# Rogue Paladin Shaman

An open-ended math puzzle for confoundingcalendar - itch.io

Status | Released |

Platforms | HTML5 |

Rating | Rated 4.5 out of 5 stars (2 total ratings) |

Author | Artless Games |

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An open-ended math puzzle for confoundingcalendar - itch.io

Status | Released |

Platforms | HTML5 |

Rating | Rated 4.5 out of 5 stars (2 total ratings) |

Author | Artless Games |

## Comments

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2 more difficulty ideas:

Medium Easy: set attributes for Rogue, Paladin, Shaman and Pay-to-win so that Pay-to-win wins all three, the rest win 1 and lose 2.

Medium Hard: set attributes for Rogue, Paladin, Shaman and All-rounder so that All-rounder ties all 3, the rest win 1, lose 1, and tie 1.

here is the abcd solution with variance:

https://www.desmos.com/calculator/ckngbupfif

type it in though, clicking the link does not work

unfortunately, four character solution requires all six colours demos has to offer.

https://www.desmos.com/calculator/qqyyxxb6iw

https://www.desmos.com/calculator/6vkfwoogbs

sorry, usage of m gave the victory to whoever had the higher hp when there was a stalemate.

I'm gonna solve it by graphing

https://www.desmos.com/calculator/iwncb9y1bs

Not the intended solution, but a really cool one. (That's what open-ended puzzle is about)

You can adjust which values are set to x and y to see the solution space for those values, given your existing values. the solution space is where three of the areas overlap (one from each pair). one should note that at usually least one of the pairs will not depend on x or y, and therefore either cover the entire screen, or not exist. m is also just a buffer, because inequalities are not graphed when one side is undefined. The bigger m is, the larger the values the function is accurate for. (do not make m or m2 negative) it sets the number of turns to beat the opponent to m*opponent's Hp. also this is just for the three character situation (and I used my original three character solution to test it), I plan to make another for the four character one. I also think the second smaller inequalities might be contained in the larger ones, and therefore be irrelevant.

There is an intended solution?

intended solution for RPS:

given the fact one always wins if they takes 0 damage, we can set up R with 1 defense, P with 2 attack and super low hp, S with 1 attack and super high hp

intended solution for ABCD:

given the fact that fight ties if both sides can't damage each other or both sides die in the same turn

we can set up A and C with high defense so that they can't hurt each other

B and D with low hp so that they otk each other

modify their stats to form a loop

I forgot the intended solution for ABCDE

ABCDE is similar to RPS, with two high hp heroes, two high defense heroes, one high attack hero, and more adjustment

my solution for RPS:

high hp for R, attack for S<(R hp)/2, defence for P>0 and <(S att)

I will use my equations to achieve the intended solution

this demonstrates how hard it is:

https://www.desmos.com/calculator/3rhphvwvip

big inequalities= turns to defeat opponent vs turns for opponent to defeat you

small inequalities asks do you block all your opponents attack while they don't block yours?

what is RPS?

rock paper scissor

It took me unbelievably long to notice it's RPS, ABCD and ABCDE.

I do think the second one is harder than the third.