hey thanks for the help:screenshot
New to programming and trying to understand the logic why the code in line 17 produces the desire result of a left-aligned pyramid. Just trying to fully understand why x= height - y - 1 gives the desired outcome.
This snippet
for (int x = height - y - 1; x < height; x++)
{
print("#");
}
does not have anything to do with alignment. The pyramid will always be left aligned without a space(" ") printing loop, which you don't have in your code and you probably don't need either.
Now let's walk through the iteration to understand what's going on
Consider the height to be 5
In the first iteration of the outer loop (i.e for (int y = 0; y < height; y++))
y = 0
x = height - y - 1 = 5 - 0 - 1 = 4
So x starts at 4 and stops before reaching height, i.e 5. So this loop will be executed exactly 1 time. Which means it'll print a singular #.
In the second iteration of the outer loop
y = 1
x = height - y - 1 = 5 - 1 - 1 = 3
So x starts at 3 and stops before reaching height, i.e 5. So this loop will be executed exactly 2 times. Which means it'll print # twice.
In the third iteration of the outer loop
y = 2
x = height - y - 1 = 5 - 2 - 1 = 2
So x starts at 2 and stops before reaching height, i.e 5. So this loop will be executed exactly 3 times. Which means it'll print # thrice.
In the fourth iteration of the outer loop
y = 3
x = height - y - 1 = 5 - 3 - 1 = 1
So x starts at 1 and stops before reaching height, i.e 5. So this loop will be executed exactly 4 times. Which means it'll print # 4 times.
In the fifth and final iteration of the outer loop
y = 4
x = height - y - 1 = 5 - 4 - 1 = 0
So x starts at 0 and stops before reaching height, i.e 5. So this loop will be executed exactly 5 times. Which means it'll print # 5 times.
So to achieve that logic, is the reasoning behind using x = height - y - 1. However there are other ways to do this too-
for (int x = 0; x < y + 1; x++)
{
print("#");
}
This will also work in the same logic, but hopefully with less confusion.
Notice, how the number of characters printed in each line matches with the line number. So the first line has 1 hash, 2nd has 2 and so on. We can deduce the line number from y. For the first line y = 0, for the second line y = 1 and so on. So we can simply add 1 to y and set that as our upper bound to print the hashes.
Related
Suppose a data like:
X y
1 5
2 6
3 1
4 7
5 3
6 8
I want to remove 3 1 and 5 3 because their difference with the previous row is more than 3. In fact, I want to draw a plot with them and want it to be smooth.
I tried
for qq = 1:size(data,1)
if data(qq,2) - data(qq-1,2) > 3
data(qq,:)=[];
end
end
However, it gives:
Subscript indices must either be real positive integers or logicals.
Moreover, I guess the size of array changes as I remove some elements.
In the end, the difference between no consecutive elements must be greater than threshold.
In practice I want to smooth the following picture where there is high fluctuate
One very simple filter from Mathematical morphology that you could try is the closing with a structuring element of size 2. It changes the value of any sample that is lower than both neighbors to the lowest of its two neighbors. Other values are not changed. Thus, it doesn't use a threshold to determine what samples are wrong, it only looks that the sample is lower than both neighbors:
y = [5, 6, 1, 7, 3, 8]; % OP's second column
y1 = y;
y1(end+1) = -inf; % enforce boundary condition
y1 = max(y1,circshift(y1,1)); % dilation
y1 = min(y1,circshift(y1,-1)); % erosion
y1 = y1(1:end-1); % undo boundary condition change
This returns y1 = [5 6 6 7 7 8].
If you want to prevent changing your signal for small deviations, you can apply your threshold as a second step:
I = y1 - y < 3;
y1(I) = y(I);
This finds the places where we changed the signal, but the change was less than the threshold of 3. At those places we write back the original value.
You have a few errors:
Your index needs to start from 2, so that you aren't trying to index 0 for a previous index.
You need to check that the absolute value of the difference is greater than 3.
Since your data matrix is changing sizes, you can't use a for loop with a fixed number of iterations. Use a while loop instead.
This should give you the results you want:
qq = 2;
while qq <= size(data, 1)
if abs(data(qq, 2) - data(qq-1, 2)) > 3,
data(qq, :) = [];
else
qq = qq+1;
end
end
Why the result is 7??? I can't find the log if that :(
#include <stdio.h>
main() {
int i, j, a = 1, b = 1;
for (i = 1; i < 4; i++)
for (j = 1; j < 3; j++)
a = a + b;
printf("a=%d", a);
}
The structure is like below
- Outer `for` loop
- inner `for` loop
- instruction
so the "instruction" (statement/block) will get executed inner for loop count times, for outer for loop count times.
What is basically says is, add the value of b to the latest value of a (in a recurring way) to get the current value of a. Now , do that for "outer" number of time, for which, do the same for "inner" number of times.
Outer for loop ==> 3 times,
inner for loop ==> 2 times
So, final value =3*2 (increment) + (initial) = (3*2)*1 + 1 = 7.
The result is 7 because b is initialized as 1 and stays 1 the whole time. The outer loop is run 3 times (1, 2, 3), the inner loop is run 2 times (1 and 2), so there are 6 runs where b is added to a (which is initialized as 1). 1 + 6 = 7.
In such scenarios, you should add a watch and debug your code line by line. I believe the shortcut is F11. Regards to why the output of your code is 7...
The inner loop runs six times. J loops twice - 1, 2, (ends when it is 3), and I loops thrice (1, 2, 3, 4 - end)... for a total of 2 X 3 = 6.
Since b is '1', you are basically adding the number 1 to a six times. Since a started with '1', the output is:
a = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7
I am trying to allocate (x, y) points to the cells of a non-uniform rectangular grid. Simply speaking, I have a grid defined as a sorted non-equidistant array
xGrid = [x1, x2, x3, x4];
and an array of numbers x lying between x1 and x4. For each x, I want to find its position in xGrid, i.e. such i that
xGrid(i) <= xi <= xGrid(i+1)
Is there a better (faster/simpler) way to do it than arrayfun(#(x) find(xGrid <= x, 1, 'last'), x)?
You are looking for the second output of histc:
[~,where] = histc(x, xGrid)
This returns the array where such that xGrid(where(i)) <= x(i) < xGrid(where(i)+1) holds.
Example:
xGrid = [2,4,6,8,10];
x = [3,5,6,9,11];
[~,where] = histc(x, xGrid)
Yields the following output:
where =
1 2 3 4 0
If you want xGrid(where(i)) < x(i) <= xGrid(where(i)+1), you need to do some trickery of negating the values:
[~,where] = histc(-x,-flip(xGrid));
where(where~=0) = numel(xGrid)-where(where~=0)
This yields:
where =
1 2 2 4 0
Because x(3)==6 is now counted for the second interval (4,6] instead of [6,8) as before.
Using bsxfun for the comparisons and exploiting find-like capabilities of max's second output:
xGrid = [2 4 6 8]; %// example data
x = [3 6 5.5 10 -10]; %// example data
comp = bsxfun(#gt, xGrid(:), x(:).'); %'// see if "x" > "xGrid"
[~, result] = max(comp, [], 1); %// index of first "xGrid" that exceeds each "x"
result = result-1; %// subtract 1 to find the last "xGrid" that is <= "x"
This approach gives 0 for values of x that lie outside xGrid. With the above example values,
result =
1 3 2 0 0
See if this works for you -
matches = bsxfun(#le,xGrid(1:end-1),x(:)) & bsxfun(#ge,xGrid(2:end),x(:))
[valid,pos] = max(cumsum(matches,2),[],2)
pos = pos.*(valid~=0)
Sample run -
xGrid =
5 2 1 6 8 9 2 1 6
x =
3 7 14
pos =
8
4
0
Explanation on the sample run -
First element of x, 3 occurs last between ...1 6 with the criteria of xGrid(i) <= xi <= xGrid(i+1) at the backend of xGrid and that 1 is at the eight position, so the first element of the output pos is 8. This continues for the second element 7, which is found between 6 and 8 and that 6 is at the fourth place in xGrid, so the second element of the output is 4. For the third element 14 which doesn't find any neighbours to satisfy the criteria xGrid(i) <= xi <= xGrid(i+1) and is therefore outputted as 0.
If x is a column this might help
xg1=meshgrid(xGrid,1:length(x));
xg2=ndgrid(x,1:length(xGrid));
[~,I]=min(floor(abs(xg1-xg2)),[],2);
or a single line implementation
[~,I]=min(floor(abs(meshgrid(xGrid,1:length(x))-ndgrid(x,1:length(xGrid)))),[],2);
Example: xGrid=[1 2 3 4 5], x=[2.5; 1.3; 1.7; 4.8; 3.3]
Result:
I =
2
1
1
4
3
consider an area with size m*n. Here the size of m and n is unknown. Now I am extracting data from each point in the area. I am scanning the area first going in the x direction till m point and the again returning to m=0 and n=1, i.e the second row. Again I scan along the x direction till the end of m. An example of the data has been shown below. Here I get value for different x,y coordinates during the scan. I can carry out operation between the first two points in x direction by
p1 = A{1}; %%reading the data from the text file
p2 = A{2};
LA=[p1 p2];
for m=1:length(y)
p= LA(m,1);
t= LA(m,2);
%%and
q=LA(m+1,1)
r=LA(m+1,2)
I want to do the same for y axis. That is I want to operate between first point in x=0 and y=1 then between x=2 and y=1 and so on. Hope you have got it.
g x y
2 0 0
3 1 0
2 2 0
4 3 0
1 4 0
2 m 0
3 0 1
2 1 1
4 2 1
5 3 1
.
.
.
.
2 m 1
now I was thinking of a logic where I will first find the size of n by counting the number of zeros
NUMX = 0;
while y((NUMX+1),:) == 0
NUMX = NUMX + 1;
end
NU= NUMX;
And then I was thinking of applying the following loop
for m=1:NU:n-1
%%and
p= LA(m,1);
t= LA(m,2);
%%and
q=LA(m+1,1)
r=LA(m+1,2)
But its showing error. Please help!!
??? Attempted to access del2(99794,:); index out of bounds because
size(del2)=[99793,1].
Here NUMX=198
Comment: The nomenclature in your question is inconsistent, making it difficult to understand what you are doing. The variable del2 you mention in the error message is nowhere to be seen.
1.) Let's start off by creating a minimal working example that illustrates the data structure and provides knowledge of the dimensions we want to retrieve later. You matrix is not m x n but m*n x 3.
The following example will set up a matrix with data similar to what you have shown in your question:
M = zeros(8,3);
for J=1:4
for I=1:2
M((J-1)*2+I,1) = rand(1);
M((J-1)*2+I,2) = I;
M((J-1)*2+I,3) = J-1;
end
end
M =
0.469 1 0
0.012 2 0
0.337 1 1
0.162 2 1
0.794 1 2
0.311 2 2
0.529 1 3
0.166 2 3
2.) Next, let's determine the number of x and y, to use the nomenclature of your question:
NUMX = 0;
while M(NUMX+1,3) == 0
NUMX = NUMX + 1;
end
NUMY = size(M,1)/NUMX;
NUMX =
2
NUMY =
4
3.) The data processing you want to do still is unclear, but here are two approaches that can be used for different means:
(a)
COUNT = 1;
for K=1:NUMX:size(M,1)
A(COUNT,1) = M(K,1);
COUNT = COUNT + 1;
end
In this case, you step through the first column of M with a step-size corresponding to NUMX. This will result in all the values for x=1:
A =
0.469
0.337
0.794
0.529
(b) You can also use NUMX and NUMY to reorder M:
for J=1:NUMY
for I=1:NUMX
NEW_M(I,J) = M((J-1)*NUMX+I,1);
end
end
NEW_M =
0.469 0.337 0.794 0.529
0.012 0.162 0.311 0.166
The matrix NEW_M now is of size m x n, with the values of constant y in the columns and the values of constant x in the rows.
Concluding remark: It is unclear how you define m and n in your code, so your specific error message cannot be resolved here.
I am trying to understand the MARIE assembly language. I don't quite understand skipcond for
doing things like <, or >, or multiply or divide.
I am taking this simple program:
x = 1
while x < 10 do
x = x +1
endwhile;
What I don't understand is how to use certain skip conditions:
Skipcond 800 if AC > 0,
Skipcond 400 if AC = 0,
Skipcond 000 if AC < 0
Now, I know I would subtract x from 10 and test using skipcond.
I am not sure which one and why. I guess if I knew how they really work maybe it would be easier to understand. Why is it used to compare to zero?
This is what I have:
100 load one
101 store x
102 subt ten
103 skipcond400 if x-10 = 0? // or skpcond000 x -10 < 0??
while x < 10 do
x = x + 1
will jump out of the loop as soon as x equals 10. If you subtract 10 from x, you'll get a negative value until x equals 10 (and the value is 0). So using skpcond000 would be wrong as it would jump out too soon. So skpcond400 is correct.
Perhaps it is easier to understand if you change the C code so it will be closer to the assembly code:
Original: while (x < 10) do
Subtract 10: while ((x - 10) < 0) do
Use != instead of <: while ((x - 10) != 0) do
Also note that you have to increase x after the condition to reproduce identical behaviour to the while loop.
This may help. There are many ways to write this but I think this is the easiest way to understand what is happening in the loop. Note: usually variables are placed at the bottom of the program.
while x<10
x = x+1
Org 100
Load One / loads accumulator = 1 from a decimal constant
Store X / initialize the var x = 1
loop, Load X / loads x into the accumulator
Subt Ten / compares x to 10
Skipcond 000 / if ac < 0 i.e. if x < 10 run rest of loop body
JUMP Endloop / if ac => 10 terminate loop
Load X / begin the Loop
ADD One / add 1 to x
Store X / store new value in X
JUMP loop / continue loop
Endloop Halt / ends loop
One = DEC 1 Constant
Ten = DEC 10 Constant
X = 0 Variable