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I'm really confused. In a book ISBN: 978-0-470-27680-8 is written:

The Euclidean distance can be generalized as a special case of a family of metrics, called Minkowski distance or L p norm, defined as, $ D(\mathbf{x}_i,{\mathbf x}_j)=\left(\sum_{l=1}^d |x_{il}-x_{jl}|^{1/p}\right)^p \tag{1} $

Is it correct? In other sources Minkowski distance is defined as:

$ \left( \color{red}{\sum_{i=1}^n} |x_{i}-y_{i}|^p\right)^{1/p} \tag{2} $

Which one is correct? (notice to powers)


J.D.'s edit: The highlighted red part was missing from a previous edit.. Originally, OP included the following two images:


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    I think it's about similarity (1st formula) and dissimilarity ($2$nd formul$a$) me$a$sures.2012-04-06

4 Answers 4

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A metric (or distance) and a norm are two different things. You can use a norm to define a metric, but not necessarily the other way around. The Minkowski metric is the metric induced by the $L_p$ norm, that is, the metric in which the distance between two vectors is the norm of their difference.

Both of these formulas describe the same family of metrics, since $p\to1/p$ transforms from one to the other. So if your question is "which of these two expressions describes the family of Minkowski metrics", the answer is both. However, if the question is "which of them gives the metric induced by the $L_p$ norm", it's the second one,

$\left(\sum_{i=1}^n|x_i-y_i|^p\right)^{1/p}\;.$

You can remember that by remembering that $L_2$ is the standard Euclidean norm, in which the exponent $2$ is on the inside and the exponent $1/2$ is on the outside.

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    @Reza: You'll have to point out what part of the answers you've already received you don't understand; otherwise there's no point in adding yet another one. Both Emre's answer and mine describe the difference between a norm and a metric as clearly as I can describe it; unless you say what part of that is unclear to you, there's nothing more I can say. Your book is simply wrong in conflating a distance and a norm.2012-04-05
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If you are asking about the difference between a metric and a norm:

A norm and a metric are two related but different things. Generally speaking, a norm is a more "vector space" concept than metrics.

A norm assigns value to a SINGLE vectors (its length) in a vector space, while metric assigns value to TWO elements (their distance) in a metric space (which is not necessarily a vector space). Of course, in a vector space, a metric system can always be induced from a norm system by defining ||x-y|| as d(x,y).

Also, a norm must have Positive Homogeneity and Translation Invariance, both of which come quite naturally for a vector space, while a metric doesn't need to.

Not sure if this answers your question?

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    Thanks a lot. Your answer is a compliant2012-04-07
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The Euclidean distance is the $L^2$ norm of the difference vector, and thus a special case as claimed. The first definition is correct; I can't read the second one.. See Wikipedia for a longer explanation.

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Both are (almost) right. And they say the same thing. Just notation adapted to the particular problem.

In

$ D(\mathbf{x}_i,{\mathbf x}_j)=\left(\sum_{l=1}^d |x_{il}-x_{jl}|^{1/p}\right)^p \tag{1} $

we have points ${\mathbf x}_i$ and ${\mathbf x}_j$ in $d$-space, where ${\mathbf x}_i$ has coordinates $x_{i1},\dots,x_{id}$ and ${\mathbf x}_j$ has coordinates $x_{j1},\dots,x_{jd}$

But in fact this one should be $ D(\mathbf{x}_i,{\mathbf x}_j)=\left(\sum_{l=1}^d |x_{il}-x_{jl}|^{p}\right)^{1/p} $
(are you sure you copied (1) correctly?)

In

$ \left( {\sum_{i=1}^n} |x_{i}-y_{i}|^p\right)^{1/p} \tag{2} $

we have points $x$ and $y$ in $n$-space, where $x$ has coordinates $x_1,\dots,x_n$ and $y$ has coordinates $y_1,\dots,y_n$.