Davies: Computer Vision, 5th edition: Solutions to selected problems 29
© E. R. Davies 2017
A1–B2–C3–D4; A2–B1; B3–C2; C4–D3; D1–A4; A3–C1; B4–D2; A5–C6; A6–C5.
The 4-element maximal clique wins over the first six 2-element cliques, leaving the
last two 2-element maximal cliques, which constitute equally probable solutions at this
level.
Thus we have an almost certain 4-feature object, and a possible 2-feature object
which is partly occluded or defective and which may be either way around.
(b) If the objects have an axis of symmetry, we will get more compatibilities, as the two
features B, C related by symmetry cannot be distinguished. The final result is that the two
objects each have twice as many maximal clique solutions:
In addition, the size of the match graph goes down from 4 6 to 3 6, saving storage
and computation, and arriving at only the meaningful answers.
(c) For the case where there are three sizes of hole, the extraneous 2-element cliques will
mostly not arise, the only one that will remain will be D1–A4. Thus the final solutions
are:
(d) However, if the further matching strategy is employed, the equal sized holes in the
image are features 1, 4, 5, and the equal holes in the template are A and D. This means
that we get just the following subset of compatibilities:
Both the 3-element cliques represent the same object, but there is only one clique to
represent the occluded object. Thus the technique locates both the objects, in spite of the
occlusion, though some means must be adopted for choosing between the two solutions
for the same object.