Box Factory is a problem in Google Code Jam 2012 Round 1C. It is similar to the Longest Common Subsequence problem, and they have given an O(n^4) solution for it. However, at the end of the analysis it says that another improvement can reduce this again to O(n^3). I am wondering what optimization can be done to the solution.
O(n^4) Algorithm
The dynamic programming approach solves for f[x][y] = the maximum number of toys that could be placed in boxes using the first x runs of boxes and the first y runs of toys.
It solves this by considering the boxes of the last type for runs between a+1 and x, and toys of the last type for runs between b+1 and y.
The O(n^4) algorithm loops over all choices for a and b, but we can simplify by only considering critical values of a and b.
O(n^3) Algorithm
The key point is that if we have a,b such that we have more boxes than toys, then there is no point changing a to get even more boxes (as this will never help us make any more products). Similarly, if we have more toys than boxes, then we can skip considering all the cases of b which would gives us even more toys.
This suggests a O(n) algorithm for the inner loop in which we trace out the boundary of a,b between having more toys and having more boxes. This is quite simple as we can just start with a=x-1, and b=y-1 and then decrease either a or b according to whether we currently have more toys or boxes. (If equal then you can decrease both.)
Each step of the algorithm decreases either a or b by 1, so this iteration will require x+y steps instead of the x*y steps of the original method.
It needs to be repeated for all values of x,y so overall the complexity is O(n^3).
Additional Improvements
A further improvement would be to store the index of the previous run of each type as this would allow several steps of the algorithm to be collapsed into a single move (because we know that our score can only improve once we work back to a run of the correct type). However, this would still be O(n^3) in the worst case (all boxes/toys of the same type).
Another practical improvement is to coalesce any runs in which the type was the same at consecutive positions, as this may significantly simplify test cases designed to expose the worst case behaviour in the previous improvement.
Related
Considering usually a sorting method products linearly sorted productions (such as "1,7,8,13,109..."), which consumes O(N) to inquiry.
Why not sort in non-linear order, consuming O(logN) or something to find element(s) by iteration or Newton method etc.? Is it expensive to make such a high-order sorted structure?
Concisely, is it a possible idea to sort results which allowed to be accessed by finding roots for ax^2 + bx + c = 0? (for contrast, usually it's finding root for ax + c = 0.) For example, we have x1 = 1, x2 = 2 as roots of a quadratic equation and just insert following xi(s). Then it is possible to use smarter ways to inquiry.
I suppose difficulty can be encountered by these aspects:
prediction of data can be rather hard. thus we cannot construct a general formula to describe well the following numbers (may be hash values).
due to the first difficulty, numbers out of certain range can be divergent. example graphed by Google:the graph. the values derived out of [-1,3] are really large, as well as rapid increment in difficulty executing the original formula.
that is actually equivalent to hash, which creates a table that contains the values. and the production rule is a formula.
the execution of a "smarter" inquiry may be expensive because of the complexity of algorithm itself.
Smarter schemes which take advantage of a known statistical distribution are typically faster by some constant. However, that still keeps them at O(log N), which is the same as a trivial binary search. The reason is that in each step, they typically narrow down the range of elements to search by a factor R > 2 , for simple binary search that's just R=2. But you need log(N)/log(R) steps to narrow it down to exactly one element.
Now whether this is a net win depends on log(R) versus the work needed at each step. A simple comparison (for binary search) takes a few cycles. As soon as you need anything more complex than +-*/ (say exp or log) to predict the location of the next element, the profit of needing less steps is gone.
So, in summary: binary search is used because each step is efficient, for many real-world distributions.
How can you compute a shortest addition chain (sac) for an arbitrary n <= 600 within one second?
Notes
This is the programming competition on codility for this month.
Addition chains are numerically very important, since they are the most economical way to compute x^n (by consecutive multiplications).
Knuth's Art of Computer Programming, Volume 2, Seminumerical Algorithms has a nice introduction to addition chains and some interesting properties, but I didn't find anything that enabled me to fulfill the strict performance requirements.
What I've tried (spoiler alert)
Firstly, I constructed a (highly branching) tree (with the start 1-> 2 -> ( 3 -> ..., 4 -> ...)) such that for each node n, the path from the root to n is a sac for n. But for values >400, the runtime is about the same as for making a coffee.
Then I used that program to find some useful properties for reducing the search space. With that, I'm able to build all solutions up to 600 while making a coffee. But for n, I need to compute all solutions up to n. Unfortunately, codility measures the class initialization's runtime, too...
Since the problem is probably NP-hard, I ended up hard-coding a lookup table. But since codility asked to construct the sac, I don't know if they had a lookup table in mind, so I feel dirty and like a cheater. Hence this question.
Update
If you think a hard-coded, full lookup table is the way to go, can you give an argument why you think a full computation/partly computed solutions/heuristics won't work?
I have just got my Golden Certificate for this problem. I will not provide a full solution because the problem is still available on the site.I will instead give you some hints:
You might consider doing a deep-first search.
There exists a minimal star-chain for each n < 12509
You need to know how prune your search space.
You need a good lower bound for the length of the chain you are looking for.
Remember that you need just one solution, not all.
Good luck.
Addition chains are numerically very important, since they are the
most economical way to compute x^n (by consecutive multiplications).
This is not true. They are not always the most economical way to compute x^n. Graham et. all proved that:
If each step in addition chain is assigned a cost equal to the product
of the numbers at that step, "binary" addition chains are shown to
minimize the cost.
Situation changes dramatically when we compute x^n (mod m), which is a common case, for example in cryptography.
Now, to answer your question. Apart from hard-coding a table with answers, you could try a Brauer chain.
A Brauer chain (aka star-chain) is an addition chain where each new element is formed as the sum of the previous element and some element (possibly the same). Brauer chain is a sac for n < 12509. Quoting Daniel. J. Bernstein:
Brauer's algorithm is often called "the left-to-right 2^k-ary method",
or simply "2^k-ary method". It is extremely popular. It is easy to
implement; constructing the chain for n is a simple matter of
inspecting the bits of n. It does not require much storage.
BTW. Does anybody know a decent C/C++ implementation of Brauer's chain computation? I'm working partially on a comparison of exponentiation times using binary and Brauer's chains for both cases: x^n and x^n (mod m).
There was a post on here recently which posed the following question:
You have a two-dimensional plane of (X, Y) coordinates. A bunch of random points are chosen. You need to select the largest possible set of chosen points, such that no two points share an X coordinate and no two points share a Y coordinate.
This is all the information that was provided.
There were two possible solutions presented.
One suggested using a maximum flow algorithm, such that each selected point maps to a path linking (source → X → Y → sink). This runs in O(V3) time, where V is the number of vertices selected.
Another (mine) suggested using the Hungarian algorithm. Create an n×n matrix of 1s, then set every chosen (x, y) coordinate to 0. The Hungarian algorithm will give you the lowest cost for this matrix, and the answer is the number of coordinates selected which equal 0. This runs in O(n3) time, where n is the greater of the number of rows or the number of columns.
My reasoning is that, for the vast majority of cases, the Hungarian algorithm is going to be faster; V is equal to n in the case where there's one chosen point for each row or column, and substantially greater for any case where there's more than that: given a 50×50 matrix with half the coordinates chosen, V is 1,250 and n is 50.
The counterargument is that there are some cases, like a 109×109 matrix with only two points selected, where V is 2 and n is 1,000,000,000. For this case, it takes the Hungarian algorithm a ridiculously long time to run, while the maximum flow algorithm is blinding fast.
Here is the question: Given that the problem doesn't provide any information regarding the size of the matrix or the probability that a given point is chosen (so you can't know for sure) how do you decide which algorithm, in general, is a better choice for the problem?
You can't, it's an imponderable.
You can only define which is better "in general" by defining what inputs you will see "in general". So for example you could whip up a probability model of the inputs, so that the expected value of V is a function of n, and choose the one with the best expected runtime under that model. But there may be arbitrary choices made in the construction of your model, so that different models give different answers. One model might choose co-ordinates at random, another model might look at the actual use-case for some program you're thinking of writing, and look at the distribution of inputs it will encounter.
You can alternatively talk about which has the best worst case (across all possible inputs with given constraints), which has the virtue of being easy to define, and the flaw that it's not guaranteed to tell you anything about the performance of your actual program. So for instance HeapSort is faster than QuickSort in the worst case, but slower in the average case. Which is faster? Depends whether you care about average case or worst case. If you don't care which case, you're not allowed to care which "is faster".
This is analogous to trying to answer the question "what is the probability that the next person you see will have an above (mean) average number of legs?".
We might implicitly assume that the next person you meet will be selected at random with uniform distribution from the human population (and hence the answer is "slightly less than one", since the mean is less than the mode average, and the vast majority of people are at the mode).
Or we might assume that your next meeting with another person is randomly selected with uniform distribution from the set of all meetings between two people, in which case the answer is still "slightly less than one", but I reckon not the exact same value as the first - one-and-zero-legged people quite possibly congregate with "their own kind" very slightly more than their frequency within the population would suggest. Or possibly they congregate less, I really don't know, I just don't see why it should be exactly the same once you take into account Veterans' Associations and so on.
Or we might use knowledge about you - if you live with a one-legged person then the answer might be "very slightly above 0".
Which of the three answers is "correct" depends precisely on the context which you are forbidding us from talking about. So we can't talk about which is correct.
Given that you don't know what each pill does, do you take the red pill or the blue pill?
If there really is not enough information to decide, there is not enough information to decide. Any guess is as good as any other.
Maybe, in some cases, it is possible to divine extra information to base the decision on. I haven't studied your example in detail, but it seems like the Hungarian algorithm might have higher memory requirements. This might be a reason to go with the maximum flow algorithm.
You don't. I think you illustrated that clearly enough. I think the proper practical solution is to spawn off both implementations in different threads, and then take the response that comes back first. If you're more clever, you can heuristically route requests to implementations.
Many algorithms require huge amounts of memory beyond the physical maximum of a machine, and in these cases, the algorithmically more ineffecient in time but efficient in space algorithm is chosen.
Given that we have distributed parallel computing, I say you just let both horses run and let the results speak for themselves.
This is a valid question, but there's no "right" answer — they are incomparable, so there's no notion of "better".
If your interest is practical, then you need to analyze the kinds of inputs that are likely to arise in practice, as well as the practical running times (constants included) of the two algorithms.
If your interest is theoretical, where worst-case analysis is often the norm, then, in terms of the input size, the O(V3) algorithm is better: you know that V ≤ n2, but you cannot polynomially bound n in terms of V, as you showed yourself. Of course the theoretical best algorithm is a hybrid algorithm that runs both and stops when whichever one of them finishes first, thus its running time would be O(min(V3,n3)).
Theoretically, they are both the same, because you actually compare how the number of operations grows when the size of the problem is increased to infinity.
The way your problem is defined, it has 2 sizes - n and number of points, so this question has no answer.
Problem description
There are different categories which contain an arbitrary amount of elements.
There are three different attributes A, B and C. Each element does have an other distribution of these attributes. This distribution is expressed through a positive integer value. For example, element 1 has the attributes A: 42 B: 1337 C: 18. The sum of these attributes is not consistent over the elements. Some elements have more than others.
Now the problem:
We want to choose exactly one element from each category so that
We hit a certain threshold on attributes A and B (going over it is also possible, but not necessary)
while getting a maximum amount of C.
Example: we want to hit at least 80 A and 150 B in sum over all chosen elements and want as many C as possible.
I've thought about this problem and cannot imagine an efficient solution. The sample sizes are about 15 categories from which each contains up to ~30 elements, so bruteforcing doesn't seem to be very effective since there are potentially 30^15 possibilities.
My model is that I think of it as a tree with depth number of categories. Each depth level represents a category and gives us the choice of choosing an element out of this category. When passing over a node, we add the attributes of the represented element to our sum which we want to optimize.
If we hit the same attribute combination multiple times on the same level, we merge them so that we can stripe away the multiple computation of already computed values. If we reach a level where one path has less value in all three attributes, we don't follow it anymore from there.
However, in the worst case this tree still has ~30^15 nodes in it.
Does anybody of you can think of an algorithm which may aid me to solve this problem? Or could you explain why you think that there doesn't exist an algorithm for this?
This question is very similar to a variation of the knapsack problem. I would start by looking at solutions for this problem and see how well you can apply it to your stated problem.
My first inclination to is try branch-and-bound. You can do it breadth-first or depth-first, and I prefer depth-first because I think it's cleaner.
To express it simply, you have a tree-walk procedure walk that can enumerate all possibilities (maybe it just has a 5-level nested loop). It is augmented with two things:
At every step of the way, it keeps track of the cost at that point, where the cost can only increase. (If the cost can also decrease, it becomes more like a minimax game tree search.)
The procedure has an argument budget, and it does not search any branches where the cost can exceed the budget.
Then you have an outer loop:
for (budget = 0; budget < ... ; budget++){
walk(budget);
// if walk finds a solution within the budget, halt
}
The amount of time it takes is exponential in the budget, so easier cases will take less time. The fact that you are re-doing the search doesn't matter much because each level of the budget takes as much or more time than all the previous levels combined.
Combine this with some sort of heuristic about the order in which you consider branches, and it may give you a workable solution for typical problems you give it.
IF that doesn't work, you can fall back on basic heuristic programming. That is, do some cases by hand, and pay attention to how you did it. Then program it the same way.
I hope that helps.
My best shot so far:
A delivery vehicle needs to make a series of deliveries (d1,d2,...dn), and can do so in any order--in other words, all the possible permutations of the set D = {d1,d2,...dn} are valid solutions--but the particular solution needs to be determined before it leaves the base station at one end of the route (imagine that the packages need to be loaded in the vehicle LIFO, for example).
Further, the cost of the various permutations is not the same. It can be computed as the sum of the squares of distance traveled between di -1 and di, where d0 is taken to be the base station, with the caveat that any segment that involves a change of direction costs 3 times as much (imagine this is going on on a railroad or a pneumatic tube, and backing up disrupts other traffic).
Given the set of deliveries D represented as their distance from the base station (so abs(di-dj) is the distance between two deliveries) and an iterator permutations(D) which will produce each permutation in succession, find a permutation which has a cost less than or equal to that of any other permutation.
Now, a direct implementation from this description might lead to code like this:
function Cost(D) ...
function Best_order(D)
for D1 in permutations(D)
Found = true
for D2 in permutations(D)
Found = false if cost(D2) > cost(D1)
return D1 if Found
Which is O(n*n!^2), e.g. pretty awful--especially compared to the O(n log(n)) someone with insight would find, by simply sorting D.
My question: can you come up with a plausible problem description which would naturally lead the unwary into a worse (or differently awful) implementation of a sorting algorithm?
I assume you're using this question for an interview to see if the applicant can notice a simple solution in a seemingly complex question.
[This assumption is incorrect -- MarkusQ]
You give too much information.
The key to solving this is realizing that the points are in one dimension and that a sort is all that is required. To make this question more difficult hide this fact as much as possible.
The biggest clue is the distance formula. It introduces a penalty for changing directions. The first thing an that comes to my mind is minimizing this penalty. To remove the penalty I have to order them in a certain direction, this ordering is the natural sort order.
I would remove the penalty for changing directions, it's too much of a give away.
Another major clue is the input values to the algorithm: a list of integers. Give them a list of permutations, or even all permutations. That sets them up to thinking that a O(n!) algorithm might actually be expected.
I would phrase it as:
Given a list of all possible
permutations of n delivery locations,
where each permutation of deliveries
(d1, d2, ...,
dn) has a cost defined by:
Return permutation P such that the
cost of P is less than or equal to any
other permutation.
All that really needs to be done is read in the first permutation and sort it.
If they construct a single loop to compare the costs ask them what the big-o runtime of their algorithm is where n is the number of delivery locations (Another trap).
This isn't a direct answer, but I think more clarification is needed.
Is di allowed to be negative? If so, sorting alone is not enough, as far as I can see.
For example:
d0 = 0
deliveries = (-1,1,1,2)
It seems the optimal path in this case would be 1 > 2 > 1 > -1.
Edit: This might not actually be the optimal path, but it illustrates the point.
YOu could rephrase it, having first found the optimal solution, as
"Give me a proof that the following convination is the most optimal for the following set of rules, where optimal means the smallest number results from the sum of all stage costs, taking into account that all stages (A..Z) need to be present once and once only.
Convination:
A->C->D->Y->P->...->N
Stage costs:
A->B = 5,
B->A = 3,
A->C = 2,
C->A = 4,
...
...
...
Y->Z = 7,
Z->Y = 24."
That ought to keep someone busy for a while.
This reminds me of the Knapsack problem, more than the Traveling Salesman. But the Knapsack is also an NP-Hard problem, so you might be able to fool people to think up an over complex solution using dynamic programming if they correlate your problem with the Knapsack. Where the basic problem is:
can a value of at least V be achieved
without exceeding the weight W?
Now the problem is a fairly good solution can be found when V is unique, your distances, as such:
The knapsack problem with each type of
item j having a distinct value per
unit of weight (vj = pj/wj) is
considered one of the easiest
NP-complete problems. Indeed empirical
complexity is of the order of O((log
n)2) and very large problems can be
solved very quickly, e.g. in 2003 the
average time required to solve
instances with n = 10,000 was below 14
milliseconds using commodity personal
computers1.
So you might want to state that several stops/packages might share the same vj, inviting people to think about the really hard solution to:
However in the
degenerate case of multiple items
sharing the same value vj it becomes
much more difficult with the extreme
case where vj = constant being the
subset sum problem with a complexity
of O(2N/2N).
So if you replace the weight per value to distance per value, and state that several distances might actually share the same values, degenerate, some folk might fall in this trap.
Isn't this just the (NP-Hard) Travelling Salesman Problem? It doesn't seem likely that you're going to make it much harder.
Maybe phrasing the problem so that the actual algorithm is unclear - e.g. by describing the paths as single-rail railway lines so the person would have to infer from domain knowledge that backtracking is more costly.
What about describing the question in such a way that someone is tempted to do recursive comparisions - e.g. "can you speed up the algorithm by using the optimum max subset of your best (so far) results"?
BTW, what's the purpose of this - it sounds like the intent is to torture interviewees.
You need to be clearer on whether the delivery truck has to return to base (making it a round trip), or not. If the truck does return, then a simple sort does not produce the shortest route, because the square of the return from the furthest point to base costs so much. Missing some hops on the way 'out' and using them on the way back turns out to be cheaper.
If you trick someone into a bad answer (for example, by not giving them all the information) then is it their foolishness or your deception that has caused it?
How great is the wisdom of the wise, if they heed not their ego's lies?