Update nearest_points.md - avoid \coloneqq and add links

izanbf1803 · web-flow · commit 5a2e34931306 · 2025-06-26T21:43:19.000+02:00
diff --git a/src/geometry/nearest_points.md b/src/geometry/nearest_points.md
@@ -183,22 +183,22 @@ Now we need to decide on how to set $d$ so that it minimizes $\Theta(\sum_{i=1}^
 
 We need $d$ to be an approximation of the minimum distance $d$, and the trick is to just sample $n$ distances randomly and choose $d$ to be the smallest of these distances. We now prove that the expected running time is linear.
 
-**Proof.** Imagine the disposition of points in squares with a particular choice of $d$, say $x$. Consider $d$ a random variable, resulting from our sampling of distances. Let's define $C(x) = \sum_{i=1}^{k(x)} n_i(x)^2$ as the cost estimation for a particular disposition when we choose $d=x$. Now, let's define $\lambda(x)$ such that $C(x) = \lambda(x) \, n$. What is the probability that such choice $x$ survives the sampling of $n$ independent distances? If a single pair among the sampled ones has distance smaller than $x$, this arrangement will be replaced by the smaller $d$. Inside a square, at least a quarter of the pairs would raise a smaller distance (imagine four subsquares in every square, and use the pigeonhole principle), so we have $\sum_{i=1}^{k} \frac{1}{4} {n_i \choose 2}$ pairs which yield a smaller final $d$. This is, approximately, $\frac{1}{8} \sum_{i=1}^{k} n_i^2 = \frac{1}{8} \lambda(x) n$. On the other hand, there are about $\frac{1}{2} n^2$ pairs that can be sampled. We have that the probability of sampling a pair with distance smaller than $x$ is at least (approximately) 
+**Proof.** Imagine the disposition of points in squares with a particular choice of $d$, say $x$. Consider $d$ a random variable, resulting from our sampling of distances. Let's define $C(x) := \sum_{i=1}^{k(x)} n_i(x)^2$ as the cost estimation for a particular disposition when we choose $d=x$. Now, let's define $\lambda(x)$ such that $C(x) = \lambda(x) \, n$. What is the probability that such choice $x$ survives the sampling of $n$ independent distances? If a single pair among the sampled ones has distance smaller than $x$, this arrangement will be replaced by the smaller $d$. Inside a square, at least a quarter of the pairs would raise a smaller distance (imagine four subsquares in every square, and use the pigeonhole principle), so we have $\sum_{i=1}^{k} \frac{1}{4} {n_i \choose 2}$ pairs which yield a smaller final $d$. This is, approximately, $\frac{1}{8} \sum_{i=1}^{k} n_i^2 = \frac{1}{8} \lambda(x) n$. On the other hand, there are about $\frac{1}{2} n^2$ pairs that can be sampled. We have that the probability of sampling a pair with distance smaller than $x$ is at least (approximately) 
 
 $$\frac{\lambda(x) n / 8}{n^2 / 2} = \frac{\lambda(x)/4}{n}$$
 
 so the probability of at least one such pair being chosen during the $n$ rounds (and therefore finding a smaller $d$) is 
 
 $$1 - \left(1 - \frac{\lambda(x)/4}{n}\right)^n \ge 1 - e^{-\lambda(x)/4}$$
 
-(we have used that $(1 + x)^n \le e^{xn}$ for any real number $x$, check https://en.wikipedia.org/wiki/Bernoulli%27s_inequality#Related_inequalities). <br> Notice this goes to $1$ exponentially as $\lambda(x)$ increases. This hints that $\lambda$ will be small usually.
+(we have used that $(1 + x)^n \le e^{xn}$ for any real number $x$, check [this Wikipedia page](https://en.wikipedia.org/wiki/Bernoulli%27s_inequality#Related_inequalities)). <br> Notice this goes to $1$ exponentially as $\lambda(x)$ increases. This hints that $\lambda$ will be small usually.
 
 
 We have shown that $\Pr(d \le x) \ge 1 - e^{-\lambda(x)/4}$, or equivalently, $\Pr(d \ge x) \le e^{-\lambda(x)/4}$. We need to know $\Pr(\lambda(d) \ge \text{something})$ to be able to estimate its expected value. We notice that $\lambda(d) \ge \lambda(x) \iff d \ge x$. This is because making the squares smaller only reduces the number of points in each square (splits the points into other squares), and this keeps reducing the sum of squares. Therefore,
 
 $$\Pr(\lambda(d) \ge \lambda(x)) = \Pr(d \ge x) \le e^{-\lambda(x)/4} \implies \Pr(\lambda(d) \ge t) \le e^{-t/4} \implies \mathbb{E}[\lambda(d)] \le \int_{0}^{+\infty} e^{-t/4} \, \mathrm{d}t = 4$$
 
-(we have used that $E[X] = \int_0^{+\infty} \Pr(X \ge x) \, \mathrm{d}x$, check https://math.stackexchange.com/a/1690829).
+(we have used that $E[X] = \int_0^{+\infty} \Pr(X \ge x) \, \mathrm{d}x$, check [the proof](https://math.stackexchange.com/a/1690829)).
 
 Finally, $\mathbb{E}[C(d)] = \mathbb{E}[\lambda(d) \, n] \le 4n$, and the expected running time is $O(n)$, with a reasonable constant factor. $\quad \blacksquare$
 
@@ -299,13 +299,13 @@ pair<pt,pt> closest_pair_of_points_rand_reals(vector<pt> P) {
 Now we introduce a different randomized algorithm which is less practical but very easy to show that it runs in expected linear time.
 
 - Permute the $n$ points randomly
-- Take $\delta \coloneqq \operatorname{dist}(p_1, p_2)$
+- Take $\delta := \operatorname{dist}(p_1, p_2)$
 - Partition the plane in squares of side $\delta/2$
 - For $i = 1,2,\dots,n$:
   - Take the square corresponding to $p_i$
   - Interate over the $25$ squares within two steps to our square in the grid of squares partitioning the plane
   - If some $p_j$ in those squares has $\operatorname{dist}(p_j, p_i) < \delta$, then
-    - Recompute the partition and squares with $\delta \coloneqq \operatorname{dist}(p_j, p_i)$
+    - Recompute the partition and squares with $\delta := \operatorname{dist}(p_j, p_i)$
     - Store points $p_1, \dots, p_i$ in the corresponding squares
   - else, store $p_i$ in the corresponding square
 - output $\delta$
