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So given 4 elements, there are 15 partitions.

The formula for d is $d((x_1, x_2),(y_1, y_2)) = [(x_1−y_1)^2+ (x_2−y_2)^2]^{1/2}.$

I've calculated the distance for one of the partitions is $\{\{(0,1),(2,1)\},\{(3,1),(5,5)\}\}$.

The distance is $2 + (2)*(5^{1/2})$.

It was calculated by taking doing {(DISTANCE between (0,1),(2,1)) + (DISTANCE between (3,1),(5,5))}

and for another partition that has 4 singleton elements: $\{ \{(0,1)\}, \{(3,1)\}, \{(5,5)\}, \{(2,1)\} \}$

Intra-block distance for this partition would go like this: $$ d((0,1),(3,1)) +d((0,1,),(5,5)) +d((0,1),(2,1)) + d((3,1),(5,5))\\ + d((3,1),(2,1)) + d((5,5),(2,1))$$

My question is, for a partition that has 3 elements that has 2 singleton pairs and 1 element that has 2 coordinates like so: $\{\{(0,1),(2,1)\},\{(3,1)\},\{(5,5)\}\}$, how do I calculate the intrablock distance for this partition.

quid
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Abhishek
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  • https://math.meta.stackexchange.com/q/5020/306553 mathjax reference for typing maths. – Siong Thye Goh Sep 01 '17 at 05:58
  • Hi, can you give me a definition or a reference to the intra-block distance? Somehow I can't find it. I'm familiar with $d(A,B)=\inf{d(a,b)~:~a\in A,~b\in B}$ which yields $d({x,\hat{x}},{y})=\min{d(x,y),~d(\hat{x},y)}$. But this might be wrong for the "intra-block distance"... – Mundron Schmidt Sep 01 '17 at 06:41
  • Show all the partitions of {(0,1),(3,1),(5,5),(2,1)}. Each element is a pair (x1, x2) ∈ R2 . A standard metric d:R2×R2→R is :d((x1, x2),(y1, y2)) = [(x1−y1)^2+ (x2−y2)^2]^0.5 (Just the l^2−norm or Euclidean distance). Create a table that calculates the intra block distances between all unique pairs. I’ve calculated one shown below. – Abhishek Sep 01 '17 at 07:01
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    You defined the metric as in your question, but you didn't defined the intra block distance. There should be a definition of the distance between sets or the distance between partitions. – Mundron Schmidt Sep 01 '17 at 07:23

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