Check predictability of names

The README to the exercise states
> The names must be random: they should not follow a predictable sequence.

Yet there are accepted [solutions](https://exercism.org/tracks/rust/exercises/robot-name/solutions/boso) that violate the requirement.

*) The linked solution can unfortunately not be commented on:
> Comments have been disabled

This new test makes sure that the most obvious kind of predictability - and probably the only one naturally used by anyone - makes tests fail.

The test cannot be run concurrently with other tests, for obvious reasons, hence the not very elegant solution using a global RwLock, which every other tests has to `read()`, in order for the new test to run exclusively.

Author: EugenDueck

exercises/practice/robot-name/tests/robot_name.rs

@@ -1,4 +1,22 @@
 use robot_name as robot;
+use std::sync::{Once, RwLock};
+
+static INIT: Once = Once::new();
+static mut PREDICTABLE_SEQ_TEST_LOCK: *const RwLock<()> = std::ptr::null();
+
+/// All tests other than names_should_not_follow_predictable_sequence need to take a read lock,
+/// so that they will not run concurrently with names_should_not_follow_predictable_sequence, who
+/// takes the write lock
+fn predictable_seq_test_lock() -> &'static RwLock<()> {
+    INIT.call_once(|| {
+        let boxed = Box::new(RwLock::new(()));
+        unsafe {
+            PREDICTABLE_SEQ_TEST_LOCK = Box::into_raw(boxed);
+        }
+    });
+
+    unsafe { &*PREDICTABLE_SEQ_TEST_LOCK }
+}
 
 fn assert_name_matches_pattern(n: &str) {
     assert!(n.len() == 5, "name is exactly 5 characters long");
@@ -14,13 +32,15 @@ fn assert_name_matches_pattern(n: &str) {
 
 #[test]
 fn name_should_match_expected_pattern() {
+    let _guard = predictable_seq_test_lock().read();
     let r = robot::Robot::new();
     assert_name_matches_pattern(r.name());
 }
 
 #[test]
 #[ignore]
 fn different_robots_have_different_names() {
+    let _guard = predictable_seq_test_lock().read();
     let r1 = robot::Robot::new();
     let r2 = robot::Robot::new();
     assert_ne!(r1.name(), r2.name(), "Robot names should be different");
@@ -29,6 +49,7 @@ fn different_robots_have_different_names() {
 #[test]
 #[ignore]
 fn many_different_robots_have_different_names() {
+    let _guard = predictable_seq_test_lock().read();
     use std::collections::HashSet;
 
     // In 3,529 random robot names, there is ~99.99% chance of a name collision
@@ -42,6 +63,7 @@ fn many_different_robots_have_different_names() {
 #[test]
 #[ignore]
 fn new_name_should_match_expected_pattern() {
+    let _guard = predictable_seq_test_lock().read();
     let mut r = robot::Robot::new();
     assert_name_matches_pattern(r.name());
     r.reset_name();
@@ -51,9 +73,38 @@ fn new_name_should_match_expected_pattern() {
 #[test]
 #[ignore]
 fn new_name_is_different_from_old_name() {
+    let _guard = predictable_seq_test_lock().read();
     let mut r = robot::Robot::new();
     let n1 = r.name().to_string();
     r.reset_name();
     let n2 = r.name().to_string();
     assert_ne!(n1, n2, "Robot name should change when reset");
 }
+
+fn name_to_num(name: &str) -> u32 {
+    let mut chars = name.chars();
+    ((chars.next().unwrap() as u32) - b'A' as u32) * 26 * 10 * 10 * 10
+        + ((chars.next().unwrap() as u32) - b'A' as u32) * 10 * 10 * 10
+        + ((chars.next().unwrap() as u32) - b'0' as u32) * 10 * 10
+        + ((chars.next().unwrap() as u32) - b'0' as u32) * 10
+        + ((chars.next().unwrap() as u32) - b'0' as u32)
+}
+
+#[test]
+#[ignore]
+fn names_should_not_follow_predictable_sequence() {
+    // needs to run exclusively, otherwise other tests may mess up predictability detection
+    let _guard = predictable_seq_test_lock().write();
+    let nums = (0..10)
+        .into_iter()
+        .map(|_| name_to_num(robot::Robot::new().name()))
+        .collect::<Vec<_>>();
+
+    // just checking the easiest kind of predictable sequences here: the difference between
+    // numerical representation of consecutive names is always the same
+    let d = nums[1] - nums[0];
+    assert!(
+        !nums.windows(2).all(|w| w[1] - w[0] == d),
+        "Name sequence is very predictable: The difference between the numerical representation of names is {d}"
+    )
+}