Problem 4. Let H : M → {0,1}" be a collision resistant hash function (CRHF). Show that truncating the output of H even by one bit does not necessarily result in a CRHF. Hint: Use H to construct another CRHF H' : M → {0, 1}", but H"(x) := H'(x)[: –1] (in Python notation) is not a CRHF.

Consider a hash table of capacity 5 that uses open addressing with linear probing. This hash table uses a hash function that takes the remainder when the hash code of a string is divided by the capacity of the hash table. (For example, "air" hashes 0.) The hashCode method has been overridden for strings, which simply returns 0 if a string begins with 'a', 1 if it begins with 'b', etc. /** Returns 0 if a String begins with 'a', 1 if it begins with 'b', etc. */ public int hashCode() { return this.charAt(0) - 'a'; } Assume that "apple", and "butter" have been inserted in this order into the hash table. "fig" is to be inserted into the hash table. (a) Which array slot/index should "fig" be placed? Answer: (b) What is the load factor after "fig" is inserted into the hash table? Answer:

In §12.2 Lamport's Hash we mentioned the notion of using only 64 bits of the hash. At each stage, 128 bits are computed, 64 bits are thrown away, and the hash of the retained 64 bits is used in the next stage. The purpose of only using 64 bits is so that a human does not need to type as long a string. Assuming the person will still only type 64 bits, does it work if hashn does a hash of all 128 bits of hashn-1, but what the person actually transmits is 64 bits of the result?

Suppose we want to store a set S of n keys in a hash table T with size m, with m = n + 2, and a hash function that behaves uniformly. Should we resolve collisions with chaining or with open- addressing? Please explain why.

If we use a hash table, there must be a slot/bucket that has at least a keys assigned to it, where a is the load factor of the current hash table. True False A binary search tree of n nodes may have a depth/height as large as 2(n). False True If A[1...n] is sorted, we can find the k th smallest element in A[1..n] in O(1) time. True False