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In the tournament selection, k = tournament size, what happens if k = population size and what happens if k = 1?
Tournament size modifies selection pressure during evolution.
In case tournament size is equal to the size of population the best individual in whole population will be picked. This will cause stuck in local optimum.
Case when tournament size is equal to one is not valid. You have to pick at least 2 individuals to compare them.
But if you mean picking one random individual as parent selection strategy ... This will cause stuck of evolution process from the begging.
Selecting right tournament size depends on population size and problem you solve.
I am familiar with the 0-1 knapsack problem and when you are given a certain number of copies from each item but I can figure out how to solve it when you are given infinite copies of each item using dynamic programming. I am trying to solve it by hand right now so I am not interested in any particular code. For example here is how I solve the 0-1 problem. How do I need to modify this if I am given an infinity amount of copies of each item?
Edit: I am aware there is a second solution to the problem containing items 1,2, and 3 with the same total value.
One possibility would be to provide a suitable number of multiplicities of the items. For item i, there can be at most
m_i := K / w_i
choices of that item, where K denotes the knapsack capacity and w_i denotes the weight of the i-th item. Furthermore, for each weight value which occurs in the instance, there is at most one item type necessary, namely the one with maximum profit with respect to the weight.
Equivalently, one could modify the evaluation of the dynamic program to reflect the different number of items to be taken, instead of just distinguishing between a choice of 0 and 1.
I'm looking for a sorting algorithm that honors a min and max range for each element1. The problem domain is a recommendations engine that combines a set of business rules (the restrictions) with a recommendation score (the value). If we have a recommendation we want to promote (e.g. a special product or deal) or an announcement we want to appear near the top of the list (e.g. "This is super important, remember to verify your email address to participate in an upcoming promotion!") or near the bottom of the list (e.g. "If you liked these recommendations, click here for more..."), they will be curated with certain position restriction in place. For example, this should always be the top position, these should be in the top 10, or middle 5 etc. This curation step is done ahead of time and remains fixed for a given time period and for business reasons must remain very flexible.
Please don't question the business purpose, UI or input validation. I'm just trying to implement the algorithm in the constraints I've been given. Please treat this as an academic question. I will endeavor to provide a rigorous problem statement, and feedback on all other aspects of the problem is very welcome.
So if we were sorting chars, our data would have a structure of
struct {
char value;
Integer minPosition;
Integer maxPosition;
}
Where minPosition and maxPosition may be null (unrestricted). If this were called on an algorithm where all positions restrictions were null, or all minPositions were 0 or less and all maxPositions were equal to or greater than the size of the list, then the output would just be chars in ascending order.
This algorithm would only reorder two elements if the minPosition and maxPosition of both elements would not be violated by their new positions. An insertion-based algorithm which promotes items to the top of the list and reorders the rest has obvious problems in that every later element would have to be revalidated after each iteration; in my head, that rules out such algorithms for having O(n3) complexity, but I won't rule out such algorithms without considering evidence to the contrary, if presented.
In the output list, certain elements will be out of order with regard to their value, if and only if the set of position constraints dictates it. These outputs are still valid.
A valid list is any list where all elements are in a position that does not conflict with their constraints.
An optimal list is a list which cannot be reordered to more closely match the natural order without violating one or more position constraint. An invalid list is never optimal. I don't have a strict definition I can spell out for 'more closely matching' between one ordering or another. However, I think it's fairly easy to let intuition guide you, or choose something similar to a distance metric.
Multiple optimal orderings may exist if multiple inputs have the same value. You could make an argument that the above paragraph is therefore incorrect, because either one can be reordered to the other without violating constraints and therefore neither can be optimal. However, any rigorous distance function would treat these lists as identical, with the same distance from the natural order and therefore reordering the identical elements is allowed (because it's a no-op).
I would call such outputs the correct, sorted order which respects the position constraints, but several commentators pointed out that we're not really returning a sorted list, so let's stick with 'optimal'.
For example, the following are a input lists (in the form of <char>(<minPosition>:<maxPosition>), where Z(1:1) indicates a Z that must be at the front of the list and M(-:-) indicates an M that may be in any position in the final list and the natural order (sorted by value only) is A...M...Z) and their optimal orders.
Input order
A(1:1) D(-:-) C(-:-) E(-:-) B(-:-)
Optimal order
A B C D E
This is a trivial example to show that the natural order prevails in a list with no constraints.
Input order
E(1:1) D(2:2) C(3:3) B(4:4) A(5:5)
Optimal order
E D C B A
This example is to show that a fully constrained list is output in the same order it is given. The input is already a valid and optimal list. The algorithm should still run in O(n log n) time for such inputs. (Our initial solution is able to short-circuit any fully constrained list to run in linear time; I added the example both to drive home the definitions of optimal and valid and because some swap-based algorithms I considered handled this as the worse case.)
Input order
E(1:1) C(-:-) B(1:5) A(4:4) D(2:3)
Optimal Order
E B D A C
E is constrained to 1:1, so it is first in the list even though it has the lowest value. A is similarly constrained to 4:4, so it is also out of natural order. B has essentially identical constraints to C and may appear anywhere in the final list, but B will be before C because of value. D may be in positions 2 or 3, so it appears after B because of natural ordering but before C because of its constraints.
Note that the final order is correct despite being wildly different from the natural order (which is still A,B,C,D,E). As explained in the previous paragraph, nothing in this list can be reordered without violating the constraints of one or more items.
Input order
B(-:-) C(2:2) A(-:-) A(-:-)
Optimal order
A(-:-) C(2:2) A(-:-) B(-:-)
C remains unmoved because it already in its only valid position. B is reordered to the end because its value is less than both A's. In reality, there will be additional fields that differentiate the two A's, but from the standpoint of the algorithm, they are identical and preserving OR reversing their input ordering is an optimal solution.
Input order
A(1:1) B(1:1) C(3:4) D(3:4) E(3:4)
Undefined output
This input is invalid for two reasons: 1) A and B are both constrained to position 1 and 2) C, D, and E are constrained to a range than can only hold 2 elements. In other words, the ranges 1:1 and 3:4 are over-constrained. However, the consistency and legality of the constraints are enforced by UI validation, so it's officially not the algorithms problem if they are incorrect, and the algorithm can return a best-effort ordering OR the original ordering in that case. Passing an input like this to the algorithm may be considered undefined behavior; anything can happen. So, for the rest of the question...
All input lists will have elements that are initially in valid positions.
The sorting algorithm itself can assume the constraints are valid and an optimal order exists.2
We've currently settled on a customized selection sort (with runtime complexity of O(n2)) and reasonably proved that it works for all inputs whose position restrictions are valid and consistent (e.g. not overbooked for a given position or range of positions).
Is there a sorting algorithm that is guaranteed to return the optimal final order and run in better than O(n2) time complexity?3
I feel that a library standard sorting algorithm could be modified to handle these constrains by providing a custom comparator that accepts the candidate destination position for each element. This would be equivalent to the current position of each element, so maybe modifying the value holding class to include the current position of the element and do the extra accounting in the comparison (.equals()) and swap methods would be sufficient.
However, the more I think about it, an algorithm that runs in O(n log n) time could not work correctly with these restrictions. Intuitively, such algorithms are based on running n comparisons log n times. The log n is achieved by leveraging a divide and conquer mechanism, which only compares certain candidates for certain positions.
In other words, input lists with valid position constraints (i.e. counterexamples) exist for any O(n log n) sorting algorithm where a candidate element would be compared with an element (or range in the case of Quicksort and variants) with/to which it could not be swapped, and therefore would never move to the correct final position. If that's too vague, I can come up with a counter example for mergesort and quicksort.
In contrast, an O(n2) sorting algorithm makes exhaustive comparisons and can always move an element to its correct final position.
To ask an actual question: Is my intuition correct when I reason that an O(n log n) sort is not guaranteed to find a valid order? If so, can you provide more concrete proof? If not, why not? Is there other existing research on this class of problem?
1: I've not been able to find a set of search terms that points me in the direction of any concrete classification of such sorting algorithm or constraints; that's why I'm asking some basic questions about the complexity. If there is a term for this type of problem, please post it up.
2: Validation is a separate problem, worthy of its own investigation and algorithm. I'm pretty sure that the existence of a valid order can be proven in linear time:
Allocate array of tuples of length equal to your list. Each tuple is an integer counter k and a double value v for the relative assignment weight.
Walk the list, adding the fractional value of each elements position constraint to the corresponding range and incrementing its counter by 1 (e.g. range 2:5 on a list of 10 adds 0.4 to each of 2,3,4, and 5 on our tuple list, incrementing the counter of each as well)
Walk the tuple list and
If no entry has value v greater than the sum of the series from 1 to k of 1/k, a valid order exists.
If there is such a tuple, the position it is in is over-constrained; throw an exception, log an error, use the doubles array to correct the problem elements etc.
Edit: This validation algorithm itself is actually O(n2). Worst case, every element has the constraints 1:n, you end up walking your list of n tuples n times. This is still irrelevant to the scope of the question, because in the real problem domain, the constraints are enforced once and don't change.
Determining that a given list is in valid order is even easier. Just check each elements current position against its constraints.
3: This is admittedly a little bit premature optimization. Our initial use for this is for fairly small lists, but we're eyeing expansion to longer lists, so if we can optimize now we'd get small performance gains now and large performance gains later. And besides, my curiosity is piqued and if there is research out there on this topic, I would like to see it and (hopefully) learn from it.
On the existence of a solution: You can view this as a bipartite digraph with one set of vertices (U) being the k values, and the other set (V) the k ranks (1 to k), and an arc from each vertex in U to its valid ranks in V. Then the existence of a solution is equivalent to the maximum matching being a bijection. One way to check for this is to add a source vertex with an arc to each vertex in U, and a sink vertex with an arc from each vertex in V. Assign each edge a capacity of 1, then find the max flow. If it's k then there's a solution, otherwise not.
http://en.wikipedia.org/wiki/Maximum_flow_problem
--edit-- O(k^3) solution: First sort to find the sorted rank of each vertex (1-k). Next, consider your values and ranks as 2 sets of k vertices, U and V, with weighted edges from each vertex in U to all of its legal ranks in V. The weight to assign each edge is the distance from the vertices rank in sorted order. E.g., if U is 10 to 20, then the natural rank of 10 is 1. An edge from value 10 to rank 1 would have a weight of zero, to rank 3 would have a weight of 2. Next, assume all missing edges exist and assign them infinite weight. Lastly, find the "MINIMUM WEIGHT PERFECT MATCHING" in O(k^3).
http://www-math.mit.edu/~goemans/18433S09/matching-notes.pdf
This does not take advantage of the fact that the legal ranks for each element in U are contiguous, which may help get the running time down to O(k^2).
Here is what a coworker and I have come up with. I think it's an O(n2) solution that returns a valid, optimal order if one exists, and a closest-possible effort if the initial ranges were over-constrained. I just tweaked a few things about the implementation and we're still writing tests, so there's a chance it doesn't work as advertised. This over-constrained condition is detected fairly easily when it occurs.
To start, things are simplified if you normalize your inputs to have all non-null constraints. In linear time, that is:
for each item in input
if an item doesn't have a minimum position, set it to 1
if an item doesn't have a maximum position, set it to the length of your list
The next goal is to construct a list of ranges, each containing all of the candidate elements that have that range and ordered by the remaining capacity of the range, ascending so ranges with the fewest remaining spots are on first, then by start position of the range, then by end position of the range. This can be done by creating a set of such ranges, then sorting them in O(n log n) time with a simple comparator.
For the rest of this answer, a range will be a simple object like so
class Range<T> implements Collection<T> {
int startPosition;
int endPosition;
Collection<T> items;
public int remainingCapacity() {
return endPosition - startPosition + 1 - items.size();
}
// implement Collection<T> methods, passing through to the items collection
public void add(T item) {
// Validity checking here exposes some simple cases of over-constraining
// We'll catch these cases with the tricky stuff later anyways, so don't choke
items.add(item);
}
}
If an element A has range 1:5, construct a range(1,5) object and add A to its elements. This range has remaining capacity of 5 - 1 + 1 - 1 (max - min + 1 - size) = 4. If an element B has range 1:5, add it to your existing range, which now has capacity 3.
Then it's a relatively simple matter of picking the best element that fits each position 1 => k in turn. Iterate your ranges in their sorted order, keeping track of the best eligible element, with the twist that you stop looking if you've reached a range that has a remaining size that can't fit into its remaining positions. This is equivalent to the simple calculation range.max - current position + 1 > range.size (which can probably be simplified, but I think it's most understandable in this form). Remove each element from its range as it is selected. Remove each range from your list as it is emptied (optional; iterating an empty range will yield no candidates. That's a poor explanation, so lets do one of our examples from the question. Note that C(-:-) has been updated to the sanitized C(1:5) as described in above.
Input order
E(1:1) C(1:5) B(1:5) A(4:4) D(2:3)
Built ranges (min:max) <remaining capacity> [elements]
(1:1)0[E] (4:4)0[A] (2:3)1[D] (1:5)3[C,B]
Find best for 1
Consider (1:1), best element from its list is E
Consider further ranges?
range.max - current position + 1 > range.size ?
range.max = 1; current position = 1; range.size = 1;
1 - 1 + 1 > 1 = false; do not consider subsequent ranges
Remove E from range, add to output list
Find best for 2; current range list is:
(4:4)0[A] (2:3)1[D] (1:5)3[C,B]
Consider (4:4); skip it because it is not eligible for position 2
Consider (2:3); best element is D
Consider further ranges?
3 - 2 + 1 > 1 = true; check next range
Consider (2:5); best element is B
End of range list; remove B from range, add to output list
An added simplifying factor is that the capacities do not need to be updated or the ranges reordered. An item is only removed if the rest of the higher-sorted ranges would not be disturbed by doing so. The remaining capacity is never checked after the initial sort.
Find best for 3; output is now E, B; current range list is:
(4:4)0[A] (2:3)1[D] (1:5)3[C]
Consider (4:4); skip it because it is not eligible for position 3
Consider (2:3); best element is D
Consider further ranges?
same as previous check, but current position is now 3
3 - 3 + 1 > 1 = false; don't check next range
Remove D from range, add to output list
Find best for 4; output is now E, B, D; current range list is:
(4:4)0[A] (1:5)3[C]
Consider (4:4); best element is A
Consider further ranges?
4 - 4 + 1 > 1 = false; don't check next range
Remove A from range, add to output list
Output is now E, B, D, A and there is one element left to be checked, so it gets appended to the end. This is the output list we desired to have.
This build process is the longest part. At its core, it's a straightforward n2 selection sorting algorithm. The range constraints only work to shorten the inner loop and there is no loopback or recursion; but the worst case (I think) is still sumi = 0 n(n - i), which is n2/2 - n/2.
The detection step comes into play by not excluding a candidate range if the current position is beyond the end of that ranges max position. You have to track the range your best candidate came from in order to remove it, so when you do the removal, just check if the position you're extracting the candidate for is greater than that ranges endPosition.
I have several other counter-examples that foiled my earlier algorithms, including a nice example that shows several over-constraint detections on the same input list and also how the final output is closest to the optimal as the constraints will allow. In the mean time, please post any optimizations you can see and especially any counter examples where this algorithm makes an objectively incorrect choice (i.e. arrives at an invalid or suboptimal output when one exists).
I'm not going to accept this answer, because I specifically asked if it could be done in better than O(n2). I haven't wrapped my head around the constraints satisfaction approach in #DaveGalvin's answer yet and I've never done a maximum flow problem, but I thought this might be helpful for others to look at.
Also, I discovered the best way to come up with valid test data is to start with a valid list and randomize it: for 0 -> i, create a random value and constraints such that min < i < max. (Again, posting it because it took me longer than it should have to come up with and others might find it helpful.)
Not likely*. I assume you mean average run time of O(n log n) in-place, non-stable, off-line. Most Sorting algorithms that improve on bubble sort average run time of O(n^2) like tim sort rely on the assumption that comparing 2 elements in a sub set will produce the same result in the super set. A slower variant of Quicksort would be a good approach for your range constraints. The worst case won't change but the average case will likely decrease and the algorithm will have the extra constraint of a valid sort existing.
Is ... O(n log n) sort is not guaranteed to find a valid order?
All popular sort algorithms I am aware of are guaranteed to find an order so long as there constraints are met. Formal analysis (concrete proof) is on each sort algorithems wikepedia page.
Is there other existing research on this class of problem?
Yes; there are many journals like IJCSEA with sorting research.
*but that depends on your average data set.
Say I have a Group data structure which contains a list of Element objects, such that each group has a unique set of elements.:
public class Group
{
public List<Element> Elements;
}
and say I have a list of populations who require certain elements, in such a way that each population has a unique set of required elements:
public class Population
{
public List<Element> RequiredElements;
}
I have an unlimited quantity of each defined Group, i.e. they are not consumed by populations.
Say I am looking at a particular Population. I want to find the best possible match of groups such that there is minimum excess elements, and no unmatched elements.
For example: I have a population which needs wood, steel, grain, and coal. The only groups available are {wood, herbs}, {steel, coal, oil}, {grain, steel}, and {herbs, meat}.
The last group - {herbs, meat} isn't required at all by my population so it isn't used. All others are needed, but herbs and oil are not required so it is wasted. Furthermore, steel exists twice in the minimum set, so one lot of steel is also wasted. The best match in this example has a wastage of 3.
So for a few hundred Population objects, I need to find the minimum wastage best match and compute how many elements are wasted.
How do I even begin to solve this? Once I have found a match, counting the wastage is trivial. Finding the match in the first place is hard. I could enumerate all possibilities but with a few thousand populations and many hundreds of groups, it's quite a task. Especially considering this whole thing sits inside each iteration of a simulated annealing algorithm.
I'm wondering whether I can formulate the whole thing as a mixed-integer program and call a solver like GLPK at each iteration.
I hope I have explained the problem correctly. I can clarify anything that's unclear.
Here's my binary program, for those of you interested...
x is the decision vector, an element of {0,1}, which says that the population in question does/doesn't receive from group i. There is an entry for each group.
b is the column vector, an element of {0,1}, which says which resources the population in question does/doesn't need. There is an entry for each resource.
A is a matrix, an element of {0,1}, which says what resources are in what groups.
The program is:
Minimise: ((Ax - b)' * 1-vector) + (x' * 1-vector);
Subject to: Ax >= b;
The constraint just says that all required resources must be satisfied. The objective is to minimise all excess and the total number of groups used. (i.e. 0 excess with 1 group used is better than 0 excess with 5 groups used).
You can formulate an integer program for each population P as follows. Use a binary variable xj to denote whether group j is chosen or not. Let A be a binary matrix, such that Aij is 1 if and only if item i is present in group j. Then the integer program is:
min Ei,j (xjAij)
s.t. Ej xjAij >= 1 for all i in P.
xj = 0, 1 for all j.
Note that you can obtain the minimum wastage by subtracting |P| from the optimal solution of the above IP.
Do you mean the Maximum matching problem?
You need to build a bipartite graph, where one of the sides is your populations and the other is groups, and edge exists between group A and population B if it have it in its set.
To find maximum edge matching you can easily use Kuhn algorithm, which is greatly described here on TopCoder.
But, if you want to find mimimum edge dominating set (the set of minimum edges that is covering all the vertexes), the problem becomes NP-hard and can't be solved in polynomial time.
Take a look at the weighted set cover problem, I think this is exactly what you described above. A basic description of the (unweighted) problem can be found here.
Finding the minimal waste as you defined above is equivalent to finding a set cover such that the sum of the cardinalities of the covering sets is minimal. Hence, the weight of each set (=a group of elements) has to be defined equal to its cardinality.
Since even the unweighted the set cover problem is NP-complete, it is not likely that an efficient algorithm for your problem instances exist. Maybe a good greedy approximation algorithm will be sufficient or your purpose? Googling weighted set cover provides several promising results, e.g. this script.
Say I have a linked list of numbers of length N. N is very large and I don’t know in advance the exact value of N.
How can I most efficiently write a function that will return k completely random numbers from the list?
There's a very nice and efficient algorithm for this using a method called reservoir sampling.
Let me start by giving you its history:
Knuth calls this Algorithm R on p. 144 of his 1997 edition of Seminumerical Algorithms (volume 2 of The Art of Computer Programming), and provides some code for it there. Knuth attributes the algorithm to Alan G. Waterman. Despite a lengthy search, I haven't been able to find Waterman's original document, if it exists, which may be why you'll most often see Knuth quoted as the source of this algorithm.
McLeod and Bellhouse, 1983 (1) provide a more thorough discussion than Knuth as well as the first published proof (that I'm aware of) that the algorithm works.
Vitter 1985 (2) reviews Algorithm R and then presents an additional three algorithms which provide the same output, but with a twist. Rather than making a choice to include or skip each incoming element, his algorithm predetermines the number of incoming elements to be skipped. In his tests (which, admittedly, are out of date now) this decreased execution time dramatically by avoiding random number generation and comparisons on each in-coming number.
In pseudocode the algorithm is:
Let R be the result array of size s
Let I be an input queue
> Fill the reservoir array
for j in the range [1,s]:
R[j]=I.pop()
elements_seen=s
while I is not empty:
elements_seen+=1
j=random(1,elements_seen) > This is inclusive
if j<=s:
R[j]=I.pop()
else:
I.pop()
Note that I've specifically written the code to avoid specifying the size of the input. That's one of the cool properties of this algorithm: you can run it without needing to know the size of the input beforehand and it still assures you that each element you encounter has an equal probability of ending up in R (that is, there is no bias). Furthermore, R contains a fair and representative sample of the elements the algorithm has considered at all times. This means you can use this as an online algorithm.
Why does this work?
McLeod and Bellhouse (1983) provide a proof using the mathematics of combinations. It's pretty, but it would be a bit difficult to reconstruct it here. Therefore, I've generated an alternative proof which is easier to explain.
We proceed via proof by induction.
Say we want to generate a set of s elements and that we have already seen n>s elements.
Let's assume that our current s elements have already each been chosen with probability s/n.
By the definition of the algorithm, we choose element n+1 with probability s/(n+1).
Each element already part of our result set has a probability 1/s of being replaced.
The probability that an element from the n-seen result set is replaced in the n+1-seen result set is therefore (1/s)*s/(n+1)=1/(n+1). Conversely, the probability that an element is not replaced is 1-1/(n+1)=n/(n+1).
Thus, the n+1-seen result set contains an element either if it was part of the n-seen result set and was not replaced---this probability is (s/n)*n/(n+1)=s/(n+1)---or if the element was chosen---with probability s/(n+1).
The definition of the algorithm tells us that the first s elements are automatically included as the first n=s members of the result set. Therefore, the n-seen result set includes each element with s/n (=1) probability giving us the necessary base case for the induction.
References
McLeod, A. Ian, and David R. Bellhouse. "A convenient algorithm for drawing a simple random sample." Journal of the Royal Statistical Society. Series C (Applied Statistics) 32.2 (1983): 182-184. (Link)
Vitter, Jeffrey S. "Random sampling with a reservoir." ACM Transactions on Mathematical Software (TOMS) 11.1 (1985): 37-57. (Link)
This is called a Reservoir Sampling problem. The simple solution is to assign a random number to each element of the list as you see it, then keep the top (or bottom) k elements as ordered by the random number.
I would suggest: First find your k random numbers. Sort them. Then traverse both the linked list and your random numbers once.
If you somehow don't know the length of your linked list (how?), then you could grab the first k into an array, then for node r, generate a random number in [0, r), and if that is less than k, replace the rth item of the array. (Not entirely convinced that doesn't bias...)
Other than that: "If I were you, I wouldn't be starting from here." Are you sure linked list is right for your problem? Is there not a better data structure, such as a good old flat array list.
If you don't know the length of the list, then you will have to traverse it complete to ensure random picks. The method I've used in this case is the one described by Tom Hawtin (54070). While traversing the list you keep k elements that form your random selection to that point. (Initially you just add the first k elements you encounter.) Then, with probability k/i, you replace a random element from your selection with the ith element of the list (i.e. the element you are at, at that moment).
It's easy to show that this gives a random selection. After seeing m elements (m > k), we have that each of the first m elements of the list are part of you random selection with a probability k/m. That this initially holds is trivial. Then for each element m+1, you put it in your selection (replacing a random element) with probability k/(m+1). You now need to show that all other elements also have probability k/(m+1) of being selected. We have that the probability is k/m * (k/(m+1)*(1-1/k) + (1-k/(m+1))) (i.e. probability that element was in the list times the probability that it is still there). With calculus you can straightforwardly show that this is equal to k/(m+1).
Well, you do need to know what N is at runtime at least, even if this involves doing an extra pass over the list to count them. The simplest algorithm to do this is to just pick a random number in N and remove that item, repeated k times. Or, if it is permissible to return repeat numbers, don't remove the item.
Unless you have a VERY large N, and very stringent performance requirements, this algorithm runs with O(N*k) complexity, which should be acceptable.
Edit: Nevermind, Tom Hawtin's method is way better. Select the random numbers first, then traverse the list once. Same theoretical complexity, I think, but much better expected runtime.
Why can't you just do something like
List GetKRandomFromList(List input, int k)
List ret = new List();
for(i=0;i<k;i++)
ret.Add(input[Math.Rand(0,input.Length)]);
return ret;
I'm sure that you don't mean something that simple so can you specify further?