Identifying Seating Arrangements at the Poker Table - Sit-and-Go Events

Note: The JavaScript code has been implemented based on the answer provided by ajrwhite. Hopefully, it proves helpful to someone.

Link: http://codepen.io/eMineiro/pen/EKrNBe

Be sure to open the CodePen console to see the examples in action.

When playing poker, player positions are defined in relation to the dealer, like this:

https://i.sstatic.net/QlktT.jpg

Blue: Small Blind and Big Blind positions

Green: Late and Dealer/Late positions

Yellow: Middle positions

Pink: Early positions

Given the following arrays:

players:[1,2,3,4,5,6,7,8,9,10];
positions:["bb","sb","btn","late","medium","medium","medium","early","early","early"];

For instance, "player1" is the "Big Blind", "player2" is the "Small Blind", and so on...

The goal is to sort the players array when changePositions(dealer) is called. For example:

changePosition(10); //indicating that "player10" is now the new Dealer

The resulting arrays should be:

players:[2,1,10,9,8,7,6,5,4,3];
positions:["bb","sb","btn","late","medium","medium","medium","early","early","early"];

If players are eliminated during the game, the "last position" in the positions array needs to be excluded along with the player. Subsequently, changePosition(X) should be called again, with X being the next non-eliminated player to the left of "player10" (current dealer).

For instance, if "player1" is eliminated, the new arrays would be:

players:[2,10,9,8,7,6,5,4,3];
positions:["bb","sb","btn","late","medium","medium","medium","early","early"];

Subsequently, changePosition(X) needs to be called again to determine the new positions. In this scenario, X=2, since "player2" is to the left of the current dealer "player10".

changePosition(2);

The resulting arrays should then be:

players:[4,3,2,10,9,8,7,6,5];
positions:["bb","sb","btn","late","medium","medium","medium","early","early"];

How can the new dealer be determined when a player gets eliminated?

Note: To address this, a function named changeNextDealer() was created. Dealing with negative indexes was not an issue since the next dealer is in a clockwise direction. Refer to the code pen link for details.

dealerArrayPosition-1; //However, when the Big Blind and Small Blind are eliminated simultaneously, a negative position is obtained.

Is there a quick way to map a negative index like -1 to the last position, or -2 to the LastPosition-1?

Note: While this specific question remains unanswered, it is not the primary focus of this discussion. It may be addressed in a separate post.

How should the changePosition(dealer) function be implemented?

Despite numerous attempts, figuring out how to accomplish this has proven challenging.

Note: A function called changePosition() has been created. It can be found in the CodePen link provided.

Answer №1

Let's say we always assign numbers 1-10 to the players. Instead of using two separate data structures to store player positions and statuses, we can simply use one indexed data structure to keep track of player state:

State 1

positions:["btn","sb","bb","early","early","early","medium","medium","medium","late"];

State 2

positions: ["late","btn","sb","bb","early","early","early","medium","medium","medium"]

State 3

positions: ["eliminated","late","btn","sb","bb","early","early","early","medium","medium"];

This setup holds the same information as using two arrays, but it is more consistent. To check a player's state, you can simply access positions[0] for player 1 or positions[9] for player 10.

With this more consistent structure, managing status changes at the end of each round should be simpler.

Removing players

After each round, update the status of an eliminated player to "eliminated". You may need a temporary array to review the previous round's results while updating information for the next round.

Treating the array as a circle

Shift the dealer chip "btn" to (currentPositionOfBtn+1)%10 - this way, the Button moves in a cycle from 1 to 2 to 3, etc. The remainder operator and total player count ensure that the Button returns to position 1 after reaching position 10.

If the player at (currentPositionOfBtn+1)%10 is eliminated, check (currentPositionOfBtn+2)%10 and continue - it's a straightforward loop to implement.

Reassigning positions

Since player positions change as eliminations occur (removing the "early" or "medium" categories), I recommend recalculating positions starting from the Dealer after each round, skipping eliminated players.

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