i want to have 10 different shuffled form of an array...but random_shuffle produce the same sequence for 10 times...
and my code is...
for(k=0;k<10;k++) {
for (l=0; l<SIZE;l++)
a[l]=l+1;
srand(time(0));
random_shuffle(a,a+SIZE); //getting the shuffled sequence
for(;i<10;i++) {
for(j=0;j<5;j++) {
rcusseq[i][j]=a[m++]; //storing the sequence in a 2d array
printf("%d\t",rcusseq[i][j]);
}
m=0;
printf("\n");
}
}
OUTPUT
5 4 2 1 3
5 4 2 1 3
5 4 2 1 3
5 4 2 1 3
5 4 2 1 3
5 4 2 1 3
5 4 2 1 3
5 4 2 1 3
5 4 2 1 3
5 4 2 1 3
Press any key to continue
You're reinitializing the random number generator each time, as your code is quick enough to run within 1 second, you get the same seed every time.
Move srand(time(0));out of the loop.
Only call srand() once in a program, e.g. at the beginning of main().
Related
In each operation we can either push an element to the end of the array or at the beginning of it
for example an array 3 2 5 1 4 6 would take 4 steps.
after first operation 2 3 5 1 4 6
after second operation 1 2 3 5 4 6
after third operation 2 3 1 4 6 5
after fourth operation 1 2 3 4 5 6
I think in the best case, the array is already sorted - 0 operations needed.
In the worst case, its sorted already, but in the opposite order (eg 6 5 4 3 2 1), you gonna need number of elements-1 operations.
I can "Vectorize" the circshift command but I'm having trouble adding dimensions to it.
See code below with working FOR loop that I'm trying to vectorize using dimensions
clear all,clf reset,tic,clc , close all
function [outMat] = vectcircshift(vectToShift,shiftVector)
%This function generates a matrix where each row is a circshift of the
%original vector from the specified interval in the shiftVector;
%
%Inputs
%vectToShift: is the original vector you want to circshift multiple times
%shiftVector: is the vector of the circshift sizes;
%
%Outputs
%outMat: is a matrix were every row is circshift by the amount in the
% shiftVector
[n,m]=size(vectToShift);
if n>m
inds=(1:n)';
i=toeplitz(flipud(inds),circshift(inds,[1 0]));
outMat=vectToShift(i(shiftVector,:));
outMat=circshift(outMat,[0,-1]); %shift to include original signal first
else
inds=1:m;
i=toeplitz(fliplr(inds),circshift(inds,[0 1]));
outMat=vectToShift(i(shiftVector,:));
outMat=circshift(outMat,[0,-1]); %shift to include original signal first
end
end
%%----Working FOR LOOP below I'm trying to vectorize.
ndim=0;
ndim_tot=[1:3] %total dimensions
for ndim=1:length(ndim_tot)
ndim=ndim+0
if ndim==1
array_sort(ndim,:)=circshift(ndim_tot,[0 ndim-1]) %start at row of sort array
else
array_sort(ndim,:)=circshift(ndim_tot,[0 mod(-ndim,length(ndim_tot))+1]) %next start of row of sort array
endif
array_sort= array_sort(ndim,:)
array_dim(:,:,ndim)=vectcircshift([1:5],array_sort)
endfor
I tired the syntax below but that logic won't work.
ndim_tot=[1:3]; %number of dimensions
array_dim2(:,:,ndim_tot)=vectcircshift([1:5],[1:3])
I get an error nonconformant arguments(op1 is 0x0x1, op2 is 3x5)
My goal is to create a multidimensional array that circshifts a signal / array and also creates and shifts it in multiple dimensions.
Example: of what the multidimensional array would look like
if I start with a signal / array a1=[1 2 3 4 5]
I'm trying to have it create.
array_dim(:,:,1)=
[
1 2 3 4 5
5 1 2 3 4
4 5 1 2 3
]
array_dim(:,:,2)=
[
5 1 2 3 4
4 5 1 2 3
1 2 3 4 5
]
array_dim(:,:,3)=
[
4 5 1 2 3
1 2 3 4 5
5 1 2 3 4
]
Please note: the the numbers won't be sequential I just used it as an example to help explain things a little easier.
PS: I'm using Octave 4.2.2
Not clear why you are shifting in mod 3, but here is a loop assignment using shift
a1=[1 2 3 4 5];
array_dim=zeros(3,5,3);
for i=0:2
array_dim(:,:,i+1)=[shift(a1,i);
shift(a1,mod(i+1,3));
shift(a1,mod(i+2,3))];
endfor
array_dim
and the output fits your example
array_dim =
ans(:,:,1) =
1 2 3 4 5
5 1 2 3 4
4 5 1 2 3
ans(:,:,2) =
5 1 2 3 4
4 5 1 2 3
1 2 3 4 5
ans(:,:,3) =
4 5 1 2 3
1 2 3 4 5
5 1 2 3 4
I am trying to construct a 2D array for an assignment. I've used a nested for loop to construct the 2D array using scanf():
int width;
int height;
scanf("%d %d",&width,&height);
int array[width][height];
for (int i=0;i<height;i++){
for (int j=0;j<width;j++){
scanf("%d",&array[i][j]);
}
}
However when I print the array, I can see that it has been constructed in a strange way, where all the numbers of the first line past a certain point are the first few numbers from the second line (instead of what they should be). The next lines after work fine.
Example:
Input:
6 2
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
The created array looks like this:
1 3 2 4 6 8 (<-- these last 4 numbers are the first 4 numbers of the second line)
2 4 6 8 0 2 (correct)
3 4 2 0 1 3 (correct)
Any ideas? Thanks a lot.
Your declaration of array
int array[width][height];
is wrong. The outer loop goes from 0 to height - 1, but array[i] can only go
from 0 to width - 1. The same applies for the inner loop. You swapped width
and height in the declaration of the array, it should be
int array[height][width];
Also note that for the matrix
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
the width is 6 and the height is 3, so the correct input should be
6 3
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
I compiled and run this code:
#include <stdio.h>
int main(void)
{
int width;
int height;
scanf("%d %d",&width,&height);
int array[height][width];
for (int i=0;i<height;i++){
for (int j=0;j<width;j++){
scanf("%d",&array[i][j]);
}
}
printf("----------------\n");
for (int i=0;i<height;i++){
for (int j=0;j<width;j++){
printf("%d ", array[i][j]);
}
printf("\n");
}
}
And the output is:
$ ./b
6 3
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
----------------
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
as you can see, now it's reading correctly. See https://ideone.com/OJjj0Y
I have a square 2d array of values, where each row is identical, and where each element of row is one bigger than the last. For example:
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
I want to filter them, such that I can make a diamond as such:
1
1 2 3
1 2 3 4 5
1 2 3 4 5 6 7
1 2 3 4 5
1 2 3
1
Notice how the first part of the array is used, no matter how many elements are to be printed on that line. Also, spacing doesn't matter. I spaced them to show the diamond.
I know how to filter the top right "chunk" out, using j-i<(j/2). This will convert the original square into:
1
1 2 3
1 2 3 4 5
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
How can I get the bottom right "chunk" to filter out also? What additional condition can I impose on the values?
Presuming you have found out and stored the length of the "side" of the square already then you could use something like below. However, if your square has an even length then it will not work (can't produce a diamond in this way from an even side length square).
The following is pseudo-code so you will need to adapt it for your language. I've also used 0-indexed arrays and presumed square is a 2D array.
for (i=0, i<length, i++)
{
for (j=0, j<Length, j++)
{
if (i < length/2)
{
if (j < length/2 AND j <= i)
print square[i][j]
}
}
else
{
if (j < length/2 AND j <= (length - i))
{
print square[i][j]
}
}
}
print newline
}
I have the following matrix:
[ 2 5 7 8 1 3 4 6 5 7 3 1;
1 1 1 1 2 2 2 2 3 3 3 3;]
The first row represents values and the second characteristic
I want to get the max value if the value in the second row is the same, i.e. their characteristic is the same. So, what I would like to have is:
[ 8 6 7], since 8 is the highest value when the second row is 1, 6 when the second row is is 2, and 7 when the second row is 3. I can do it with a loop, but I would like vectorized solution, and if possible of course, in one line.
accumarray does exactly what you want
x=[ 2 5 7 8 1 3 4 6 5 7 3 1; 1 1 1 1 2 2 2 2 3 3 3 3;]
accumarray(x(2,:)',x(1,:)',[],#max)