Over a million developers have joined DZone.
{{announcement.body}}
{{announcement.title}}

Neo4j: Find the Midpoint Between Two Lat/Longs [Code Snippet]

DZone's Guide to

Neo4j: Find the Midpoint Between Two Lat/Longs [Code Snippet]

In this post, look at how to quickly determine the quickest route between two points on a map using Neo4j.

· Database Zone
Free Resource

Whether you work in SQL Server Management Studio or Visual Studio, Redgate tools integrate with your existing infrastructure, enabling you to align DevOps for your applications with DevOps for your SQL Server databases. Discover true Database DevOps, brought to you in partnership with Redgate.

2016 10 31 06 06 00

Over the last couple of weekends I’ve been playing around with some transport data and I wanted to run the A* algorithm to find the quickest route between two stations.

The A* algorithm takes an estimateEvaluator as one of its parameters and the evaluator looks at lat/longs of nodes to work out whether a path is worth following or not. I therefore needed to add lat/longs for each station and I found it surprisingly hard to find this location date for all the points in the dataset.

Luckily I tend to have the lat/longs for two points either side of a station so I can work out the midpoint as an approximation for the missing one.

I found an article which defines a formula we can use to do this and there’s a StackOverflow post which has some Java code that implements the formula.

I wanted to find the midpoint between Surrey Quays station (51.4931963543,-0.0475185810) and a point further south on the train line (51.47908,-0.05393950). I wrote the following Cypher query to calculate this point:

WITH 51.4931963543 AS lat1, -0.0475185810 AS lon1, 
     51.47908 AS lat2 , -0.05393950 AS lon2

WITH radians(lat1) AS rlat1, radians(lon1) AS rlon1, 
     radians(lat2) AS rlat2, radians(lon2) AS rlon2, 
     radians(lon2 - lon1) AS dLon

WITH rlat1, rlon1, rlat2, rlon2, 
     cos(rlat2) * cos(dLon) AS Bx, 
     cos(rlat2) * sin(dLon) AS By

WITH atan2(sin(rlat1) + sin(rlat2), 
           sqrt( (cos(rlat1) + Bx) * (cos(rlat1) + Bx) + By * By )) AS lat3,
     rlon1 + atan2(By, cos(rlat1) + Bx) AS lon3

RETURN degrees(lat3) AS midLat, degrees(lon3) AS midLon
╒═════════════════╤═════════════════════╕
│midLat           │midLon               │
╞═════════════════╪═════════════════════╡
│51.48613822097523│-0.050729537454086385│
└─────────────────┴─────────────────────┘

The Google Maps screenshot on the right-hand side shows the initial points at the top and bottom and the midpoint in between. It’s not perfect; ideally, I’d like the midpoint to be on the track, but I think it’s good enough for the purposes of the algorithm.

Now I need to go and fill in the lat/longs for my location-less stations!

It’s easier than you think to extend DevOps practices to SQL Server with Redgate tools. Discover how to introduce true Database DevOps, brought to you in partnership with Redgate

Topics:
cypher ,algorithm ,neo4j ,database

Published at DZone with permission of Mark Needham, DZone MVB. See the original article here.

Opinions expressed by DZone contributors are their own.

{{ parent.title || parent.header.title}}

{{ parent.tldr }}

{{ parent.urlSource.name }}