Can the DBSCAN algorithm create a cluster with less than minPts?

Question

I've just written the DBSCAN algorithm and I am wondering if the DBSCAN algorithm can allow a cluster where the number of points in the cluster is less than the minPts parameter used.

I've been using http://people.cs.nctu.edu.tw/~rsliang/dbscan/testdatagen.html to verify my implementation and it seems to work fine, just running into this issue.

I'm running some simulations against a sample data set and I've been using a minPts of 3. The DBSCAN algorithm will often create clusters of 2 points (never 1 though) from the data set. Is this by design or have I messed up the implementation?

Some sample data, eps = 0.1, minPts = 3.

 0.307951851891331 0.831249445598223
 0.0223402371734102 0.352948855307395
 0.780763753587736 0.691021379870838
 0.950537940464233 0.849805725668467
 0.66559538881555 0.603627873865714
 0.983049284658883 0.320016804300256
 0.710854941844407 0.646746252033276
 0.404260418566065 0.610378857986247
 0.740377815785062 0.899680181825385
 0.430522446721104 0.597713506593236
 0.0365937198682659 0.109160974206944
 0.378702778545536 0.115744969861463
 0.765229786171219 0.568206346858389
 0.760991609078362 0.59582572271853
 0.970256112036414 0.480310371834929
 0.110018607280226 0.541528500403058
 0.679553015939683 0.951676915377228
 0.730563320094051 0.806108465793593
 0.30542559935964 0.500680956757013
 0.740971321585109 0.670210885196091
 0.877572476806851 0.221948942738561
 0.882196086404005 0.674841667374057
 0.808923079077584 0.740714808339586
 0.935197343553974 0.438659039064617
 0.283511740287539 0.271373094185895
 0.0740317893559261 0.602333299630477
 0.30702819223843 0.0683579570932118
 0.31839294653311 0.198790877684388
 0.452546667052687 0.906595267311947
 0.587719069136176 0.212557406729347
 0.930029770792476 0.354712217745703
 0.879549613632052 0.185285016980621
 0.493609266585488 0.441520784255825
 0.640463788360573 0.759178026467179
 0.916182931939225 0.598151952772472

Output clusters:

(Cluster: 1 { 0.780763753587736,0.691021379870838 }, { 0.66559538881555,0.603627873865714 }, { 0.710854941844407,0.646746252033276 }, { 0.765229786171219,0.568206346858389 }, { 0.760991609078362,0.59582572271853 }, { 0.740971321585109,0.670210885196091 }, { 0.882196086404005,0.674841667374057 }, { 0.808923079077584,0.740714808339586 }, { 0.916182931939225,0.598151952772472 })
(Cluster: 2 { 0.983049284658883,0.320016804300256 }, { 0.970256112036414,0.480310371834929 }, { 0.935197343553974,0.438659039064617 }, { 0.930029770792476,0.354712217745703 })
(Cluster: 3 { 0.404260418566065,0.610378857986247 }, { 0.430522446721104,0.597713506593236 })
(Cluster: 4 { 0.740377815785062,0.899680181825385 }, { 0.679553015939683,0.951676915377228 }, { 0.730563320094051,0.806108465793593 })
(Cluster: 5 { 0.378702778545536,0.115744969861463 }, { 0.30702819223843,0.0683579570932118 })
(Cluster: 6 { 0.110018607280226,0.541528500403058 }, { 0.0740317893559261,0.602333299630477 })
(Cluster: 7 { 0.877572476806851,0.221948942738561 }, { 0.879549613632052,0.185285016980621 })
(Cluster: 8 { 0.283511740287539,0.271373094185895 }, { 0.31839294653311,0.198790877684388 })

Answer 1

Yes.

A cluster in DBSCAN is only guaranteed to consists of at least 1 core point.

Since border points that belong to more than 1 cluster will be "randomly" (usually: first-come) assigned to one of the clusters, a core point may not be able to retain/get all its neighbors. Thus, it may be smaller than minPts.

One dimensional example: minPts=4, epsilon=1:

0.0 0.5 1.0     2.0     3.0 3.5 4.0
 |-------X-------|                   Core point at 1.0, Cluster 1.
                 |-------X-------|   Core point at 3.0, Cluster 2.

All other points are not core points. Since the point 2.0 is not a core point, it does not connect the two clusters. One of the clusters will have 4 members, the other only 3.

The closest thing to a reference implementation probably is the DBSCAN implementation in ELKI. Maybe you should check that your results match the results of ELKI. Because in your data set, I believe you do have some programming error, too.