Related
I'm trying to find a name for my problem, so I don't have to re-invent wheel when coding an algorithm which solves it...
I have say 2,000 binary (row) vectors and I need to pick 500 from them. In the picked sample I do column sums and I want my sample to be as close as possible to a pre-defined distribution of the column sums. I'll be working with 20 to 60 columns.
A tiny example:
Out of the vectors:
110
010
011
110
100
I need to pick 2 to get column sums 2, 1, 0. The solution (exact in this case) would be
110
100
My ideas so far
one could maybe call this a binary multidimensional knapsack, but I did not find any algos for that
Linear Programming could help, but I'd need some step by step explanation as I got no experience with it
as exact solution is not always feasible, something like simulated annealing brute force could work well
a hacky way using constraint solvers comes to mind - first set the constraints tight and gradually loosen them until some solution is found - given that CSP should be much faster than ILP...?
My concrete, practical (if the approximation guarantee works out for you) suggestion would be to apply the maximum entropy method (in Chapter 7 of Boyd and Vandenberghe's book Convex Optimization; you can probably find several implementations with your favorite search engine) to find the maximum entropy probability distribution on row indexes such that (1) no row index is more likely than 1/500 (2) the expected value of the row vector chosen is 1/500th of the predefined distribution. Given this distribution, choose each row independently with probability 500 times its distribution likelihood, which will give you 500 rows on average. If you need exactly 500, repeat until you get exactly 500 (shouldn't take too many tries due to concentration bounds).
Firstly I will make some assumptions regarding this problem:
Regardless whether the column sum of the selected solution is over or under the target, it weighs the same.
The sum of the first, second, and third column are equally weighted in the solution (i.e. If there's a solution whereas the first column sum is off by 1, and another where the third column sum is off by 1, the solution are equally good).
The closest problem I can think of this problem is the Subset sum problem, which itself can be thought of a special case of Knapsack problem.
However both of these problem are NP-Complete. This means there are no polynomial time algorithm that can solve them, even though it is easy to verify the solution.
If I were you the two most arguably efficient solution of this problem are linear programming and machine learning.
Depending on how many columns you are optimising in this problem, with linear programming you can control how much finely tuned you want the solution, in exchange of time. You should read up on this, because this is fairly simple and efficient.
With Machine learning, you need a lot of data sets (the set of vectors and the set of solutions). You don't even need to specify what you want, a lot of machine learning algorithms can generally deduce what you want them to optimise based on your data set.
Both solution has pros and cons, you should decide which one to use yourself based on the circumstances and problem set.
This definitely can be modeled as (integer!) linear program (many problems can). Once you have it, you can use a program such as lpsolve to solve it.
We model vector i is selected as x_i which can be 0 or 1.
Then for each column c, we have a constraint:
sum of all (x_i * value of i in column c) = target for column c
Taking your example, in lp_solve this could look like:
min: ;
+x1 +x4 +x5 >= 2;
+x1 +x4 +x5 <= 2;
+x1 +x2 +x3 +x4 <= 1;
+x1 +x2 +x3 +x4 >= 1;
+x3 <= 0;
+x3 >= 0;
bin x1, x2, x3, x4, x5;
If you are fine with a heuristic based search approach, here is one.
Go over the list and find the minimum squared sum of the digit wise difference between each bit string and the goal. For example, if we are looking for 2, 1, 0, and we are scoring 0, 1, 0, we would do it in the following way:
Take the digit wise difference:
2, 0, 1
Square the digit wise difference:
4, 0, 1
Sum:
5
As a side note, squaring the difference when scoring is a common method when doing heuristic search. In your case, it makes sense because bit strings that have a 1 in as the first digit are a lot more interesting to us. In your case this simple algorithm would pick first 110, then 100, which would is the best solution.
In any case, there are some optimizations that could be made to this, I will post them here if this kind of approach is what you are looking for, but this is the core of the algorithm.
You have a given target binary vector. You want to select M vectors out of N that have the closest sum to the target. Let's say you use the eucilidean distance to measure if a selection is better than another.
If you want an exact sum, have a look at the k-sum problem which is a generalization of the 3SUM problem. The problem is harder than the subset sum problem, because you want an exact number of elements to add to a target value. There is a solution in O(N^(M/2)). lg N), but that means more than 2000^250 * 7.6 > 10^826 operations in your case (in the favorable case where vectors operations have a cost of 1).
First conclusion: do not try to get an exact result unless your vectors have some characteristics that may reduce the complexity.
Here's a hill climbing approach:
sort the vectors by number of 1's: 111... first, 000... last;
use the polynomial time approximate algorithm for the subset sum;
you have an approximate solution with K elements. Because of the order of elements (the big ones come first), K should be a little as possible:
if K >= M, you take the M first vectors of the solution and that's probably near the best you can do.
if K < M, you can remove the first vector and try to replace it with 2 or more vectors from the rest of the N vectors, using the same technique, until you have M vectors. To sumarize: split the big vectors into smaller ones until you reach the correct number of vectors.
Here's a proof of concept with numbers, in Python:
import random
def distance(x, y):
return abs(x-y)
def show(ls):
if len(ls) < 10:
return str(ls)
else:
return ", ".join(map(str, ls[:5]+("...",)+ls[-5:]))
def find(is_xs, target):
# see https://en.wikipedia.org/wiki/Subset_sum_problem#Pseudo-polynomial_time_dynamic_programming_solution
S = [(0, ())] # we store indices along with values to get the path
for i, x in is_xs:
T = [(x + t, js + (i,)) for t, js in S]
U = sorted(S + T)
y, ks = U[0]
S = [(y, ks)]
for z, ls in U:
if z == target: # use the euclidean distance here if you want an approximation
return ls
if z != y and z < target:
y, ks = z, ls
S.append((z, ls))
ls = S[-1][1] # take the closest element to target
return ls
N = 2000
M = 500
target = 1000
xs = [random.randint(0, 10) for _ in range(N)]
print ("Take {} numbers out of {} to make a sum of {}", M, xs, target)
xs = sorted(xs, reverse = True)
is_xs = list(enumerate(xs))
print ("Sorted numbers: {}".format(show(tuple(is_xs))))
ls = find(is_xs, target)
print("FIRST TRY: {} elements ({}) -> {}".format(len(ls), show(ls), sum(x for i, x in is_xs if i in ls)))
splits = 0
while len(ls) < M:
first_x = xs[ls[0]]
js_ys = [(i, x) for i, x in is_xs if i not in ls and x != first_x]
replace = find(js_ys, first_x)
splits += 1
if len(replace) < 2 or len(replace) + len(ls) - 1 > M or sum(xs[i] for i in replace) != first_x:
print("Give up: can't replace {}.\nAdd the lowest elements.")
ls += tuple([i for i, x in is_xs if i not in ls][len(ls)-M:])
break
print ("Replace {} (={}) by {} (={})".format(ls[:1], first_x, replace, sum(xs[i] for i in replace)))
ls = tuple(sorted(ls[1:] + replace)) # use a heap?
print("{} elements ({}) -> {}".format(len(ls), show(ls), sum(x for i, x in is_xs if i in ls)))
print("AFTER {} splits, {} -> {}".format(splits, ls, sum(x for i, x in is_xs if i in ls)))
The result is obviously not guaranteed to be optimal.
Remarks:
Complexity: find has a polynomial time complexity (see the Wikipedia page) and is called at most M^2 times, hence the complexity remains polynomial. In practice, the process is reasonably fast (split calls have a small target).
Vectors: to ensure that you reach the target with the minimum of elements, you can improve the order of element. Your target is (t_1, ..., t_c): if you sort the t_js from max to min, you get the more importants columns first. You can sort the vectors: by number of 1s and then by the presence of a 1 in the most important columns. E.g. target = 4 8 6 => 1 1 1 > 0 1 1 > 1 1 0 > 1 0 1 > 0 1 0 > 0 0 1 > 1 0 0 > 0 0 0.
find (Vectors) if the current sum exceed the target in all the columns, then you're not connecting to the target (any vector you add to the current sum will bring you farther from the target): don't add the sum to S (z >= target case for numbers).
I propose a simple ad hoc algorithm, which, broadly speaking, is a kind of gradient descent algorithm. It seems to work relatively well for input vectors which have a distribution of 1s “similar” to the target sum vector, and probably also for all “nice” input vectors, as defined in a comment of yours. The solution is not exact, but the approximation seems good.
The distance between the sum vector of the output vectors and the target vector is taken to be Euclidean. To minimize it means minimizing the sum of the square differences off sum vector and target vector (the square root is not needed because it is monotonic). The algorithm does not guarantee to yield the sample that minimizes the distance from the target, but anyway makes a serious attempt at doing so, by always moving in some locally optimal direction.
The algorithm can be split into 3 parts.
First of all the first M candidate output vectors out of the N input vectors (e.g., N=2000, M=500) are put in a list, and the remaining vectors are put in another.
Then "approximately optimal" swaps between vectors in the two lists are done, until either the distance would not decrease any more, or a predefined maximum number of iterations is reached. An approximately optimal swap is one where removing the first vector from the list of output vectors causes a maximal decrease or minimal increase of the distance, and then, after the removal of the first vector, adding the second vector to the same list causes a maximal decrease of the distance. The whole swap is avoided if the net result is not a decrease of the distance.
Then, as a last phase, "optimal" swaps are done, again stopping on no decrease in distance or maximum number of iterations reached. Optimal swaps cause a maximal decrease of the distance, without requiring the removal of the first vector to be optimal in itself. To find an optimal swap all vector pairs have to be checked. This phase is much more expensive, being O(M(N-M)), while the previous "approximate" phase is O(M+(N-M))=O(N). Luckily, when entering this phase, most of the work has already been done by the previous phase.
from typing import List, Tuple
def get_sample(vects: List[Tuple[int]], target: Tuple[int], n_out: int,
max_approx_swaps: int = None, max_optimal_swaps: int = None,
verbose: bool = False) -> List[Tuple[int]]:
"""
Get a sample of the input vectors having a sum close to the target vector.
Closeness is measured in Euclidean metrics. The output is not guaranteed to be
optimal (minimum square distance from target), but a serious attempt is made.
The max_* parameters can be used to avoid too long execution times,
tune them to your needs by setting verbose to True, or leave them None (∞).
:param vects: the list of vectors (tuples) with the same number of "columns"
:param target: the target vector, with the same number of "columns"
:param n_out: the requested sample size
:param max_approx_swaps: the max number of approximately optimal vector swaps,
None means unlimited (default: None)
:param max_optimal_swaps: the max number of optimal vector swaps,
None means unlimited (default: None)
:param verbose: print some info if True (default: False)
:return: the sample of n_out vectors having a sum close to the target vector
"""
def square_distance(v1, v2):
return sum((e1 - e2) ** 2 for e1, e2 in zip(v1, v2))
n_vec = len(vects)
assert n_vec > 0
assert n_out > 0
n_rem = n_vec - n_out
assert n_rem > 0
output = vects[:n_out]
remain = vects[n_out:]
n_col = len(vects[0])
assert n_col == len(target) > 0
sumvect = (0,) * n_col
for outvect in output:
sumvect = tuple(map(int.__add__, sumvect, outvect))
sqdist = square_distance(sumvect, target)
if verbose:
print(f"sqdist = {sqdist:4} after"
f" picking the first {n_out} vectors out of {n_vec}")
if max_approx_swaps is None:
max_approx_swaps = sqdist
n_approx_swaps = 0
while sqdist and n_approx_swaps < max_approx_swaps:
# find the best vect to subtract (the square distance MAY increase)
sqdist_0 = None
index_0 = None
sumvect_0 = None
for index in range(n_out):
tmp_sumvect = tuple(map(int.__sub__, sumvect, output[index]))
tmp_sqdist = square_distance(tmp_sumvect, target)
if sqdist_0 is None or sqdist_0 > tmp_sqdist:
sqdist_0 = tmp_sqdist
index_0 = index
sumvect_0 = tmp_sumvect
# find the best vect to add,
# but only if there is a net decrease of the square distance
sqdist_1 = sqdist
index_1 = None
sumvect_1 = None
for index in range(n_rem):
tmp_sumvect = tuple(map(int.__add__, sumvect_0, remain[index]))
tmp_sqdist = square_distance(tmp_sumvect, target)
if sqdist_1 > tmp_sqdist:
sqdist_1 = tmp_sqdist
index_1 = index
sumvect_1 = tmp_sumvect
if sumvect_1:
tmp = output[index_0]
output[index_0] = remain[index_1]
remain[index_1] = tmp
sqdist = sqdist_1
sumvect = sumvect_1
n_approx_swaps += 1
else:
break
if verbose:
print(f"sqdist = {sqdist:4} after {n_approx_swaps}"
f" approximately optimal swap{'s'[n_approx_swaps == 1:]}")
diffvect = tuple(map(int.__sub__, sumvect, target))
if max_optimal_swaps is None:
max_optimal_swaps = sqdist
n_optimal_swaps = 0
while sqdist and n_optimal_swaps < max_optimal_swaps:
# find the best pair to swap,
# but only if the square distance decreases
best_sqdist = sqdist
best_diffvect = diffvect
best_pair = None
for i0 in range(M):
tmp_diffvect = tuple(map(int.__sub__, diffvect, output[i0]))
for i1 in range(n_rem):
new_diffvect = tuple(map(int.__add__, tmp_diffvect, remain[i1]))
new_sqdist = sum(d * d for d in new_diffvect)
if best_sqdist > new_sqdist:
best_sqdist = new_sqdist
best_diffvect = new_diffvect
best_pair = (i0, i1)
if best_pair:
tmp = output[best_pair[0]]
output[best_pair[0]] = remain[best_pair[1]]
remain[best_pair[1]] = tmp
sqdist = best_sqdist
diffvect = best_diffvect
n_optimal_swaps += 1
else:
break
if verbose:
print(f"sqdist = {sqdist:4} after {n_optimal_swaps}"
f" optimal swap{'s'[n_optimal_swaps == 1:]}")
return output
from random import randrange
C = 30 # number of columns
N = 2000 # total number of vectors
M = 500 # number of output vectors
F = 0.9 # fill factor of the target sum vector
T = int(M * F) # maximum value + 1 that can be appear in the target sum vector
A = 10000 # maximum number of approximately optimal swaps, may be None (∞)
B = 10 # maximum number of optimal swaps, may be None (unlimited)
target = tuple(randrange(T) for _ in range(C))
vects = [tuple(int(randrange(M) < t) for t in target) for _ in range(N)]
sample = get_sample(vects, target, M, A, B, True)
Typical output:
sqdist = 2639 after picking the first 500 vectors out of 2000
sqdist = 9 after 27 approximately optimal swaps
sqdist = 1 after 4 optimal swaps
P.S.: As it stands, this algorithm is not limited to binary input vectors, integer vectors would work too. Intuitively I suspect that the quality of the optimization could suffer, though. I suspect that this algorithm is more appropriate for binary vectors.
P.P.S.: Execution times with your kind of data are probably acceptable with standard CPython, but get better (like a couple of seconds, almost a factor of 10) with PyPy. To handle bigger sets of data, the algorithm would have to be translated to C or some other language, which should not be difficult at all.
Suppose I have a real number. I want to approximate it with something of the form a+sqrt(b) for integers a and b. But I don't know the values of a and b. Of course I would prefer to get a good approximation with small values of a and b. Let's leave it undefined for now what is meant by "good" and "small". Any sensible definitions of those terms will do.
Is there a sane way to find them? Something like the continued fraction algorithm for finding fractional approximations of decimals. For more on the fractions problem, see here.
EDIT: To clarify, it is an arbitrary real number. All I have are a bunch of its digits. So depending on how good of an approximation we want, a and b might or might not exist. Brute force is naturally not a particularly good algorithm. The best I can think of would be to start adding integers to my real, squaring the result, and seeing if I come close to an integer. Pretty much brute force, and not a particularly good algorithm. But if nothing better exists, that would itself be interesting to know.
EDIT: Obviously b has to be zero or positive. But a could be any integer.
No need for continued fractions; just calculate the square-root of all "small" values of b (up to whatever value you feel is still "small" enough), remove everything before the decimal point, and sort/store them all (along with the b that generated it).
Then when you need to approximate a real number, find the radical whose decimal-portion is closet to the real number's decimal-portion. This gives you b - choosing the correct a is then a simple matter of subtraction.
This is actually more of a math problem than a computer problem, but to answer the question I think you are right that you can use continued fractions. What you do is first represent the target number as a continued fraction. For example, if you want to approximate pi (3.14159265) then the CF is:
3: 7, 15, 1, 288, 1, 2, 1, 3, 1, 7, 4 ...
The next step is create a table of CFs for square roots, then you compare the values in the table to the fractional part of the target value (here: 7, 15, 1, 288, 1, 2, 1, 3, 1, 7, 4...). For example, let's say your table had square roots for 1-99 only. Then you would find the closest match would be sqrt(51) which has a CF of 7: 7,14 repeating. The 7,14 is the closest to pi's 7,15. Thus your answer would be:
sqrt(51)-4
As the closest approximation given a b < 100 which is off by 0.00016. If you allow larger b's then you could get a better approximation.
The advantage of using CFs is that it is faster than working in, say, doubles or using floating point. For example, in the above case you only have to compare two integers (7 and 15), and you can also use indexing to make finding the closest entry in the table very fast.
This can be done using mixed integer quadratic programming very efficiently (though there are no run-time guarantees as MIQP is NP-complete.)
Define:
d := the real number you wish to approximate
b, a := two integers such that a + sqrt(b) is as "close" to d as possible
r := (d - a)^2 - b, is the residual of the approximation
The goal is to minimize r. Setup your quadratic program as:
x := [ s b t ]
D := | 1 0 0 |
| 0 0 0 |
| 0 0 0 |
c := [0 -1 0]^T
with the constraint that s - t = f (where f is the fractional part of d)
and b,t are integers (s is not)
This is a convex (therefore optimally solvable) mixed integer quadratic program since D is positive semi-definite.
Once s,b,t are computed, simply derive the answer using b=b, s=d-a and t can be ignored.
Your problem may be NP-complete, it would be interesting to prove if so.
Some of the previous answers use methods that are of time or space complexity O(n), where n is the largest “small number” that will be accepted. By contrast, the following method is O(sqrt(n)) in time, and O(1) in space.
Suppose that positive real number r = x + y, where x=floor(r) and 0 ≤ y < 1. We want to approximate r by a number of the form a + √b. If x+y ≈ a+√b then x+y-a ≈ √b, so √b ≈ h+y for some integer offset h, and b ≈ (h+y)^2. To make b an integer, we want to minimize the fractional part of (h+y)^2 over all eligible h. There are at most √n eligible values of h. See following python code and sample output.
import math, random
def findb(y, rhi):
bestb = loerror = 1;
for r in range(2,rhi):
v = (r+y)**2
u = round(v)
err = abs(v-u)
if round(math.sqrt(u))**2 == u: continue
if err < loerror:
bestb, loerror = u, err
return bestb
#random.seed(123456) # set a seed if testing repetitively
f = [math.pi-3] + sorted([random.random() for i in range(24)])
print (' frac sqrt(b) error b')
for frac in f:
b = findb(frac, 12)
r = math.sqrt(b)
t = math.modf(r)[0] # Get fractional part of sqrt(b)
print ('{:9.5f} {:9.5f} {:11.7f} {:5.0f}'.format(frac, r, t-frac, b))
(Note 1: This code is in demo form; the parameters to findb() are y, the fractional part of r, and rhi, the square root of the largest small number. You may wish to change usage of parameters. Note 2: The
if round(math.sqrt(u))**2 == u: continue
line of code prevents findb() from returning perfect-square values of b, except for the value b=1, because no perfect square can improve upon the accuracy offered by b=1.)
Sample output follows. About a dozen lines have been elided in the middle. The first output line shows that this procedure yields b=51 to represent the fractional part of pi, which is the same value reported in some other answers.
frac sqrt(b) error b
0.14159 7.14143 -0.0001642 51
0.11975 4.12311 0.0033593 17
0.12230 4.12311 0.0008085 17
0.22150 9.21954 -0.0019586 85
0.22681 11.22497 -0.0018377 126
0.25946 2.23607 -0.0233893 5
0.30024 5.29150 -0.0087362 28
0.36772 8.36660 -0.0011170 70
0.42452 8.42615 0.0016309 71
...
0.93086 6.92820 -0.0026609 48
0.94677 8.94427 -0.0024960 80
0.96549 11.95826 -0.0072333 143
0.97693 11.95826 -0.0186723 143
With the following code added at the end of the program, the output shown below also appears. This shows closer approximations for the fractional part of pi.
frac, rhi = math.pi-3, 16
print (' frac sqrt(b) error b bMax')
while rhi < 1000:
b = findb(frac, rhi)
r = math.sqrt(b)
t = math.modf(r)[0] # Get fractional part of sqrt(b)
print ('{:11.7f} {:11.7f} {:13.9f} {:7.0f} {:7.0f}'.format(frac, r, t-frac, b,rhi**2))
rhi = 3*rhi/2
frac sqrt(b) error b bMax
0.1415927 7.1414284 -0.000164225 51 256
0.1415927 7.1414284 -0.000164225 51 576
0.1415927 7.1414284 -0.000164225 51 1296
0.1415927 7.1414284 -0.000164225 51 2916
0.1415927 7.1414284 -0.000164225 51 6561
0.1415927 120.1415831 -0.000009511 14434 14641
0.1415927 120.1415831 -0.000009511 14434 32761
0.1415927 233.1415879 -0.000004772 54355 73441
0.1415927 346.1415895 -0.000003127 119814 164836
0.1415927 572.1415909 -0.000001786 327346 370881
0.1415927 911.1415916 -0.000001023 830179 833569
I do not know if there is any kind of standard algorithm for this kind of problem, but it does intrigue me, so here is my attempt at developing an algorithm that finds the needed approximation.
Call the real number in question r. Then, first I assume that a can be negative, in that case we can reduce the problem and now only have to find a b such that the decimal part of sqrt(b) is a good approximation of the decimal part of r. Let us now write r as r = x.y with x being the integer and y the decimal part.
Now:
b = r^2
= (x.y)^2
= (x + .y)^2
= x^2 + 2 * x * .y + .y^2
= 2 * x * .y + .y^2 (mod 1)
We now only have to find an x such that 0 = .y^2 + 2 * x * .y (mod 1) (approximately).
Filling that x into the formulas above we get b and can then calculate a as a = r - b. (All of these calculations have to be carefully rounded of course.)
Now, for the time being I am not sure if there is a way to find this x without brute forcing it. But even then, one can simple use a simple loop to find an x good enough.
I am thinking of something like this(semi pseudo code):
max_diff_low = 0.01 // arbitrary accuracy
max_diff_high = 1 - max_diff_low
y = r % 1
v = y^2
addend = 2 * y
x = 0
while (v < max_diff_high && v > max_diff_low)
x++;
v = (v + addend) % 1
c = (x + y) ^ 2
b = round(c)
a = round(r - c)
Now, I think this algorithm is fairly efficient, while even allowing you to specify the wished accuracy of the approximation. One thing that could be done that would turn it into an O(1) algorithm is calculating all the x and putting them into a lookup table. If one only cares about the first three decimal digits of r(for example), the lookup table would only have 1000 values, which is only 4kb of memory(assuming that 32bit integers are used).
Hope this is helpful at all. If anyone finds anything wrong with the algorithm, please let me know in a comment and I will fix it.
EDIT:
Upon reflection I retract my claim of efficiency. There is in fact as far as I can tell no guarantee that the algorithm as outlined above will ever terminate, and even if it does, it might take a long time to find a very large x that solves the equation adequately.
One could maybe keep track of the best x found so far and relax the accuracy bounds over time to make sure the algorithm terminates quickly, at the possible cost of accuracy.
These problems are of course non-existent, if one simply pre-calculates a lookup table.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
how to get uniformed random between a, b by a known uniformed random function RANDOM(0,1)
In the book of Introduction to algorithms, there is an excise:
Describe an implementation of the procedure Random(a, b) that only makes calls to Random(0,1). What is the expected running time of your procedure, as a function of a and b? The probability of the result of Random(a,b) should be pure uniformly distributed, as Random(0,1)
For the Random function, the results are integers between a and b, inclusively. For e.g., Random(0,1) generates either 0 or 1; Random(a, b) generates a, a+1, a+2, ..., b
My solution is like this:
for i = 1 to b-a
r = a + Random(0,1)
return r
the running time is T=b-a
Is this correct? Are the results of my solutions uniformly distributed?
Thanks
What if my new solution is like this:
r = a
for i = 1 to b - a //including b-a
r += Random(0,1)
return r
If it is not correct, why r += Random(0,1) makes r not uniformly distributed?
Others have explained why your solution doesn't work. Here's the correct solution:
1) Find the smallest number, p, such that 2^p > b-a.
2) Perform the following algorithm:
r=0
for i = 1 to p
r = 2*r + Random(0,1)
3) If r is greater than b-a, go to step 2.
4) Your result is r+a
So let's try Random(1,3).
So b-a is 2.
2^1 = 2, so p will have to be 2 so that 2^p is greater than 2.
So we'll loop two times. Let's try all possible outputs:
00 -> r=0, 0 is not > 2, so we output 0+1 or 1.
01 -> r=1, 1 is not > 2, so we output 1+1 or 2.
10 -> r=2, 2 is not > 2, so we output 2+1 or 3.
11 -> r=3, 3 is > 2, so we repeat.
So 1/4 of the time, we output 1. 1/4 of the time we output 2. 1/4 of the time we output 3. And 1/4 of the time we have to repeat the algorithm a second time. Looks good.
Note that if you have to do this a lot, two optimizations are handy:
1) If you use the same range a lot, have a class that computes p once so you don't have to compute it each time.
2) Many CPUs have fast ways to perform step 1 that aren't exposed in high-level languages. For example, x86 CPUs have the BSR instruction.
No, it's not correct, that method will concentrate around (a+b)/2. It's a binomial distribution.
Are you sure that Random(0,1) produces integers? it would make more sense if it produced floating point values between 0 and 1. Then the solution would be an affine transformation, running time independent of a and b.
An idea I just had, in case it's about integer values: use bisection. At each step, you have a range low-high. If Random(0,1) returns 0, the next range is low-(low+high)/2, else (low+high)/2-high.
Details and complexity left to you, since it's homework.
That should create (approximately) a uniform distribution.
Edit: approximately is the important word there. Uniform if b-a+1 is a power of 2, not too far off if it's close, but not good enough generally. Ah, well it was a spontaneous idea, can't get them all right.
No, your solution isn't correct. This sum'll have binomial distribution.
However, you can generate a pure random sequence of 0, 1 and treat it as a binary number.
repeat
result = a
steps = ceiling(log(b - a))
for i = 0 to steps
result += (2 ^ i) * Random(0, 1)
until result <= b
KennyTM: my bad.
I read the other answers. For fun, here is another way to find the random number:
Allocate an array with b-a elements.
Set all the values to 1.
Iterate through the array. For each nonzero element, flip the coin, as it were. If it is came up 0, set the element to 0.
Whenever, after a complete iteration, you only have 1 element remaining, you have your random number: a+i where i is the index of the nonzero element (assuming we start indexing on 0). All numbers are then equally likely. (You would have to deal with the case where it's a tie, but I leave that as an exercise for you.)
This would have O(infinity) ... :)
On average, though, half the numbers would be eliminated, so it would have an average case running time of log_2 (b-a).
First of all I assume you are actually accumulating the result, not adding 0 or 1 to a on each step.
Using some probabilites you can prove that your solution is not uniformly distibuted. The chance that the resulting value r is (a+b)/2 is greatest. For instance if a is 0 and b is 7, the chance that you get a value 4 is (combination 4 of 7) divided by 2 raised to the power 7. The reason for that is that no matter which 4 out of the 7 values are 1 the result will still be 4.
The running time you estimate is correct.
Your solution's pseudocode should look like:
r=a
for i = 0 to b-a
r+=Random(0,1)
return r
As for uniform distribution, assuming that the random implementation this random number generator is based on is perfectly uniform the odds of getting 0 or 1 are 50%. Therefore getting the number you want is the result of that choice made over and over again.
So for a=1, b=5, there are 5 choices made.
The odds of getting 1 involves 5 decisions, all 0, the odds of that are 0.5^5 = 3.125%
The odds of getting 5 involves 5 decisions, all 1, the odds of that are 0.5^5 = 3.125%
As you can see from this, the distribution is not uniform -- the odds of any number should be 20%.
In the algorithm you created, it is really not equally distributed.
The result "r" will always be either "a" or "a+1". It will never go beyond that.
It should look something like this:
r=0;
for i=0 to b-a
r = a + r + Random(0,1)
return r;
By including "r" into your computation, you are including the "randomness" of all the previous "for" loop runs.
Is there a good algorithm to split a randomly generated number into three buckets, each with constraints as to how much of the total they may contain.
For example, say my randomly generated number is 1,000 and I need to split it into buckets a, b, and c.
These ranges are only an example. See my edit for possible ranges.
Bucket a may only be between 10% - 70% of the number (100 - 700)
Bucket b may only be between 10% - 50% of the number (100 - 500)
Bucket c may only be between 5% - 25% of the number (50 - 250)
a + b + c must equal the randomly generated number
You want the amounts assigned to be completely random so there's just as equal a chance of bucket a hitting its max as bucket c in addition to as equal a chance of all three buckets being around their percentage mean.
EDIT: The following will most likely always be true: low end of a + b + c < 100%, high end of a + b + c > 100%. These percentages are only to indicate acceptable values of a, b, and c. In a case where a is 10% while b and c are their max (50% and 25% respectively) the numbers would have to be reassigned since the total would not equal 100%. This is the exact case I'm trying to avoid by finding a way to assign these numbers in one pass.
I'd like to find a way to pick these number randomly within their range in one pass.
The problem is equivalent to selecting a random point in an N-dimensional object (in your example N=3), the object being defined by the equations (in your example):
0.1 <= x <= 0.7
0.1 <= y <= 0.5
0.05 <= z <= 0.25
x + y + z = 1 (*)
Clearly because of the last equation (*) one of the coordinates is redundant, i.e. picking values for x and y dictates z.
Eliminating (*) and one of the other equations leaves us with an (N-1)-dimensional box, e.g.
0.1 <= x <= 0.7
0.1 <= y <= 0.5
that is cut by the inequality
0.05 <= (1 - x - y) <= 0.25 (**)
that derives from (*) and the equation for z. This is basically a diagonal stripe through the box.
In order for the results to be uniform, I would just repeatedly sample the (N-1)-dimensional box, and accept the first sampled point that fulfills (**). Single-pass solutions might end up having biased distributions.
Update: Yes, you're right, the result is not uniformly distributed.
Let's say your percent values are natural numbers (if this assumption is wrong, you don't have to read further :) In that case I don't have a solution).
Let's define an event e as a tuple of 3 values (percentage of each bucket): e = (pa, pb, pc). Next, create all possible events en. What you have here is a tuple space consisting of a discrete number of events. All of the possible events should have the same possibility to occur.
Let's say we have a function f(n) => en. Then, all we have to do is take a random number n and return en in a single pass.
Now, the problem remains to create such a function f :)
In pseudo code, a very slow method (just for illustration):
function f(n) {
int c = 0
for i in [10..70] {
for j in [10..50] {
for k in [5..25] {
if(i + j + k == 100) {
if(n == c) {
return (i, j, k) // found event!
} else {
c = c + 1
}
}
}
}
}
}
What you have know is a single pass solution, but problem is only moved away. The function f is very slow. But you can do better: I think you can calculate everything a bit faster if you set your ranges correctly and calculate offsets instead of iterating through your ranges.
Is this clear enough?
First of all you probably have to adjust your ranges. 10% in bucket a is not possible, since you can't get condition a+b+c = number to hold.
Concerning your question: (1) Pick a random number for bucket a inside your range, then (2) update the range for bucket b with minimum and maximum percentage (you should only narrow the range). Then (3) pick a random number for bucket b. In the end c should be calculated that your condition holds (4).
Example:
n = 1000
(1) a = 40%
(2) range b [35,50], because 40+35+25 = 100%
(3) b = 45%
(4) c = 100-40-45 = 15%
Or:
n = 1000
(1) a = 70%
(2) range b [10,25], because 70+25+5 = 100%
(3) b = 20%
(4) c = 100-70-20 = 10%
It is to check whether all the events are uniformly distributed. If that should be a problem you might want to randomize the range update in step 2.
UPDATE:
Combinatorics and unranking was eventually what I needed.
The links below helped alot:
http://msdn.microsoft.com/en-us/library/aa289166(v=vs.71).aspx
http://www.codeproject.com/Articles/21335/Combinations-in-C-Part-2
The Problem
Given a list of N symbols say {0,1,2,3,4...}
And NCr combinations of these
eg. NC3 will generate:
0 1 2
0 1 3
0 1 4
...
...
1 2 3
1 2 4
etc...
For the ith combination (i = [1 .. NCr]) I want to determine Whether a symbol (s) is part of it.
Func(N, r, i, s) = True/False or 0/1
eg. Continuing from above
The 1st combination contains 0 1 2 but not 3
F(N,3,1,"0") = TRUE
F(N,3,1,"1") = TRUE
F(N,3,1,"2") = TRUE
F(N,3,1,"3") = FALSE
Current approaches and tibits that might help or be related.
Relation to matrices
For r = 2 eg. 4C2 the combinations are the upper (or lower) half of a 2D matrix
1,2 1,3 1,4
----2,3 2,4
--------3,4
For r = 3 its the corner of a 3D matrix or cube
for r = 4 Its the "corner" of a 4D matrix and so on.
Another relation
Ideally the solution would be of a form something like the answer to this:
Calculate Combination based on position
The nth combination in the list of combinations of length r (with repitition allowed), the ith symbol can be calculated
Using integer division and remainder:
n/r^i % r = (0 for 0th symbol, 1 for 1st symbol....etc)
eg for the 6th comb of 3 symbols the 0th 1st and 2nd symbols are:
i = 0 => 6 / 3^0 % 3 = 0
i = 1 => 6 / 3^1 % 3 = 2
i = 2 => 6 / 3^2 % 3 = 0
The 6th comb would then be 0 2 0
I need something similar but with repition not allowed.
Thank you for following this question this far :]
Kevin.
I believe your problem is that of unranking combinations or subsets.
I will give you an implementation in Mathematica, from the package Combinatorica, but the Google link above is probably a better place to start, unless you are familiar with the semantics.
UnrankKSubset::usage = "UnrankKSubset[m, k, l] gives the mth k-subset of set l, listed in lexicographic order."
UnrankKSubset[m_Integer, 1, s_List] := {s[[m + 1]]}
UnrankKSubset[0, k_Integer, s_List] := Take[s, k]
UnrankKSubset[m_Integer, k_Integer, s_List] :=
Block[{i = 1, n = Length[s], x1, u, $RecursionLimit = Infinity},
u = Binomial[n, k];
While[Binomial[i, k] < u - m, i++];
x1 = n - (i - 1);
Prepend[UnrankKSubset[m - u + Binomial[i, k], k-1, Drop[s, x1]], s[[x1]]]
]
Usage is like:
UnrankKSubset[5, 3, {0, 1, 2, 3, 4}]
{0, 3, 4}
Yielding the 6th (indexing from 0) length-3 combination of set {0, 1, 2, 3, 4}.
There's a very efficient algorithm for this problem, which is also contained in the recently published:Knuth, The Art of Computer Programming, Volume 4A (section 7.2.1.3).
Since you don't care about the order in which the combinations are generated, let's use the lexicographic order of the combinations where each combination is listed in descending order. Thus for r=3, the first 11 combinations of 3 symbols would be: 210, 310, 320, 321, 410, 420, 421, 430, 431, 432, 510. The advantage of this ordering is that the enumeration is independent of n; indeed it is an enumeration over all combinations of 3 symbols from {0, 1, 2, …}.
There is a standard method to directly generate the ith combination given i, so to test whether a symbol s is part of the ith combination, you can simply generate it and check.
Method
How many combinations of r symbols start with a particular symbol s? Well, the remaining r-1 positions must come from the s symbols 0, 1, 2, …, s-1, so it's (s choose r-1), where (s choose r-1) or C(s,r-1) is the binomial coefficient denoting the number of ways of choosing r-1 objects from s objects. As this is true for all s, the first symbol of the ith combination is the smallest s such that
∑k=0s(k choose r-1) ≥ i.
Once you know the first symbol, the problem reduces to finding the (i - ∑k=0s-1(k choose r-1))-th combination of r-1 symbols, where we've subtracted those combinations that start with a symbol less than s.
Code
Python code (you can write C(n,r) more efficiently, but this is fast enough for us):
#!/usr/bin/env python
tC = {}
def C(n,r):
if tC.has_key((n,r)): return tC[(n,r)]
if r>n-r: r=n-r
if r<0: return 0
if r==0: return 1
tC[(n,r)] = C(n-1,r) + C(n-1,r-1)
return tC[(n,r)]
def combination(r, k):
'''Finds the kth combination of r letters.'''
if r==0: return []
sum = 0
s = 0
while True:
if sum + C(s,r-1) < k:
sum += C(s,r-1)
s += 1
else:
return [s] + combination(r-1, k-sum)
def Func(N, r, i, s): return s in combination(r, i)
for i in range(1, 20): print combination(3, i)
print combination(500, 10000000000000000000000000000000000000000000000000000000000000000)
Note how fast this is: it finds the 10000000000000000000000000000000000000000000000000000000000000000th combination of 500 letters (it starts with 542) in less than 0.5 seconds.
I have written a class to handle common functions for working with the binomial coefficient, which is the type of problem that your problem falls under. It performs the following tasks:
Outputs all the K-indexes in a nice format for any N choose K to a file. The K-indexes can be substituted with more descriptive strings or letters. This method makes solving this type of problem quite trivial.
Converts the K-indexes to the proper index of an entry in the sorted binomial coefficient table. This technique is much faster than older published techniques that rely on iteration. It does this by using a mathematical property inherent in Pascal's Triangle. My paper talks about this. I believe I am the first to discover and publish this technique, but I could be wrong.
Converts the index in a sorted binomial coefficient table to the corresponding K-indexes.
Uses Mark Dominus method to calculate the binomial coefficient, which is much less likely to overflow and works with larger numbers.
The class is written in .NET C# and provides a way to manage the objects related to the problem (if any) by using a generic list. The constructor of this class takes a bool value called InitTable that when true will create a generic list to hold the objects to be managed. If this value is false, then it will not create the table. The table does not need to be created in order to perform the 4 above methods. Accessor methods are provided to access the table.
There is an associated test class which shows how to use the class and its methods. It has been extensively tested with 2 cases and there are no known bugs.
To read about this class and download the code, see Tablizing The Binomial Coeffieicent.
This class can easily be applied to your problem. If you have the rank (or index) to the binomial coefficient table, then simply call the class method that returns the K-indexes in an array. Then, loop through that returned array to see if any of the K-index values match the value you have. Pretty straight forward...