Synonym finder algorithm - php

Question

I think example will be much better than loooong description :)
Let's assume we have an array of arrays:
("Server1", "Server_1", "Main Server", "192.168.0.3")
("Server_1", "VIP Server", "Main Server")
("Server_2", "192.168.0.4")
("192.168.0.3", "192.168.0.5")
("Server_2", "Backup")
Each line contains strings which are synonyms. And as a result of processing of this array I want to get this:
("Server1", "Server_1", "Main Server", "192.168.0.3", "VIP Server", "192.168.0.5")
("Server_2", "192.168.0.4", "Backup")
So I think I need a kind of recursive algorithm. Programming language actually doesn't matter — I need only a little help with idea in general. I'm going to use php or python.
Thank you!

Answer 1

This problem can be reduced to a problem in graph theory where you find all groups of connected nodes in a graph.
An efficient way to solve this problem is doing a "flood fill" algorithm, which is essentially a recursive breath first search. This wikipedia entry describes the flood fill algorithm and how it applies to solving the problem of finding connected regions of a graph.
To see how the original question can be made into a question on graphs: make each entry (e.g. "Server1", "Server_1", etc.) a node on a graph. Connect nodes with edges if and only if they are synonyms. A matrix data structure is particularly appropriate for keeping track of the edges, provided you have enough memory. Otherwise a sparse data structure like a map will work, especially since the number of synonyms will likely be limited.
Server1 is Node #0
Server_1 is Node #1
Server_2 is Node #2
Then edge[0][1] = edge[1][0] = 1, indicated that there is an edge between nodes #0 and #1 ( which means that they are synonyms ). While edge[0][2] = edge[2][0] = 0, indicating that Server1 and Server_2 are not synonyms.
Complexity Analysis
Creating this data structure is pretty efficient because a single linear pass with a lookup of the mapping of strings to node numbers is enough to crate it. If you store the mapping of strings to node numbers in a dictionary then this would be a O(n log n) step.
Doing the flood fill is O(n), you only visit each node in the graph once. So, the algorithm in all is O(n log n).

Answer 2

Introduce integer marking, which indicates synonym groups. On start one marks all words with different marks from 1 to N.
Then search trough your collection and if you find two words with indexes i and j are synonym, then remark all of words with marking i and j with lesser number of both. After N iteration you get all groups of synonyms.
It is some dirty and not throughly efficient solution, I believe one can get more performance with union-find structures.

Answer 3

Edit: This probably is NOT the most efficient way of solving your problem. If you are interested in max performance (e.g., if you have millions of values), you might be interested in writing more complex algorithm.
PHP, seems to be working (at least with data from given example):
$data = array(
array("Server1", "Server_1", "Main Server", "192.168.0.3"),
array("Server_1", "VIP Server", "Main Server"),
array("Server_2", "192.168.0.4"),
array("192.168.0.3", "192.168.0.5"),
array("Server_2", "Backup"),
);
do {
$foundSynonyms = false;
foreach ( $data as $firstKey => $firstValue ) {
foreach ( $data as $secondKey => $secondValue ) {
if ( $firstKey === $secondKey ) {
continue;
}
if ( array_intersect($firstValue, $secondValue) ) {
$data[$firstKey] = array_unique(array_merge($firstValue, $secondValue));
unset($data[$secondKey]);
$foundSynonyms = true;
break 2; // outer foreach
}
}
}
} while ( $foundSynonyms );
print_r($data);
Output:
Array
(
[0] => Array
(
[0] => Server1
[1] => Server_1
[2] => Main Server
[3] => 192.168.0.3
[4] => VIP Server
[6] => 192.168.0.5
)
[2] => Array
(
[0] => Server_2
[1] => 192.168.0.4
[3] => Backup
)
)

Answer 4

This would yield lower complexity then the PHP example (Python 3):
a = [set(("Server1", "Server_1", "Main Server", "192.168.0.3")),
set(("Server_1", "VIP Server", "Main Server")),
set(("Server_2", "192.168.0.4")),
set(("192.168.0.3", "192.168.0.5")),
set(("Server_2", "Backup"))]
b = {}
c = set()
for s in a:
full_s = s.copy()
for d in s:
if b.get(d):
full_s.update(b[d])
for d in full_s:
b[d] = full_s
c.add(frozenset(full_s))
for k,v in b.items():
fsv = frozenset(v)
if fsv in c:
print(list(fsv))
c.remove(fsv)

Answer 5

I was looking for a solution in python, so I came up with this solution. If you are willing to use python data structures like sets
you can use this solution too. "It's so simple a cave man can use it."
Simply this is the logic behind it.
foreach set_of_values in value_collection:
alreadyInSynonymSet = false
foreach synonym_set in synonym_collection:
if set_of_values in synonym_set:
alreadyInSynonymSet = true
synonym_set = synonym_set.union(set_of_values)
if not alreadyInSynonymSet:
synonym_collection.append(set(set_of_values))
vals = (
("Server1", "Server_1", "Main Server", "192.168.0.3"),
("Server_1", "VIP Server", "Main Server"),
("Server_2", "192.168.0.4"),
("192.168.0.3", "192.168.0.5"),
("Server_2", "Backup"),
)
value_sets = (set(value_tup) for value_tup in vals)
synonym_collection = []
for value_set in value_sets:
isConnected = False # If connected to a term in the graph
print(f'\nCurrent Value Set: {value_set}')
for synonyms in synonym_collection:
# IF two sets are disjoint, they don't have common elements
if not set(synonyms).isdisjoint(value_set):
isConnected = True
synonyms |= value_set # Appending elements of new value_set to synonymous set
break
# If it's not related to any other term, create a new set
if not isConnected:
print ('Value set not in graph, adding to graph...')
synonym_collection.append(value_set)
print('\nDone, Completed Graphing Synonyms')
print(synonym_collection)
This will have a result of
Current Value Set: {'Server1', 'Main Server', '192.168.0.3', 'Server_1'}
Value set not in graph, adding to graph...
Current Value Set: {'VIP Server', 'Main Server', 'Server_1'}
Current Value Set: {'192.168.0.4', 'Server_2'}
Value set not in graph, adding to graph...
Current Value Set: {'192.168.0.3', '192.168.0.5'}
Current Value Set: {'Server_2', 'Backup'}
Done, Completed Graphing Synonyms
[{'VIP Server', 'Main Server', '192.168.0.3', '192.168.0.5', 'Server1', 'Server_1'}, {'192.168.0.4', 'Server_2', 'Backup'}]