I am struck very badly in a problem. my problem goes this way; I need to find permutations of n objects(there could be repetitions) such a way that every permutation differs from the other by atleast k objects.
For ex: if there are 5 objects a,b,c,d,e and each permutation differs by 2 or more object and if aabcd is a permutation then I cannot have aabdd as a permutation as both differ by just one object.
If anyone can point me out a general formula or procedure to solve this problem, I'd be highly grateful
Thanks for your time and consideration of this request
--Ady
This sounds like it's related to Conway's Lexicode thereom. I heard him give a lecture about it once. It was quite entertaining.
http://www.dpmms.cam.ac.uk/seminars/Kuwait/abstracts/L25.pdf
Related
I have a problem that I have a number of questions about. First, I'm mostly looking for help describing and understanding the problem at hand. Solutions are always welcome, but most importantly I could use some advice from someone more experienced than I. Now, to the problem at hand:
I have a set of orders that each require some number of items. I also have several groupings of items that each contain some number of some items (call them groups). The goal is to find a subset of the orders that can be fulfilled using as few groups as possible and where the total number of items contained within the orders is between n and N.
Edit: The constraints on the number of items contained in the orders (n and N) are chosen independently.
To me at least, that's a really complicated way of saying the problem so I've been trying to re-phrase it as a knapsack problem (I suspect this might reduce to a subset-sum). To help my conceptual understanding of this I've started using the following definitions:
First, lets say that a dimension exists for each possible item, and somethings 'length' in that dimension is the number of that particular type of item it either has or requires.
From this, an order becomes an 'n-dimensional object' where its value in each dimension corresponds to the number of that item that it requires.
In addition, a group can be seen as an 'n-dimensional box' that has space in each dimension corresponding to the number of items it provides.
An objects value is equal to the sum of its length in all dimensions.
Boxes can be combined.
Given the above I've rephrased the problem to this:
What is the smallest combination of boxes that can hold a combination of items with value between n and N.
Question #1: Is this a correct/useful way to express the problem? Does it seem like I've missed anything obvious?
As I see it, since there are two combinations that I'm looking for I need to break the problem into two parts. So far I think breaking the problem up like this is a good step:
How many objects can box (or combination of boxes) X hold?
Check all (or preferably some small subset of) the possible combinations of boxes and pick the 'best'.
That makes it a little more manageable, but I'm still struggling with the details.
Question #2: Solved To solve the first part I think it's appropriate to say that the cost of an object is equal to the sum of its length in all dimensions, so is it's value. That places me into a subset-sum problem, right? Obviously it's a special case, but does this problem have a name?
Question #3: Solved I've been looking into subset-sum solutions a lot, but I don't understand how to apply them to something like this in multiple dimensions. I assume it's been done before, but I'm unsure where to start my research. Could someone either describe the principles at work or point me in a research direction?
Edit: After looking at everyone's feedback and digging into the terms I think I've found a good algorithm I can implement to solve part 1. Since I will have a very large number of dimensions compared to the number of items it looks like using a 'primal effective capacity heuristic (PECH)' will be a good fit. I'd be interested in hearing someones thoughts about it if they have experience with such an algorithm.
Question #4: For the second part, performance is a concern and I doubt it will be realistic to brute force it. So I intend to treat all combinations of boxes as a really big tree of solutions. The idea is to compute part 1 for all combinations of M-1 boxes where M is the total number of boxes. Somehow determine the 'best' couple box combinations from that set and do the same to their child nodes on the tree. Does this sound like it would help me arrive at something close to optimal? How would I choose the 'best' box combinations?
Thanks for reading! Suggestions for edits and clarifications are welcome.
While writing code, i found the following problem, to state it in a simple way:
Partition an array of floats X in array A and B such that the difference between the sum of the values in A and the sum of values of B is minimized
This was part of an investigation I was doing, but I can't find a way to efficiently perform this operation.
Edit:
To answer to those who believe this is from a math contest like PE, SPOJ or homework, it is not. I just had curiosity about this when i was trying to partition an already factorized number p in the set of factors a and b such that b=a+1. If we take logs from both sides, we can show this problem is equivalent to minimize a diference of sums, but that is where i have got stuck.
Just a first simple idea. Use dynamic programming methods.
I assume that this problem can be transformed to knapsack problem. You need to pick items from X (there'll be array A) to maximize sum but don't exceed (sumX - sumA) value (there'll be sum of items from array B). For algorithm to solve knapsack problem by dynamic programming approach look at wiki e.g.
This solution can be wrong, btw... but even if it'll work I'm more than sure that more efficient, elegant and short solutions exist.
I have this problem in my textbook:
Given a group of n items, each with a distinct value V(i), what is the best way to divide the items into 3 groups so the group with the highest value is minimIzed? Give the value of this largest group.
I know how to do the 2 pile variant of this problem: it just requires running the knapsack algorithm backwards on the problem. However, I am pretty puzzled as how to solve this problem. Could anyone give me any pointers?
Answer: Pretty much the same thing as the 0-1 knapsack, although 2D
Tough homework problem. This is essentially the optimization version of the 3-partition problem.
http://en.wikipedia.org/wiki/3-partition_problem
It is closely related to bin packing, partition, and subset-sum (and, as you noted, knapsack). However, it happens to be strongly NP-Complete, which makes it a harder than its cousins. Anyway, I suggest you start by looking at dynamic programming solutions to the related problems (I'd start with partition, but find a non-wikipedia explanation of the DP solution).
Update: I apologize. I have mislead you. The 3-partition problem splits the input into sets of 3, not 3 sets. The rest of what I said still applies, but with the renewed hope that your variant isn't strongly np-complete.
Let f[i][j][k] denotes whether it is possible to have value j in the first set and value k in the second set, with the first i items.
So we have f[i][j][k] = f[i-1][j-v[i]][k] or f[i-1][j][k-v[i]].
and initially we have f[0][0][0] = True.
for every f[i][j][k] = True, update your answer depends on how you defines fairly.
I don't know about "The Best" mathematically speaking, but one obvious approach would be to build a population of groups initially with one item in each group. Then, for as long as you have more groups than the desired number of final groups, extract the two groups with the lowest values and combine them into a new group that you add back into the collection. This is similar to how Huffman compression trees are built.
Example:
1 3 7 9 10
becomes
4(1+3) 7 9 10
becomes
9 10 11(1+3+7)
I have 'n' number of amounts (non-negative integers). My requirement is to determine an optimal set of amounts so that the sum of the combination is less than or equal to a given fixed limit and the total is as large as possible. There is no limit to the number of amounts that can be included in the optimal set.
for sake of example: amounts are 143,2054,546,3564,1402 and the given limit is 5000.
As per my understanding the knapsack problem has 2 attributes for each item (weight and value). But the problem stated above has only one attribute (amount). I hope that would make things simpler? :)
Can someone please help me with the algorithm or source code for solving this?
this is still an NP-hard problem, but if you want to (or have to) to do something like that, maybe this topic helps you out a bit:
find two or more numbers from a list of numbers that add up towards a given amount
where i solved it like this and NikiC modified it to be faster. only difference: that one was about getting the exact amount, not "as close as possible", but that would be only some small changes in code (and you'll have to translate it into the language you're using).
take a look at the comments in my code to understand what i'm trying to do, wich is, in short form:
calculating all possible combinations of the given parts and sum them up
if the result is the amount i'm looking for, save the solution to an array
at least, sort all possible solutions to get the one using the least parts
so you'll have to change:
save a solution if it's lower than the amount you're looking for
sort solutions by total amount instead of number of used parts
The book "Knapsack Problems" By Hans Kellerer, Ulrich Pferschy and David Pisinger calls this The Subset Sum Problem and dedicates an entire chapter (Ch 4) to it. The chapter is very comprehensive and covers algorithms as well as computational results.
Even though this problem is a special case of the knapsack problem, it is still NP-hard.
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?