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Posted on Aug 17, 2019

Daily Challenge #43 - Boardgame Fight Resolver

#challenge

Today's challenge is a smaller part of a larger idea.

You are creating a board game, similar to a mix between Fire Emblem and Chess. The game features four unique pieces: Swordsman, Cavalry, Archer, and Pikeman. Each piece has its own advantages and weaknesses in combat against other pieces.

You must write a function fightResolve that takes the attacking and defending pieces as input parameters and returns the winning piece.

The outcome of the fight between two pieces depends on which piece attacks, the type of the attacking piece and the type of the defending piece.

Archers > Swordsmen > Pikemen > Cavalry > Archers

Archers always win against swordsmen, swordsmen always win against pikemen, pikemen always win against cavalry and cavalry always win against archers.

If a matchup occurs that was not previously mentioned (for example Archers vs Pikemen) the attacker will always win.

This challenge comes from user Brysen on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge for a future post? Email yo+challenge@dev.to with your suggestions!

Top comments (19)

Alvaro Montoro • Aug 17 '19 • Edited

CSS

Add the attributes data-piece1 and data-piece2 to any element, and it will announce the winner:

[data-piece1][data-piece2]::before {
  content: attr(data-piece1) ' vs ' attr(data-piece2) ': ';
}

/* By default, piece 1 will win */
[data-piece1][data-piece2]::after {
  content: attr(data-piece1) ' wins!';
}

/* Only in some particular cases, piece 2 will win */
[data-piece2="archers"i][data-piece1="swordsmen"i]::after,
[data-piece2="swordsmen"i][data-piece1="pikemen"i]::after,
[data-piece2="pikemen"i][data-piece1="cavalry"i]::after,
[data-piece2="cavalry"i][data-piece1="archers"i]::after { 
  content: attr(data-piece2) ' wins!'
}

In CSS Selectors Level 4, there is an option to make some selectors case-insensitive (adding an i at the end as displayed above). And it is fairly well supported (haven't tried on IE). So it doesn't matter if the user writes "archers" or "Archers" or "aRcHeRs", all of them will be matched.

Basti Ortiz • Aug 17 '19

This is by far the most clever CSS I have ever seen. I applaud you.

reko91 • Aug 17 '19

Mad

Giorgio Bertolotti • Aug 18 '19

Okay close it all, we have a winner!

Avalander • Aug 17 '19

Look ma, pattern matching!

fightResolve :: String -> String -> String
fightResolve "Swordsman" "Archer"  = "Archer"
fightResolve "Pikeman" "Swordsman" = "Swordsman"
fightResolve "Cavalry" "Pikeman"   = "Pikeman"
fightResolve "Archer" "Cavalry"    = "Cavalry"
fightResolve attacker _            = attacker

Josh • Aug 19 '19 • Edited

Ahh, pattern matches. So elegance, much match ( *-*)

Kona Arctic • Aug 17 '19 • Edited



#!/bin/bash

# ( attack , defend )
# archer: 1; swordsmen: 2; pikemen: 3; cavalry: 4;
function fightResolve {
    echo "$1$2" | grep -qEe '(21|32|43|14)' - &&
        return $2
    return $1
}

willsmart • Aug 17 '19 • Edited

I'm surprised no-one has used the age old xor-the-second-bit-of-the-first-characters-and-xor-the-fifth-bit-of-first-and-second-characters trick 😉

In C:

const char *defenderWins(const char *attacker, const char *defender) {
    return (((attacker[0] ^ defender[0]) & 2) | ((attacker[1] ^ defender[0]) & 0x10)) == 0x12;
}

Clearly, this doesn't really need any testing, but here's an exhaustive dump, err, full set of trials, just to be pedantic

#include <stdio.h>

const char *winner(const char *attacker, const char *defender) {
    return (((attacker[0] ^ defender[0]) & 2) | ((attacker[1] ^ defender[0]) & 0x10)) == 0x12 ? defender : attacker;
}

int main()
{
    const char *teams[] = { "archers","swordsmen", "pikemen", "cavalry" };
    const size_t teamCount=sizeof(teams)/sizeof(*teams);

    for (int attackeri = 0; attackeri<teamCount; ++attackeri) {
        for (int defenderi = 0; defenderi<teamCount; ++defenderi) {
            const char *attacker = teams[attackeri];
            const char *defender = teams[defenderi];
            printf("%s -> %s : %s wins\n", attacker, defender, winner(attacker,defender));
        }
    }
    return 0;
}

prints:

archers -> archers : archers wins
archers -> swordsmen : archers wins
archers -> pikemen : archers wins
archers -> cavalry : cavalry wins
swordsmen -> archers : archers wins
swordsmen -> swordsmen : swordsmen wins
swordsmen -> pikemen : swordsmen wins
swordsmen -> cavalry : swordsmen wins
pikemen -> archers : pikemen wins
pikemen -> swordsmen : swordsmen wins
pikemen -> pikemen : pikemen wins
pikemen -> cavalry : pikemen wins
cavalry -> archers : cavalry wins
cavalry -> swordsmen : cavalry wins
cavalry -> pikemen : pikemen wins
cavalry -> cavalry : cavalry wins

Avalander • Aug 18 '19

Care to explain how it works? I don't get it 😅

willsmart • Aug 18 '19

Sure thing. It's a "don't look behind the curtain" kind of thing 🧙‍♂️
here's the rest of the code...

const a = [
  'archers',
  'swordsmen', 'pikemen', 'cavalry'
]
let i1 = 2,
  i2 = 1,
  i3 = 3,
  i4 = 3,
  i5 = 2,
  i6 = 2,
  i7 = 2,
  i8 = 2

winner_mank = (s1, s2) => ((s1.charCodeAt(i1) >> i3) ^ ((s2.charCodeAt(i2)) >> i4)) & ((s1.charCodeAt(i5) >> i7) ^ ((s2.charCodeAt(i6)) >> i8)) & 1 ? s2 : s1

winner_good = (s1, s2) => a.indexOf(s1) == (a.indexOf(s2) + 1) % 4 ? s2 : s1

verify = () => {
  for (const s1 of a) {
    for (const s2 of a) {
      if (winner_good(s1, s2) != winner_mank(s1, s2)) return
    }
  }
  console.log({
    i1,
    i2,
    i3,
    i4,
    i5,
    i6,
    i7,
    i8
  })
  debugger
  return true
}

for (i1 = 0; i1 < 7; i1++) {
  for (i2 = 0; i2 < 7; i2++) {
    for (i3 = 0; i3 < 8; i3++) {
      for (i4 = 0; i4 < 8; i4++) {
        for (i5 = 0; i5 < 7; i5++) {
          for (i6 = 0; i6 < 7; i6++) {
            for (i7 = 0; i7 < 8; i7++) {
              for (i8 = 0; i8 < 8; i8++) {
                verify()
              }
            }
          }
        }
      }
    }
  }
}

winner_mank is reducible to the one posted when i1=i3=0, i2=i4=1, i5=1,i7=0, i6=i8=4. And I felt it was may as well be in C.
I was lying about the age old trick thing. Upside is it's really really quick, downsides are all the other things 😂

Jay • Aug 18 '19

Rust Solution: Playground

fn fight_resolver(defender: Class, attacker: Class) -> Class {
    match (defender, attacker) {
        (Class::Swordsmen, Class::Archer)
        | (Class::Pikemen, Class::Swordsmen)
        | (Class::Cavalry, Class::Pikemen)
        | (Class::Archer, Class::Cavalry) => defender,
        (_, att) => att,
    }
}

E. Choroba • Aug 22 '19

Perl solution, using a hash of known fight results.

#!/usr/bin/perl
use warnings;
use strict;

{   my %stronger = (
        archers   => 'swordsmen',
        swordsmen => 'pikemen',
        pikemen   => 'cavalry',
        cavalry   => 'archers');
    sub fight_resolve {
        my ($attacker, $defender) = @_;
        ($stronger{$defender} // "") eq $attacker ? $defender : $attacker
    }
}

use Test::More tests => 16;

is fight_resolve('archers', 'archers'    ), 'archers';
is fight_resolve('archers', 'swordsmen'  ), 'archers';
is fight_resolve('archers', 'pikemen'    ), 'archers';
is fight_resolve('archers', 'cavalry'    ), 'cavalry';
is fight_resolve('swordsmen', 'archers'  ), 'archers';
is fight_resolve('swordsmen', 'swordsmen'), 'swordsmen';
is fight_resolve('swordsmen', 'pikemen'  ), 'swordsmen';
is fight_resolve('swordsmen', 'cavalry'  ), 'swordsmen';
is fight_resolve('pikemen', 'archers'    ), 'pikemen';
is fight_resolve('pikemen', 'swordsmen'  ), 'swordsmen';
is fight_resolve('pikemen', 'pikemen'    ), 'pikemen';
is fight_resolve('pikemen', 'cavalry'    ), 'pikemen';
is fight_resolve('cavalry', 'archers'    ), 'cavalry';
is fight_resolve('cavalry', 'swordsmen'  ), 'cavalry';
is fight_resolve('cavalry', 'pikemen'    ), 'pikemen';
is fight_resolve('cavalry', 'cavalry'    ), 'cavalry';

Oleksii Filonenko • Aug 23 '19

Rust:

enum Piece {
    Archer,
    Swordsman,
    Pikeman,
    Cavalry,
}

fn fight_resolve<'a>(attacker: &'a Piece, defender: &'a Piece) -> &'a Piece {
    use Piece::*;

    match (attacker, defender) {
        (Swordsman, Archer) | (Pikeman, Swordsman) | (Cavalry, Pikeman) | (Archer, Cavalry) => {
            defender
        }
        (attacker, _) => attacker,
    }
}

La blatte • Aug 19 '19

How bout some G O L F

f=(a,d)=>(t='ASPC',(t.indexOf(d[0])-t.indexOf(a[0])+1)%4?a:d)

If I understood the problem correctly, the defender only wins if it is the "previous" element on the list. Hence, I just store a string with the initials in the correct order, check the initial index. If the defender is just before the attacker, it wins.

r=['Archers', 'Swordsmen', 'Pikemen', 'Cavalry'];
s='',r.forEach(e=>r.forEach(k=>s+=`${e} VS ${k}: ${f(e,k)}\n`)),s 


"Archers VS Archers: Archers
Archers VS Swordsmen: Archers
Archers VS Pikemen: Archers
Archers VS Cavalry: Cavalry
Swordsmen VS Archers: Archers
Swordsmen VS Swordsmen: Swordsmen
Swordsmen VS Pikemen: Swordsmen
Swordsmen VS Cavalry: Swordsmen
Pikemen VS Archers: Pikemen
Pikemen VS Swordsmen: Swordsmen
Pikemen VS Pikemen: Pikemen
Pikemen VS Cavalry: Pikemen
Cavalry VS Archers: Cavalry
Cavalry VS Swordsmen: Cavalry
Cavalry VS Pikemen: Pikemen
Cavalry VS Cavalry: Cavalry
"

Chris Achard • Aug 17 '19

I put a front end on it with React:

I kept the actual method super simple though with a bunch of if statements! Though there are a whole bunch of more clever ways to do it :)

Seiei Miyagi • Aug 17 '19

ruby <3

Swordsman, Cavalry, Archer, Pikeman = :swardsman, :cavalry, :archer, :pikeman

def fightResolve(attacking_piece, defending_piece)
  case [attacking_piece, defending_piece]
  when [Swordsman, Archer],
       [Pikeman, Swordsman],
       [Cavalry, Pikeman],
       [Archer, Cavalry]
    defending_piece
  else
    attacking_piece
  end
end

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Daily Challenge #43 - Boardgame Fight Resolver

Top comments (19)

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