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Recently I needed to do weighted random selection of elements from a list, both with and without replacement. While there are well known and good algorithms for unweighted selection, and some for weighted selection without replacement (such as modifications of the resevoir algorithm), I couldn't find any good algorithms for weighted selection with replacement. I also wanted to avoid the resevoir method, as I was selecting a significant fraction of the list, which is small enough to hold in memory.
Does anyone have any suggestions on the best approach in this situation? I have my own solutions, but I'm hoping to find something more efficient, simpler, or both.
One of the fastest ways to make many with replacement samples from an unchanging list is the alias method. The core intuition is that we can create a set of equal-sized bins for the weighted list that can be indexed very efficiently through bit operations, to avoid a binary search. It will turn out that, done correctly, we will need to only store two items from the original list per bin, and thus can represent the split with a single percentage.
Let's us take the example of five equally weighted choices, (a:1, b:1, c:1, d:1, e:1)
To create the alias lookup:
Normalize the weights such that they sum to 1.0. (a:0.2 b:0.2 c:0.2 d:0.2 e:0.2) This is the probability of choosing each weight.
Find the smallest power of 2 greater than or equal to the number of variables, and create this number of partitions, |p|. Each partition represents a probability mass of 1/|p|. In this case, we create 8 partitions, each able to contain 0.125.
Take the variable with the least remaining weight, and place as much of it's mass as possible in an empty partition. In this example, we see that a fills the first partition. (p1{a|null,1.0},p2,p3,p4,p5,p6,p7,p8) with (a:0.075, b:0.2 c:0.2 d:0.2 e:0.2)
If the partition is not filled, take the variable with the most weight, and fill the partition with that variable.
Repeat steps 3 and 4, until none of the weight from the original partition need be assigned to the list.
For example, if we run another iteration of 3 and 4, we see
(p1{a|null,1.0},p2{a|b,0.6},p3,p4,p5,p6,p7,p8) with (a:0, b:0.15 c:0.2 d:0.2 e:0.2) left to be assigned
At runtime:
Get a U(0,1) random number, say binary 0.001100000
bitshift it lg2(p), finding the index partition. Thus, we shift it by 3, yielding 001.1, or position 1, and thus partition 2.
If the partition is split, use the decimal portion of the shifted random number to decide the split. In this case, the value is 0.5, and 0.5 < 0.6, so return a.
Here is some code and another explanation, but unfortunately it doesn't use the bitshifting technique, nor have I actually verified it.
A simple approach that hasn't been mentioned here is one proposed in Efraimidis and Spirakis. In python you could select m items from n >= m weighted items with strictly positive weights stored in weights, returning the selected indices, with:
import heapq
import math
import random
def WeightedSelectionWithoutReplacement(weights, m):
elt = [(math.log(random.random()) / weights[i], i) for i in range(len(weights))]
return [x[1] for x in heapq.nlargest(m, elt)]
This is very similar in structure to the first approach proposed by Nick Johnson. Unfortunately, that approach is biased in selecting the elements (see the comments on the method). Efraimidis and Spirakis proved that their approach is equivalent to random sampling without replacement in the linked paper.
Here's what I came up with for weighted selection without replacement:
def WeightedSelectionWithoutReplacement(l, n):
"""Selects without replacement n random elements from a list of (weight, item) tuples."""
l = sorted((random.random() * x[0], x[1]) for x in l)
return l[-n:]
This is O(m log m) on the number of items in the list to be selected from. I'm fairly certain this will weight items correctly, though I haven't verified it in any formal sense.
Here's what I came up with for weighted selection with replacement:
def WeightedSelectionWithReplacement(l, n):
"""Selects with replacement n random elements from a list of (weight, item) tuples."""
cuml = []
total_weight = 0.0
for weight, item in l:
total_weight += weight
cuml.append((total_weight, item))
return [cuml[bisect.bisect(cuml, random.random()*total_weight)] for x in range(n)]
This is O(m + n log m), where m is the number of items in the input list, and n is the number of items to be selected.
I'd recommend you start by looking at section 3.4.2 of Donald Knuth's Seminumerical Algorithms.
If your arrays are large, there are more efficient algorithms in chapter 3 of Principles of Random Variate Generation by John Dagpunar. If your arrays are not terribly large or you're not concerned with squeezing out as much efficiency as possible, the simpler algorithms in Knuth are probably fine.
It is possible to do Weighted Random Selection with replacement in O(1) time, after first creating an additional O(N)-sized data structure in O(N) time. The algorithm is based on the Alias Method developed by Walker and Vose, which is well described here.
The essential idea is that each bin in a histogram would be chosen with probability 1/N by a uniform RNG. So we will walk through it, and for any underpopulated bin which would would receive excess hits, assign the excess to an overpopulated bin. For each bin, we store the percentage of hits which belong to it, and the partner bin for the excess. This version tracks small and large bins in place, removing the need for an additional stack. It uses the index of the partner (stored in bucket[1]) as an indicator that they have already been processed.
Here is a minimal python implementation, based on the C implementation here
def prep(weights):
data_sz = len(weights)
factor = data_sz/float(sum(weights))
data = [[w*factor, i] for i,w in enumerate(weights)]
big=0
while big<data_sz and data[big][0]<=1.0: big+=1
for small,bucket in enumerate(data):
if bucket[1] is not small: continue
excess = 1.0 - bucket[0]
while excess > 0:
if big==data_sz: break
bucket[1] = big
bucket = data[big]
bucket[0] -= excess
excess = 1.0 - bucket[0]
if (excess >= 0):
big+=1
while big<data_sz and data[big][0]<=1: big+=1
return data
def sample(data):
r=random.random()*len(data)
idx = int(r)
return data[idx][1] if r-idx > data[idx][0] else idx
Example usage:
TRIALS=1000
weights = [20,1.5,9.8,10,15,10,15.5,10,8,.2];
samples = [0]*len(weights)
data = prep(weights)
for _ in range(int(sum(weights)*TRIALS)):
samples[sample(data)]+=1
result = [float(s)/TRIALS for s in samples]
err = [a-b for a,b in zip(result,weights)]
print(result)
print([round(e,5) for e in err])
print(sum([e*e for e in err]))
The following is a description of random weighted selection of an element of a
set (or multiset, if repeats are allowed), both with and without replacement in O(n) space
and O(log n) time.
It consists of implementing a binary search tree, sorted by the elements to be
selected, where each node of the tree contains:
the element itself (element)
the un-normalized weight of the element (elementweight), and
the sum of all the un-normalized weights of the left-child node and all of
its children (leftbranchweight).
the sum of all the un-normalized weights of the right-child node and all of
its chilren (rightbranchweight).
Then we randomly select an element from the BST by descending down the tree. A
rough description of the algorithm follows. The algorithm is given a node of
the tree. Then the values of leftbranchweight, rightbranchweight,
and elementweight of node is summed, and the weights are divided by this
sum, resulting in the values leftbranchprobability,
rightbranchprobability, and elementprobability, respectively. Then a
random number between 0 and 1 (randomnumber) is obtained.
if the number is less than elementprobability,
remove the element from the BST as normal, updating leftbranchweight
and rightbranchweight of all the necessary nodes, and return the
element.
else if the number is less than (elementprobability + leftbranchweight)
recurse on leftchild (run the algorithm using leftchild as node)
else
recurse on rightchild
When we finally find, using these weights, which element is to be returned, we either simply return it (with replacement) or we remove it and update relevant weights in the tree (without replacement).
DISCLAIMER: The algorithm is rough, and a treatise on the proper implementation
of a BST is not attempted here; rather, it is hoped that this answer will help
those who really need fast weighted selection without replacement (like I do).
This is an old question for which numpy now offers an easy solution so I thought I would mention it. Current version of numpy is version 1.2 and numpy.random.choice allows the sampling to be done with or without replacement and with given weights.
Suppose you want to sample 3 elements without replacement from the list ['white','blue','black','yellow','green'] with a prob. distribution [0.1, 0.2, 0.4, 0.1, 0.2]. Using numpy.random module it is as easy as this:
import numpy.random as rnd
sampling_size = 3
domain = ['white','blue','black','yellow','green']
probs = [.1, .2, .4, .1, .2]
sample = rnd.choice(domain, size=sampling_size, replace=False, p=probs)
# in short: rnd.choice(domain, sampling_size, False, probs)
print(sample)
# Possible output: ['white' 'black' 'blue']
Setting the replace flag to True, you have a sampling with replacement.
More info here:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.random.choice.html#numpy.random.choice
We faced a problem to randomly select K validators of N candidates once per epoch proportionally to their stakes. But this gives us the following problem:
Imagine probabilities of each candidate:
0.1
0.1
0.8
Probabilities of each candidate after 1'000'000 selections 2 of 3 without replacement became:
0.254315
0.256755
0.488930
You should know, those original probabilities are not achievable for 2 of 3 selection without replacement.
But we wish initial probabilities to be a profit distribution probabilities. Else it makes small candidate pools more profitable. So we realized that random selection with replacement would help us – to randomly select >K of N and store also weight of each validator for reward distribution:
std::vector<int> validators;
std::vector<int> weights(n);
int totalWeights = 0;
for (int j = 0; validators.size() < m; j++) {
int value = rand() % likehoodsSum;
for (int i = 0; i < n; i++) {
if (value < likehoods[i]) {
if (weights[i] == 0) {
validators.push_back(i);
}
weights[i]++;
totalWeights++;
break;
}
value -= likehoods[i];
}
}
It gives an almost original distribution of rewards on millions of samples:
0.101230
0.099113
0.799657
Suppose you are given an unsorted list of positive integers, and you wish to order them in a manner such that the elements alternate as: (less than preceding element), (greater than preceding element), (less than preceding element), etc... The very first element in the output list may ignore the rule. So for example, suppose your list was: 1,4,9,2,7,5,3,8,6.
One correct output would be...
1,9,2,8,3,7,4,6,5
Another would be...
3,4,2,7,5,6,1,9,8
Assume that the list contains no duplicates, is arbitrarily large, and is not already sorted.
What is the most processing efficient algorithm to achieve this?
Now, the standard approach would be to simply sort the list in ascending order first, and then peel elements from the ends of the list in alternation. However, I'd like to know: Is there a more time-efficient way to do this without first sorting the list?
My reason for asking: (read this only if you care)
Apparently this is a question my sister's boyfriend poses to people at job interviews out in San Francisco. My sister asked me the question, and I immediately came up with the standard response. That's what everyone answers. However, apparently one girl came up with a completely different solution that does not require sorting the list, and it appears to work. My sister couldn't explain to me this solution, but the idea has been confounding me since last night. I'd appreciate any help! Thanks!
You can do this in O(n) by placing each element in turn at the end, or at the penultimate position based on a comparison with the current last element.
For example,
1,4,9,2,7,5,3,8,6
Place 1 at end, current list [1]
4>1 true so place 4 at end, current list [1,4]
9<4 false so place 9 at penultimate position [1,9,4]
2>4 false so place 2 at penultimate [1,9,2,4]
7<4 false so place 7 at penultimate [1,9,2,7,4]
5>4 true so place 5 at end [1,9,2,7,4,5]
3<5 true so place 3 at end [1,9,2,7,4,5,3]
8>3 true so place 8 at end [1,9,2,7,4,5,3,8]
6<8 true so place 6 at end [1,9,2,7,4,5,3,8,6]
Note that the equality tests alternate, and that we place at the end if the equality is true, or at the penultimate position if it is not true.
Example Python Code
A=[1,4,9,2,7,5,3,8,6]
B=[]
for i,a in enumerate(A):
if i==0 or (i&1 and a>B[-1]) or (i&1==0 and a<B[-1]):
B.insert(i,a)
else:
B.insert(i-1,a)
print B
One solution is this. Given in Pseudocode.
Assuming, nums has at least two elements and all elements in nums are distinct.
nums = [list of numbers]
if nums[0] < nums[1]: last_state = INCREASING else: last_state = DECREASING
for i = 2 to len(nums - 1):
if last_state = INCREASING:
if nums[i] > nums[i-1]:
swap (nums[i], nums[i-1])
last_state = DECREASING
else
if nums[i] < nums[i-1]:
swap (nums[i], nums[i-1])
last_state = INCREASING
Proof of correctness:
After each loop iteration, elements upto index i in nums remain alternating and last_state is represent the order of i th and i-1 th elements.
Note that a swapping happens only if last 3 items considered are in order. (Increasing or Decreasing) Therefore, if we swapped ith element with i-1 th element, the order of i-2 th element and i-1th element will not change.
For example, I have the numbers 46,47,54,58,60, and 66. I want to make group them in such a way as to make the largest possible group sizes. Numbers get grouped if their values are within plus or minus 10 (inclusive). So, depending on which number you start with, for this example there can be three possible outcomes (shown in the image).
What I want is the second possible outcome, which would occur if you started with 54, as the numbers within 44 to 64 would be grouped, leaving 66 by itself, and creating the largest group (5 items).
I realize I could easily brute force this example, but I really have a long list of numbers and it needs to do this across thousands of numbers.. Can anyone tell me about algorithms I should be reading about or give me suggestions?
You can simply sort the array first. Then for every i th number you can do a binary search to find the right most number that's within ith number + 20 range, let the position of such right most index is X. You have to find the largest (X-i+1) for all ith numbers and we are done :)
Runtime analysis: Runtime for this algorithm will be O(NlgN), where N is the number of items in the original array.
A better solution: Let's assume we have the array ar[] and ar[] has N items.
sort ar[] in non decreasing order
set max_result = 0, set cur_index = 0, i=0
increase i while i
set max_result to max(max_result,i-cur_index+1)
set cur_index=cur_index+1
if cur_index
Runtime Analysis: O(N), where N is the number of items in the array ar[] as cur_index will iterate through the array exactly once and i will iterate just once too.
Correctness: as the array is sorted in non decreasing order, if i < j and j < k and ar[i]+20 > ar[k] then ar[j]+20 > ar[k] too. So we don't need to check for these items those are already checked for previous item.
This is what I wanted to do. Sorry I didn't explain myself very well. Each iteration finds the largest possible group, using the numbers that are left after removing the previous largest group. Matlab code:
function out=groupNums(y)
d=10;
out=[];
if length(y)==1
out=y;
return
end
group=[];
for i=1:length(y)
group{i}=find(y<=y(i)+d & y>=y(i)-d);
end
[~,idx]=max(cellfun(#length,group));
out=[out,{y(group{idx})}];
y(group{idx})=[];
out=[out,groupNums(y)];
How can I generate a random number that is in the range (1,n) but not in a certain list (i,j)?
Example: range is (1,500), list is [1,3,4,45,199,212,344].
Note: The list may not be sorted
Rejection Sampling
One method is rejection sampling:
Generate a number x in the range (1, 500)
Is x in your list of disallowed values? (Can use a hash-set for this check.)
If yes, return to step 1
If no, x is your random value, done
This will work fine if your set of allowed values is significantly larger than your set of disallowed values:if there are G possible good values and B possible bad values, then the expected number of times you'll have to sample x from the G + B values until you get a good value is (G + B) / G (the expectation of the associated geometric distribution). (You can sense check this. As G goes to infinity, the expectation goes to 1. As B goes to infinity, the expectation goes to infinity.)
Sampling a List
Another method is to make a list L of all of your allowed values, then sample L[rand(L.count)].
The technique I usually use when the list is length 1 is to generate a random
integer r in [1,n-1], and if r is greater or equal to that single illegal
value then increment r.
This can be generalised for a list of length k for small k but requires
sorting that list (you can't do your compare-and-increment in random order). If the list is moderately long, then after the sort you can start with a bsearch, and add the number of values skipped to r, and then recurse into the remainder of the list.
For a list of length k, containing no value greater or equal to n-k, you
can do a more direct substitution: generate random r in [1,n-k], and
then iterate through the list testing if r is equal to list[i]. If it is
then set r to n-k+i (this assumes list is zero-based) and quit.
That second approach fails if some of the list elements are in [n-k,n].
I could try to invest something clever at this point, but what I have so far
seems sufficient for uniform distributions with values of k much less than
n...
Create two lists -- one of illegal values below n-k, and the other the rest (this can be done in place).
Generate random r in [1,n-k]
Apply the direct substitution approach for the first list (if r is list[i] then set r to n-k+i and go to step 5).
If r was not altered in step 3 then we're finished.
Sort the list of larger values and use the compare-and-increment method.
Observations:
If all values are in the lower list, there will be no sort because there is nothing to sort.
If all values are in the upper list, there will be no sort because there is no occasion on which r is moved into the hazardous area.
As k approaches n, the maximum size of the upper (sorted) list grows.
For a given k, if more value appear in the upper list (the bigger the sort), the chance of getting a hit in the lower list shrinks, reducing the likelihood of needing to do the sort.
Refinement:
Obviously things get very sorty for large k, but in such cases the list has comparatively few holes into which r is allowed to settle. This could surely be exploited.
I might suggest something different if many random values with the same
list and limits were needed. I hope that the list of illegal values is not the
list of results of previous calls to this function, because if it is then you
wouldn't want any of this -- instead you would want a Fisher-Yates shuffle.
Rejection sampling would be the simplest if possible as described already. However, if you didn't want use that, you could convert the range and disallowed values to sets and find the difference. Then, you could choose a random value out of there.
Assuming you wanted the range to be in [1,n] but not in [i,j] and that you wanted them uniformly distributed.
In Python
total = range(1,n+1)
disallowed = range(i,j+1)
allowed = list( set(total) - set(disallowed) )
return allowed[random.randrange(len(allowed))]
(Note that this is not EXACTLY uniform since in all likeliness, max_rand%len(allowed) != 0 but this will in most practical applications be very close)
I assume that you know how to generate a random number in [1, n) and also your list is ordered like in the example above.
Let's say that you have a list with k elements. Make a map(O(logn)) structure, which will ensure speed if k goes higher. Put all elements from list in map, where element value will be the key and "good" value will be the value. Later on I'll explain about "good" value. So when we have the map then just find a random number in [1, n - k - p)(Later on I'll explain what is p) and if this number is in map then replace it with "good" value.
"GOOD" value -> Let's start from k-th element. It's good value is its own value + 1, because the very next element is "good" for us. Now let's look at (k-1)th element. We assume that its good value is again its own value + 1. If this value is equal to k-th element then the "good" value for (k-1)th element is k-th "good" value + 1. Also you will have to store the largest "good" value. If the largest value exceed n then p(from above) will be p = largest - n.
Of course I recommend you this only if k is big number otherwise #Timothy Shields' method is perfect.
I have a collection of objects, each of which has a weight and a value. I want to pick the pair of objects with the highest total value subject to the restriction that their combined weight does not exceed some threshold. Additionally, I am given two arrays, one containing the objects sorted by weight and one containing the objects sorted by value.
I know how to do it in O(n2) but how can I do it in O(n)?
This is a combinatorial optimization problem, and the fact the values are sorted means you can easily try a branch and bound approach.
I think that I have a solution that works in O(n log n) time and O(n) extra space. This isn't quite the O(n) solution you wanted, but it's still better than the naive quadratic solution.
The intuition behind the algorithm is that we want to be able to efficiently determine, for any amount of weight, the maximum value we can get with a single item that uses at most that much weight. If we can do this, we have a simple algorithm for solving the problem: iterate across the array of elements sorted by value. For each element, see how much additional value we could get by pairing a single element with it (using the values we precomputed), then find which of these pairs is maximum. If we can do the preprocessing in O(n log n) time and can answer each of the above queries in O(log n) time, then the total time for the second step will be O(n log n) and we have our answer.
An important observation we need to do the preprocessing step is as follows. Our goal is to build up a structure that can answer the question "which element with weight less than x has maximum value?" Let's think about how we might do this by adding one element at a time. If we have an element (value, weight) and the structure is empty, then we want to say that the maximum value we can get using weight at most "weight" is "value". This means that everything in the range [0, max_weight - weight) should be set to value. Otherwise, suppose that the structure isn't empty when we try adding in (value, weight). In that case, we want to say that any portion of the range [0, weight) whose value is less than value should be replaced by value.
The problem here is that when we do these insertions, there might be, on iteration k, O(k) different subranges that need to be updated, leading to an O(n2) algorithm. However, we can use a very clever trick to avoid this. Suppose that we insert all of the elements into this data structure in descending order of value. In that case, when we add in (value, weight), because we add the elements in descending order of value, each existing value in the data structure must be higher than our value. This means that if the range [0, weight) intersects any range at all, those ranges will automatically be higher than value and so we don't need to update them. If we combine this with the fact that each range we add always spans from zero to some value, the only portion of the new range that could ever be added to the data structure is the range [weight, x), where x is the highest weight stored in the data structure so far.
To summarize, assuming that we visit the (value, weight) pairs in descending order of value, we can update our data structure as follows:
If the structure is empty, record that the range [0, value) has value "value."
Otherwise, if the highest weight recorded in the structure is greater than weight, skip this element.
Otherwise, if the highest weight recorded so far is x, record that the range [weight, x) has value "value."
Notice that this means that we are always splitting ranges at the front of the list of ranges we have encountered so far. Because of this, we can think about storing the list of ranges as a simple array, where each array element tracks the upper endpoint of some range and the value assigned to that range. For example, we might track the ranges [0, 3), [3, 9), and [9, 12) as the array
3, 9, 12
If we then needed to split the range [0, 3) into [0, 1) and [1, 3), we could do so by prepending 1 to he list:
1, 3, 9, 12
If we represent this array in reverse (actually storing the ranges from high to low instead of low to high), this step of creating the array runs in O(n) time because at each point we just do O(1) work to decide whether or not to add another element onto the end of the array.
Once we have the ranges stored like this, to determine which of the ranges a particular weight falls into, we can just use a binary search to find the largest element smaller than that weight. For example, to look up 6 in the above array we'd do a binary search to find 3.
Finally, once we have this data structure built up, we can just look at each of the objects one at a time. For each element, we see how much weight is left, use a binary search in the other structure to see what element it should be paired with to maximize the total value, and then find the maximum attainable value.
Let's trace through an example. Given maximum allowable weight 10 and the objects
Weight | Value
------+------
2 | 3
6 | 5
4 | 7
7 | 8
Let's see what the algorithm does. First, we need to build up our auxiliary structure for the ranges. We look at the objects in descending order of value, starting with the object of weight 7 and value 8. This means that if we ever have at least seven units of weight left, we can get 8 value. Our array now looks like this:
Weight: 7
Value: 8
Next, we look at the object of weight 4 and value 7. This means that with four or more units of weight left, we can get value 7:
Weight: 7 4
Value: 8 7
Repeating this for the next item (weight six, value five) does not change the array, since if the object has weight six, if we ever had six or more units of free space left, we would never choose this; we'd always take the seven-value item of weight four. We can tell this since there is already an object in the table whose range includes remaining weight four.
Finally, we look at the last item (value 3, weight 2). This means that if we ever have weight two or more free, we could get 3 units of value. The final array now looks like this:
Weight: 7 4 2
Value: 8 7 3
Finally, we just look at the objects in any order to see what the best option is. When looking at the object of weight 2 and value 3, since the maximum allowed weight is 10, we need tom see how much value we can get with at most 10 - 2 = 8 weight. A binary search over the array tells us that this value is 8, so one option would give us 11 weight. If we look at the object of weight 6 and value 5, a binary search tells us that with five remaining weight the best we can do would be to get 7 units of value, for a total of 12 value. Repeating this on the next two entries doesn't turn up anything new, so the optimum value found has value 12, which is indeed the correct answer.
Hope this helps!
Here is an O(n) time, O(1) space solution.
Let's call an object x better than an object y if and only if (x is no heavier than y) and (x is no less valuable) and (x is lighter or more valuable). Call an object x first-choice if no object is better than x. There exists an optimal solution consisting either of two first-choice objects, or a first-choice object x and an object y such that only x is better than y.
The main tool is to be able to iterate the first-choice objects from lightest to heaviest (= least valuable to most valuable) and from most valuable to least valuable (= heaviest to lightest). The iterator state is an index into the objects by weight (resp. value) and a max value (resp. min weight) so far.
Each of the following steps is O(n).
During a scan, whenever we encounter an object that is not first-choice, we know an object that's better than it. Scan once and consider these pairs of objects.
For each first-choice object from lightest to heaviest, determine the heaviest first-choice object that it can be paired with, and consider the pair. (All lighter objects are less valuable.) Since the latter object becomes lighter over time, each iteration of the loop is amortized O(1). (See also searching in a matrix whose rows and columns are sorted.)
Code for the unbelievers. Not heavily tested.
from collections import namedtuple
from operator import attrgetter
Item = namedtuple('Item', ('weight', 'value'))
sentinel = Item(float('inf'), float('-inf'))
def firstchoicefrombyweight(byweight):
bestsofar = sentinel
for x in byweight:
if x.value > bestsofar.value:
bestsofar = x
yield (x, bestsofar)
def firstchoicefrombyvalue(byvalue):
bestsofar = sentinel
for x in byvalue:
if x.weight < bestsofar.weight:
bestsofar = x
yield x
def optimize(items, maxweight):
byweight = sorted(items, key=attrgetter('weight'))
byvalue = sorted(items, key=attrgetter('value'), reverse=True)
maxvalue = float('-inf')
try:
i = firstchoicefrombyvalue(byvalue)
y = i.next()
for x, z in firstchoicefrombyweight(byweight):
if z is not x and x.weight + z.weight <= maxweight:
maxvalue = max(maxvalue, x.value + z.value)
while x.weight + y.weight > maxweight:
y = i.next()
if y is x:
break
maxvalue = max(maxvalue, x.value + y.value)
except StopIteration:
pass
return maxvalue
items = [Item(1, 1), Item(2, 2), Item(3, 5), Item(3, 7), Item(5, 8)]
for maxweight in xrange(3, 10):
print maxweight, optimize(items, maxweight)
This is similar to Knapsack problem. I will use naming from it (num - weight, val - value).
The essential part:
Start with a = 0 and b = n-1. Assuming 0 is the index of heaviest object and n-1 is the index of lightest object.
Increase a til objects a and b satisfy the limit.
Compare current solution with best solution.
Decrease b by one.
Go to 2.
Update:
It's the knapsack problem, except there is a limit of 2 items. You basically need to decide how much space you want for the first object and how much for the other. There is n significant ways to split available space, so the complexity is O(n). Picking the most valuable objects to fit in those spaces can be done without additional cost.