Algorithm to distribute elements evenly among a population

In the software I am doing I have three different categories:

    1- Variety A
    2- Variety B
    3- Variety C
    4- Variety D
    5- Variety E
    6- Variety F
    7- Variety G
    1- Type A
    2- Type B
    3- Type C
    4- Type D
    1- Form A
    2- Form B
    3- Form C
    4- Form D
    5- Form E

With this categories I need to produce combinations of Maize+Fertilizer+Management. So I get 140 combinations. I need to distribute the 140 combinations into 200 recipients randomly (some will repeat of course) but balanced (to avoid to many of one type of combination)

I saw this post but I might have a different case.

Is there an algorithm to do this?

  • 1
    $\begingroup$ Welcome to CS.SE! Can you define more precisely what you mean by "balanced"? What condition does the distribution have to satisfy? $\endgroup$
    – D.W.
    Aug 1, 2017 at 5:46
  • $\begingroup$ Choose a random order over the 140 combinations and over the 200 recipients, and then assign recipient $i$ combination $i \bmod 140$. $\endgroup$ Aug 1, 2017 at 6:24
  • $\begingroup$ @D.W. Hello, if you have 140 combinations and 200 recipients inevitably some combinations will be repeated. If the assignation is random then a combination could be assigned too many times. Balanced here is who to ensure that a combination is not assigned to many times $\endgroup$
    – QLands
    Aug 22, 2017 at 17:29
  • $\begingroup$ @YuvalFilmus I'm a programmer so can you explain a bit more your solution? $\endgroup$
    – QLands
    Aug 22, 2017 at 17:31
  • 1
    $\begingroup$ Choose a random permutation P on the 140 combinations and a random permutation Q on the 200 recipients. For each $1 \leq I \leq 200$, assign to recipient $Q[i]$ the combination $P[i \bmod 140]$. $\endgroup$ Aug 22, 2017 at 18:34

1 Answer 1


Traditionally, this kind of problem is described as a fractional factorial design problem. However, your problem also can be elegantly solved via a probabilistic / statistical method that exploits the power of low discrepancy sequences.

For additional context, this method can be applied to very similar questions as yours, such as:

Low Discrepancy Sequences

Low discrepancy sequences, (often described as quasirandom sequences) are numbers that generally appear random, but have an underlying structure to them that makes them more evenly distributed and so have less 'gaps' and 'clumping'.

There are many ways to construct low discrepancy sequences, and they can be generalised to arbitrarily high dimensions, too. For example, below is a picture of the first 100 terms of five different low discrepancy sequences in two dimensional and how they compare to simple random sampling. Note that all of them are substantially more evenly distributed than the conventional uniform random sampling.

enter image description here

Further details can be found at the Wikipedia article: "Halton Sequence".

Equidistributed Allocation

A post I wrote, "The unreasonable effectiveness of quasirandom sequences" , describes how to easily create an efficient extensible $d$-dimensional low discrepancy sequences. (For this particular case, we need a $d=3$ dimensional sequence.)

Define an $n$ x $3$ dimensional array, where each of the $n=200$ rows represents a different trial, and each of the components of each 3-vector defines which type of variety to use for each of the $d=3$ categories.

For $d=3$, there is one magical number involved which is $ \phi_3 = 1.2207440846... $ (see the article if you are interested in what is so special about this number).

Now simply define for all integer $n$, a 3-dimensional vector $\pmb{t}_n$ such that: $$ \mathbf{t}_n = \pmb{s} + n \pmb{\alpha} \; (\textrm{mod} \; 1),  \quad n=1,2,3,... $$ $$ \textrm{where} \quad \pmb{\alpha} =(\frac{1}{\phi_3}, \frac{1}{\phi_3^2},\frac{1}{\phi_3^3}).$$

In nearly all instances, the choice of initial seed $\pmb{s} $ does not change the key characteristics, and so for reasons of obvious simplicity, $\pmb{s} =\pmb{0}$ is the usual choice. (I have included it in this definition, because if you wish to create another sampling distribution, all you need to do is re-run the code but with a different starting seed!)

Specifically for $d=3$; with $\phi_3 = 1.2207440846$ and $\pmb{s}= (0.5,0.5,0.5) $, the first 5 terms of this sequence are:

  1. (0.319173, 0.171044, 0.0497005)
  2. (0.138345, 0.842087, 0.599401)
  3. (0.957518, 0.513131, 0.149101)
  4. (0.77669, 0.184174, 0.698802)
  5. (0.595863, 0.855218, 0.248502) ...

Now consider the first component $x_n$ of each vector $t_n$, to represent the first category ("Maize"). (ie $x_1 = 0.319173$). If

  • $ 0 \leq x < 1/7$, then select Maize, option A
  • $ 1/7 \leq x < 2/7$, then select Maize, option B
  • $ 2/7 \leq x < 3/7$, then select Maize, option C
  • ...
  • $ 6/7 \leq x < 1$, then select Maize, option G

Similarly, consider the second component $y_n$ of each vector $t_n$, to represent the second category ("Fertilizer"). If

  • $ 0 \leq x < 1/4$, then select Fertilizer, option A
  • $ 1/4 \leq x < 2/4$, then select Fertilizer, option B
  • $ 2/4 \leq x < 3/4$, then select Fertilizer, option C
  • $ 3/4 \leq x < 1$, then select Fertilizer, option G


The beautiful result of this low discrepancy method is that, for any $n$, the number of varieties allocated in each category will not differ by more than 1.

Said another way, the number of samples for every variety and every category, will never be more than 1 away to the ideal equal allocation method.


Python code to create the low discrepancy sequences is

# Use Newton-Rhapson-Method
def gamma(d):
    for i in range(20):
        x = x-(pow(x,d+1)-x-1)/((d+1)*pow(x,d)-1)
    return x


g = gamma(d)
alpha = np.zeros(d)                 
s=0.5  # initial seed.

for j in range(d):
    alpha[j] = pow(1/g,j+1) %1
z = np.zeros((n, d))    
for i in range(n):
    z = (s + alpha*(i+1)) %1


Hope that helps!


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