📄 assignmentsuboptimal1.m
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function [assignment, cost] = assignmentsuboptimal1(distMatrix)
%ASSIGNMENTSUBOPTIMAL1 Compute suboptimal assignment
% ASSIGNMENTSUBOPTIMAL1(DISTMATRIX) computes a suboptimal assignment
% (minimum overall costs) for the given rectangular distance or cost
% matrix, for example the assignment of tracks (in rows) to observations
% (in columns). The result is a column vector containing the assigned
% column number in each row (or 0 if no assignment could be done).
%
% [ASSIGNMENT, COST] = ASSIGNMENTSUBOPTIMAL1(DISTMATRIX) returns the
% assignment vector and the overall cost.
%
% The algorithm is designed for distance matrices with many forbidden and
% singly validated assignments, (rows or columns containing only one
% finite element). The algorithm first searches the matrix for singly
% validated columns and rejects all assignments with multiply validated
% row. Afterwards, singly validated rows are searched and assignments to
% multiply validated columns are rejected. Then, for each row that
% validates only with singly validated columns (and the other way
% around), the minimum elements is chosen and the assignment is made. If
% there are still assignments open, the minimum element in the distMatrix
% is searched and the corresponding assignment is made.
%
% In scenarios without any forbidden assignments, the algorithm reduceds
% to the last step is will give the same result as ASSIGNMENTOPTIMAL2. If
% there are only some assignments forbidden, the algorithm will perform
% poorly because singly validated assignments are preferred.
%
% The last step can still be optimized, see the comments in
% ASSIGNMENTOPTIMAL2.
%
% Markus Buehren
% Last modified 30.01.2008
% initialize
[nOfRows, nOfColumns] = size(distMatrix);
nOfValidObservations = zeros(nOfRows,1);
nOfValidTracks = zeros(1,nOfColumns);
assignment = zeros(nOfRows,1);
cost = 0;
% compute number of validations for each track
for row=1:nOfRows
nOfValidObservations(row) = length(find(isfinite(distMatrix(row,:))));
end
if any(nOfValidObservations < nOfColumns)
if all(nOfValidObservations == 0)
return
end
repeatSteps = 1;
while repeatSteps
repeatSteps = 0;
% step 1: reject assignments of multiply validated tracks to singly validated observations
for col=1:nOfColumns
index = isfinite(distMatrix(:,col));
nOfValidTracks(col) = length(find(index));
if any(nOfValidObservations(index) == 1)
index = index & (nOfValidObservations > 1);
if any(index)
distMatrix(index, col) = inf;
nOfValidObservations(index) = nOfValidObservations(index) - 1;
nOfValidTracks(col) = nOfValidTracks(col) - length(find(index));
repeatSteps = 1;
end
end
end
% step 2: reject assignments of multiply validated observations to singly validated tracks
if nOfColumns > 1
for row=1:nOfRows
index = isfinite(distMatrix(row,:));
if any(nOfValidTracks(index) == 1)
index = index & (nOfValidTracks > 1);
if any(index)
distMatrix(row, index) = inf;
nOfValidTracks(index) = nOfValidTracks(index) - 1;
nOfValidObservations(row) = nOfValidObservations(row) - length(find(index));
repeatSteps = 1;
end
end
end
end
end % while repeatSteps
%disp(sprintf('xx = %d', xx));
% for each multiply validated track that validates only with singly validated
% observations, choose the observation with minimum distance
for row=1:nOfRows
if nOfValidObservations(row) > 1
index = isfinite(distMatrix(row,:));
if all(nOfValidTracks(index) == 1)
[minDist, col] = min(distMatrix(row,:));
assignment(row) = col;
cost = cost + minDist;
distMatrix(row,:) = inf;
distMatrix(:,col) = inf;
end
end
end
% for each multiply validated observation that validates only with singly validated
% track, choose the track with minimum distance
for col=1:nOfColumns
if nOfValidTracks(col) > 1
index = isfinite(distMatrix(:,col));
if all(nOfValidObservations(index) == 1)
[minDist, row] = min(distMatrix(:,col));
assignment(row) = col;
cost = cost + minDist;
distMatrix(row,:) = inf;
distMatrix(:,col) = inf;
end
end
end
end
% now, recursively search for the minimum element and do the assignment
while 1
% find minimum distance observation-to-track pair
[minDist, index1] = min(distMatrix, [], 1);
[minDist, index2] = min(minDist);
row = index1(index2);
col = index2;
if isfinite(minDist)
% make the assignment
assignment(row) = col;
cost = cost + minDist;
distMatrix(row, :) = inf;
distMatrix(:, col) = inf;
else
break
end
end
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