I have a following problem Im trying to solve.
I have hundreds of particles with their corresponding chemical composition (elements with their weight percentages).
As an example, here are some made-up simplified particles:
Particle 1 - S (32%), K (25%), C (43%)
Particle 2 - S (33%), K (12%), C (15%), O (40%)
Particle 3 - Ti (18%), S (72%)
Particle 4 - Ti (10%), S (79%), K (12%)
In reality there are hundreds of them, some of them quite different to one another, some of them quite similar. As you can see, some particles do not have certain elements (i.e. they could be used as 0%).
What I would try to achieve is perform a cluster analysis, that would group the particles into groups with similar particles and give me some averages in terms of that cluster element composition.
I was looking at how cluster analysis works, but usually it only uses 2 parameters, whereas I have many elements for each particle and I want to take into account more than just one element for each particle while clustering it. I am not so much interested in the exact match in terms of all the elements contained. In other words, if for example some 2 particles were quite similar except that one contained one extra element in a very small quantity, that would be ok too. Very low percentages are sometimes caused by background noise when measuring it.
Once I know which strategy to use I would ideally use R to do it. But giving me just a hint as to how to go about it, or a link, would be enough.
Related
I am trying to solve an optimization which looks similar to a knapsack-problem. The setting is the following:
I am having a pool of ~80,000 players of which I want to build the cheapest squad of exactly 11 players. Each player has multiple attributes, the main position he is playing in, nation, club, league and rating.
The players not only need to be selected but also assigned to a position in the formation:
Stating the following problem:
The first constraint is a minimum rating of the squad, which can simply be formulated as a linear constraint. The second and third constraint make sure that exactly one player is selected for each position and each player can only be selected once.
There are several other linear constrains that can occur like a minimum amount of players from one nation or at most three players from a specific club etc.
The chemistry of a squad is a non-linear constraint with a step function.
A players individual chemistry is the product of his position & link bonus.
The position bonus is defined by what the players main position is and where in the formation he is placed in. A central defender placed in the according position gets 3 points, used as a striker he gets 0 points. The bonuses can be seen in the next table.
This part of the constraint still can be formulated linearly. The link bonus is the non linear component. Each position/node in the formation/graph has a weight between [0-3], two adjacent players have a weight of 1 if they are from the same nation, league or club. Sharing two attributes is a weight of 2 and for three respectively. The bonus for a specific position is the average of all edges multiplied by a factor 3.
This bonus is plugged into a step function, which can be seen in the next figure (mapping values between [0-1] to 0.9 etc.). The link bonus is multiplied by the position bonus and capped to 10. The team chemistry is defined as the sum of the individual player chemistries.
I implemented it as described with miniZinc solving it with the osicbc solver, but even for a player pool of ~100 players this is not really feasible to compute, depending on the additional constraints.
Now I am looking for an implementation that can approximate the solution. I was thinking about a simulated annealing or genetic algorithm. However, due to this chemistry constraint these approaches produce a lot of invalid solutions, wandering around in the dark.
Does anyone have an approach that might be applicable to my problem?
I am using the finite difference scheme to find gradients.
Lets say i have 2 outputs (y1,y2) and 1 input (x) in a single component. And in advance I know that the sensitivity of y1 with respect to x is not same as the sensitivity of y2 to x. And thus i could potentially have two different steps for those as in ;
self.declare_partials(of=y1, wrt=x, method='fd',step=0.01, form='central')
self.declare_partials(of=y2, wrt=x, method='fd',step=0.05, form='central')
There is nothing that stops me (algorithmically) but it is not clear what would openmdao gradient calculation exactly do in this case?
does it exchange information from the case where the steps are different by looking at the steps ratios or simply treating them independently and therefore doubling computational time ?
I just tested this, and it does the finite difference twice with the two different step sizes, and only saves the requested outputs for each step. I don't think we could do anything with the ratios as you suggested, as the reason for using different stepsizes to resolve individual outputs is because you don't trust the accuracy of the outputs at the smaller (or large) stepsize.
This is a fair question about the effect of the API. In typical FD applications you would get only 1 function call per design variable for forward and backward difference and 2 function calls for central difference.
However in this case, you have asked for two different step sizes for two different outputs, both with central difference. So here, you'll end up with 4 function calls to compute all the derivatives. dy1_dx will be computed using the step size of .01 and dy2_dx will be computed with a step size of .05.
There is no crosstalk between the two different FD calls, and you do end up with more function calls than you would have if you just specified a single step size via:
self.declare_partials(of='*', wrt=x, method='fd',step=0.05, form='central')
If the cost is something you can bear, and you get improved accuracy, then you could use this method to get different step sizes for different outputs.
I am making a roguelike where the setting is open world on a procedurally generated planet. I want the distribution of each biome to be organic. There are 5 different biomes. Is there a way to organically distribute them without a huge complicated algorithm? I want the amount of space each biome takes up to be nearly equal.
I have worked with cellular automata before when I was making the terrain generators for each biome. There were 2 different states for each tile there. Is there an efficient way to do 5?
I'm using python 2.5, although specific code isn't necessary. Programming theory on it is fine.
If the question is too open ended, are there any resources out there that I could look at for this kind of problem?
You can define a cellular automaton on any cell state space. Just formulate the cell update function as F:Q^n->Q where Q is your state space (here Q={0,1,2,3,4,5}) and n is the size of your neighborhood.
As a start, just write F as a majority rule, that is, 0 being the neutral state, F(c) should return the value in 1-5 with the highest count in the neighborhood, and 0 if none is present. In case of equality, you may pick one of the max at random.
As an initial state, start with a configuration with 5 relatively equidistant cells with the states 1-5 (you may build them deterministically through a fixed position that can be shifted/mirrored, or generate these points randomly).
When all cells have a value different than 0, you have your map.
Feel free to improve on the update function, for example by applying the rule with a given probability.
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I have a large sales database of a 'home and construction' retail.
And I need to know who are the electricians, plumbers, painters, etc. in the store.
My first approach was to select the articles related to a specialty (wires [article] is related to an electrician [specialty], for example) And then, based on customer sales, know who the customers are.
But this is a lot of work.
My second approach is to make a cluster segmentation first, and then discover which cluster belong to a specialty. (this is a lot better because I would be able to discover new segments)
But, how can I do that? What type of clustering should I occupy? Kmeans, fuzzy? What variables should I take to that model? Should I use PCA to know how many cluster to search?
The header of my data (simplified):
customer_id | transaction_id | transaction_date | item_article_id | item_group_id | item_category_id | item_qty | sales_amt
Any help would be appreciated
(sorry my english)
You want to identify classes of customers based on what they buy (I presume this is for marketing reasons). This calls for a clustering approach. I will talk you through the entire setup.
The clustering space
Let us first consider what exactly you are clustering: either orders or customers. In either case, the way you characterize the items and the distances between them is the same. I will discuss the basic case for orders first, and then explain the considerations that apply to clustering by customers instead.
For your purpose, an order is characterized by what articles were purchased, and possibly also how many of them. In terms of a space, this means that you have a dimension for each type of article (item_article_id), for example the "wire" dimension. If all you care about is whether an article is bought or not, each item has a coordinate of either 0 or 1 in each dimension. If some order includes wire but not pipe, then it has a value of 1 on the "wire" dimension and 0 on the "pipe" dimension.
However, there is something to say for caring about the quantities. Perhaps plumbers buy lots of glue while electricians buy only small amounts. In that case, you can set the coordinate in each dimension to the quantity of the corresponding article (presumably item_qty). So suppose you have three articles, wire, pipe and glue, then an order described by the vector (2, 3, 0) includes 2 wire, 3 pipe and 0 glue, while an order described by the vector (0, 1, 4) includes 0 wire, 1 pipe and 4 glue.
If there is a large spread in the quantities for a given article, i.e. if some orders include order of magnitude more of some article than other orders, then it may be helpful to work with a log scale. Suppose you have these four orders:
2 wire, 2 pipe, 1 glue
3 wire, 2 pipe, 0 glue
0 wire, 100 pipe, 1 glue
0 wire, 300 pipe, 3 glue
The former two orders look like they may belong to electricians while the latter two look like they belong to plumbers. However, if you work with a linear scale, order 3 will turn out to be closer to orders 1 and 2 than to order 4. We fix that by using a log scale for the vectors that encode these orders (I use the base 10 logarithm here, but it does not matter which base you take because they differ only by a constant factor):
(0.30, 0.30, 0)
(0.48, 0.30, -2)
(-2, 2, 0)
(-2, 2.48, 0.48)
Now order 3 is closest to order 4, as we would expect. Note that I have used -2 as a special value to indicate the absence of an article, because the logarithm of 0 is not defined (log(x) tends to negative infinity as x tends to 0). -2 means that we pretend that the order included 1/100th of the article; you could make the special value more or less extreme, depending on how much weight you want to give to the fact that an article was not included.
The input to your clustering algorithm (regardless of which algorithm you take, see below) will be a position matrix with one row for each item (order or customer), one column for each dimension (article), and either the presence (0/1), amount, or logarithm of the amount in each cell, depending on which you choose based on the discussion above. If you cluster by customers, you can simply sum the amounts from all orders that belong to that customer before you calculate what goes into each cell of your position matrix (if you use the log scale, sum the amounts before taking the logarithm).
Clustering by orders rather than by customers gives you more detail, but also more noise. Customers may be consistent within an order but not between them; perhaps a customer sometimes behaves like a plumber and sometimes like an electrician. This is a pattern that you will only find if you cluster by orders. You will then find how often each customer belongs to each cluster; perhaps 70% of somebody's orders belong to the electrician type and 30% belong to the plumber type. On the other hand, a plumber may only buy pipe in one order and then only buy glue in the next order. Only if you cluster by customers and sum the amounts of their orders, you get a balanced view of what each customer needs on average.
From here on I will refer to your position matrix by the name my.matrix.
The clustering algorithm
If you want to be able to discover new customer types, you probably want to let the data speak for themselves as much as possible. A good old fashioned
hierarchical clustering with complete linkage (CLINK) may be an appropriate choice in this case. In R, you simply do hclust(dist(my.matrix)) (this will use the Euclidean distance measure, which is probably good enough in your case). It will join closely neighbouring items or clusters together until all items are categorized in a hierarchical tree. You can treat any branch of the tree as a cluster, observe typical article amounts for that branch and decide whether that branch represents a customer segment by itself, should be split in sub-branches, or joined with a sibling branch instead. The advantage is that you find the "full story" of which items and clusters of items are most similar to each other and how much. The disadvantage is that the outcome of the algorithm does not tell you where to draw the borders between your customer segments; you can cut up the clustering tree in many ways, so it's up to your interpretation how you want to identify your customer types.
On the other hand, if you are comfortable fixing the number of clusters (k) beforehand, k-means is a very robust way to get just any segmentation of your customers in k distinct types. In R, you would do kmeans(my.matrix, k). For marketing purposes, it may be sufficient to have (say) 5 different profiles of customers that you make custom advertisement for, rather than treating all customers the same. With k-means you don't explore all of the diversity that is present in your data, but you might not need to do so anyway.
If you don't want to fix the number of clusters beforehand, but you also don't want to manually decide where to draw the borders between the segments afterwards, there is a third possibility. You start with the k-means algorithm, where you let it generate an amount of cluster centers that is much larger than the number of clusters that you hope to end up with (for example, if you hope to end up with somewhere about 10 clusters, let the k-means algorithm look for 200 clusters). Then, use the mean shift algorithm to further cluster the resulting centers. You will end up with a smaller number of compact clusters. The approach is explained in more detail by James Li over here. You can use the mean shift algorithm in R with the ms function from the LPCM package, see this documentation.
About using PCA
PCA will not tell you how many clusters you need. PCA answers a different question: which variables seem to represent a common underlying (hidden) factor. In a sense, it is a way to cluster variables, i.e. properties of entities, not to cluster the entities themselves. The number of principal components (common underlying factors) is not indicative of the number of clusters needed. PCA can still be interesting if you want to learn something about the predictive value of each article about a customer's interests.
Sources
Michael J. Crawley, 2005. Statistics. An Introduction using R.
Gerry P. Quinn and Michael J. Keough, 2002. Experimental Design and Data Analysis for Biologists.
Wikipedia: hierarchical clustering, k-means, mean shift, PCA
I have a collection of 15M (Million) DAGs (directed acyclic graphs - directed hypercubes actually) that I would like to remove isomorphisms from. What is the common algorithm for this? Each graph is fairly small, a hybercube of dimension N where N is 3 to 6 (for now) resulting in graphs of 64 nodes each for N=6 case.
Using networkx and python, I implemented it like this which works for small sets like 300k (Thousand) just fine (runs in a few days time).
def isIsomorphicDuplicate(hcL, hc):
"""checks if hc is an isomorphism of any of the hc's in hcL
Returns True if hcL contains an isomorphism of hc
Returns False if it is not found"""
#for each cube in hcL, check if hc could be isomorphic
#if it could be isomorphic, then check if it is
#if it is isomorphic, then return True
#if all comparisons have been made already, then it is not an isomorphism and return False
for saved_hc in hcL:
if nx.faster_could_be_isomorphic(saved_hc, hc):
if nx.fast_could_be_isomorphic(saved_hc, hc):
if nx.is_isomorphic(saved_hc, hc):
return True
return False
One better way to do it would be to convert each graph to its canonical ordering, sort the collection, then remove the duplicates. This bypasses checking each of the 15M graphs in a binary is_isomophic() test, I believe the above implementation is something like O(N!N) (not taking isomorphic time into account) whereas a clean convert all to canonical ordering and sort should take O(N) for the conversion + O(log(N)N) for the search + O(N) for the removal of duplicates. O(N!N) >> O(log(N)N)
I found this paper on Canonical graph labeling, but it is very tersely described with mathematical equations, no pseudocode: "McKay's Canonical Graph Labeling Algorithm" - http://www.math.unl.edu/~aradcliffe1/Papers/Canonical.pdf
tldr: I have an impossibly large number of graphs to check via binary isomorphism checking. I believe the common way this is done is via canonical ordering. Do any packaged algorithms or published straightforward to implement algorithms (i.e. have pseudocode) exist?
Here is a breakdown of McKay ’ s Canonical Graph Labeling Algorithm, as presented in the paper by Hartke and Radcliffe [link to paper].
I should start by pointing out that an open source implementation is available here: nauty and Traces source code.
Ok, let's do this! Unfortunately this algorithm is heavy in graph theory, so we need some terms. First I will start by defining isomorphic and automorphic.
Isomorphism:
Two graphs are isomorphic if they are the same, except that the vertices are labelled differently. The following two graphs are isomorphic.
Automorphic:
Two graphs are automorphic if they are completely the same, including the vertex labeling. The following two graphs are automorphic. This seems trivial, but turns out to be important for technical reasons.
Graph Hashing:
The core idea of this whole thing is to have a way to hash a graph into a string, then for a given graph you compute the hash strings for all graphs which are isomorphic to it. The isomorphic hash string which is alphabetically (technically lexicographically) largest is called the "Canonical Hash", and the graph which produced it is called the "Canonical Isomorph", or "Canonical Labelling".
With this, to check if any two graphs are isomorphic you just need to check if their canonical isomporphs (or canonical labellings) are equal (ie are automorphs of each other). Wow jargon! Unfortuntately this is even more confusing without the jargon :-(
The hash function we are going to use is called i(G) for a graph G: build a binary string by looking at every pair of vertices in G (in order of vertex label) and put a "1" if there is an edge between those two vertices, a "0" if not. This way the j-th bit in i(G) represents the presense of absence of that edge in the graph.
McKay ’ s Canonical Graph Labeling Algorithm
The problem is that for a graph on n vertices, there are O( n! ) possible isomorphic hash strings based on how you label the vertices, and many many more if we have to compute the same string multiple times (ie automorphs). In general we have to compute every isomorph hash string in order to find the biggest one, there's no magic sort-cut. McKay's algorithm is a search algorithm to find this canonical isomoprh faster by pruning all the automorphs out of the search tree, forcing the vertices in the canonical isomoprh to be labelled in increasing degree order, and a few other tricks that reduce the number of isomorphs we have to hash.
(1) Sect 4: the first step of McKay's is to sort vertices according to degree, which prunes out the majority of isomoprhs to search, but is not guaranteed to be a unique ordering since there may be more than one vertex of a given degree. For example, the following graph has 6 vertices; verts {1,2,3} have degree 1, verts {4,5} have degree 2 and vert {6} has degree 3. It's partial ordering according to vertex degree is {1,2,3|4,5|6}.
(2) Sect 5: Impose artificial symmetry on the vertices which were not distinguished by vertex degree; basically we take one of the groups of vertices with the same degree, and in turn pick one at a time to come first in the total ordering (fig. 2 in the paper), so in our example above, the node {1,2,3|4,5|6} would have children { {1|2,3|4,5|6}, {2|1,3|4,5|6}}, {3|1,2|4,5|6}} } by expanding the group {1,2,3} and also children { {1,2,3|4|5|6}, {1,2,3|5|4|6} } by expanding the group {4,5}. This splitting can be done all the way down to the leaf nodes which are total orderings like {1|2|3|4|5|6} which describe a full isomorph of G. This allows us to to take the partial ordering by vertex degree from (1), {1,2,3|4,5|6}, and build a tree listing all candidates for the canonical isomorph -- which is already a WAY fewer than n! combinations since, for example, vertex 6 will never come first. Note that McKay evaluates the children in a depth-first way, starting with the smallest group first, this leads to a deeper but narrower tree which is better for online pruning in the next step. Also note that each total ordering leaf node may appear in more than one subtree, there's where the pruning comes in!
(3) Sect. 6: While searching the tree, look for automorphisms and use that to prune the tree. The math here is a bit above me, but I think the idea is that if you discover that two nodes in the tree are automorphisms of each other then you can safely prune one of their subtrees because you know that they will both yield the same leaf nodes.
I have only given a high-level description of McKay's, the paper goes into a lot more depth in the math, and building an implementation will require an understanding of this math. Hopefully I've given you enough context to either go back and re-read the paper, or read the source code of the implementation.
This is indeed an interesting problem.
I would approach it from the adjacency matrix angle. Two isomorphic graphs will have adjacency matrices where the rows / columns are in a different order. So my idea is to compute for each graph several matrix properties which are invariant to row/column swaps, off the top of my head:
numVerts, min, max, sum/mean, trace (probably not useful if there are no reflexive edges), norm, rank, min/max/mean column/row sums, min/max/mean column/row norm
and any pair of isomorphic graphs will be the same on all properties.
You could make a hash function which takes in a graph and spits out a hash string like
string hashstr = str(numVerts)+str(min)+str(max)+str(sum)+...
then sort all graphs by hash string and you only need to do full isomorphism checks for graphs which hash the same.
Given that you have 15 million graphs on 36 nodes, I'm assuming that you're dealing with weighted graphs, for unweighted undirected graphs this technique will be way less effective.
This is an interesting question which I do not have an answer for! Here is my two cents:
By 15M do you mean 15 MILLION undirected graphs? How big is each one? Any properties known about them (trees, planar, k-trees)?
Have you tried minimizing the number of checks by detecting false positives in advance? Something includes computing and comparing numbers such as vertices, edges degrees and degree sequences? In addition to other heuristics to test whether a given two graphs are NOT isomorphic. Also, check nauty. It may be your way to check them (and generate canonical ordering).
If all your graphs are hypercubes (like you said), then this is trivial: All hypercubes with the same dimension are isomorphic, hypercubes with different dimension aren't. So run through your collection in linear time and throw each graph in a bucket according to its number of nodes (for hypercubes: different dimension <=> different number of nodes) and be done with it.
since you mentioned that testing smaller groups of ~300k graphs can be checked for isomorphy I would try to split the 15M graphs into groups of ~300k nodes and run the test for isomorphy on each group
say: each graph Gi := VixEi (Vertices x Edges)
(1) create buckets of graphs such that the n-th bucket contains only graphs with |V|=n
(2) for each bucket created in (1) create subbuckets such that the (n,m)-th subbucket contains only graphs such that |V|=n and |E|=m
(3) if the groups are still too large, sort the nodes within each graph by their degrees (meaning the nr of edges connected to the node), create a vector from it and distribute the graphs by this vector
example for (3):
assume 4 nodes V = {v1, v2, v3, v4}. Let d(v) be v's degree with d(v1)=3, d(v2)=1, d(v3)=5, d(v4)=4, then find < := transitive hull ( { (v2,v1), (v1,v4), (v4,v3) } ) and create a vector depening on the degrees and the order which leaves you with
(1,3,4,5) = (d(v2), d(v1), d(v4), d(v3)) = d( {v2, v1, v4, v3} ) = d(<)
now you have divided the 15M graphs into buckets where each bucket has the following characteristics:
n nodes
m edges
each graph in the group has the same 'out-degree-vector'
I assume this to be fine grained enough if you are expecting not to find too many isomorphisms
cost so far: O(n) + O(n) + O(n*log(n))
(4) now, you can assume that members inside each bucket are likely to be isomophic. you can run your isomorphic-check on the bucket and only need to compare the currently tested graph against all representants you have already found within this bucket. by assumption there shouldn't be too many, so I assume this to be quite cheap.
at step 4 you also can happily distribute the computation to several compute nodes, which should really speed up the process
Maybe you can just use McKay's implementation? It is found here now: http://pallini.di.uniroma1.it/
You can convert your 15M graphs to the compact graph6 format (or sparse6) which nauty uses and then run the nauty tool labelg to generate the canonical labels (also in graph6 format).
For example - removing isomorphic graphs from a set of random graphs:
#gnp.py
import networkx as nx
for i in range(100000):
graph = nx.gnp_random_graph(10,0.1)
print nx.generate_graph6(graph,header=False)
[nauty25r9]$ python gnp.py > gnp.g6
[nauty25r9]$ cat gnp.g6 |./labelg |sort |uniq -c |wc -l
>A labelg
>Z 10000 graphs labelled from stdin to stdout in 0.05 sec.
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