I was solving the common coin change problem and it makes sense to solve using DP, since the greedy approach won't work here. Eg. For amount 12, if coins = {3,5} then greedy algorithm to use minimum coins will use two 5-coins and then fail. So far so good.
Now I came across another problem where we are given an integer which we want to split into a sum of a maximum number of unique positive even integers.
This problem looked a lot like the coin change problem (as in for both problems we have to add upto a sum using the given denominations - repeating is okay in coins problem and not okay in this one), however I found out that this can be solved using a greedy approach.
I can of-course try and write down different use cases and figure this out, but is there any way of intuitively coming to this conclusion? Asking for this problem, and also in general.
Related
General problem
I have a set of short strings, each of different length with minimum X > 0 and maximum Y. What is an algorithm which will optimally fit together these short strings to make long strings of length M, where M >> Y? Optimal would be defined as the greatest number of long strings with lengths closest to M as possible.
Details
I am writing a tweet creator to practice javascript. I have a list of greetings and a list of account names. I want my program to create tweets such that each tweet has one greeting and the rest of the characters are used for account names. Each tweet has a limit of 140 characters.
Hello! #person1 #acc2 #mygoodfriend3 ...
Of course, each account has a different number of characters. I want each tweet to use up as many of the 140 characters as possible by optimally selecting combinations of account names.
I am pretty certain there is a class of problems / algorithm that is known to solve this problem but I can't remember it.
This kind of problem is called a knapsack problem, and an exact (or exactly optimal) solution is famous for being intractable.
However, there are reasonable approximate solvers, as well as a "pseudo-polynomial time" dynamic-programming algorithm.
I think it is related to the knapsack problem; it is a particular case of "multiple linear bin packing problem". Both are NP-hard problems. Here you can find a greedy algorithm for the linear bin packing problem, but the multiple case is much harder. A constraint programming language/library would help to solve these kind of problems.
There's an algorithm currently driving me crazy.
I've seen quite a few variations of it, so I'll just try to explain the easiest one I can think about.
Let's say I have a project P:
Project P is made up of 4 sub projects.
I can solve each of those 4 in two separate ways, and each of those modes has a specific cost and a specific time requirement:
For example (making it up):
P: 1 + 2 + 3 + 4 + .... n
A(T/C) Ta1/Ca1 Ta2/Ca2 etc
B(T/C) Tb1/Cb1 etc
Basically I have to find the combination that of those four modes which has the lowest cost. And that's kind of easy, the problem is: the combination has to be lower than specific given time.
In order to find the lowest combination I can easily write something like:
for i = 1 to n
aa[i] = min(aa[i-1],ba[i-1]) + value(a[i])
bb[i] = min(bb[i-1],ab[i-1]) + value(b[i])
ba[i] = min(bb[i-1],ab[i-1]) + value(b[i])
ab[i] = min(aa[i-1],ba[i-1]) + value(a[i])
Now something like is really easy and returns the correct value every time, the lowest at the last circle is gonna be the correct one.
Problem is: if min returns modality that takes the last time, in the end I'll have the fastest procedure no matter the cost.
If if min returns the lowest cost, I'll have the cheapest project no matter the amount of time taken to realize it.
However I need to take both into consideration: I can do it easily with a recursive function with O(2^n) but I can't seem to find a solution with dynamic programming.
Can anyone help me?
If there are really just four projects, you should go with the exponential-time solution. There are only 16 different cases, and the code will be short and easy to verify!
Anyway, the I'm pretty sure the problem you describe is the knapsack problem, which is NP-hard. So, there will be no exact solution that's sub-exponential unless P=NP. However, depending on what "n" actually is (is it 4 in your case? or the values of the time and cost?) there may be a pseudo-polynomial time solution. The Wikipedia article contains descriptions of these.
I think the best way to form this question is with an example...so, the actual reason I decided to ask about this is because of because of Problem 55 on Project Euler. In the problem, it asks to find the number of Lychrel numbers below 10,000. In an imperative language, I would get the list of numbers leading up to the final palindrome, and push those numbers to a list outside of my function. I would then check each incoming number to see if it was a part of that list, and if so, simply stop the test and conclude that the number is NOT a Lychrel number. I would do the same thing with non-lychrel numbers and their preceding numbers.
I've done this before and it has worked out nicely. However, it seems like a big hassle to actually implement this in Haskell without adding a bunch of extra arguments to my functions to hold the predecessors, and an absolute parent function to hold all of the numbers that I need to store.
I'm just wondering if there is some kind of tool that I'm missing here, or if there are any standards as a way to do this? I've read that Haskell kind of "naturally caches" (for example, if I wanted to define odd numbers as odds = filter odd [1..], I could refer to that whenever I wanted to, but it seems to get complicated when I need to dynamically add elements to a list.
Any suggestions on how to tackle this?
Thanks.
PS: I'm not asking for an answer to the Project Euler problem, I just want to get to know Haskell a bit better!
I believe you're looking for memoizing. There are a number of ways to do this. One fairly simple way is with the MemoTrie package. Alternatively if you know your input domain is a bounded set of numbers (e.g. [0,10000)) you can create an Array where the values are the results of your computation, and then you can just index into the array with your input. The Array approach won't work for you though because, even though your input numbers are below 10,000, subsequent iterations can trivially grow larger than 10,000.
That said, when I solved Problem 55 in Haskell, I didn't bother doing any memoization whatsoever. It turned out to just be fast enough to run (up to) 50 iterations on all input numbers. In fact, running that right now takes 0.2s to complete on my machine.
This is a problem appeared in today's Pacific NW Region Programming Contest during which no one solved it. It is problem B and the complete problem set is here: http://www.acmicpc-pacnw.org/icpc-statements-2011.zip. There is a well-known O(n^2) algorithm for LCS of two strings using Dynamic Programming. But when these strings are extended to rings I have no idea...
P.S. note that it is subsequence rather than substring, so the elements do not need to be adjacent to each other
P.S. It might not be O(n^2) but O(n^2lgn) or something that can give the result in 5 seconds on a common computer.
Searching the web, this appears to be covered by section 4.3 of the paper "Incremental String Comparison", by Landau, Myers, and Schmidt at cost O(ne) < O(n^2), where I think e is the edit distance. This paper also references a previous paper by Maes giving cost O(mn log m) with more general edit costs - "On a cyclic string to string correcting problem". Expecting a contestant to reproduce either of these papers seems pretty demanding to me - but as far as I can see the question does ask for the longest common subsequence on cyclic strings.
You can double the first and second string and then use the ordinary method, and later wrap the positions around.
It is a good idea to "double" the strings and apply the standard dynamic programing algorithm. The problem with it is that to get the optimal cyclic LCS one then has to "start the algorithm from multiple initial conditions". Just one initial condition (e.g. setting all Lij variables to 0 at the boundaries) will not do in general. In practice it turns out that the number of initial states that are needed are O(N) in number (they span a diagonal), so one gets back to an O(N^3) algorithm.
However, the approach does has some virtue as it can be used to design efficient O(N^2) heuristics (not exact but near exact) for CLCS.
I do not know if a true O(N^2) exist, and would be very interested if someone knows one.
The CLCS problem has quite interesting properties of "periodicity": the length of a CLCS of
p-times reapeated strings is p times the CLCS of the strings. This can be proved by adopting a geometric view off the problem.
Also, there are some additional benefits of the problem: it can be shown that if Lc(N) denotes the averaged value of the CLCS length of two random strings of length N, then
|Lc(N)-CN| is O(\sqrt{N}) where C is Chvatal-Sankoff's constant. For the averaged length L(N) of the standard LCS, the only rate result of which I know says that |L(N)-CN| is O(sqrt(Nlog N)). There could be a nice way to compare Lc(N) with L(N) but I don't know it.
Another question: it is clear that the CLCS length is not superadditive contrary to the LCS length. By this I mean it is not true that CLCS(X1X2,Y1Y2) is always greater than CLCS(X1,Y1)+CLCS(X2,Y2) (it is very easy to find counter examples with a computer).
But it seems possible that the averaged length Lc(N) is superadditive (Lc(N1+N2) greater than Lc(N1)+Lc(N2)) - though if there is a proof I don't know it.
One modest interest in this question is that the values Lc(N)/N for the first few values of N would then provide good bounds to the Chvatal-Sankoff constant (much better than L(N)/N).
As a followup to mcdowella's answer, I'd like to point out that the O(n^2 lg n) solution presented in Maes' paper is the intended solution to the contest problem (check http://www.acmicpc-pacnw.org/ProblemSet/2011/solutions.zip). The O(ne) solution in Landau et al's paper does NOT apply to this problem, as that paper is targeted at edit distance, not LCS. In particular, the solution to cyclic edit distance only applies if the edit operations (add, delete, replace) all have unit (1, 1, 1) cost. LCS, on the other hand, is equivalent to edit distances with (add, delete, replace) costs (1, 1, 2). These are not equivalent to each other; for example, consider the input strings "ABC" and "CXY" (for the acyclic case; you can construct cyclic counterexamples similarly). The LCS of the two strings is "C", but the minimum unit-cost edit is to replace each character in turn.
At 110 lines but no complex data structures, Maes' solution falls towards the upper end of what is reasonable to implement in a contest setting. Even if Landau et al's solution could be adapted to handle cyclic LCS, the complexity of the data structure makes it infeasible in a contest setting.
Last but not least, I'd like to point out that an O(n^2) solution DOES exist for CLCS, described here: http://arxiv.org/abs/1208.0396 At 60 lines, no complex data structures, and only 2 arrays, this solution is quite reasonable to implement in a contest setting. Arriving at the solution might be a different matter, though.
I am aware that languages like Prolog allow you to write things like the following:
mortal(X) :- man(X). % All men are mortal
man(socrates). % Socrates is a man
?- mortal(socrates). % Is Socrates mortal?
yes
What I want is something like this, but backwards. Suppose I have this:
mortal(X) :- man(X).
man(socrates).
man(plato).
man(aristotle).
I then ask it to give me a random X for which mortal(X) is true (thus it should give me one of 'socrates', 'plato', or 'aristotle' according to some random seed).
My questions are:
Does this sort of reverse inference have a name?
Are there any languages or libraries that support it?
EDIT
As somebody below pointed out, you can simply ask mortal(X) and it will return all X, from which you can simply pick a random one from the list. What if, however, that list would be very large, perhaps in the billions? Obviously in that case it wouldn't do to generate every possible result before picking one.
To see how this would be a practical problem, imagine a simple grammar that generated a random sentence of the form "adjective1 noun1 adverb transitive_verb adjective2 noun2". If the lists of adjectives, nouns, verbs, etc. are very large, you can see how the combinatorial explosion is a problem. If each list had 1000 words, you'd have 1000^6 possible sentences.
Instead of the deep-first search of Prolog, a randomized deep-first search strategy could be easyly implemented. All that is required is to randomize the program flow at choice points so that every time a disjunction is reached a random pole on the search tree (= prolog program) is selected instead of the first.
Though, note that this approach does not guarantees that all the solutions will be equally probable. To guarantee that, it is required to known in advance how many solutions will be generated by every pole to weight the randomization accordingly.
I've never used Prolog or anything similar, but judging by what Wikipedia says on the subject, asking
?- mortal(X).
should list everything for which mortal is true. After that, just pick one of the results.
So to answer your questions,
I'd go with "a query with a variable in it"
From what I can tell, Prolog itself should support it quite fine.
I dont think that you can calculate the nth solution directly but you can calculate the n first solutions (n randomly picked) and pick the last. Of course this would be problematic if n=10^(big_number)...
You could also do something like
mortal(ID,X) :- man(ID,X).
man(X):- random(1,4,ID), man(ID,X).
man(1,socrates).
man(2,plato).
man(3,aristotle).
but the problem is that if not every man was mortal, for example if only 1 out of 1000000 was mortal you would have to search a lot. It would be like searching for solutions for an equation by trying random numbers till you find one.
You could develop some sort of heuristic to find a solution close to the number but that may affect (negatively) the randomness.
I suspect that there is no way to do it more efficiently: you either have to calculate the set of solutions and pick one or pick one member of the superset of all solutions till you find one solution. But don't take my word for it xd