1. Open Addressing: Quadratic Probing
ECE Data Structures and Algorithms
Open Addressing: Quadratic Probing
Douglas Wilhelm Harder, M.Math. LEL
Department of Electrical and Computer Engineering
University of Waterloo
Copyright © 2006 by Douglas Wilhelm Harder. All rights reserved.

2. Outline Problems with linear problem and primary clustering
Outline of quadratic probing
insertions, searching
restrictions
deletions
weaknesses

3. Quadratic Probing
Primary clustering occurs with linear probing because the same linear pattern:
if a bin is inside a cluster, then the next bin must either:
also be in that cluster, or
expand the cluster
Instead of searching forward in a linear fashion, consider searching forward using a quadratic function

4. Quadratic Probing Suppose that an element should appear in bin h:
if bin h is occupied, then check the following sequence of bins:
h + 12, h + 22, h + 32, h + 42, h + 52, ...
h + 1, h + 4, h + 9, h + 16, h + 25, ...
For example, with M = 17:

5. Quadratic Probing
If one of h + i2 falls into a cluster, this does not imply the next one will

6. Quadratic Probing
For example, suppose an element was to be inserted in bin 23 in a hash table with 31 bins
The sequence in which the bins would be checked is:
23, 24, 27, 1, 8, 17, 28, 10, 25, 11, 30, 20, 12, 6, 2, 0

7. Quadratic Probing
Even if two bins are initially close, the sequence in which subsequent bins are checked varies greatly
Again, with M = 31 bins, compare the first 16 bins which are checked starting with 22 and 23:
22 22, 23, 26, 0, 7, 16, 27, 9, 24, 10, 29, 19, 11, 5, 1, 30
23 23, 24, 27, 1, 8, 17, 28, 10, 25, 11, 30, 20, 12, 6, 2, 0

8. Quadratic Probing
Thus, quadratic probing solves the problem of primary clustering
Unfortunately, there is a second problem which must be dealt with
Suppose we have M = 8 bins:
12 ≡ 1, 22 ≡ 4, 32 ≡ 1
In this case, we are checking bin h + 1 twice having checked only one other bin

9. Quadratic Probing Unfortunately, there is no guarantee that
h + i2 mod M
will cycle through 0, 1, ..., M – 1
Solution:
require that M be prime
in this case, h + i2 mod M for i = 0, ..., (M – 1)/2 will cycle through exactly (M + 1)/2 values before repeating

10. Quadratic Probing Example with M = 11: With M = 13: With M = 17:
0, 1, 4, 9, 16 ≡ 5, 25 ≡ 3, 36 ≡ 3
With M = 13:
0, 1, 4, 9, 16 ≡ 3, 25 ≡ 12, 36 ≡ 10, 49 ≡ 10
With M = 17:
0, 1, 4, 9, 16, 25 ≡ 8, 36 ≡ 2, 49 ≡ 15, 64 ≡ 13, 81 ≡ 13

11. Quadratic Probing Thus, quadratic probing avoids primary clustering
Unfortunately, we are not guaranteed that we will use all the bins
In reality, if the hash function is reasonable, this is not a significant problem until l approaches 1

12. Quadratic Probing
For example, with a hash table with M = 19 using quadratic probing, insert the following random 3-digit numbers:
086, 198, 466, 709, 973, 981, 374,
766, 473, 342, 191, 393, 300, 011,
538, 913, 220, 844, 565
using the number modulo 19 to be the initial bin

13. Quadratic Probing The first two fall into their correct bin:
086 → 10, 198 → 8
The next already causes a collision:
466 → 10 → 11
The next four cause no collisons:
709 → 6, 973 → 4, 981 → 12, 374 → 13
Then another collision:
766 → 6 → 7

14. Quadratic Probing
At this point, two clusters have appeared and the load factor is l = 0.42

15. Quadratic Probing The next three also go into their appropriat bin:
473 → 17, 342 → 0, 191 → 1
Then there is one more collision
393 → 13 → 14
and 300 falls into its correct bin:
300 → 15

16. Quadratic Probing
With five more insertions, the load factor is is l = 0.68 with one large cluster:

17. Quadratic Probing At this point, insertions become more tedious:
011 → 11 → 12 → 15 → 1 → 8 → 17 → 9
538 → 6 → 7 → 10 → 15 → 3
913 → 1 → 2
220 → 11 → ⋅⋅⋅ → 9 → 3 → 18
844 → 8 → 9 → 12 → 17 → 5

18. Quadratic Probing
To show how quadratic probing works, consider the addition of 583, starting in bin 6:
The first four bins all fall within the same cluster, however, the fifth bin checked falls far outside the cluster

19. Quadratic Probing
At this point, the array is almost full (bin 16 is open) and the load factor is l = 0.95
If we try to add the last number, 565 the sequence of bins checked is
14 → 15 → 18 → 4 → 11 → 1 → 12 → 6 → 2 → 0
which does not hit bin 16

20. Quadratic versus Linear Probing
We can compare the number of probes required with that of linear probing:
086 → 10, → 8
466 → 10 → → 6
973 → → 12
374 → → 6 → 7
473 → → 0
191 → → 13 → 14
300 → → 11 → 12 → 13 → 14 → 15 → 16
538 → 6 → 7 → 8 → → 1 → 2
220 → 11 → 12 → 13 → 14 → 15 → 16 → 17 → 18
844 → 8 → 9 → 10 → 11 → 12 → 13 → 14 → 15 → 16 → 17 → 18 → 0 → 1 → 2 → 3
565 → 14 → 15 → 16 → 17 → 18 → 0 → 1 → 2 → 3 → 4 → 5

21. Quadratic Probing: Deletions
We have seen how we can perform insertions – next is deletions
With linear probing, if we deleted the contents of a bin, we had to search ahead to determine if any nodes had to be moved back
easy with linear probing; we simply moved from bin to bin until an empty bin was located

22. Quadratic Probing: Deletions
The nonlinear probing associated with quadratic probing does not allow us to do this efficiently
For example, suppose we delete 466 which is currently in bin 11:
The two other entries which pass through bin 11 were 011 and 220
We cannot (efficiently) find these entries

23. Quadratic Probing: Deletions
Solution:
associate with each bin a field which is either EMPTY, OCCUPIED, or DELETED

24. Quadratic Probing: Deletions
Initially, all bins are initially marked EMPTY
When a bin is filled, it is marked OCCUPIED
If a bin is emptied (as a result of a remove), it is marked DELETED
Note that a bin which is marked as being DELETED may once again be filled (and hence marked OCCUPIED)

25. Quadratic Probing: Deletions
For example, given a hash table with M = 11 bins, enter the values
Bin 1 2 3 4 5 6 7 8 9 10
Entry
Flag E

26. Quadratic Probing: Deletions
The first three are straight-forward:
135 → → → 4
Bin 1 2 3 4 5 6 7 8 9 10
Entry
135
246
909
Flag E O

27. Secondary Clustering
The phenomenon of primary clustering will not occur with quadratic probing
However, if multiple items all hash to the same initial bin, the same sequence of numbers will be followed
This is termed secondary clustering
The effect is less significant than that of primary clustering

28. Secondary Clustering
Secondary clustering may be a problem if the hash function does not produce an even distribution of entries
One solution to secondary is double hashing: associating with each element an initial bin (defined by one hash function) and a skip (defined by a second hash function)

29. Example Insert the 6 elements 3, 107, 9, 119, 35, 112
into an initially empty hash table of size 11 using quadratic hashing
Let the hash function be the number modulo 11

30. Example: Insertion 14, 107, 31, 119, 35, 112
The first three fall into bins 3, 8, and 9, respectively 1 2 3 4 5 6 7 8 9 10 14
107 31

31. Example: Insertion 14, 107, 31, 118, 35, 112
118 also falls into bin 8 (occupied)
Thus, we check:
8 + 1 = 9 - occupied
8 + 4 = 1 - unoccupied 1 2 3 4 5 6 7 8 9 10
118 14
107 31

32. Example: Insertion 14, 107, 31, 118, 35, 112
34 falls into bin 1 which is occupied, thus we check:
1 + 1 = 2 - unoccupied 1 2 3 4 5 6 7 8 9 10
118 34 14
107 31

33. Example: Insertion 14, 107, 31, 118, 35, 112
112 falls into bin 2 which is now occupied, thus we check:
2 + 1 = 3 - occupied
2 + 4 = 6 - unoccupied 1 2 3 4 5 6 7 8 9 10
118 34 14
112
107 31

34. Example: Insertion At this point, the hash table is over half full
We are no longer guaranteed that the insertion of a new element may be possible
Solution: increase the size of the table (perhaps only after failing)
Problem: the new size must, too, be prime 1 2 3 4 5 6 7 8 9 10
118 34 14
112
107 31

35. Example: Removal
To remove an element, we must simply mark it as deleted
In our example, removing 118, we begin in bin 8, and continue to check 9, and then 1
Mark that bin as having had an element deleted: 1 2 3 4 5 6 7 8 9 10
DEL 34 14
112
107 31

36. Example: Finding
To find an element we start by checking the bin it should have initially been in, and then begin checking following quadratic probing until either:
we find it, or
we find a bin which is neither occupied or deleted 1 2 3 4 5 6 7 8 9 10
DEL 34 14
112
107 31

37. Example: Finding We find 14 in bin 3
We don’t find 34 in bin 1 (marked as deleted), so we check bin = 2, and find it 1 2 3 4 5 6 7 8 9 10
DEL 34 14
112
107 31

38. Example: Finding We search for 19 in bin 8 Not finding it, we check:
= occupied
= deleted
= occupied
= occupied
= unoccupied: not found 1 2 3 4 5 6 7 8 9 10
DEL 34 14
112
107 31

39. Usage Notes
These slides are made publicly available on the web for anyone to use
If you choose to use them, or a part thereof, for a course at another institution, I ask only three things:
that you inform me that you are using the slides,
that you acknowledge my work, and
that you alert me of any mistakes which I made or changes which you make, and allow me the option of incorporating such changes (with an acknowledgment) in my set of slides
Sincerely,
Douglas Wilhelm Harder, MMath