Approach
I am maintaining two stacks, one for inorder traversal and the other for reverse inorder traversal, but can't seem to figure out a way to print both 2 10, and 5 7. How should I go about this ? I want to have a space complexity of O(height) and time complexity of O(nodes). I have seen articles on leetcode, geeksforgeeks etc. but am not able to get hold of this example
The main question is connected with extracting the contact pressure from .odb file.
The issue is described in three facts written below:
Imagine that we have simple 3D contact model in Abaqus/CAE
1.If we make a plot of CPRESS on a deformed shape in visualisation module, we'll get a one value of CPRESS for each node. The same (one value for one node) we will get if we request XYdata field output for all frames. And this all seems to be ok, because as far as I know Abaqus CAE use averaging for surface output (CPRESS) to make it possible to request as nodal output.
2.If we use "Probe values" instrument to examine CPRESS value in node, we'll get four values for one node. It still seems to be ok, because, i suppose, it shows the values befor averaging.
3.If we request CPRESS value from command window using this script:
odb.steps['step_name'].frames[frame_number].fieldOutputs['CPRESS'].getSubset(region='node_path').values
length of this vector of CPRESS values in a single node may be from 1 to 6 depending on a chosen node. And the quantity of CPRESS valuse got using this method have no connection with the quantity got using method 2.
So the trick is that I can't inderstand how the vector of CPRESS in node is forming.
Found very little information about this topic in Abaqus Manual.
Hope somebody may help)
Probe Values, extacts the CPRESS values for the whole element. It shows the face number and its node IDs toghether with their corresponding values.
I have products with different details in different attributes and I need to develop an algorithm to find the most similar to the one I'm trying to find.
For example, if a product has:
Weight: 100lb
Color: Black, Brown, White
Height: 10in
Conditions: new
Others can have different colors, weight, etc. Then I need to do a search where the most similar return first. For example, if everything matches but the color is only Black and White but not Brown, it's a better match than another product that is only Black but not White or Brown.
I'm open to suggestions as the project is just starting.
One approach, for example, I could do is restrict each attribute (weight, color, size) a limited set of option, so I can build a binary representation. So I have something like this for each product:
Colors Weight Height Condition
00011011000 10110110 10001100 01
Then if I do an XOR between the product's binary representation and my search, I can calculate the number of set bits to see how similar they are (all zeros would mean exact match).
The problem with this approach is that I cannot index that on a database, so I would need to read all the products to make the comparison.
Any suggestions on how I can approach this? Ideally I would like to have something I can index on a database so it's fast to query.
Further question: also if I could use different weights for each attribute, it would be awesome.
You basically need to come up with a distance metric to define the distance between two objects. Calculate the distance from the object in question to each other object, then you can either sort by minimum distance or just select the best.
Without some highly specialized algorithm based on the full data set, the best you can do is a linear time distance comparison with every other item.
You can estimate the nearest by keeping sorted lists of certain fields such as Height and Weight and cap the distance at a threshold (like in broad phase collision detection), then limit full distance comparisons to only those items that meet the thresholds.
What you want to do is a perfect use case for elasticsearch and other similar search oriented databases. I don't think you need to hack with bitmasks/etc.
You would typically maintain your primary data in your existing database (sql/cassandra/mongo/etc..anything works), and copy things that need searching to elasticsearch.
What are you talking about very similar to BK-trees. BK-tree constructs search tree with some metric associated with keys of this tree. Most common use of this tree is string corrections with Levenshtein or Damerau-Levenshtein distances. This is not static data structure, so it supports future insertions of elements.
When you search exact element (or insert element), you need to look through nodes of this tree and go to links with weight equal to distance between key of this node and your element. if you want to find similar objects, you need to go to several nodes simultaneously that supports your wishes of constrains of distances. (Maybe it can be even done with A* to fast finding one most similar object).
Simple example of BK-tree (from the second link)
BOOK
/ \
/(1) \(4)
/ \
BOOKS CAKE
/ / \
/(2) /(1) \(2)
/ | |
BOO CAPE CART
Your metric should be Hamming distance (count of differences between bit representations of two objects).
BUT! is it good to compare two integers as count of different bits in their representation? With Hamming distance HD(10000, 00000) == HD(10000, 10001). I.e. difference between numbers 16 and 0, and 16 and 17 is equal. Is it really what you need?
BK-tree with details:
https://hamberg.no/erlend/posts/2012-01-17-BK-trees.html
https://nullwords.wordpress.com/2013/03/13/the-bk-tree-a-data-structure-for-spell-checking/
Thinking about it, I thought the time complexity for insertion and search for any data structure should be the same, because to insert, you first have to search for the location you want to insert, and then you have to insert.
According to here: http://bigocheatsheet.com/, for a linked list, search is linear time but insertion is constant time. I understand how searching is linear (start from the front, then keep going through the nodes on the linked list one after another until you find what you are searching for), but how is insertion constant time?
Suppose I have this linked list:
1 -> 5 -> 8 -> 10 -> 8
and I want to insert the number 2 after the number 8, then would I have to first search for the number 8 (search is linear time), and then take an extra 2 steps to insert it (so, insertion is still linear time?)?
#insert y after x in python
def insert_after(x, y):
search_for(y)
y.next = x.next
x.next = y
Edit: Even for a doubly linked list, shouldn't it still have to search for the node first (which is linear time), and then insert?
So if you already have a reference to the node you are trying to insert then it is O(1). Otherwise, it is search_time + O(1). It is a bit misleading but on wikipedia there is a chart explains it a bit better:
Contrast this to a dynamic array, which, if you want to insert at the beginning is: Θ(n).
Just for emphasis: The website you reference is referring to the actual act of inserting given we already know where we want to insert.
Time to insert = Time to set three pointers = O(3) = constant time.
Time to insert the data is not the same as time to insert the data at a particular location. The time asked is the time to insert the data only.
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.