Discrete and Computational Geometry |
by Satyan Devadoss and Joseph O'Rourke |
Chap |
Page |
Location |
Old |
New |
Explanation |
Corrector |
Date |
---|---|---|---|---|---|---|---|
1 |
7 |
Cor. 1.9 |
has at least two ears. | has at least two ears with non-adjacent tips. | This stronger statement is needed for the induction step, and is used in the exercises. | Giovanni Viglietta |
27 Jun 2011 |
1 |
11 |
Thm.1.20 |
let P be any polygon with n labeled ordered vertices...let Q be a convex polygon also with n vertices | let P be any polygon with n+2 labeled ordered vertices...let Q be a convex polygon also with n+2 vertices | P and Q should have n+2 vertices. | Mario Lopez |
6 Sep 2013 |
1 |
22 |
Ex. 1.48 |
The claim in the exercise is false. | Will delete or modify in 2nd Ed. | Student found counterexample | Lori Ziegelmeier & Caleb Williams | 9 Oct 2023 |
1 |
27-28 |
Line-2 |
More precisely, let f : R → Q ... | [to be written] | Our attempt here to simplify Dehn's proof is incorrect. The function f should be defined on R / (π · Q), and we need to define a basis for the Q-linear vector space. We will be writing up a revsion of the proof. Meanwhile, consult Proofs from THE BOOK (Aigner, Ziegler), Chap. 7. | JOR & SLD |
22 Nov 2011 |
1 |
30 |
Example 1.64 |
could be the identity function on...= 3√2 f(arctan √2) | could be the constant function 1 on ...= 3√2 | The identity function fails to satisfy Property 1 on p.28. | JOR |
6 Nov 2011 |
3 |
83 |
Thm. 3.49 |
and A within and C outside the circle | and A within and C outside the circle where A, B, C lie on the same side of the line through P and Q | Assumption needed. | Chris Atkinson |
27 Sep 2012 |
5 |
132,4 |
Figs. 5.12,14 |
Bottom edge of R in 5.12, and top edge of -R in 5.14. Convolution curve is missing a loop at the upper reflex vertex. | The topmost blue edge of -R in Fig.5.14 should have a larger slope so that it is steeper than the nearly parallel red edge incident to the top corner of P1. For a sketch of the repaired figure, see Fig5.14.jpg. | The nearly parallel edges was an accident. The convolution should continue ccw cycling around -R until the outgoing P1 edge direction is reached. | Günter Rote |
17 May 2011 |
5 |
137 |
Line+3 |
The many convolution cycles in Figure 5.17 | The many convolution loops in Figure 5.17 | There is just one convolution cycle, which self-crosses in many "loops." As Günter points out, there can in fact be several convolution cycles that require new starting points to detect. | Günter Rote |
17 May 2011 |
6 |
157 |
Exer. 6.2. |
Show that each face of a convex polyhedron must | Show that each face of a convex polyhedron without coplanar faces must | Temporary definition permits coplanar faces, but then could partition a convex face into nonconvex subfaces. | Giovanni Viglietta |
12 Mar 2011 |
6 |
177 |
Thm. 6.35 |
with corresponding faces congruent, then | with corresponding faces congruent and similarly arranged around each vertex, then | Here is an example of Connelly showing two incongruent polyhedra composed of congruent faces, but the correspondence between the two does not map the angles around each vertex identically. | Robert Connelly |
26 Aug 2011 |
Chap |
Page |
Location |
Old |
New |
Explanation |
Corrector |
Date |
---|---|---|---|---|---|---|---|
1 |
19 |
Line+2 of bottom para |
In 1992, Raimund Seidel constructed | In 1982, Raimund Seidel and William Thurston independently constructed | Typo on date; found 2nd reference. | Giovanni Viglietta; JOR |
12 Mar 2011 |
1 |
12 |
Line-5 |
in any of its triangulation. | in any of its triangulations. | Typo | Jennifer Sadler |
14 Sep 2011 |
1 |
13 |
Exer. 1.26 |
Should be starred. | ∗ Exercise 1.26. | It is a hard exercise! | Giovanni Viglietta |
27 Jun 2011 |
1 |
14 |
Exer. 1.28 |
Are there are polygons | Are there polygons | Typo | Giovanni Viglietta; Justin Iwerks |
27 Jun 2011 |
1 |
18 |
Thm. 1.38 |
⌈n/2⌉ guards are needed | ⌈n/2⌉ boundary guards are needed | (In contrast to Exer. 1.40's "anywhere in the plane.") | Giovanni Viglietta |
27 Jun 2011 |
1 |
26 |
Line+8 of Sec. 1.5 |
Max Dehn a few years later. | Max Dehn within a year. | The publication is: "Über den Rauminhalt", Mathematische Annalen 55 (1901), no. 3, pages 465–478. | Giovanni Viglietta |
27 Jun 2011 |
1 |
26 |
Line+3 of Definition |
n1 . n2 = cos θ | n1 . n2 = – cos θ | The minus sign is needed because these are outward-pointing normals. | Giovanni Viglietta |
27 Jun 2011 |
1 |
30 |
Line+1 after Thm 1.67 |
This, along with the Dehn-Hadwiger theorem, show that | This, along with the Dehn-Hadwiger theorem, shows that | Grammar. | Giovanni Viglietta |
27 Jun 2011 |
1 |
24,25, 252 |
several locations |
Bolyai-Gerwein | Bolyai-Gerwien | Spelling: ei vs. ie | SLD |
17 Mar 2016 |
2 |
35 |
Part II, after 1st displayed equation |
conv(S') ⊂ conv(S) [twice] | conv(S') ⊆conv(S) | It could be that pn is in S'. | Giovanni Viglietta |
27 Jun 2011 |
2 |
35 |
Fig. 2.5 caption. |
Tangent line lines transitioning | Tangent line transitioning | Typo | Giovanni Viglietta |
27 Jun 2011 |
2 |
43 |
Line-3 |
, which is n. | , which is n-1. | ~Typo. One point to all others. | Giovanni Viglietta |
27 Jun 2011 |
2 |
43 |
Line+6 |
makes the largest angle | makes the smallest angle | Typo | Lori Ziegelmeier | 9 Oct 2023 |
2 |
51 |
Line+3 of Sec. 2.7 |
a convex polyhedra | a convex polyhedron | Typo | Giovanni Viglietta |
27 Jun 2011 |
2 |
53 |
Line+7 |
all but one of them extends | all but one of them extend | Grammar | Giovanni Viglietta |
27 Jun 2011 |
3 |
65 |
Uns. Prob. 11 |
an exponential number of triangles | an exponential number of triangulations | Typo. | Dianna Xu |
24 Feb 2013 |
3 |
60 |
Line-1 above box |
Notice that both ... affects | Notice that both ... affect | Grammar | Giovanni Viglietta |
27 Jun 2011 |
3 |
69 |
Line+2 of para. before Cor. 3.24 |
A quantity that give | A quantity that gives | Typo | Giovanni Viglietta |
27 Jun 2011 |
3 |
68 |
Line-1 |
which by our induction hypothesis | which by our induction hypothesis establishes the theorem. | Printing error (last three words of sentence missing). | Justin Iwerks |
15 Aug 2011 |
3 |
75 |
2nd para., Line+9 |
based on the six ways a hexagon may be cut into two quadrilaterals | based on the three ways a hexagon may be cut into two quadrilaterals | ~Typo | Justin Iwerks |
15 Aug 2011 |
3 |
96 |
Unsolved Prob. 18, parenthetical sentence |
for all points sets | for all point sets | Typo | Justin Iwerks |
19 Aug 2011 |
4 |
102 |
Line+3 after Exer. 4.8 |
not all these bisectors becomes Voronoi edges | not all these bisectors become Voronoi edges | Typo | Justin Iwerks |
27 Aug 2011 |
4 |
103 |
Proof of Theorem 4.12 |
so the number of notes of G is n | so the number of faces of G is n | ~Typo | Uuganbaatar Ninjbat |
2 Jan 2017 |
4 |
104 |
Line +2 |
Substituting e = (v+1)+b-2 | Substituting e = (v+1)+n-2 | Typo | Uuganbaatar Ninjbat |
3 Jan 2017 |
4 |
110 |
Theorem 4.26 |
We should have mentioned explicitly that it is easy to construct Vor(S) from Del(S) and vice versa. | Uuganbaatar Ninjbat |
2 Jan 2017 |
||
4 |
115 |
Fig. 4.12 caption |
The point set in identical | The point set is identical | Typo | Justin Iwerks |
27 Aug 2011 |
4 |
117 |
4th reference |
Otfried Schwartzkopf | Otfried Cheong | (He changed his name.) | Justin Iwerks |
27 Aug 2011 |
5 |
133 |
Definition eqn for α * β |
x ∈ A, y ∈ B | x ∈ α, y ∈ β | Typo. | Dianna Xu |
3 Nov 2015 |
6 |
181 |
Line-2 |
... + 6 f7 + ... | ... + 6 f6 + ... | Typo: f7 → f6 | SLD |
30 Mar 2016 |
7 |
207 |
Motion Planning Box heading |
Voronoi Diagram Algorithm | Path Finding Algorithm | ~Typo. | SLD |
28 Feb 2011 |