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We have the test image with M rows and N columns as
f(x,y), for x∈ [1,M] and y∈ [1,N ]. The horizontal absolute
difference value of a pixel is defined by
D (x, y) = |f (x, y +1) − f (x, y −1)|.
need help in how to implement it in matlab
This will generate same size matrix, that you need:
mat1 = [zeros(2,size(f,2)); f];% adds 2 rows of zeros to begining
mat2 = [f;zeros(2,size(f,2))]; %adds 2 row of zeros to the end
Dd = mat1-mat2;
D = Dd(2:((size(Dd,1)-1)),:);%crop Dd matrix to size(f)
D = abs( f(1:end-1,:) - f(2:end,:) );
check out diff command as well. Note that D has 1 row less than f.
aux = abs(diff(f,[],2));
D = max(aux(:,1:end-1), aux(:,2:end));
For example: given
f = [3 5 6 4
2 5 4 3
8 9 3 1];
the result is
>> D
D =
2 2
3 1
6 6
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matrix m:
1 2 3
4 5 6
7 8 9
output: (matrix r)
1 3 6
5 12 21
12 27 45
How we get the results is:
staring index = (0,0)
for example,
element at the (1,1) position of the result matrix would be,
r[1][1] = m[0][0] + m[0][1] + m[1][0] + m[1][1]
sum of the elements inside the red box:
element at the (2,1) position of the result matrix would be,
r[2][1] = matrix[0][0] + m[0][1] + m[1][0] + m[1][1] + m[2][0] + m[2][1]
sum of the elements inside the red box:
One important observation here is that for i > 0 and j > 0:
r[i][j] = m[i][j] + r[i-1][j] + r[i][j-1]
^ ^
When i == 0 or j == 0 then just drop the terms from the above expression that become invalid.
So:
r[0][0] = m[0][0]
And:
r[0][1] = m[0][1] + r[0][0]
If you continue with the first row from left to right and then the next rows in the same fashion, you'll always have the information needed to calculate r[i][j].
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Given any 3 digits N as an input,
we can only use digits from 1 to 9, in such a way that order never breaks and any repetition of number.
for example
If N = 150.
123 + 4 + 5 - 6 + 7 + 8 + 9 = 150
We can combine digits and insert '-' and '+' operations to get the desired N value.
Line up the numbers: 1 2 3 4 5 6 7 8 9.
There are 8 spaces between these numbers. Each space could be a '+' or a '-' or a blank (joining the digits together).
Thus, there are 3 ** 8 i.e., 6561 different possible combinations of operations you could use.
That's small enough to just try all of them in a loop and check which one works.
You can do it like this(this code is written by python):
N = 150
digit_from, digit_to = 1, 9 ### from 1 to 9 ###
def find(N, pos, equation, num, coff):
if pos > digit_to:
if N - coff * num == 0:
print(equation)
else:
find( N-coff*num, pos+1, equation+'+'+str(pos), pos, 1 ) ### plus ###
find( N-coff*num, pos+1, equation+'-'+str(pos), pos, -1 ) ### minus ###
find( N, pos+1, equation+str(pos), num*10+pos, coff ) ### blank ###
return
find( N, digit_from + 1, '150 = 1', digit_from, 1 )
Result(In case of N = 150 and digits between 1 to 9):
150 = 1+23+45-6+78+9
150 = 1+234+5+6-7-89
150 = 1-2+3-4+56+7+89
150 = 12+3+45-6+7+89
150 = 123+4+5-6+7+8+9
150 = 123+45+6-7-8-9
150 = 123-4-56+78+9
Thanks
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Euler Totient Function finds the number of numbers from 1 to n that are relatively prime to n.
For example--
phi(1) = 1
phi(2) = 1
phi(3) = 2
phi(4) = 2
phi(15) = 8
phi(16) = 8
phi(20) = 8
phi(24) = 8
. . . . . . . .
phi(n) = x
Here 15, 16, 20, 24 have 8 numbers between 1 to them that are relatively prime. But 15 is the smallest number with number of co-primes = 8.
How to find 15 if we input 8
Find a general algorithm for finding the smallest n such that phi(n) = x for an arbitrary x where x <= 2*10^9
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Given an integer x in the interval [1 .. 6], I am looking for two mathematical functions y1 and y2 so that:
y1(x) ∈ [1 .. 6], y2(x) ∈ [1 .. 6]
y1(x) ≠ y2(x) ≠ x
y1(x) and y2(x) are integers
I tried y1(x) = 7-x and y2(x) = (1+x)%6 where % is the remainder or modulo operation.
That solution does not work for x=6. I get y1(x) = y2(x) = 1, which does not fulfills the condition 2. Neither for x=3 and x=5.
Does anyone sees a working solution?
You can use for example:
y1=(x % 6) +1
y2=((x+1) % 6) +1
Functions as table:
x y1 y2
1 2 3
2 3 4
3 4 5
4 5 6
5 6 1
6 1 2
Technically, y1=1+((x+1) %6) and y2=(1+(x+2) %6) both satisfy your request.
I guess though you were thinking about something with a unified distribution of some sort (which is usually the motivation for such attempts...).
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We have this matrix of 4x4:
a b c d
e f g h
1 2 3 4
5 6 7 8
By transposing the matrix we get:
a e 1 5
b f 2 6
c g 3 7
d h 4 8
My question is:
What matrix do we get by "transposing column 2 with row 4?"
I need to understand the operation in itself, what does it imply/mean? I never thought of "transposing a column with a line".
AFAIK, It means you are to swap column 2 and row 4, instead of column 1 with row 1 and column2 with row 2 etc.
The code is basically the same as a full transposition, except you only have one column/row
Matrix transposition is a mathematical operation in which a matrix's rows become its columns. From a mathematical perspective, there's no real benefit to transposing only one row in a M x N matrix, but the code to transpose one row is not much different than transposing an entire matrix.
The matrix you get after the transposition would be:
a b 1 d
e f 2 h
c g 3 7
5 6 4 8