I'm reading the book of Computer networks by Andrew S. Tanenbaum. I'm confused about the hamming distance in this paragraph. Why does it say the Hamming distance is 3 in this construction? Thanks.
You are confusing between the figure and the written part.They are 2 different examples.In the diagram example of (11,7) Hamming code is given,while the example above it is discussing about (11,29).
The Hamming distance between two strings of equal length is the number
of positions at which the corresponding symbols are different.
Hamming distance of (11,29) will be 3 as 11=01011 and 29=11101
Reach out if you want something else.
Cheers
Related
here the question
1001110
1110101
1010011
0011011
How can calculate hamming distance for this example. Can you explain pls for more than 2 inputs.
Do you mean the minimum Hamming distance? The minimum Hamming distance of a code is the smallest Hamming distance between a pair of codewords. In your example code, which contains four codewords, the two closest codewords are the last two codewords, and they differ in exactly two positions. Hence, the minimum Hamming distance of this code is 2.
In dynamical networks, one may calculate the Hamming distance to compare the similarity between two graphs, can anyone explain how?
Assuming that the Hamming distance of two graphs have equal edge density, what is the difference between Hamming distance and expected Hamming distance between two independent Erdos-Renyi random graphs? How does the later arise?
The Hamming distance measures the minimum number of substitutions required to change (transform) one mathematical 'object' (i.e. strings or binary) into another.
So in network theory it can be defined as a the number of different connections between two networks (it can be formulated also for not equally-sized networks and for weighted or directed graphs). In a simple case in which you have two Erdos-Renyi networks (the adjacency matrix has 1 if the node pair is connected and 0 if not) the distance is mathematically defined as follows:
The values that are subtracted are the two adjacency matrix. If you take two Erdos-Renyi networks with wiring probability of 0.5 and compute the hamming distance between them you should get a value around 0.5. I generated different Erdos-Renyi graph and their Hamming distances produced a Gaussian curve around 0.5 (as we can expect; see below).
If it is needed I can give you the code I used.
How can I find the minimum Hamming Distance for the above?
I understand the string comparison idea and putting it into a table based on C0, C1, C2, etc but I'm not sure how to group the code above. Any suggestions? Thank you in advance.
Generally, to find the minimum Hamming distance you have to compute the Hamming distance of each pair of code words and then take the minimum of these. For special cases, e.g. linear codes there are theorems for quicker determination of the minimum Hamming distance (https://en.wikipedia.org/wiki/Linear_code).
In your example, the eye spots several adjacent code word pairs differing only in one bit, so as Egor wrote, the minimum Hamming distance is 1.
MY DATA
I have a matrix Median that contains three qualities, Speed, Angle & Acceleration, in virtual 3D space. Each set of qualities belongs to an individual person, termed Class.
Speed<-c(18,21,25,19)
Angle<-c(90,45,90,120)
Acceleration<-c(4,5,9,4)
Class<-c("Nigel","Paul","Kelly","Steve")
Median = data.frame(Class,Speed,Angle,Acceleration)
mm = as.matrix(Median)
In the example above, Nigel's Speed, Angle and Acceleration qualities would be (18,90,4).
MY PROBLEM
I wish to know the euclidean distance between each individual person/class. For example, the euclidean distance between Nigel and Paul, Nigel and Kelly etc. I then wish to display the results in a dendrogram, as a result of hierarchical clustering.
WHAT I HAVE (UNSUCCESSFULLY) ATTEMPTED
I first used hc = hclust(dist(mm)) then plot(hc) although this results in a dendrogram of Speed only. It seems the function pdist() can compute distance between two matrices of observations, but I have three matrices. Is this possible in R? I am new to the language and have found a similar question in MATLAB here Calculating Euclidean distance of pairs of 3D points in matlab but how do I write this in R code?
Many thanks.
When you transform your data.frame into a matrix, all values become characters, I don't think that is what you want... (moreover, you're trying to compute distance with the "class" names as one of the variables...)
The best would be to put your "Class" as row.names and then compute your distances and hclust :
mm<-Median[,-1]
row.names(mm)<-Median[,1]
Then you can compute the euclidean distances between Class with
dist(mm,method="euclidean") :
> dist(mm,method="euclidean")
Nigel Paul Kelly
Paul 45.110974
Kelly 8.602325 45.354162
Steve 30.016662 75.033326 31.000000
Finally, perform your hierarchical classification :
hac<-hclust(dist(mm,method="euclidean"))
and plot(hac,hang=-1) to display the dendrogram.
I need to calculate the trace of a matrix to the power of 3 and 4 and it needs to be as fast as it can get.
The matrix here is an adjacency matrix of a simple graph, therefore it is square, symmetric, its entries are always 1 or 0 and the diagonal elements are always 0.
Optimization is trivial for the trace of the matrix to the power of 2:
We only need the diagonal entries (i,i) for the trace, skip all others
As the matrix is symmetric these entries are just the entries of the i-th row squared and summed up
And as the entries are just 1 or 0 the square-operation can be skipped
Another idea I found on wikipedia was summing up all elements of the Hadamard product, i.e. entry-wise multiplication, but I don't know how to extend this method to the power of 3 and 4.
See http://en.wikipedia.org/wiki/Trace_(linear_algebra)#Properties
Maybe I'm just blind but I can't think of a simple solution.
In the end I need a C++ implementation, but I think that's not important to the question.
Thanks in advance for any help.
The trace is the sum of the eigenvalues and the eigenvalues of a matrix power are just the eigenvalues to that power.
That is, if l_1,...,l_n are the eigenvalues of your matrix then trace(M^p) = 1_1^p + l_2^p +...+l_n^p.
Depending on your matrix you may want to go with computing the eigenvalues and then summing. If your matrix has low rank (or can be well approximated with a low rank matrix) you can compute the eigenvalues very cheaply (a partial eigendecomposition has complexity O(n*k^2) where k is the rank).
Edit: You mention in the comments that it's 1600x1600 in which case finding all the eigenvalues should be no problem. Here's one of many C++ codes that you can use for this http://code.google.com/p/redsvd/
Ok, I just figured this one out myself.
The important thing I did not know was this:
If A is the adjacency matrix of the directed or undirected graph G, then the matrix An (i.e., the matrix product of n copies of A) has an interesting interpretation: the entry in row i and column j gives the number of (directed or undirected) walks of length n from vertex i to vertex j. This implies, for example, that the number of triangles in an undirected graph G is exactly the trace of A^3 divided by 6.
(Copied from http://en.wikipedia.org/wiki/Adjacency_matrix#Properties)
Retrieving the number of paths of a given length from node i to i for all n nodes can essentially be done in O(n) when dealing with sparse graphs and using adjacency lists instead of matrices.
Nevertheless, thanks for your answers!