#$3.$ Ex$.6\mathrm{.}6\mathrm{.}6$ Show the result of Exercise $6\mathrm{.}6\mathrm{.}4($see below$)$ assuming collisions are handled by quadratic

probing, up to the point where the method fails because no empty slot is found.

$($a$)$ Draw the $11-$item hash table resulting from hashing the keys $12,44,13,88,23,94,11,39,20,16,$ and $5,$ using the hash function

$h(i)=(2i+5)mod11$ and assuming collisions are handled by chaining.

Hash function: $h(i)=(2(i)+5)$ mod $11$

Key:

Hash value:

Slot:

$12,h(12)=(2(12)+5)mod11=$

$44,h(44)=(2(44)+5)mod11=$

$13,h(13)=(2(13)+5)mod11=$

$88,h(88)=(2(88)+5)mod11=$

$23,h(23)=(2(23)+5)mod11=$

$94,h(94)=(2(94)+5)mod11=$

$11,h(11)=(2(11)+5)mod11=$

$39,h(39)=(2(39)+5)mod11=$

$20,h(20)=(2(20)+5)mod11=$

$16,h(16)=(2(16)+5)mod11=$

$5,h(5)=(2(5)+5)mod11=$

#$3$ Part $2$

What is the result of Exercise $6\mathrm{.}6\mathrm{.}4($see below$)$ assuming collisions are handled by double hashing

using a secondary hash function $h^{\text{'}}(k)=7-(kmod7)?$

$($a$)$ Draw the $11-$item hash table resulting from hashing the keys $12,44,13,88,23,94,11,39,20,16,$ and $5,$ using the hash function

$h(i)=(2i+5)mod11$ and assuming collisions are handled by chaining.