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Best way to create a large,iterative database of frequency count, of binary numbers varying by 1 bit?

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Forrest
Forrest 2013-11-8
关闭： MATLAB Answer Bot 2021-8-20
So, I have a general question, not necessarily looking for code, but just a general idea of how to accomplish this task in the best way possible.
To start, I will have a string, that will be several thousands bits long, of simple 1's and 0's. What I want to do, is for each new bit (1 or 0) added to my string, take the last n bits off the end, and store that sequence into a database.
If the string is a NEW/unique string not seen before, I'd like to add it to the database. If it is a repeated sequence, I'd like to add to the count.
What I initially did, since these are binary numbers, using n-bits, was to create a matrix of all possible permutations, where the possible events would be 2^n. So if n==3, I would have 000,001,010,011,100,101,110,111.
And run through my big string 1 bit at a time, look at the latest n bits on the end, and compare to each value in my list and update the count of each occurrence.
What is important, is that I am able to compare the frequency of the sequences that differ by the last bit. So if n==3, I would like to compare for example, the frequency/percentage of occurrence of:
  • 00 0 vs 00 1
  • 01 0 vs 01 1
  • 10 0 vs 10 1
  • 11 0 vs 11 1
For now I'm not too concerned with efficiency, just a basic working framework in place.
So in general I am wondering if there is a better way to accomplish what I am trying to do other than the way I just described above?
Thanks!

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Walter Roberson
Walter Roberson 2013-11-8
Let B be the bit values in numeric form (0 vs 1), already arranged n wide per row
IDX = 2.^(n-1:-1:0).' * B + 1; %matrix multiplication not .*
numents = length(IDX);
counts = accumarray(IDX(:), 1, [1, 2^n]);
Now counts(K) is the number of occurrences of the bit sequence whose decimal equivalent is K-1.
The marginals are now
T = 2 * (0:(2^(n-1)-1));
marginals = counts([T + 1, T + 2]);
I did not convert the marginal counts into ratios or relative percentages because of the possibility that one or both might be 0.
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Forrest
Forrest 2013-11-12
Ok..... so it took me a while, but finally got a structure in place that will at least grab and create each string. It's not a string actually, but just a numeric value of each string as you suggested.
Now I'm onto figuring out how to compare values that differ by that last bit. I am attaching a zip, that has the original script, along with the input string array. There are 3, each one can be used by 'commenting' it out at the top of the script. Values are stored in the variable "kArray."
So your suggestion is the convert each string to an integer and calculate %s that way?
Forrest
Forrest 2013-11-12
Have the computations of the percentages that differ by the last bit working. It is probably EXTREMELY computationally inefficient though as I loop through to count the individual occurrences.
A string of length 4 happens pretty quickly, taking approximately 24 seconds.
By contrast, a 10 string takes 1570 seconds! I will want to do this for strings up to a length of 20. What is worse is that I will want to generate reports of strings from 6-20 all at once. So I can't imagine how bad that will be? :)

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