I am working on error control in WMSNs. I want to transmit a video through binary symmetric channel with error probability p. So I have frames (images) in each gop which are shown by a matrix.
Each matrix element have decimal value which might be positive or negative. As explained here I need to convert this whole matrix to a binary stream. I used
reshape(dec2bin(typecast(b,'uint8'),8).',1,[])
for converting elements to binary streams but I cannot get back the exact number using
typecast(uint8(bin2dec(reshape(m,8,[]).')),'double').
On the other side, I think for getting the right bit error rate, I have to convert the whole matrix to just one bit stream which I'm not sure how to do that. And convert them to a matrix of measured values of a image again.
I think you need
m = reshape(dec2bin(typecast(b(:),'uint8'),8).',1,[]);
Note that this reads the matrix in Matlab's standard, column-major order (down, then across).
Then you can convert back with
b_recovered = reshape(typecast(uint8(bin2dec(reshape(m,8,[]).')),'double'),size(b));
Since typecast converts data types without changing the underlying data, this process entails no loss of accuracy. For example,
>> b = randn(2,3)
b =
-0.241247174335006 0.540703471823211 0.526269662140438
0.908207564087271 -0.507829312416083 -1.067884765919437
>> m = reshape(dec2bin(typecast(b(:),'uint8'),8).',1,[])
m =
101100101011100000000010111110100010111111100001110011101011111101001010100000100011011101001111000010010001000011101101001111110010100100001111000010100101111001110001010011011110000100111111001011101101111100011000010000100010001101000000111000001011111100010010101001010111100001111001001100111101011111100000001111110101110001010100000110000101011000001110000101101111000110111111
>> b_recovered = reshape(typecast(uint8(bin2dec(reshape(m,8,[]).')),'double'),size(b))
b_recovered =
-0.241247174335006 0.540703471823211 0.526269662140438
0.908207564087271 -0.507829312416083 -1.067884765919437
>> b==b_recovered
ans =
2×3 logical array
1 1 1
1 1 1
>>
Related
I have two images to fuse. Images are
A
B
I have fused them and got this image fused
Now I want difference between fused image F and the image B.
I have execute the code, but not getting desirable results
I am getting this image -> difference normalized difference,
But I want this -> Required
Values of difference image are normalized to the range of 0 to 1.
The code used is
difference=F-B;
figure,imshow(difference);
normImage = mat2gray(difference);
figure,imshow(normImage);
Please anyone help. Thank you.
Using:
R = mat2gray(im2double(F)-im2double(B));
My result is:
To see why conversion to double is important, look at an area of the image where B(y,x) > F(y,x), such as (343, 280) in your sample images.
>> F(343,280)
ans = 32
>> B(343,280)
ans = 107
Mathematically, we'd expect 32-107 to equal -75, but:
>> F(343,280) - B(343,280)
ans = 0
This is because both F and B are arrays of uint8:
>> class(F)
ans = uint8
>> class(B)
ans = uint8
As an unsigned integer, uint8 can't take a negative value, so any attempt to assign a negative value to a uint8 variable results in 0. Since both operands are uint8, the result is uint8. Trying to cast that value to a double after it has already been clamped to be with in the range of 0-255 would simply result in a double variable with a value of 0. (The same thing also happens at the upper end of the range. Try uint8(444).)
Casting F and B to a signed type (one big enough to the range -max to +max, or -255 to 255 in this case) will take care of the math problem:
>> int16(F(343,280)) - int16(B(343,280))
ans = -75
For images, though, casting to double feels more natural and gives you more precision than integers when you're doing calculations and rescaling. Plus, there's this handy im2double function we can use that not only casts the array to doubles, but rescales everything to be between 0 and 1:
>> Fd = im2double(F);
>> Fd(343,280)
ans = 0.1255 % 32.0/255.0
>> Bd = im2double(B);
>> Bd(343,280)
ans = 0.4196 % 107.0/255.0
But now when we try to subtract the two, we actually get a negative value as expected:
>> Fd(343,280) - Bd(343,280)
ans = -0.2941 % -75.0/255.0
So, im2double(F)-im2double(B) gives us double values between -1.0 and 1.0. mat2gray takes care of scaling those values back to a range of 0.0 to 1.0 for display.
Note: I chose the coordinates (343,280) very carefully because that's where F-B is most negative. If you're curious about how the conversion happens and what values get scaled to what, you can also have a look at (53,266).
I am writing a program where I need to delete duplicate points stored in a matrix. The problem is that when it comes to check whether those points are in the matrix, MATLAB can't recognize them in the matrix although they exist.
In the following code, intersections function gets the intersection points:
[points(:,1), points(:,2)] = intersections(...
obj.modifiedVGVertices(1,:), obj.modifiedVGVertices(2,:), ...
[vertex1(1) vertex2(1)], [vertex1(2) vertex2(2)]);
The result:
>> points
points =
12.0000 15.0000
33.0000 24.0000
33.0000 24.0000
>> vertex1
vertex1 =
12
15
>> vertex2
vertex2 =
33
24
Two points (vertex1 and vertex2) should be eliminated from the result. It should be done by the below commands:
points = points((points(:,1) ~= vertex1(1)) | (points(:,2) ~= vertex1(2)), :);
points = points((points(:,1) ~= vertex2(1)) | (points(:,2) ~= vertex2(2)), :);
After doing that, we have this unexpected outcome:
>> points
points =
33.0000 24.0000
The outcome should be an empty matrix. As you can see, the first (or second?) pair of [33.0000 24.0000] has been eliminated, but not the second one.
Then I checked these two expressions:
>> points(1) ~= vertex2(1)
ans =
0
>> points(2) ~= vertex2(2)
ans =
1 % <-- It means 24.0000 is not equal to 24.0000?
What is the problem?
More surprisingly, I made a new script that has only these commands:
points = [12.0000 15.0000
33.0000 24.0000
33.0000 24.0000];
vertex1 = [12 ; 15];
vertex2 = [33 ; 24];
points = points((points(:,1) ~= vertex1(1)) | (points(:,2) ~= vertex1(2)), :);
points = points((points(:,1) ~= vertex2(1)) | (points(:,2) ~= vertex2(2)), :);
The result as expected:
>> points
points =
Empty matrix: 0-by-2
The problem you're having relates to how floating-point numbers are represented on a computer. A more detailed discussion of floating-point representations appears towards the end of my answer (The "Floating-point representation" section). The TL;DR version: because computers have finite amounts of memory, numbers can only be represented with finite precision. Thus, the accuracy of floating-point numbers is limited to a certain number of decimal places (about 16 significant digits for double-precision values, the default used in MATLAB).
Actual vs. displayed precision
Now to address the specific example in the question... while 24.0000 and 24.0000 are displayed in the same manner, it turns out that they actually differ by very small decimal amounts in this case. You don't see it because MATLAB only displays 4 significant digits by default, keeping the overall display neat and tidy. If you want to see the full precision, you should either issue the format long command or view a hexadecimal representation of the number:
>> pi
ans =
3.1416
>> format long
>> pi
ans =
3.141592653589793
>> num2hex(pi)
ans =
400921fb54442d18
Initialized values vs. computed values
Since there are only a finite number of values that can be represented for a floating-point number, it's possible for a computation to result in a value that falls between two of these representations. In such a case, the result has to be rounded off to one of them. This introduces a small machine-precision error. This also means that initializing a value directly or by some computation can give slightly different results. For example, the value 0.1 doesn't have an exact floating-point representation (i.e. it gets slightly rounded off), and so you end up with counter-intuitive results like this due to the way round-off errors accumulate:
>> a=sum([0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1]); % Sum 10 0.1s
>> b=1; % Initialize to 1
>> a == b
ans =
logical
0 % They are unequal!
>> num2hex(a) % Let's check their hex representation to confirm
ans =
3fefffffffffffff
>> num2hex(b)
ans =
3ff0000000000000
How to correctly handle floating-point comparisons
Since floating-point values can differ by very small amounts, any comparisons should be done by checking that the values are within some range (i.e. tolerance) of one another, as opposed to exactly equal to each other. For example:
a = 24;
b = 24.000001;
tolerance = 0.001;
if abs(a-b) < tolerance, disp('Equal!'); end
will display "Equal!".
You could then change your code to something like:
points = points((abs(points(:,1)-vertex1(1)) > tolerance) | ...
(abs(points(:,2)-vertex1(2)) > tolerance),:)
Floating-point representation
A good overview of floating-point numbers (and specifically the IEEE 754 standard for floating-point arithmetic) is What Every Computer Scientist Should Know About Floating-Point Arithmetic by David Goldberg.
A binary floating-point number is actually represented by three integers: a sign bit s, a significand (or coefficient/fraction) b, and an exponent e. For double-precision floating-point format, each number is represented by 64 bits laid out in memory as follows:
The real value can then be found with the following formula:
This format allows for number representations in the range 10^-308 to 10^308. For MATLAB you can get these limits from realmin and realmax:
>> realmin
ans =
2.225073858507201e-308
>> realmax
ans =
1.797693134862316e+308
Since there are a finite number of bits used to represent a floating-point number, there are only so many finite numbers that can be represented within the above given range. Computations will often result in a value that doesn't exactly match one of these finite representations, so the values must be rounded off. These machine-precision errors make themselves evident in different ways, as discussed in the above examples.
In order to better understand these round-off errors it's useful to look at the relative floating-point accuracy provided by the function eps, which quantifies the distance from a given number to the next largest floating-point representation:
>> eps(1)
ans =
2.220446049250313e-16
>> eps(1000)
ans =
1.136868377216160e-13
Notice that the precision is relative to the size of a given number being represented; larger numbers will have larger distances between floating-point representations, and will thus have fewer digits of precision following the decimal point. This can be an important consideration with some calculations. Consider the following example:
>> format long % Display full precision
>> x = rand(1, 10); % Get 10 random values between 0 and 1
>> a = mean(x) % Take the mean
a =
0.587307428244141
>> b = mean(x+10000)-10000 % Take the mean at a different scale, then shift back
b =
0.587307428244458
Note that when we shift the values of x from the range [0 1] to the range [10000 10001], compute a mean, then subtract the mean offset for comparison, we get a value that differs for the last 3 significant digits. This illustrates how an offset or scaling of data can change the accuracy of calculations performed on it, which is something that has to be accounted for with certain problems.
Look at this article: The Perils of Floating Point. Though its examples are in FORTRAN it has sense for virtually any modern programming language, including MATLAB. Your problem (and solution for it) is described in "Safe Comparisons" section.
type
format long g
This command will show the FULL value of the number. It's likely to be something like 24.00000021321 != 24.00000123124
Try writing
0.1 + 0.1 + 0.1 == 0.3.
Warning: You might be surprised about the result!
Maybe the two numbers are really 24.0 and 24.000000001 but you're not seeing all the decimal places.
Check out the Matlab EPS function.
Matlab uses floating point math up to 16 digits of precision (only 5 are displayed).
I have two geotiff images (saying "A" and "B") imported in Matlab as matrices with Geotiffread. One has different values, while the second has only 0 and 255s.
What I'd like to do is replacing all the 255s with the values inside the other image (or matrix), according to their positions.
A and B differs in size, but they have the same projections.
I tried this:
A (A== 255)= B;
the output is the error:
??? In an assignment A(:) = B, the number of elements in A and B must be the same.
Else, I also tried with the logical approach:
if A== 255
A= B;
end
and nothing happens.
Is there a way to replace the values of A with values of B according to a specific value and the position in the referenced space?
As darthbith put in his comment, you need to make sure that the number of entries you want to replace is the same as the number values you are putting in.
By doing A(A==255)=B you are trying to put the entire matrix B into the subset of A that equals 255.
However, if, as you said, the projections are the same, you can simply do A(A==255) = B(A==255), under the assumption that B is larger or the same size as A.
Some sample code to provide a proof of concept.
A = randi([0,10],10,10);
B = randi([0,4],15,15);
C = A % copy original A matrix for comparison later
A(A==5) = B(A==5); % replace values
C==A % compare original and new
This example code creates two matrices, A is a 10x10 and B is a 15x15 and replaces all values that equal 5 in A with the corresponding values in B. This is shown to be true by doing C==A which shows where the new matrix and the old matrix vary, proving replacement did happen.
It seems to me that you are trying to mask an image with a binary mask. You can do this:
BW = im2bw(B,0.5);
A=A.*BW;
hope it helps
Try A(A==255) = B(A==255). The error is telling you that when you try to assign values to the elements of an array, you cannot give it any more or fewer values than you are trying to assign.
Also, regarding the if statement: if A==255 means the same as if all(A==255), as in, if any elements of A are not 255, false is returned. You can check this at the command line.
If you're really desperate, you can use a pair of nested for loops to achieve this (assuming A and B are the same size and shape):
[a,b] = size(A);
for ii = 1:a
for jj = 1:b
if A(ii,jj) == 255
A(ii,jj) = B(ii,jj);
end
end
end
Lets have matrix A say A = magic(100);. I have seen 2 ways of computing sum of all elements of matrix A.
sumOfA = sum(sum(A));
Or
sumOfA = sum(A(:));
Is one of them faster (or better practise) then other? If so which one is it? Or are they both equally fast?
It seems that you can't make up your mind about whether performance or floating point accuracy is more important.
If floating point accuracy were of paramount accuracy, then you would segregate the positive and negative elements, sorting each segment. Then sum in order of increasing absolute value. Yeah, I know, its more work than anyone would do, and it probably will be a waste of time.
Instead, use adequate precision such that any errors made will be irrelevant. Use good numerical practices about tests, etc, such that there are no problems generated.
As far as the time goes, for an NxM array,
sum(A(:)) will require N*M-1 additions.
sum(sum(A)) will require (N-1)*M + M-1 = N*M-1 additions.
Either method requires the same number of adds, so for a large array, even if the interpreter is not smart enough to recognize that they are both the same op, who cares?
It is simply not an issue. Don't make a mountain out of a mole hill to worry about this.
Edit: in response to Amro's comment about the errors for one method over the other, there is little you can control. The additions will be done in a different order, but there is no assurance about which sequence will be better.
A = randn(1000);
format long g
The two solutions are quite close. In fact, compared to eps, the difference is barely significant.
sum(A(:))
ans =
945.760668102446
sum(sum(A))
ans =
945.760668102449
sum(sum(A)) - sum(A(:))
ans =
2.72848410531878e-12
eps(sum(A(:)))
ans =
1.13686837721616e-13
Suppose you choose the segregate and sort trick I mentioned. See that the negative and positive parts will be large enough that there will be a loss of precision.
sum(sort(A(A<0),'descend'))
ans =
-398276.24754782
sum(sort(A(A<0),'descend')) + sum(sort(A(A>=0),'ascend'))
ans =
945.7606681037
So you really would need to accumulate the pieces in a higher precision array anyway. We might try this:
[~,tags] = sort(abs(A(:)));
sum(A(tags))
ans =
945.760668102446
An interesting problem arises even in these tests. Will there be an issue because the tests are done on a random (normal) array? Essentially, we can view sum(A(:)) as a random walk, a drunkard's walk. But consider sum(sum(A)). Each element of sum(A) (i.e., the internal sum) is itself a sum of 1000 normal deviates. Look at a few of them:
sum(A)
ans =
Columns 1 through 6
-32.6319600960983 36.8984589766173 38.2749084367497 27.3297721091922 30.5600109446534 -59.039228262402
Columns 7 through 12
3.82231962760523 4.11017616179294 -68.1497901792032 35.4196443983385 7.05786623564426 -27.1215387236418
Columns 13 through 18
When we add them up, there will be a loss of precision. So potentially, the operation as sum(A(:)) might be slightly more accurate. Is it so? What if we use a higher precision for the accumulation? So first, I'll form the sum down the columns using doubles, then convert to 25 digits of decimal precision, and sum the rows. (I've displayed only 20 digits here, leaving 5 digits hidden as guard digits.)
sum(hpf(sum(A)))
ans =
945.76066810244807408
Or, instead, convert immediately to 25 digits of precision, then summing the result.
sum(hpf(A(:))
945.76066810244749807
So both forms in double precision were equally wrong here, in opposite directions. In the end, this is all moot, since any of the alternatives I've shown are far more time consuming compared to the simple variations sum(A(:)) or sum(sum(A)). Just pick one of them and don't worry.
Performance-wise, I'd say both are very similar (assuming a recent MATLAB version). Here is quick test using the TIMEIT function:
function sumTest()
M = randn(5000);
timeit( #() func1(M) )
timeit( #() func2(M) )
end
function v = func1(A)
v = sum(A(:));
end
function v = func2(A)
v = sum(sum(A));
end
the results were:
>> sumTest
ans =
0.0020917
ans =
0.0017159
What I would worry about is floating-point issues. Example:
>> M = randn(1000);
>> abs( sum(M(:)) - sum(sum(M)) )
ans =
3.9108e-11
Error magnitude increases for larger matrices
i think a simple way to understand is apply " tic_ toc "function in first and last of your code.
tic
A = randn(5000);
format long g
sum(A(:));
toc
but when you used randn function ,elements of it are random and time of calculation can
different in each cycle CPU calculation .
This better you used a unique matrix whit so large elements to compare time of calculation.
How to calculate the entropy of a file? (Or let's just say a bunch of bytes)
I have an idea, but I'm not sure that it's mathematically correct.
My idea is the following:
Create an array of 256 integers (all zeros).
Traverse through the file and for each of its bytes,
increment the corresponding position in the array.
At the end: Calculate the "average" value for the array.
Initialize a counter with zero,
and for each of the array's entries:
add the entry's difference
to "average" to the counter.
Well, now I'm stuck. How to "project" the counter result in such a way
that all results would lie between 0.0 and 1.0? But I'm sure,
the idea is inconsistent anyway...
I hope someone has better and simpler solutions?
Note: I need the whole thing to make assumptions on the file's contents:
(plaintext, markup, compressed or some binary, ...)
At the end: Calculate the "average" value for the array.
Initialize a counter with zero,
and for each of the array's entries:
add the entry's difference to "average" to the counter.
With some modifications you can get Shannon's entropy:
rename "average" to "entropy"
(float) entropy = 0
for i in the array[256]:Counts do
(float)p = Counts[i] / filesize
if (p > 0) entropy = entropy - p*lg(p) // lgN is the logarithm with base 2
Edit:
As Wesley mentioned, we must divide entropy by 8 in order to adjust it in the range 0 . . 1 (or alternatively, we can use the logarithmic base 256).
A simpler solution: gzip the file. Use the ratio of file sizes: (size-of-gzipped)/(size-of-original) as measure of randomness (i.e. entropy).
This method doesn't give you the exact absolute value of entropy (because gzip is not an "ideal" compressor), but it's good enough if you need to compare entropy of different sources.
To calculate the information entropy of a collection of bytes, you'll need to do something similar to tydok's answer. (tydok's answer works on a collection of bits.)
The following variables are assumed to already exist:
byte_counts is 256-element list of the number of bytes with each value in your file. For example, byte_counts[2] is the number of bytes that have the value 2.
total is the total number of bytes in your file.
I'll write the following code in Python, but it should be obvious what's going on.
import math
entropy = 0
for count in byte_counts:
# If no bytes of this value were seen in the value, it doesn't affect
# the entropy of the file.
if count == 0:
continue
# p is the probability of seeing this byte in the file, as a floating-
# point number
p = 1.0 * count / total
entropy -= p * math.log(p, 256)
There are several things that are important to note.
The check for count == 0 is not just an optimization. If count == 0, then p == 0, and log(p) will be undefined ("negative infinity"), causing an error.
The 256 in the call to math.log represents the number of discrete values that are possible. A byte composed of eight bits will have 256 possible values.
The resulting value will be between 0 (every single byte in the file is the same) up to 1 (the bytes are evenly divided among every possible value of a byte).
An explanation for the use of log base 256
It is true that this algorithm is usually applied using log base 2. This gives the resulting answer in bits. In such a case, you have a maximum of 8 bits of entropy for any given file. Try it yourself: maximize the entropy of the input by making byte_counts a list of all 1 or 2 or 100. When the bytes of a file are evenly distributed, you'll find that there is an entropy of 8 bits.
It is possible to use other logarithm bases. Using b=2 allows a result in bits, as each bit can have 2 values. Using b=10 puts the result in dits, or decimal bits, as there are 10 possible values for each dit. Using b=256 will give the result in bytes, as each byte can have one of 256 discrete values.
Interestingly, using log identities, you can work out how to convert the resulting entropy between units. Any result obtained in units of bits can be converted to units of bytes by dividing by 8. As an interesting, intentional side-effect, this gives the entropy as a value between 0 and 1.
In summary:
You can use various units to express entropy
Most people express entropy in bits (b=2)
For a collection of bytes, this gives a maximum entropy of 8 bits
Since the asker wants a result between 0 and 1, divide this result by 8 for a meaningful value
The algorithm above calculates entropy in bytes (b=256)
This is equivalent to (entropy in bits) / 8
This already gives a value between 0 and 1
I'm two years late in answering, so please consider this despite only a few up-votes.
Short answer: use my 1st and 3rd bold equations below to get what most people are thinking about when they say "entropy" of a file in bits. Use just 1st equation if you want Shannon's H entropy which is actually entropy/symbol as he stated 13 times in his paper which most people are not aware of. Some online entropy calculators use this one, but Shannon's H is "specific entropy", not "total entropy" which has caused so much confusion. Use 1st and 2nd equation if you want the answer between 0 and 1 which is normalized entropy/symbol (it's not bits/symbol, but a true statistical measure of the "entropic nature" of the data by letting the data choose its own log base instead of arbitrarily assigning 2, e, or 10).
There 4 types of entropy of files (data) of N symbols long with n unique types of symbols. But keep in mind that by knowing the contents of a file, you know the state it is in and therefore S=0. To be precise, if you have a source that generates a lot of data that you have access to, then you can calculate the expected future entropy/character of that source. If you use the following on a file, it is more accurate to say it is estimating the expected entropy of other files from that source.
Shannon (specific) entropy H = -1*sum(count_i / N * log(count_i / N))
where count_i is the number of times symbol i occured in N.
Units are bits/symbol if log is base 2, nats/symbol if natural log.
Normalized specific entropy: H / log(n)
Units are entropy/symbol. Ranges from 0 to 1. 1 means each symbol occurred equally often and near 0 is where all symbols except 1 occurred only once, and the rest of a very long file was the other symbol. The log is in the same base as the H.
Absolute entropy S = N * H
Units are bits if log is base 2, nats if ln()).
Normalized absolute entropy S = N * H / log(n)
Unit is "entropy", varies from 0 to N. The log is in the same base as the H.
Although the last one is the truest "entropy", the first one (Shannon entropy H) is what all books call "entropy" without (the needed IMHO) qualification. Most do not clarify (like Shannon did) that it is bits/symbol or entropy per symbol. Calling H "entropy" is speaking too loosely.
For files with equal frequency of each symbol: S = N * H = N. This is the case for most large files of bits. Entropy does not do any compression on the data and is thereby completely ignorant of any patterns, so 000000111111 has the same H and S as 010111101000 (6 1's and 6 0's in both cases).
Like others have said, using a standard compression routine like gzip and dividing before and after will give a better measure of the amount of pre-existing "order" in the file, but that is biased against data that fits the compression scheme better. There's no general purpose perfectly optimized compressor that we can use to define an absolute "order".
Another thing to consider: H changes if you change how you express the data. H will be different if you select different groupings of bits (bits, nibbles, bytes, or hex). So you divide by log(n) where n is the number of unique symbols in the data (2 for binary, 256 for bytes) and H will range from 0 to 1 (this is normalized intensive Shannon entropy in units of entropy per symbol). But technically if only 100 of the 256 types of bytes occur, then n=100, not 256.
H is an "intensive" entropy, i.e. it is per symbol which is analogous to specific entropy in physics which is entropy per kg or per mole. Regular "extensive" entropy of a file analogous to physics' S is S=N*H where N is the number of symbols in the file. H would be exactly analogous to a portion of an ideal gas volume. Information entropy can't simply be made exactly equal to physical entropy in a deeper sense because physical entropy allows for "ordered" as well disordered arrangements: physical entropy comes out more than a completely random entropy (such as a compressed file). One aspect of the different For an ideal gas there is a additional 5/2 factor to account for this: S = k * N * (H+5/2) where H = possible quantum states per molecule = (xp)^3/hbar * 2 * sigma^2 where x=width of the box, p=total non-directional momentum in the system (calculated from kinetic energy and mass per molecule), and sigma=0.341 in keeping with uncertainty principle giving only the number of possible states within 1 std dev.
A little math gives a shorter form of normalized extensive entropy for a file:
S=N * H / log(n) = sum(count_i*log(N/count_i))/log(n)
Units of this are "entropy" (which is not really a unit). It is normalized to be a better universal measure than the "entropy" units of N * H. But it also should not be called "entropy" without clarification because the normal historical convention is to erringly call H "entropy" (which is contrary to the clarifications made in Shannon's text).
For what it's worth, here's the traditional (bits of entropy) calculation represented in C#:
/// <summary>
/// returns bits of entropy represented in a given string, per
/// http://en.wikipedia.org/wiki/Entropy_(information_theory)
/// </summary>
public static double ShannonEntropy(string s)
{
var map = new Dictionary<char, int>();
foreach (char c in s)
{
if (!map.ContainsKey(c))
map.Add(c, 1);
else
map[c] += 1;
}
double result = 0.0;
int len = s.Length;
foreach (var item in map)
{
var frequency = (double)item.Value / len;
result -= frequency * (Math.Log(frequency) / Math.Log(2));
}
return result;
}
Is this something that ent could handle? (Or perhaps its not available on your platform.)
$ dd if=/dev/urandom of=file bs=1024 count=10
$ ent file
Entropy = 7.983185 bits per byte.
...
As a counter example, here is a file with no entropy.
$ dd if=/dev/zero of=file bs=1024 count=10
$ ent file
Entropy = 0.000000 bits per byte.
...
There's no such thing as the entropy of a file. In information theory, the entropy is a function of a random variable, not of a fixed data set (well, technically a fixed data set does have an entropy, but that entropy would be 0 — we can regard the data as a random distribution that has only one possible outcome with probability 1).
In order to calculate the entropy, you need a random variable with which to model your file. The entropy will then be the entropy of the distribution of that random variable. This entropy will equal the number of bits of information contained in that random variable.
If you use information theory entropy, mind that it might make sense not to use it on bytes. Say, if your data consists of floats you should instead fit a probability distribution to those floats and calculate the entropy of that distribution.
Or, if the contents of the file is unicode characters, you should use those, etc.
Calculates entropy of any string of unsigned chars of size "length". This is basically a refactoring of the code found at http://rosettacode.org/wiki/Entropy. I use this for a 64 bit IV generator that creates a container of 100000000 IV's with no dupes and a average entropy of 3.9. http://www.quantifiedtechnologies.com/Programming.html
#include <string>
#include <map>
#include <algorithm>
#include <cmath>
typedef unsigned char uint8;
double Calculate(uint8 * input, int length)
{
std::map<char, int> frequencies;
for (int i = 0; i < length; ++i)
frequencies[input[i]] ++;
double infocontent = 0;
for (std::pair<char, int> p : frequencies)
{
double freq = static_cast<double>(p.second) / length;
infocontent += freq * log2(freq);
}
infocontent *= -1;
return infocontent;
}
Re: I need the whole thing to make assumptions on the file's contents:
(plaintext, markup, compressed or some binary, ...)
As others have pointed out (or been confused/distracted by), I think you're actually talking about metric entropy (entropy divided by length of message). See more at Entropy (information theory) - Wikipedia.
jitter's comment linking to Scanning data for entropy anomalies is very relevant to your underlying goal. That links eventually to libdisorder (C library for measuring byte entropy). That approach would seem to give you lots more information to work with, since it shows how the metric entropy varies in different parts of the file. See e.g. this graph of how the entropy of a block of 256 consecutive bytes from a 4 MB jpg image (y axis) changes for different offsets (x axis). At the beginning and end the entropy is lower, as it part-way in, but it is about 7 bits per byte for most of the file.
Source: https://github.com/cyphunk/entropy_examples. [Note that this and other graphs are available via the novel http://nonwhiteheterosexualmalelicense.org license....]
More interesting is the analysis and similar graphs at Analysing the byte entropy of a FAT formatted disk | GL.IB.LY
Statistics like the max, min, mode, and standard deviation of the metric entropy for the whole file and/or the first and last blocks of it might be very helpful as a signature.
This book also seems relevant: Detection and Recognition of File Masquerading for E-mail and Data Security - Springer
Here's a Java algo based on this snippet and the invasion that took place during the infinity war
public static double shannon_entropy(File file) throws IOException {
byte[] bytes= Files.readAllBytes(file.toPath());//byte sequence
int max_byte = 255;//max byte value
int no_bytes = bytes.length;//file length
int[] freq = new int[256];//byte frequencies
for (int j = 0; j < no_bytes; j++) {
int value = bytes[j] & 0xFF;//integer value of byte
freq[value]++;
}
double entropy = 0.0;
for (int i = 0; i <= max_byte; i++) {
double p = 1.0 * freq[i] / no_bytes;
if (freq[i] > 0)
entropy -= p * Math.log(p) / Math.log(2);
}
return entropy;
}
usage-example:
File file=new File("C:\\Users\\Somewhere\\In\\The\\Omniverse\\Thanos Invasion.Log");
int file_length=(int)file.length();
double shannon_entropy=shannon_entropy(file);
System.out.println("file length: "+file_length+" bytes");
System.out.println("shannon entropy: "+shannon_entropy+" nats i.e. a minimum of "+shannon_entropy+" bits can be used to encode each byte transfer" +
"\nfrom the file so that in total we transfer atleast "+(file_length*shannon_entropy)+" bits ("+((file_length*shannon_entropy)/8D)+
" bytes instead of "+file_length+" bytes).");
output-example:
file length: 5412 bytes
shannon entropy: 4.537883805240875 nats i.e. a minimum of 4.537883805240875 bits can be used to encode each byte transfer
from the file so that in total we transfer atleast 24559.027153963616 bits (3069.878394245452 bytes instead of 5412 bytes).
Without any additional information entropy of a file is (by definition) equal to its size*8 bits. Entropy of text file is roughly size*6.6 bits, given that:
each character is equally probable
there are 95 printable characters in byte
log(95)/log(2) = 6.6
Entropy of text file in English is estimated to be around 0.6 to 1.3 bits per character (as explained here).
In general you cannot talk about entropy of a given file. Entropy is a property of a set of files.
If you need an entropy (or entropy per byte, to be exact) the best way is to compress it using gzip, bz2, rar or any other strong compression, and then divide compressed size by uncompressed size. It would be a great estimate of entropy.
Calculating entropy byte by byte as Nick Dandoulakis suggested gives a very poor estimate, because it assumes every byte is independent. In text files, for example, it is much more probable to have a small letter after a letter than a whitespace or punctuation after a letter, since words typically are longer than 2 characters. So probability of next character being in a-z range is correlated with value of previous character. Don't use Nick's rough estimate for any real data, use gzip compression ratio instead.