I am designing a tournament scheduling algorithm but i am facing a problem that 2 same player may match again within the same tournament.
I am not sure are there an official name for this type of tournament, will be appreciate if someone can let me know it name too.
Requirement:
- Winner get 3 points, loser get 0 points, and 1 points each player for draw
- Player will play against another player who have similar score in each round
- The tournament end when the Max(score) have only 1 player
What I tried to do:
1. Initial the data in db as follow
tournamentId playerId score
1 A 0
1 B 0
......
1 Z 0
back end select the db by tournamentId=1, order by score and do some shuffling for the same score players.
take out players 2 by 2 as the first round schedule.
using the example in 1. I will have A vs B, C vs D and so on.
Update the score after all games done and start from 2 again until Max(score) have 1 player.
My Problem:
For my implementation, there will be a possibility that a player will play with the same player again.
For example:
round 1 : A vs B
...
round 4 : if both of them have the same score, they may meet again
How can I modify my algorithm to prevent this issue?
You have to introduce another condition for this special case. For example, if the players have already played against each other, you pick the most suitable next candidate, which didn't play against one of those players, even if he has a lower score than both. Actually, you can have a look at the swiss system tournament. I think it is quite close to what you want to achieve.
Related
I have been working on the algorithm for this problem, but can't figure it out. The problem is below:
In a tournament with X player, each player is betting on the outcomes of basketball matches in the NBA.
Guessing the correct match outcome earns a player 3 points, guessing the MVP of the match earns 1 point and guessing both wrong - 0 points.
The algorithm needs to be able to determine if a certain player can't reach the number 1 spot in this betting game.
For example, let's say there are a total of 30 games in the league, so the max points a player can get for guessing right is (3+1)*30=120.
In the table below you can see players X,Y and Z.
Player X guessed correctly so far 20 matches so he have 80 points.
Players Y and Z have 26 and 15 points, and since there are only 10 matches left, even if they guess correctly all the remaining 10 it would not be enough to reach the number 1 spot.
Therefore, the algorithm determined that they are eliminated from the game.
Team
Points
Points per match
Total Games
Max Points possible
Games left
Points Available
Eliminated?
X
80
0-L 1-MVP 3-W
30
120
10
0-40
N
Y
26
0-L 1-MVP 3-W
30
120
10
0-40
Y
Z
15
0-L 1-MVP 3-W
30
120
10
0-40
Y
The baseball elimination problem seems to be the most similar to this problem, but it's not exactly it.
How should I build the reduction of the maximum-flow problem to suit this problem?
Thank you.
I don't get why you are looking at very complex max-flow algorithms. Those might be needed for very complex things (especially when pairings lead to zero-sum results and order/remaining parings start to matter -> !much! harder to do worst-case analysis).
Maybe the baseball problem you mention is one of those (did not check it). But your use-case sounds trivial.
1. Get current leader score LS
2. Get remaining matches N
3. For each player P
4. Get current player score PS
5. Eliminate iff PS + 3 * N < LS
(assumes parallel progress: standings always synced to all players P have played M games
-> easy to generalize though)
This is simple. Given your description there is nothing preventing us from asumming worst-case performance from every other player aka it's a valid scenario that every other player guesses wrong for all upcoming guesses -> score S of player P can stay at S for all remaining games.
Things might quickly change to NP-hard decision-problems when there are more complex side-constraints (e.g. statistical distributions / expectations)
I have a requirement to represent ladder results, that were previously kept in an excel sheet, on a website.
The players on the ladder don't have scores. If player A beat player B, player A's going to be placed above player B in the ladder.
My problem is the following:
Let's say I have 5 players, ranked as follows:
Player 1
Player 2
Player 3
Player 4
Player 5
If Player 5 beat Player 1, Player 5 now moves on the top of the ladder. If the player order is decided on the ranking number I have to update the ranking number of all players between those who actually played which just sounds wrong to me.
I had an idea of assigning scores to each player, if Player A beats Player B then Player A gets Player B's score + 1, then the ranking is score in descending order. This works reasonably well up to the point when players end up on the same score. I can get around that too with adding a "LastUpdated" field in the database so whichever score got last updated goes below the other (the LastUpdated player hasn't actually beaten the player on the same score, they were just moved up there). This then fails at the point when a person does multiple adjustments on the same day so multiple players end up on the same score and the ranking order purely depends on which player got updated last. This is a problem as there's a requirement for the admin to be able to move players up and down freely. It also makes it impossible to add a player between players that have the same score (the winning player would have to jump the whole group).
My last resort would be to get the current ranking order from the database, put it all in a table, allow the admin to freely move players wherever they want, then have a single save operation that would get the current ranking of all the players and update everybody. This sounds horribly wrong though.
There are about 160 players in the ladder so the operation wouldn't be particularly lengthy, just sounds wrong.
So, what would the best way to change ranking order and get it saved in the database be?
If your database support exact numeric data types, it can be easily solved by having the ranking column an exact numeric.
Then when player 5 beat player 2, all you have to do is set player 5's rank between player 2's and player 1's. since it's an exact numeric your limit is the precision limit of the data type of choice.
Sample data:
id name rank
1. Player 1 1.0
2. Player 2 2.0
3. Player 3 3.0
4. Player 4 4.0
5. Player 5 5.0
Now player 5 beat player 2. so all you have to do is set it's rank to 1.5, that's between player 2's rank and the first rank higher then player 2's.
If player 4 would beat player 1, I would set it's rank to 0.5, since player 1 has no one ranked higher.
I have a specific number of teams. I want every team to play exactly 4 matches with 4 different opponents at 4 specified times.
The complication arises from that no team can play 2 different matches at the same time.
For example if team 1 is playing like this
team1 vs team2, team1 vs team3, team1 vs team4, team1 vs team5
then team2 already has the first time slot occupied so team2 can play like this
(team2 vs team1), team2 vs team3, team2 vs team4, team2 vs team5
But here the problem arises, team3 will play in the second time slot with team1 and team2 and this can not be done.
I do not know what this algorithm can be called, but I am searching for algorithm to implement this.
I made a search to find round-robin and other tournament like matching algorithm and also the marriage problem, but I think my problem is different. Please correct me if I am wrong.
Any help is greatly appreciated.
I have concluded there is no solution if the number of teams is odd. Let N be the number of teams. We need a total of N*4/2 matches, four matches per team but each match counts for two teams. To do N*2 matches in four time slots we must average N/2 matches per slot. We can do at most floor(N/2) matches at a time. If N is odd, floor(N/2) < N/2.
Would a solution that only works for even N, if it exists, be useful?
A simple algorithm:
Round 1
1 2 3 4
8 7 6 5
Then rotate places 2-8...
Round 2
1 8 2 3
7 6 5 4
Round 3
1 7 8 2
6 5 4 3
http://en.wikipedia.org/wiki/Round-robin_tournament#Scheduling_algorithm
(If there is an odd number, one can extend the this by adding a dummy pairing to indicate a bye, but then as Patricia Shanahan notes, not every team would play every round. So an even number of teams and at least six teams are needed to fit the requirements.)
This algorithm will work for any number of teams.
Suppose there are 6 teams in a tournament.
This solution basically tells you how to fill the 6x6 matrix, and each entry in the matrix is the match between the teams in the row vs column.
Consider few rules in the algorithm
1. Increment value in the matrix from left to right and top to bottom. mat[0][0] = 1
2. Whenever i == j, then fill the matrix at [n-1][j] instead at [i][j]. Basically, no entry will be there at i==j
3. And whenever entry in the matrix reaches to 6, make it 1
We will follow this rule and start filling the matrix from [0][0] column-wise.
Means first we will fill every row of 0th column then move to 1st column and so.
- At [0][0], apply rule 2. So fill mat[n-1][0] = 0
- At mat[1][0], fill next number ie 2 and similarly for [2][0], [3][0], [4][0]
- And now column 1, starts with value 2
- mat[1][0] = 2;
- At mat[1][1] apply rule 2, fill last row of current column ie mat[n-1][1] = 3
If you want that every team should play only one game with other team, use lower triangle.
And if you want that every team should play 2 games with other teams, one home and away use both lower and upper triangle.
Hope you guys understand my solution.
Cheers
I have a data set of players' skill ranking, age and sex and would like to create evenly matched teams.
Teams will have the same number of players (currently 8 teams of 12 players).
Teams should have the same or similar male to female ratio.
Teams should have similar age curve/distribution.
I would like to try this in Haskell but the choice of coding language is the least important aspect of this problem.
This is a bin packing problem, or a multi-dimensional knapsack problem. Björn B. Brandenburg has made a bin packing heuristics library in Haskell that you may find useful.
You need something like...
data Player = P { skill :: Int, gender :: Bool, age :: Int }
Decide on a number of teams n (I'm guessing this is a function of the total number of players).
Find the desired total skill per team:
teamSkill n ps = sum (map skill ps) / n
Find the ideal gender ratio:
genderRatio ps = sum (map (\x -> if gender x then 1 else 0)) / length ps
Find the ideal age variance (you'll want the Math.Statistics package):
ageDist ps = pvar (map age ps)
And you must assign the three constraints some weights to come up with a scoring for a given team:
score skillW genderW ageW team = skillW * sk + genderW * g + ageW * a
where (sk, (g, a)) = (teamSkill 1 &&& genderRatio &&& ageDist) team
The problem reduces to the minimization of the difference in scores between teams. A brute force approach will take time proportional to Θ(nk−1). Given the size of your problem (8 teams of 12 players each), this translates to about 6 to 24 hours on a typical modern PC.
EDIT
An approach that may work well for you (since you don't need an exact solution in practise) is simulated annealing, or continual improvement by random permutation:
Pick teams at random.
Get a score for this configuration (see above).
Randomly swap players between two or more teams.
Get a score for the new configuration. If it's better than the previous one, keep it and recurse to step 3. Otherwise discard the new configuration and try step 3 again.
When the score has not improved for some fixed number of iterations (experiment to find the knee of this curve), stop. It's likely that the configuration you have at this point will be close enough to the ideal. Run this algorithm a few times to gain confidence that you have not hit on some local optimum that is considerably worse than ideal.
Given the number of players per team and the gender ration (which you can easily compute). The remaining problem is called n-partition problem, which is unfortunately NP-complete and thus very hard to solve exactly. You will have to use approximative or heuristic allgorithms (evolutionary algorithms), if your problem size is too big for a brute force solution. A very simple approximation would be sorting by age and assign in an alternating way.
Assign point values to the skill levels, gender, and age
Assign the sum of the points for each criteria to each player
Sort players by their calculated point value
Assign the next player to the first team
Assign players to the second team until it has >= total points than the first team or the team reaches the maximum players.
Perform 5 for each team, looping back to the first team, until all players are assigned
You can tweak the skill level, gender, and age point values to change the distribution of each.
Lets say you have six players (for a simple example). We can use the same algorithm which pairs opponents in single-elimination tournaments and adapt that to generate "even" teams based on any criteria you choose.
First rank your players best-to-worst. Don't take this too literally. You want a list of players sorted by the criteria you wish to separate them.
Why?
Let's look at single elimination tournaments for a second. The idea of using an algorithm to generate optimal single-elimination matches is to avoid the problem of the "top players" meeting too soon in the tournament. If top players meet too soon, one of the top players will be eliminated early on, making the tournament less interesting. We can use this "optimal" pairing to generate teams in which the "top" players are spread out evenly across the teams. Then spread out the the second top players, etc, etc.
So list you players by the criteria you want them separated: men first, then women... sorted by age second. We get (for example):
Player 1: Male - 18
Player 2: Male - 26
Player 3: Male - 45
Player 4: Female - 18
Player 5: Female - 26
Player 6: Female - 45
Then we'll apply the single-elimination algorithm which uses their "rank" (which is just their player number) to create "good match ups".
The single-elimination tournament generator basically works like this: take their rank (player number) and reverse the bits (binary). This new number you come up with become their "slot" in the tournament.
Player 1 in binary (001), reversed becomes 100 (4 decimal) = slot 4
Player 2 in binary (010), reversed becomes 010 (2 decimal) = slot 2
Player 3 in binary (011), reversed becomes 110 (6 decimal) = slot 6
Player 4 in binary (100), reversed becomes 001 (1 decimal) = slot 1
Player 5 in binary (101), reversed becomes 101 (5 decimal) = slot 5
Player 6 in binary (110), reversed becomes 011 (3 decimal) = slot 3
In a single-elimination tournament, slot 1 plays slot 2, 3-vs-4, 5-vs-6. We're going to uses these "pair ups" to generate optimal teams.
Looking at the player number above, ordered by their "slot number", here is the list we came up with:
Slot 1: Female - 18
Slot 2: Male - 26
Slot 3: Female - 45
Slot 4: Male - 18
Slot 5: Female - 26
Slot 6: Male - 45
When you split the slots up into teams (two or more) you get the players in slot 1-3 vs players in slot 4-6. That is the best/optimal grouping you can get.
This technique scales very well with many more players, multiple criteria (just group them together correctly), and multiple teams.
Idea:
Sort players by skill
Assign best players in order (i.e.: team A: 1st player, team B: 2nd player, ...)
Assign worst players in order
Loop on 2
Evaluate possible corrections and perform them (i.e.: if team A has a total skill of 19 with a player with skill 5 and team B has a total skill of 21 with a player with skill 4, interchange them)
Evaluate possible corrections on gender distribution and perform them
Evaluate possible corrections on age distribution and perform them
Almost trivial approach for two teams:
Sort all player by your skill/rank assessment.
Assign team A the best player.
Assign team B the next two best players
Assign team A the next two best players
goto 3
End when you're out of players.
Not very flexible, and only works on one column ranking, so it won't try to get similar gender or age profiles. But it does make fair well matched teams if the input distribution is reasonably smooth. Plus it doesn't always end with team A have the spare player when there are an odd number.
Well,
My answer is not about scoring strategies of teams/players because all the posted are good, but I would try a brute force or a random search approach.
I don't think it's worth create a genetic algorithm.
Regards.
I have a project for school where I have to come up with an algorithm for scheduling 4 teams to play volleyball on one court, such that each team gets as close to the same amount of time as possible to play.
If you always have the winners stay in and rotate out the loser, then the 4th ranked team will never play and the #1 team always will.
The goal is to have everybody play the same amount of time.
The simplest answer is team 1 play team 2, then team 3 play team 4 and keep switching, but then team 1 never gets to play team 3 or 4 and so on.
So I'm trying to figure out an algorithm that will allow everybody to play everybody else at some point without having one team sit out a lot more than any other team.
Suggestions?
How about this: Make a hashtable H of size NC2, in this case, 6. It looks like:
H[12] = 0
H[13] = 0
H[14] = 0
H[23] = 0
H[24] = 0
H[34] = 0
I am assuming it would be trivial to generate the keys.
Now to schedule a game, scan through the hash and pick the key with the lowest value (one pass). The teams denoted by the key play the game and you increment the value by one.
EDIT:
To add another constraint that no team should wait too long, make another hash W:
W[1] = 0
W[2] = 0
W[3] = 0
W[4] = 0
After every game increment the W value for the team that did not play, by one.
Now when picking up the least played team if there are more than one team combo with low play score, take help from this hash to determine which team must play next.
well you should play 1-2 3-4, 1-3 2-4, 1-4 2-3 and then start all over again.
If there are N teams and you want all pairs of them to play once, then there are "N choose 2" = N*(N-1)/2 games you need to run.
To enumerate them, just put the teams in an ordered list and have the first team play every other team, then have the second team play all the teams below it in the list, and so on. If you want to spread the games out so teams have similar rest intervals between games, then see Knuth.
Check out the wikipedia entry on round robin scheduling.
pretend it's a small sports league, and repeat the "seasons"...
(in most sports leagues in Europe, all teams play against all other teams a couple of times during a season)
The REQUIREMENTS for the BALANCED ROUND ROBIN algorithm, for the Team championship scheduling may be found here:
Constellation Algorithm - Balanced Round Robin
The requirements of the algorithm can be defined by these four constraints:
1) All versus all
Each team must meet exactly once, and once only, the other teams in the division/ league.
If the division is composed of n teams, the championship takes place in the n-1 rounds.
2) Alternations HOME / AWAY rule
The sequence of alternations HOME / AWAY matches for every teams in the division league, should be retained if possible. For any team in the division league at most once in the sequence of consecutive matches HAHA, occurs the BREAK of the rhythm, i.e. HH or AA match in the two consecutive rounds.
3) The rule of the last slot number
The team with the highest slot number must always be positioned in the last row of the grid. For each subsequent iteration the highest slot number of grid alternates left and right position; left column (home) and right (away).
The system used to compose the league schedule is "counter-clockwise circuit." In the construction of matches in one round of the championship, a division with an even number of teams. If in a division is present odd number of teams, it will be inserted a BYE/Dummy team in the highest slot number of grid/ring.
4) HH and AA non-terminal and not initial
Cadence HH or AA must never happen at the beginning or at the end of the of matches for any team in the division.