cluster analysis - Finding connected components in an array

I have an $w\times h$ array with non-negative integers.

I wish to partition the set of coordinates $(c,r)$ with $1\leq c \leq w$ and $1\leq r \leq h$, in subsets, so that each subset corresponds to a cluster of the same value in the array.

Two coordinates are neighbouring if they either are on top of another, or one is down-left of the other. This corresponds to finding clusters in a parallelogram.

belisarius was nice to provide a picture:

I have some code to find one connected cluster in an array (I call it gt):

FloodFillExtractTile[gt_, {sr_, sc_}] := 
 Module[{r, c, toExplore, visited = {}},
  toExplore = {{sr, sc}};

  While[Length@toExplore > 0,
   (* Pop *)
   {r, c} = Last@toExplore;
   toExplore = Most[toExplore];
   AppendTo[visited, {r, c}];
   (* Down-left *)

   If[c > 1 && r < h && 
     gt[[r + 1, c - 1]] == gt[[r, c]] && ! 
      MemberQ[visited, {r + 1, c - 1}],

    AppendTo[toExplore, {r + 1, c - 1}];
    ];
   (* Down-right *)

   If[ r < h && 
     gt[[r + 1, c]] == gt[[r, c]] && ! MemberQ[visited, {r + 1, c}],
    AppendTo[toExplore, {r + 1, c}];
    ];
   (* Up-right *)


   If[r > 1 && c < w && 
     gt[[r - 1, c + 1]] == gt[[r, c]] && ! 
      MemberQ[visited, {r - 1, c + 1}],
    AppendTo[toExplore, {r - 1, c + 1}];
    ];
   (* Up-left *)

   If[ r > 1 && 
     gt[[r - 1, c]] == gt[[r, c]] && ! MemberQ[visited, {r - 1, c}],
    AppendTo[toExplore, {r - 1, c}];

    ];
   ];
  Return@visited;
];

Then FloodFillExtractTile[gtp, {1, 1}] gives the component connected to the upper left hand corner. However, this method feels ugly, and extending it to all components feels even uglier.

I was looking at Gather, but the problem is that it wants all points in a cluster to be equal, see for example Gather dependency on list order?

Edit: So this is the type of arrays I am looking at. The $6$, the $3$'s, the $2$'s and the $0$'s are in one component respectively, but there are two components with $1$'s.

Being neighbours means being adjacent down-left, down-right, up-left and up-right.

Now, the rows is stored just like regular rows in a rectangular matrix, so that is why this translates to a bit strange criteria for being neighbours.

This is the best code I have so far, first extract points with the same value, then do connected-component analysis on those parts.

GetGTTiles[gtp_] := 
  Module[{testSame, testEdge, h, w, pts, sameClusters, getEdges, 
    tiles},
   {h, w} = Dimensions[gtp];
   pts = Join @@ Table[{r, c}, {r, h}, {c, w}];
   testSame[{r1_, c1_}, {r2_, c2_}] := (gtp[[r1, c1]] == 
      gtp[[r2, c2]]);
   testEdge[{r1_, c1_}, {r2_, 
      c2_}] := (gtp[[r1, c1]] == 

       gtp[[r2, c2]]) &&
     ((c1 == c2 && 
         Abs[r1 - r2] <= 1) || (c1 == c2 - 1 && 
         r1 == r2 + 1) || (c1 == c2 + 1 && r1 == r2 - 1));
   sameClusters = Gather[pts, testSame];
   getEdges[clust_] := 
    Join @@ Outer[If[testEdge[#1, #2], #1 -> #2, Sequence @@ {}] &, 
      clust, clust, 1];
   tiles = 
    Join @@ (ConnectedComponents[Graph@getEdges[#]] & /@ sameClusters);

   Return@tiles;
   ];

This is the output I expect for the example given (5 clusters found):

{{{1, 1}}, {{1, 2}, {2, 1}}, 
{{1, 3}, {2, 3}, {3, 2}, {4, 1}, {3, 1}, {2, 2}}, 
{{1, 4}}, 
{{2, 4}, {3, 4}, {4, 4}, {5, 4}, {5, 3}, {4,3}}, 
{{3, 3}, {4, 2}, {5, 2}, {5, 1}}}

EDIT: So this is the final code, based on belisarius solution:

GTTiles[gtp_List] := Module[{fromEuclidean, toEuclidean,
    getOneTile, elements, elmPos, pts, tile, tiles},

   (* This is used to changefrom different coordinate systems. *)

   fromEuclidean[{r_, c_}] := {r, (c - r)/2 + 1};
   toEuclidean[{r_, c_}] := {r, 2 c + r - 2};

   getOneTile[pts_List, maxDist_?NumericQ] := Module[{f},

     f = Nearest[pts];
     FixedPoint[
      Union@Flatten[f[#, {Infinity, maxDist}] & /@ #, 1] &, {First@
        pts}]];

   elements = Union @@ gtp;
   elmPos = (toEuclidean /@ Position[gtp, #]) & /@ elements;
   (* This is really strange code. *)
   tiles = Flatten[Flatten[
        Reap[NestWhile[Complement[#,

             Sow@getOneTile[#, N@Sqrt@2]] &, #, # != {} &]][[2]], 
        1] & /@ elmPos, 1];
   tiles = Map[fromEuclidean, tiles, {2}];
   Return@tiles;
   ];

Answer

l1 = {{6, 3, 2, 1}, {3, 2, 2, 0}, {2, 2, 1, 0}, {2, 1, 0, 0}, {1, 1, 0, 0}};
l = Riffle[#, ""] & /@ l1;
els = Union @@ l;
par = MapIndexed[PadRight[PadLeft[#1, #2[[1]] + Length@#1 - 1, ""], 

                          Length@l + Length@#1 - 1, ""] &, l];
eachElm = Position[par, #] & /@ els;
getOneCluster[pts_List, maxDist_?NumericQ] :=
  Module[{f},
   f = Nearest[pts];
   FixedPoint[Union@Flatten[f[#, {Infinity, maxDist}] & /@ #, 1] &, {First@pts}]];
clusters = 
  Flatten[Flatten[
      Reap[NestWhile[
         Complement[#, 

           Sow@getOneCluster[#, N@Sqrt@2]] &, #, # != {} &]][[2]], 1] & /@ eachElm, 1];
Grid[par, 
 ItemStyle -> {Automatic, Automatic, Flatten@MapIndexed[#1 -> Hue[#2[[1]]/3] &, 
                                                       clusters, {2}]}]