Colonizer is a tool developed to better analyze the popular board game, Settler of Catan. It can help (beginner) players make better decisions and understand their common gameplay mistakes such as placing a suboptimal initial settlement, or performing an unfavorable trade using economic theories.
Colonizer keeps track of statistics to help a player make better decisions. Currently, I am only focusing on the initial settlement placements, but eventually I'd hope that Colonizer can become a bot that can make decisions on its own for the whole duration of the game.
This app is built in Python and uses tkinter for UI/UX.
Games should be loaded using the GameBoards.xlsx file, 1 game per row.
For example, if we are trying to load this specific game (that I actually played on Colonist) as game 1 :
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In the GameBoards file, column A will be the (sequential) ID of the games that we can load.
- For this game, we will put 1 in column A
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Columns B:T will be the resource tiles, which should be recorded from left to right, then row by row from top to bottom. Note the resources need to be named as Lumber, Brick, Sheep, Wheat, Rock, and Desert.
- For this game, the resource tiles should be recorded as:
- (row 1) lumber, brick, rock,
- (row 2) sheep, lumber, sheep, wheat,
- (row 3) rock, sheep, brick, desert, sheep,
- (row 4) wheat, lumber, brick, rock,
- (row 5) lumber, wheat, wheat
- For this game, the resource tiles should be recorded as:
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Columns U:AC will be for the ports, which should be recorded clockwise, starting from the port attached to the first resource tile in the upper left corner. Note that general ports (3 for 1 ports) should be recorded as desert ports.
- For this game, it will be recorded as: rock, desert, desert, wheat, lumber, desert, desert, sheep, brick
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In the game of Catan, the rules actually specified how the dice numbers should be assigned. The sequence is [5, 2, 6, 3, 8, 10, 9, 12, 11, 4, 8, 10, 9, 4, 5, 6, 3, 11], and needs to be assigned starting from a randomly chosen cornered resource tile, and assigned in either direction (clockwise or counter-clockwise), from the outer-most ring to the most central resource hex.
- First, look at where the 5 is, it will be on one of the 6 corner hex tiles, unless the first tile is a desert, then look for a 5 next to the desert. Going clockwise, the "5" Starting Location Offset will be either 0, 2, 4, 6, 8, or 10.
- For this game, the first 5 dice sequence begins at position 0, we will record 0 in column AD
- Just for reference:
- Offset of 2 is currently taken by rock 8
- Offset of 4 is currently taken by sheep 11
- Offset of 6 is currently taken by wheat 9
- Offset of 8 is currently taken by lumber 8
- Offset of 10 is currently taken by rock 6
- Next, look at where the 2 is next to the initial 5, unless it is a desert, then go to the next tile. Which direction does the 2 follow the 5? It will be either clockwise or counter-clockwise.
- For this game, it's counter-clockwise, so we will record False for ClockwiseDice in column AE.
- First, look at where the 5 is, it will be on one of the 6 corner hex tiles, unless the first tile is a desert, then look for a 5 next to the desert. Going clockwise, the "5" Starting Location Offset will be either 0, 2, 4, 6, 8, or 10.
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Colonizer will check to make sure the board is a valid board (having the correct number of tiles, etc). And the dice number will be automatically assigned based on the initial position and the direction of assignment.
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Update the gameID in
ColonizerMain.pyto the game that we want to study. Or set it as 0 and then latest game will be chosen instead.
- For this game, it'll be 1.
- If everything is set up correctly, you should get the following board.
We can compare how well each player setup their initial settlements using a few different methods to evaluate their performance.
Let's supposed that the players placed their initial settlements like this:
Methodologies:
- Getting the Most Resources: Players are ranked by their total resource production ability. Here is a sample calculation for Orange:
- 10 brick has probability of 0.0833
- 8 rock has probability of 0.1389
- 3 sheep has probability of 0.0556
- 3 wheat has probability of 0.0556
- 5 lumber has probability of 0.1111
- 8 lumber has probability of 0.1389
- Total probability (score) = 0.583
| Orange's Rank (and Score) | Black's Rank (and Score) | Blue's Rank (and Score) | Red's Rank (and Score) |
|---|---|---|---|
| 1 (0.583) | 4 (0.500) | 3 (0.528) | 2 (0.556) |
- Getting the Most & Diverse Resources (Sharpe Ratio): Players are ranked by their total resource production ability, divided by the standard deviation between their resource productions. Here is a sample calculation for Orange:
- Total lumber production of 0.1111 + 0.1389 = 0.2500
- Total brick production of 0.0833
- Total sheep production of 0.0556
- Total wheat production of 0.0556
- Total rock production of 0.1389
- Mean of production = 0.1167
- Standard deviation of production = 0.0733
- Sharpe ratio (score) = 0.1167 / 0.0733 = 1.592
| Orange's Rank (and Score) | Black's Rank (and Score) | Blue's Rank (and Score) | Red's Rank (and Score) |
|---|---|---|---|
| 2 (1.592) | 1 (2.654) | 4 (1.440) | 3 (1.534) |
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Getting the Most Rare Resources by Tiles Available: Players are ranked by their resource production ability; but resource values are weighted depending on how many resource tiles are available. For example, because there are always 4 lumber tiles available, but only 3 brick tiles, 4 lumbers are worth the same as 3 bricks. Here is a sample calculation for Orange:
- Because there are 4 lumber tiles, 3 brick tiles, 4 sheep tiles, 4 wheat tiles, and 3 rock ties
- The average resource value is (4 + 3 + 4 + 4 + 3) / 5 = 3.6
- The relative lumber value is 3.6/4
- The relative brick value is 3.6/3 (more valuable than lumber)
- The relative sheep value is 3.6/4 (just as valuable as lumber)
- The relative wheat value is 3.6/4
- The relative rock value is 3.6/3
- Total lumber production of (0.1111 + 0.1389) * 3.6/4 = 0.2250
- Total brick production of 0.0833 * 3.6/3 = 0.1000
- Total sheep production of 0.0556 * 3.6/4 = 0.0500
- Total wheat production of 0.0556 * 3.6/4 = 0.0500
- Total rock production of 0.1389 * 3.6/3 = 0.1667
- Total production score = 0.2550 + 0.1000 + 0.0500 + 0.0500 + 0.1667 = 0.592
Orange's Rank (and Score) Black's Rank (and Score) Blue's Rank (and Score) Red's Rank (and Score) 1 (0.592) 4 (0.508) 3 (0.542) 2 (0.542) - Because there are 4 lumber tiles, 3 brick tiles, 4 sheep tiles, 4 wheat tiles, and 3 rock ties
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Getting the Most Rare Resources by Tiles' Probability: Players are ranked by their resource production ability; but resource values are weighted depending on how the board is set up randomly. For example, if there are less rock tiles available, like the example game above, then rocks are worth more. Here is a sample calculation for Orange:
- We need to calculate how scarce each resource is:
- The scarcity of lumber is 0.1111 + 0.1111 + 0.1111 + 0.1389 = 0.4722
- The 4 values are the probability of each lumber tile, which are 5 (11.11%), 9 (11.11%), 5 (11.11%), and 8 (8.33%)
- The scarcity of brick is 0.0833 + 0.0556 + 0.1389 = 0.2778
- The scarcity of sheep is 0.0278 + 0.0556 + 0.0833 + 0.0556 = 0.2222
- The scarcity of wheat is 0.0833 + 0.0556 + 0.0833 + 0.1111 = 0.3333
- The scarcity of rock is 0.1389 + 0.1389 + 0.0278 = 0.3056
- The scarcity of lumber is 0.1111 + 0.1111 + 0.1111 + 0.1389 = 0.4722
- Calculate the average value of resources, which will always be 0.3222
- (0.4722 + 0.2778 + 0.2222 + 0.3333 + 0.3056)/5 = 0.3222
- We calculate the relative value of each resource:
- The relative lumber value is 0.3222/0.4722
- The relative brick value is 0.3222/0.2778
- The relative sheep value is 0.3222/0.2222
- The relative wheat value is 0.3222/0.3333
- The relative rock value is 0.3222/0.3056
- Lastly, for Orange:
- Total lumber production of (0.1111 + 0.1389) * 0.3222/0.4722 = 0.1706
- Total brick production of 0.0833 * 0.3222/0.2778 = 0.0967
- Total sheep production of 0.0556 * 0.3222/0.2222 = 0.0806
- Total wheat production of 0.0556 * 0.3222/0.3333 = 0.0537
- Total rock production of 0.1389 * 0.3222/0.3056 = 0.1465
- Total production score = 0.1706 + 0.0967 + 0.0806 + 0.0537 + 0.1465 = 0.548
- We need to calculate how scarce each resource is:
| Orange's Rank (and Score) | Black's Rank (and Score) | Blue's Rank (and Score) | Red's Rank (and Score) |
|---|---|---|---|
| 2 (0.548) | 3 (0.542) | 4 (0.475) | 1 (0.554) |
- Getting the Most Rarely Produced Resources: Players are ranked by their resource production ability; but resource values are weighted depending on what resources are occupied by all players. For example, if rocks are rare (like the example game), but still occupied more than another resource (even after taking probability into account), rock's value will be driven down.
- We need to calculate how scarce each resource is based on what all players produce in the economy:
- The scarcity of lumber is 3 * 0.1111 + 1 * 0.1389 + 3 * 0.1111 = 0.8056
- The value is calculated because there are 3 settlements on 5 lumber, 1 settlement on 8 lumber, and 3 settlements on 9 lumber
- The scarcity of brick is 0.3889
- The scarcity of sheep is 0.3056
- The scarcity of wheat is 0.2778
- The scarcity of rock is 0.4167
- The scarcity of lumber is 3 * 0.1111 + 1 * 0.1389 + 3 * 0.1111 = 0.8056
- Calculate the average value of resources
- (0.8056 + 0.3889 + 0.3056 + 0.2778 + 0.4167)/5 = 0.4389
- We calculate the relative value of each resource:
- The relative lumber value is 0.4389/0.8056
- The relative brick value is 0.4389/0.3889
- The relative sheep value is 0.4389/0.3056
- The relative wheat value is 0.4389/0.2778
- The relative rock value is 0.4389/0.4167
- Lastly, for Orange:
- Total lumber production of (0.1111 + 0.1389) * 0.4389/0.8056 = 0.1362
- Total brick production of 0.0833 * 0.4389/0.3889 = 0.0940
- Total sheep production of 0.0556 * 0.4389/0.3056 = 0.0798
- Total wheat production of 0.0556 * 0.4389/0.2778 = 0.0878
- Total rock production of 0.1389 * 0.4389/0.4167 = 0.1463
- Total production score = 0.1362 + 0.0940 + 0.0798 + 0.0878 + 0.1463 = 0.544
- We need to calculate how scarce each resource is based on what all players produce in the economy:
| Orange's Rank (and Score) | Black's Rank (and Score) | Blue's Rank (and Score) | Red's Rank (and Score) |
|---|---|---|---|
| 3 (0.544) | 2 (0.555) | 4 (0.494) | 1 (0.574) |
Some functionalities are already built in the tool, with many more to be expanded.
- Debug Mode: Very useful when some formulas aren't working as expected, shows the coordinates for each of the hex tiles, building number, and road number
- Dice Stats: Dice rolls can be tracked, and compared against the empirical distribution to see if the specific roll is "hot" or "cold"
- Resource Tracking:
- When the dice is rolled, resources are automatically distributed
- When trading occurs, right click to increment a resource for a specific player, and left click to decrement a resource for a specific player
- Development Cards Tracking: The functionality is not yet robust, but there is basic framework to set up how many cards are bought (and hidden), how many cards are played, and how many cards are known and unplayed (your own cards)
- Building Tracking: Settlements are represented with a thin ring around the building, and cities are represented with a thick ring around the building
- Useful Statistics: There are also some useful statistics built within the app
- Resources: from Resource Tracking
- Development Cards Tracking* from Development Cards Tracking
- Dice rolls: from Dice stats
- Economic power: based on resource production, the bank's economic power is the total resource production power summed over all players
- Building location value: taking into consideration of how the board is set up to calculate how much each building location is worth
In Catan, the most important goal is to win. In fact, it does not matter how you win, such as how many resources you need or how many turns it takes for one player to win. As long as the player is the first agent to get 10 victory points (or more), a winner is declared.
In a side analysis, WaysToWin.py explores all the pathways to victory that a player can take to achieve this and becomes the winner. This is actually a combinatorial optimization problem, or the Knapsack problem, where a player needs to "fill their bags" with a combination of "items" that are worth points.
There are in fact, 142 total ways to win the game (142 different game ends result for a player to win). Some are trivial, some are very specific scenarios where it is not at all likely to achieve in a real game.
The result of this analysis is not yet considered in the main Colonizer app. However, it is the first step that I have taken to understand how to win a game in Catan. We can consider further expanding the result of this analysis, such as the probability of drawing development cards, or how some pathways are known to be unachievable at any point of the game.
Clearly, there is a lot to expand, with many directions to focus on, but for now, there is now a basic framework that is easy to work with.
Pull requests are very welcomed. Feel free to improve any part of the code, and make this a learning experience for everyone interested in getting involved.
While this is my first major project using Python, I really would appreciate any feedback that you may have. Feedback, both good or bad will help me learn and get better.
Contributions are welcome as well. I love cooperating with people and get new ideas!