I have seen in some codes that people define a variable and assign values like 1e-8 or 1e5.
for example
const int MAXN = 1e5 + 123;
What are these numbers? I couldn't find any thing on the web...
1e5 is a number expressed using scientific notation and it means 1 multiplied by 10 to the 5th power (the e meaning 'exponent')
so 1e5 equals 1*100000and is equal to 100000, the three notations are interchangeable meaning the same.
1e5 means 1 × 105.
Similarly, 12.34e-9 means 12.34 × 10−9.
Generally, AeB means A × 10B.
this is scientific notation for 10^5 = 100000
1e5 is 100000. 5 stand for the amount of zeros you add in behind that number. For example, lets say I have 1e7. I would put 7 zeros behind 1 so it will become 10,000,000. But lets say that the number is 1.234e6. You would still add 6 zeros at the end of the number so it's 1.234000000, but since there is that decimal, you would have to move it to the right 6 times since it's e6.
The values like:
1e-8 or 1e5
means;
1e-8 = 1 * 10^(-8)
And
1e5 = 1 * 10
Related
Here is an example of convolution given:
I have two questions here:
Why is the vector 𝑥 padded with two 0s on each side? As, the length of kernel ℎ is 3. If 𝑥 is padded with one 0 on each side, the middle element of convolution output would be within the range of the length of 𝑥, why not one 0 on each side?
Explain the following output to me:
>> x = [1, 2, 1, 3];
>> h = [2, 0, 1];
>> y = conv(x, h, 'valid')
y =
3 8
>>
What is valid doing here in the context of the previously shown mathematics on vectors 𝑥 and ℎ?
I can't speak as to the amount of zero padding that is proper .... That being said, any zero padding is making up data that is not there. This isn't necessarily wrong, but you should be aware that the values computing this information may be biased. Sometimes you care about this, sometimes you don't. Introducing 1 zero (in this case) would leave the middle kernel value always in the data, but why should that be a stopping criteria? Importantly, adding on 2 zeros still leaves one multiplication of values that are actually present in the data and the kernel (the x[0]*h[0] and x[3]*h[2] - using 0 based indexing). Adding on a 3rd zero (or more) would just yield zeros in the output since 3 is the length of the kernel. In other words zero padding will always yield an output that is partially based on the actual data (but not completely) for any zero padding from n=1 to n = length(h)-1 (in this case either 1 or 2).
Even though zero padding with length 2 or 1 still has multiplications based on real data, some values are summed over "fake" data (those multiplied with a padded zero). In this case Matlab gives you 3 options for how you want the data returned. First, you can get the full convolution, which includes values that are biased because they include adding in 0 values that aren't really in the data. Alternatively you can get same, which means the length of the output is the length of the data y = [4 3 8 1]. This corresponds to 1 zero but note that for longer kernels you could technically get other lengths between full and same, Matlab just doesn't return those for you.
Finally, and probably most important to understand out of all this, you have the valid option. In your example only 2 samples of the output are computed from summations that occur only from multiplications over real data (i.e. from multiplying samples of the kernel with samples from x and not from zeros). More specifically:
y[2] = h[2]*x[0] + h[1]*x[1] + h[2]*x[2] = 3 //0 based indexing like example
y[3] = h[2]*x[1] + h[1]*x[2] + h[2]*x[3] = 8
Note none of the other y values are computed with only h and x, they all involve a padded zero which is not necessarily indicative of the real data. For example:
y[4] = h[2]*x[2] + h[1]*x[3] + h[2]*0 <= padded zero
Im looking for a way to divide a certain matrix elements with its lowest common divisor.
for example, I have vectors
[0,0,0; 2,4,2;-2,0,8]
I can tell the lowest common divisor is 2, so the matrix after the division will be
[0,0,0;1,2,1;-1,0,4]
What is the built in method that can compute this?
Thanks in advance
p.s. I personally do not like using loops for this computation, it seems like there is built in computation that can perform matrix element division.
Since you don't like loops, how about recursive functions?
iif = #(varargin) varargin{2 * find([varargin{1:2:end}], 1, 'first')}();
gcdrec=#(v,gcdr) iif(length(v)==1,v, ...
v(1)==1,1, ...
length(v)==2,#()gcd(v(1),v(2)), ...
true,#()gcdr([gcd(v(1),v(2)),v(3:end)],gcdr));
mygcd=#(v)gcdrec(v(:)',gcdrec);
A=[0,0,0; 2,4,2;-2,0,8];
divisor=mygcd(A);
A=A/divisor;
The first function iif will define an inline conditional construct. This allows to define a recursive function, gcdrec, to find the greatest common divisor of your array. This iif works like this: it tests whether the first argument is true, if it is, then it returns the second argument. Otherwise it tests the third argument, and if that's true, then it returns the fourth, and so on. You need to protect recursive functions and sometimes other quantities appearing inside it with #(), otherwise you can get errors.
Using iif the recursive function gcdrec works like this:
if the input vector is a scalar, it returns it
else if the first component of the vector is 1, there's no chance to recover, so it returns 1 (allows quick return for large matrices)
else if the input vector is of length 2, it returns the greatest common divisor via gcd
else it calls itself with a shortened vector, in which the first two elements are substituted with their greatest common divisor.
The function mygcd is just a front-end for convenience.
Should be pretty fast, and I guess only the recursion depth could be a problem for very large problems. I did a quick timing check to compare with the looping version of #Adriaan, using A=randi(100,N,N)-50, with N=100, N=1000 and N=5000 and tic/toc.
N=100:
looping 0.008 seconds
recursive 0.002 seconds
N=1000:
looping 0.46 seconds
recursive 0.04 seconds
N=5000:
looping 11.8 seconds
recursive 0.6 seconds
Update: interesting thing is that the only reason that I didn't trip the recursion limit (which is by default 500) is that my data didn't have a common divisor. Setting a random matrix and doubling it will lead to hitting the recursion limit already for N=100. So for large matrices this won't work. Then again, for small matrices #Adriaan's solution is perfectly fine.
I also tried to rewrite it to half the input vector in each recursive step: this indeed solves the recursion limit problem, but it is very slow (2 seconds for N=100, 261 seconds for N=1000). There might be a middle ground somewhere, where the matrix size is large(ish) and the runtime's not that bad, but I haven't found it yet.
A = [0,0,0; 2,4,2;-2,0,8];
B = 1;
kk = max(abs(A(:))); % start at the end
while B~=0 && kk>=0
tmp = mod(A,kk);
B = sum(tmp(:));
kk = kk - 1;
end
kk = kk+1;
This is probably not the fastest way, but it will do for now. What I did here is initialise some counter, B, to store the sum of all elements in your matrix after taking the mod. the kk is just a counter which runs through integers. mod(A,kk) computes the modulus after division for each element in A. Thus, if all your elements are wholly divisible by 2, it will return a 0 for each element. sum(tmp(:)) then makes a single column out of the modulo-matrix, which is summed to obtain some number. If and only if that number is 0 there is a common divisor, since then all elements in A are wholly divisible by kk. As soon as that happens your loop stops and your common divisor is the number in kk. Since kk is decreased every count it is actually one value too low, thus one is added.
Note: I just edited the loop to run backwards since you are looking for the Greatest cd, not the Smallest cd. If you'd have a matrix like [4,8;16,8] it would stop at 2, not 4. Apologies for that, this works now, though both other solutions here are much faster.
Finally, dividing matrices can be done like this:
divided_matrix = A/kk;
Agreed, I don't like the loops either! Let's kill them -
unqA = unique(abs(A(A~=0))).'; %//'
col_extent = [2:max(unqA)]'; %//'
B = repmat(col_extent,1,numel(unqA));
B(bsxfun(#gt,col_extent,unqA)) = 0;
divisor = find(all(bsxfun(#times,bsxfun(#rem,unqA,B)==0,B),2),1,'first');
if isempty(divisor)
out = A;
else
out = A/divisor;
end
Sample runs
Case #1:
A =
0 0 0
2 4 2
-2 0 8
divisor =
2
out =
0 0 0
1 2 1
-1 0 4
Case #2:
A =
0 3 0
5 7 6
-5 0 21
divisor =
1
out =
0 3 0
5 7 6
-5 0 21
Here's another approach. Let A be your input array.
Get nonzero values of A and take their absolute value. Call the resulting vector B.
Test each number from 1 to max(B), and see if it divides all entries of B (that is, if the remainder of the division is zero).
Take the largest such number.
Code:
A = [0,0,0; 2,4,2;-2,0,8]; %// data
B = nonzeros(abs(A)); %// step 1
t = all(bsxfun(#mod, B, 1:max(B))==0, 1); %// step 2
result = find(t, 1, 'last'); %// step 3
For my experiment I have 20 categories which contain 9 pictures each. I want to show these pictures in a pseudo-random sequence where the only constraint to randomness is that one image may not be followed directly by one of the same category.
So I need something similar to
r = randi([1 20],1,180);
just with an added constraint of two numbers not directly following each other. E.g.
14 8 15 15 7 16 6 4 1 8 is not legitimate, whereas
14 8 15 7 15 16 6 4 1 8 would be.
An alternative way I was thinking of was naming the categories A,B,C,...T, have them repeat 9 times and then shuffle the bunch. But there you run into the same problem I think?
I am an absolute Matlab beginner, so any guidance will be welcome.
The following uses modulo operations to make sure each value is different from the previous one:
m = 20; %// number of categories
n = 180; %// desired number of samples
x = [randi(m)-1 randi(m-1, [1 n-1])];
x = mod(cumsum(x), m) + 1;
How the code works
In the third line, the first entry of x is a random value between 0 and m-1. Each subsequent entry represents the change that, modulo m, will give the next value (this is done in the fourth line).
The key is to choose that change between 1 and m-1 (not between 0 and m-1), to assure consecutive values will be different. In other words, given a value, there are m-1 (not m) choices for the next value.
After the modulo operation, 1 is added to to transform the range of resulting values from 0,...,m-1 to 1,...,m.
Test
Take all (n-1) pairs of consecutive entries in the generated x vector and count occurrences of all (m^2) possible combinations of values:
count = accumarray([x(1:end-1); x(2:end)].', 1, [m m]);
imagesc(count)
axis square
colorbar
The following image has been obtained for m=20; n=1e6;. It is seen that all combinations are (more or less) equally likely, except for pairs with repeated values, which never occur.
You could look for the repetitions in an iterative manner and put new set of integers from the same group [1 20] only into those places where repetitions have occurred. We continue to do so until there are no repetitions left -
interval = [1 20]; %// interval from where the random integers are to be chosen
r = randi(interval,1,180); %// create the first batch of numbers
idx = diff(r)==0; %// logical array, where 1s denote repetitions for first batch
while nnz(idx)~=0
idx = diff(r)==0; %// logical array, where 1s denote repetitions for
%// subsequent batches
rN = randi(interval,1,nnz(idx)); %// new set of random integers to be placed
%// at the positions where repetitions have occured
r(find(idx)+1) = rN; %// place ramdom integers at their respective positions
end
Say I have matrices A and B. I want to create a third matrix C where
A = [1,0,0,1]
B = [1,0,1,0]
C = [11, 00, 01, 10]
Is there such a function in Matlab? If not how would I go about creating a function that does this?
Edit: C is not literal numbers. They are concatenated values of A,B element-wise.
Edit2: The actual problem I am dealing with is I have 10 large matrices of the size [x,y] where x,y > 1000. The elements in these matrices all have 0s and 1s. No other number. What I need to accomplish is have the element [x1,y1] in matrix 1 to be appended to the element in [x1,y1] of matrix 2. and then that value to be appended to [x1,y1] of matrix 3.
Another example:
A = [1,1,1,1;
0,0,0,0]
B = [0,0,0,0;
1,1,1,1]
C = [1,0,1,0;
0,1,0,1]
And I need a matrix D where
D = [101, 100, 101, 101; 010, 011, 010, 011]
I recommend that you avoid manipulating binary numbers as strings where possible. It seems tempting, and there are cases where matlab provides a more elegant solution if you treat a binary number as a string, but you cannot perform binary arithmetic on strings. You can always work with integers as if they were binary (they are stored as bits in your machine after all), and then just display them as binary numbers using dec2bin when necessary.
A = [1,0,0,1]
B = [1,0,1,0]
C = bitshift(A,1)+B;
display(dec2bin(C));
In the other case you show in your question you could use:
A = [1,1,1,1; 0,0,0,0];
B = [0,0,0,0; 1,1,1,1];
C = [1,0,1,0; 0,1,0,1];
D = bitshift(A,2) + bitshift(B,1) + C;
You can also convert an arbitrary length row vector of zeros and ones into its decimal equivalent by defining this simple function:
mat2dec = #(x) x*2.^(size(x,2)-1:-1:0)';
This will also work for matrices too. For example
>> M = [0 0 1; 0 1 1; 0 1 0; 1 1 0; 1 1 1; 1 1 0; 1 0 0];
>> dec2bin(mat2dec(M))
ans =
001
011
010
110
111
110
100
In my experience, treating binary numbers as strings obfuscates your code and is not very flexible. For example, try adding two binary "strings" together. You have to use bin2dec every time, so why not just leave the numbers as numbers until you want to display them? You have already run into some of the issues caused by strings of different lengths too. You will be amazed how one simple change can break everything when treating numbers as strings. The worst part is that an algorithm may work great for one set of data and not for another. If all I test with is two-bit binary numbers and a three-bit number somehow sneaks its way in, I may not see an error, but my results will be inexplicably incorrect. I realize that this is a very subjective issue, and I think that I definitely stand in the minority on StackOverflow, so take it for what it's worth.
It depends how you want the output formatted. You could apply bitshift to the numerical values and convert to binary:
>> b = dec2bin(bitshift(A,1)+B)
b =
11
00
01
10
For a general matrix Digits:
>> Digits
Digits =
1 0 0 0
0 1 0 0
1 0 1 1
1 1 0 0
1 0 0 0
1 0 1 1
1 1 0 0
1 0 1 0
0 1 1 1
0 1 1 1
>> Ddec = D*(2.^(size(D,2)-1:-1:0))'; % dot product
>> dec2bin(Ddec)
ans =
1000
0100
1011
1100
1000
1011
1100
1010
0111
0111
Another way to write that is dec2bin(sum(D .* repmat(2.^(size(D,2)-1:-1:0),size(D,1),1),2)).
For your larger problem, with 10 large matrixes (say M1, M2, ..., M10), you can build the starting Digits matrix by:
Digits = [M1(:) M2(:) M3(:) M4(:) M5(:) M6(:) M7(:) M8(:) M9(:) M10(:)];
If that is reverse the order of the digits, just do Digits = fliplr(Digits);.
If you would rather not reshape anything you can compute the matrix decimal values from the matrices of digits as follows:
M = cat(3,A,B,C);
Ddec = sum(bsxfun(#times,M,permute(2.^(size(M,3)-1:-1:0),[1 3 2])),3)
I see some quite extensive answers so perhaps this is thinking too simple, but how about just this assuming you have vectors of ones and zeros representing your binary numbers:
A = [1,0,0,1];
B = [1,0,1,0];
C = 10*A + B
This should give you the numbers you want, you may want to add leading zeros. Of course this method can easily be expanded to append multiple matrices, you just need to make sure there is a factor (base) 10 between them before adding.
I'm using the matrix as an initial population for multiobjective optimization using NSGA-II in matlab. The size of my chromosome vector,(C), is 1x192 and each gene must be within the range 0<=gene<=40 and the genes must be integers. The rule is that the sum of groupings of 6 genes must be less or equal to 40.that is:
sum(reshape(6,[]))<=40
I've use the following code but it outputs either an all-zero population matrix(population matrix=vertical concatenation of 500 chromosomes) or a matrix that does not satisfy the rule:
X=zeros(500,192);
while i<501
r=randi(40,6,32);
if nnz(((sum(r))./40)>1)==0
X(i,:)=reshape(r,1,[]);
i=i+1;
clear r;
else
clear r;
end
end
It is also taking forever to exit the while loop.
What am I doing wrong here? Is there another way of doing the above?
I've also tried this:
i=1;
while i<17500
r=randi([1,40],6,1);
s=sum(r);
if s<=40
X(:,i)=r;
i=i+1;
else
clear r;
end
end
X=unique(X','rows')';
A=X(:,randperm(size(X,2)));
A=X(randperm(size(X,1)),:);
The above tries to create random columns that will be reshaped to the population matrix. But the numbers are repeating; i.e in the 17500(16448 after removing duplicate columns) columns there is no occurrence of the numbers 37 and 40. Is there any way I can optimize the spread of the generated random numbers?
#0x90
I have a vector,called 'chromosome', of size 1x192 and each successive group of 6 members(called phenotype) must sum to 40 or less. To make it clearer:
That is, each P must be an integer in the range 0 to 40 inclusive and the sum at each phenotype must be <=40. I need 500 chromosomes like this.
I hope it makes sense now. ><
You should use randi([min,max],n,m). randint is going to be deprecated.
>> r = randi([1,4],3,2)
r =
3 3
2 2
4 4