Given a bunch of numbers, I am trying to determine whether there is a "clump" anywhere where numbers are very densely packed.
To make things more precise, I thought I'd ask a more specific problem: given a set of numbers, I would like to determine whether there is a subset of size n which has a standard deviation <= s. If there are many such subsets, I'd like to find the subset with the lowest standard deviation.
So question #1 : does this formal problem definition effectively capture the intuitive concept of a "clump" of densely packed numbers?
EDIT: I don't actually care about determining which numbers belong to this "clump", I'm much more interested in determining where the clump is centred, which is why I think that specifying n in advance is okay. But feel free to correct me!
And question #2 : assuming it does, what is the best way to go about implementing something like this (in particular, I want a solution with lowest time complexity)? So far I think I have a solution that runs in n log n:
First, note that the lowest-standard-deviation-possessing subset of a given size must consist of consecutive numbers. So step 1 is sort the numbers (this is n log n)
Second, take the first n numbers and compute their standard deviation. If our array of numbers is 0-based, then the first n numbers are [0, n-1]. To get standard deviation, compute s1 and s2 as follows:
s1 = sum of numbers
s2 = sum of squares of numbers
Then, wikipedia says that the standard deviation is sqrt(n*s2 - s1^2)/n. Record this value as the highest standard deviation seen so far.
Find the standard deviation of [1, n], [2, n+1], [3, n+2] ... until you hit the the last n numbers. To do each computation takes only constant time if you keep track of s1 and s2 running totals: for example, to get std dev of [1, n], just subtract the 0th element from the s1 and s2 totals and add the nth element, then recalculate standard deviation. This means that the entire standard deviation calculating portion of the algorithm takes linear time.
So total time complexity n log n.
Is my assessment right? Is there a better way to do this? I really need this to run fast on fairly large sets, so the faster the better! Space is less of an issue (I think).
Having been working recently on a similar problem, both the definition of the clumps and the proposed implementation seem reasonable.
Another reasonable definition would be to find the minimum of all the ranges of n numbers. Thus, given that the list of numbers x is sorted, one would just find the minimum of x[n]-x[1], x[n+1]-x[2], etc. This would be slightly quicker than finding the standard deviation because it would avoid the multiplications and square roots. Indeed, you can avoid the square roots even when looking for the lowest standard deviation by finding the minimum variance (the square of the standard deviation), rather than the sd itself.
A caution would be that the location of the biggest clump might be quite sensitive to the choice of n. If there is an a priori reason to select a particular n, that won't be a problem. If not, however, it might require some experimentation to select the value of n that fairly reliably finds the clumps you are looking for, whether you are selecting by range or by standard deviation. Some ideas on this can be found in Chapter 6 of the online book ABC of EDA.
Related
The naive binary search is a very efficient algorithm: you take the midpoint of your high and low points in a sorted array and adjust your high or low point accordingly. Then you recalculate your endpoint and iterate until you find your target value (or you don't, of course.)
Now, quite clearly, if you don't use the midpoint, you introduce some risk to the system. Let's say you shift your search target away from the midpoint and you create two sides - I'll call them a big side and small side. (It doesn't matter whether the shift is toward high or low, because it would be symmetrical.) The risk is that if you miss, your search space is bigger than it would be: you've got to search the big side which is bigger. But the reward is that if you hit your search space is smaller.
It occurs to me that the number of spaces being risked vs rewarded is the same, and (without patterns, which I'm assuming there are none) the likelihood of an element being higher and lower than the midpoint is equal. So the risk is that it falls between the new target and the midpoint.
Now because the number of spaces affects the search space, and the search space is measured logrithmically, it seems to me if I used, let's say 1/4 and 3/4 for our search spaces, I've cut the log of the small space in half, where the large space has only gone up in by about .6 or .7.
So with all this in mind: is there a more efficient way of performing a binary search than just using the midpoint?
Let's agree that the search key is equally likely to be at position in the array—otherwise, we'd want to design an algorithm based on our special knowledge of the location. So all we can choose is where to split the array each time. If we choose a number 0 < x < 1 and split the array there, the chance that it's on the left is x and the chance that it's on the right is 1-x. In the first case we shorten the array by a factor of x and in the second by a factor of 1-x. If we did this many times we'd have a product of many of these factors, and so the 'right' average to use here is the geometric mean. In that sense, the average decrease per step is x with weight x and 1-x with weight 1-x, for a total of x^x * (1-x)^(1-x).
So when is this minimized? If this were the math stackexchange, we'd take derivatives (with the product rule, chain rule, and exponent rule), set them to zero, and solve. But this is stackoverflow, so instead we graph it:
You can see that the further you get from 1/2, the worse you get. For a better understanding I recommend information theory or calculus which have interesting and complementary perspectives on this.
Question after BIG edition :
I need to built a ranking using genetic algorithm, I have data like this :
P(a>b)=0.9
P(b>c)=0.7
P(c>d)=0.8
P(b>d)=0.3
now, lets interpret a,b,c,d as names of football teams, and P(x>y) is probability that x wins with y. We want to build ranking of teams, we lack some observations P(a>d),P(a>c) are missing due to lack of matches between a vs d and a vs c.
Goal is to find ordering of team names, which the best describes current situation in that four team league.
If we have only 4 teams than solution is straightforward, first we compute probabilities for all 4!=24 orderings of four teams, while ignoring missing values we have :
P(abcd)=P(a>b)P(b>c)P(c>d)P(b>d)
P(abdc)=P(a>b)P(b>c)(1-P(c>d))P(b>d)
...
P(dcba)=(1-P(a>b))(1-P(b>c))(1-P(c>d))(1-P(b>d))
and we choose the ranking with highest probability. I don't want to use any other fitness function.
My question :
As numbers of permutations of n elements is n! calculation of probabilities for all
orderings is impossible for large n (my n is about 40). I want to use genetic algorithm for that problem.
Mutation operator is simple switching of places of two (or more) elements of ranking.
But how to make crossover of two orderings ?
Could P(abcd) be interpreted as cost function of path 'abcd' in assymetric TSP problem but cost of travelling from x to y is different than cost of travelling from y to x, P(x>y)=1-P(y<x) ? There are so many crossover operators for TSP problem, but I think I have to design my own crossover operator, because my problem is slightly different from TSP. Do you have any ideas for solution or frame for conceptual analysis ?
The easiest way, on conceptual and implementation level, is to use crossover operator which make exchange of suborderings between two solutions :
CrossOver(ABcD,AcDB) = AcBD
for random subset of elements (in this case 'a,b,d' in capital letters) we copy and paste first subordering - sequence of elements 'a,b,d' to second ordering.
Edition : asymetric TSP could be turned into symmetric TSP, but with forbidden suborderings, which make GA approach unsuitable.
It's definitely an interesting problem, and it seems most of the answers and comments have focused on the semantic aspects of the problem (i.e., the meaning of the fitness function, etc.).
I'll chip in some information about the syntactic elements -- how do you do crossover and/or mutation in ways that make sense. Obviously, as you noted with the parallel to the TSP, you have a permutation problem. So if you want to use a GA, the natural representation of candidate solutions is simply an ordered list of your points, careful to avoid repitition -- that is, a permutation.
TSP is one such permutation problem, and there are a number of crossover operators (e.g., Edge Assembly Crossover) that you can take from TSP algorithms and use directly. However, I think you'll have problems with that approach. Basically, the problem is this: in TSP, the important quality of solutions is adjacency. That is, abcd has the same fitness as cdab, because it's the same tour, just starting and ending at a different city. In your example, absolute position is much more important that this notion of relative position. abcd means in a sense that a is the best point -- it's important that it came first in the list.
The key thing you have to do to get an effective crossover operator is to account for what the properties are in the parents that make them good, and try to extract and combine exactly those properties. Nick Radcliffe called this "respectful recombination" (note that paper is quite old, and the theory is now understood a bit differently, but the principle is sound). Taking a TSP-designed operator and applying it to your problem will end up producing offspring that try to conserve irrelevant information from the parents.
You ideally need an operator that attempts to preserve absolute position in the string. The best one I know of offhand is known as Cycle Crossover (CX). I'm missing a good reference off the top of my head, but I can point you to some code where I implemented it as part of my graduate work. The basic idea of CX is fairly complicated to describe, and much easier to see in action. Take the following two points:
abcdefgh
cfhgedba
Pick a starting point in parent 1 at random. For simplicity, I'll just start at position 0 with the "a".
Now drop straight down into parent 2, and observe the value there (in this case, "c").
Now search for "c" in parent 1. We find it at position 2.
Now drop straight down again, and observe the "h" in parent 2, position 2.
Again, search for this "h" in parent 1, found at position 7.
Drop straight down and observe the "a" in parent 2.
At this point note that if we search for "a" in parent one, we reach a position where we've already been. Continuing past that will just cycle. In fact, we call the sequence of positions we visited (0, 2, 7) a "cycle". Note that we can simply exchange the values at these positions between the parents as a group and both parents will retain the permutation property, because we have the same three values at each position in the cycle for both parents, just in different orders.
Make the swap of the positions included in the cycle.
Note that this is only one cycle. You then repeat this process starting from a new (unvisited) position each time until all positions have been included in a cycle. After the one iteration described in the above steps, you get the following strings (where an "X" denotes a position in the cycle where the values were swapped between the parents.
cbhdefga
afcgedbh
X X X
Just keep finding and swapping cycles until you're done.
The code I linked from my github account is going to be tightly bound to my own metaheuristics framework, but I think it's a reasonably easy task to pull the basic algorithm out from the code and adapt it for your own system.
Note that you can potentially gain quite a lot from doing something more customized to your particular domain. I think something like CX will make a better black box algorithm than something based on a TSP operator, but black boxes are usually a last resort. Other people's suggestions might lead you to a better overall algorithm.
I've worked on a somewhat similar ranking problem and followed a technique similar to what I describe below. Does this work for you:
Assume the unknown value of an object diverges from your estimate via some distribution, say, the normal distribution. Interpret your ranking statements such as a > b, 0.9 as the statement "The value a lies at the 90% percentile of the distribution centered on b".
For every statement:
def realArrival = calculate a's location on a distribution centered on b
def arrivalGap = | realArrival - expectedArrival |
def fitness = Σ arrivalGap
Fitness function is MIN(fitness)
FWIW, my problem was actually a bin-packing problem, where the equivalent of your "rank" statements were user-provided rankings (1, 2, 3, etc.). So not quite TSP, but NP-Hard. OTOH, bin-packing has a pseudo-polynomial solution proportional to accepted error, which is what I eventually used. I'm not quite sure that would work with your probabilistic ranking statements.
What an interesting problem! If I understand it, what you're really asking is:
"Given a weighted, directed graph, with each edge-weight in the graph representing the probability that the arc is drawn in the correct direction, return the complete sequence of nodes with maximum probability of being a topological sort of the graph."
So if your graph has N edges, there are 2^N graphs of varying likelihood, with some orderings appearing in more than one graph.
I don't know if this will help (very brief Google searches did not enlighten me, but maybe you'll have more success with more perseverance) but my thoughts are that looking for "topological sort" in conjunction with any of "probabilistic", "random", "noise," or "error" (because the edge weights can be considered as a reliability factor) might be helpful.
I strongly question your assertion, in your example, that P(a>c) is not needed, though. You know your application space best, but it seems to me that specifying P(a>c) = 0.99 will give a different fitness for f(abc) than specifying P(a>c) = 0.01.
You might want to throw in "Bayesian" as well, since you might be able to start to infer values for (in your example) P(a>c) given your conditions and hypothetical solutions. The problem is, "topological sort" and "bayesian" is going to give you a whole bunch of hits related to markov chains and markov decision problems, which may or may not be helpful.
I have a process, similar to tineye that generates perceptual hashes, these are 32bit ints.
I intend to store these in a sql database (maybe a nosql db) in the future
However, I'm stumped at how I would be able to retrieve records based on the similarity of hashes.
Any Ideas?
A common approach (at least common to me) is to divide your hash bit string in several chunks and query on these chunks for an exact match. This is a "pre-filter" step. You then can perform a bitwise hamming distance computation on the returned results which should be only a smaller subset of your overall dataset. This can be done using data files or SQL tables.
So in simple terms: Say you have a bunch of 32 bits hashes in a DB and that you want to find every hash that are within a 4 bits hamming distance of your "query" hash:
Create a table with four columns: each will contain an 8 bits (as a string or int) slice of the 32 bits hashes, islice 1 to 4.
Slice your query hash the same way in qslice 1 to 4.
Query this table such that any of qslice1=islice1 or qslice2=islice2 or qslice3=islice3 or qslice4=islice4. This gives you every DB hash that are within 3 bits (4 - 1) of the query hash. It may contain more results, and this is why there is a step 4.
For each returned hash, compute the exact hamming distance pair-wise with you query hash (reconstructing the index-side hash from the four slices)
The number of operations in step 4 should be much less than a full pair-wise hamming computation of your whole table.
This approach was first described afaik by Moses Charikar in its "simhash" seminal paper and the corresponding Google patent:
APPROXIMATE NEAREST NEIGHBOR SEARCH IN HAMMING SPACE
[...]
Given bit vectors consisting of d bits each, we choose
N = O(n 1/(1+ ) ) random permutations of the bits. For each
random permutation σ, we maintain a sorted order O σ of
the bit vectors, in lexicographic order of the bits permuted
by σ. Given a query bit vector q, we find the approximate
nearest neighbor by doing the following:
For each permutation σ, we perform a binary search on O σ to locate the two bit vectors closest to q (in the lexicographic order obtained by bits permuted by σ). We now search in each of the sorted orders O σ examining elements above and below the position returned by the binary search in order of the length of the longest prefix that matches q.
Monika Henziger expanded on this in her paper "Finding near-duplicate web pages: a large-scale evaluation of algorithms":
3.3 The Results for Algorithm C
We partitioned the bit string of each page into 12 non-
overlapping 4-byte pieces, creating 20B pieces, and computed the C-similarity of all pages that had at least one
piece in common. This approach is guaranteed to find all
pairs of pages with difference up to 11, i.e., C-similarity 373,
but might miss some for larger differences.
NB: C-similarity is the same as the Hamming distance: The Hamming distance is the number of positions at which the corresponding bits differ while C-similarity is the number of positions at which the corresponding bits agree.
This is also explained in the paper Detecting Near-Duplicates for Web Crawling by Gurmeet Singh Manku, Arvind Jain, and Anish Das Sarma:
THE HAMMING DISTANCE PROBLEM
Definition: Given a collection of f -bit fingerprints and a
query fingerprint F, identify whether an existing fingerprint
differs from F in at most k bits. (In the batch-mode version
of the above problem, we have a set of query fingerprints
instead of a single query fingerprint)
[...]
Intuition: Consider a sorted table of 2 d f -bit truly random fingerprints. Focus on just the most significant d bits
in the table. A listing of these d-bit numbers amounts to
“almost a counter” in the sense that (a) quite a few 2 d bit-
combinations exist, and (b) very few d-bit combinations are
duplicated. On the other hand, the least significant f − d
bits are “almost random”.
Now choose d such that |d − d| is a small integer. Since
the table is sorted, a single probe suffices to identify all fingerprints which match F in d most significant bit-positions.
Since |d − d| is small, the number of such matches is also
expected to be small. For each matching fingerprint, we can
easily figure out if it differs from F in at most k bit-positions
or not (these differences would naturally be restricted to the
f − d least-significant bit-positions).
The procedure described above helps us locate an existing
fingerprint that differs from F in k bit-positions, all of which
are restricted to be among the least significant f − d bits of
F. This takes care of a fair number of cases. To cover all
the cases, it suffices to build a small number of additional
sorted tables, as formally outlined in the next Section.
PS: Most of these fine brains are/were associated with Google at some level or some time for these, FWIW.
To find hamming distance, you can just use bitwise addition and subtraction (& and ~ on the integers) in order to compute these.
SQL isn't made for this sort of processing. The comparisons on large data sets get very messy, and will not have the speed of a query that utilizes the strength of the system. That said, I've done similar things.
This will give you individual differences, which would need to be run on the full data set and ordered, which is messy at best. If you want it to run faster, you will need to use strategies like indexing by "region," or finding natural groupings in your data. There are umbrella clustering strategies, and similar - there is a lot of literature. It will, however, be messy in most traditional Database systems.
David's discussion is correct, but if you don't have a lot of data, check out Hamming distance on binary strings in SQL
I have read that it has something to do with time, also you get from including time.h, so I assumed that much, but how does it work exactly? Also, does it have any tendencies towards odd or even numbers or something like that? And finally is there something with better distribution in the C standard library or the Foundation framework?
Briefly:
You use time.h to get a seed, which is an initial random number. C then does a bunch of operations on this number to get the next random number, then operations on that one to get the next, then... you get the picture.
rand() is able to touch on every possible integer. It will not prefer even or odd numbers regardless of the input seed, happily. Still, it has limits - it repeats itself relatively quickly, and in almost every implementation only gives numbers up to 32767.
C does not have another built-in random number generator. If you need a real tough one, there are many packages available online, but the Mersenne Twister algorithm is probably the most popular pick.
Now, if you are interested on the reasons why the above is true, here are the gory details on how rand() works:
rand() is what's called a "linear congruential generator." This means that it employs an equation of the form:
xn+1 = (*a****xn + ***b*) mod m
where xn is the nth random number, and a and b are some predetermined integers. The arithmetic is performed modulo m, with m usually 232 depending on the machine, so that only the lowest 32 bits are kept in the calculation of xn+1.
In English, then, the idea is this: To get the next random number, multiply the last random number by something, add a number to it, and then take the last few digits.
A few limitations are quickly apparent:
First, you need a starting random number. This is the "seed" of your random number generator, and this is where you've heard of time.h being used. Since we want a really random number, it is common practice to ask the system what time it is (in integer form) and use this as the first "random number." Also, this explains why using the same seed twice will always give exactly the same sequence of random numbers. This sounds bad, but is actually useful, since debugging is a lot easier when you control the inputs to your program
Second, a and b have to be chosen very, very carefully or you'll get some disastrous results. Fortunately, the equation for a linear congruential generator is simple enough that the math has been worked out in some detail. It turns out that choosing an a which satisfies *a***mod8 = 5 together with ***b* = 1 will insure that all m integers are equally likely, independent of choice of seed. You also want a value of a that is really big, so that every time you multiply it by xn you trigger a the modulo and chop off a lot of digits, or else many numbers in a row will just be multiples of each other. As a result, two common values of a (for example) are 1566083941 and 1812433253 according to Knuth. The GNU C library happens to use a=1103515245 and b=12345. A list of values for lots of implementations is available at the wikipedia page for LCGs.
Third, the linear congruential generator will actually repeat itself because of that modulo. This gets to be some pretty heady math, but the result of it all is happily very simple: The sequence will repeat itself after m numbers of have been generated. In most cases, this means that your random number generator will repeat every 232 cycles. That sounds like a lot, but it really isn't for many applications. If you are doing serious numerical work with Monte Carlo simulations, this number is hopelessly inadequate.
A fourth much less obvious problem is that the numbers are actually not really random. They have a funny sort of correlation. If you take three consecutive integers, (x, y, z), from an LCG with some value of a and m, those three points will always fall on the lattice of points generated by all linear combinations of the three points (1, a, a2), (0, m, 0), (0, 0, m). This is known as Marsaglia's Theorem, and if you don't understand it, that's okay. All it means is this: Triplets of random numbers from an LCG will show correlations at some deep, deep level. Usually it's too deep for you or I to notice, but its there. It's possible to even reconstruct the first number in a "random" sequence of three numbers if you are given the second and third! This is not good for cryptography at all.
The good part is that LCGs like rand() are very, very low footprint. It typically requires only 32 bits to retain state, which is really nice. It's also very fast, requiring very few operations. These make it good for noncritical embedded systems, video games, casual applications, stuff like that.
PRNGs are a fascinating topic. Wikipedia is always a good place to go if you are hungry to learn more on the history or the various implementations that are around today.
rand returns numbers generated by a pseudo-random number generator (PRNG). The sequence of numbers it returns is deterministic, based on the value with which the PRNG was initialized (by calling srand).
The numbers should be distributed such that they appear somewhat random, so, for example, odd and even numbers should be returned at roughly the same frequency. The actual implementation of the random number generator is left unspecified, so the actual behavior is specific to the implementation.
The important thing to remember is that rand does not return random numbers; it returns pseudo-random numbers, and the values it returns are determined by the seed value and the number of times rand has been called. This behavior is fine for many use cases, but is not appropriate for others (for example, rand would not be appropriate for use in many cryptographic applications).
How does rand() work?
http://en.wikipedia.org/wiki/Pseudorandom_number_generator
I have read that it has something to
do with time, also you get from
including time.h
rand() has nothing at all to do with the time. However, it's very common to use time() to obtain the "seed" for the PRNG so that you get different "random" numbers each time your program is run.
Also, does it have any tendencies
towards odd or even numbers or
something like that?
Depends on the exact method used. There's one popular implementation of rand() that alternates between odd and even numbers. So avoid writing code like rand() % 2 that depends on the lowest bit being random.
I have read through various papers on the 'Balls and Bins' problem and it seems that if a hash function is working right (ie. it is effectively a random distribution) then the following should/must be true if I hash n values into a hash table with n slots (or bins):
Probability that a bin is empty, for large n is 1/e.
Expected number of empty bins is n/e.
Probability that a bin has k balls is <= 1/ek! (corrected).
Probability that a bin has at least k collisions is <= ((e/k)**k)/e (corrected).
These look easy to check. But the max-load test (the maximum number of collisions with high probability) is usually stated vaguely.
Most texts state that the maximum number of collisions in any bin is O( ln(n) / ln(ln(n)) ).
Some say it is 3*ln(n) / ln(ln(n)). Other papers mix ln and log - usually without defining them, or state that log is log base e and then use ln elsewhere.
Is ln the log to base e or 2 and is this max-load formula right and how big should n be to run a test?
This lecture seems to cover it best, but I am no mathematician.
http://pages.cs.wisc.edu/~shuchi/courses/787-F07/scribe-notes/lecture07.pdf
BTW, with high probability seems to mean 1 - 1/n.
That is a fascinating paper/lecture-- makes me wish I had taken some formal algorithms class.
I'm going to take a stab at some answers here, based on what I've just read from that, and feel free to vote me down. I'd appreciate a correction, though, rather than just a downvote :) I'm also going to use n and N interchangeably here, which is a big no-no in some circles, but since I'm just copy-pasting your formulae, I hope you'll forgive me.
First, the base of the logs. These numbers are given as big-O notation, not as absolute formulae. That means that you're looking for something 'on the order of ln(n) / ln(ln(n))', not with an expectation of an absolute answer, but more that as n gets bigger, the relationship of n to the maximum number of collisions should follow that formula. The details of the actual curve you can graph will vary by implementation (and I don't know enough about the practical implementations to tell you what's a 'good' curve, except that it should follow that big-O relationship). Those two formulae that you posted are actually equivalent in big-O notation. The 3 in the second formula is just a constant, and is related to a particular implementation. A less efficient implementation would have a bigger constant.
With that in mind, I would run empirical tests, because I'm a biologist at heart and I was trained to avoid hard-and-fast proofs as indications of how the world actually works. Start with N as some number, say 100, and find the bin with the largest number of collisions in it. That's your max-load for that run. Now, your examples should be as close as possible to what you expect actual users to use, so maybe you want to randomly pull words from a dictionary or something similar as your input.
Run that test many times, at least 30 or 40. Since you're using random numbers, you'll need to satisfy yourself that the average max-load you're getting is close to the theoretical 'expectation' of your algorithm. Expectation is just the average, but you'll still need to find it, and the tighter your std dev/std err about that average, the more you can say that your empirical average matches the theoretical expectation. One run is not enough, because a second run will (most likely) give a different answer.
Then, increase N, to say, 1000, 10000, etc. Increase it logarithmically, because your formula is logarithmic. As your N increases, your max-load should increase on the order of ln(n) / ln(ln(n)). If it increases at a rate of 3*ln(n) / ln(ln(n)), that means that you're following the theory that they put forth in that lecture.
This kind of empirical test will also show you where your approach breaks down. It may be that your algorithm works well for N < 10 million (or some other number), but above that, it starts to collapse. Why could that be? Maybe you have some limitation to 32 bits in your code without realizing it (ie, using a 'float' instead of a 'double'), or some other implementation detail. These kinds of details let you know where your code will work well in practice, and then as your practical needs change, you can modify your algorithm. Maybe making the algorithm work for very large datasets makes it very inefficient for very small ones, or vice versa, so pinpointing that tradeoff will help you further characterize how you could adapt your algorithm to particular situations. Always a useful skill to have.
EDIT: a proof of why the base of the log function doesn't matter with big-O notation:
log N = log_10 (N) = log_b (N)/log_b (10)= (1/log_b(10)) * log_b(N)
1/log_b(10) is a constant, and in big-O notation, constants are ignored. Base changes are free, which is why you're encountering such variation in the papers.
Here is a rough start to the solution of this problem involving uniform distributions and maximum load.
Instead of bins and balls or urns or boxes or buckets or m and n, people (p) and doors (d) will be used as designations.
There is an exact expected value for each of the doors given a certain number of people. For example, with 5 people and 5 doors, the expected maximum door is exactly 1.2864 {(1429-625) / 625} above the mean (p/d) and the minimum door is exactly -0.9616 {(24-625) / 625} below the mean. The absolute value of the highest door's distance from the mean is a little larger than the smallest door's because all of the people could go through one door, but no less than zero can go through one of the doors. With large numbers of people (p/d > 3000), the difference between the absolute value of the highest door's distance from the mean and the lowest door's becomes negligible.
For an odd number of doors, the center door is essentially zero and is not scalable, but all of the other doors are scalable from certain values representing p=d. These rounded values for d=5 are:
-1.163 -0.495 0* 0.495 1.163
* slowly approaching zero from -0.12
From these values, you can compute the expected number of people for any count of people going through each of the 5 doors, including the maximum door. Except for the middle ordered door, the difference from the mean is scalable by sqrt(p/d).
So, for p=50,000 and d=5:
Expected number of people going through the maximum door, which could be any of the 5 doors, = 1.163 * sqrt(p/d) + p/d.
= 1.163 * sqrt(10,000) + 10,000 = 10,116.3
For p/d < 3,000, the result from this equation must be slightly increased.
With more people, the middle door slowly becomes closer and closer to zero from -0.11968 at p=100 and d=5. It can always be rounded up to zero and like the other 4 doors has quite a variance.
The values for 6 doors are:
-1.272 -0.643 -0.202 0.202 0.643 1.272
For 1000 doors, the approximate values are:
-3.25, -2.95, -2.79 … 2.79, 2.95, 3.25
For any d and p, there is an exact expected value for each of the ordered doors. Hopefully, a good approximation (with a relative error < 1%) exists. Some professor or mathematician somewhere must know.
For testing uniform distribution, you will need a number of averaged ordered sessions (750-1000 works well) rather than a greater number of people. No matter what, the variances between valid sessions are great. That's the nature of randomness. Collisions are unavoidable. *
The expected values for 5 and 6 doors were obtained by sheer brute force computation using 640 bit integers and averaging the convergence of the absolute values of corresponding opposite doors.
For d=5 and p=170:
-6.63901 -2.95905 -0.119342 2.81054 6.90686
(27.36099 31.04095 33.880658 36.81054 40.90686)
For d=6 and p=108:
-5.19024 -2.7711 -0.973979 0.734434 2.66716 5.53372
(12.80976 15.2289 17.026021 18.734434 20.66716 23.53372)
I hope that you may evenly distribute your data.
It's almost guaranteed that all of George Foreman's sons or some similar situation will fight against your hash function. And proper contingent planning is the work of all good programmers.
After some more research and trial-and-error I think I can provide something part way to to an answer.
To start off, ln and log seem to refer to log base-e if you look into the maths behind the theory. But as mmr indicated, for the O(...) estimates, it doesn't matter.
max-load can be defined for any probability you like. The typical formula used is
1-1/n**c
Most papers on the topic use
1-1/n
An example might be easiest.
Say you have a hash table of 1000 slots and you want to hash 1000 things. Say you also want to know the max-load with a probability of 1-1/1000 or 0.999.
The max-load is the maximum number of hash values that end up being the same - ie. collisions (assuming that your hash function is good).
Using the formula for the probability of getting exactly k identical hash values
Pr[ exactly k ] = ((e/k)**k)/e
then by accumulating the probability of exactly 0..k items until the total equals or exceeds 0.999 tells you that k is the max-load.
eg.
Pr[0] = 0.37
Pr[1] = 0.37
Pr[2] = 0.18
Pr[3] = 0.061
Pr[4] = 0.015
Pr[5] = 0.003 // here, the cumulative total is 0.999
Pr[6] = 0.0005
Pr[7] = 0.00007
So, in this case, the max-load is 5.
So if my hash function is working well on my set of data then I should expect the maxmium number of identical hash values (or collisions) to be 5.
If it isn't then this could be due to the following reasons:
Your data has small values (like short strings) that hash to the same value. Any hash of a single ASCII character will pick 1 of 128 hash values (there are ways around this. For example you could use multiple hash functions, but slows down hashing and I don't know much about this).
Your hash function doesn't work well with your data - try it with random data.
Your hash function doesn't work well.
The other tests I mentioned in my question also are helpful to see that your hash function is running as expected.
Incidentally, my hash function worked nicely - except on short (1..4 character) strings.
I also implemented a simple split-table version which places the hash value into the least used slot from a choice of 2 locations. This more than halves the number of collisions and means that adding and searching the hash table is a little slower.
I hope this helps.