Implementing Colour Refinement by RheyaM · Pull Request #905 · digraphs/Digraphs

RheyaM · 2026-03-11T15:20:54Z

Implementation of DigraphColourRefinement. Takes a digraph and returns a colouring (A list where value at i^th position is the colour of the i^th node), such that all nodes with the same colour have the same configuration of coloured neighbours. That is, if two nodes have the same colour, they will have the same number of neighbours of colour x, and colour y, etc. DigraphColourRefinement considers out and in neighbours separately.

If two digraphs produce two different colouring under Colour Refinement, we can know that they are not isomorphic.

Now includes optimisation step

… and without optimisation

…marking

james-d-mitchell · 2026-03-18T13:47:04Z

+        Label="for a digraph"/>
+  <Returns>A list of lists of integers.</Returns>
+  <Description>
+    Colour Refinement is a method of colouring a digraph such that for a colour, every 


Inappropriate capitalisation.

james-d-mitchell · 2026-03-18T13:47:44Z

+  <Description>
+    Colour Refinement is a method of colouring a digraph such that for a colour, every 
+    node with that colouring has an identical configuration of coloured neighbours. That is, 
+    all nodes of colour <A>q</A> have the same number of neighbours of colour <A>x</A>, and colour y</A>, etc.


Suggested change

all nodes of colour <A>q</A> have the same number of neighbours of colour <A>x</A>, and colour y</A>, etc.

all nodes of colour <A>q</A> have the same number of neighbours of colour <A>x</A>, and colour <A>y</A>, etc.

but also this should probably say: "all nodes with the same colour have equal numbers of nodes of every colour"

james-d-mitchell · 2026-03-18T13:48:09Z

+    Colour Refinement is a method of colouring a digraph such that for a colour, every 
+    node with that colouring has an identical configuration of coloured neighbours. That is, 
+    all nodes of colour <A>q</A> have the same number of neighbours of colour <A>x</A>, and colour y</A>, etc.
+    <C>DigraphColourRefinement</C> considers the out neighbours and in neighbours of a node separately.


This should be expanded to say exactly how they are considered separately.

james-d-mitchell · 2026-03-18T13:48:56Z

+    all nodes of colour <A>q</A> have the same number of neighbours of colour <A>x</A>, and colour y</A>, etc.
+    <C>DigraphColourRefinement</C> considers the out neighbours and in neighbours of a node separately.
+    <P/>
+    This involves recolouring the digraph each iteration until it is 'refined'. It returns the colouring as a


What is This? What iterations are you referring to? What does 'refined' mean in this context? This is too imprecise.

james-d-mitchell · 2026-03-18T13:49:43Z

+    <P/>
+    This involves recolouring the digraph each iteration until it is 'refined'. It returns the colouring as a
+    list where the value at the ith position is the colour of node i. For two digraphs with different colourings, 
+    we can be sure that they are not isomorphic. However, identical colourings for two digraphs does not necessarily 


I'm not sure I believe this, the colouring as returned by this function depends on the order the vertices appear in the digraph, but isomorphism does not.

james-d-mitchell · 2026-03-18T13:50:18Z

+  j, P, v, Out, In, Sets, pair, current, currentPair, newSet, colour;
+
+  # Or just remove loops?
+  if not DigraphNrLoops(D) = 0 then


Suggested change

if not DigraphNrLoops(D) = 0 then

if DigraphHasLoops(D) then

james-d-mitchell · 2026-03-18T13:51:41Z

+  C := [];
+  for v in DigraphVertices(D) do
+    C[v] := 1;
+  od;


Suggested change

C := [];

for v in DigraphVertices(D) do

C[v] := 1;

od;

C := List(DigraphNrVertices(D), x -> 1);

mtorpey

A few suggestions for changes here! Great work, and this stuff shouldn't take too much longer.

mtorpey · 2026-04-01T16:05:58Z

+        largest := 1;
+        for j in [1 .. Length(colour_cell)] do
+          if Length(colour_cell[j]) > Length(colour_cell[largest]) then
+            largest := j;
+          fi;
+        od;


I think this might be replacable with largest := PositionMaximum(colour_cell, Length). Could you try this?

Done!

mtorpey · 2026-04-01T16:19:47Z

Hi @RheyaM! Well done on the quick fixes.

Once you've tried that PositionMaximum fix above, it's just about resolving that "canonical" thing. If you can have a quick think about that and then leave a comment here summarising to what extent it's canonical, and what "canonical" means, I think we can merge this.

Joseph-Edwards

I've had quite a thorough look at this now, and it seems to me like the colouring that get's returned is canonical; as far as I can tell, we don't use any information of the vertex labels when deciding which cells/colours to refine with respect to. If we could mention this in the documentation (and also do what Michael suggested in his comment) then I would be happy for this to be merged.

Nice work @RheyaM!

azure-pipelines · 2026-04-15T13:37:41Z

Azure Pipelines: 1 pipeline(s) were filtered out due to trigger conditions.

azure-pipelines · 2026-04-15T13:43:44Z

Azure Pipelines: 1 pipeline(s) were filtered out due to trigger conditions.

mtorpey

Sorry, nearly done

mtorpey · 2026-04-15T15:29:47Z

    colour of node i. The time complexity of this algorithm is <M>O(n^2 log n)</M>.
    <P/>
+    Because the labels of the vertices are not used in any capacity during the refinement process (I.e. to determine which cells 
+    to refine), the colouring produced is canonical. This means that for two isomorphic digraphs, they would produce the same colouring. 


Sorry to keep fussing, but one more thing that should be made clear here.

The colouring is canonical, but my understanding is that the output of the function is not canonical (because it's a list that uses the ordering of vertices). This should be made clear, because as written it sounds like isomorphic digraphs have the same output.

Have I got this right

I added an explanation for this, but I'm not sure if it comes across clearly so I included two isomorphic digraphs in the examples and their 'identical' colourings (I just swapped vertices 1 and 8, and the output swaps the colours of those vertices as well). Is that okay/does it make sense?

mtorpey · 2026-04-15T15:29:59Z

    <C>DigraphColourRefinement</C> returns the colouring as a list where the value at the ith position is the 
    colour of node i. The time complexity of this algorithm is <M>O(n^2 log n)</M>.
    <P/>
+    Because the labels of the vertices are not used in any capacity during the refinement process (I.e. to determine which cells 


Lower case i.e.

…efinement

rm387 added 5 commits February 25, 2026 14:13

ColourRefinement non-optimised method

a9431e7

Tests added for DigraphColourRefinement()

d965a01

DigraphColourRefinement considering out and in neighbours separately,…

57ce239

… and without optimisation

Adjusting output of DigraphColourRefinement and moving code to oper.tst

4935b0a

Created separate optimised DigraphColourRefinement() method for bench…

c9af0ab

…marking

james-d-mitchell reviewed Mar 18, 2026

View reviewed changes

Comment thread doc/oper.xml

james-d-mitchell reviewed Mar 18, 2026

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rm387 added 6 commits March 24, 2026 22:52

Optimisation fixed

182c019

Optimisation fixed (real this time)

b0f491b

Removing unoptimised code + updating .xml files

f33b4cb

Updated Colour Refinement tests

99920ba

Small fixes to .xml file

4bef5bb

Overlong line >:(

f6b9267

RheyaM marked this pull request as ready for review March 25, 2026 15:26

Cleaning up Colour Refinement

42c8b89

mtorpey mentioned this pull request Apr 1, 2026

Uses of colour refinement #916

Open

mtorpey previously requested changes Apr 1, 2026

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Changes made to ColourRefinement + tests added

16eaa7b

mtorpey reviewed Apr 1, 2026

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Joseph-Edwards reviewed Apr 1, 2026

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Using PositionMaximum and updated explanations on canonical colouring

912ec56

Trailing whitespace :(

8e0e5b4

mtorpey requested changes Apr 15, 2026

View reviewed changes

Updating oper.xml to explain canonical colouring vs output of ColourR…

7e37cbe

…efinement

	all nodes of colour <A>q</A> have the same number of neighbours of colour <A>x</A>, and colour y</A>, etc.
	all nodes of colour <A>q</A> have the same number of neighbours of colour <A>x</A>, and colour <A>y</A>, etc.

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RheyaM commented Mar 11, 2026 •

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