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The hash function: h(x) = |3x + 1| mod M a bucket array of capacity N a set of o

ID: 3824243 • Letter: T

Question

The hash function: h(x) = |3x + 1| mod M a bucket array of capacity N a set of objects with keys: 12, 44, 13, 88, 23, 94, 11, 39, 20, 16, 5 (to input from left to right) Write the hash table where M = N = 11 and collisions are handled using separate chaining. Write the hash table where M = N = 11 and collisions are handled using linear probing. Would a size N for the bucket array exist, such that no collisions would happen with the hash function hash function h(x) = |2x + 5| mod 11 and the keys above?

Explanation / Answer

A

M=N=11

Using separate chaining

Element 12

( 12*3+1 ) mod 11=4

Element 44

44*3+1 = 133 mod 11=1

Element 13.

(13 *3 + 1) mod 11=7

Element 88.

(88 *3 + 1) mod 11=1

Element 23.

(23 *3 + 1) mod 11=4

Element 94.

(94 *3 + 1) mod 11=8

Element 11.

(11 *3 + 1) mod 11=1

Element 39.

(39 *3 + 1) mod 11=8

Element 20.

(20 *3 + 1) mod 11=6

Element 16.

(16 *3 + 1) mod 11=5

Element 5

(5 *3 + 1) mod 11=5

a:

Separate chaining

0

1

2

3

4

5

6

7

8

9

10

44

88

11

12

23

16

5

20

13

39

94

B

M=N=11

Using Linear probing

Element 12

( 12*3+1 ) mod 11=4

Element 44

44*3+1 = 133 mod 11=1

Element 13.

(13 *3 + 1) mod 11=7

Element 88.

(88 *3 + 1) mod 11=1

Element 23.

(23 *3 + 1) mod 11=4

Element 94.

(94 *3 + 1) mod 11=8

Element 11.

(11 *3 + 1) mod 11=1

Element 39.

(39 *3 + 1) mod 11=8

Element 20.

(20 *3 + 1) mod 11=6

Element 16.

(16 *3 + 1) mod 11=5

Element 5

(5 *3 + 1) mod 11=5

a:

Linear probing

0

1

2

3

4

5

6

7

8

9

10

6

44

88

11

12

23

20

13

94

39

16

C

Element 12

( 12*2+1 ) mod 11=3

Element 44

44*2+1 = 133 mod 11=1

Element 13.

(13 *2 + 1) mod 11=5

Element 88.

(88 *2 + 1) mod 11=1

Element 23.

(23 *2 + 1) mod 11=3

Element 94.

(94 *2 + 1) mod 11=8

Element 11.

(11 *2 + 1) mod 11=1

Element 39.

(39 *2 + 1) mod 11=2

Element 20.

(20*2 + 1) mod 11=8

Element 16.

(16 *2 + 1) mod 11=0

Element 5

(5 *2 + 1) mod 11=0

So cleary no matter what would be size of N collision will occur as value of |2x+1| mod 11 are same for few elements.

0

1

2

3

4

5

6

7

8

9

10

44

88

11

12

23

16

5

20

13

39

94