This blackbox puzzle can be roughly divided into three steps.

**Step 1: Deciphering the letters**

After a bit of playing around, we found out that the comparison function seems to only care about the "multiset of case-insensitive letters" in the string.

Attempting to identify the "smallest" letter sets gives an inequality chain of

A < B < AA < C < AB < D < AAA < BB < AC < E < AAB < F < .....

And we can recognize them to be in the same pattern of prime factorizations of small integers.

This implies that the letters represents primes in order (A=2,B=3,C=5,...,Z=101), and these primes are meant to be multiplied together.

This is further confirmed by the flavortext of "primary particles" and "basic building blocks of natural things" (referring to the description of primes as basic building blocks of natural numbers).

**Step 2: Identifying the collision function**

Once the letters and strings are deciphered, we can write the given collision table in a purely numerical form:

2 | 3 | 5 | 7 | 11 | 13 | 17 | 19 | 23 | 29 | 31 | |
---|---|---|---|---|---|---|---|---|---|---|---|

2 | 2 | 9 | 20 | 54 | 77 | 135 | 170 | 252 | 405 | 464 | |

3 | 7 | 18 | 52 | 75 | 133 | 168 | 250 | 403 | 462 | ||

5 | 11 | 45 | 68 | 126 | 161 | 243 | 396 | 455 | |||

7 | 34 | 57 | 115 | 150 | 232 | 385 | 444 | ||||

11 | 23 | 81 | 116 | 198 | 351 | 410 | |||||

13 | 58 | 93 | 175 | 328 | 387 | ||||||

17 | 35 | 117 | 270 | 329 | |||||||

19 | 82 | 235 | 294 | ||||||||

23 | 153 | 212 | |||||||||

27 | 59 | ||||||||||

31 |

Either by noticing the constant differences between different rows or by performing a polynomial fit, one concludes that the "collision" function is given by f(p,q)=|p(p-1)/2-q(q-1)/2|.

This allows us to build a full two-letter collision table.

**Step 3: Solving the cycles**

Each collision in the cycle results in a set of letters, where one goes onward into the next collision and the others form the "observations".

The key to identifying the missing letters is noticing that several given observations has a very few number of possibilities.

For example, in cycle 1, there is only one possible two-letter collision that gives ?BBS as the outcome, namely DðŸ’¥R=BBBS (or f(7,61)=3^3*37). Using this as a starting point we can fully determine cycle 1.

Cycles 2 and 3 are similarly solved; a common starting point is that ?AABBK can only be produced by either IðŸ’¥T=AAABBK or EðŸ’¥W=AABBBK, and in both cycles the next collision reveals EðŸ’¥W is the only possibility, and we can work from this point.

The completed collision cycles are shown as below (starting from top and going clockwise):

Cycle 1

? Letters | F | R | O | M | T | U | R | T | W | I | G |

Letters in the cycle | A | D | B | E | G | J | Z | C | E | K | P |

Collision outcome | DE | BBBS | ADDE | BBCG | BBBBJ | AEZ | AACDI | BBCCE | AABBBK | AAP | ABBBI |

Cycle 2

? Letters | T | O | A | R | C | E | U | S |

Letters in the cycle | U | F | Q | H | V | W | B | D |

Collision outcome | EF | GQ | ABBCH | BDV | LW | AABBBK | BCCCD | ABCU |

Cycle 3

? Letters | S | I | X | T | H | R | E | E |

Letters in the cycle | A | G | F | Z | C | I | W | B |

Collision outcome | ACFG | BBF | AHZ | BBBCH | DI | H | AABBBK | AAF |

The ? letters spell the clue phrase **FROMTURTWIG TOARCEUS SIXTHREE**, or **FROM TURTWIG TO ARCEUS (6 3)**.

Turtwig and Arceus are the beginning and end in the list of the generation IV Pokemons. This fact along with the enumeration (6 3) pinpoints the final answer of FOURTH GEN.