Merge pull request 'Added second threading exercise.' (#65) from threading2 into main

Reviewed-on: https://codeberg.org/ziglings/exercises/pulls/65

README.md

@@ -219,7 +219,6 @@ Zig Core Language
 * [X] Bit manipulation
 * [X] Working with C
 * [X] Threading
-* [ ] Interfaces part 2
 
 Zig Standard Library
 

build.zig

@@ -1117,6 +1117,10 @@ const exercises = [_]Exercise{
         \\Zig is cool!
         ,
     },
+    .{
+        .main_file = "105_threading2.zig",
+        .output = "PI ≈ 3.14159265",
+    },
     .{
         .main_file = "999_the_end.zig",
         .output =

exercises/105_threading2.zig

@@ -0,0 +1,107 @@
+//
+// Now that we are familiar with the principles of multithreading, we
+// boldly venture into a practical example from mathematics.
+// We will determine the circle number PI with sufficient accuracy.
+//
+// There are different methods for this, and some of them are several
+// hundred years old. For us, the dusty procedures are surprisingly well
+// suited to our exercise. Because the mathematicians of the time didn't
+// have fancy computers with which we can calculate something like this
+// in seconds today.
+// Whereby, of course, it depends on the accuracy, i.e. how many digits
+// after the decimal point we are interested in.
+// But these old procedures can still be tackled with paper and pencil,
+// which is why they are easier for us to understand.
+// At least for me. ;-)
+//
+// So let's take a mental leap back a few years.
+// Around 1672 (if you want to know and read about it in detail, you can
+// do so on Wikipedia, for example), various mathematicians once again
+// discovered a method of approaching the circle number PI.
+// There were the Scottish mathematician Gregory and the German
+// mathematician Leibniz, and even a few hundred years earlier the Indian
+// mathematician Madhava. All of them independently developed the same
+// formula, which was published by Leibnitz in 1682 in the journal
+// "Acta Eruditorum".
+// This is why this method has become known as the "Leibnitz series",
+// although the other names are also often used today.
+// We will not go into the formula and its derivation in detail, but
+// will deal with the series straight away:
+//
+//        4     4     4     4     4
+//  PI = --- - --- + --- - --- + --- ...
+//        1     3     5     7     9
+//
+// As you can clearly see, the series starts with the whole number 4 and
+// approaches the circle number by subtracting and adding smaller and
+// smaller parts of 4. Pretty much everyone has learned PI = 3.14 at school,
+// but very few people remember other digits, and this is rarely necessary
+// in practice. Because either you don't need the precision, or you use a
+// calculator in which the number is stored as a very precise constant.
+// But at some point this constant was calculated and we are doing the same
+// now.The question at this point is, how many partial values do we have
+// to calculate for which accuracy?
+//
+// The answer is chewing, to get 8 digits after the decimal point we need
+// 1,000,000,000 partial values. And for each additional digit we have to
+// add a zero.
+// Even fast computers - and I mean really fast computers - get a bit warmer
+// on the CPU when it comes to really many diggits. But the 8 digits are
+// enough for us for now, because we want to understand the principle and
+// nothing more, right?
+//
+// As we have already discovered, the Leibnitz series is a series with a
+// fixed distance of 2 between the individual partial values. This makes
+// it easy to apply a simple loop to it, because if we start with n = 1
+// (which is not necessarily useful now) we always have to add 2 in each
+// round.
+// But wait! The partial values are alternately added and subtracted.
+// This could also be achieved with one loop, but not very elegantly.
+// It also makes sense to split this between two CPUs, one calculates
+// the positive values and the other the negative values. And so we can
+// simply start two threads and add everything up at the end and we're
+// done.
+// We just have to remember that if only the positive or negative values
+// are calculated, the distances are twice as large, i.e. 4.
+//
+// So that the whole thing has a real learning effect, the first thread
+// call is specified and you have to make the second.
+// But don't worry, it will work out. :-)
+//
+const std = @import("std");
+
+pub fn main() !void {
+    const count = 1_000_000_000;
+    var pi_plus: f64 = 0;
+    var pi_minus: f64 = 0;
+
+    {
+        // First thread to calculate the plus numbers.
+        const handle1 = try std.Thread.spawn(.{}, thread_pi, .{ &pi_plus, 5, count });
+        defer handle1.join();
+
+        // Second thread to calculate the minus numbers.
+        ???
+        
+    }
+    // Here we add up the results.
+    std.debug.print("PI ≈ {d:.8}\n", .{4 + pi_plus - pi_minus});
+}
+
+fn thread_pi(pi: *f64, begin: u64, end: u64) !void {
+    var n: u64 = begin;
+    while (n < end) : (n += 4) {
+        pi.* += 4 / @as(f64, @floatFromInt(n));
+    }
+}
+// If you wish, you can increase the number of loop passes, which
+// improves the number of digits.
+//
+// But be careful:
+// In order for parallel processing to really show its strengths,
+// the compiler must be given the "-O ReleaseFast" flag when it
+// is created. Otherwise the debug functions slow down the speed
+// to such an extent that seconds become minutes during execution.
+//
+// And you should remove the formatting restriction in "print",
+// otherwise you will not be able to see the additional diggits.

patches/patches/105_threading2.patch

@@ -0,0 +1,13 @@
+--- exercises/105_threading2.zig	2024-03-23 16:35:14.754540802 +0100
++++ answers/105_threading2.zig	2024-03-23 16:38:00.577539733 +0100
+@@ -81,8 +81,8 @@
+         defer handle1.join();
+ 
+         // Second thread to calculate the minus numbers.
+-        ???
+-        
++        const handle2 = try std.Thread.spawn(.{}, thread_pi, .{ &pi_minus, 3, count });
++        defer handle2.join();
+     }
+     // Here we add up the results.
+     std.debug.print("PI ≈ {d:.8}\n", .{4 + pi_plus - pi_minus});