2019-03-10 03:01:48 8 Comments

Take a random walk in the plane from the origin, each step of unit length in a uniformly random direction.

. How many steps on average until the path self-intersects?Q

My simulations suggest ~$8.95$ steps.

^{ Red: origin. Top: the $8$-th step self-intersects. Bottom: the $11$-th step self-intersects. (Not to same scale.) }

I suspect this is known in the SAW literature (SAW=Self-Avoiding Walk), but I am not finding this explicit number.

Related: self-avoidance time of random walk.

*Added*. Here is a histogram of the number of steps to self-intersection.

^{ $10000$ random trials. }

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## 3 comments

## @Matt F. 2019-03-22 19:43:20

We can get an upper bound of $15.52$ by considering just the cases of interesction of next-to-consecutive steps, which we call quick intersections.

Let $\alpha$ be the angle between step $AB$ and step $BC$, and let $u=\alpha/\pi$. So $u$ is (initially) uniformly distributed between $0$ and $1$. Then, as Gerhard Paseman and Bill Bradley determined, the probability that steps $AB$ and $CD$ intersect is $$ P(\text{intersection|u})= \begin{cases} \frac14-\frac u4\ \ \text{ if } u\le \frac13\\ \frac12-u\ \ \text{ if } \frac13\le u \le \frac12\\ \phantom{1-2}0 \ \ \text{ if } \frac12 \le u\\ \end{cases} $$ So the (initial) overall probability of an intersection is $\int P(\text{intersection}|u)du=1/12$.

This means that for the next step, we have $$ P(u|\text{no intersection last time})=\frac{12}{11} \begin{cases} \frac34+\frac u4\ \ \text{ if } u\le \frac13\\ \frac12 + u\, \ \ \text{ if } \frac13\le u \le \frac12\\ \phantom{1-2}1 \ \ \text{ if } \frac12 \le u\\ \end{cases} $$ $$ \begin{align} P(\text{intersection|no intersection last time}) &=\frac{12}{11}\int_{u=0}^1 \begin{cases} (\frac34+\frac u4)(\frac14-\frac u4)\ \ \text{ if } u\le \frac13\\ (\frac12+u)(\frac12-u)\ \ \, \text{ if } \frac13\le u \le \frac12\\ \ \ \phantom{1-2}1(0)\ \ \ \ \ \ \ \ \ \ \ \text{ if } \frac12 \le u \end{cases} du\\ &=\frac{29}{396} \end{align} $$ So the probability of no quick intersection after $n$ steps is $$\frac{11}{12}\left(\frac{367}{396}\right)^{n-3}$$ and the expected number of steps until a quick intersection is $$ 3\left(\frac{1}{12}\right)+\sum_{n=4}^\infty n\left( \frac{11}{12}\left(\frac{367}{396}\right)^{n-4}\! - \frac{11}{12}\left(\frac{367}{396}\right)^{n-3} \right)=\frac{450}{29} \simeq 15.52$$

## @Joseph O'Rourke 2019-03-23 00:11:15

Nice analysis, MattF!

## @Bill Bradley 2019-03-20 15:37:53

(Thanks for noticing my earlier error, Gerhard!)

Here's a loose but very simple upper bound. Consider three consecutive unit steps, which might look like this:

We take $\alpha\in[0, \pi]$ and $\beta\in[0,2\pi)$. (By symmetry we can reflect $\alpha\in (\pi, 2\pi)$ back to $\pi-\alpha$.) Then to overlap, we need $\alpha < \pi/2$, $\beta<\pi/2$ and $\alpha + \beta < \pi$ The first two events are independent, with probabilities 1/2 and 1/4.

For the third event, consider a case where the three segments just barely cross, e.g.:

If we decrease the angles $\alpha, \beta$, the steps will still intersect, so it's sufficient to consider this case.

Because the bottom and left edges are both length one, the upper and rightmost angles are both $\beta$. Since we have a triangle, $\alpha+2\beta=\pi$, so the probability of this event is 2/3. (There's a symmetric case where the roles of $\alpha$ and $\beta$ are reversed.) It's possibly clearer to see this by examining a phase space diagram ("blue" = "line segments intersect"):

The blue region takes up 2/3rds of the (uniformly sampled) space. Conditioning on everything, the probability of overlap is 1/2 * 1/4 * 2/3 = 1/12

We can apply this argument to multiple non-overlapping pairs of angles. Therefore, the expected number of steps until self-intersection is upper-bounded by $$ \leq 1 + 2\times 12 = 25$$

## @Gerhard Paseman 2019-03-20 15:49:25

Something does not look right. Do you want to find it, or shall I expand on my comments above? Gerhard "Pretty Sure About One Twelfth" Paseman, 2019.03.20.

## @Gerhard Paseman 2019-03-21 23:55:43

Looks good now. I like the picture and the estimate. Gerhard "Blue Is A Nice Color" Paseman, 2019.03.21.

## @Gerhard Paseman 2019-03-20 17:03:41

Inspired by Bill Bradley's post, I've decided to lower bound the probability of an intersection which (using the method suggested in Bill's post) gives a slightly less weak upper bound for the expected length of a random walk.

Using Bill's picture and rescaling so that angles are measured as nonnegative fractions of 2pi, we see one domain of self intersection is when $\alpha \lt \beta$ and $\alpha + 2\beta \lt 1/2$. This region is a triangle of base 1/4, and height 1/6, with an area of 1/48. Missing the measure 0 set where $\alpha=\beta$, we reflect across that line, and then once more across the line $\alpha+\beta=1$ to get the other three regions having an intersection, giving the probability of self intersection of three-step paths to be 4/48=1/12.

Now let's extend this to paths with four steps. The number of cases to determine the probability exactly is more than I care to tackle, so I will cheat and underestimate it. I do this by considering just two cases: that an intersection occurs in the first three steps, or that it does not but occurs in the last three steps.

If we considered these cases as disjoint, we would get a probability of 1/12 + 1/12 = 1/6. They aren't, so we will consider the case that $\beta \leq 1/4$ and $\alpha \gt 1/4$ or more visibly use 3/4 of the chance that the last three steps self-intersect. This gives a lower bound of (1/12) times (1+3/4), or 7/48, which gives 1/7 as a weak lower bound.

Using this and Bill's estimating method gives 1 + 7*(4-1)=22 as a weak upper bound on the number of steps needed to self-intersect. If we could improve the lower bound from 1/7 to 1/6 for self intersection of a four step path, this would lead to an upper bound of 19 instead of 22.

But if we can cheat once, we can cheat again. The first cheat went from two neighboring angles to three, so we will go from three to five and lower bound the crossing probability of six segments as (7/4) times the (lower bound of the) probability of four segments. We share the middle angle so that we can nicely separate the domains of the middle angle turning sharp left or sharp right. We do a similar cut and trim and end up with a lower bound of 49/192 =1/4+1/192 for the probability of 6 segments crossing. This gives a slightly better upper bound of 1 +4*5 or 21 for the step length of a walk that does not cross itself.

To justify the cheat, one has to do a lot more work to make sure that the probability grows by 7/4 at this step. However, I won't do that. I use the second cheat instead to motivate the idea that 22 is an upper bound that can be improved with a simple argument, and hope to inspire someone to come up with a simple and more honest and precise estimate.

Gerhard "Long As It's Not Graded" Paseman, 2019.03.20.

## @Gerhard Paseman 2019-03-20 17:21:51

I would like to point out that many of my signatures encode some double meaning. In this particular case, I want to make it clear that this analysis fails on other manifolds. In particular, Joseph's problem makes sense on a torus or pseudo sphere, even on non-oriented manifolds or graphs with a rigid geometric structure. I encourage people to consider the version for a torus. Gerhard "Promises To Not Grade Work" Paseman, 2019.03.20.

## @Gerhard Paseman 2019-03-20 18:20:11

The version of Joseph's question on a compact space is likely in the graph theory literature. Consider a cover of the space by uniform mostly disjoint regions, and make a graph with a region a vertex and a line joining two regions if a step takes one from one region to another. One can now approximate a random walk by a transition matrix, or use the theory of random walks in graphs to compute a desired expectation. Thus the actual expectation can be considered as a limit of expectations using various partitions of the space. Gerhard "Still Interested In Torus Version" Paseman, 2019.03.20.