I am lost as to how to increment this vector. I know that the value from each number squared increase by odd numbers starting from 3. From 1^2 to 2^2 we have a space of three, from 2^2 to 3^2 we have a space of 5 in between and then 7 in between for 3^2 to 4^2 and 9 in between 4^2 and 5^2 and so on and so forth. But I just can't think of how I would write those increments for a general case as I have to do in this given problem.
You cannot define d in the a:d:b vector with one formula because it changes constantly. Therefore, you need to define your vector as [1 3 5 7 ... 2n+1] and square it.
(1:2:2*n+1).^2
ans =
1 9 25 49 81 121
Related
I have 3 sets of array each contains 12 elements of same type
a=[1 1 1 1 1 1 1 1 1 1 1];
b=[2 2 2 2 2 2 2 2 2 2 2];
c=[3 3 3 3 3 3 3 3 3 3 3];
I have to find how many ways it can be picked up if I need to pickup 12 items at a time
here, 1 1 2 is same as 2 1 1
I found this link Generating all combinations with repetition using MATLAB.
Ho this can be done in matlab within reasonable time.
is that way correct
abc=[a b c];
allcombs=nmultichoosek(abc,12);
combs=unique(allcombs,'rows');
If you only need to find the number of ways to select the items, then using generating functions is a way to very efficiently compute that, even for fairly large values of N and k.
If you are not familiar with generating functions, you can read up on the math background here:
http://mathworld.wolfram.com/GeneratingFunction.html
and here:
http://math.arizona.edu/~faris/combinatoricsweb/generate.pdf
The solution hinges on the fact that the number of ways to choose k items from 36, with each of 3 items repeated 12 times, can be determined from the product of the generating function:
g(x) = 1 + x + x^2 + x^3 + ... + x^12
with itself 3 times. The 12 comes from the fact the elements are repeated 12 times (NOT from the fact you are choosing 12), and multiplying by itself 3 times is because there are three different sets of elements. The number of ways to choose 12 elements is then just the coefficient of the power of x^12 in this product of polynomials (try it for smaller examples if you want to prove to yourself that it works).
The great thing about that is that MATLAB has a simple function conv for multiplying polynomials:
>> g = ones(1,13); %% array of 13 ones, for a 12th degree polynomial with all `1` coefficents
>> prod = conv(g, conv(g, g)); %% multiply g by itself 3 times, as a polynomial
>> prod(13)
ans =
91
So there are 91 ways to select 12 elements from your list of 36. If you want to select 11 elements, that's z(12) = 78. If you want to select 13 elements, that's z(14) = 102.
Finally, if you had different numbers of elements in the sets, say 10 1's, 12 2's, and 14 3's, then you would have 3 distinct polynomials of the same form, 1 + x + x^2 + ..., with degrees 10, 12 and 14 respectively. Inspecting the coefficient of the degree k term again gives you the number of ways to choose k elements from this set.
Basically the sum function calculate the sum of the columns, that is to say if we have a 4x4 matrix we would get a 1X4 vector
A = magic(4)
A =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
sum(A)
ans =
34 34 34 34
But if I want to get the Summation of the rows then i have 2 methods, the first is to get the transpose of the matrix then get the summation of the transposed matrix,and finally get the transpose of the result...., The Second method is to use dimension argument for the Sum function "sum(A, 2)"
A = magic(4)
A =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
sum(A,2)
ans =
34
34
34
34
The problem is here I cannot understand how this is done, If anyone could please tell me the idea/concept behind this method,
It's hard to tell exactly how sum internally works, but we can guess it does something similar to this.
Matlab stores matrices (or N-dimensional arrays) in memory using column-major order. This means the order for the elements in memory for a 3 x 4 matrix is
1 4 7 10
2 5 8 11
3 6 9 12
So it first stores element (1,1), then (1,2), then (13), then (2,1), ...
In fact, this is the order you use when you apply linear indexing (that is, index a matrix with a single number). For example, let
A = [7 8 6 2
9 0 3 5
6 3 2 1];
Then A(4) gives 8.
With this in mind, it's easy to guess that what sum(A,1) does is traverse elements consecutively: A(1)+A(2)+A(3) to obtain the sum of the first column, then A(4)+A(5)+A(6) to sum the second column, etc. In contrast, sum(A,2) proceeds in steps of size(A,1) (3 in this example): A(1)+A(4)+A(7)+A(10) to compute the sum of the first row, etc.
As a side note, this is probably related with the observed fact that sum(A,1) is faster than sum(A,2).
I'm really not sure what you are asking. sum takes two inputs, the first of which is a multidimensional array A, say.
Now let's take sA = size(A), and d between 1 and ndims(A).
To understand what B = sum(A,d) does, first we find out what the size of B is.
That's easy, sB = sA; sB(d) = 1;. So in a way, it will "reduce" the size of A along dimension d.
The rest is trivial: every element in B is the sum of elements in A along dimension d.
Basically, sum(A) = sum(A,1) which outputs the sum of the columns in the matrix. 1 indicates the columns. So, sum(A,2) outputs the sum of the rows in the matrix. 2 indicating the rows. More than that, the sum command will output the entire matrix because there is only 2 dimensions (rows and columns)
I have a matrix of 2d lets assume the values of the matrix
a =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
17 24 1 8 15
11 18 25 2 9
This matrix is going to be divided into three different matrices randomly let say
b =
17 24 1 8 15
23 5 7 14 16
c =
4 6 13 20 22
11 18 25 2 9
d =
10 12 19 21 3
17 24 1 8 15
How can i know the index of the vectors in matrix d for example in the original matrix a,note that the values of the matrix can be duplicated.
for example if i want to know the index of {10 12 19 21 3} in matrix a?
or the index of {17 24 1 8 15} in matrix a,but for this one should return only on index value?
I would appreciate it so much if you can help me with this. Thank you in advance
You can use ismember with the 'rows' option. For example:
tf = ismember(a, c, 'rows')
Should produce:
tf =
0
0
1
0
0
1
To get the indices of the rows, you can apply find on the result of ismember (note that it's redundant if you're planning to use this vector for matrix indexing). Here find(tf) return the vector [3; 6].
If you want to know the number of the row in matrix a that matches a single vector, you either use the method explained and apply find, or use the second output parameter of ismember. For example:
[tf, loc] = ismember(a, [10 12 19 21 3], 'rows')
returns loc = 4 for your example. Note that here a is the second parameter, so that the output variable loc would hold a meaningful result.
Handling floating-point numbers
If your data contains floating point numbers, The ismember approach is going to fail because floating-point comparisons are inaccurate. Here's a shorter variant of Amro's solution:
x = reshape(c', size(c, 2), 1, []);
tf = any(all(abs(bsxfun(#minus, a', x)) < eps), 3)';
Essentially this is a one-liner, but I've split it into two commands for clarity:
x is the target rows to be searched, concatenated along the third dimension.
bsxfun subtracts each row in turn from all rows of a, and the magnitude of the result is compared to some small threshold value (e.g eps). If all elements in a row fall below it, mark this row as "1".
It depends on how you build those divided matrices. For example:
a = magic(5);
d = a([2 1 2 3],:);
then the matching rows are obviously: 2 1 2 3
EDIT:
Let me expand on the idea of using ismember shown by #EitanT to handle floating-point comparisons:
tf = any(cell2mat(arrayfun(#(i) all(abs(bsxfun(#minus, a, d(i,:)))<1e-9,2), ...
1:size(d,1), 'UniformOutput',false)), 2)
not pretty but works :) This would be necessary for comparisons such as: 0.1*3 == 0.3
(basically it compares each row of d against all rows of a using an absolute difference)
a=[2 3 6 7 2 1 0.01 6 8 10 12 15 18 9 6 5 4 2].
Here is an array i need to extract the exact values where the increasing and decreasing trend starts.
the output for the array a will be [2(first element) 2 6 9]
a=[2 3 6 7 2 1 0.01 6 8 10 12 15 18 9 6 5 4 2].
^ ^ ^ ^
| | | |
Kindly help me to get the result in MATLAB for any similar type of array..
You just have to find where the sign of the difference between consecutive numbers changes.
With some common sense and the functions diff, sign and find, you get this solution:
a = [2 3 6 7 2 1 0.01 6 8 10 12 15 18 9 6 5 4 2];
sda = sign(diff(a));
idx = [1 find(sda(1:end-1)~=sda(2:end))+2 ];
result = a(idx);
EDIT:
The sign function messes things up when there are two consecutive numbers which are the same, because sign(0) = 0, which is falsely identified as a trend change. You'd have to filter these out. You can do this by first removing the consecutive duplicates from the original data. Since you only want the values where the trend change starts, and not the position where it actually starts, this is easiest:
a(diff(a)==0) = [];
This is a great place to use the diff function.
Your first step will be to do the following:
B = [0 diff(a)]
The reason we add the 0 there is to keep the matrix the same length because of the way the diff function works. It will start with the first element in the matrix and then report the difference between that and the next element. There's no leading element before the first one so is just truncates the matrix by one element. We add a zero because there is no change there as it's the starting element.
If you look at the results in B now it is quite obvious where the inflection points are (where you go from positive to negative numbers).
To pull this out programatically there are a number of things you can do. I tend to use a little multiplication and the find command.
Result = find(B(1:end-1).*B(2:end)<0)
This will return the index where you are on the cusp of the inflection. In this case it will be:
ans =
4 7 13
I'm sure there's an easy answer to this, but I'm not really sure what to search for. I have an array, M, of D dimensions, where D is constrained to be 1 <= D <= 5, and a vector of length D, X. I'd like to use D as an address within M and increment the value at that address, so if D were [1 2 3], I would want to increment M(1,2,3). I know I can do it like so:
if D == 1
M(X(1)) = M(X(1)) + 1;
end
if D == 2
M(X(1), X(2)) = M(X(1), X(2)) + 1;
end
But it's really ugly and I have to imagine there's a simpler, less clumsy way. Thanks!
You can use the function sub2ind to convert the address vector D to the corresponding dimensions in M. However, this would require that you store D as a cell and not a vector. The following example should help.
A=magic(5);%# just a test matrix
A=
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
d={3,4};%we need the element at row 3, column 4
indx=sub2ind(size(A),d{:});%# get the index corresponding to the subscript 3,4
A(indx)
ans=
20
You can also directly index it into the matrix A as A(sub2ind(size(A),d{:})), without having to create a separate variable.
You can also use num2cell to convert the vector to a cell. This might be a better option, as you might want to store D as a vector for other purposes. So the corresponding line becomes
indx=sub2ind(size(A),num2cell(d));