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I am looking for a way to generate a de Bruijn sequence iteratively instead of with recursion. My goal is to generate it character by character.
I found some example code in Python for generating de Bruijn sequences and translated it into Rust. I am not yet able to comprehend this technique well enough to create my own method.
Translated into Rust:
fn gen(sequence: &mut Vec<usize>, a: &mut [usize], t: usize, p: usize, k: usize, n: usize) {
if t > n {
if n % p == 0 {
for x in 1..(p + 1) {
sequence.push(a[x])
}
}
} else {
a[t] = a[t - p];
gen(sequence, a, t + 1, p, k, n);
for x in (a[t - p] + 1)..k {
a[t] = x;
gen(sequence, a, t + 1, t, k, n);
}
}
}
fn de_bruijn<T: Clone>(alphabet: &[T], n: usize) -> Vec<T> {
let k = alphabet.len();
let mut a = vec![0; n + 1];
let vecsize = k.checked_pow(n as u32).unwrap();
let mut sequence = Vec::with_capacity(vecsize);
gen(&mut sequence, &mut a, 1, 1, k, n);
sequence.into_iter().map(|x| alphabet[x].clone()).collect()
}
However this is not able to generate iteratively - it goes through a whole mess of recursion and iteration which is impossible to untangle into a single state.
Consider this approach:
Choose the first (lexicographically) representative from every necklace class
Here is Python code for generation of representatives for (binary) necklaces containing d ones (it is possible to repeat for all d values). Sawada article link
Sort representatives in lexicographic order
Make periodic reduction for every representative (if possible): if string is periodic s = p^m like 010101, choose 01
To find the period, it is possible to use string doubling or z-algorithm (I expect it's faster for compiled languages)
Concatenate reductions
Example for n=3,k=2:
Sorted representatives: 000, 001, 011, 111
Reductions: 0, 001, 011, 1
Result: 00010111
The same essential method (with C code) is described in Jörg Arndt's book "Matters Computational", chapter 18
A similar way is mentioned in the wiki
An alternative construction involves concatenating together, in
lexicographic order, all the Lyndon words whose length divides n
You might look for effective way to generate appropriate Lyndon words
I am not familiar with Rust, so I programmed and tested it in Python. Since the poster translated the version in the question from a Python program, I hope it will not be a big issue.
# the following function treats list a as
# k-adic number with n digtis
# and increments this number returning
# the index of the leftmost digit changed
def increment_a7(a, k, n):
digit= n-1
a[digit]+= 1
while a[digit] >= k and digit> 0:
#a[digit]= 0
a[digit]= a[0]+1
a[digit-1]+= 1
digit-= 1
return digit
# the following function adds a to the sequence
# and takes into account, that the beginning of a
# could overlap with the end of sequence
# in that case, it just removes the overlapping digits
# from a before adding the remaining digits to sequence
def append_to_sequence(sequence, a, n):
# here we can assume safely, that a
# does not overlap completely with sequence[-n:]
i= -1
for i in range(n-1, -1, -1):
found= True
# check if the last i digits in sequence
# overlap with the first i digits in a
for j in range(i):
if a[j] != sequence[-i+j]:
# no, they don't overlap
found= False
break
if found:
# yes they overlap, so no need to
# continue the check with a smaller i
break
# now we can just append everything from
# digit i (digit 0 - i-1 are swallowed)
sequence.extend(a[i:])
return n-i
# during the operation we have to keep track of
# the k-adic numbers a, that already occured in
# the sequence. We store them in a set called used
# everytime we add something to the sequence
# we have to update it and add one entry for each
# digit inserted
def update_used(sequence, used, n, num_inserted):
l= len(sequence)
for i in range(num_inserted):
used.add(tuple(sequence[-n-i:l-i]))
# the main work is done in the following function
# it creates and returns the generated sequence
def gen4(k, n):
a= [0]*n
sequence= a[:]
used= set()
# create a fake sequence to add the segments obtained by the cyclic nature
fake= ([k-1] * (n-1))
for i in range(n-1):
fake.append(0)
update_used(fake, used, n, 1)
update_used(sequence, used, n, 1)
valid= True
while valid:
# a is still a valid k-adic number
# this means the generation process
# has not ended
# so construct a new number from the n-1
# last digits of sequence
# followed by a zero
a= sequence[-n+1:]
a.append(0)
while valid and tuple(a) in used:
# the constructed k-adict number a
# was already used, so increment it
# and try again
increment_a(a, k, n)
valid= a[0]<k
if valid:
# great, the number is still valid
# and is not jet part of the sequence
# so add it after removing the overlapping
# digits and update the set with the segments
# we already used
num_inserted= append_to_sequence(sequence, a, n)
update_used(sequence, used, n, num_inserted)
return sequence
I tested the code above by generating some sequences with the original version of gen and this one using the same parameters. For all sets of parameters I tested, the result was the same in both versions.
Please note that this code is less efficient than the original version, especially if the sequence gets long. I guess the costs of the set operations has a non-linear influence on the runtime.
If you like, you can improve it further such as by using a more efficient way to store the used segments. Instead of operating on the k-adic representation (the a-list), you could use a multidimensional array instead.
I am trying to solve a problem where I have pairs like:
A C
B F
A D
D C
F E
E B
A B
B C
E D
F D
and I need to group them in groups of 3 where I must have a triangule of matching from that list. Basically I need a result if its possible or not to group a collection.
So the possible groups are (ACD and BFE), or (ABC and DEF) and this collection is groupable since all letters can be grouped in groups of 3 and no one is left out.
I made a script where I can achieve this for small ammounts of input but for big ammounts it gets too slow.
My logic is:
make nested loop to find first match (looping untill I find a match)
> remove 3 elements from the collection
> run again
and I do this until I am out of letters. Since there can be different combinations I run this multiple times starting on different letters until I find a match.
I can understand that this gives me loops in order at least N^N and can get too slow. Is there a better logic for such problems? can a binary tree be used here?
This problem can be modeled as a graph Clique cover problem. Every letter is a node and every pair is an edge and you want to partition the graph into vertex-disjoint cliques of size 3 (triangles). If you want the partitioning to be of minimum cardinality then you want a minimum clique cover.
Actually this would be a k-clique cover problem, because in the clique cover problem you can have cliques of arbitrary/different sizes.
As Alberto Rivelli already stated, this problem is reducible to the Clique Cover problem, which is NP-hard.
It is also reducible to the problem of finding a clique of particular/maximum size. Maybe there are others, not NP-hard problems to which your particular case could be reduced to, but I didn't think of any.
However, there do exist algorithms which can find the solution in polynomial time, although not always for worst cases. One of them is Bron–Kerbosch algorithm, which is known by far to be the most efficient algorithm for finding the maximum clique and can find a clique in the worst case of O(3^(n/3)). I don't know the size of your inputs, but I hope it will be sufficient for your problem.
Here is the code in Python, ready to go:
#!/usr/bin/python3
# #by DeFazer
# Solution to:
# stackoverflow.com/questions/40193648/algorithm-to-group-items-in-groups-of-3
# Input:
# N P - number of vertices and number of pairs
# P pairs, 1 pair per line
# Output:
# "YES" and groups themselves if grouping is possible, and "NO" otherwise
# Input example:
# 6 10
# 1 3
# 2 6
# 1 4
# 4 3
# 6 5
# 5 2
# 1 2
# 2 3
# 5 4
# 6 4
# Output example:
# YES
# 1-2-3
# 4-5-6
# Output commentary:
# There are 2 possible coverages: 1-2-3*4-5-6 and 2-5-6*1-3-4.
# If required, it can be easily modified to return all possible groupings rather than just one.
# Algorithm:
# 1) List *all* existing triangles (1-2-3, 1-3-4, 2-5-6...)
# 2) Build a graph where vertices represent triangles and edges connect these triangles with no common... vertices. Sorry for ambiguity. :)
# 3) Use [this](en.wikipedia.org/wiki/Bron–Kerbosch_algorithm) algorithm (slightly modified) to find a clique of size N/3.
# The grouping is possible if such clique exists.
N, P = map(int, input().split())
assert (N%3 == 0) and (N>0)
cliquelength = N//3
pairs = {} # {a:{b, d, c}, b:{a, c, f}, c:{a, b}...}
# Get input
# [(0, 1), (1, 3), (3, 2)...]
##pairlist = list(map(lambda ab: tuple(map(lambda a: int(a)-1, ab)), (input().split() for pair in range(P))))
pairlist=[]
for pair in range(P):
a, b = map(int, input().split())
if a>b:
b, a = a, b
a, b = a-1, b-1
pairlist.append((a, b))
pairlist.sort()
for pair in pairlist:
a, b = pair
if a not in pairs:
pairs[a] = set()
pairs[a].add(b)
# Make list of triangles
triangles = []
for a in range(N-2):
for b in pairs.get(a, []):
for c in pairs.get(b, []):
if c in pairs[a]:
triangles.append((a, b, c))
break
def no_mutual_elements(sortedtupleA, sortedtupleB):
# Utility function
# TODO: if too slow, can be improved to O(n) since tuples are sorted. However, there are only 9 comparsions in case of triangles.
return all((a not in sortedtupleB) for a in sortedtupleA)
# Make a graph out of that list
tgraph = [] # if a<b and (b in tgraph[a]), then triangles[a] has no common elements with triangles[b]
T = len(triangles)
for t1 in range(T):
s = set()
for t2 in range(t1+1, T):
if no_mutual_elements(triangles[t1], triangles[t2]):
s.add(t2)
tgraph.append(s)
def connected(a, b):
if a > b:
b, a = a, b
return (b in tgraph[a])
# Finally, the magic algorithm!
CSUB = set()
def extend(CAND:set, NOT:set) -> bool:
# while CAND is not empty and there is no vertex in NOT connected to *all* vertexes in CAND
while CAND and all((any(not connected(n, c) for c in CAND)) for n in NOT):
v = CAND.pop()
CSUB.add(v)
newCAND = {c for c in CAND if connected(c, v)}
newNOT = {n for n in NOT if connected(n, v)}
if (not newCAND) and (not newNOT) and (len(CSUB)==cliquelength): # the last condition is the algorithm modification
return True
elif extend(newCAND, newNOT):
return True
else:
CSUB.remove(v)
NOT.add(v)
if extend(set(range(T)), set()):
print("YES")
# If the clique itself is not needed, it's enough to remove the following 2 lines
for a, b, c in [triangles[c] for c in CSUB]:
print("{}-{}-{}".format(a+1, b+1, c+1))
else:
print("NO")
If this solution is still too slow, perphaps it may be more efficient to solve the Clique Cover problem instead. If that's the case, I can try to find a proper algorithm for it.
Hope that helps!
Well i have implemented the job in JS where I feel most confident. I also tried with 100000 edges which are randomly selected from 26 letters. Provided that they are all unique and not a point such as ["A",A"] it resolves in like 90~500 msecs. The most convoluted part was to obtain the nonidentical groups, those without just the order of the triangles changing. For the given edges data it resolves within 1 msecs.
As a summary the first reduce stage finds the triangles and the second reduce stage groups the disconnected ones.
function getDisconnectedTriangles(edges){
return edges.reduce(function(p,e,i,a){
var ce = a.slice(i+1)
.filter(f => f.some(n => e.includes(n))), // connected edges
re = []; // resulting edges
if (ce.length > 1){
re = ce.reduce(function(r,v,j,b){
var xv = v.find(n => e.indexOf(n) === -1), // find the external vertex
xe = b.slice(j+1) // find the external edges
.filter(f => f.indexOf(xv) !== -1 );
return xe.length ? (r.push([...new Set(e.concat(v,xe[0]))]),r) : r;
},[]);
}
return re.length ? p.concat(re) : p;
},[])
.reduce((s,t,i,a) => t.used ? s
: (s.push(a.map((_,j) => a[(i+j)%a.length])
.reduce((p,c,k) => k-1 ? p.every(t => t.every(n => c.every(v => n !== v))) ? (c.used = true, p.push(c),p) : p
: [p].every(t => t.every(n => c.every(v => n !== v))) ? (c.used = true, [p,c]) : [p])),s)
,[]);
}
var edges = [["A","C"],["B","F"],["A","D"],["D","C"],["F","E"],["E","B"],["A","B"],["B","C"],["E","D"],["F","D"]],
ps = 0,
pe = 0,
result = [];
ps = performance.now();
result = getDisconnectedTriangles(edges);
pe = performance.now();
console.log("Disconnected triangles are calculated in",pe-ps, "msecs and the result is:");
console.log(result);
You may generate random edges in different lengths and play with the code here
Given that I have an initially empty graph, to which will, incrementally, be added edges (one by one), what would be the best way to detect and identify appearing cycles?
Would I have to check for cycles in the entire graph every time a new edge is added? This approach doesn't take advantage of computations already made. Is there an algorithm that I still haven't found?
Thanks.
You can use the Quick Union Find algorithm here to 1st check whether the two ends of the new edge are already connected.
EDIT:
As noted in the comment, this solution works only for an undirected graph. For a directed graph, following link may help https://cs.stackexchange.com/questions/7360/directed-union-find.
implemented the algo that #Abhishek linked in python ..
q=QFUF(10)
In [131]: q.is_cycle([(0,1),(1,3),(2,3),(0,3)])
Out[131]: True
In [132]: q.is_cycle([0,1,4,3,1])
Out[132]: True
only cares about first two elements of the tuple
In [134]: q.is_cycle([(0,1,0.3),(1,3,0.7),(2,3,0.4),(0,3,0.2)])
Out[134]: True
here
import numpy as np
class QFUF:
""" Detect cycles using Union-Find algorithm """
def __init__(self, n):
self.n = n
self.reset()
def reset(self): self.ids = np.arange(self.n)
def find(self, a): return self.ids[a]
def union(self, a, b):
#assign cause can be updated in the for loop
aid = self.ids[a]
bid = self.ids[b]
if aid == bid : return
for x in range(self.n) :
if self.ids[x] == aid : self.ids[x] = bid
#given next ~link/pair check if it forms a cycle
def step(self, a, b):
# print(f'{a} => {b}')
if self.find(a) == self.find(b) : return True
self.union(a, b)
return False
def is_cycle(self, seq):
self.reset()
#if not seq of pairs, make them
if not isinstance(seq[0], tuple) :
seq = zip(seq[:-1], seq[1:])
for tpl in seq :
a,b = tpl[0], tpl[1]
if self.step(a, b) : return True
return False
Currently I'm reading "The Algorithm Design Manual" by Skiena (well, beginning to read)
He asks a problem he calls the "Movie Scheduling Problem":
Problem: Movie Scheduling Problem
Input: A set I of n intervals on the line.
Output: What is the largest subset of mutually non-overlapping intervals which can
be selected from I?
Example: (Each dashed line is a movie, you want to find a set with the highest quantity of movies)
----a---
-----b---- -----c--- ---d---
-----e--- -------f---
--g-- --h--
The algorithm I thought of to solve it was like this:
I could throw out the "worst offender" (intersects with the most other movies) until there are no worst offenders (zero intersections). The only problem I see is that if there is a tie (say two different movies each intersect with 3 other movies) could it matter which one I throw out?
Basically I'm wondering how I go about turning the idea into "math" and how to prove it correct/incorrect.
The algorithm is incorrect. Let's consider the following example:
Counterexample
|----F----| |-----G------|
|-------D-------| |--------E--------|
|-----A------| |------B------| |------C-------|
You can see that there is a solution of size at least 3 because you can pick A, B and C.
Firstly, let's count, for each interval the number of intersections:
A = 2 [F, D]
B = 4 [D, F, E, G]
C = 2 [E, G]
D = 3 [A, B, F]
E = 3 [B, C, G]
F = 3 [A, B, D]
G = 3 [B, C, E]
Now consider a run of your algorithm. In the first step we delete B because it intersects with the most number of invervals and we get:
|----F----| |-----G------|
|-------D-------| |--------E--------|
|-----A------| |------C-------|
It's easy to see that now from {A, D, F} you can choose only one, because each pair intersects. The same case with {G, E, C}, so after deleting B, you can choose at most one from {A, D, F} and at most one from {G, E, C}, to get the total of 2, which is smaller than the size of {A, B, C}.
The conclusion is, that after deleting B which intersects with the most number of invervals, you can't get the maximum number of nonintersecting movies.
Correct solution
The problem is very well known and one solution is to pick the interval which ends first, delete all intervals intersecting with it and continue until there are no intervals to examine. This is an example of a greedy method and you can find or develop a proof that it's correct.
This looks like a dynamic programming problem to me:
Define the following functions:
sched(t) = best schedule starting at time t
next(t) = set of movies that start next after time t
len(m) = length of movie m
next returns a set because there may be more than one movie that starts at the same time.
then sched should be defined as follows:
sched(t) = max { 1 + sched(t + len(m)), sched(t+1) } where m in next(t)
This recursive function selects a movie m from next(t) and compares the largest possible sets that either include or don't include m.
Invoke sched with the time of your first movie and you will get the size of the optimal set. Getting the optimal set itself just requires a little extra logic to remember which movies you select at each invocation.
I think this recursive (as opposed to iterative) algorithm runs in O(n^2) if you use memoization, where n is the number of movies.
It's correct, but I'd have to consult my algorithms textbook to give you an explicit proof, but hopefully this algorithm makes intuitive sense why it is correct.
# go through the database and create a 2-D matrix indexed a..h by a..h. Set each
# element of the matrix to 1 if the row index movie overlaps the column index movie.
mtx = []
for i in range(8):
column = []
for j in range(8):
column.append(0)
mtx.append(column)
# b <> e
mtx[1][4] = 1
mtx[4][1] = 1
# e <> g
mtx[4][6] = 1
mtx[6][4] = 1
# e <> c
mtx[4][2] = 1
mtx[2][4] = 1
# c <> a
mtx[2][0] = 1
mtx[0][2] = 1
# c <> f
mtx[2][5] = 1
mtx[5][2] = 1
# c <> g
mtx[2][6] = 1
mtx[6][2] = 1
# c <> h
mtx[2][7] = 1
mtx[7][2] = 1
# d <> f
mtx[3][5] = 1
mtx[5][3] = 1
# a <> f
mtx[0][5] = 1
mtx[5][0] = 1
# a <> d
mtx[0][3] = 1
mtx[3][0] = 1
# a <> h
mtx[0][7] = 1
mtx[7][0] = 1
# g <> e
mtx[4][7] = 1
mtx[7][4] = 1
# print out contstraints
for line in mtx:
print line
# keep track of which movies are still allowed
allowed = set(range(8))
# loop through in greedy fashion, picking movie that throws out the least
# number of other movies at each step
best = 8
while best > 0:
best_col = None
best_lost = set()
best = 8 # score if move does not overlap with any other
# each step, only try movies still allowed
for col in allowed:
lost = set()
for row in range(8):
# keep track of other movies eliminated by this selection
if mtx[row][col] == 1:
lost.add(row)
# this was the best of all the allowed choices so far
if len(lost) < best:
best_col = col
best_lost = lost
best = len(lost)
# there was a valid selection, process
if best_col > 0:
print 'watch movie: ', str(unichr(best_col+ord('a')))
for row in best_lost:
# now eliminate the other movies you can't now watch
if row in allowed:
print 'throwing out: ', str(unichr(row+ord('a')))
allowed.remove(row)
# also throw out this movie from the allowed list (can't watch twice)
allowed.remove(best_col)
# this is just a greedy algorithm, not guaranteed optimal!
# you could also iterate through all possible combinations of movies
# and simply eliminate all illegal possibilities (brute force search)
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.