My math classes are far behind, and I'm currently struggling to find a decent solution to a problem I'm having: I have a tree, in which nodes are actions, and are "weighted" according to multiple criteria : the cost of said action, the time it will take, the necessary resources, the disturbance, etc...
And I want to to find in this tree the path that minimizes both the cost AND the time for example, or the disturbance AND cost AND time, etc. My problem is that I have no idea on how to do it, except by coming up with a global cost function F(cost, time, resources,...), and apply a regular tree traversal algorithm using the result from F(...) as my only weight.
But then, how do I come up with F ? Something like "F(cost, time, resources) = a * cost + b * time + c * resources" feels very "unprofessional"...
Edit:
I wanted to avoid the word "summing" as I wasn't sure it was really the way to go, but in essence, that's what I'm doing: computing a total cost for each "path" or "branch" that goes from that top node, to one of the leafs, and choosing the "path" or "branch" that minimizes the cost. The problem being that each node has a weight based on the time necessary, on financial cost, on resource usage, etc.
So it seems inevitable to have to come up with a formula, as Stephan says, that will reduce all these parameters to one global cost, per node, which I can then sum across nodes as I travel down the tree, and pick the path that minimizes the total cost.
So I guess my question really is, it there a methodology to choose that function ?
Thanks for your answers and comments, it's starting to be a bit more clear in my head now.
Let's say we have four pairs (x, y) like (1, 4), (1, 5), (2, 3), (3, 3). Now you want to minimize "both x and y". What do you mean? If you minimize first x and then y you end up with (1, 4). If you minimize first y and then x you find (2, 3).
Unless you choose a global cost function F(x, y), like in your observation, I can't see any meaning of "both". (Anyway, once F is chosen, there may still be multiple minimum points.) By the way, in my opinion a linear combination (the positive multipliers a,b,c being understood as "weights") is not "unprofessional" at all, at least if you have no idea of what a more suitable cost function could be.
Edit:
So it seems inevitable to have to come up with a formula, as Stephan says, that will reduce all these parameters to one global cost, per node, which I can then sum across nodes as I travel down the tree, and pick the path that minimizes the total cost.
Caution. Indeed this strategy makes sense only if F is linear. Surely cost, time, resources etc. are additive functions, in the sense that time(node1 -> node2 -> node3) = time(node1) + time(node2) + time(node3), but in general this is not the case for F, unless it is linear. (i.e. F(cost(node1 -> node2)) = F(cost(node1) + cost(node2)) != F(cost(node1)) + F(cost(node2)). )
If you choose a nonlinear global cost function, the right strategy is to compute, for each node, the total cost, total time, total resources from the root to that node, and compute (then minimize) F only for the terminal nodes.
Coming up with F is the most important thing. If I can give you 6 cost and 5 time or 5 time and 6 cost, which do you prefer? Your cost function needs to take that into consideration. There's no algorithm that's going to solve that problem for you, unfortunately. I denied that for a week before I sat down and wrote F for an optimization application I was working on. Worst case, leave parameters for the user to tinker with.
Why wouldn't a normal graph search algorithm like A* work?
For the path-cost function, you could use the running sum of the relevant criteria. The distance to the goal is more difficult...
It could be the distance to the nearest leaf, pre-computed for all or some nodes, although that sounds awfully expensive. Depending on the structure of your tree, you could come up with a cheaper under-estimate - if it's perfectly balanced, for example, it's trivial.
Related
I am working on a Traveling salesman problem and can't figure how to solve it. The problem contains ten workers, ten workplaces where they should be driven to and one car driving them one by one. There is a cost of $1.5 per km. Also, all the nodes (both workers and workplaces) are positioned in a 10*10 matrix and the distance between each block in the matrix is 1 km.
The problem should be solved using AMPL.
I have already calculated the distances between each coordinate in excel and have copy pasted the matrix to the dat.file in AMPL.
This is my mod.file so far (without the constrains):
param D > 0;
param D > 0;
set A = 1..W cross 1..D;
var x{A}; # 1 if the route goes from person p to work d,
# 0 otherwise
param cost;
param distance;
minimize Total_Cost:
sum {(w,d) in A} cost * x[w,d];
OK, so your route looks like: start-worker 1-job 1-worker 2-job 2-worker 3-job-3-...-job 10-end (give or take start & end points, depending on how you formulate the problem.
That being the case, the "worker n-job n" parts of your route are predetermined. You don't need to include "worker n-job n" costs in the route optimisation, because there's no choice about those parts of the route (though you do need to remember them for calculating total cost, of course).
So what you have here is really a basic TSP with 10 "destinations" (each representing a single worker and their assigned job) but with an asymmetric cost matrix (because cost of travel from job i to worker j isn't the same as cost of travel from job j to worker i).
If you already have an implementation for the basic TSP, it should be easy to adapt. If not, then you need to write one and make that small change for an asymmetric cost matrix. I've seen two different approaches to this in AMPL.
2-D decision matrix with subtour elimination
Decision variable x{1..10,1..10} is defined as: x[i,j] = 1 if the route goes from job i to job j, and 0 otherwise. Constraints require that every row and column of this matrix has exactly one 1.
The challenging part with this approach is preventing subtours (i.e. the "route" produced is actually two or more separate cycles instead of one large cycle). It sounds like your current attempt is at this stage.
One solution to the problem of subtours is an iterative approach:
Write an implementation that includes all requirements except for subtour prevention.
Solve with this implementation.
Check the resulting solution for subtours.
If no subtours are found, return the solution and end.
If you do find subtours, add a constraint which prevents that particular subtour. (Identify the arcs involved in the subtour, and set a constraint which implies they can't all be selected.) Then go to #2.
For a small exercise you may be able to do the subtour elimination by hand. For a larger exercise, or if your lecturer doesn't like that approach, you can create a .run that automates it. See Bob Fourer's post of 31/7/2013 in this thread for an example of implementation.
3-D decision matrix with time dimension
Under this approach, you set up a decision variable x{1..10,1..10,1..10} where x[i,j,t] = 1 if the route goes from job i to worker j at time t, and 0 otherwise. Your constraints then require that the route goes to and from each job/worker combination exactly once, that if it goes to worker i at time t then it must go from job i at time t+1 (excepting first/last issues), that it's doing exactly one thing at time t, and that the endpoint at time 10 matches the startpoint at time 1 (assuming you want a circuit).
This prevents subtours, because it forces a route that starts at some point at time 1, returns to that point at time 10, and doesn't visit any other point more than once - meaning that it has to go through all of them exactly once.
I know that my question is general, but I'm new to AI area.
I have an experiment with some parameters (almost 6 parameters). Each one of them is independent one, and I want to find the optimal solution for maximum or minimum the output function. However, if I want to do it in traditional programming technique it will take much time since i will use six nested loops.
I just want to know which AI technique to use for this problem? Genetic Algorithm? Neural Network? Machine learning?
Update
Actually, the problem could have more than one evaluation function.
It will have one function that we should minimize it (Cost)
and another function the we want to maximize it (Capacity)
Maybe another functions can be added.
Example:
Construction a glass window can be done in a million ways. However, we want the strongest window with lowest cost. There are many parameters that affect the pressure capacity of the window such as the strength of the glass, Height and Width, slope of the window.
Obviously, if we go to extreme cases (Largest strength glass, with smallest width and height, and zero slope) the window will be extremely strong. However, the cost for that will be very high.
I want to study the interaction between the parameters in specific range.
Without knowing much about the specific problem it sounds like Genetic Algorithms would be ideal. They've been used a lot for parameter optimisation and have often given good results. Personally, I've used them to narrow parameter ranges for edge detection techniques with about 15 variables and they did a decent job.
Having multiple evaluation functions needn't be a problem if you code this into the Genetic Algorithm's fitness function. I'd look up multi objective optimisation with genetic algorithms.
I'd start here: Multi-Objective optimization using genetic algorithms: A tutorial
First of all if you have multiple competing targets the problem is confused.
You have to find a single value that you want to maximize... for example:
value = strength - k*cost
or
value = strength / (k1 + k2*cost)
In both for a fixed strength the lower cost wins and for a fixed cost the higher strength wins but you have a formula to be able to decide if a given solution is better or worse than another. If you don't do this how can you decide if a solution is better than another that is cheaper but weaker?
In some cases a correctly defined value requires a more complex function... for example for strength the value could increase up to a certain point (i.e. having a result stronger than a prescribed amount is just pointless) or a cost could have a cap (because higher than a certain amount a solution is not interesting because it would place the final price out of the market).
Once you find the criteria if the parameters are independent a very simple approach that in my experience is still decent is:
pick a random solution by choosing n random values, one for each parameter within the allowed boundaries
compute target value for this starting point
pick a random number 1 <= k <= n and for each of k parameters randomly chosen from the n compute a random signed increment and change the parameter by that amount.
compute the new target value from the translated solution
if the new value is better keep the new position, otherwise revert to the original one.
repeat from 3 until you run out of time.
Depending on the target function there are random distributions that work better than others, also may be that for different parameters the optimal choice is different.
Some time ago I wrote a C++ code for solving optimization problems using Genetic Algorithms. Here it is: http://create-technology.blogspot.ro/2015/03/a-genetic-algorithm-for-solving.html
It should be very easy to follow.
I understand the basics of minimax and alpha-beta pruning. In all the literature, they talk about the time complexity for the best case is O(b^(d/2)) where b = branching factor and d = depth of the tree, and the base case is when all the preferred nodes are expanded first.
In my example of the "best case", I have a binary tree of 4 levels, so out of the 16 terminal nodes, I need to expand at most 7 nodes. How does this relate to O(b^(d/2))?
I don't understand how they come to O(b^(d/2)).
O(b^(d/2)) correspond to the best case time complexity of alpha-beta pruning. Explanation:
With an (average or constant) branching factor of b, and a search
depth of d plies, the maximum number of leaf node positions evaluated
(when the move ordering is pessimal) is O(bb...*b) = O(b^d) – the
same as a simple minimax search. If the move ordering for the search
is optimal (meaning the best moves are always searched first), the
number of leaf node positions evaluated is about O(b*1*b*1*...*b) for
odd depth and O(b*1*b*1*...*1) for even depth, or O(b^(d/2)). In the
latter case, where the ply of a search is even, the effective
branching factor is reduced to its square root, or, equivalently, the
search can go twice as deep with the same amount of computation.
The explanation of b*1*b*1*... is that all the first player's moves
must be studied to find the best one, but for each, only the best
second player's move is needed to refute all but the first (and best)
first player move – alpha–beta ensures no other second player moves
need be considered.
Put simply, you "skip" every two level:
O describes the limiting behavior of a function when the argument tends towards a particular value or infinity, so in your case comparing precisely O(b^(d/2)) with small values of b and d doesn't really make sense.
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 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.