126 lines
3.7 KiB
Markdown
126 lines
3.7 KiB
Markdown
# Chapter 5
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## Linear Probing
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When a collision occurs, we go through the hash table and find the next available slot.
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## Quadratic Probing
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Going through a hash table and finding the next available slot by adding a quadratic value to the current index.
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$(H + c1*i + c2*i^2) \mod \textit{(table size)}$
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Where $H$ is the hash value, $c1$ and $c2$ are constants, and $i$ is the number of times we've probed.
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For instance:
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| Index | Value |
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|-------|-------|
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| 0 | 20 |
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| 1 | 41 |
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| 2 | null |
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| 3 | null |
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| 4 | null |
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- $c1 = 2$
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- $c2 = 4$
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So running `insert(40)` would result in:
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- $H = 40 \bmod 5 = 0$
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- $i = 1$
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- $(0 + (2 * 1) + (4 * 1^2)) \bmod 5$ -> $(0 + 2 + 4) \bmod 5 = 1$
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- $i = 2$ - retrying because there's already a value at 1
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- $(0 + (2 * 2) + (4 * 2^2)) \bmod 5$ -> $(0 + 4 + 16) \bmod 5 = 0$
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- $i = 3$ - retrying because there's already a value at 0
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- $(0 + (2 * 3) + (4 * 3^2)) \bmod 5$ -> $(0 + 6 + 36) \bmod 5 = 2$
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- There is no value at 2, so we can insert 40 there.
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| Index | Value |
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|-------|-------|
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| 0 | 20 |
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| 1 | 41 |
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| 2 | 40 |
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| 3 | null |
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| 4 | null |
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And a script to do it automatically:
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```python
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import sys
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size = int(sys.argv[-5]) # size of the table
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num = int(sys.argv[-4]) # number to hash
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c1 = int(sys.argv[-3]) # c1
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c2 = int(sys.argv[-2]) # c2
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i = int(sys.argv[-1]) # number of times we've probed
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h = num % size # hash value
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print((h + (c1 * i) + (c2 * i * i)) % size)
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```
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## Double Hashing
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Another collision resolution thing, where we use a second hash function to find the next available slot. The formula is:
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$(h1(key) + i * h2(key)) \bmod (table size)$
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Where $h1$ is the first hash function, $h2$ is the second hash function, and $i$ is the number of times we've probed.
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For instance:
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| Index | Value |
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|-------|-------|
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| 0 | 20 |
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| 1 | 41 |
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| 2 | null |
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| 3 | null |
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| 4 | null |
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Assuming:
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- $h1() = h \bmod 5$
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- $h2() = (i * h \bmod 3)$ - idk if there's any logic to this, but it's just an example
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So running `insert(40)` would result in:
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- $h1(40) = 40 \bmod 5 = 0$ - using the second function because there's already a value at 0
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- $h2(40) = (1 * 40) \bmod 3 = 1$;  $i = 0 + (1 * 1) = 1$ - there's already a value at 1
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- $h2(40) = (2 * 40) \bmod 3 = 2$;  $i = 0 + (2 * 2) = 4$ - there's no value at 4, so we can insert 40 there.
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## Common Hash Functions
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### Mid-Square hashing
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1. Square the key
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2. Take the middle $R$ digits - $R$ must be greater than or equal to $log_{10}(\text{table size})$
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3. Use that as the hash value - if it's above the number of slots, we can use the modulo of the table size. e.g., if the table size is 100, and the hash value is 123, we can use 23.
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#### Binary
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Usually mid-square hashing is done using binary, since that's what computers work with and is faster. It works the same, but using base 2:
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1. Square the key
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2. Take the middle $R$ bits - $R$ must be greater than or equal to $log_2(\text{table size})$
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3. Use that as the hash value - if it's above the number of slots, we can use the modulo of the table size.
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### Multiplicative string hashing
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Multiplicative string hashing in Python-like pseudocode:
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```py
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multiplicativeStringHash(string key) {
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string_hash = initial_value # using Bernstein's hash, initial_value is 5381
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hash_multiplier = 33 # using Bernstein's hash, hash_multiplier is 33
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for character in key {
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string_hash *= hash_multiplier
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strChar += ascii(character) # returns the ASCII number for the character
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}
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return string_hash % table_size
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}
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```
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Bernstein's hash uses an initial value of 5381 and a hash multiplier of 33, and works well for hashing short English strings.
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