Question: Suppose h 1 : {0,1} 2m {0,1} m is a collision resistant hash function. (1) Define h 2 : {0,1} 4m {0,1} m as follows:

Suppose h1: {0,1}2m {0,1}m is a collision resistant hash function.

(1) Define h2: {0,1}4m {0,1}m as follows:

a. Write x{0,1}4m as x=x1||x2, where x1, x2 {0,1}2m.

b. Define h2(x)=h1(h1(x1)||h1(x2)).

Prove that h2 is collision resistant.

(2) For an integer i 2, define a hash function hi: {0,1}Suppose h1: {0,1}2m {0,1}m is a collision resistant hash function. (1) Defineh2: {0,1}4m {0,1}m as follows: a. Write x{0,1}4m as x=x1||x2, where x1,(2^i)m {0,1}m recursively from hi-1, as follows:

a. Write x {0,1}x2 {0,1}2m. b. Define h2(x)=h1(h1(x1)||h1(x2)). Prove that h2 is collision resistant. (2)For an integer i 2, define a hash function hi: {0,1}(2^i)m {0,1}m(2^i)m as x=x1||x2, where x1, x2 {0,1}recursively from hi-1, as follows: a. Write x {0,1}(2^i)m as x=x1||x2, wherex1, x2 {0,1}(2^i)m b. Define hi(x)=h1(hi-1(x1)||hi-1(x2)). Prove that hi is collision resistant.(2^i)m

b. Define hi(x)=h1(hi-1(x1)||hi-1(x2)).

Prove that hi is collision resistant.

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