Let's look at this example:

simple.data = data.frame(
  x = c(0,1,0.5),
  y = c(0,0,0.9)

par(mfrow = c(1,2))
plot(simple.data, xlab = "Dimension 1", ylab = "Dimension 2")
text(simple.data[1,], labels = 1, pos = 3)
text(simple.data[2,], labels = 2, pos = 3)
text(simple.data[3,], labels = 3, pos = 1)

eucl = dist(simple.data, method = "euclidean")

#         1        2
# 2 1.000000         
# 3 1.029563 1.029563

agglo = hclust(eucl, method = "centroid")

#          1        2
# 2 1.000000         
# 3 0.779563 0.779563

par(mar = c(2,4,1,1))
plot(as.dendrogram(agglo), main = "Dendrogram", ylab = "Height", ylim = c(0,1))

enter image description here

  1. I know that the distances between clusters can be computed using the Lance und Williams Formula, e.g. $$D(A \cup B,C)=\alpha_1 d(A,C)+\alpha_2 d(B,C)+ \beta d(A,B) + \gamma |d(A,C)-d(B,C)|.$$ with $\alpha_1 = \tfrac{|A|}{|A|+|B|}, \alpha_2 = \tfrac{|B|}{|A|+|B|}, \beta = \tfrac{|A||B|}{(|A|+|B|)^2}, \gamma = 0$ for centroid linkage (see also https://de.wikipedia.org/wiki/Hierarchische_Clusteranalyse#Lance_und_Williams_Formel).

  2. I also know that, in R, the dendrogram and the function cophenetic() computes the distances between two clusters with this formula, e.g. after merging the two closest points (in example above: point 1 and point 2), the distance between the cluster that consists of point 1 and 2 and the second cluster that only consists of point 3 is, according to the Lance und Williams Formula with $\alpha_1 = \alpha_2 = 0.5, \gamma = 0$ and $\beta = 0.25$: 0.5*1.029563 + 0.5*1.029563 - 0.25*1 = 0.779563. Therefore, the dendrogram shows a merge of those two clusters at "Height" 0.779563.

  3. However, since "in centroid method, the distance between two clusters is the distance between the two cluster centroids", I would have computed this distance differently, namely:

    • compute centroid of point 1 and 2, which is at (0.5, 0)
    • "centroid" of point 3 is point 3 itself (located at (0.5, 0.9), see plot).
    • euclidean distance between the two cluster centroids is therefore sqrt((0.5-0.5)^2+(0-0.9)^2) = 0.9, which is not the same distance as computed with the Lance und Williams Formula.

So, my questions is: Why do we use the Lance und Williams Formula (e.g. the 0.779563) to plot the dendrogram and do not use "the distance between the two cluster centroids", which is 0.9?

  • 1
    $\begingroup$ I think you need to re calculate your tree. This dengrogram is not drawn well. It has a reverse branch. Have a look at the end of this page: link Also check Dendrogram and some related areas in MATLAB website. They have great detailed explanations which if I remember correctly, your answer is also there. Sorry, I only know these through MATLAB, not R. $\endgroup$
    – PM0087
    Jun 6, 2016 at 14:01
  • 2
    $\begingroup$ Centroid method (and also median, Ward) expect that the input distances are squared euclidean, their L-W formula implies it. Either square your euclidean distances or tell the program that they aren't square (if there's such an option). $\endgroup$
    – ttnphns
    Jun 6, 2016 at 15:52
  • $\begingroup$ It seems that if I use squared euclidean distance agglo = hclust(eucl^2, method = "centroid"), I still have to take the root of the cophenetic distances sqrt(cophenetic(agglo)) to get 0.9. However, I did not see a possibility how to do this in the dendrogram... $\endgroup$
    – Giuseppe
    Jun 7, 2016 at 6:51
  • $\begingroup$ Well this does what I want: agglo = hclust(dist(simple.data, method = "euclidean")^2, method = "centroid");agglo$height = sqrt(agglo$height);plot(as.dendrogram(agglo), ylim = c(0,1)). I have to say that for interpretation purposes this is really strange: In the first step, the $y$ axis in the dendrogram can be interpreted as the euclidean distance between the two points. However, when we merge clusters, the $y$ axis of the dendrogram does not show the euclidean distances... $\endgroup$
    – Giuseppe
    Jun 7, 2016 at 7:17
  • 1
    $\begingroup$ Once again, by points. L-W formula for centroid method is formulated w.r.t. squared euclidean distance (s.e.d.); it is done for convenience and speed. Centroid method needs s.e.d. as the input distance matrix. Logically, s.e.d. is what should be plotted on a dendrogram. Whether it is possible to input nonsquared e.d. (and still get the right result!) and/or to plot dendrogram with root taken from the plotted linkage coefficients - depends entirely on your function additional options, these are features that might exist. $\endgroup$
    – ttnphns
    Jun 7, 2016 at 7:41

1 Answer 1


According to the book Introduction to Information Retrieval. Christopher D. Manning, Prabhakar Raghavan and Hinrich Schütze:

centroid clustering is not monotonic. So-called inversions can occur: Similarity can increase during clustering as in the example in Figure 17.12, where we define similarity as negative distance.

this seems this is an typical behavior.

Clearly, 1 and 2 are the closest points at distance 1. So that is the best possible merge here.

The squared Euclidean distance of the centroids is $0^2+0.9^2=0.81$.

With Lance-Williams: The squared Euclidean distance is $d(1,3)=d(2,3)=0.9^2+0.5^2 = 1.06$, while d(1,2)=1. So we get $$\frac121.06+\frac121.06-\frac141=0.81.$$


  1. The result is correct - if you use squared Euclidean distances. I don't think you can use the (efficient!) Lance-Williams approach with other distances. Plus, the centroid makes most sense with squared errors, too.
  2. Inversions in centroid linkage do occur. Even if you would use non-squared Euclidean distance. With regular Euclidean distance, you would still merge 1,2 at distance 1, and then merge 1,2,3 at $\sqrt{0.81}=0.9$; which is less than 1.

But don't ask me for a proof that the Lance-Williams and "direct" definition are equivalent. Apparently the proof can only work for squared Euclidean? I guess it is similar to the derivation of Ward's, which appears to be the "properly weighted" version of centroid linkage.


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