Seeking Neo4J Cypher query for long but (nearly) unique paths. Returns any nodes connected by an outgoing relationship to the Person node with the name property set to Oliver Stone. will be the based on the highest value of V1,V2. Using your programming language's form of infinity for d and k will give you all paths§. Construct NEO4J Cypher Graph Query to show intermediate nodes in path. Trying to find all possible paths (even if up to length of 20) in a graph of millions of nodes is likely to cause a memory overflow. Improve this question. Find centralized, trusted content and collaborate around the technologies you use most. Since you said you don't want all pairs with multiple paths, you can adjust the limit (I've set it to 2 in. Matching multiple independent (or dependent) paths. Then I want a path of length at most 4 between A and B, having at least one node in common with a path from A to C and a path from B to. I have two labels (A and B) let us say we have the following connectivity. problems with: Dijkstra, shortestPath and allShortestPaths: Only returns the shortest path or paths with the shortest length. Sorted by: 0. With small reusable samples, for less time-consuming labs. What the above query is doing: The variable length 1. The code for your all pairs shortest path is correct, it is just a very expensive algorithm. In order to find the shortest path between to nodes based on their IDs, this cypher-query does the trick: MATCH (martin:RoadNode), (oliver:RoadNode), p = shortestPath ( (martin)- [*. Get all paths between two nodes via specific node Neo4j Graph Platform migrated Reid (Reid) September 26, 2022, 9:20am 1 We have a Neo4J database that tracks company Supply for companies. For example,I want to query allShortestPaths between 3 nodes(A,B,C),it means i want to query: 1. Dijkstra Source-Target Shortest Path. The variable-length relationship pattern [*0. v(6) x. 15]- (oliver)) WHERE id (martin) = 16814 AND id (oliver) = 16820 RETURN p. Note: In our graph database nodes can have multi direction. dijkstra. The variable-length relationship pattern [*0. The output should include: A . This query will find all paths of length up to 10 between nodes c1 and c2. sourceNode: source, You can specify a target node, which looks like your end node, so you can find a path between two nodes. Path finding algorithms find the shortest path between two or more nodes or evaluate the availability and quality of paths. BFS + Reverse DFS Added: My current algorithm is a similarity matrix, where I have one path on the x-axis and another on the y-axis. I enclosed here an example with with few nodes just as an illustrations. To do this efficiently, do the following: MATCH p= (parent: Root )- [: HAS_CHILD * 1. START n=node (*),m=node (*) MATCH p=n- [r*1. targetId = 5678 MATCH p = shortestPath ( (source)- [*]- (target)) return p; If this query runs too long, try limiting the maximum path length searched. MATCH (x:Test)- [r:CONNECTED_TO*]-> (z:Test) RETURN x, r, z. All queries return invalid input or no rows. BTW when trying to find this query, I found the following : When trying to filter the collection of nodes(p), without the unwind, the identifier is not treated as a Node but as a collection, I don't know if it is the expected behavior : The following query will fail saying that the identifier s is a collection and a Node is expected Create an index on the Name property of nodes with the X label like this: MATCH p= ( (n:X {Name:"Node1"}) - [*1. would find the shortest path from start to end based on the number of relationships in. 1. EDIT. Line 2 aggregates the node and relationship collections using the collect method to reduce the data from all the paths to one collection to meet your requirement. Two customers have an account in the "Money Bank". spanningTree (start, {terminatorNodes: [end]}) YIELD path return path. For the Start node. We have a Neo4J database that tracks company Supply for companies. How can I find a path between given two nodes in neo4j Java API, the product of all weights of the path is maximum in all paths between two nodes, what can I do? In my graph database, there are two elements: one is node, another is relationship, and all have a name property, but the relationship have an extra property: weight (double type, and. So i need to get the particular entire path of a node named "Trinity" and. All relationships are directional. In your example, it looks like these two nodes are connected via a path length of 8. The general structure is `````(Company)-[:Supply_For]->(Company)` A company can have multiple rels at the same time and each company can have multiple company supply rels. all path between two nodes in neo4j with TraversalDescription returns incorrect paths. both. In this category, Dijkstra’s algorithm is the most well known. EDIT: After getting more information, here's a use case using the. The path choice between a node and direct candidates C1,C2. it finds the end of the chain). The general structure is `````(Company)-[:Supply_For]->(Company)` A company can have multiple rels at the same time and each company can have multiple company supply rels. 5]->b // match paths (best to provide maximum lengths to prevent queries from running away) where not (a-->b. This allows the target node to be the same as the root or leaf node. Is there probably a better way to re-write the query ? Also it is not clear to me if the memory allocated to the. I want to find all shortest paths between "Money Bank" and "ACME Inc. paths Lets assume there are 2 shortest path of equal distance between two given nodes. ] matches relationship lengths of 0 or more (where 0 means there is actually no relationship and the 2 "endpoints" are the same node). This rule should apply from the start node S till the end node E. :Style) which are connected somehow to a specific set of nodes (e. ]-m WITH count (p) AS totalPaths,n,m WHERE totalPaths>1 RETURN n,m,totalPaths LIMIT 2. This feature is in the alpha tier. Here's two. The query won't be as elegant, but it should work. So, the 1-st step is to do: td = td. Expecting below results, how to do this? First A & C Path count: 2 Path1: BAD (from A. Why would you want to extract the genre property when you need to find shortest paths between nodes. Create a database connection by creating a driver instance. Get all Routes between two nodes neo4j Get all Routes between two nodes neo4j neo4j cypher 10,840 Solution 1 In cypher, to get all paths between two nodes not linked by a relationship, and sort by a total in a weight, you can use the reduce function introduced in 1. @GarethRees Assume there is a polynomial time (NOT pseudo polynomial) algorithm for kth shortest simple path between two nodes. my question is I want to find the path via two given nodes id, I tried as below pattern: To start off, you can just prepend a zero to the cost array. Regarding the APOC you could try the following: MATC. 5 ]->b // match paths (best to provide maximum lengths to prevent queries from running away) where not ( a. That is, say the persons are A, B and C. Path finding algorithms find the path between two or more nodes or evaluate the availability and quality of paths. . Trying to find all possible paths (even if up to length of 20) in a graph of millions of nodes is likely to cause a memory overflow. Table 6. 1. Related nodes The symbol -- means related to, without regard to type or direction of the relationship. users and groups can see red nodes (there is no 'see' relation to non-red nodes) The db schema is empty because your nodes don't have any labels. But all I get as response from the REST API is a . 10 will find all paths (there should be only one), for any Parent-Child path that spans at most 10 hops. Querying the shortest path for a defined start and end node this approach would create n start nodes and m. spanningTree (start, {terminatorNodes: [end]}) YIELD path return path. I have been looking into the APOC- and the GDSL-libraries, but. The query plan for the new query is the following: I was expecting a greater speedup though. Relation between nodes (e. I have a scenario where I have more than 2 random nodes. 2. RELATIONSHIP_GLOBAL); desc. It has the following use cases: Finding directions between physical locations. Collect them into a list. path. However, I have noticed that it is not always the case that the number of paths or the longest path play a role in the performance. I also want to consider the type of relationship between the nodes. Since your graph has cycles, the "longest path" between 2 nodes can be infinitely long. shortestPath. I want is A - [Supply_FOR*0. Pathfinding has a long history and is considered to be one of the classical. This query will return the longest path between the 2 endpoints: MATCH path = (:Node {Id: 1})- [:Link*]-> (:Node {Id: 4}) RETURN path ORDER BY LENGTH (path) DESC LIMIT 1; Well, you are specifically asking in your query to find the shortest path. By clicking Accept, you consent to the use of cookies. You mean like a persistent path ID? This isn't directly featured, but you can do it in a Query in Cypher. All relationships are directional. For example, you could think that I have the git history of a complex project. match p = (n{name:"Node1"})-[:Route*1. name IN {names} WITH collect(n) as nodes UNWIND nodes as n UNWIND nodes as m WITH * WHERE id(n) < id(m. sourceId = 1234 AND target. path. Improve this question. I have figured out how to find a path of length n, though it isn't very fast: MATCH path= (start:Node {name:'nodeA'})- [:CONNECTS_TO*19. This is the query. In your first image, you had :Station nodes and you had a :Goto relationship, all other node labels and relationship types were lowercase. ), b=node (. I want to find paths between a particular node and set of nodes. All relationships are directional. I have a complex graph as shown in image below : Here every relationship has a type value. g. . I am using the graph. coll. Identify paths between two nodes in neo4j. Using a gremlin script and neo4j I try to find all paths between two nodes, descending at most 10 levels down. 0. My TraversalDescription cannot be used to find more than 1 path because of default Uniqueness. Since you already have the relationships between the nodes that force the types to be what you want (for example :CD will always connect a :C node to a 😄 node) you can leave the labels of the nodes out, and. Try this: MATCH (a) WHERE id(a) = 801 CALL apoc. I need to get all possible paths connecting all three nodes. 1. v(2) y = g. You can use a variable length relationship to return all such paths. The WHERE clause is needed to filter the paths to only those where the leaf child nodes have no outgoing :HAS_CHILD relationships (i. This visual presentation of the Neo4j graph algorithms is focused on quick understanding and less implementation details. Each node can be described with a JSON object that has: {'id':'id', 'outbound':['id1','id2','id3']} } So if I had these nodes in the database: 4. . 0. This is the most common usage, and web mapping. Share. So as result I want to get all paths from A to B except this one via node of type X, for example A -> C (type different than X) -> B - ok, A -> D. 3]-(b:PERSON{name: 'EK'}) RETURN paths This will return paths of length up to 3, change it as you need. Internally, Neo4j will use a fast bidirectional breadth-first search algorithm if the predicates can be evaluated whilst searching for the path. You should always use shortestPath for finding the span of minute nodes; matching on (startMinute)-[:NEXT*]->(endMinute) without wrapping it in shortestPath is trying to find all paths of any length between the two nodes, so it's exhaustive and takes much longer, whereas shortestPath can stop as soon as it's found the path. The query returns 1,927,493 tuples in the form (source_node, target_node, count_of_paths). But this is problematic because every :Address node can have multiple :LEADS_TO relationships. Planning shortest paths in Cypher can lead to different query plans depending on the predicates that need to be evaluated. The following Cypher query using a variable-length pattern matching, nodes (), unwind and collect () should work. MATCH (b:Person)-[:RELATES*]->(a:Person)<-[:RELATES*]-(c:Person) where ID(b) = 7 and ID(c) = 50 RETURN a a -> b -> c 1. There are a couple of approaches. Assuming these have a common label (I'll use :Node), this should be easy enough: MATCH path = (start:Node - 18936 This website uses cookies. For example, I would like the query to return foo in the following example, because both relationships to foo have prop = false. 7. To check whether you have loops on node 0 execute: 2 Answers. Weighted Shortest Path MATCH (c1:City {name: "London"})-[ edge_list:ROAD_TO *wShortest 10 (e, n | e. Neo4j Find all paths back to the same node. Get all Routes between two nodes neo4j. paths3. Each tuple counts only a few paths (for most of them less than 10) Note that when adding two more relations to the same query, the execution time is increased to approximately two hours. 20]-> (end:Node {name:'nodeB'}) RETURN path LIMIT 1. expandConfig(c1, { relationshipFilter: "Supply_FOR", minLevel: 0, maxLevel: 2 }) YIELD path RETURN path. 9: The Shortest Path algorithm calculates the shortest (weighted) path between a pair of nodes. We have a Neo4J database that tracks company Supply for companies. . I am using the following syntax from Cypher to find the shortest node. For example, I would like the query to return foo in the following example, because both relationships to foo have prop = false. neo4j. 0. the allShortestPaths between A and C . I'm trying to draw graph with all connections (paths) between two nodes except paths with specific type of node. neo4j. Something like this, though it is really going to take long depending on the size of your graph-. This query will find all paths of length up to 10 between nodes c1 and c2. Let's say I need to find all the path of lenght 2 between Keanu Reeves and Laurence Fishburne. Unfortunately, the graphs that are involved tend to be rather large, such that normal cypher queries would be to slow to perform this kind of operation. Ideally you would want the shortest path algorithm itself to check that each relationship in the path has the same value for x, but alternatively you can go one relationship out from your start node and check that the shortest path shares an x value with that first relationship. If there are several paths of the same length i want all of them. Regarding the APOC you could try the following: MATC. If you need that all relationships between n and n1 have a property called RelationLabel that CONTAINS the value "may_be_treat", then you can use the ALL function in conjunction. Given a couple of Persons, say 3 or 4 of them, I want to find all paths between all of them of length 4, and I just want simple paths. The above query works and returns all the results, but does not display the relationship between the User and the Repository if there is an intermediate node. 2]-(B) it's more like finding a path between A and B and here sometime allowed another one node C, In general, three patterns includi. Therefore, the two nodes that are furthest will have longest shortest path between them. Lets say I have 3 places from 2 different sources. When the direction of a relationship is of interest, it is shown by using -→ or ←- . g. object. The more nodes and relationships you can filter out the more efficient the search will be. Database contains Sr manager as root node. For example, I would like the query to return foo in the following example, because both relationships to foo have prop = false. You should find the source and target first, and then invoke shortestpath: MATCH (source:example_nodes), (target:example_nodes) WHERE source. java. The general structure is `````(Company)-[:Supply_For]->(Company)` A company can have multiple rels at the same time and each company can have multiple company supply rels. . What you can do is break it down into smaller segments. , do you want to avoid revisiting the same node, or revisiting the same relationship, or use some other approach?). Weighted Shortest Path MATCH (c1:City {name: "London"})-[ edge_list:ROAD_TO *wShortest 10 (e, n | e. For example, I would like the query to return foo in the following example, because both relationships to foo have prop = false. If you want to restrict level to say 3 then CALL apoc. I want to store the nodes in a database and have the database execute a search that shows me all paths between two nodes. Unfortunately, in my case, I may have multiple nodes in between of start and end node and these between nodes may vary for another set of data. This is easier if you are using subqueries (Neo4j 4. Node1 and Node2 are not of the same type but i will try your suggestion by creating labels and indexes on the nodes. Combining results with Cypher. You seem to have a different definition of "distance" that may be difficult to calculate (or whose value can be very ambiguous) -- given that nodes can have any number of relationships of various types between them, and that the same node might appear multiple times in the same. 6. targetId = 5678 MATCH p = shortestPath ( (source)- [*]- (target)) return p; If this query runs too long, try limiting the maximum path length searched. Since log{(3/2)n!} is polynomial in n, both encoding the number and the number of repeats. I want to print all path present form parent node to child node for entire trees present in graph. 10 ]-> (child: Node ) WHERE NOT (child)- [: HAS_CHILD ]-> () RETURN p What the above query is doing: The variable length 1. Internally, Neo4j will use a fast bidirectional breadth-first search algorithm if the predicates can be evaluated whilst searching for the path. 15]- (oliver)) WHERE id (martin) = 16814 AND id (oliver) = 16820 RETURN p. I would like to handle multiple paths between 2 nodes in neo4j. But this is problematic because every :Address node can have multiple :LEADS_TO relationships. In order to find the shortest path between to nodes based on their IDs, this cypher-query does the trick: MATCH (martin:RoadNode), (oliver:RoadNode), p = shortestPath ( (martin)- [*. For example, you could think that I have the git history of a complex project. StartNode - Verb - Adverb - End node StartNode - Noun- Verb - adjective- End node. k is the number of paths to find. NODE_PATH); But now we don't have a stop condition and traversal returns not only shortest paths. , do you want to avoid revisiting the same node, or revisiting the same relationship, or use some other approach?). // 2,153 and 485 are nodeid. ] matches relationship lengths of 0 or more (where 0 means there is actually no relationship and the 2 "endpoints" are the same node). 0. cypher query to find paths between set of nodes takes a very long time. To start off, you can just prepend a zero to the cost array. Between two vertices can be more relations with different weight. I'm using a graph to get routes by bus and bike between two stops. Neo4j similar paths. I have two labels (A and B) let us say we have the following connectivity. Cart Facebook Twitter Pinterest YouTube RSS Find all nodes with a specific label: Query MATCH (movie:Movie) RETURN movie. For your example (which has relationships pointing in both directions), this query using an undirected variable length relationship should work: MATCH p= (:Foo {id: 'A'})- [*]- (:Foo {id: 'B'}) RETURN p. cypher query for shortest paths between set of nodes. Neo4j query for complete paths. . Neo4j is case sensitive, you need to keep your cases consistent. I'm trying to find a generic way to search for a node or set of nodes which does not have a link to a another node or set of nodes. java. I am pretty sure there exist paths with length of 5. . If you want somthing persistent, you can always use an index, so create a Relationship index that would store the Relationship s of Path 1 under the key/value of Path:1. NODE_GLOBAL: once discovered, the end node will never be visited again. What the above query is doing: The variable length 1. thanks for your reply. The relationships between the nodes have the property "Distance". The Shortest Path algorithm calculates the shortest (weighted) path between a pair of nodes. In cypher, to get all paths between two nodes not linked by a relationship, and sort by a total in a weight, you can use the reduce function introduced in 1. Under the Hood. Improve this answer. sum function. I'm using a graph to get routes by bus and bike between two stops. One more query on similar lines, I have a relationship property called health for each, while finding the path between each A and C, I also need to find the overall health. it finds the end of the chain). The query returns 1,927,493 tuples in the form (source_node, target_node, count_of_paths). ArrayIndexOutOfBoundsException: -1 Here is the script: x = g. It returns exactly 3 results. B1'} ) // get all nodes (departments) from path with nodes (p) as deps // unwind deps. The shortestPath function in Cypher does not take into account accumulating of relationship properties, so this: MATCH (start:Point {title: 'Some Point 1'}), (end:Point {title: 'Some Point 5'}) MATCH p=shortestPath ( (start)- [:distance*]-> (end)) RETURN p. If you want somthing persistent, you can always use an index, so create a Relationship index that would store the Relationship s of Path 1 under the key/value of Path:1. g. (bar0) - [ {prop:false}]-> (foo) <- [ {prop:false}]- (bar1) To do this efficiently, do the following: MATCH p= (parent: Root )- [: HAS_CHILD * 1. So to be clear, lets say I want to find K best paths between two nodes until a maximum length M. Next, I create a list of array that are used to calculate the cumulative sum of cost along the way using the apoc. enter image description here thank you for your help. If there are several paths of the same length i want all of them. For example, I would like the query to return foo in the following example, because both relationships to foo have prop = false. I want is A - [Supply_FOR*0. 1 Answer. Unwind the list twice, once for every side of the path. problems with: Dijkstra, shortestPath and allShortestPaths: Only returns the shortest path or paths with the shortest length. my question is I want to find the path via two given nodes id, I tried as below pattern: We also know that for any two nodes, the given minimum path length will be two (i. . ID ='3007' AND to. Creating your first node and relationship using Cypher. The relationship also has a property "altitude_difference" between the two nodes it connects: (altitude-EndNode minus altitude-StartNode). You. Without exporting to a file, the query runs in 32 seconds. 2 Answers. I start to learn neo4j. , two -[:PART_OF]- relationships,. For example, I would like the query to return foo in the following example, because both relationships to foo have prop = false. Path finding algorithms find the path between two or more nodes or evaluate the availability and quality of paths. Here's a screenshot of the results and what I mean visually: Note that for some reason, where there is a single hop between the User and a Repository, the relationship is shown. In a neo4j database, I have created 4 node types: A, B, C and D each one with 1500 nodes. For example, If node "fcproc" is given, I expect to get a path("itest->bkqa->fcproc"); If node "itest" is given, I expect to get 2 paths("itest->bkqa->fcproc", "itest. object. match and return the paths. v(2) y = g. This rule should apply from the start node S till the end node E. title Returns all the nodes with the Movie label in the database. g. I'm trying to draw graph with all connections (paths) between two nodes except paths with specific type of node. Cypher: Find any path between nodes. The relationships have a property "length_km" which I can use as a weight property. This returns only 6 possible paths. ), b=node (. shortestPath. Then I want a path of length at most 4 between A and B, having at least one node in common with a path from A to C and a path from B to C. Data Modeling and Leveraging with Neo4j. Remove inverse pairs by id comparison. I have a graph with this hierarchy, so I want to find all nodes and relationships connected to a specific node. so is it possible to return also the distance of the shortest path in an efficient way? –Find all paths between two nodes. We have a tree of nodes. 5. 2] ->(c)<-[Supply_FOR*0. 1 Answer. 7. MATCH (a:Address)- [:LEADS_TO]-> (aa:Milestone) and similar for the end node and start the query from there. MATCH p=(o{value:"a"})-[r*]-(x{value:"b"}) WHERE has(r. We have the concept of endNodes, where you can supply a list of possible nodes that you want to expand to and end upon, and we can also supply a limit for the number of paths to. MATCH (: Person {name: 'Oliver Stone' })--> (movie) RETURN movie. The WHERE clause is needed to filter the paths to only those where the leaf child nodes have no outgoing :HAS_CHILD relationships (i. return count (*) so you don't have to employ the pattern matcher just check the relationship-existence. . I want to understand all paths between any two nodes (all pairs of nodes) Restrict the paths to a maximum of length 10, but should also get paths if found, length 4 etc. For the Start node. both. 0. Find path between nodes through cypher. For example, I would like the query to return foo in the following example, because both relationships to foo have prop = false. Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type. Trek & mountaineering routing. VDOMDHTMLtml> How to efficiently count all paths between two nod. coll. Well, that skeleton code is a bit weird anyhow. dijkstra. weight) total_weight ]-(c2:City {name: "Paris"}) RETURN *; The above query will find the shortest path of length up to 10 nodes between nodes c1 and c2. Create a database connection by creating a driver instance. The Shortest Path algorithm calculates the shortest (weighted) path between a pair of nodes. In the same graph, the relationship between A and B1 is having health "BAD". I read the apoc path finding algorithms, didn't find which algo meets my requirement. I have tried in many ways but I am not able to success. Finding the Shortest Path between two nodes of a graph in Neo4j using CQL and Python: From a Python program import the GraphDatabase module, which is available through installing Neo4j Python driver. will be the based on the highest value of V1,V2. Can anybody recommend a client? Or alternatively, is there a way using the REST api to check if. This was a problem of changing case. Length of the shortest path merely returns the number of hops, and not the actual distance between node x and node y (start and end nodes). Result. ", but I don't want to consider the path (Company)-(Customer)-(Bank) in my search for the shortest paths. maxmemory to 5000. e. Since there are at most (3/2)n! such paths, you can do binary search and find if there is a simple path of length n. For example Path: I am currently working with neo4j and I need to find path between 2 nodes in large graph. Without exporting to a file, the query runs in 32 seconds. lang. (bar0) - [ {prop:false}]-> (foo) <- [ {prop:false}]- (bar1) 1 I'm trying to retrieve nodes where none of the relationships to that node have a specific property value. title. I want to find all paths with a given node. spanningTree(a, {}) YIELD path return path Returns all available levels, nodes and relationships. hope you can help me with this, so basically I'm looking for a way to Find all paths till the first node of certain label (the query should explore all the paths from a node and stop when a node with a specific label is found). 4]-> (m:X {Name:"Node2"})) RETURN p; If you do the above, then your query should be faster. Shortest path finding in Cypher ® and how it is planned. And once you have calculated all the paths, you only return the 10 shortest. Run START n=node (*) RETURN n to get the ids of node 27 and 0 and then run your query. Each tuple counts only a few paths (for most of them less than 10). Example there are two shortest path in graph: 1) A - B -. Is there any other way to get all nodes and relations between two nodes without using the APOC procedure? Here is query with apoc procedure: MATCH (start:temp {Name:"Joel"}), (end:temp {Name:"Jack"}) CALL apoc. How do we find out paths in neo4j which will cover all the selected set of nodes. Pathfinding has a long history and is considered to be one of the classical. I used these code: TraversalDescription desc = Traversal. We have recently started using Neo4j database and I am using the below code (uses GraphAlgoFactory API) for finding all the possible paths to reach from one Node to another Node : Node node1 = db. (bar0) - [ {prop:false}]-> (foo) <- [ {prop:false}]- (bar1) 1 I'm trying to retrieve nodes where none of the relationships to that node have a specific property value. g. Note, however, that variable length relationship. 4. Here is a picture of the data under consideration. I have figured out how to find a path of length n, though it isn't very fast: MATCH path=(start:Node {name:'nodeA'})-[:CONNECTS_TO*19. 4]-> (m:X {Name:"Node2"})) RETURN p; If you do the above, then your query should be faster. Path finding algorithms find the shortest path between two or more nodes or evaluate the availability and quality of paths. e. sum function. Thanks in advance! Also, normally a single path step is considered a "distance" of 1. This cypher returned a row for every "person" to "edge" relationships. (bar0) - [ {prop:false}]-> (foo) <- [ {prop:false}]- (bar1) 1 I'm trying to retrieve nodes where none of the relationships to that node have a specific property value. Match p = (n {ConceptID: 'C0000039'})- [*]- (m) WHERE ALL (r IN relationships (p). There are a few hundred thousand nodes in this chain. For example I'm looking paths between A and B except connection via node of type X. The variable-length relationship pattern [*0. Some of the nodes are marked as red. It don't solve my problem because I don't know path length (it depends of graph). 1 source 45 gives me 2 places that are in-fact duplicates of each other, and 1 source 55 gives me 1 correct identifier. v(6) x. weight) total_weight ]-(c2:City {name: "Paris"}) RETURN *; The above query will find the shortest path of length up to 10 nodes between nodes c1 and c2. 9: start a =node (. loop(10){!it. Sorted by: 0. Trek & mountaineering routing - Stack Overflow. You can return the path if you need all the relationships and nodes in between. This blog post explains the nuisances of parallel processing and database recovery, how modern computer architectures can be utilized to better serve the needs of the user in the case of recovery, and how Memgraph specifically deals with this issue. Without exporting to a file, the query runs in 32 seconds. 1. START startnode=node (2), group = node (153 ,485. The nodes can be entered in any order, like x64->Linux->Oracle or Oracle->Linux->10.