Met een beetje google-fu heb ik de volgende matlab code gevonden, niet zo gelikt als het orgineel maar het punt is duidelijk
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%% define the batman equation
BatmanEq = @(x,y) ...
((x/7)^2*sqrt(abs(abs(x)-3)/(abs(x)-3)) + (y/3)^2*sqrt(abs(y+3*sqrt(33)/7)/(y+3*sqrt(33)/7))-1) ...
* (abs(x/2)-((3*sqrt(33)-7)/112)*x^2 -3 +sqrt(1-(abs(abs(x)-2)-1)^2) -y) ...
* (9*sqrt(abs((abs(x)-1)*(abs(x)-0.75))/((1-abs(x))*(abs(x)-0.75))) -8*abs(x) -y) ...
* (3*abs(x) + 0.75*sqrt(abs((abs(x)-0.75)*(abs(x)-0.5))/((0.75-abs(x))*(abs(x)-0.5))) -y) ...
* (2.25*sqrt(abs((x-0.5)*(x+0.5))/((0.5-x)*(0.5+x))) -y) ...
* (6*sqrt(10)/7 + (1.5-0.5*abs(x))*sqrt(abs(abs(x)-1)/(abs(x)-1)) -6*sqrt(10)/14*sqrt(4-(abs(x)-1)^2) -y);
%% Calculate X- and Y-array defining the Batman logo
% set x-points
xx = (-7.1:0.01:7.1);
% init y line arrays
yy_pos = nan(size(xx));
yy_neg = nan(size(xx));
% set the range of y values to scan for the line
y_scan = 0:0.01:3;
% init Batman equation value arrays
EqValue = zeros(size(y_scan));
% Find y-values for each x-point
for ii = 1:length(xx)
% POSITIVE Y VALUES:
% calculate equation values for a vertical line of (x,y)-coordinates
% Note that EqValues have also imaginary part!
for jj = 1:length(y_scan)
EqValue(jj) = BatmanEq(xx(ii), y_scan(jj));
end
% Find y-coordinate around which the abs(EqValue) minimizes.
[~, mini] = min(abs(EqValue));
% list indices around that minimum
ind = max(mini-2, 1):min(mini+2,length(y_scan));
% Although the batman logo "looks" continuous, it has breaks where it is
% not defined (0 over 0 e.g. at x==3). We cannot interpolate in those
% places.
if any(isnan(EqValue(ind)))
yy_pos(ii) = nan;
else
% Use interpolation to find the Eq==0 point more accurately.
% Spline interpolation works best here:
% Spline interpolation likely finds also the points where eq gets
% values [+,0,+] or [-,0,-] (as oppose to only [-,0,+] or [+,0,-]
% found by the other methods). However, spline does not perform
% well with sharp corners. Thus interp of abs(EqValue) will not
% work as most zero points are in sharp corners thus using
% real(EqValue) is preferred.
yy_pos(ii) = interp1(real(EqValue(ind)), y_scan(ind), 0, 'spline');
% Interpolation using only real part of the eq value, may lead to
% totally false values in some circumstances. Thus we must make sure
% that absolute value of the Eq value is lower than treshold.
if abs(BatmanEq(xx(ii), yy_pos(ii))) > 1
yy_pos(ii) = nan;
end
end
% NEGATIVE Y VALUES:
% Exactly same as with positive but:
% 'y_scan' -> '-y_scan')
% 'yy_pos' -> 'yy_neg')
for jj = 1:length(y_scan)
EqValue(jj) = BatmanEq(xx(ii), -y_scan(jj));
end
[~, mini] = min(abs(EqValue));
ind = max(mini-2, 1):min(mini+2,length(y_scan));
if any(isnan(EqValue(ind)))
yy_neg(ii) = nan;
else
yy_neg(ii) = interp1(real(EqValue(ind)), -y_scan(ind), 0, 'spline');
if abs(BatmanEq(xx(ii), yy_neg(ii))) > 1
yy_neg(ii) = nan;
end
end
end
% build plot arrays and ignore nan values
nan_filter_pos = find(~isnan(yy_pos));
nan_filter_neg = find(~isnan(yy_neg));
xx_plot = [xx(nan_filter_pos), xx(nan_filter_neg(end:(-1):1))];
yy_plot = [yy_pos(nan_filter_pos), yy_neg(nan_filter_neg(end:(-1):1))];
%% plot Batman logo
% init figure
clf
axis equal
% set background to yellow
set(gca, 'Color', 'y')
% patch the Batman logo
patch(xx_plot, yy_plot, [0,0,0], ...
'EdgeColor', 'r', ...
'LineWidth', 4, ...
'LineStyle', ':')
. En bovendien dat het voorschrift uit de openingspost, klopt.
. Maar alle delen apart geven wel wat ze moeten geven. Helaas ben ik zo dom geweest om niet steeds de 'code' bij te houden. Hierdoor kan ik er maar eentje geven: