Making card configurations in Sagemath

asked 2023-06-21 02:13:05 +0200

Luisz
11 ●2

I'm following SageMath's tutorial on Combinatorics (can't post link here cause I have low karma) and one of the exercizes is to calculate the set of Four of a Kind hands (in card games a hand containing four cards of the same value is called a four of a kind and a Hand is a subset of 5 cards). Each card has a Value and a Suit.

I want to generate the hand in a structured way like {(1,"Hearts"), (1,"Spades"), (1,"Clubs"), (1,"Diamonds"), (5,"Clubs") )}, so my idea was to create all the cards first and then join them in a set, like so:

Suits = Set(["Hearts", "Diamonds", "Spades", "Clubs"])
Values = Set([2, 3, 4, 5, 6, 7, 8, 9, 10,
              "Jack", "Queen", "King", "Ace"])
Cards = cartesian_product([Values, Suits])

pick = Values.random_element()
card1 = Subsets(cartesian_product([Set([pick]), Suits]),1)
card2 = Subsets(cartesian_product([Set([pick]), Suits]),1)
card3 = Subsets(cartesian_product([Set([pick]), Suits]),1)
card4 = Subsets(cartesian_product([Set([pick]), Suits]),1)
card5 = Subsets(Cards,1)

FourOfaKind = cartesian_product([card1,card2,card3,card4,card5])

FourOfaKind.cardinality()

FourOfaKind.random_element()

However, I realised I'm picking Suits with reposition, which is not what I want!

How can i tell Sage Math that I don't want to pick the choice of the previous card? If i instantiate a Suit with Suit.random_element() each time I fix a card, then that won't be counted as part of the construction of the Suit/{whatever suit is instantiated} for the following cards, which puts me in quite a pickle! I'm sure there's some method to do this but I've been at hours for this and the Sage documentation is huge!

Any help would be really appreciated!

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Tutorial link: https://doc.sagemath.org/html/en/refe...

John Palmieri ( 2023-06-21 22:30:13 +0200 )edit

The exercise gives a hint: use Arrangements(Values, 2) to get two values, and then choose a suit for the first value. Then the hand will be (1st value + suit, 4 of a kind of 2nd value).

John Palmieri ( 2023-06-21 22:35:10 +0200 )edit

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answered 2023-06-21 22:43:27 +0200

John Palmieri

8551 ●19 ●71 ●186 http://www.math.washin...

Using what you have so far:

Suits = Set(["Hearts", "Diamonds", "Spades", "Clubs"])
Values = Set([2, 3, 4, 5, 6, 7, 8, 9, 10, "Jack", "Queen", "King", "Ace"])
X = Arrangements(Values, 2)  # two values: first will be the singleton, second will be 4 of a kind

Now we can construct the hands in several ways. Using a loop:

Hands = []
for a in X:
    for s in Suits:
        first_card = (a[0], s)
        rest = [(a[1], Suits[i]) for i in range(4)]
        hand = [first_card] + rest
        Hands.append(hand)

Or more briefly using list comprehension:

Hands = [[(a[0], s)] + [(a[1], Suits[i]) for i in range(4)] for a in X for s in Suits]

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So, essentialy, we build the entire set of Hands from scratch! (computational bruteforce) Brilliant! Thank you very much for the answers, I found it very enlightening indeed!

Luisz ( 2023-06-24 11:00:39 +0200 )edit

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answered 2023-06-21 14:39:21 +0200

Max Alekseyev

5733 ●5 ●37 ●123

Here is a way:

 FourOfaKind = [ {(v,s) for s in Suits} | {(v2,s2)} for v,v2 in Permutations(Values,2) for s2 in Suits ]

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