculation and want to know what letter it is our goal is to nd the unique value of b such that 194 E b mod 26 with 0 S b lt 26 We can determine that 194E 12 mod 26 We can verify this because 194 E 12 182 and 182 26x7 The easy way to nd b when the starting value is greater than 26 is to divide 26 into the number When we divide 26 into 194 it goes in 7 times leaving a remainder of 12 which is our desired value Consider a second example 85 E b mod 26 with 0 S b lt 26 By dividing we nd that 85 E 7 mod 26 since 85 E 7 78 and 78 26x3 but we also see that we haven t gotten the desired value of b either We can simply add 26 to 7 to do that since adding or subtracting multiples of 26 will create other values equivalent to the original Thus we have 85 2722677 19 mod 26 Now let s look at some rules with mod if aEbmodnthenachcmodn if a E b mod n then ac E bc mod n and ac E bc mod cn but this latter fact is rarely used if a E b mod n then ak E bk mod n if a E b mod n and c E d mod n then ac E bd mod n and ac E bd mod n These are fairly straightforward to apply However division rules are tricky since we are now dealing with integers If we have a situation such as 3a E 16 mod 26 we deal with it by multiplying through by the inverse of 3 mod 26 which is 9 to yield the following equation 93a E 916 mod 26 27a E 144 mod 26 a E 14 mod 26 Here is a list of the inverses mod 26 UILth I l 7 5 ll l9 17 23 25 9 1 Note 1 is an inverse of itself as is 25 The rest are pairs so 3 is the inverse of9 and 9 is the inverse of 3 mod 26 etc But what about an equation like 4a E 14 mod 26 or 4a E 7 mod 26 This literally means 4a 7 l4 26c for some int c 4a 7 7 26c for some int c 2a77l3cso 74a726c 2a E 7 mod 13 is all we can 7 22a E 13c which is impossible ascertain the following above since 7 is NOT divisible by 2 implies that a E 10 mod 13 which can be determined by multiplying through by 7 Ifwe nd that a E 10 mod 13 that means that a E 10 mod 26 or a E 23 mod 26 We can see this because if a E 10 l3c for some integer c then setting c 0 1 shows that a could be 10 or 23 Setting c 2 shows that a could be 36 but 36 is equivalent to 10 mod 26 This information is relevant in the following situations 1 Solving for the inverse of a matrix 2 Solving for a key in a known plaintext attack on the Hill cipher For the former if it is known that the matrix does have an inverse then there will be a unique solution that satis es all of the given equations To take an example from the 3 l a b l 0 notes chapter 4 when solving the equation 6 SI d 0 1Jmod 26 we found c that c E 8 mod 26 We could have used that and substituted into the equation 6a 5c E 0 mod 26 yielding 6a 58 E 0 mod 26 6a E 40 mod 26 6a E 12 mod 26 3a E 6 mod 13 a E 2 mod 13 which means a E 2 mod 26 or a E 15 mod 26 Which of these two is correct can only be ascertained by plugging into the other relevant equation 3a c E 1 mod 26 For 2 it may be the case that the equations formed don t provide a unique solution for the key This was illustrated in the notes for chapter 4 Here we can na1row the key down to a few options and from there we can simply try out all of the candidates Diffie Hellman Key Exchange Motivation for K e Exchange Algorithm The fastest encryption schemes are symmetric encryption schemes In order to use these the two communicators Alice amp Bob would have to meet in a secure location BUT this sort of defeats the purpose because their whole goal is to communicate when securely when they aren39t in the same place A modern day example of why it would be useful to exchange a key without meeting deals with an online purchase You are making the purchase online so that you DON39T have to go to the store to meet with the vendor So you don t really want to go to the store to just exchange a secret key either You want to be able to do that from the comfort of your own home Thus that is the underlying problem how do two users exchange a secret key without a secure communication channel The first solution to this problem was the DiffieHellman Key Exchange What39s interesting about this algorithm is that neither user actually gets to choose the key But at the end of the algorithm both users have calculated the same key which is not easy for an eavesdropper to calculate In order to understand why the DiffieHellman Key Exchange is difficult it is important to understand the Discrete Log Problem Discrete Log Problem The discrete exponentiation problem is as follows Given a base a an exponent b and a modulus p calculate c such that ab E c mod p and 0 S c lt p It turns out that this problem is fairly easy and can be calculated quotquicklyquot using fast exponentiation The discrete log problem is the inverse problem Given a base a a result c 0 S c lt p and a modulus p calculate the exponent b such that ab E c mod p It turns out that no one has found a quick way to solve this problem To get an intuition as to why this is the case try picking different values of a and p and listing out each successive power of a mod p What you will find is that there is no discemable pattern for the list of numbers created Thus given a number on the list it s very difficult to predict where it appears on the list Here s a concrete example Given a 2 b 7 and p 37 calculate c 27 128 and 128 E 17 mod 37 Given a 2 c l7 and p 37 calculate b Try each value of b until you nd one that works For large prime numbers this is too slow Diffze Hellman K 6 Exchange Alice and Bob agree on two values a large prime number p and a generator g l lt g lt p It39s better if g is an actual generator meaning that when you raise it to the 1st 2nd 3rd pl powers you get all different answers These values are known to everyone In secret Alice picks a value a with l lt a lt p In secret Bob picks a value b with l lt b lt p Alice calculates ga mod p call this fa and sends it to Bob Bob calculates gb mod p call this fb and sends it to Alice Note that fa and fb are also known by everyone In secret Alice computes fba mod p 7 this is the exchanged key In secret Bob computes fab mod p 7 this is again the exchanged key Why does this work fb E gb E g mod p Similarly fa E ga E ga mod p Here s a concrete example Letp37andg 13 Let Alice pick a 10 Alice calculates 1310 mod 37 which is 4 and sends that to Bob Let Bob pick b 7 Bob calculates 137 mod 27 which is 32 and sends that to Alice Note 6 and 7 are secret to Alice and Bob respectively but both 4 and 32 are known by all Alice receives 32 and calculates 3210 mod 37 which is 30 the secret key Bob receives 4 and calculates 47 mod 37 which is 30 the same secret key Note that neither Alice nor Bob chose 30 but that they ended up with that secret key anyway Furthermore note that even with knowing p 37 g l3 fa 4 and fb 32 it is difficult to ascertain the secret key 30 without doing a brute force check In particular if the discrete log problem were easy this scheme could be broken Consider the following If an adversary saw that fa 4 p 37 and g l3 and could solve the discrete log problem then they could calculate that 1310 E 4 mod 37 Once they had this value 10 then they could take fb 32 and then calculate 3210 mod 37 to arrive at 30 Thus the Dif eHellman Key Exchange is only as secure as the Discrete Log problem


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