I'll get straight to it. I'm working on an web or phone app that is responsible for scheduling. I want students to input courses they took, and I give them possible combinations of courses they should take that fits their requirements.
However, let's say there's 150 courses that fits their requirements and they're looking for 3 courses. That would be 150C3 combinations, right?.
Would it be feasible to run something like this in browser or a mobile device?
First of all you need a smarter algorithm which can prune the search tree. Also, if you are doing this for the same set of courses over and over again, doing the computation on the server would be better, and perhaps precomputing a feasible data structure can reduce the execution time of the queries. For example, you can create a tree where each sub-tree under a node contains nodes that are 'compatible'.
Sounds to me like you're viewing this completely wrong. At most institutions there are 1) curriculum requirements for graduation, and 2) prerequisites for many requirements and electives. This isn't a pure combinatorial problem, it's a dependency tree. For instance, if Course 201, Course 301, and Course 401 are all required for the student's major, higher numbers have the lower numbered ones as prereqs, and the student is a Junior, you should be strongly recommending that Course 201 be taken ASAP.
Yay, mathematics I think I can handle!
If there are 150 courses, and you have to choose 3, then the amount of possibilities are (150*149*148)/(3*2) (correction per jerry), which is certainly better than 150 factorial which is a whole lot more zeros ;)
Now, you really don't want to build an array that size, and you don't have to! All web languages have the idea of randomly choosing an element in an array, so you get an element in an array and request 3 random unique entries from it.
While the potential course combinations is very large, based on your post I see no reason to even attempt to calculate them. This task of random selection of k items from n-sized list is delightfully trivial even for old, slow devices!
Is there any particular reason you'd need to calculate all the potential course combinations, instead of just grab-bagging one random selection as a suggestion? If not, problem solved!
Option 1 (Time\Space costly): let the user on mobile phone browse the list of (150*149*148) possible choices, page by page, the processing is done at the server-side.
Option 2 (Simple): instead of the (150*149*148)-item decision tree, provide a 150-item bag, if he choose one item from the bag, remove it from the bag.
Option 3 (Complex): expand your decision tree (possible choices) using a dependency tree (parent course requires child courses) and the list of course already taken by the student, and his track\level.
As far as I know, most educational systems use the third option, which requires having a profile for the student.
Related
I am looking to create a large list of items that allows for easy insertion of new items and for easily changing the position of items within that list. When updating the position of an item, I want to change as few fields as possible regarding the order of items.
After some research, I found that Jira's Lexorank algorithm fulfills all of these needs. Each story in Jira has a 'rank-field' containing a string which is built up of 3 parts: <bucket>|<rank>:<sub-rank>. (I don't know whether these parts have actual names, this is what I will call them for ease of reference)
Examples of valid rank-fields:
0|vmis7l:hl4
0|i000w8:
0|003fhy:zzzzzzzzzzzw68bj
When dragging a card above 0|vmis7l:hl4, the new card will receive rank 0|vmis7l:hl2, which means that only the rank-field for this new card needs to be updated while the entire list can always be sorted on this rank-field. This is rather clever, and I can't imagine that Lexorank is the only algorithm to use this.
Is there a name for this method of sorting used in the sub-rank?
My question is related to the creation of new cards in Jira. Each new card starts with an empty sub-rank, and the rank is always chosen such that the new card is located at the bottom of the list. I've created a bunch of new stories just to see how the rank would change, and it seems that the rank is always incremented by 8 (in base-36).
Does anyone know more specifically how the rank for new cards is generated? Why is it incremented by 8?
I can only imagine that after some time (270 million cards) there are no more ranks to generate, and the system needs to recalculate the rank-field of all cards to make room for additional ranks.
Are there other triggers that require recalculation of all rank-fields?
I suppose the bucket plays a role in this recalculation. I would like to know how?
We are talking about a special kind of indexing here. This is not sorting; it is just preparing items to end up in a certain order in case someone happens to sort them (by whatever sorting algorithm). I know that variants of this kind of indexing have been used in libraries for decades, maybe centuries, to ensure that books belonging together but lacking a common title end up next to each other in the shelves, but I have never heard of a name for it.
The 8 is probably chosen wisely as a compromise, maybe even by analyzing typical use cases. Consider this: If you choose a small increment, e. g. 1, then all tickets will have ranks like [a, b, c, …]. This will be great if you create a lot of tickets (up to 26) in the correct order because then your rank fields keep small (one letter). But as soon as you move a ticket between two other tickets, you will have to add a letter: [a, b] plus a new ticket between them: [a, an, b]. If you expect to have this a lot, you better leave gaps between the ranks: [a, i, q, …], then an additional ticket can get a single letter as well: [a, e, i, q, …]. But of course if you now create lots of tickets in the correct order right in the beginning, you quickly run out of letters: [a, i, q, y, z, za, zi, zq, …]. The 8 probably is a good value which allows for enough gaps between the tickets without increasing the need for many letters too soon. Keep in mind that other scenarios (maybe not Jira tickets which are created manually) might make other values more reasonable.
You are right, the rank fields get recalculated now and then, Lexorank calls this "balancing". Basically, balancing takes place in one of three occasions: ① The ranks are exhausted (largest value reached), ② the ranks are due to user-reranking of tickets too close together ([a, b, i] and something is supposed to go in between a and b), and ③ a balancing is triggered manually in the management page. (Actually, according to the presentation, Lexorank allows for up to three letter ranks, so "too close together" can be something like aaa and aab but the idea is the same.)
The <bucket> part of the rank is increased during balancing, so a messy [0|a, 0|an, 0|b] can become a nice and clean [1|a, 1|i, 1|q] again. The brownbag presentation about Lexorank (as linked by #dandoen in the comments) mentions a round-robin use of <buckets>, so instead of a constant increment (0→1→2→3→…) a 2 is increased modulo 3, so it will turn back to 0 after the 2 (0→1→2→0→…). When comparing the ranks, the sorting algorithm can consider a 0 "greater" than a 2 (it will not be purely lexicographical then, admitted). If now the balancing algorithm works backwards (reorder the last ticket first), this will keep the sorting order intact all the time. (This is just a side aspect, that's why I keep the explanation small, but if this is interesting, ask, and I will elaborate on this.)
Sidenote: Lexorank also keeps track of minimum and maximum values of the ranks. For the functioning of the algorithm itself, this is not necessary.
Here is the facts first.
In the game of bridge there are 4
players named North, South, East and
West.
All 52 cards are dealt with 13 cards
to each player.
There is a Honour counting systems.
Ace=4 points, King=3 points, Queen=2
points and Jack=1 point.
I'm creating a "Card dealer" with constraints where for example you might say that the hand dealt to north has to have exactly 5 spades and between 13 to 16 Honour counting points, the rest of the hands are random.
How do I accomplish this without affecting the "randomness" in the best way and also having effective code?
I'm coding in C# and .Net but some idea in Pseudo code would be nice!
Since somebody already mentioned my Deal 3.1, I'd like to point out some of the optimizations I made in that code.
First of all, to get the most flexibly constraints, I wanted to add a complete programming language to my dealer, so you could generate whole libraries of constraints with different types of evaluators and rules. I used Tcl for that language, because I was already learning it for work, and, in 1994 when Deal 0.0 was released, Tcl was the easiest language to embed inside a C application.
Second, I needed the constraint language to run fairly fast. The constraints are running deep inside the loop. Quite a lot of code in my dealer is little optimizations with lookup tables and the like.
One of the most surprising and simple optimizations was to not deal cards to a seat until a constraint is checked on that seat. For example, if you want north to match constraint A and south to match constraint B, and your constraint code is:
match constraint A to north
match constraint B to south
Then only when you get to the first line do you fill out the north hand. If it fails, you reject the complete deal. If it passes, next fill out the south hand and check its constraint. If it fails, throw out the entire deal. Otherwise, finish the deal and accept it.
I found this optimization when doing some profiling and noticing that most of the time was spent in the random number generator.
There is one fancy optimization, which can work in some instances, call "smart stacking."
deal::input smartstack south balanced hcp 20 21
This generates a "factory" for the south hand which takes some time to build but which can then very quickly fill out the one hand to match this criteria. Smart stacking can only be applied to one hand per deal at a time, because of conditional probability problems. [*]
Smart stacking takes a "shape class" - in this case, "balanced," a "holding evaluator", in this case, "hcp", and a range of values for the holding evaluator. A "holding evaluator" is any evaluator which is applied to each suit and then totaled, so hcp, controls, losers, and hcp_plus_shape, etc. are all holding evalators.
For smartstacking to be effective, the holding evaluator needs to take a fairly limited set of values. How does smart stacking work? That might be a bit more than I have time to post here, but it's basically a huge set of tables.
One last comment: If you really only want this program for bidding practice, and not for simulations, a lot of these optimizations are probably unnecessary. That's because the very nature of practicing makes it unworthy of the time to practice bids that are extremely rare. So if you have a condition which only comes up once in a billion deals, you really might not want to worry about it. :)
[Edit: Add smart stacking details.]
Okay, there are exactly 8192=2^13 possible holdings in a suit. Group them by length and honor count:
Holdings(length,points) = { set of holdings with this length and honor count }
So
Holdings(3,7) = {AK2, AK3,...,AKT,AQJ}
and let
h(length,points) = |Holdings(length,points)|
Now list all shapes that match your shape condition (spades=5):
5-8-0-0
5-7-1-0
5-7-0-1
...
5-0-0-8
Note that the collection of all possible hand shapes has size 560, so this list is not huge.
For each shape, list the ways you can get the total honor points you are looking for by listing the honor points per suit. For example,
Shape Points per suit
5-4-4-0 10-3-0-0
5-4-4-0 10-2-1-0
5-4-4-0 10-1-2-0
5-4-4-0 10-0-3-0
5-4-4-0 9-4-0-0
...
Using our sets Holdings(length,points), we can compute the number of ways to get each of these rows.
For example, for the row 5-4-4-0 10-3-0-0, you'd have:
h(5,10)*h(4,3)*h(4,0)*h(0,0)
So, pick one of these rows at random, with relative probability based on the count, and then, for each suit, choose a holding at random from the correct Holdings() set.
Obviously, the wider the range of hand shapes and points, the more rows you will need to pre-compute. A little more code, you can still do this with some cards pre-determined - if you know where the ace of spades or west's whole hand or whatever.
[*] In theory, you can solve these conditional probability issues for smart stacking with multiple hands, but the solution to the problem would make it effective only for extremely rare types of deals. That's because the number of rows in the factory's table is roughly the product of the number of rows for stacking one hand times the number of rows for stacking the other hand. Also, the h() table has to be keyed on the number of ways of dividing the n cards amongst hand 1, hand 2, and other hands, which changes the number of values from roughly 2^13 to 3^13 possible values, which is about two orders of magnitude bigger.
Since the numbers are quite small here, you could just take the heuristic approach: Randomly deal your cards, evaluate the constraints and just deal again if they are not met.
Depending on how fast your computer is, it might be enough to do this:
Repeat:
do a random deal
Until the board meets all the constraints
As with all performance questions, the thing to do is try it and see!
edit I tried it and saw:
done 1000000 hands in 12914 ms, 4424 ok
This is without giving any thought to optimisation - and it produces 342 hands per second meeting your criteria of "North has 5 spades and 13-16 honour points". I don't know the details of your application but it seems to me that this might be enough.
I would go for this flow, which I think does not affect the randomness (other than by pruning solutions that do not meet constraints):
List in your program all possible combinations of "valued" cards whose total Honour points count is between 13 and 16. Then pick randomly one of these combinations, removing the cards from a fresh deck.
Count how many spades you already have among the valued cards, and pick randomly among the remaining spades of the deck until you meet the count.
Now pick from the deck as much non-spades, non-valued cards as you need to complete the hand.
Finally pick the other hands among the remaining cards.
You can write a program that generates the combinations of my first point, or simply hardcode them while accounting for color symmetries to reduce the number of lines of code :)
Since you want to practise bidding, I guess you will likely be having various forms of constraints (and not just 1S opening, as I guess for this current problem) coming up in the future. Trying to come up with the optimal hand generation tailored to the constraints could be a huge time sink and not really worth the effort.
I would suggest you use rejection sampling: Generate a random deal (without any constraints) and test if it satisfies your constraints.
In order to make this feasible, I suggest you concentrate on making the random deal generation (without any constraints) as fast as you can.
To do this, map each hand to a 12byte integer (the total number of bridge hands fits in 12 bytes). Generating a random 12 byte integer can be done in just 3, 4 byte random number calls, of course since the number of hands is not exactly fitting in 12 bytes, you might have a bit of processing to do here, but I expect it won't be too much.
Richard Pavlicek has an excellent page (with algorithms) to map a deal to a number and back.
See here: http://www.rpbridge.net/7z68.htm
I would also suggest you look at the existing bridge hand dealing software (like Deal 3.1, which is freely available) too. Deal 3.1 also supports doing double dummy analysis. Perhaps you could make it work for you without having to roll one of your own.
Hope that helps.
Given a list of (say) songs, what's the best way to determine their relative "popularity"?
My first thought is to use Google Trends. This list of songs:
Subterranean Homesick Blues
Empire State of Mind
California Gurls
produces the following Google Trends report: (to find out what's popular now, I restricted the report to the last 30 days)
http://s3.amazonaws.com/instagal/original/image001.png?1275516612
Empire State of Mind is marginally more popular than California Gurls, and Subterranean Homesick Blues is far less popular than either.
So this works pretty well, but what happens when your list is 100 or 1000 songs long? Google Trends only allows you to compare 5 terms at once, so absent a huge round-robin, what's the right approach?
Another option is to just do a Google Search for each song and see which has the most results, but this doesn't really measure the same thing
Excellent question - one song by Britney Spears, might be phenomenally popular for 2 months then (thankfully) forgotten, while another song by Elvis might have sustained popularity for 30 years. How do you quantitatively distinguish the two? We know we want to think that sustained popularity is more important than a "flash in the pan", but how to get this result?
First, I would normalize around the release date - Subterranean Homesick Blues might be unpopular now (not in my house, though), but normalizing back to 1965 might yield a different result.
Since most songs climb in popularity, level off, then decline, let's choose the area when they level off. One might assume that during that period, that the two series are stationary, uncorrelated, and normally distributed. Now you can just apply a test to determine if the means are different.
There's probably less restrictive tests to determine the magnitude of difference between two time series, but I haven't run across them yet.
Anyone?
You could search for the item on Twitter and see how many times it is mentioned. Or look it up on Amazon to see how many people have reviewed it and what rating they gave it. Both Twitter and Amazon have APIs.
There is an unoffical google trends api. See http://zoastertech.com/projects/googletrends/index.php?page=Getting+Started I have not used it but perhaps it is of some help.
I would certainly treat Google's API of "restricted".
In general, comparison functions used for sorting algorithms are very "binary":
input: 2 elements
output: true/false
Here you have:
input: 5 elements
output: relative weights of each element
Therefore you will only need a linear number of calls to the API (whereas sorting usually requires O(N log N) calls to comparison functions).
You will need exactly ceil( (N-1)/4 ) calls. That you can parallelize, though do read the user guide closely as for the number of requests you are authorized to submit.
Then, once all of them are "rated" you can have a simple sort in local.
Intuitively, in order to gather them properly you would:
Shuffle your list
Pop the 5 first elements
Call the API
Insert them sorted in the result (use insertion sort here)
Pick up the median
Pop the 4 first elements (or less if less are available)
Call the API with the median and those 4 first
Go Back to Insert until your run out of elements
If your list is 1000 songs long, that 250 calls to the API, nothing too scary.
I'm study recommendation engines, and I went through the paper that defines how Google News generates recommendations to users for news items which might be of their interest, based on collaborative filtering.
One interesting technique that they mention is Minhashing. I went through what it does, but I'm pretty sure that what I have is a fuzzy idea and there is a strong chance that I'm wrong. The following is what I could make out of it :-
Collect a set of all news items.
Define a hash function for a user. This hash function returns the index of the first item from the news items which this user viewed, in the list of all news items.
Collect, say "n" number of such values, and represent a user with this list of values.
Based on the similarity count between these lists, we can calculate the similarity between users as the number of common items. This reduces the number of comparisons a lot.
Based on these similarity measures, group users into different clusters.
This is just what I think it might be. In Step 2, instead of defining a constant hash function, it might be possible that we vary the hash function in a way that it returns the index of a different element. So one hash function could return the index of the first element from the user's list, another hash function could return the index of the second element from the user's list, and so on. So the nature of the hash function satisfying the minwise independent permutations condition, this does sound like a possible approach.
Could anyone please confirm if what I think is correct? Or the minhashing portion of Google News Recommendations, functions in some other way? I'm new to internal implementations of recommendations. Any help is appreciated a lot.
Thanks!
I think you're close.
First of all, the hash function first randomly permutes all the news items, and then for any given person looks at the first item. Since everyone had the same permutation, two people have a decent chance of having the same first item.
Then, to get a new hash function, rather than choosing the second element (which would have some confusing dependencies on the first element), they choose a whole new permutation and take the first element again.
People who happen to have the same hash value 2-4 times (that is, the same first element in 2-4 permutations) are put together in a cluster. This algorithm is repeated 10-20 times, so that each person gets put into 10-20 clusters. Finally, recommendations are given based (the small number of) other people in the 10-20 clusters. Since all this work is done by hashing, people are put directly into buckets for their clusters, and large numbers of comparisons aren't needed.
We have an auto-complete list that's populated when an you send an email to someone, which is all well and good until the list gets really big you need to type more and more of an address to get to the one you want, which goes against the purpose of auto-complete
I was thinking that some logic should be added so that the auto-complete results should be sorted by some function of most recently contacted or most often contacted rather than just alphabetical order.
What I want to know is if there's any known good algorithms for this kind of search, or if anyone has any suggestions.
I was thinking just a point system thing, with something like same day is 5 points, last three days is 4 points, last week is 3 points, last month is 2 points and last 6 months is 1 point. Then for most often, 25+ is 5 points, 15+ is 4, 10+ is 3, 5+ is 2, 2+ is 1. No real logic other than those numbers "feel" about right.
Other than just arbitrarily picked numbers does anyone have any input? Other numbers also welcome if you can give a reason why you think they're better than mine
Edit: This would be primarily in a business environment where recentness (yay for making up words) is often just as important as frequency. Also, past a certain point there really isn't much difference between say someone you talked to 80 times vs say 30 times.
Take a look at Self organizing lists.
A quick and dirty look:
Move to Front Heuristic:
A linked list, Such that whenever a node is selected, it is moved to the front of the list.
Frequency Heuristic:
A linked list, such that whenever a node is selected, its frequency count is incremented, and then the node is bubbled towards the front of the list, so that the most frequently accessed is at the head of the list.
It looks like the move to front implementation would best suit your needs.
EDIT: When an address is selected, add one to its frequency, and move to the front of the group of nodes with the same weight (or (weight div x) for courser groupings). I see aging as a real problem with your proposed implementation, in that it requires calculating a weight on each and every item. A self organizing list is a good way to go, but the algorithm needs a bit of tweaking to do what you want.
Further Edit:
Aging refers to the fact that weights decrease over time, which means you need to know each and every time an address was used. Which means, that you have to have the entire email history available to you when you construct your list.
The issue is that we want to perform calculations (other than search) on a node only when it is actually accessed -- This gives us our statistical good performance.
This kind of thing seems similar to what is done by firefox when hinting what is the site you are typing for.
Unfortunately I don't know exactly how firefox does it, point system seems good as well, maybe you'll need to balance your points :)
I'd go for something similar to:
NoM = Number of Mail
(NoM sent to X today) + 1/2 * (NoM sent to X during the last week)/7 + 1/3 * (NoM sent to X during the last month)/30
Contacts you did not write during the last month (it could be changed) will have 0 points. You could start sorting them for NoM sent in total (since it is on the contact list :). These will be showed after contacts with points > 0
It's just an idea, anyway it is to give different importance to the most and just mailed contacts.
If you want to get crazy, mark the most 'active' emails in one of several ways:
Last access
Frequency of use
Contacts with pending sales
Direct bosses
Etc
Then, present the active emails at the top of the list. Pay attention to which "group" your user uses most. Switch to that sorting strategy exclusively after enough data is collected.
It's a lot of work but kind of fun...
Maybe count the number of emails sent to each address. Then:
ORDER BY EmailCount DESC, LastName, FirstName
That way, your most-often-used addresses come first, even if they haven't been used in a few days.
I like the idea of a point-based system, with points for recent use, frequency of use, and potentially other factors (prefer contacts in the local domain?).
I've worked on a few systems like this, and neither "most recently used" nor "most commonly used" work very well. The "most recent" can be a real pain if you accidentally mis-type something once. Alternatively, "most used" doesn't evolve much over time, if you had a lot of contact with somebody last year, but now your job has changed, for example.
Once you have the set of measurements you want to use, you could create an interactive apoplication to test out different weights, and see which ones give you the best results for some sample data.
This paper describes a single-parameter family of cache eviction policies that includes least recently used and least frequently used policies as special cases.
The parameter, lambda, ranges from 0 to 1. When lambda is 0 it performs exactly like an LFU cache, when lambda is 1 it performs exactly like an LRU cache. In between 0 and 1 it combines both recency and frequency information in a natural way.
In spite of an answer having been chosen, I want to submit my approach for consideration, and feedback.
I would account for frequency by incrementing a counter each use, but by some larger-than-one value, like 10 (To add precision to the second point).
I would account for recency by multiplying all counters at regular intervals (say, 24 hours) by some diminisher (say, 0.9).
Each use:
UPDATE `addresslist` SET `favor` = `favor` + 10 WHERE `address` = 'foo#bar.com'
Each interval:
UPDATE `addresslist` SET `favor` = FLOOR(`favor` * 0.9)
In this way I collapse both frequency and recency to one field, avoid the need for keeping a detailed history to derive {last day, last week, last month} and keep the math (mostly) integer.
The increment and diminisher would have to be adjusted to preference, of course.