Let's say that I want to create a fictional product called g.
I know that:
a+b=c
x+y=z
and finally that
c+z=g
So clearly if I start off with products
a,b,x,y
I can create g in three steps:
a+b=c
x+y=z
c+z=g
So a naive algorithm for reaching a goal could be:
For each component required to make the goal (here c and z), recursively find a cause and effect tuple that can create that component.
But there are snags with that algorithm.
For example, let's say that my cause and effect tuples are:
a+b=c
x+y+c=z (NOTE THE EXTRA 'c' REQUIRED!!)
c+z=g
Now when I run my naive algorithm I will do
a+b=c
x+y+c=z (Using up the 'c' I created in the previous step)
c+z=g (Uh oh! I can't do this because I don't have the 'c' any more)
It seems like quite a basic area of research - how we can combine known causes and effects to reach a goal - so I suspect that work must have been done on it, but I've looked around and couldn't find anything and I don't really know where to look now.
Many thanks for any assistance!
Assuming that using a product consumes one item of it, which can then be replaced by producing a second item of that product, I would model this by giving each product a cost and working out how to minimize the cost of the final product. In this case I think this is the same as minimizing the costs of every product, because minimizing the cost of an input never increases the cost of any output. You end up with loads of equations like
a=min(b+c, d+e, f+g)
where a is the cost of a product that can be produced in alternative ways, one way consuming units with cost of b and c, another way consuming units with costs of d and e, another way consuming units with costs of f and g, and so on. There may be cycles in the associated graph.
One way to solve such a problem would be to start by assigning the cost infinity to all products not originally provided as inputs (with costs) and then repeatedly reducing costs where equations show a way of calculating a cost less than the current cost, keeping track of re-calculations caused by inputs not yet considered or reductions in costs. At each stage I would consider the consequences of the smallest input or recalculated value available, with ties broken by a second component which amounts to a tax on production. The outputs produced from a calculation are always at least as large as any input, so newly produced values are always larger than the recalculated value considered, and the recalculated value considered at each stage never decreases, which should reduce repeated recalculation.
Another way would be to turn this into a linear program and throw it at a highly optimized guaranteed polynomial time (at least in practice) linear programming solver.
a = min(b+c, d+e, f+g)
becomes
a = b+c-x
a = d+e-y
a = f+g-z
x >= 0
y >= 0
z >= 0
minimize sum(x+y+z+....)
Related
Say you have a warehouse with fragile goods (f.e. vegetables or fruits), and you can only take out a container with vegetables once. If you move them twice, they'll rot too fast and cant be sold anymore.
So if you give a value to every container of vegetables (depending on how long they'll still be fresh), you want to sell the lowest value first. And when a client asks a certain weight, you want to deliver a good service, and give the exact weight (so you need to take some extra out of your warehouse, and throw the extra bit away after selling).
I don't know if this problem has a name, but I would consider this the dual form of the knapsack problem. In the knapsack problem, you want to maximise the value and limit the weight to a maximum. While here you want to minimise the value and limit the weight to a minimum.
You can easily see this duality by treating the warehouse as the knapsack, and optimising the warehouse for the maximum value and limited weight to a maximum of the current weight minus what the client asks.
However, many practical algorithms on solving the knapsack problem rely on the assumption that the weight you can carry is small compared to the total weight you can chose from. F.e. the dynamic programming 0/1 solution relies on looping until you reach the maximum weight, and the FPTAS solution guarantees to be correct within a factor of (1-e) of the total weight (but a small factor of a huge value can still make a pretty big difference).
So both have issues when the wanted weight is big.
As such, I wondered if anyone studied the "dual knapsack problem" already (if some literature can be found around it), or if there's some easy modification to the existing algorithms that I'm missing.
The usual pseudopolynomial DP algorithm for solving knapsack asks, for each i and w, "What is the largest total value I can get from the first i items if I use at most w capacity?"
You can instead ask, for each i and w, "What is the smallest total value I can get from the first i items if I use at least w capacity?" The logic is almost identical, except that the direction of the comparison is reversed, and you need a special value to record the possibility that even taking all i of the first i items cannot reach w capacity -- infinity works for this, since you want this value to lose against any finite value when they are compared with min().
I'm trying to come up with a fast and reasonably optimal algorithm to solve the following TSP/hamiltonian-path-like problem:
A delivery vehicle has a number of pickups and dropoffs it needs to
perform:
For each delivery, the pickup needs to come before the
dropoff.
The vehicle is quite small and the packages vary in size.
The total carriage cannot exceed some upper bound (e.g. 1 cubic
metre). Each delivery has a deadline.
The planner can run mid-route, so the vehicle will begin with a number of jobs already picked up and some capacity already taken up.
A near-optimal solution should minimise the total cost (for simplicity, distance) between each waypoint. If a solution does not exist because of the time constraints, I need to find a solution that has the fewest number of late deliveries. Some illustrations of an example problem and a non-optimal, but valid solution:
I am currently using a greedy best first search with backtracking bounded to 100 branches. If it fails to find a solution with on-time deliveries, I randomly generate as many as I can in one second (the most computational time I can spare) and pick the one with the fewest number of late deliveries. I have looked into linear programming but can't get my head around it - plus I would think it would be inappropriate given it needs to be run very frequently. I've also tried algorithms that require mutating the tour, but the issue is mutating a tour nearly always makes it invalid due to capacity constraints and precedence. Can anyone think of a better heuristic approach to solving this problem? Many thanks!
Safe Moves
Here are some ideas for safely mutating an existing feasible solution:
Any two consecutive stops can always be swapped if they are both pickups, or both deliveries. This is obviously true for the "both deliveries" case; for the "both pickups" case: if you had room to pick up A, then pick up B without delivering anything in between, then you have room to pick up B first, then pick up A. (In fact a more general rule is possible: In any pure-delivery or pure-pickup sequence of consecutive stops, the stops can be rearranged arbitrarily. But enumerating all the possibilities might become prohibitive for long sequences, and you should be able to get most of the benefit by considering just pairs.)
A pickup of A can be swapped with any later delivery of something else B, provided that A's original pickup comes after B was picked up, and A's own delivery comes after B's original delivery. In the special case where the pickup of A is immediately followed by the delivery of B, they can always be swapped.
If there is a delivery of an item of size d followed by a pickup of an item of size p, then they can be swapped provided that there is enough extra room: specifically, provided that f >= p, where f is the free space available before the delivery. (We already know that f + d >= p, otherwise the original schedule wouldn't be feasible -- this is a hint to look for small deliveries to apply this rule to.)
If you are starting from purely randomly generated schedules, then simply trying all possible moves, greedily choosing the best, applying it and then repeating until no more moves yield an improvement should give you a big quality boost!
Scoring Solutions
It's very useful to have a way to score a solution, so that they can be ordered. The nice thing about a score is that it's easy to incorporate levels of importance: just as the first digit of a two-digit number is more important than the second digit, you can design the score so that more important things (e.g. deadline violations) receive a much greater weight than less important things (e.g. total travel time or distance). I would suggest something like 1000 * num_deadline_violations + total_travel_time. (This assumes of course that total_travel_time is in units that will stay beneath 1000.) We would then try to minimise this.
Managing Solutions
Instead of taking one solution and trying all the above possible moves on it, I would instead suggest using a pool of k solutions (say, k = 10000) stored in a min-heap. This allows you to extract the best solution in the pool in O(log k) time, and to insert new solutions in the same time.
You could initially populate the pool with randomly generated feasible solutions; then on each step, you would extract the best solution in the pool, try all possible moves on it to generate child solutions, and insert any child solutions that are better than their parent back into the pool. Whenever the pool doubles in size, pull out the first (i.e. best) k solutions and make a new min-heap with them, discarding the old one. (Performing this step after the heap grows to a constant multiple of its original size like this has the nice property of leaving the amortised time complexity unchanged.)
It can happen that some move on solution X produces a child solution Y that is already in the pool. This wastes memory, which is unfortunate, but one nice property of the min-heap approach is that you can at least handle these duplicates cheaply when they arrive at the front of the heap: all duplicates will have identical scores, so they will all appear consecutively when extracting solutions from the top of the heap. Thus to avoid having duplicate solutions generate duplicate children "down through the generations", it suffices to check that the new top of the heap is different from the just-extracted solution, and keep extracting and discarding solutions until this holds.
A note on keeping worse solutions: It might seem that it could be worthwhile keeping child solutions even if they are slightly worse than their parent, and indeed this may be useful (or even necessary to find the absolute optimal solution), but doing so has a nasty consequence: it means that it's possible to cycle from one solution to its child and back again (or possibly a longer cycle). This wastes CPU time on solutions we have already visited.
You are basically combining the Knapsack Problem with the Travelling Salesman Problem.
Your main problem here seems to be actually the Knapsack Problem, rather then the Travelling Salesman Problem, since it has the one hard restriction (maximum delivery volume). Maybe try to combine the solutions for the Knapsack Problem with the Travelling Salesman.
If you really only have one second max for calculations a greedy algorithm with backtracking might actually be one of the best solutions that you can get.
I am trying to develop an algorithm to select a subset of activities from a larger list. If selected, each activity uses some amount of a fixed resource (i.e. the sum over the selected activities must stay under a total budget). There could be multiple feasible subsets, and the means of choosing from them will be based on calculating the opportunity cost of the activities not selected.
EDIT: There are two reasons this is not the 0-1 knapsack problem:
Knapsack requires integer values for the weights (i.e. resources consumed) whereas my resource consumption (i.e. mass in the knapsack parlance) is a continuous variable. (Obviously it's possible to pick some level of precision and quantize the required resources, but my bin size would have to be very small and Knapsack is O(2^n) in W.
I cannot calculate the opportunity cost a priori; that is, I can't evaluate the fitness of each one independently, although I can evaluate the utility of a given set of selected activities or the marginal utility from adding an additional task to an existing list.
The research I've done suggests a naive approach:
Define the powerset
For each element of the powerset, calculate it's utility based on the items not in the set
Select the element with the highest utility
However, I know there are ways to speed up execution time and required memory. For example:
fully enumerating a powerset is O(2^n), but I don't need to fully enumerate the list because once I've found a set of tasks that exceeds the budget I know that any set that adds more tasks is infeasible and can be rejected. That is if {1,2,3,4} is infeasible, so is {1,2,3,4} U {n}, where n is any one of the tasks remaining in the larger list.
Since I'm just summing duty the order of tasks doesn't matter (i.e. if {1,2,3} is feasible, so are {2,1,3}, {3,2,1}, etc.).
All I need in the end is the selected set, so I probably only need the best utility value found so far for comparison purposes.
I don't need to keep the list enumerations, as long as I can be sure I've looked at all the feasible ones. (Although I think keeping the duty sum for previously computed feasible sub-sets might speed run-time.)
I've convinced myself a good recursion algorithm will work, but I can't figure out how to define it, even in pseudo-code (which probably makes the most sense because it's going to be implemented in a couple of languages--probably Matlab for prototyping and then a compiled language later).
The knapsack problem is NP-complete, meaning that there's no efficient way of solving the problem. However there's a pseudo-polynomial time solution using dynamic programming. See the Wikipedia section on it for more details.
However if the maximum utility is large, you should stick with an approximation algorithm. One such approximation scheme is to greedily select items that have the greatest utility/cost. If the budget is large and the cost of each item is small, then this can work out very well.
EDIT: Since you're defining the utility in terms of items not in the set, you can simply redefine your costs. Negate the cost and then shift everything so that all your values are positive.
As others have mentioned, you are trying to solve some instance of the Knapsack problem. While theoretically, you are doomed, in practice you may still do a lot to increase the performance of your algorithm. Here are some (wildly assorted) ideas:
Be aware of Backtracking. This corresponds to your observation that once you crossed out {1, 2, 3, 4} as a solution, {1, 2, 3, 4} u {n} is not worth looking at.
Apply Dynamic Programming techniques.
Be clear about your actual requirements:
Maybe you don't need the best set? Will a good one do? I am not aware if there is an algorithm which provides a good solution in polynomial time, but there might well be.
Maybe you don't need the best set all the time? Using randomized algorithms you can solve some NP-Problems in polynomial time with the risk of failure in 1% (or whatever you deem "safe enough") of all executions.
(Remember: It's one thing to know that the halting problem is not solvable, but another to build a program that determines whether "hello world" implementations will run indefinetly.)
I think the following iterative algorithm will traverse the entire solution set and store the list of tasks, the total cost of performing them, and the opportunity cost of the tasks not performed.
It seems like it will execute in pseudo-polynomial time: polynomial in the number of activities and exponential in the number of activities that can fit within the budget.
ixCurrentSolution = 1
initialize empty set solution {
oc(ixCurrentSolution) = opportunity cost of doing nothing
tasklist(ixCurrentSolution) = empty set
costTotal(ixCurrentSolution) = 0
}
for ixTask = 1:cActivities
for ixSolution = 1:ixCurrentSolution
costCurrentSolution = costTotal(ixCurrentSolution) + cost(ixTask)
if costCurrentSolution < costMax
ixCurrentSolution++
costTotal(ixCurrentSolution) = costCurrentSolution
tasklist(ixCurrentSolution) = tasklist(ixSolution) U ixTask
oc(ixCurrentSolution) = OC of tasks not in tasklist(ixCurrentSolution)
endif
endfor
endfor
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?