1) The document presents Lemma 1 which establishes that the length of the face-crossing path segment between oi and oi+1 is at most the length of the corresponding surface path segment between pi and pi+1 plus one divided by k+1 times the length of the longest edge of the crossed face, where k is the number of cut-vertices along each edge of the face. 2) It provides a proof sketch that divides each edge of the face into intervals of length l/(k+1) since there are k evenly placed cut-vertices along each edge. 3) Case 1 of the proof is illustrated, where pi and pi+1 are located on cut-vertices of adjacent edges

1. A
pi+1
oi+1
oi
pi
B
C
(a) Case 1
Figure 1: Cases 1 and k = 2.
Lemma 1 (1/2 Lower Bound) Let Π(s, t) and ΠF C (s, t) be the the shortest surface path and
corresponding surface face-crossing path between source s and destination t on tetrahedral
surface P, respectively. Further, let (oi , oi+1 ) denote the segment of path ΠF C (s, t) that crosses
face fi and (pi , pi+1 ) denote the segment of path Π(s, t) that crosses face fi . Then,
(oi , oi+1 ) −
l
≤ (pi , pi+1 ),
k+1
where l is the length of the longest edge of face fi and k is the number of cut-vertices placed
along each edge of the face.
Proof Sketch: Each edge of the face fi has k evenly placed points on the edge, hence it divides
l
the edge into intervals of size k+1 .
Case 1: pi and pi+1 are located on cut-vertices of adjacent edges This case is illustrated
l
l
in Figure 1(a). In this case, |(pi , oi )| ≤ 1 k+1 and |(pi+1 , oi+1 )| ≤ 1 k+1 . Combining, the two
2
2
l
and applying triangle inequality as in the previous case we get, (oi , oi+1 ) − k+1 ≤ |(pi , pi+1 )|.
Note, that we must take into account the differences for both pi and pi+1 to their corresponding
oi and oi+1 .
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