From 3bd158c665f7f464ed9e0d5b02a408eedbf4d610 Mon Sep 17 00:00:00 2001
From: nham <hamann.nick@gmail.com>
Date: Fri, 8 Aug 2014 15:37:06 -0400
Subject: [PATCH 1/4] Add example of estimating pi using Monte Carlo simulation
 to std::rand

---
 src/libstd/rand/mod.rs | 45 ++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 45 insertions(+)

diff --git a/src/libstd/rand/mod.rs b/src/libstd/rand/mod.rs
index 40d8f80171c4..06b9b464ebde 100644
--- a/src/libstd/rand/mod.rs
+++ b/src/libstd/rand/mod.rs
@@ -70,6 +70,51 @@
 //! println!("{}", tuple)
 //! ```
 //!
+//! ## Monte Carlo estimation of pi
+//!
+//! For this example, imagine we have a square with sides of length 2 and a unit
+//! circle, both centered at the origin. Since the area of a unit circle is pi,
+//! the ratio
+//!
+//!     (area of unit circle) / (area of square)
+//!
+//! is equal to pi / 4. So if we sample many points randomly from the square,
+//! roughly pi / 4 of them should be inside the circle.
+//!
+//! We can use the above fact to estimate the value of pi: pick many points in the
+//! square at random, calculate the fraction that fall within the circle, and
+//! multiply this fraction by 4.
+//!
+//! ```
+//! use std::rand;
+//! use std::rand::distributions::{IndependentSample, Range};
+//!
+//! fn dist(x: f64, y: f64) -> f64 {
+//!    (x*x + y*y).sqrt()
+//! }
+//!
+//! fn main() {
+//!    let between = Range::new(-1f64, 1.);
+//!    let mut rng = rand::task_rng();
+//!
+//!    let total = 1_000_000u;
+//!    let mut in_circle = 0u;
+//!
+//!    for _ in range(0u, total) {
+//!        let a = between.ind_sample(&mut rng);
+//!        let b = between.ind_sample(&mut rng);
+//!        if dist(a, b) <= 1. {
+//!            in_circle += 1;
+//!        }
+//!    }
+//!
+//!    // prints something close to 3.14159...
+//!    println!("{}", 4. * (in_circle as f64) / (total as f64));
+//! }
+//! ```
+//!
+//! ## Monty Hall Problem
+//!
 //! This is a simulation of the [Monty Hall Problem][]:
 //!
 //! > Suppose you're on a game show, and you're given the choice of three doors:

From 9315d841c7f2041e3b6960a5f5400747b1989647 Mon Sep 17 00:00:00 2001
From: nham <hamann.nick@gmail.com>
Date: Fri, 8 Aug 2014 21:02:51 -0400
Subject: [PATCH 2/4] Remove the dist function; it is more efficient to compare
 squared distances

---
 src/libstd/rand/mod.rs | 6 +-----
 1 file changed, 1 insertion(+), 5 deletions(-)

diff --git a/src/libstd/rand/mod.rs b/src/libstd/rand/mod.rs
index 06b9b464ebde..b9066c2f338f 100644
--- a/src/libstd/rand/mod.rs
+++ b/src/libstd/rand/mod.rs
@@ -89,10 +89,6 @@
 //! use std::rand;
 //! use std::rand::distributions::{IndependentSample, Range};
 //!
-//! fn dist(x: f64, y: f64) -> f64 {
-//!    (x*x + y*y).sqrt()
-//! }
-//!
 //! fn main() {
 //!    let between = Range::new(-1f64, 1.);
 //!    let mut rng = rand::task_rng();
@@ -103,7 +99,7 @@
 //!    for _ in range(0u, total) {
 //!        let a = between.ind_sample(&mut rng);
 //!        let b = between.ind_sample(&mut rng);
-//!        if dist(a, b) <= 1. {
+//!        if a*a + b*b <= 1. {
 //!            in_circle += 1;
 //!        }
 //!    }

From 348132196a1a25336d2df6b6e0989540cdc8f7e1 Mon Sep 17 00:00:00 2001
From: nham <hamann.nick@gmail.com>
Date: Thu, 21 Aug 2014 02:08:17 -0400
Subject: [PATCH 3/4] Surround formula in a 'notrust' code block

---
 src/libstd/rand/mod.rs | 2 ++
 1 file changed, 2 insertions(+)

diff --git a/src/libstd/rand/mod.rs b/src/libstd/rand/mod.rs
index b9066c2f338f..00abd84d5c27 100644
--- a/src/libstd/rand/mod.rs
+++ b/src/libstd/rand/mod.rs
@@ -76,7 +76,9 @@
 //! circle, both centered at the origin. Since the area of a unit circle is pi,
 //! the ratio
 //!
+//! ```notrust
 //!     (area of unit circle) / (area of square)
+//! ```
 //!
 //! is equal to pi / 4. So if we sample many points randomly from the square,
 //! roughly pi / 4 of them should be inside the circle.

From 86587224d0b99c69da01ebfaf98052694036302b Mon Sep 17 00:00:00 2001
From: nham <hamann.nick@gmail.com>
Date: Thu, 21 Aug 2014 02:42:15 -0400
Subject: [PATCH 4/4] Use unicode pi symbol in pi estimation example.
 Additional tweaks

---
 src/libstd/rand/mod.rs | 14 +++++++-------
 1 file changed, 7 insertions(+), 7 deletions(-)

diff --git a/src/libstd/rand/mod.rs b/src/libstd/rand/mod.rs
index 00abd84d5c27..0739d91b5840 100644
--- a/src/libstd/rand/mod.rs
+++ b/src/libstd/rand/mod.rs
@@ -70,20 +70,20 @@
 //! println!("{}", tuple)
 //! ```
 //!
-//! ## Monte Carlo estimation of pi
+//! ## Monte Carlo estimation of π
 //!
 //! For this example, imagine we have a square with sides of length 2 and a unit
-//! circle, both centered at the origin. Since the area of a unit circle is pi,
-//! the ratio
+//! circle, both centered at the origin. Since the area of a unit circle is π,
+//! we have:
 //!
 //! ```notrust
-//!     (area of unit circle) / (area of square)
+//!     (area of unit circle) / (area of square) = π / 4
 //! ```
 //!
-//! is equal to pi / 4. So if we sample many points randomly from the square,
-//! roughly pi / 4 of them should be inside the circle.
+//! So if we sample many points randomly from the square, roughly π / 4 of them
+//! should be inside the circle.
 //!
-//! We can use the above fact to estimate the value of pi: pick many points in the
+//! We can use the above fact to estimate the value of π: pick many points in the
 //! square at random, calculate the fraction that fall within the circle, and
 //! multiply this fraction by 4.
 //!