My brother, Teal Knight, came up with a logic puzzle and I'm posting it on his behalf:
The notorious villain, "
Cheese Shogun", is at large, and The Resistance needs to send a party of Heroes to stop him! He has stolen an artifact called the "Dark Mirror", which has X charges (3 by default), and when a charge is used on a Hero, a Dark Copy of that Hero is made - so strong that it can wipe the entire party singlehandedly!
However, The Resistance has a trick up their sleeve - For each Hero they send, they can imbue that Hero with the ability to Counter one or more other specific Heroes. (For example, the 1st Hero might be a Dragon with a Sword of Elf Slaying, the 2nd Hero might be an Elf with a Sword of Demon Slaying, and the 3rd Hero might be a Demon with a Sword of Dragon Slaying).
After assembling a team of Heroes, repeat the following steps over and over:
1) The Villain uses one Dark Mirror charge to make a Dark Copy of a Hero.
2) The Dark Copy kills all Heroes that it can counter (for example, the Elf Dark Copy slays the Demon)
3) If a Hero is still alive that can counter the Dark Copy, then the Dark Copy dies (for example, the Dragon slays the Elf Dark Copy)
4) If the Dark Copy is still alive, the entire party dies and the Villain wins.
5) If the Dark Mirror is out of charges, the Villain loses as the remaining party members overwhelm him.
6) GOTO 1.
For each value of X charges, what is the minimum number of heroes N The Resistance must send to ensure the Villain cannot win, and what setup of counters do those N heroes have?
For X = 1, it's very simple - just a rock paper scissors triangle of counters. But for X = 2 and beyond, it gets harder, because each hero needs to be countered by at least 2 other heroes (otherwise you would copy the counter to that hero's hero's one counter, then copy that hero, and win that way), and those two counters need to not be countered by the same hero (otherwise you would copy that hero's two counters' shared counter, then copy that hero, and win that way). And the more charges, the more ways there are to fail.
