%------------------------------------------------
% 2011/01/02 version 1.0
%
% [xxx, yyy] = LatexFilterHullCurve(x, y, NumberOfPoints)
%
%------------------------------------------------
% output:
% xxx, yyy: filtered data
% input:
% x, y: data to be filtered
% tol: tolerance
% author: Thomas Koenig
% date: 2011/01/02
%
% Copyright 2010 Thomas König, Alexander Michel
%
% This file is part of NumericPlots.
%
% NumericPlots is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
% any later version.
%
% NumericPlots is distributed in the hope that it will be useful,
% but WITHOUT ANY WARRANTY; without even the implied warranty of
% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
% GNU General Public License for more details.
%
% You should have received a copy of the GNU General Public License
% along with NumericPlots. If not, see .
function [xxx, yyy] = LatexFilterMinMax(x, y, NumberOfPoints)
% fprintf('LatexFilter: Input - %i ...', size(x,2));
if (NumberOfPoints - round(NumberOfPoints))~=0
error('NumberOfPoints must be an integer.');
end
if NumberOfPoints<3
error('NumberOfPoints<3 does not make much sense.');
end
if NumberOfPoints>size(x, 2)/2
% This would mean, that the filtered data would have more points than
% the input data. This filter is not meant to do that.
error('NumberOfPoints>length(x)');
end
IntLength = floor(size(x, 2)/NumberOfPoints);
% The last NrLongInt intervals are enlarged by one to get all points in
% x.
NrLongInt = size(x,2)-(IntLength*NumberOfPoints);
IntLengthLong = IntLength+1;
xxx = nan(size(x, 1), NumberOfPoints*2+2);
yyy = nan(size(y, 1), NumberOfPoints*2+2);
% always use first and last value
xxx(:,1) = x(:,1);
yyy(:,1) = y(:,1);
xxx(:,end) = x(:,end);
yyy(:,end) = y(:,end);
% Search maximum and minimum in each interval of length IntLength
for k=0:NumberOfPoints-NrLongInt-1
[mi, mi_idx] = min(y(:,(k*IntLength+1):(k+1)*IntLength), [], 2);
[ma, ma_idx] = max(y(:,(k*IntLength+1):(k+1)*IntLength), [], 2);
if mi_idx