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Athanasios Paraskevopoulos
Athanasios Paraskevopoulos
Last activity on 14 Mar 2024

Integral Ghost

Can you solve it?
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s = ['M','A','T','L','A','B']
9%
char([77,65,84,76,65,66])
7%
"MAT" + "LAB"
21%
upper(char('matlab' - '0' + 48))
17%
fliplr("BALTAM")
17%
rot90(rot90('BALTAM'))
30%
2929 votes
The File Exchange team is thrilled to introduce a more streamlined approach to working with GitHub and File Exchange - the MATLAB and Simulink Integration for GitHub!
Key Enhancements:
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Action Required for File Exchange Contributors!
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Several of the colormaps are great for a 256 color surface plot, but aren't well optimized for extracting m colors for plotting several independent lines. The issue is that many colormaps have start/end colors that are too similar or are suboptimal colors for lines. There are certainly many workarounds for this, but it would be a great quality of life to adjust that directly when calling this.
Example:
x = linspace(0,2*pi,101)';
y = [1:6].*cos(x);
figure; plot(x,y,'LineWidth',2); grid on; axis tight;
And now if I wanted to color these lines, I could use something like turbo(6) or gray(6) and then apply it using colororder.
colororder(turbo(6))
But my issue is that the ends of the colormap are too similar. For other colormaps, you may get lines that are too light to be visible against the white background. There are plenty of workarounds, with my preference being to create extra colors and truncate that before using colororder.
cmap = turbo(8); cmap = cmap(2:end-1,:); % Truncate the end colors
figure; plot(x,y,'LineWidth',2); grid on; axis tight;
colororder(cmap)
I think it would be really awesome to add some name-argument input pair to these colormaps that can specify the range you want so this could even be done inside the colororder calling if desired. An example of my proposed solution would look something like this:
cmap = turbo(6,'Range',[0.1 0.8]); % Proposed idea to add functionality
Where in this scenario, the resulting colormap would be 6 equally spaced colors that range from 10% to 80% of the total color range. This would be especially nice because you could more quickly modify the range of colors, or you could set the limits regardless of whether you need to plot 3, 6, or 20 lines.
Big congratulations to @VBBV for achieving the remarkable milestone of 3,000 reputation points, earning the prestigious title of Editor within our community.
This achievement is a testament to @VBBV's exceptional contributions and steadfast commitment to the community. These efforts have also been endorsed by fellow top contributors, underscoring the value and impact of @VBBV's expertise.
Welcome to the Editors' Club, @VBBV – we are excited to witness and support your continued journey and influence within our community!
eye(3) - diag(ones(1,3))
11%
0 ./ ones(3)
9%
cos(repmat(pi/2, [3,3]))
16%
zeros(3)
20%
A(3, 3) = 0
32%
mtimes([1;1;0], [0,0,0])
12%
3009 votes
2 x 2 행렬의 행렬식은
  • 행렬의 두 row 벡터로 정의되는 평행사변형의 면적입니다.
  • 물론 두 column 벡터로 정의되는 평행사변형의 면적이기도 합니다.
  • 좀 더 정확히는 signed area입니다. 면적이 음수가 될 수도 있다는 뜻이죠.
  • 행렬의 두 행(또는 두 열)을 맞바꾸면 행렬식의 부호도 바뀌고 면적의 부호도 바뀌어야합니다.
일반적으로 n x n 행렬의 행렬식은
  • 각 row 벡터(또는 각 column 벡터)로 정의되는 N차원 공간의 평행면체(?)의 signed area입니다.
  • 제대로 이해하려면 대수학의 개념을 많이 가지고 와야 하는데 자세한 설명은 생략합니다.(=저도 모른다는 뜻)
  • 더 자세히 알고 싶으시면 수학하는 만화의 '넓이 이야기' 편을 추천합니다.
  • 수학적인 정의를 알고 싶으시면 위키피디아를 보시면 됩니다.
  • 이렇게 생겼습니다. 좀 무섭습니다.
아래 코드는...
  • 2 x 2 행렬에 대해서 이것을 수식 없이 그림만으로 증명하는 과정입니다.
  • gif 생성에는 ScreenToGif를 사용했습니다. (gif 만들기엔 이게 킹왕짱인듯)
Determinant of 2 x 2 matrix is...
  • An area of a parallelogram defined by two row vectors.
  • Of course, same one defined by two column vectors.
  • Precisely, a signed area, which means area can be negative.
  • If two rows (or columns) are swapped, both the sign of determinant and area change.
More generally, determinant of n x n matrix is...
  • Signed area of parallelepiped defined by rows (or columns) of the matrix in n-dim space.
  • For a full understanding, a lot of concepts from abstract algebra should be brought, which I will not write here. (Cuz I don't know them.)
  • For a mathematical definition of determinant, visit wikipedia.
  • A little scary, isn't it?
The code below is...
  • A process to prove the equality of the determinant of 2 x 2 matrix and the area of parallelogram.
  • ScreenToGif is used to generate gif animation (which is, to me, the easiest way to make gif).
% 두 점 (a, b), (c, d)의 좌표
a = 4;
b = 1;
c = 1;
d = 3;
% patch 색 pre-define
lightgreen = [144, 238, 144]/255;
lightblue = [169, 190, 228]/255;
lightorange = [247, 195, 160]/255;
% animation params.
anim_Nsteps = 30;
% create window
figure('WindowStyle','docked')
ax = axes;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
ax.XTick = [];
ax.YTick = [];
hold on
ax.XLim = [-.4, a+c+1];
ax.YLim = [-.4, b+d+1];
% create ad-bc patch
area = patch([0, a, a+c, c], [0, b, b+d, d], lightgreen);
p_ab = plot(a, b, 'ko', 'MarkerFaceColor', 'k');
p_cd = plot(c, d, 'ko', 'MarkerFaceColor', 'k');
p_ab.UserData = text(a+0.1, b, '(a, b)', 'FontSize',16);
p_cd.UserData = text(c+0.1, d-0.2, '(c, d)', 'FontSize',16);
area.UserData = text((a+c)/2-0.5, (b+d)/2, 'ad-bc', 'FontSize', 18);
pause
%% Is this really ad-bc?
area.UserData.String = 'ad-bc...?';
pause
%% fade out ad-bc
fadeinout(area, 0)
area.UserData.Visible = 'off';
pause
%% fade in ad block
rect_ad = patch([0, a, a, 0], [0, 0, d, d], lightblue, 'EdgeAlpha', 0, 'FaceAlpha', 0);
uistack(rect_ad, 'bottom');
fadeinout(rect_ad, 1, t_pause=0.003)
draw_gridline(rect_ad, ["23", "34"])
rect_ad.UserData = text(mean(rect_ad.XData), mean(rect_ad.YData), 'ad', 'FontSize', 20, 'HorizontalAlignment', 'center');
pause
%% fade-in bc block
rect_bc = patch([0, c, c, 0], [0, 0, b, b], lightorange, 'EdgeAlpha', 0, 'FaceAlpha', 0);
fadeinout(rect_bc, 1, t_pause=0.0035)
draw_gridline(rect_bc, ["23", "34"])
rect_bc.UserData = text(b/2, c/2, 'bc', 'FontSize', 20, 'HorizontalAlignment', 'center');
pause
%% slide ad block
patch_slide(rect_ad, ...
[0, 0, 0, 0], [0, b, b, 0], t_pause=0.004)
draw_gridline(rect_ad, ["12", "34"])
pause
%% slide ad block
patch_slide(rect_ad, ...
[0, 0, d/(d/c-b/a), d/(d/c-b/a)],...
[0, 0, b/a*d/(d/c-b/a), b/a*d/(d/c-b/a)], t_pause=0.004)
draw_gridline(rect_ad, ["14", "23"])
pause
%% slide bc block
uistack(p_cd, 'top')
patch_slide(rect_bc, ...
[0, 0, 0, 0], [d, d, d, d], t_pause=0.004)
draw_gridline(rect_bc, "34")
pause
%% slide bc block
patch_slide(rect_bc, ...
[0, 0, a, a], [0, 0, 0, 0], t_pause=0.004)
draw_gridline(rect_bc, "23")
pause
%% slide bc block
patch_slide(rect_bc, ...
[d/(d/c-b/a), 0, 0, d/(d/c-b/a)], ...
[b/a*d/(d/c-b/a), 0, 0, b/a*d/(d/c-b/a)], t_pause=0.004)
pause
%% finalize: fade out ad, bc, and fade in ad-bc
rect_ad.UserData.Visible = 'off';
rect_bc.UserData.Visible = 'off';
fadeinout([rect_ad, rect_bc, area], [0, 0, 1])
area.UserData.String = 'ad-bc';
area.UserData.Visible = 'on';
%% functions
function fadeinout(objs, inout, options)
arguments
objs
inout % 1이면 fade-in, 0이면 fade-out
options.anim_Nsteps = 30
options.t_pause = 0.003
end
for alpha = linspace(0, 1, options.anim_Nsteps)
for i = 1:length(objs)
switch objs(i).Type
case 'patch'
objs(i).FaceAlpha = (inout(i)==1)*alpha + (inout(i)==0)*(1-alpha);
objs(i).EdgeAlpha = (inout(i)==1)*alpha + (inout(i)==0)*(1-alpha);
case 'constantline'
objs(i).Alpha = (inout(i)==1)*alpha + (inout(i)==0)*(1-alpha);
end
pause(options.t_pause)
end
end
end
function patch_slide(obj, x_dist, y_dist, options)
arguments
obj
x_dist
y_dist
options.anim_Nsteps = 30
options.t_pause = 0.003
end
dx = x_dist/options.anim_Nsteps;
dy = y_dist/options.anim_Nsteps;
for i=1:options.anim_Nsteps
obj.XData = obj.XData + dx(:);
obj.YData = obj.YData + dy(:);
obj.UserData.Position(1) = mean(obj.XData);
obj.UserData.Position(2) = mean(obj.YData);
pause(options.t_pause)
end
end
function draw_gridline(patch, where)
ax = patch.Parent;
for i=1:length(where)
v1 = str2double(where{i}(1));
v2 = str2double(where{i}(2));
x1 = patch.XData(v1);
x2 = patch.XData(v2);
y1 = patch.YData(v1);
y2 = patch.YData(v2);
if x1==x2
xline(x1, 'k--')
else
fplot(@(x) (y2-y1)/(x2-x1)*(x-x1)+y1, [ax.XLim(1), ax.XLim(2)], 'k--')
end
end
end
Hannah
Hannah
Last activity on 8 Feb 2024

Flipbook Mini Hack

I have been procrastinating on schoolwork by looking at all the amazing designs created in the last MATLAB Flipbook Mini Hack! They are just amazing. The voting is over but what are y'all's personal favorites? Mine is the flapping butterfly, it is for sure a creation I plan to share with others in the future!
Mike Croucher
Mike Croucher
Last activity on 29 Jan 2024

Pong Wars in MATLAB

One of my colleauges, Michio, recently posted an implementation of Pong Wars in MATLAB
Making me wonder about variations. What might the resulting patterns look with differing numbers of balls? Different physics etc?
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I recently wrote about the new ODE solution framework in MATLAB over the The MATLAB Blog The new solution framework for Ordinary Differential Equations (ODEs) in MATLAB R2023b » The MATLAB Blog - MATLAB & Simulink (mathworks.com)
This was a very popular post at the time - many thousands of views. Clearly everyone cares about ODEs in MATLAB.
This made me wonder. If you could wave a magic wand, what ODE functionality would you have next and why?
sky
sky
Last activity on 18 Jan 2024

Relative pose in 2D: problem 1

I'm having problem in its test 6 ... passing 5/6 what would be the real issue..
am wring Transformation matrix correct.. as question said SSW should be 202.5 degree...
so what is the issue..
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Matt J
Matt J
Last activity on 29 Jan 2024

3D polyshapes

Is there a reason for TMW not to invest in 3D polyshapes? Is the mathematical complexity of having all the same operations in 3D (union, intersection, subtract,...) prohibitive?
Image Analyst
Image Analyst
Last activity on 16 Feb 2024

My favorite sport to watch is

American style football
12%
Soccer / football
39%
baseball
5%
basketball
12%
tennis or golf
7%
rugby, track, cricket, racing, etc.
26%
3712 votes
Congratulations, @Cris LaPierre for achieving 10K reputation points.
You reached this milestone by providing valuable contribution to the community since you started answering questions in Since September 2018.
You provided 3984 answers and received 1142 votes. You are ranked #24 in the community. Thank you for your contribution to the community and please keep up the good track record!
MATLAB Central Team