For an assignment, I have used iPython to create the dendrogram below, using Ward's method and Euclidean distance, from the following data: $$a=(0,0)$$ $$b=(1,2)$$ $$c=(3,4)$$ $$d=(4,1)$$ $$e=(2,2)$$ enter image description here where dist({a},{b,e}) = 2.88, and dist({a,b,e},{c,d})=4.27.

How are these values derived? I've tried using the recursive method given here, but am not getting the same results.

Any help muchos appreciated.

I know that dist(a,b)=$\sqrt{5}$, dist(a,e)=$\sqrt{8}$, so using the formula on the wikipedia page, I have dist({a},{b,e})=$\frac{2}{3}\sqrt{5} + \frac{2}{3}\sqrt{8} - \frac{1}{3} = 3.04$. Where am I going wrong?

  • $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ Dec 15, 2016 at 14:26
  • $\begingroup$ done - btw this isn't a homework question, it's a question I've put together myself for a quiz $\endgroup$
    – Harr
    Dec 15, 2016 at 14:35
  • $\begingroup$ The Talk page at the wikipedia article discusses all the necessary details. Please see also in stats.stackexchange.com/a/217742/3277. And remember Ward's method operates on squared distances. $\endgroup$
    – ttnphns
    Dec 15, 2016 at 15:17

2 Answers 2


Having banged my head on the wall for the last 2 hours on this, I feel your pain.

The result is the square root of the increase in within-cluster sum of squares (vs. cluster means), multiplied by $\sqrt{2}$ for some reason.

$$ \sqrt{ 2 \left( \sum_{i \in (C_1 \cup C_2)} \lVert \vec x_i - \bar x_{C_1 \cup C_2} \rVert^2 - \sum_{i \in (C_1)} \lVert \vec x_i - \bar x_{C_1} \rVert^2 - \sum_{i \in (C_2)} \lVert \vec x_i - \bar x_{C_2} \rVert^2 \right) } $$

For $C_1 = \{ a, b, e \}, C_2 = \{ c,d \}$ the result is $4.27$ as you require.

As for how exactly they are derived, depends on the specifics of the algorithm in use.

  • 1
    $\begingroup$ Michael, the Talk page of Wikipedia's Ward method article explains that sqrt(2) $\endgroup$
    – ttnphns
    Dec 19, 2018 at 16:05

The right is: $$dist(\{a\},\{b,e\})=\left(\frac23\right)\cdot 5+\left(\frac23\right)\cdot 8–\frac13\cdot 1=\frac{25}3$$ Then $$dist(\{a\},\{b,e\})=\sqrt{\frac{25}3}=2.88$$


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