In the first step I generated a sequence of bits (0,1)..
I used a randi command x = randi([0 1],1,3) to generate random bits
I stuck with these 2 steps :
Divide sequence by 3 bits into 8 levels
[000, 001, 010, 011, 100, 101, 110, 111]
For each quantum level assigns amplitude value from the range [-2, 2]
I won't provide the full source code to leave a bit of the homework for you, but I will give you some hints:
randi() is creating a sequence of 0 and 1 floating point numbers
Look at the documentation of function bitpack. This allows you to pack your bits from array elements into a single byte. Be aware that you need to provide an 8 element array of "bits" to fill a byte. User 'uint8' as the class argument.
before passing the array of floating point numbers to bitpack you have to convert it to a logical array by using the logical() function.
look at the documentation of linspace() to create an array with 8 elements containing your equally spaces amplitude values
lookup the amplitude value in this array for each "digital" value.
Related
I am essentially trying to generate all possible nxm matrices. I have seen some codes in R and Python that kind of does this with a single function, but I cant find anything similar for matlab :(
here's a way to do that in matlab:
n=4;
m=3;
for c=0:2^(n*m)-1
A(:,c+1)=str2double(split(dec2bin(c,n*m),'',1));
end
A(1,:)=[]; A(end,:)=[];
A=reshape(A,n,m,[]);
the logic, it doesnt matter if it is an nxm array or a 1D vector of length nxm, we can later reshape the vector to an nxm array (last line). Then, the # of possible permutation for a binary string of Length L is just all the decimal # from 0 to 2^L-1 transformed to binary string of length nxm. This is what the dec2bin(...) function does. Then we get a long string (for example '010001010111') that needs to be split to individual elements('0','1',...) , so we can later convert this strings to numbers using str2double. The split(...) function does that but generates NaN at the edges, so we get rid of them in the row before the last one...
so for this example I chose n=4, m=3, which generates just 2^(nxm)=4096 possible array...
you can see them if if you try A(:,:,j) with j being a # from 1 to 4096
I'm new to Matlab and for an assignment my professor is having the class write (complete really) a custom Matlab function for generating a histogram from a set of data. Essentially a new vector is being created, L which is being updated with the information from a 2D matrix M. The first column of L contains the information from M(i,j) and in a second column contains the count (total) of M(i,j) in the data set. I'm in need of some direction as to how to proceed next.
Below is where I'm at thus far:
function L = hist_count(M)
L = [ [0:255' zeros(256,1) ];
for i = 1:size(M,1)
for j = 1:size(M,2)
L(double(M(i,j))+1,2) = <<finish code here>>;
end
end
figure;
plot(L(:1),L(:2));
The <<finish code here>> section is where I'm stuck. I understand everything up to the point where I need to update L with the information.
Assistance is appreciated.
Note: Your initialization of your histogram L has the brackets mismatched.
Remove the second [ bracket in the code. In addition, the creation of the 0:255 vector is incorrect. Doing 0:255' transposes the single constant of 255, which means that it will still create a horizontal vector of 0:255 which will make the code fail. You should surround the creation of this vector with parantheses, then transpose that result. Therefore:
L = [ (0:255)' zeros(256,1) ];
Now onto your actual problem. Judging by your initialization of the histogram, there are 256 possible values so your input is most likely of type uint8, which means that the values in your data will only be from [0-255] in steps of 1. Recall that a histogram records the total number of times you see a value. In this case, you have a two column matrix where the first column tells you the value you want to examine and the second column tells you how many times you see that value in your data. Therefore, each row tells you which value you are examining in your data as well as how many times you have seen that value in your data. Note that the counts are all initialized to zero, so the logic is that every time you see a value, you need to access the right row corresponding to the data point, then increment that value by 1.
Therefore, the line is simply just accessing the current count and adding 1 to it... you then store it back:
L(double(M(i,j))+1,2) = L(double(M(i,j))+1,2) + 1;
M(i,j) is the value found at location (i,j) in your 2D data. The last question you have is why cast the intensity to double and add 1? You cast to double because the input may be an integer type. This means that any values that are beyond the dynamic range of the type will get saturated. Because your input is type uint8, any values beyond 255 will saturate to 255. In MATLAB, we index into rows and columns of a matrix starting at 1 and because the values will potentially start at value 0, this corresponds to row 1 of your histogram so you have to offset by 1. When we get to the most extreme case of value 255 for type uint8 for example, adding 1 to this using the native uint8 will saturate to 255, which means that the values of 254 and 255 get lumped into the same bin. Therefore, you must convert to some type that extends beyond the limits of uint8 and then you add by 1 to avoid saturation. double is usually done here as a default as it has higher precision than uint8, but any type that is higher than uint8 in precision is suitable.
I have a matrix 1x5000 with numbers. Now I am interested in getting values from the matrix in different positions, more precisely in six different places of the matrix. The places should be based on the length, these are the numbers I want to get out:
Number in 1/6 of the matrix length
Number in 2/6 of the matrix length
Number in 3/6 of the matrix length
Number in 4/6 of the matrix length
Number in 5/6 of the matrix length
Number in 6/6 of the matrix length
These values could be printed out in another matrix, so assume the matrix is 1x5000, 3/6 would give the number in the middle of the matrix. I am new in Matlab and therefore the help is much appreciated!
Since your question is unclear I can try to give you an example.
First of all you can use numel function to get matrix's size.
It's easy to get necessary element in Matlab: you can address directly to any element if you know its number (index). So:
x(100) returns 100th element.
Now you got size and know what to do. Last moment - what to do if numel(x) / 6 return non integer?
You can use rounding functions: ceil, floor or round.
index = ceil(numel(x)/6) %if you want NEXT element always
result = x(index)
Next step: there are a lot of ways to divide data. For example now you have just 6 numbers (1/6, 2/6 and so on) but what if there are 1000 of them? You can't do it manually. So you can use for loop, or you can use matrix of indexes or perfect comment Stewie Griffin.
My example:
divider = [6 5 4 3 2 1] % lets take 1/6 1/5 1/4 1/3 1/2 and 1/1
ind = ceil( numel(x)./divider)
res = x(ind)
The colon notation in MATLAB provides an easy way to extract a range of elements from v:
v(3:7) %Extract the third through the seventh elements
You could either manually input range or use a function to convert fractions into suitable ranges
Given a positive integer n, I want to generate all possible n bit combinations in matlab.
For ex : If n=3, then answer should be
000
001
010
011
100
101
110
111
How do I do it ?
I want to actually store them in matrix. I tried
for n=1:2^4
r(n)=dec2bin(n,5);
end;
but that gave error "In an assignment A(:) = B, the number of elements in A and B must be the same.
Just loop over all integers in [0,2^n), and print the number as binary. If you always want to have n digits (e.g. insert leading zeros), this would look like:
for ii=0:2^n-1,
fprintf('%0*s\n', n, dec2bin(ii));
end
Edit: there are a number of ways to put the results in a matrix. The easiest is to use
x = dec2bin(0:2^n-1);
which will produce an n-by-2^n matrix of type char. Each row is one of the bit strings.
If you really want to store strings in each row, you can do this:
x = cell(1, 2^n);
for ii=0:2^n-1,
x{ii} = dec2bin(ii);
end
However, if you're looking for efficient processing, you should remember that integers are already stored in memory in binary! So, the vector:
x = 0 : 2^n-1;
Contains the binary patterns in the most memory efficient and CPU efficient way possible. The only trade-off is that you will not be able to represent patterns with more than 32 of 64 bits using this compact representation.
This is a one-line answer to the question which gives you a double array of all 2^n bit combinations:
bitCombs = dec2bin(0:2^n-1) - '0'
So many ways to do this permutation. If you are looking to implement with an array counter: set an array of counters going from 0 to 1 for each of the three positions (2^0,2^1,2^2). Let the starting number be 000 (stored in an array). Use the counter and increment its 1st place (2^0). The number will be 001. Reset the counter at position (2^0) and increase counter at 2^1 and go on a loop till you complete all the counters.
I am trying to compare an array of doubles to a scalar double for equality, but equality is never recognized under certain circumstances. I suspect that this has to do with the way the double is represented (e.g., 1.0 vs 1.00), but I can't figure it out.
For instance, I have generated an array composed of thousands of double values, the last few of which at some instant in time are given by
10.6000
-11.0000
10.2000
22.6000
3.4000
If I test for equality to 10.2 (or 10.2000) by the command array==10.2 (or array=10.2000), I return an array of 0s. If I place the values shown into an array manually (e.g., array=[10.6000 -11.0000 10.2000 22.6000 3.4000]), then the command is successful (i.e., array==10.2 returns 0 0 1 0 0). Could someone please explain why the equality succeeds if I input the values manually, but fails if the array is generated in the context of a program? I am able to rectify the comparison failure by using an approximate rather than an exact comparison (e.g., (array<10.20001) & (array>10.19999)), but this seems unsatisfying.
Edit: The values in the array are generated by iterative addition or subtraction of a constant double (e.g., 0.2). The modulus of this array by 0.2 should therefore be everywhere equal to 0. In fact, the modulus of each element is equal to either 0 or 0.2, as shown below for the above sequence of numbers in the array:
mod(array,0.2)
...
0.2000
0
0.2000
0.2000
0
Again, if the values are placed in an array manually and the modulus is taken, the expected value of all 0s is obtained.
The reason is MATLAB truncated the numbers in the array to preserve only 4 digits after the decimal point when displaying,. That is, the real value of your array may be [10.600000000001, -10.99999999999, ...]. You are right, this is due to the internal representation of floating-point numbers in computer, which may cause tiny errors in the computations.
Personally I think there are two solutions, one is approximate matching like you did, while the other is to round the array up first (say, with this tool from FileExchange), and then do exact matching.
Something is probably in single precision somewhere, and double elsewhere. The binary representation of e.g. 10.2 in each is different as they terminate after a different number of bits. Thus they are different:
>> if (single(10.2) == 10.2) disp('honk'); end
>> if (single(10.2) == single(10.2)) disp('honk'); end
honk
You'll need to check for equality within some small difference:
eps = 0.001;
result = abs(array-10.2) < eps;
You can find the precision used in an array using whos:
>> whos A
Name Size Bytes Class Attributes
A 1x2 8 single
Create a MATLAB function file that will accept a modulo values (from 3 to 9; that is Z3 to Z9) and
will output the least possible value describe by the modulo conditions.
Sample Simulation:
Z = [ 3 4 5 ]; % Modulo Z3, Z4, and Z5
r = [ 2 1 4 ]; % Remainder Values
Least Possible Value is 29.
The Z inputs must be an array matrix ... where in you can type any number from 3 to 9 .... and you can type 3,4,5,6,7,8,9 in any order, in any pairings or groupings ...
The r inputs should be equal to the number of z inputs too...
the output should yield the least possible value though modulo conditions...