*
If the two pixels x and w are deleted first, then
q cannot be deleted by Lemma 1. This is a
contradiction as q ∈ D.
– Lastly, let us assume that the three pixels q, x,
and w are deleted at a time (i.e., in the same
subiteration). In this case
*
x is not SI-0-deletable, since it is not a 0-
border pixel (i.e., pixel q is black);
*
x is not SI-1-deletable, since it is not a 1-
border pixel (i.e., pixel p is black);
*
x is not SI-2-deletable, since it is not a 2-
border pixel (i.e., pixel w is black).
Thus all the three pixels q, x, and w need to be
SI-3-deletable. Consequently, pixels d, e, and
f are all white. It can be readily seen, by The-
orem 1, pixel q is simple iff all the three pixels
a, b, and c are also white. In this case, q is an
endpixel of type 2. Thus q is not SI-3-deletable,
and we arrive at a contradiction. (Similarly, by
Theorem 1, pixel w is simple iff all the three
pixels g, h, and i are also white. Since w is
an endpixel of type 2, it is not SI-3-deletable.
Thus we arrive at a contradiction.)
Since a non-SI-0-deletable border pixel may not
be SI-i-deletable (i = 0,1, 2,3) in the remaining
subiterations of the thinning process, this theorem
holds.
4.2 Examining Algorithm SF
Similarly to algorithm SI, we can state that algorithm
SF fulfills the 1-attempt property:
Theorem 5. Algorithm SF is 1-attempt.
Proof. Without loss of generality, we can assume that
a border pixel p to be examined is in S
0
.
We need to verify that if p ∈ B in picture
(Z
2
,8, 4,B) is non-SF-0-deletable in the actual subit-
eration, it remains non-SF-0-deletable.
Let us assume indirectly that p is SF-0-deletable
for B\ D, where D ⊆ B\ {p} is a set of deleted pixels.
Since pixel p is a non-SF-0-deletable border pixel
for B, at least one of the last two conditions of Defini-
tion 4 is violated. Consequently,
• p is an endpixel of type 1 for B, or
• p is non-simple for B.
It is obvious that if p is an endpixel of type 1 for
B (i.e., the first point holds), p remains an endpixel of
type 1 for B \ D for any D ⊆ B\ {p}. Thus only the
second point is to be examined (i.e., p is non-simple
for B).
According to our assumption, p is deleted in a re-
maining thinning phase, thus it becomes simple for
B \ D. It means that N(p) ∩ D is a simplifier set of
deleted pixels.
Since p is a non-simple pixel for B, at least one
template shown in Fig. 3 matches it. By Proposition
7 and Proposition 9 we can ignore the five templates
T
0
, T
1
, T
2
, T
3
, and T
5
. Thus T
4
is the only template to
be investigated.
If p is matched by T
4
, by Proposition 8, ex-
actly one of the two sets of pixels {q,w,x} ∩ B and
{s,t,u} ∩ B is a subset of all possible simplifier sets
associated to p. Consequently,one of {q, w,x}∩B and
{s,t,u} ∩ B is to be completely deleted. Without loss
of generality, we can assume that ({q,w, x} ∩ B) ⊆ D.
It is known that black pixel x ∈ D (see Fig. 3), thus
the following four cases depicted in Fig. 10 are to be
checked:
• q 6∈ D,w 6∈ D (see Fig. 10a):
By Lemma 2, pixel x cannot be deleted. This is a
contradiction as x ∈ D.
• q 6∈ D,w ∈ D (see Fig. 10b):
By Proposition 2, x and w belong to different sub-
fields. Thus these two pixels cannot be deleted in
the same subiteration.
– If x is deleted first, w cannot be deleted by
Lemma 1. This is a contradiction as w ∈ D.
– If w is deleted first, x cannot be deleted by
Lemma 2. This is a contradiction as x ∈ D.
• q ∈ D,w 6∈ D (see Fig. 10c):
Similarly to the previous case, by Proposition 2, x
and q belong to different subfields. Thus these two
pixels cannot be deleted in the same subiteration.
– If x is deleted first, q cannot be deleted by
Lemma 1. This is a contradiction as q ∈ D.
– If q is deleted first, x cannot be deleted by
Lemma 2. This is a contradiction as x ∈ D.
• q ∈ D,w ∈ D (see Fig. 10d):
By Proposition 2, x and q are in different sub-
fields, x and w are also in different subfields.
(Note that q and w are in the same subfield, see
Fig. 1b.) Then the following points are to be
checked:
– If just one of the two pixels q and w is deleted
first, then we get the previously examined cases
shown in Fig. 10b and Fig. 10c, respectively.
Thus we arrive at a contradiction.
– If x is deleted first, then both pixels q and w
cannot be deleted by Lemma 1. This is a con-
tradiction as q ∈ D and w ∈ D.
Since a non-SF-0-deletable border pixel in S
0
may
not be SF-i-deletable (i = 0,1,2, 3) in the remaining
subiterations, the proof by contradiction is completed.