Attacks on hash functions. Birthday attacks and Multicollisions
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1 Attacks on hash functions Birthday attacks and Multicollisions
2 Birthday Attack Basics In a group of 23 people, the probability that there are at least two persons on the same day in the same month is greater than ½. Proof: The probability that none of the 23 people has the same birthday is: Thus, > 1/2.
3 Strong Collision Resistance Complexity Upper Bound Complexity upper bound of breaking strong collision resistance Let H be a cryptographic hash function with output length N. Then H will only have at most 2 N different outputs. Q: Is 2 N the complexity upper bound of breaking strong collision resistance? A: No. We can use birthday attack to reduce the complexity to 2 N/2 with a success rate of over 50%. Birthday attack for collisions for hash functions: From a basket of n balls of different colors, pick k (k<n) balls uniformly and independently at random and record their colors. The probability that least one ball that is picked more than once is The probability at least ½ if. Complexity upper bound of SHA-1: For the probability is at least ½ that there is a collision. In the case of SHA-1 k is approximately 2 160/2 = 2 80
4 Yuval s birthday attack Let H be n-bit hash function. Input: Legitimate message m 1 and forgery message m 2. Output: m 1, m 2 resulting from minor modifications of m 1 and m 2 with H(m 1 )=H(m 2 ). 1. Generate t = 2 n/2 minor modifications m 1 of m 1 2. Hash each such message and store the hash values. 3. Generate minor modifications m 2 of m 2, computing H(m 2 ) for each and checking for matches with any m 1 from the previous step; continue until match is found. A match can be expected after about t candidates of m 2
5 Multicollisions Basic facts about hash functions An iterated hash function H is built by iterating a basic compression function f. Figure 1. Compression function f in SHA-1
6 Basic facts about hash functions The hashing process works as follows: Pad the original message M and split it into blocks m 1,..., m t. Set H 0 to the initial value IV. For i from 1 to t, let H i = f(h i 1, m i ). Output H(M) = H t. Note: The initial value H 0 in SHA-1 is A B C D E.
7 Multicollisions A k-collision is a r-tuple of messages M (1),, M (k) such that H(M (1) ) = = H(M (k) ). Finding an k-collision could be done by hashing about 2 n (k 1)/k messages.
8 Constructing Multicollisions Assume that the size of the message blocks m i is bigger than the size of the hash values. Suppose that the output of the function H, has n bits. Use a birthday attack to find blocks m 0 and m 0 such that f(iv, m 0 )= f(iv, m 0 ). In approximately 2 n/2 steps one can find such m 0 and m 0.
9 Let h 1 = f(iv, m 0 ). Constructing Multicollisions (cont.) Use second birthday attack to find m 1 and m 1 with f(h 1, m 1 ) = f(h 1, m 1 ). In approximately another 2 n/2 steps one can find such m 1 and m 1. Let h i = f(h i-1,m i-1 ). Use a birthday attack to find m i and m i with f(h i, m i ) = f(h i, m i ). In approximately another 2 n/2 steps one can find such m i and m i. This process is continued until we have blocks m 0,m 0,m 1,m 1,...,m t 1,m t 1 where t is some integer.
10 Constructing Multicollisions (cont.) Construct the following messages: There are 2 t such messages. m 0 m 1 m t-1 m 0 m 1 m t-1 m 0 m 1 m t-1 m 0 m 1 m t-1.. m 0 m 1 m t-1 m 0 m 1 m t-1 m 0 m 1 m t-1
11 Multicollisions (cont.) Each of the 2 t messages has the same hash value. Proof: At each calculation h i = f(h i-1,m i-1 ) the same value h i is obtained whether m = m i 1 or m = m i 1. Therefore, the output of the function f during each step of the hash algorithm is independent of whether an m i 1 or an m i 1 is used. Therefore, the final output of the hash algorithm is the same for all messages. Therefore, we have have 2 t -collision. Note: After t2 n/2 steps there is a 2 t - collision.
12 Concatenation and Multicollisions Suppose we have two hash functions H 1 and H 2. Concatenate H(M) = H 1 (M) H 2 (M) Question: Is H(M) stronger hash function than either H 1 or H 2 individually? Answer: No.
13 Concatenation and Multicollisions Suppose the output of H i has n i bits. Assume that H 1 is calculated by an iterative algorithm. In n 2 n 1 2 n 1 /2 steps, we can find 2 n 2 /2 messages that all have the same hash value for H 1. Compute the value of H 2 for each of these 2 n 2 /2 messages. By the birthday attack there will be a match among these values of H 2. They all have the same H 1 value, so there will be collision for H 1 H 2.
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