Combinatoric commanders

For the discussion of math. Duh.

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fealuinix
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Combinatoric commanders

Postby fealuinix » Thu Dec 29, 2016 5:59 am UTC

Imagine five colors on a wheel--white, blue, black, red, and green, looping back to white.  I have 20 cards with some of those colors on each.  5 have four of the colors, each missing a different one.  5 have two colors; 1 of each possible combination of colors next to each other on the wheel (white-blue, blue-black, etc.)  10 have two colors that are not next to each other; 2 distinct cards for each possible color combination.  How many ways can I make 5 piles of cards where each pile has exactly four colors, each pile is missing a different color, and each pile consists of either 1 four-color card, or 2 two-color cards?

Some of you may recognize that what I'm really asking is "If I have one of each of the 2016 MTG Commander decks, how many ways can I swap around the commanders for all five decks?"

Lothario O'Leary
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Joined: Fri Feb 05, 2016 3:39 pm UTC

Re: Combinatoric commanders

Postby Lothario O'Leary » Thu Dec 29, 2016 7:52 am UTC

I tried to do it by listing the possible cases, but it's really complicated. I'd try a program, but I'm not that good of a programmer.
It looks like the answer is somewhere around 20000, however.

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PM 2Ring
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Location: Mid north coast, NSW, Australia

Re: Combinatoric commanders

Postby PM 2Ring » Thu Dec 29, 2016 12:23 pm UTC

I'll represent the colors by their initials, except I'll use the printers' convention of using K for black, so our colors in color-wheel order are WBKRG.

The 5 four-color cards are

BKRG WKRG WBRG WBKG WBKR

The 15 two-color cards are

WB BK KR RG GW WK KW WR RW BR RB BG GB KG GK

For each color there are 7 two-color pairs that do not contain that color. Here are the pairs that don't contain W:

BK+RG
BR+GK
BR+KG
KR+BG
KR+GB
RB+GK
RB+KG

Thus for a given color there are 1 + 7 = 8 different ways to make a pile that doesn't contain that color since we can either choose the appropriate four-colour card or one of the 7 two-colour pairs.

Now things start to get tricky. :)

Here's some Python code that does a brute force search of the 85 = 32768 combinations looking for ones that don't reuse any two-color cards. It finds 5915 different ways to make the piles.

This code runs on Python 3 or Python 2.7; with slight modifications it can run on older versions of Python 2.

Code: Select all

from __future__ import print_function
from itertools import combinations, product

colors = 'WBKRG'
colorset = set(colors)
print(colors)

# Four-color cards
col4 = [colors[:i] + colors[i+1:] for i in range(len(colors))]
print(*col4)
groups = {k: [(u,)] for k, u in zip(colors, col4)}

# Two-color adjacents
col2 = [u + v for u, v in zip(colors, colors[1:] + colors[:1])]

# Add the two-color non-adjacents
for u, v in combinations(colors, 2):
    a, b = u + v, v + u
    if not (a in col2 or b in col2):
        col2.extend((a, b))

print(*col2)

pairs = []
for u, v in combinations(col2, 2):
    k = colorset - set(u + v)
    if len(k) == 1:
        pairs.append((k.pop(), (u, v)))

pairs.sort()

for k, t in pairs:
    groups[k].append(t)

for k in colors:
    print(k, groups[k])

# Generate valid piles
good = 0
for piles in product(*[groups[k] for k in colors]):
    used = set()
    for t in piles:
        if len(t) == 2:
            s = set(t)
            if not used.intersection(s):
                used.update(s)
            else:
                break
    else:
        #print(piles)
        good += 1

print('\nNumber of ways to make the piles =', good)

Output:

Code: Select all

WBKRG
BKRG WKRG WBRG WBKG WBKR
WB BK KR RG GW WK KW WR RW BR RB BG GB KG GK
W [('BKRG',), ('BK', 'RG'), ('BR', 'GK'), ('BR', 'KG'), ('KR', 'BG'), ('KR', 'GB'), ('RB', 'GK'), ('RB', 'KG')]
B [('WKRG',), ('KR', 'GW'), ('RG', 'KW'), ('RG', 'WK'), ('RW', 'GK'), ('RW', 'KG'), ('WR', 'GK'), ('WR', 'KG')]
K [('WBRG',), ('GW', 'BR'), ('GW', 'RB'), ('RW', 'BG'), ('RW', 'GB'), ('WB', 'RG'), ('WR', 'BG'), ('WR', 'GB')]
R [('WBKG',), ('BK', 'GW'), ('KW', 'BG'), ('KW', 'GB'), ('WB', 'GK'), ('WB', 'KG'), ('WK', 'BG'), ('WK', 'GB')]
G [('WBKR',), ('BK', 'RW'), ('BK', 'WR'), ('KW', 'BR'), ('KW', 'RB'), ('WB', 'KR'), ('WK', 'BR'), ('WK', 'RB')]

Number of ways to make the piles = 5915

To get this code to actually list all of the piles just uncomment the #print(piles) line.

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Sizik
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Re: Combinatoric commanders

Postby Sizik » Thu Dec 29, 2016 2:46 pm UTC

PM 2Ring wrote:I'll represent the colors by their initials, except I'll use the printers' convention of using K for black, so our colors in color-wheel order are WBKRG.

The official convention is to use B for black and U for blue.
gmalivuk wrote:
King Author wrote:If space (rather, distance) is an illusion, it'd be possible for one meta-me to experience both body's sensory inputs.
Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses.

Demki
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Joined: Fri Nov 30, 2012 9:29 pm UTC

Re: Combinatoric commanders

Postby Demki » Thu Dec 29, 2016 3:54 pm UTC

Sizik wrote:
PM 2Ring wrote:I'll represent the colors by their initials, except I'll use the printers' convention of using K for black, so our colors in color-wheel order are WBKRG.

The official convention is to use B for black and U for blue.

PM 2Ring refers to the CMYK color model, where K stands for Key and is the black color.
In fact, B is quite commonly used for Blue, as in the RGB color model

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WibblyWobbly
Can't Get No
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Re: Combinatoric commanders

Postby WibblyWobbly » Thu Dec 29, 2016 4:03 pm UTC

Demki wrote:
Sizik wrote:
PM 2Ring wrote:I'll represent the colors by their initials, except I'll use the printers' convention of using K for black, so our colors in color-wheel order are WBKRG.

The official convention is to use B for black and U for blue.

PM 2Ring refers to the CMYK color model, where K stands for Key and is the black color.
In fact, B is quite commonly used for Blue, as in the RGB color model

Sizik is referring to the long-standing convention in M:tG. Which is more or less the way WotC did it way back in the long long ago (at least, since I started, which was during the Revised/Third Ed. era)

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PM 2Ring
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Location: Mid north coast, NSW, Australia

Re: Combinatoric commanders

Postby PM 2Ring » Fri Dec 30, 2016 7:58 am UTC

Sizik wrote:
PM 2Ring wrote:I'll represent the colors by their initials, except I'll use the printers' convention of using K for black, so our colors in color-wheel order are WBKRG.

The official convention is to use B for black and U for blue.


Ah, ok. I'm not really into Pokémon I don't know much about M:tG.

If you want to run my code using that convention just change the color assignment line from

Code: Select all

colors = 'WBKRG'

to

Code: Select all

colors = 'WUBRG'


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