The problem is:
Initially, the sequence is empty. There are n queries and 4 types of queries:
Add(x): add x to the sequence, if there is already x in the sequence, still add x.
Remove(x): remove x from the sequence 1 times.
Xor(x): replace all N of the sequence with N xor x.
Sum(K): find sum of the k smallest elements in the sequence.
0 <= x, n, K <= 10^5
For each query sum(x), output the sum of the x smallest elements in the sequence.
Input:
7
Add(4) // A[] = {4}
Remove(3) // A[] = {4}
Add(2) // A[] = {4, 2}
Sum(2) // A[] = {4, 2} => Output: 6
Xor(2) // A[] = {4^2, 2^2} = {6, 0}
Sum(1) // A[] = {6, 0} => Output: 0
Sum(2) // A[] = {6, 0} => Output: 6
I solved the problem with the following way:
Use a vector A to hold the sequence of numbers, and an array Count[] where Count[x] is the number of occurrences of x in A. Initially A is empty, and every Count[x] = 0.
For each Add(x) query, I add x to A, and Count[x] = Count[x]+1
For each Remove(x) query, if Count[x] = 0 then skip, otherwise, remove x from A and Count[x] = Count[x]-1
For each Xor(x) query, replace every A[i] with A[i]^x
For each Sum(x) query, sort A in ascending value, take the sum of the first x numbers
It seems that my way has a complexity of O(n^2), so for n <= 100000 the above algorithm cannot work. Is there a better way to solve this problem? Thanks a lot.
My code can run well in n <= 5000. Here is it:
int Count[100001];
vector<int> A;
void Add(int x) {
A.push_back(x);
Count[x] = Count[x]+1;
}
void Remove(int x) {
if (Count[x] == 0) return;
Count[x] = Count[x]-1;
auto Find = find(A.begin(), A.end(), x);
A.erase(Find);
}
void Xor(int x) {
for (int& i : A)
i = i^x;
}
int Sum(int x) {
int Num = 0, S = 0;
for (int i : A) {
if (Num + 1 > x) return S;
S = S + i; Num = Num + 1;
}
return S;
}
I'll describe a data structure that supports Add(x)/Remove(x)/Count()/SumXorWith(x) (returns the sum of all elements xor x; doesn't modify the sequence) and then sketch how to extend it to a full solution where each operation is O(log^2 n) (taking n to be both the number of operations and the upper bound on the values).
First observe that Count and SumXorWith can be used to count, for each bit position, how many numbers have that position set (e.g., for the low order bit, it's (Count() + SumXorWith(0) - SumXorWith(1)) / 2). Conversely, it's enough to maintain these counts. In pseudocode:
*** Variables, initially zero:
count : int
bit_count : int[17]
*** Operations:
Add(x):
increment count
for j from 0 to 16, add the j'th bit of x to bit_count[j]
Remove(x):
decrement count
for j from 0 to 16, subtract the j'th bit of x from bit_count[j]
Count():
return count
SumXorWith(x):
return the sum for j from 0 to 16 of
2**j * (if j'th bit of x = 0 then bit_count[j] else count - bit_count[j])
To extend this data structure to handle Xor(x)/Sum(), we could just take count - bit_count for each bit set in x, but for efficiency (that we'll need later), there's a trick. The idea is that we store the sequence xor cum_xor. More pseudocode:
*** Additional variable, initially zero
cum_xor : int
*** Operations:
Add(x): super.Add(x xor cum_xor)
Remove(x): super.Remove(x xor cum_xor)
Xor(x): cum_xor <- cum_xor xor x
Count(): return super.Count()
Sum(): return super.SumXorWith(cum_xor)
Finally, we need to handle Sum(x), with selection. This is, frankly, the tedious part. We set up a height-17 (ceiling of log2(100000)) trie on big-endian bit patterns, with one of the data structures above at each node of the trie. To Add/Remove, we descend the trie, doing Add/Remove at each node. Xor we handle as before, by updating cum_xor. Sum(x) is the trickiest, of course. Starting at the root of the trie, we examine the current node. If it has at most x elements, just sum it. Otherwise, its "favored" child is the one that agrees with cum_xor, and its "disfavored" child is the one that disagrees. If the favored child has at least x elements, then we can operate recursively on it and ignore the disfavored child. Otherwise, we sum the whole favored child and operate recursively on the disfavored child, decreasing x by the number of elements in the favored child.
(For maximum practical efficiency, we'd want something with higher fan-out than the trie and likely the naive implementation near the leaves, but this is as simple as I can make it and likely fast enough.)
I am trying to solve a problem from codility
"Even sums"
but am unable to do so. Here is the question below.
Even sums is a game for two players. Players are given a sequence of N positive integers and take turns alternately. In each turn, a player chooses a non-empty slice (a subsequence of consecutive elements) such that the sum of values in this slice is even, then removes the slice and concatenates the remaining parts of the sequence. The first player who is unable to make a legal move loses the game.
You play this game against your opponent and you want to know if you can win, assuming both you and your opponent play optimally. You move first.
Write a function:
string solution(vector< int>& A);
that, given a zero-indexed array A consisting of N integers, returns a string of format "X,Y" where X and Y are, respectively, the first and last positions (inclusive) of the slice that you should remove on your first move in order to win, assuming you have a winning strategy. If there is more than one such winning slice, the function should return the one with the smallest value of X. If there is more than one slice with the smallest value of X, the function should return the shortest. If you do not have a winning strategy, the function should return "NO SOLUTION".
For example, given the following array:
A[0] = 4 A[1] = 5 A[2] = 3 A[3] = 7 A[4] = 2
the function should return "1,2". After removing a slice from positions 1 to 2 (with an even sum of 5 + 3 = 8), the remaining array is [4, 7, 2]. Then the opponent will be able to remove the first element (of even sum 4) or the last element (of even sum 2). Afterwards you can make a move that leaves the array containing just [7], so your opponent will not have a legal move and will lose. One of possible games is shown on the following picture
Note that removing slice "2,3" (with an even sum of 3 + 7 = 10) is also a winning move, but slice "1,2" has a smaller value of X.
For the following array:
A[0] = 2 A[ 1 ] = 5 A[2] = 4
the function should return "NO SOLUTION", since there is no strategy that guarantees you a win.
Assume that:
N is an integer within the range [1..100,000]; each element of array A is an integer within the range [1..1,000,000,000]. Complexity:
expected worst-case time complexity is O(N); expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments). Elements of input arrays can be modified.
I have found a solution online in python.
def check(start, end):
if start>end:
res = 'NO SOLUTION'
else:
res = str(start) + ',' + str(end)
return res
def trans( strr ):
if strr =='NO SOLUTION':
return (-1, -1)
else:
a, b = strr.split(',')
return ( int(a), int(b) )
def solution(A):
# write your code in Python 2.7
odd_list = [ ind for ind in range(len(A)) if A[ind]%2==1 ]
if len(odd_list)%2==0:
return check(0, len(A)-1)
odd_list = [-1] + odd_list + [len(A)]
res_cand = []
# the numbers at the either end of A are even
count = odd_list[1]
second_count = len(A)-1-odd_list[-2]
first_count = odd_list[2]-odd_list[1]-1
if second_count >= count:
res_cand.append( trans(check( odd_list[1]+1, len(A)-1-count )))
if first_count >= count:
res_cand.append( trans(check( odd_list[1]+count+1, len(A)-1 )))
twosum = first_count + second_count
if second_count < count <= twosum:
res_cand.append( trans(check( odd_list[1]+(first_count-(count-second_count))+1, odd_list[-2] )))
###########################################
count = len(A)-1-odd_list[-2]
first_count = odd_list[1]
second_count = odd_list[-2]-odd_list[-3]-1
if first_count >= count:
res_cand.append( trans(check( count, odd_list[-2]-1 )))
if second_count >= count:
res_cand.append( trans(check( 0, odd_list[-2]-count-1)) )
twosum = first_count + second_count
if second_count < count <= twosum:
res_cand.append( trans(check( count-second_count, odd_list[-3])) )
res_cand = sorted( res_cand, key=lambda x: (-x[0],-x[1]) )
cur = (-1, -2)
for item in res_cand:
if item[0]!=-1:
cur = item
return check( cur[0], cur[1] )
This code works and I am unable to understand the code and flow of one function to the the other. However I don't understand the logic of the algorithm. How it has approached the problem and solved it. This might be a long task but can anybody please care enough to explain me the algorithm. Thanks in advance.
So far I have figured out that the number of odd numbers are crucial to find out the result. Especially the index of the first odd number and the last odd number is needed to calculate the important values.
Now I need to understand the logic behind the comparison such as "if first_count >= count" and if "second_count < count <= twosum".
Update:
Hey guys I found out the solution to my question and finally understood the logic of the algorithm.
The idea lies behind the symmetry of the array. We can never win the game if the array is symmetrical. Here symmetrical is defined as the array where there is only one odd in the middle and equal number of evens on the either side of that one odd.
If there are even number of odds we can directly win the game.
If there are odd number of odds we should always try to make the array symmetrical. That is what the algorithm is trying to do.
Now there are two cases to it. Either the last odd will remain or the first odd will remain. I will be happy to explain more if you guys didn't understand it. Thanks.
Given a number n of x digits. How to remove y digits in a way the remaining digits results in the greater possible number?
Examples:
1)x=7 y=3
n=7816295
-8-6-95
=8695
2)x=4 y=2
n=4213
4--3
=43
3)x=3 y=1
n=888
=88
Just to state: x > y > 0.
For each digit to remove: iterate through the digits left to right; if you find a digit that's less than the one to its right, remove it and stop, otherwise remove the last digit.
If the number of digits x is greater than the actual length of the number, it means there are leading zeros. Since those will be the first to go, you can simply reduce the count y by a corresponding amount.
Here's a working version in Python:
def remove_digits(n, x, y):
s = str(n)
if len(s) > x:
raise ValueError
elif len(s) < x:
y -= x - len(s)
if y <= 0:
return n
for r in range(y):
for i in range(len(s)):
if s[i] < s[i+1:i+2]:
break
s = s[:i] + s[i+1:]
return int(s)
>>> remove_digits(7816295, 7, 3)
8695
>>> remove_digits(4213, 4, 2)
43
>>> remove_digits(888, 3, 1)
88
I hesitated to submit this, because it seems too simple. But I wasn't able to think of a case where it wouldn't work.
if x = y we have to remove all the digits.
Otherwise, you need to find maximum digit in first y + 1 digits. Then remove all the y0 elements before this maximum digit. Then you need to add that maximum to the answer and then repeat that task again, but you need now to remove y - y0 elements now.
Straight forward implementation will work in O(x^2) time in the worst case.
But finding maximum in the given range can be done effectively using Segment Tree data structure. Time complexity will be O(x * log(x)) in the worst case.
P. S. I just realized, that it possible to solve in O(x) also, using the fact, that exists only 10 digits (but the algorithm maybe a little bit complicated). We need to find the minimum in the given range [L, R], but the ranges in this task will "change" from left to the right (L and R always increase). And we just need to store 10 pointers to the digits (1 per digit) to the first position in the number such that position >= L. Then to find the minimum, we need to check only 10 pointers. To update the pointers, we will try to move them right.
So the time complexity will be O(10 * x) = O(x)
Here's an O(x) solution. It builds an index that maps (i, d) to j, the smallest number > i such that the j'th digit of n is d. With this index, one can easily find the largest possible next digit in the solution in O(1) time.
def index(digits):
next = [len(digits)+1] * 10
for i in xrange(len(digits), 0, -1):
next[ord(digits[i-1])-ord('0')] = i-1
yield next[::-1]
def minseq(n, y):
n = str(n)
idx = list(index(n))[::-1]
i, r = 0, []
for ry in xrange(len(n)-y):
i = next(j for j in idx[i] if j <= y+ry) + 1
r.append(n[i - 1])
return ''.join(r)
print minseq(7816295, 3)
print minseq(4213, 2)
Pseudocode:
Number.toDigits().filter (sortedSet (Number.toDigits()). take (y))
Imho you don't need to know x.
For efficiency, Number.toDigits () could be precalculated
digits = Number.toDigits()
digits.filter (sortedSet (digits).take (y))
Depending on language and context, you either output the digits and are done or have to convert the result into a number again.
Working Scala-Code for example:
def toDigits (l: Long) : List [Long] = if (l < 10) l :: Nil else (toDigits (l /10)) :+ (l % 10)
val num = 734529L
val dig = toDigits (num)
dig.filter (_ > ((dig.sorted).take(2).last))
A sorted set is a set which is sorted, which means, every element is only contained once and then the resulting collection is sorted by some criteria, for example numerical ascending. => 234579.
We take two of them (23) and from that subset the last (3) and filter the number by the criteria, that the digits have to be greater than that value (3).
Your question does not explicitly say, that each digit is only contained once in the original number, but since you didn't give a criterion, which one to remove in doubt, I took it as an implicit assumption.
Other languages may of course have other expressions (x.sorted, x.toSortedSet, new SortedSet (num), ...) or lack certain classes, functions, which you would have to build on your own.
You might need to write your own filter method, which takes a pedicate P, and a collection C, and returns a new collection of all elements which satisfy P, P being a Method which takes one T and returns a Boolean. Very useful stuff.
There is known Random(0,1) function, it is a uniformed random function, which means, it will give 0 or 1, with probability 50%. Implement Random(a, b) that only makes calls to Random(0,1)
What I though so far is, put the range a-b in a 0 based array, then I have index 0, 1, 2...b-a.
then call the RANDOM(0,1) b-a times, sum the results as generated idx. and return the element.
However since there is no answer in the book, I don't know if this way is correct or the best. How to prove that the probability of returning each element is exactly same and is 1/(b-a+1) ?
And what is the right/better way to do this?
If your RANDOM(0, 1) returns either 0 or 1, each with probability 0.5 then you can generate bits until you have enough to represent the number (b-a+1) in binary. This gives you a random number in a slightly too large range: you can test and repeat if it fails. Something like this (in Python).
def rand_pow2(bit_count):
"""Return a random number with the given number of bits."""
result = 0
for i in xrange(bit_count):
result = 2 * result + RANDOM(0, 1)
return result
def random_range(a, b):
"""Return a random integer in the closed interval [a, b]."""
bit_count = math.ceil(math.log2(b - a + 1))
while True:
r = rand_pow2(bit_count)
if a + r <= b:
return a + r
When you sum random numbers, the result is not longer evenly distributed - it looks like a Gaussian function. Look up "law of large numbers" or read any probability book / article. Just like flipping coins 100 times is highly highly unlikely to give 100 heads. It's likely to give close to 50 heads and 50 tails.
Your inclination to put the range from 0 to a-b first is correct. However, you cannot do it as you stated. This question asks exactly how to do that, and the answer utilizes unique factorization. Write m=a-b in base 2, keeping track of the largest needed exponent, say e. Then, find the biggest multiple of m that is smaller than 2^e, call it k. Finally, generate e numbers with RANDOM(0,1), take them as the base 2 expansion of some number x, if x < k*m, return x, otherwise try again. The program looks something like this (simple case when m<2^2):
int RANDOM(0,m) {
// find largest power of n needed to write m in base 2
int e=0;
while (m > 2^e) {
++e;
}
// find largest multiple of m less than 2^e
int k=1;
while (k*m < 2^2) {
++k
}
--k; // we went one too far
while (1) {
// generate a random number in base 2
int x = 0;
for (int i=0; i<e; ++i) {
x = x*2 + RANDOM(0,1);
}
// if x isn't too large, return it x modulo m
if (x < m*k)
return (x % m);
}
}
Now you can simply add a to the result to get uniformly distributed numbers between a and b.
Divide and conquer could help us in generating a random number in range [a,b] using random(0,1). The idea is
if a is equal to b, then random number is a
Find mid of the range [a,b]
Generate random(0,1)
If above is 0, return a random number in range [a,mid] using recursion
else return a random number in range [mid+1, b] using recursion
The working 'C' code is as follows.
int random(int a, int b)
{
if(a == b)
return a;
int c = RANDOM(0,1); // Returns 0 or 1 with probability 0.5
int mid = a + (b-a)/2;
if(c == 0)
return random(a, mid);
else
return random(mid + 1, b);
}
If you have a RNG that returns {0, 1} with equal probability, you can easily create a RNG that returns numbers {0, 2^n} with equal probability.
To do this you just use your original RNG n times and get a binary number like 0010110111. Each of the numbers are (from 0 to 2^n) are equally likely.
Now it is easy to get a RNG from a to b, where b - a = 2^n. You just create a previous RNG and add a to it.
Now the last question is what should you do if b-a is not 2^n?
Good thing that you have to do almost nothing. Relying on rejection sampling technique. It tells you that if you have a big set and have a RNG over that set and need to select an element from a subset of this set, you can just keep selecting an element from a bigger set and discarding them till they exist in your subset.
So all you do, is find b-a and find the first n such that b-a <= 2^n. Then using rejection sampling till you picked an element smaller b-a. Than you just add a.
Selecting without any weights (equal probabilities) is beautifully described here.
I was wondering if there is a way to convert this approach to a weighted one.
I am also interested in other approaches as well.
Update: Sampling without replacement
If the sampling is with replacement, you can use this algorithm (implemented here in Python):
import random
items = [(10, "low"),
(100, "mid"),
(890, "large")]
def weighted_sample(items, n):
total = float(sum(w for w, v in items))
i = 0
w, v = items[0]
while n:
x = total * (1 - random.random() ** (1.0 / n))
total -= x
while x > w:
x -= w
i += 1
w, v = items[i]
w -= x
yield v
n -= 1
This is O(n + m) where m is the number of items.
Why does this work? It is based on the following algorithm:
def n_random_numbers_decreasing(v, n):
"""Like reversed(sorted(v * random() for i in range(n))),
but faster because we avoid sorting."""
while n:
v *= random.random() ** (1.0 / n)
yield v
n -= 1
The function weighted_sample is just this algorithm fused with a walk of the items list to pick out the items selected by those random numbers.
This in turn works because the probability that n random numbers 0..v will all happen to be less than z is P = (z/v)n. Solve for z, and you get z = vP1/n. Substituting a random number for P picks the largest number with the correct distribution; and we can just repeat the process to select all the other numbers.
If the sampling is without replacement, you can put all the items into a binary heap, where each node caches the total of the weights of all items in that subheap. Building the heap is O(m). Selecting a random item from the heap, respecting the weights, is O(log m). Removing that item and updating the cached totals is also O(log m). So you can pick n items in O(m + n log m) time.
(Note: "weight" here means that every time an element is selected, the remaining possibilities are chosen with probability proportional to their weights. It does not mean that elements appear in the output with a likelihood proportional to their weights.)
Here's an implementation of that, plentifully commented:
import random
class Node:
# Each node in the heap has a weight, value, and total weight.
# The total weight, self.tw, is self.w plus the weight of any children.
__slots__ = ['w', 'v', 'tw']
def __init__(self, w, v, tw):
self.w, self.v, self.tw = w, v, tw
def rws_heap(items):
# h is the heap. It's like a binary tree that lives in an array.
# It has a Node for each pair in `items`. h[1] is the root. Each
# other Node h[i] has a parent at h[i>>1]. Each node has up to 2
# children, h[i<<1] and h[(i<<1)+1]. To get this nice simple
# arithmetic, we have to leave h[0] vacant.
h = [None] # leave h[0] vacant
for w, v in items:
h.append(Node(w, v, w))
for i in range(len(h) - 1, 1, -1): # total up the tws
h[i>>1].tw += h[i].tw # add h[i]'s total to its parent
return h
def rws_heap_pop(h):
gas = h[1].tw * random.random() # start with a random amount of gas
i = 1 # start driving at the root
while gas >= h[i].w: # while we have enough gas to get past node i:
gas -= h[i].w # drive past node i
i <<= 1 # move to first child
if gas >= h[i].tw: # if we have enough gas:
gas -= h[i].tw # drive past first child and descendants
i += 1 # move to second child
w = h[i].w # out of gas! h[i] is the selected node.
v = h[i].v
h[i].w = 0 # make sure this node isn't chosen again
while i: # fix up total weights
h[i].tw -= w
i >>= 1
return v
def random_weighted_sample_no_replacement(items, n):
heap = rws_heap(items) # just make a heap...
for i in range(n):
yield rws_heap_pop(heap) # and pop n items off it.
If the sampling is with replacement, use the roulette-wheel selection technique (often used in genetic algorithms):
sort the weights
compute the cumulative weights
pick a random number in [0,1]*totalWeight
find the interval in which this number falls into
select the elements with the corresponding interval
repeat k times
If the sampling is without replacement, you can adapt the above technique by removing the selected element from the list after each iteration, then re-normalizing the weights so that their sum is 1 (valid probability distribution function)
I know this is a very old question, but I think there's a neat trick to do this in O(n) time if you apply a little math!
The exponential distribution has two very useful properties.
Given n samples from different exponential distributions with different rate parameters, the probability that a given sample is the minimum is equal to its rate parameter divided by the sum of all rate parameters.
It is "memoryless". So if you already know the minimum, then the probability that any of the remaining elements is the 2nd-to-min is the same as the probability that if the true min were removed (and never generated), that element would have been the new min. This seems obvious, but I think because of some conditional probability issues, it might not be true of other distributions.
Using fact 1, we know that choosing a single element can be done by generating these exponential distribution samples with rate parameter equal to the weight, and then choosing the one with minimum value.
Using fact 2, we know that we don't have to re-generate the exponential samples. Instead, just generate one for each element, and take the k elements with lowest samples.
Finding the lowest k can be done in O(n). Use the Quickselect algorithm to find the k-th element, then simply take another pass through all elements and output all lower than the k-th.
A useful note: if you don't have immediate access to a library to generate exponential distribution samples, it can be easily done by: -ln(rand())/weight
I've done this in Ruby
https://github.com/fl00r/pickup
require 'pickup'
pond = {
"selmon" => 1,
"carp" => 4,
"crucian" => 3,
"herring" => 6,
"sturgeon" => 8,
"gudgeon" => 10,
"minnow" => 20
}
pickup = Pickup.new(pond, uniq: true)
pickup.pick(3)
#=> [ "gudgeon", "herring", "minnow" ]
pickup.pick
#=> "herring"
pickup.pick
#=> "gudgeon"
pickup.pick
#=> "sturgeon"
If you want to generate large arrays of random integers with replacement, you can use piecewise linear interpolation. For example, using NumPy/SciPy:
import numpy
import scipy.interpolate
def weighted_randint(weights, size=None):
"""Given an n-element vector of weights, randomly sample
integers up to n with probabilities proportional to weights"""
n = weights.size
# normalize so that the weights sum to unity
weights = weights / numpy.linalg.norm(weights, 1)
# cumulative sum of weights
cumulative_weights = weights.cumsum()
# piecewise-linear interpolating function whose domain is
# the unit interval and whose range is the integers up to n
f = scipy.interpolate.interp1d(
numpy.hstack((0.0, weights)),
numpy.arange(n + 1), kind='linear')
return f(numpy.random.random(size=size)).astype(int)
This is not effective if you want to sample without replacement.
Here's a Go implementation from geodns:
package foo
import (
"log"
"math/rand"
)
type server struct {
Weight int
data interface{}
}
func foo(servers []server) {
// servers list is already sorted by the Weight attribute
// number of items to pick
max := 4
result := make([]server, max)
sum := 0
for _, r := range servers {
sum += r.Weight
}
for si := 0; si < max; si++ {
n := rand.Intn(sum + 1)
s := 0
for i := range servers {
s += int(servers[i].Weight)
if s >= n {
log.Println("Picked record", i, servers[i])
sum -= servers[i].Weight
result[si] = servers[i]
// remove the server from the list
servers = append(servers[:i], servers[i+1:]...)
break
}
}
}
return result
}
If you want to pick x elements from a weighted set without replacement such that elements are chosen with a probability proportional to their weights:
import random
def weighted_choose_subset(weighted_set, count):
"""Return a random sample of count elements from a weighted set.
weighted_set should be a sequence of tuples of the form
(item, weight), for example: [('a', 1), ('b', 2), ('c', 3)]
Each element from weighted_set shows up at most once in the
result, and the relative likelihood of two particular elements
showing up is equal to the ratio of their weights.
This works as follows:
1.) Line up the items along the number line from [0, the sum
of all weights) such that each item occupies a segment of
length equal to its weight.
2.) Randomly pick a number "start" in the range [0, total
weight / count).
3.) Find all the points "start + n/count" (for all integers n
such that the point is within our segments) and yield the set
containing the items marked by those points.
Note that this implementation may not return each possible
subset. For example, with the input ([('a': 1), ('b': 1),
('c': 1), ('d': 1)], 2), it may only produce the sets ['a',
'c'] and ['b', 'd'], but it will do so such that the weights
are respected.
This implementation only works for nonnegative integral
weights. The highest weight in the input set must be less
than the total weight divided by the count; otherwise it would
be impossible to respect the weights while never returning
that element more than once per invocation.
"""
if count == 0:
return []
total_weight = 0
max_weight = 0
borders = []
for item, weight in weighted_set:
if weight < 0:
raise RuntimeError("All weights must be positive integers")
# Scale up weights so dividing total_weight / count doesn't truncate:
weight *= count
total_weight += weight
borders.append(total_weight)
max_weight = max(max_weight, weight)
step = int(total_weight / count)
if max_weight > step:
raise RuntimeError(
"Each weight must be less than total weight / count")
next_stop = random.randint(0, step - 1)
results = []
current = 0
for i in range(count):
while borders[current] <= next_stop:
current += 1
results.append(weighted_set[current][0])
next_stop += step
return results
In the question you linked to, Kyle's solution would work with a trivial generalization.
Scan the list and sum the total weights. Then the probability to choose an element should be:
1 - (1 - (#needed/(weight left)))/(weight at n). After visiting a node, subtract it's weight from the total. Also, if you need n and have n left, you have to stop explicitly.
You can check that with everything having weight 1, this simplifies to kyle's solution.
Edited: (had to rethink what twice as likely meant)
This one does exactly that with O(n) and no excess memory usage. I believe this is a clever and efficient solution easy to port to any language. The first two lines are just to populate sample data in Drupal.
function getNrandomGuysWithWeight($numitems){
$q = db_query('SELECT id, weight FROM theTableWithTheData');
$q = $q->fetchAll();
$accum = 0;
foreach($q as $r){
$accum += $r->weight;
$r->weight = $accum;
}
$out = array();
while(count($out) < $numitems && count($q)){
$n = rand(0,$accum);
$lessaccum = NULL;
$prevaccum = 0;
$idxrm = 0;
foreach($q as $i=>$r){
if(($lessaccum == NULL) && ($n <= $r->weight)){
$out[] = $r->id;
$lessaccum = $r->weight- $prevaccum;
$accum -= $lessaccum;
$idxrm = $i;
}else if($lessaccum){
$r->weight -= $lessaccum;
}
$prevaccum = $r->weight;
}
unset($q[$idxrm]);
}
return $out;
}
I putting here a simple solution for picking 1 item, you can easily expand it for k items (Java style):
double random = Math.random();
double sum = 0;
for (int i = 0; i < items.length; i++) {
val = items[i];
sum += val.getValue();
if (sum > random) {
selected = val;
break;
}
}
I have implemented an algorithm similar to Jason Orendorff's idea in Rust here. My version additionally supports bulk operations: insert and remove (when you want to remove a bunch of items given by their ids, not through the weighted selection path) from the data structure in O(m + log n) time where m is the number of items to remove and n the number of items in stored.
Sampling wihout replacement with recursion - elegant and very short solution in c#
//how many ways we can choose 4 out of 60 students, so that every time we choose different 4
class Program
{
static void Main(string[] args)
{
int group = 60;
int studentsToChoose = 4;
Console.WriteLine(FindNumberOfStudents(studentsToChoose, group));
}
private static int FindNumberOfStudents(int studentsToChoose, int group)
{
if (studentsToChoose == group || studentsToChoose == 0)
return 1;
return FindNumberOfStudents(studentsToChoose, group - 1) + FindNumberOfStudents(studentsToChoose - 1, group - 1);
}
}
I just spent a few hours trying to get behind the algorithms underlying sampling without replacement out there and this topic is more complex than I initially thought. That's exciting! For the benefit of a future readers (have a good day!) I document my insights here including a ready to use function which respects the given inclusion probabilities further below. A nice and quick mathematical overview of the various methods can be found here: Tillé: Algorithms of sampling with equal or unequal probabilities. For example Jason's method can be found on page 46. The caveat with his method is that the weights are not proportional to the inclusion probabilities as also noted in the document. Actually, the i-th inclusion probabilities can be recursively computed as follows:
def inclusion_probability(i, weights, k):
"""
Computes the inclusion probability of the i-th element
in a randomly sampled k-tuple using Jason's algorithm
(see https://stackoverflow.com/a/2149533/7729124)
"""
if k <= 0: return 0
cum_p = 0
for j, weight in enumerate(weights):
# compute the probability of j being selected considering the weights
p = weight / sum(weights)
if i == j:
# if this is the target element, we don't have to go deeper,
# since we know that i is included
cum_p += p
else:
# if this is not the target element, than we compute the conditional
# inclusion probability of i under the constraint that j is included
cond_i = i if i < j else i-1
cond_weights = weights[:j] + weights[j+1:]
cond_p = inclusion_probability(cond_i, cond_weights, k-1)
cum_p += p * cond_p
return cum_p
And we can check the validity of the function above by comparing
In : for i in range(3): print(i, inclusion_probability(i, [1,2,3], 2))
0 0.41666666666666663
1 0.7333333333333333
2 0.85
to
In : import collections, itertools
In : sample_tester = lambda f: collections.Counter(itertools.chain(*(f() for _ in range(10000))))
In : sample_tester(lambda: random_weighted_sample_no_replacement([(1,'a'),(2,'b'),(3,'c')],2))
Out: Counter({'a': 4198, 'b': 7268, 'c': 8534})
One way - also suggested in the document above - to specify the inclusion probabilities is to compute the weights from them. The whole complexity of the question at hand stems from the fact that one cannot do that directly since one basically has to invert the recursion formula, symbolically I claim this is impossible. Numerically it can be done using all kind of methods, e.g. Newton's method. However the complexity of inverting the Jacobian using plain Python becomes unbearable quickly, I really recommend looking into numpy.random.choice in this case.
Luckily there is method using plain Python which might or might not be sufficiently performant for your purposes, it works great if there aren't that many different weights. You can find the algorithm on page 75&76. It works by splitting up the sampling process into parts with the same inclusion probabilities, i.e. we can use random.sample again! I am not going to explain the principle here since the basics are nicely presented on page 69. Here is the code with hopefully a sufficient amount of comments:
def sample_no_replacement_exact(items, k, best_effort=False, random_=None, ε=1e-9):
"""
Returns a random sample of k elements from items, where items is a list of
tuples (weight, element). The inclusion probability of an element in the
final sample is given by
k * weight / sum(weights).
Note that the function raises if a inclusion probability cannot be
satisfied, e.g the following call is obviously illegal:
sample_no_replacement_exact([(1,'a'),(2,'b')],2)
Since selecting two elements means selecting both all the time,
'b' cannot be selected twice as often as 'a'. In general it can be hard to
spot if the weights are illegal and the function does *not* always raise
an exception in that case. To remedy the situation you can pass
best_effort=True which redistributes the inclusion probability mass
if necessary. Note that the inclusion probabilities will change
if deemed necessary.
The algorithm is based on the splitting procedure on page 75/76 in:
http://www.eustat.eus/productosServicios/52.1_Unequal_prob_sampling.pdf
Additional information can be found here:
https://stackoverflow.com/questions/2140787/
:param items: list of tuples of type weight,element
:param k: length of resulting sample
:param best_effort: fix inclusion probabilities if necessary,
(optional, defaults to False)
:param random_: random module to use (optional, defaults to the
standard random module)
:param ε: fuzziness parameter when testing for zero in the context
of floating point arithmetic (optional, defaults to 1e-9)
:return: random sample set of size k
:exception: throws ValueError in case of bad parameters,
throws AssertionError in case of algorithmic impossibilities
"""
# random_ defaults to the random submodule
if not random_:
random_ = random
# special case empty return set
if k <= 0:
return set()
if k > len(items):
raise ValueError("resulting tuple length exceeds number of elements (k > n)")
# sort items by weight
items = sorted(items, key=lambda item: item[0])
# extract the weights and elements
weights, elements = list(zip(*items))
# compute the inclusion probabilities (short: π) of the elements
scaling_factor = k / sum(weights)
π = [scaling_factor * weight for weight in weights]
# in case of best_effort: if a inclusion probability exceeds 1,
# try to rebalance the probabilities such that:
# a) no probability exceeds 1,
# b) the probabilities still sum to k, and
# c) the probability masses flow from top to bottom:
# [0.2, 0.3, 1.5] -> [0.2, 0.8, 1]
# (remember that π is sorted)
if best_effort and π[-1] > 1 + ε:
# probability mass we still we have to distribute
debt = 0.
for i in reversed(range(len(π))):
if π[i] > 1.:
# an 'offender', take away excess
debt += π[i] - 1.
π[i] = 1.
else:
# case π[i] < 1, i.e. 'save' element
# maximum we can transfer from debt to π[i] and still not
# exceed 1 is computed by the minimum of:
# a) 1 - π[i], and
# b) debt
max_transfer = min(debt, 1. - π[i])
debt -= max_transfer
π[i] += max_transfer
assert debt < ε, "best effort rebalancing failed (impossible)"
# make sure we are talking about probabilities
if any(not (0 - ε <= π_i <= 1 + ε) for π_i in π):
raise ValueError("inclusion probabilities not satisfiable: {}" \
.format(list(zip(π, elements))))
# special case equal probabilities
# (up to fuzziness parameter, remember that π is sorted)
if π[-1] < π[0] + ε:
return set(random_.sample(elements, k))
# compute the two possible lambda values, see formula 7 on page 75
# (remember that π is sorted)
λ1 = π[0] * len(π) / k
λ2 = (1 - π[-1]) * len(π) / (len(π) - k)
λ = min(λ1, λ2)
# there are two cases now, see also page 69
# CASE 1
# with probability λ we are in the equal probability case
# where all elements have the same inclusion probability
if random_.random() < λ:
return set(random_.sample(elements, k))
# CASE 2:
# with probability 1-λ we are in the case of a new sample without
# replacement problem which is strictly simpler,
# it has the following new probabilities (see page 75, π^{(2)}):
new_π = [
(π_i - λ * k / len(π))
/
(1 - λ)
for π_i in π
]
new_items = list(zip(new_π, elements))
# the first few probabilities might be 0, remove them
# NOTE: we make sure that floating point issues do not arise
# by using the fuzziness parameter
while new_items and new_items[0][0] < ε:
new_items = new_items[1:]
# the last few probabilities might be 1, remove them and mark them as selected
# NOTE: we make sure that floating point issues do not arise
# by using the fuzziness parameter
selected_elements = set()
while new_items and new_items[-1][0] > 1 - ε:
selected_elements.add(new_items[-1][1])
new_items = new_items[:-1]
# the algorithm reduces the length of the sample problem,
# it is guaranteed that:
# if λ = λ1: the first item has probability 0
# if λ = λ2: the last item has probability 1
assert len(new_items) < len(items), "problem was not simplified (impossible)"
# recursive call with the simpler sample problem
# NOTE: we have to make sure that the selected elements are included
return sample_no_replacement_exact(
new_items,
k - len(selected_elements),
best_effort=best_effort,
random_=random_,
ε=ε
) | selected_elements
Example:
In : sample_no_replacement_exact([(1,'a'),(2,'b'),(3,'c')],2)
Out: {'b', 'c'}
In : import collections, itertools
In : sample_tester = lambda f: collections.Counter(itertools.chain(*(f() for _ in range(10000))))
In : sample_tester(lambda: sample_no_replacement_exact([(1,'a'),(2,'b'),(3,'c'),(4,'d')],2))
Out: Counter({'a': 2048, 'b': 4051, 'c': 5979, 'd': 7922})
The weights sum up to 10, hence the inclusion probabilities compute to: a → 20%, b → 40%, c → 60%, d → 80%. (Sum: 200% = k.) It works!
Just one word of caution for the productive use of this function, it can be very hard to spot illegal inputs for the weights. An obvious illegal example is
In: sample_no_replacement_exact([(1,'a'),(2,'b')],2)
ValueError: inclusion probabilities not satisfiable: [(0.6666666666666666, 'a'), (1.3333333333333333, 'b')]
b cannot appear twice as often as a since both have to be always be selected. There are more subtle examples. To avoid an exception in production just use best_effort=True, which rebalances the inclusion probability mass such that we have always a valid distribution. Obviously this might change the inclusion probabilities.
I used a associative map (weight,object). for example:
{
(10,"low"),
(100,"mid"),
(10000,"large")
}
total=10110
peek a random number between 0 and 'total' and iterate over the keys until this number fits in a given range.