# Examples of non symmetric distances

It is well known that the symmetric property is $d(x,y)=d(y,x)$ is not necessary in the definition of distance if the triangle inequality is carefully stated. On the other hand there are examples of functions satifying

(1) $d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$

(2) $d(x,y)\leq d(x,z)+d(z,y)$

which are not symmetric: in the three point space $(a,b,c)$ take the non-zero values of $d$ as $1=d(a,b)=d(c,b)$, $2=d(b,a)=d(b,c)=d(a,c)=d(c,a)$.

Do you know other examples of "non symmetric distances"? Are there examples on the real numbers, etc.? Are there examples of spaces were every function satisfying (1) and (2) is symmetric?

• I'm not sure I understand your last question. What is a space? Do you mean a set? In that case, any of several constructions shows that any set with at least 2 elements admits an asymmetric metric. – Qiaochu Yuan Feb 23 '11 at 17:29
• I think it's worth mentioning that such spaces are sometimes called "quasimetric spaces" en.wikipedia.org/wiki/Metric_%28mathematics%29#Quasimetrics or "asymmetric metric spaces". (When you google for these phrases, you will find quite a few scientific papers using them.) The papers Wilson: On quasi-metric spaces scholar.google.com/scholar?cites=6007326225178920620 and Kelly: Bitopological spaces scholar.google.com/scholar?cites=17692636850287560255 are often mentioned as references. – Martin Sleziak Mar 29 '11 at 17:20
• I lived in a part of Toronto where most of the residential streets were one-way ( which tended to channel most of the traffic onto the 2-way main streets). So the minimum driving distance $d(x,y)$ from $x$ to $y$ was often not equal to $d(y,x).$ – DanielWainfleet Aug 30 '18 at 1:40

A classic example is the circle $S^1$ with metric the length of the shortest clockwise path between $x$ and $y$, but let me say some things in general.

Lawvere once made the point that this is really a much more natural definition of a metric space, since it allows them to be interpreted as a type of enriched category. The triangle inequality then becomes a consequence of composition of morphisms, which is extremely reasonable if you think of distances in a metric space as measuring "the best way to get from $a$ to $b$": clearly the best way to get from $a$ to $b$ to $c$ is at most as good at the best way to get from $a$ to $c$, and this is precisely the (asymmetric) triangle inequality.

On the other hand, there is no reason in general that the best way to get from $a$ to $b$ has to look anything like the best way to get from $b$ to $a$. This is the sense in which the symmetry requirement is unnatural. There is also no reason in general that it should be possible to get from $a$ to $b$ at all! This is the sense in which the requirement that distances be finite is unnatural. Finally, there is no reason in general that it should not be possible to instantaneously get from $a$ to $b$ (in other words, that $a$ and $b$ be isomorphic); this is the sense in which the requirement that distances be positive-definite is unnatural.

Here is how to use that idea to generate a large class of examples. Let $(M, d)$ be a metric space and let $h : M \to \mathbb{R}$ be a function. Define a new metric $$d'(x, y) = \begin{cases} d(x, y) + h(y) - h(x) \text{ if } h(y) \ge h(x) \\\ d(x, y) \text{ otherwise} \end{cases}.$$

Intuitively, $h$ is a potential (e.g. a height if one is thinking of gravitational potential), and the new metric $d'$ penalizes you for going against the potential (e.g. going uphill).

The directed graph example given by mjqxxxx is also a good illustration of this philosophy about metric spaces.

• In the third paragraph, if your geometric intuition is getting in the way pretend that a and b are not points in some space but possible states of a physical system and d(a, b) measures the minimal amount of energy that needs to be put into the system to get it from state a to state b. – Qiaochu Yuan Feb 23 '11 at 17:33

Most distances in real life are going to be more or less asymmetric due to one-way roads, going uphill resp. downhill, different public transportation schedules, congestion, etc.

• and don't forget friendship closeness. I might not be my close friend's close friend. – MaudPieTheRocktorate Jul 5 '18 at 7:49

The Hausdorff distance is a symmetric version of a natural non-symmetric distance.

For any directed graph $G=(V,E)$, we may define a non-symmetric distance as follows: for $x,y\in V$, the directed graph distance is the length of the shortest directed path from $x$ to $y$, or $\infty$ if there is no such path. This clearly satisfies the triangle inequality as stated above. An even more general family of examples is generated by allowing a positive weight, in which case we may take the distance to be the smallest weight of any directed path from $x$ to $y$ (or, if $G$ is infinite, the greatest lower bound of all such weights; in this case we must additionally require all edge weights to be bounded away from zero).

This may or may not be what you're looking for, but I thought the idea was fun, so I'll share:

This will be a non-symmetric metric on $\mathbb{Z}^+$:

Suppose that one is only allowed to move left one unit, and in addition, one is allowed to jump right from one power of two to the next. I.e., to find a sequence of steps to get from 5 to 11, one must travel as follows:

$$5 \to 4 \to8 \to16 \to15 \to 14 \to13 \to12\to11.$$

For a (positive) integer $n$, define $T(n)$ to be largest power of $2$ which is less than or equal to n. Similarly, define $S(n)$ to be the smallest power of $2$ greater than or equal to $n$. Then, if $|\cdot|$ denotes the usual absolute value, we have:

$$d(x,y) = \left\{ \begin{array}{lcc} |x - y| & & x \geq y \\ |x - T(x)| + \log_2(S(y)/T(x)) + |S(y) - y| & & x < y \end{array} \right.$$

If $x \leq y$, one just moves left $|y - x|$ units. If not, one moves from $x$ to the largest power of $x$ less than or equal to $x$. One then jumps $\log_2(S(y)) - \log_2(T(x))$ powers of $2$, and finally moves the remaining $|S(y) - y|$ units. It is not hard to show that this a metric (It clear satisfies condition (i), and to show it satisfies (ii), one just needs to consider the case of $d(x,y)$ when $x > y$).

This metric came to mind in lieu of Cauchy's Proof of the Arithmetic-Geometric mean inequality (http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proof_by_Cauchy).

Let $P(n)$ denote the statement that the AG-mean inequality is true for $n$ numbers. Cauchy's idea was to prove the inequality by first inducting along powers of two (i.e., P($2^k$) is true implies P($2^{k+1}$) is true. Then, he showed that $P(n)$ is true implies $P(n-1)$ is true.

Assuming that we know the arithmetic-geometric mean inequality is true for some step $n$, $d(n,m)$ is the number of steps needed using Cauchy's proof to show $P(m)$ is true, assuming that we know $P(n)$ is true.

What about the following distance function on the reals?

$$\begin{array}{rcl} d(x,y) & = 1, & \mbox{if} \;\; x<y \\ & = 2, & \mbox{if} \;\; x>y \\ & = 0, & \mbox{if} \;\; x=y \end{array}$$

A bit artificial, but it seems to fulfill the requirements, unless I overlooked something.

• Nice! It seems to be possible to generalize your answer as follows: let $X$ be a set with at least two elements $a$ and $b$. Let $d(a,b)=1$ and $d(x,y)=2$ for all other pairs of distinct elements of $X$ (and let $d(x,x)=0$). If this works, every space with at leas two elements has a non symmetric distance. – user7416 Feb 23 '11 at 17:57
• If you want a generalization, see Qiaochu's answer. I think it is more deserving of being attributed the tick mark. – Raskolnikov Feb 23 '11 at 18:01

$d(x,y)=\begin{cases}x-y, \mbox{ if } x\geq y\\1, \mbox{ otherwise }\end{cases}$ will do!

The Kullback-Leibler divergence is another example too. Defined between two probability distributions $$p(x)$$ & $$q(x)$$ it takes the form, $$D_{KL}(p||q) = \int p(x)\log\left(\frac{p(x)}{q(x)}\right)dx$$ It can easily be seen that $D_{KL}(p||q)\neq D_{KL}(q||p)$\$.

I came here looking for something else and was surprised that nobody has mentioned distances induced by convex bodies (these are classically called Minkowski distances, and "convex distance functions" in the Computational Geometry literature).

Let $$K$$ be an arbitrary convex body (= compact convex set with non-empty interior) in $$\mathbb R^d$$ with the origin in its interior. You can define the $$K$$-norm $$||v||_K$$ of a vector $$v$$ to be the unique positive $$\lambda\in [0,\infty)$$ with $$\lambda v \in \partial K$$, and the $$K$$-distance between $$a$$ and $$b$$ to be $$||b-a||_K$$.

If $$K$$ is centrally symmetric then these are a norm and a distance in the usual sense (the Euclidean distance and all the $$L_p$$ distances are examples).

If $$K$$ is not centrally symmetric then this distance is not symmetric ($$||-v||\ne -||v||$$), but it still satisfies the asymmetric triangle inequality.