I have rather large dataset (20mln nodes, 200mln edges), simplest shortestPath queries finish in milliseconds, everything is great.
But... I need to allow shortestPath to have ZERO or ONE relation of type 999 and it can be only the first from the start node.
So, my query became like this:
MATCH (one:Obj{oid:'startID'})-[r1*0..1]-(b:Obj)
WHERE all(rel in r1 where rel.val = 999)
WITH one, b
MATCH (two:Obj{oid:'endID'}), path=shortestPath((one) -[0..21]-(two))
WHERE ALL (x IN RELATIONSHIPS(path)
WHERE (x.val > -1 and x.val<101) or (x.val=999 or x.val=998)) return path
it runs in milliseconds when there's a short path (up to 2-4), but can take 5 or 20 seconds for paths like 5++. Maybe I've composed inefficient query?
This question will be bountied when available.
Some of your requirements are a bit unclear to me, so I'll reiterate my understanding and offer a solution.
You want to inspect the shortest paths between a start and end node.
The paths returned should have ZERO or ONE relationship with a val of 999. If it's ONE relationship with that value, it should be the first.
Here's an attempt based on that logic:
MATCH (start:Obj {oid:'startID'}),
(end:Obj {oid:'endID'}),
path=shortestPath((start)-[1..21]->(end))
WITH path, relationships(path) AS rels
WHERE all(r IN relationships WHERE r.val != 999)
OR (relationships[0].val = 999
AND all(r IN relationships[1..] WHERE r.val != 999))
RETURN path
I haven't had a chance to test on actual data, but hopefully this logic and approach at least point you in the right direction.
Also note: it's possible the entire WHERE clause at the end could be reduced to:
WHERE all(r IN relationships[1..] WHERE r.val != 999)
Meaning you don't even need to check the first relationship.
Related
I'm using neo4j as a graph database to store user's connections detail into this. here I want to show the level of one user with respect to another user in their connections like Linkedin. for example- first layer connection, second layer connection, third layer and above the third layer shows 3+. but I don't know how this happens using neo4j. i searched for this but couldn't find any solution for this. if anybody knows about this then please help me to implement this functionality.
To find the shortest "connection level" between 2 specific people, just get the shortest path and add 1:
MATCH path = shortestpath((p1:Person)-[*..]-(p2:Person))
WHERE p1.id = 1 AND p2.id = 2
RETURN LENGTH(path) + 1 AS level
NOTE: You may want to put a reasonable upper bound on the variable-length relationship pattern (e.g., [*..6]) to avoid having the query taking too long or running out of memory in a large DB). You should probably ignore very distant connections anyway.
it would be something like this
// get all persons (or users)
MATCH (p:Person)
// create a set of unique combinations , assuring that you do
// not do double work
WITH COLLECT(p) AS personList
UNWIND personList AS personA
UNWIND personList AS personB
WITH personA,personB
WHERE id(personA) < id(personB)
// find the shortest path between any two nodes
MATCH path=shortestPath( (personA)-[:LINKED_TO*]-(personB) )
// return the distance ( = path length) between the two nodes
RETURN personA.name AS nameA,
personB.name AS nameB,
CASE WHEN length(path) > 3 THEN '3+'
ELSE toString(length(path))
END AS distance
I´m having serious performance issues with a job that is running everyday and I think i cannot improve the algorithm; so I´m gonnga explain you what is the problem to solve and the algorithm we have, and maybe you have some other ideas to solve the problem better.
So the problem we have to solve is:
There is a set of Rules, ~ 120.000 Rules.
Every rule has a set of combinations of Codes. Codes are basically strings. So we have ~8 combinations per rule. Example of a combination: TTAAT;ZZUHH;GGZZU;WWOOF;SSJJW;FFFOLL
There is a set of Objects, ~800 objects.
Every object has a set of ~200 codes.
We have to check for every Rule, if there is at least one Combination of Codes that is fully contained in the Objects. It means =>
loop in Rules
Loop in Combinations of the rule
Loop in Objects
every code of the combination found in the Object? => create relationship rule/object and continue with the next object
end of loop
end of loop
end of loop
For example, if we have the Rule with this combination of two codes: HHGGT; ZZUUF
And let´s say we have an object with this codes: HHGGT; DHZZU; OIJUH; ZHGTF; HHGGT; JUHZT; ZZUUF; TGRFE; UHZGT; FCDXS
Then we create a relationship between the Object and the Rule because every code of the combination of the rule is contained in the codes of the object => this is what the algorithm has to do.
As you can see this is quite expensive, because we need 120.000 x 8 x 800 = 750 millions of times in the worst-case scenario.
This is a simplified scenario of the real problem; actually what we do in the loops is a little bit more complicated, that´s why we have to reduce this somehow.
I tried to think in a solution but I don´t have any ideas!
Do you see something wrong here?
Best regards and thank you for the time :)
Something like this might work better if I'm understanding correctly (this is in python):
RULES = [
['abc', 'def',],
['aaa', 'sfd',],
['xyy', 'eff',]]
OBJECTS = [
('rrr', 'abc', 'www', 'def'),
('pqs', 'llq', 'aaa', 'sdr'),
('xyy', 'hjk', 'fed', 'eff'),
('pnn', 'rrr', 'mmm', 'qsq')
]
MapOfCodesToObjects = {}
for obj in OBJECTS:
for code in obj:
if (code in MapOfCodesToObjects):
MapOfCodesToObjects[code].add(obj)
else:
MapOfCodesToObjects[code] = set({obj})
RELATIONS = []
for rule in RULES:
if (len(rule) == 0):
continue
if (rule[0] in MapOfCodesToObjects):
ValidObjects = MapOfCodesToObjects[rule[0]]
else:
continue
for i in range(1, len(rule)):
if (rule[i] in MapOfCodesToObjects):
codeObjects = MapOfCodesToObjects[rule[i]]
else:
ValidObjects = set()
break
ValidObjects = ValidObjects.intersection(codeObjects)
if (len(ValidObjects) == 0):
break
for vo in ValidObjects:
RELATIONS.append((rule, vo))
for R in RELATIONS:
print(R)
First you build a map of codes to objects. If there are nObj objects and nCodePerObj codes on average per object, this takes O(nObj*nCodePerObj * log(nObj*nCodePerObj).
Next you iterate through the rules and look up each code in each rule in the map you built. There is a relation if a certain object occurs for every code in the rule, i.e. if it is in the set intersection of the objects for every code in the rule. Since hash lookups have O(1) time complexity on average, and set intersection has time complexity O(min of the lengths of the 2 sets), this will take O(nRule * nCodePerRule * nObjectsPerCode), (note that is nObjectsPerCode, not nCodePerObj, the performance gets worse when one code is included in many objects).
For a new POC I have the following use case:
For a given node do a 3-Level expand but apply also a filter on all expanded node (means I want to filter all resulting nodes for certain properties)
Test set:
nodes: ~ 17 Mio
edges: ~ 40 Mio
Properties: ~ 2650 Mio
My first solution looks like this:
MATCH path=(startNode:Entity {id:'RVDJRcV_yfXbG0-syGKp3Q..'})-[*..3]-(endNode:Entity)
WITH path
WHERE ALL (n IN nodes(path)[1..]
WHERE n.key = '1' AND n.domain = 'facebook.com' AND n.investigationID='any')
RETURN path
LIMIT 100
This does the job, but it is not very fast. Avg. query times in my test set are 2-3 seconds but with many timeouts (time > 30 seconds). I assume the problem is the path handling and that my node has lots of properties...
Explain plan:
Variant 1: i removed the "with path"
Solution:
Based on the tip that i should avoid [1..] in the query
MATCH path=(startNode:Entity {id:'v-jXIO7kozAa35gMUpUkvg..'})-[*..3]-(endNode:Entity)
WHERE ALL (n IN nodes(path)
WHERE n=startNode OR (n.key = '1' AND n.domain = 'facebook.com' AND n.investigationID='any'))
RETURN path
LIMIT 100
While you can filter during expansion with variable-length paths, Cypher currently can't apply that filter during expansion when you're working with a slice of the list instead of the whole list. It will fall back to doing the full var-length expansion, and then applying the filter to all results found.
We need to only use ALL (n IN nodes(path) ..., we can't use the slice of the path.
To do this, we need to add one more predicate within the all() function. Since the start node probably doesn't meet the current predicate, we'll create an exception for it:
MATCH path=(startNode:Entity {id:'RVDJRcV_yfXbG0-syGKp3Q..'})-[*..3]-(endNode:Entity)
WHERE ALL (n IN nodes(path)[1..]
WHERE n=startNode OR (n.key = '1' AND n.domain = 'facebook.com' AND n.investigationID='any'))
RETURN path
LIMIT 100
I'm thinking about an algorithm that will create X most unique concatenations of Y parts, where each part can be one of several items. For example 3 parts:
part #1: 0,1,2
part #2: a,b,c
part #3: x,y,z
And the (random, one case of some possibilities) result of 5 concatenations:
0ax
1by
2cz
0bz (note that '0by' would be "less unique " than '0bz' because 'by' already was)
2ay (note that 'a' didn't after '2' jet, and 'y' didn't after 'a' jet)
Simple BAD results for next concatenation:
1cy ('c' wasn't after 1, 'y' wasn't after 'c', BUT '1'-'y' already was as first-last
Simple GOOD next result would be:
0cy ('c' wasn't after '0', 'y' wasn't after 'c', and '0'-'y' wasn't as first-last part)
1az
1cx
I know that this solution limit possible results, but when all full unique possibilities will gone, algorithm should continue and try to keep most avaible uniqueness (repeating as few as possible).
Consider real example:
Boy/Girl/Martin
bought/stole/get
bottle/milk/water
And I want results like:
Boy get milk
Martin stole bottle
Girl bought water
Boy bought bottle (not water, because of 'bought+water' and not milk, because of 'Boy+milk')
Maybe start with a tree of all combinations, but how to select most unique trees first?
Edit: According to this sample data, we can see, that creation of fully unique results for 4 words * 3 possibilities, provide us only 3 results:
Martin stole a bootle
Boy bought an milk
He get hard water
But, there can be more results requested. So, 4. result should be most-available-uniqueness like Martin bought hard milk, not Martin stole a water
Edit: Some start for a solution ?
Imagine each part as a barrel, wich can be rotated, and last item goes as first when rotates down, first goes as last when rotating up. Now, set barells like this:
Martin|stole |a |bootle
Boy |bought|an |milk
He |get |hard|water
Now, write sentences as We see, and rotate first barell UP once, second twice, third three and so on. We get sentences (note that third barell did one full rotation):
Boy |get |a |milk
He |stole |an |water
Martin|bought|hard|bootle
And we get next solutions. We can do process one more time to get more solutions:
He |bought|a |water
Martin|get |an |bootle
Boy |stole |hard|milk
The problem is that first barrel will be connected with last, because rotating parallel.
I'm wondering if that will be more uniqe if i rotate last barrel one more time in last solution (but the i provide other connections like an-water - but this will be repeated only 2 times, not 3 times like now). Don't know that "barrels" are good way ofthinking here.
I think that we should first found a definition for uniqueness
For example, what is changing uniqueness to drop ? If we use word that was already used ? Do repeating 2 words close to each other is less uniqe that repeating a word in some gap of other words ? So, this problem can be subjective.
But I think that in lot of sequences, each word should be used similar times (like selecting word randomly and removing from a set, and after getting all words refresh all options that they can be obtained next time) - this is easy to do.
But, even if we get each words similar number od times, we should do something to do-not-repeat-connections between words. I think, that more uniqe is repeating words far from each other, not next to each other.
Anytime you need a new concatenation, just generate a completely random one, calculate it's fitness, and then either accept that concatenation or reject it (probabilistically, that is).
const C = 1.0
function CreateGoodConcatenation()
{
for (rejectionCount = 0; ; rejectionCount++)
{
candidate = CreateRandomConcatination()
fitness = CalculateFitness(candidate) // returns 0 < fitness <= 1
r = GetRand(zero to one)
adjusted_r = Math.pow(r, C * rejectionCount + 1) // bias toward acceptability as rejectionCount increases
if (adjusted_r < fitness)
{
return candidate
}
}
}
CalculateFitness should never return zero. If it does, you might find yourself in an infinite loop.
As you increase C, less ideal concatenations are accepted more readily.
As you decrease C, you face increased iterations for each call to CreateGoodConcatenation (plus less entropy in the result)
I work in a consulting organization and am most of the time at customer locations. Because of that I rarely meet my colleagues. To get to know each other better we are going to arrange a dinner party. There will be many small tables so people can have a chat. In order to talk to as many different people as possible during the party, everybody has to switch tables at some interval, say every hour.
How do I write a program that creates the table switching schedule? Just to give you some numbers; in this case there will be around 40 people and there can be at most 8 people at each table. But, the algorithm needs to be generic of course
heres an idea
first work from the perspective of the first person .. lets call him X
X has to meet all the other people in the room, so we should divide the remaining people into n groups ( where n = #_of_people/capacity_per_table ) and make him sit with one of these groups per iteration
Now that X has been taken care of, we will consider the next person Y
WLOG Y be a person X had to sit with in the first iteration itself.. so we already know Y's table group for that time-frame.. we should then divide the remaining people into groups such that each group sits with Y for every consecutive iteration.. and for each iteration X's group and Y's group have no person in common
.. I guess, if you keep doing something like this, you will get an optimal solution (if one exists)
Alternatively you could crowd source the problem by giving each person a card where they could write down the names of all the people they got dine with.. and at the end of event, present some kind of prize to the person with the most names in their card
This sounds like an application for genetic algorithm:
Select a random permutation of the 40 guests - this is one seating arrangement
Repeat the random permutation N time (n is how many times you are to switch seats in the night)
Combine the permutations together - this is the chromosome for one organism
Repeat for how ever many organisms you want to breed in one generation
The fitness score is the number of people each person got to see in one night (or alternatively - the inverse of the number of people they did not see)
Breed, mutate and introduce new organisms using the normal method and repeat until you get a satisfactory answer
You can add in any other factors you like into the fitness, such as male/female ratio and so on without greatly changing the underlying method.
Why not imitate real world?
class Person {
void doPeriodically() {
do {
newTable = random (numberOfTables);
} while (tableBusy(newTable))
switchTable (newTable)
}
}
Oh, and note that there is a similar algorithm for finding a mating partner and it's rumored to be effective for those 99% of people who don't spend all of their free time answering programming questions...
Perfect Table Plan
You might want to have a look at combinatorial design theory.
Intuitively I don't think you can do better than a perfect shuffle, but it's beyond my pre-coffee cognition to prove it.
This one was very funny! :D
I tried different method but the logic suggested by adi92 (card + prize) is the one that works better than any other I tried.
It works like this:
a guy arrives and examines all the tables
for each table with free seats he counts how many people he has to meet yet, then choose the one with more unknown people
if two tables have an equal number of unknown people then the guy will choose the one with more free seats, so that there is more probability to meet more new people
at each turn the order of the people taking seats is random (this avoid possible infinite loops), this is a "demo" of the working algorithm in python:
import random
class Person(object):
def __init__(self, name):
self.name = name
self.known_people = dict()
def meets(self, a_guy, propagation = True):
"self meets a_guy, and a_guy meets self"
if a_guy not in self.known_people:
self.known_people[a_guy] = 1
else:
self.known_people[a_guy] += 1
if propagation: a_guy.meets(self, False)
def points(self, table):
"Calculates how many new guys self will meet at table"
return len([p for p in table if p not in self.known_people])
def chooses(self, tables, n_seats):
"Calculate what is the best table to sit at, and return it"
points = 0
free_seats = 0
ret = random.choice([t for t in tables if len(t)<n_seats])
for table in tables:
tmp_p = self.points(table)
tmp_s = n_seats - len(table)
if tmp_s == 0: continue
if tmp_p > points or (tmp_p == points and tmp_s > free_seats):
ret = table
points = tmp_p
free_seats = tmp_s
return ret
def __str__(self):
return self.name
def __repr__(self):
return self.name
def Switcher(n_seats, people):
"""calculate how many tables and what switches you need
assuming each table has n_seats seats"""
n_people = len(people)
n_tables = n_people/n_seats
switches = []
while not all(len(g.known_people) == n_people-1 for g in people):
tables = [[] for t in xrange(n_tables)]
random.shuffle(people) # need to change "starter"
for the_guy in people:
table = the_guy.chooses(tables, n_seats)
tables.remove(table)
for guy in table:
the_guy.meets(guy)
table += [the_guy]
tables += [table]
switches += [tables]
return switches
lst_people = [Person('Hallis'),
Person('adi92'),
Person('ilya n.'),
Person('m_oLogin'),
Person('Andrea'),
Person('1800 INFORMATION'),
Person('starblue'),
Person('regularfry')]
s = Switcher(4, lst_people)
print "You need %d tables and %d turns" % (len(s[0]), len(s))
turn = 1
for tables in s:
print 'Turn #%d' % turn
turn += 1
tbl = 1
for table in tables:
print ' Table #%d - '%tbl, table
tbl += 1
print '\n'
This will output something like:
You need 2 tables and 3 turns
Turn #1
Table #1 - [1800 INFORMATION, Hallis, m_oLogin, Andrea]
Table #2 - [adi92, starblue, ilya n., regularfry]
Turn #2
Table #1 - [regularfry, starblue, Hallis, m_oLogin]
Table #2 - [adi92, 1800 INFORMATION, Andrea, ilya n.]
Turn #3
Table #1 - [m_oLogin, Hallis, adi92, ilya n.]
Table #2 - [Andrea, regularfry, starblue, 1800 INFORMATION]
Because of the random it won't always come with the minimum number of switch, especially with larger sets of people. You should then run it a couple of times and get the result with less turns (so you do not stress all the people at the party :P ), and it is an easy thing to code :P
PS:
Yes, you can save the prize money :P
You can also take look at stable matching problem. The solution to this problem involves using max-flow algorithm. http://en.wikipedia.org/wiki/Stable_marriage_problem
I wouldn't bother with genetic algorithms. Instead, I would do the following, which is a slight refinement on repeated perfect shuffles.
While (there are two people who haven't met):
Consider the graph where each node is a guest and edge (A, B) exists if A and B have NOT sat at the same table. Find all the connected components of this graph. If there are any connected components of size < tablesize, schedule those connected components at tables. Note that even this is actually an instance of a hard problem known as Bin packing, but first fit decreasing will probably be fine, which can be accomplished by sorting the connected components in order of biggest to smallest, and then putting them each of them in turn at the first table where they fit.
Perform a random permutation of the remaining elements. (In other words, seat the remaining people randomly, which at first will be everyone.)
Increment counter indicating number of rounds.
Repeat the above for a while until the number of rounds seems to converge.