Q$6$ Your turn to Hash!

$5$ Points

Grading comment:

Assume you are hashing a set $($unknown length$)$ of randomly generated Strings, into a HashTable with a a size of $50($this means your HashTable has $50$ buckets$).$ You can also assume that Strings are iterable and Characters utilize the Java native hash function.

Examine the two hashCode$()$ implementations below. In general, which hashCode$()$ option will result in a greater number of collisions for all Strings? Justify your answer in $1-2$ sentences describing how each hashCode may generate a range of possible hashCode values.

Assume that when we call this in our hash functions, we are referring to each String object we are hashing.

Option $1$:

This code utilizes Java's implementation of hashCode$()$ for Characters which returns the unique int value associated with each character based on its assigned ASCII value. Ex: $\text{'}$a$\text{'}$ returns $97,\text{'}$A$\text{'}$ returns $65.$

public int hashCode$1()\{$

Iterator iterator $=$ this.iterator$()$;

int result $=0$;

int i $=0$;

while $($iterator$.$hasNext$())\{$

result $+=$ iterator.next$().$hashCode$()$;

i$++$;

$\}$

return result;

$\}$

Option $2$:

This code utilizes Java's implementation of hashCode$()$ for Strings which is the following $($value is an array of the characters within the array$)$:

$//$Java$\text{'}$s String hashCode implementation

public static int hashCode$($byte$[]$ value$)\{$

int h $=0$;

int length $=$ value.length;

for $($int i $=0$; i $<$ length; i$++)\{$

h $=31*$ h $+$ getChar$($value$,$ i$)$;

$\}$

return h;

$\}$

public int hashCode$2()\{$

return this.hashCode$().$

$\}$