In order to bridge two edge loops having an unequal number of edges the diamond pattern can be used. Video tutorials by Jonathan Williamson.

Are there more patterns that preserve the quads?

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  1. Select 3 vertices (Image 1)
  2. Merge at Center
  3. Select another 3 vertices connected to the ones selected in step 1.
  4. Merge the in 3. selected vertices. (Image 2)
  5. Remove the superfluous (single edge) edge loop (Image 3, to illustrate how the edge flow is redirected)

1 Answer 1


Usually one can catch such topologycal tricks by watching the vertices (poles) neighbour count. In your example there is a cube which has one vertex with 5 neighbour and another with 3. The following example shows one where these poles are neighbours. As Hount House notet in his comment, sometimes they are called N-poles and E-poles.


When avoiding triangles, you can reduce or increase edge amount somewhere by the cost of doing it somewhere else. In your example on the same side you loose one and another edge. In my example I added a new edge in the right side, and had to redirect the loop Flow to the top. When simply removing an edge loop, you take one edge from one side and one from the front side.

Generall, when using only quads on manifold meshes, every sub-mesh of it (for example strangly subdivided planes like these) must have even number of border edges (only one face connected). So a hole with odd number of edges cannot be filled with quads.

Maybe I am getting too wild now. I am currently working on a trees retopology, where you can see, that I mostly only have the usual poles or E-poles.


A similar example to yours (two edges taken from bottom side), but with another loop Flow:


Must note the tutorial in this related question. I Recommend to watch all episode.

  • $\begingroup$ You have also poles with 3 neighbors. Someone once coined the phrase E-poles and N-poles because E has 5 and N has 3 strokes. It's the old 'cube on a subdivided plane' topology where you have E-poles at the base and N-poles at the top. $\endgroup$ Sep 19, 2013 at 1:55

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