and

[knowclue]

I will have to do some research to give you a good example of the hash table portion, but assuming mod means the modulus operator then mode71 is not much different than mod 10. The modulo operator returns the remainder after a division so taking the first number for example, 4371 mod 10 would be 1 (4371/10 = 431 r. 1) and 4371 mod 7 would be 3 (4371/7 = 624 r 3)

[closed account Sz6592yv]

I figured it out. Here are the values that I got for both mod 10 and mod 7.

mod 10: 4371->1, 1323 ->3, 6173->3, 4199->9, 9679->9, 1989->9
mod 7: 3,0,6,6,5,1

[knowclue]

Looks good. As for drawing out the hash tables that will depend on which collision method you are using. Linear probing and chaining are the most common. I believe that chaining is the common method that is taught at universities (it was at mine). which looks something like this.

using mod 10 on 4320, 4350, and 4351

0[]-->[4320]-->[4350]
1[]-->[4351]
2[]
3[]
4[]
5[]
6[]
7[]
8[]
9[]

Essentially you used a pointer to the next value if they map to the same one.

Once you know which collision method you use, you can draw the table.

[closed account Sz6592yv]

Sorry the values for the h(x) = 7-(x mod 7) are : 4,7, 1,1 2, 6

[closed account Sz6592yv]

how does rehashing works???

[knowclue]

Try searching for it on google. Here are 2 results that show up:

https://courses.cs.washington.edu/courses/cse373/12sp/lectures/05-09-hashing3/18-hashing3.pdf

https://secweb.cs.odu.edu/~zeil/cs361/web/website/Lectures/rehashing/page/rehashing.html

[closed account Sz6592yv]

Here is what I have:

x mod 10:
0
1 ->[ 4371]
2
3 ->[1323] ->[6173]
4
5
6
7
8
9 ->[4199] ->[9679] ->[1989]

7-(x mod 7):
0
1->[4199] -> [6173]
2->[9679]
3
4->[4371]
5
6 ->[1989]
7->[1323]
8
9

Is this how I'm suppose do?
How do I apply the rehashing for 7-(x mod 7)?