I have project that I need to display a leaderboard of the top 20, and if the user not in the leaderboard they will appear in the 21st place with their current ranking.
Is there efficient way to this?
I am using Cloud Firestore as a database. I believe it was mistake to choose it instead of MongoDB but I am in the middle of the project so I must do it with Cloud Firestore.
The app will be use by 30K users. Is there any way to do it without getting all the 30k users?
this.authProvider.afs.collection('profiles', ref => ref.where('status', '==', 1)
.where('point', '>', 0)
.orderBy('point', 'desc').limit(20))
This is code I did to get the top 20 but what will be the best practice for getting current logged in user rank if they are not in the top 20?
Finding an arbitrary player's rank in leaderboard, in a manner that scales is a common hard problem with databases.
There are a few factors that will drive the solution you'll need to pick, such as:
Total Number players
Rate that individual players add scores
Rate that new scores are added (concurrent players * above)
Score range: Bounded or Unbounded
Score distribution (uniform, or are their 'hot scores')
Simplistic approach
The typical simplistic approach is to count all players with a higher score, eg SELECT count(id) FROM players WHERE score > {playerScore}.
This method works at low scale, but as your player base grows, it quickly becomes both slow and resource expensive (both in MongoDB and Cloud Firestore).
Cloud Firestore doesn't natively support count as it's a non-scalable operation. You'll need to implement it on the client-side by simply counting the returned documents. Alternatively, you could use Cloud Functions for Firebase to do the aggregation on the server-side to avoid the extra bandwidth of returning documents.
Periodic Update
Rather than giving them a live ranking, change it to only updating every so often, such as every hour. For example, if you look at Stack Overflow's rankings, they are only updated daily.
For this approach, you could schedule a function, or schedule App Engine if it takes longer than 540 seconds to run. The function would write out the player list as in a ladder collection with a new rank field populated with the players rank. When a player views the ladder now, you can easily get the top X + the players own rank in O(X) time.
Better yet, you could further optimize and explicitly write out the top X as a single document as well, so to retrieve the ladder you only need to read 2 documents, top-X & player, saving on money and making it faster.
This approach would really work for any number of players and any write rate since it's done out of band. You might need to adjust the frequency though as you grow depending on your willingness to pay. 30K players each hour would be $0.072 per hour($1.73 per day) unless you did optimizations (e.g, ignore all 0 score players since you know they are tied last).
Inverted Index
In this method, we'll create somewhat of an inverted index. This method works if there is a bounded score range that is significantly smaller want the number of players (e.g, 0-999 scores vs 30K players). It could also work for an unbounded score range where the number of unique scores was still significantly smaller than the number of players.
Using a separate collection called 'scores', you have a document for each individual score (non-existent if no-one has that score) with a field called player_count.
When a player gets a new total score, you'll do 1-2 writes in the scores collection. One write is to +1 to player_count for their new score and if it isn't their first time -1 to their old score. This approach works for both "Your latest score is your current score" and "Your highest score is your current score" style ladders.
Finding out a player's exact rank is as easy as something like SELECT sum(player_count)+1 FROM scores WHERE score > {playerScore}.
Since Cloud Firestore doesn't support sum(), you'd do the above but sum on the client side. The +1 is because the sum is the number of players above you, so adding 1 gives you that player's rank.
Using this approach, you'll need to read a maximum of 999 documents, averaging 500ish to get a players rank, although in practice this will be less if you delete scores that have zero players.
Write rate of new scores is important to understand as you'll only be able to update an individual score once every 2 seconds* on average, which for a perfectly distributed score range from 0-999 would mean 500 new scores/second**. You can increase this by using distributed counters for each score.
* Only 1 new score per 2 seconds since each score generates 2 writes
** Assuming average game time of 2 minute, 500 new scores/second could support 60000 concurrent players without distributed counters. If you're using a "Highest score is your current score" this will be much higher in practice.
Sharded N-ary Tree
This is by far the hardest approach, but could allow you to have both faster and real-time ranking positions for all players. It can be thought of as a read-optimized version of of the Inverted Index approach above, whereas the Inverted Index approach above is a write optimized version of this.
You can follow this related article for 'Fast and Reliable Ranking in Datastore' on a general approach that is applicable. For this approach, you'll want to have a bounded score (it's possible with unbounded, but will require changes from the below).
I wouldn't recommend this approach as you'll need to do distributed counters for the top level nodes for any ladder with semi-frequent updates, which would likely negate the read-time benefits.
Final thoughts
Depending on how often you display the leaderboard for players, you could combine approaches to optimize this a lot more.
Combining 'Inverted Index' with 'Periodic Update' at a shorter time frame can give you O(1) ranking access for all players.
As long as over all players the leaderboard is viewed > 4 times over the duration of the 'Periodic Update' you'll save money and have a faster leaderboard.
Essentially each period, say 5-15 minutes you read all documents from scores in descending order. Using this, keep a running total of players_count. Re-write each score into a new collection called scores_ranking with a new field players_above. This new field contains the running total excluding the current scores player_count.
To get a player's rank, all you need to do now is read the document of the player's score from score_ranking -> Their rank is players_above + 1.
One solution not mentioned here which I'm about to implement in my online game and may be usable in your use case, is to estimate the user's rank if they're not in any visible leaderboard because frankly the user isn't going to know (or care?) whether they're ranked 22,882nd or 22,838th.
If 20th place has a score of 250 points and there are 32,000 players total, then each point below 250 is worth on average 127 places, though you may want to use some sort of curve so as they move up a point toward bottom of the visible leaderboard they don't jump exactly 127 places each time - most of the jumps in rank should be closer to zero points.
It's up to you whether you want to identify this estimated ranking as an estimation or not, and you could add some a random salt to the number so it looks authentic:
// Real rank: 22,838
// Display to user:
player rank: ~22.8k // rounded
player rank: 22,882nd // rounded with random salt of 44
I'll be doing the latter.
Alternative perspective - NoSQL and document stores make this type of task overly complex. If you used Postgres this is pretty simple using a count function. Firebase is tempting because it's easy to get going with but use cases like this are when relational databases shine. Supabase is worth a look https://supabase.io/ similar to firebase so you can get going quickly with a backend but its opensource and built on Postgres so you get a relational database.
A solution that hasn't been mentioned by Dan is the use of security rules combined with Google Cloud Functions.
Create the highscore's map. Example:
highScores (top20)
Then:
Give the users write/read access to highScores.
Give the document/map highScores the smallest score in a property.
Let the users only write to highScores if his score > smallest score.
Create a write trigger in Google Cloud Functions that will activate when a new highScore is written. In that function, delete the smallest score.
This looks to me the easiest option. It is realtime as well.
You could do something with cloud storage. So manually have a file that has all the users' scores (in order), and then you just read that file and find the position of the score in that file.
Then to write to the file, you could set up a CRON job to periodically add all documents with a flag isWrittenToFile false, add them all to the file (and mark them as true). That way you won't eat up your writes. And reading a file every time the user wants to view their position is probably not that intensive. It could be done from a cloud function.
2022 Updated and Working Answer
To solve the problem of having a leaderboards with user and points, and to know your position in this leaderboards in an less problematic way, I have this solution:
1) You should create your Firestorage Document like this
In my case, I have a document perMission that has for each user a field, with the userId as property and the respective leaderboard points as value.
It will be easier to update the values inside my Javascript code.
For example, whenever an user completed a mission (update it's points):
import { doc, setDoc, increment } from "firebase/firestore";
const docRef = doc(db, 'leaderboards', 'perMission');
setDoc(docRef, { [userId]: increment(1) }, { merge: true });
The increment value can be as you want. In my case I run this code every time the user completes a mission, increasing the value.
2) To get the position inside the leaderboards
So here in your client side, to get your position, you have to order the values and then loop through them to get your position inside the leaderboards.
Here you can also use the object to get all the users and its respective points, ordered. But here I am not doing this, I am only interested in my position.
The code is commented explaining each block.
// Values coming from the database.
const leaderboards = {
userId1: 1,
userId2: 20,
userId3: 30,
userId4: 12,
userId5: 40,
userId6: 2
};
// Values coming from your user.
const myUser = "userId4";
const myPoints = leaderboards[myUser];
// Sort the values in decrescent mode.
const sortedLeaderboardsPoints = Object.values(leaderboards).sort(
(a, b) => b - a
);
// To get your specific position
const myPosition = sortedLeaderboardsPoints.reduce(
(previous, current, index) => {
if (myPoints === current) {
return index + 1 + previous;
}
return previous;
},
0
);
// Will print: [40, 30, 20, 12, 2, 1]
console.log(sortedLeaderboardsPoints);
// Will print: 4
console.log(myPosition);
You can now use your position, even if the array is super big, the logic is running in the client side. So be careful with that. You can also improve the client side code, to reduce the array, limit it, etc.
But be aware that you should do the rest of the code in your client side, and not Firebase side.
This answer is mainly to show you how to store and use the database in a "good way".
I have a unique problem and I'm not aware of any algorithm that can help me. Maybe someone on here does.
I have a dataset compiled from many different sources (teams). One field in particular is called "type". Here are some example values for type:
aple, apples, appls, ornge, fruits, orange, orange z, pear,
cauliflower, colifower, brocli, brocoli, leeks, veg, vegetables.
What I would like to be able to do is to group them together into e.g. fruits, vegetables, etc.
Put another way I have multiple spellings of various permutations of a parent level variable (fruits or vegetables in this example) and I need to be able to group them as best I can.
The only other potentially relevant feature of the data is the team that entered it, assuming some consistency in the way each team enters their data.
So, I have several million records of multiple spellings and short spellings (e.g. apple, appls) and I want to group them together in some way. In this example by fruits and vegetables.
Clustering would be challenging since each entry is most often 1 or two words, making it tricky to calculate a distance between terms.
Short of creating a massive lookup table created by a human (not likely with millions of rows), is there any approach I can take with this problem?
You will need to first solve the spelling problem, unless you have Google scale data that could allow you to learn fixing spelling with Google scale statistics.
Then you will still have the problem that "Apple" could be a fruit or a computer. Apple and "Granny Smith" will be completely different. You best guess at this second stage is something like word2vec trained on massive data. Then you get high dimensional word vectors, and can finally try to solve the clustering challenge, if you ever get that far with decent results. Good luck.
Is there a way I can match players using Google Play Game Service based on individual players' skill level in the game?
I have locally stored the player level of each player, and want a player to be matched to his/her closest ranked player.
For example: a player ranked 10 (beginner) should be paired with the closest ranked player available (e.g. 5 to 15) instead of an expert level 100 player, so that we can have a balanced competition.
There are two variables that can be set to influence the match making:
First, you can set a variant of the game using RoomConfig.Builder.setVariant(). This method takes an non-negative value indicating the type of match. The variant specified needs to match exactly with other participants in order for auto-matching to take place. I suppose you could be strict in your match making and use the variant as the player's level. In this way, players would only be matched with players of the same level. An alternative would be to group levels together in a range, for example levels 1-5 could play each other, likewise group 6-8, etc.
The second variable is the exclusiveBitMask. This is passed in when calling RoomConfig.CreateAutoMatchCritera(). This method takes the min and max number of players to match, and the exclusiveBitMask. This mask when logically AND'ed with the other players will equal 0. This is used for things like role based games (need to have 1 offense and 1 defense). One possible use of this would be to mask out high level vs. low level capabilities so there is no outrageous mismatch.
I think Clayton Wilkinson's answer is all correct and I voted it up.
But, I imagine the OP is hoping for some way to do skill-based matching without splitting the player-base into segments.
Sadly, the answer is no you can't. The choices are to use some other matchmaking system or split up your player base. If you choose to split your player base then you need a lot of concurrent users to avoid making your players wait a long time.
On a recent title we rolled our own matchmaking service based on Raknet because we wanted more nuanced matchmaking. It's a lot of hassle though, and GPGS is pretty great otherwise, so skill based matching would have to be a very high priority before you consider abandoning GPGS.
While studying for Interviews, this question came to my mind.
I am planning to design a parking lot and I am assuming the following things:
It has multiple levels. Each Level has 2 rows.
The vehicle type could be small, compact and large.
Each row at a certain level has multiple parking spots.
Each parking spot has red/green light indicator( Red- No space, Green- Free space)
Also each parking spot will have size small, compact and large.
6(optional). Also want to add handicapped person situation.
These are some of the assumptions that I could come up with(Not sure if they are enough or I need more).
I was thinking of designing the system in such a way that as soon as the Vehicle enters the Parking entrance.
He should be given information that where is the nearest vacant spot available(For example- Level 3, Row 2, Spot # 10)
I would like to know how to go about designing such a system? I have seen many other designs but none achieves this i guess.
Not sure what you want to achieve through point 6, but the rest are pretty simple to achieve with an OO design, through the usual abstraction, inheritance and polymorphism principles.
You can have an interface called AvailabilityIndicator which has a boolean method isAvailable(), which represents the light indicator (the bulb will show red if isAvailable() is false, and green if true).
You can have an abstract class called ParkingSlot, which implements AvailabilityIndicator.
This could have the level, row and spot number in it.
You can have 3 classes LargeParkingSlot, CompactParkingSlot and SmallParkingSlot which extend ParkingSlot. (Not that your functionality actually needs this, unless the different parking slots have different behaviours or data you want to model, but since you mentioned you wanted an OO approach I mentioned it, otherwise a simple slotType parameter in ParkingSlot would do.)
Then its a question of when the vehicle arrives, checking what type of slot it needs and looking up which are the available ones that match. You might want to put them in a Map data structure which maps each slot type to the list of available ones, so that when one is taken it is removed and put in a separate unavailable list, for fast lookup of available slots when a vehicle arrives.
I'm currently developing an iOS app which shows realtime data, now I receive 33,265 timingpoints from the API, which are all stops. So stops on opposite sides of the street are counted as 2, bus stations which have multiple platforms are also counted as many times as there are platforms.
Now, this is confusing on a map. You'd want all data for example a bus station on one screen and don't browse past 10 platforms to get the bus you'd like to take. So how can I group these annotation, which have the same name, and often are near or overlapping each other?
You can find an example of the JSON results from the API here: http://pastebin.com/RiKS4G0Q
Just make a new entity Location and have a to-one relationship to each stop (reverse is to-many, of course). Now one stop can share a location and you can present the dat in an appropriate way. During import, you could decide to create a new location if the coordinates are close enough to each other (and maybe the stop names correspond).