Page 409 - Introduction to AI Robotics

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is the
.Since dontknow
regions: three for O and one for dontknow 11 Localization and Map Making
ccupied
set = fOccupied;Empty g, the intersection of dontknow and O is
ccupied
.
T
he
O c c u p i e dsubregions of a square of unit 1.0 area can be projected onto a
numberline of length one. This means that the “area” of belief mass created
by the intersections of the two belief functions forms a third belief function!
The belief mass for Occupied is taken by summing the areas of each of the
three subregions, as seen in Fig. 11.6c. Therefore, the orthogonal sum is:
ccupied
B 1 e = l m(O ) :4; = 0
m(E ) :0 = 0
mpty
m(dontknow ) :6 = 0
ccupied
B 2 e = l m(O ) :6; = 0
mpty
m(E ) :0; = 0
m(dontknow ) :4 = 0
B 3 = eB l 1 eB l 2 e = l m(O ) :76 ; = 0
ccupied
mpty
m(E ) :0; = 0
m(dontknow ) :24 = 0

Now consider the case where the two readings are contradictory:

mpty
B 1 e = l m(O ) :4; (E m) = :0; (dontknow m =) :6 = 0 0 0
ccupied
B 2 e = l m(O ) :0; (E m) = :6; (dontknow m =) :4 = 0 0 0
ccupied
mpty
As shown in Fig. 11.7, there is now a region where the intersection of Oc-
cupied and Empty occurs. Since these are mutually exclusive propositions,
the resulting set intersection is the empty set ;. The emergence of a subre-
gion associated with ; is a problem. Recall from the definition of a Shafer
belief function that no belief mass can be assigned to ;. But if the area of 0.24
associated with ; is simply left out, the resulting combined belief function
will not equal 1.0.
Dempster’s rule solves this problem by normalizing the belief function; the
area for ; is distributed equally to each of the non-empty areas. Each area
gets a little bigger and they now all sum to 1.0. The normalization can be
carried out by noticing that the belief mass for a particular proposition C
is really the sum of all k areas with set C divided by the total area of valid