t
A brief analysis of packing regular pentagons in
the plane
Wöden Kusner1
Technische Universität Graz
ESI
October 2014
1
Supported by an FWF special research program, NSF grant 1104102 and a fellowship from the University of Pittsburgh.
Wöden Kusner TU Graz
1 / 35
t
Abstract
It is a still an open problem to describe the densest packing of the
plane by regular pentagons. Here, the general problem is described
near a conjectured optimum by a nonlinear programming problem.
Under certain assumptions, that optimum can be certified locally via a
linear programming problem. This method can be used to
computationally prove* the local optimality of the conjectured globally
optimal pentagon packing.
Wöden Kusner TU Graz
2 / 35
Motivation
t
Density
A packing of a region X ⊆ Rn by objects Pi ⊆ X is a countable family
P = {Pi }i∈I with disjoint interiors.
The upper density ρ+ of a packing P in Rn will be defined as
ρ+ (P) = lim sup
r →∞
Vol(r Bn ∩ P)
,
Vol(r Bn )
where Bn is the unit n-ball centered at 0.
Remark
The packing density is bounded by the density of the densest “cell” in
some “nice decomposition” of the packing.
Wöden Kusner TU Graz
3 / 35
Pentagon Packing
t
Pentagons
The best lower bound for the density of pentagon packings and the
conjectured maximal density
configuration is shown. This packing
√
has a density of (5 − 5)/3 = 0.92131 . . . .
Wöden Kusner TU Graz
4 / 35
Pentagon Packing
t
Pentagons
For circles, saturated packings and the existence of Delaunay
triangulations can guide a proof of Thue’s theorem.
This doesn’t work for pentagons. The conjectured best packing is not
a even critical point for density inside a Delaunay triangle.
Wöden Kusner TU Graz
5 / 35
Pentagon Packing
t
Pentagons
For four pentagons there is a pairing that gives a local result.
Wöden Kusner TU Graz
6 / 35
Pentagon Packing
t
Nonlinear Program
The problem of finding the densest packing of pentagons can be
phrased as the nonlinear program: Maximize the density of a
configuration of pentagons subject to the condition that the
pentagons do not intersect.
Parameterize the configuration space of 4 pentagons not with
independent pentagons, but rather as a coupled system.
Wöden Kusner TU Graz
7 / 35
Pentagon Packing
t
Objective
The objective function for the nonlinear program is defined in terms of
the areas and corresponds to the normalized density function on a
neighborhood of the conjectured optimal configuration. The objective
is made up of constituent functions which are the areas of various
triangular regions of the pentagons and the area of the convex hull of
{C1 , C2 , C3 , C4 }. The objective function may be written as
4
X
Area(Hull({C1 , C2 , C3 , C4 }) ∩ Pi )
i=1
Wöden Kusner TU Graz
Area(Hull({C1 , C2 , C3 , C4 }))
√
5− 5
−
.
3
8 / 35
Pentagon Packing
t
Objective
Wöden Kusner TU Graz
9 / 35
Pentagon Packing
t
Objective
1
r(1+
√5
√5
1 (1−√5)+ 1 (−1+√5)− 1
2 )( 5 −
5
√5
(1+ 4
4
4
8
8 )(1+√5))(r 8 − 8 −x1)
r
2(−
−x8+ 1
√5
4 (−1+√5)+1)
5−
8
8
v
r(1+
u
√5
√5
u
1 (1−√5)+ 1 (−1+√5)− 1
2 )( 5 −
5
u
√5
(1+ 4
4
4
8
8 )(1+√5))(r 8 − 8 −x1)
u
v
u
r(1+
r
(−
−x8+ 1
u
√5
√5
√5
4 (−1+√5)+1)2
u
u
1
1
2 )( 5 −
5
u
5−
u
√5
√5
√5
√5
(1+ 1
u
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(r 8 − 8 −x1)
u
8
8
2 )( 5 −
5
5
u1−
u
r
√5
+x8+ 1
r(1+
√5
√5
√5
8
8 )(r 8 − 8 −x1)+x7−r 8 + 8 )2+(
4 (1−√5)−1)2 u
t(−r(1+
u
5
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
(1+ 1
8− 8
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(r 8 − 8 −x1)
u
5
5
2 )( 5 −
1 (1−√5)−1)2
u
r
(−r(1+ √5
+x8+ 4
√5
8
8 )(r 8 − 8 −x1)+x7−r 8 + 8 )2+(
t
5
8− 8
+
√
+2
2((
5−2√5
1
1
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
r(1+
r
√5
√5
2 )( 5 −
5
√5
8
8 )(1+√5))(x1+r 8 − 8 )
√5
5−
8
8
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
2 )( 5 −
5
5
2 )( 5 −
1
u
r
√5
−x2+x5+r(1+ √5
√5
8
8 )(x1+r 8 − 8 )+x4+r 8 + 8 )2+(
8
8 )+ 4 (1−√5))2
t(−x1+r(1+
5
8− 8
√
1
√5
√5
√5
√5
2 )( 5 −
1
π
2 )( 5 −
5
5
π
√5
−x2+x5+r(1+ √5
8
8 )+ 4 (1−√5)) cos( 5 −x3)+(−x1+r(1+
8
8 )(x1+r 8 − 8 )+x4+r 8 + 8 ) sin( 5 −x3))
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
2 )( 5 −
1
π
2 )( 5 −
5
5
π
u
r
√5
((
−x2+x5+r(1+ √5
√5
8
8 )+ 4 (1−√5)) cos( 5 −x3)+(−x1+r(1+
8
8 )(x1+r 8 − 8 )+x4+r 8 + 8 ) sin( 5 −x3))2
u
u
5−
u
8
8
π
u1−
cos2( π
r(1+
u
√5
√5
5 −x3)+sin2( 5 −x3)u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
5
5
2 )( 5 −
2 )( 5 −
1
π
π
u
r
((−x1+r(1+ √5
−x2+x5+r(1+ √5
√5
8
8 )(x1+r 8 − 8 )+x4+r 8 + 8 )2+(
8
8 )+ 4 (1−√5))2)(cos2( 5 −x3)+sin2( 5 −x3))
t
5
8− 8
+
√
+2
2((
5−2√5
1
1
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
r(1+
r
√5
√5
2 )( 5 −
5
√5
8
8 )(1+√5))(r 8 − 8 −x1)
√5
5−
8
8
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(r 8 − 8 −x1)
u
2 )( 5 −
5
5
2 )( 5 −
1
u
r
√5
−x2+x8+r(1+ √5
√5
8
8 )(r 8 − 8 −x1)−x1+x7−r 8 + 8 )2+(
8
8 )+ 4 (1−√5))2
t(−r(1+
5
8− 8
1
√5
√5
√5
√5
2 )( 5 −
1
π
2 )( 5 −
5
5
π
√5
−x2+x8+r(1+ √5
8
8 )+ 4 (1−√5)) cos(x3+ 5 )−(−r(1+
8
8 )(r 8 − 8 −x1)−x1+x7−r 8 + 8 ) sin(x3+ 5 ))
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(r 8 − 8 −x1)
u
2 )( 5 −
1
π
2 )( 5 −
5
5
π
u
r
√5
((
−x2+x8+r(1+ √5
√5
8
8 )+ 4 (1−√5)) cos(x3+ 5 )−(−r(1+
8
8 )(r 8 − 8 −x1)−x1+x7−r 8 + 8 ) sin(x3+ 5 ))2
u
u
5−
u
8
8
π u1−
cos2(x3+ π
r(1+
u
√5
√5
5 )+sin2(x3+ 5 )u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(r 8 − 8 −x1)
u
5
5
2 )( 5 −
2 )( 5 −
1
π
π
u
r
((−r(1+ √5
−x2+x8+r(1+ √5
√5
8
8 )(r 8 − 8 −x1)−x1+x7−r 8 + 8 )2+(
8
8 )+ 4 (1−√5))2)(cos2(x3+ 5 )+sin2(x3+ 5 ))
t
5
8− 8
+
√
+2
2((−
1
1
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
r(1+
5−2√5
r
√5
√5
2 )( 5 −
5
√5
8
8 )(1+√5))(x1+r 8 − 8 )
√5
5−
8
8
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
2 )( 5 −
5
5
1 (−1+√5)+1)2
u
r
√5
−x5+ 4
√5
8
8 )(x1+r 8 − 8 )−x4−r 8 + 8 )2+(−
t(−r(1+
5
8− 8
√
1
1
2 (− 4
r
5(3+√5) 2
x1
5−√5
−
1
4
q
3+√5 2
x
5−√5 1
2
+ 2q 5−√5
x21 −
1
2
q
1
2 (3
+
√5)x
2 x1
+ x5 x1 +
1
2
q
1
2 (1
+
2
√5
)( 58
−
√5
8
)(3 +
√5)x
1
−
√
1
8
1
+
√5
√5
√5
π
1
1
2 )( 5 −
5
5
π
√5
−x5+ 1
4 (−1+√5)+1)(sin( 10 −x6)+ 4 (−1+√5)+ 4 (1−√5))−(−r(1+
8
8 )(x1+r 8 − 8 )−x4−r 8 + 8 ) cos( 10 −x6))
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
π
1
1
2 )( 5 −
5
5
π
u
r
√5
((−
−x5+ 1
√5
4 (−1+√5)+1)(sin( 10 −x6)+ 4 (−1+√5)+ 4 (1−√5))−(−r(1+
8
8 )(x1+r 8 − 8 )−x4−r 8 + 8 ) cos( 10 −x6))2
u
u
5−
u
8
8
π −x6)+(sin( π −x6)+ 1 (−1+√5)+ 1 (1−√5))2 u1−
cos2( 10
r(1+
u
√5
√5
10
4
4
u
2 )( 5 −
5
1 (1−√5)+ 1 (−1+√5)− 1
u
√5
√5
√5
√5
(1+ 4
4
4
8
8 )(1+√5))(x1+r 8 − 8 )
u
5
5
2 )( 5 −
1 (−1+√5)+1)2)(cos2( π −x6)+(sin( π −x6)+ 1 (−1+√5)+ 1 (1−√5))2)
u
r
((−r(1+ √5
−x5+ 4
√5
8
8 )(x1+r 8 − 8 )−x4−r 8 + 8 )2+(−
10
10
4
4
t
5
q
5
2 (3
+
√5)x
1
+
3
8
q
1
2 (3
+
√5)x
1
−
√5x
4
1
+
x1
4
−
1
2
q
1
2 (5
+
√5)x
2
−
1
8
q(5
−
√5)(3
+
√5)x
2
− x2 x4 +
q(1
8− 8
+
2
√5
)( 58
−
√5
8
)x4 + x4 +
1
2
q
1
2 (1
+
2
√5
)( 58
−
√5
8
)(5 +
√5)
+
1
2
q
1
2 (5
+
√5)
+
1
8
q(1
√
+2
2((−
1 (1−√5)+ 1 (−1+√5)− 1
(1+ 4
4
4
r(1+
5−2√5
r
√5
√5
2 )( 5 −
5
√5
8
8 )(1+√5))(x1+r 8 − 8 )
√5
5−
8
8
v
r(1+
u
√5
√5
u
1 (1−√5)+ 1 (−1+√5)− 1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 4
4
4
8
8 )(1+√5))(x1+r 8 − 8 )
u
2 )( 5 −
5
5
2 )( 5 −
u
r
√5
√5
+x2−x5+ 1
√5
8
8 )(x1+r 8 − 8 )−x4−r 8 + 8 )2+(−
4 (−1+√5)−r(1+
8
8 ))2
t(x1−r(1+
5
8− 8
+
2
√5
)(5
−
√5)( 5
8
−
√5
8
)(3 +
√5)
+
1
8
q(5
−
√5)(3
+
√5))
+ 12 (− 14
r
5(3+√5) 2
x1
5−√5
−
1
4
q
3+√5 2
x
5−√5 1
2
+ 2q 5−√5
x21 +
1
2
q
1
2 (3
+
√5)x
2 x1
− x8 x1 −
√
1
2
1
1
2 (1
+
2
√5
)( 85
−
√5
8
)(3 +
√5)x
1
+
1
8
q
5
2 (3
+
√5)x
1
+
√5
√5
√5
√5
2 )( 5 −
π
1
1
2 )( 5 −
5
5
π
√5
√5
+x2−x5+ 1
4 (−1+√5)−r(1+
8
8 ))(sin( 10 −x6)+ 4 (−1+√5)+ 4 (1−√5))−(x1−r(1+
8
8 )(x1+r 8 − 8 )−x4−r 8 + 8 ) cos( 10 −x6))
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
2 )( 5 −
π
1
1
2 )( 5 −
5
5
π
u
r
√5
√5
((−
+x2−x5+ 1
√5
4 (−1+√5)−r(1+
8
8 ))(sin( 10 −x6)+ 4 (−1+√5)+ 4 (1−√5))−(x1−r(1+
8
8 )(x1+r 8 − 8 )−x4−r 8 + 8 ) cos( 10 −x6))2
u
u
5−
u
8
8
π −x6)+(sin( π −x6)+ 1 (−1+√5)+ 1 (1−√5))2 u1−
cos2( 10
r(1+
u
√5
√5
10
4
4
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
5
5
2 )( 5 −
2 )( 5 −
π
π
1
1
u
r
√5
((x1−r(1+ √5
+x2−x5+ 1
√5
8
8 )(x1+r 8 − 8 )−x4−r 8 + 8 )2+(−
4 (−1+√5)−r(1+
8
8 ))2)(cos2( 10 −x6)+(sin( 10 −x6)+ 4 (−1+√5)+ 4 (1−√5))2)
t
5
q
1
−
3
8
q
1
2 (3
+
√5)x
1
+
√5x
4
1
−
x1
4
−
1
2
q
8− 8
1
2 (5
+
√5)x
2
−
1
8
q(5
−
√5)(3
+
√5)x
2
+ x2 x7 −
q(1
+
2
√5
)( 85
−
√5
8
)x7 − x7 +
1
2
q
1
2 (1
+
2
√5
)( 58
−
√5
8
)(5 +
√5)
+
1
2
√
+2
q
+
√5)
+
r(1+
√5
√5
2 )( 5 −
5
√5
8
8 )(1+√5))(r 8 − 8 −x1)
1 (−1+√5)+1)(sin(x9+ π )+ 1 (−1+√5)+ 1 (1−√5)))
r
−x8+ 4
√5
10
4
4
5−
8
8
v
r(1+
u
√5
√5
u
1 (1−√5)+ 1 (−1+√5)− 1
2 )( 5 −
5
u
√5
√5
√5
√5
(1+ 4
4
4
8
8 )(1+√5))(r 8 − 8 −x1)
u
2 )( 5 −
5
5
π
π
1
1
u
r
((r(1+ √5
−x8+ 1
√5
8
8 )(r 8 − 8 −x1)−x7+r 8 + 8 ) cos(x9+ 10 )+(−
4 (−1+√5)+1)(sin(x9+ 10 )+ 4 (−1+√5)+ 4 (1−√5)))2
u
u
5−
u
8
8
π )+(sin(x9+ π )+ 1 (−1+√5)+ 1 (1−√5))2 u1−
cos2(x9+ 10
r(1+
u
√5
√5
10
4
4
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(r 8 − 8 −x1)
u
5
5
2 )( 5 −
1 (−1+√5)+1)2)(cos2(x9+ π )+(sin(x9+ π )+ 1 (−1+√5)+ 1 (1−√5))2)
u
r
((r(1+ √5
−x8+ 4
√5
8
8 )(r 8 − 8 −x1)−x7+r 8 + 8 )2+(−
10
10
4
4
t
5
1
1
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
2 )( 5 −
5
5
π
2((r(1+ √5
8
8 )(r 8 − 8 −x1)−x7+r 8 + 8 ) cos(x9+ 10 )+(−
5−2√5
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(r 8 − 8 −x1)
u
2 )( 5 −
5
5
u
r
√5
−x8+ 1
√5
8
8 )(r 8 − 8 −x1)−x7+r 8 + 8 )2+(−
4 (−1+√5)+1)2
t(r(1+
5
8− 8
1
2 (5
1
8
q(1
+
2
√5
)(5
−
√5)( 5
8
−
√5
8
)(3 +
√5)
+
1
8
q(5
√
8− 8
−
√5)(3
+
√5))
+
√
+2
1
r(1+
√5
√5
2 )( 5 −
5
√5
√5
8
8 )(1+√5))(r 8 − 8 −x1)
1 (−1+√5)−r(1+ √5
2 )( 5 −
π
1
1
r
+x2−x8+ 4
√5
8
8 ))(sin(x9+ 10 )+ 4 (−1+√5)+ 4 (1−√5)))
5−
8
8
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(r 8 − 8 −x1)
u
2 )( 5 −
5
5
π
1 (−1+√5)−r(1+ √5
2 )( 5 −
π
1
1
u
r
((r(1+ √5
+x2−x8+ 4
√5
8
8 )(r 8 − 8 −x1)+x1−x7+r 8 + 8 ) cos(x9+ 10 )+(−
8
8 ))(sin(x9+ 10 )+ 4 (−1+√5)+ 4 (1−√5)))2
u
u
5−
u
8
8
π )+(sin(x9+ π )+ 1 (−1+√5)+ 1 (1−√5))2 u1−
cos2(x9+ 10
r(1+
u
√5
√5
10
4
4
u
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(r 8 − 8 −x1)
u
5
5
2 )( 5 −
1 (−1+√5)−r(1+ √5
2 )( 5 −
π
π
1
1
u
r
((r(1+ √5
+x2−x8+ 4
√5
8
8 )(r 8 − 8 −x1)+x1−x7+r 8 + 8 )2+(−
8
8 ))2)(cos2(x9+ 10 )+(sin(x9+ 10 )+ 4 (−1+√5)+ 4 (1−√5))2)
t
5
8− 8
1
1
√5
√5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
2 )( 5 −
5
5
π
2((r(1+ √5
8
8 )(r 8 − 8 −x1)+x1−x7+r 8 + 8 ) cos(x9+ 10 )+(−
5−2√5
v
r(1+
u
√5
√5
u
1 (1−√5)+ 1 (−1+√5)− 1
2 )( 5 −
5
u
√5
√5
√5
√5
√5
(1+ 4
4
4
8
8 )(1+√5))(r 8 − 8 −x1)
u
2 )( 5 −
5
5
1 (−1+√5)−r(1+ √5
2 )( 5 −
u
r
√5
+x2−x8+ 4
√5
8
8 )(r 8 − 8 −x1)+x1−x7+r 8 + 8 )2+(−
8
8 ))2
t(r(1+
5
8− 8
√
+
√
+2
5−2√5
1 √
2
2q1− √5
+2 5−2√5
+
1
r(1+
√5
√5
1
1
2 )( 5 −
5
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
r
2(−
−x5+ 1
√5
4 (−1+√5)+1)
5−
8
8
v
r(1+
u
√5
√5
u
1
1
2 )( 5 −
5
u
√5
(1+ 1
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
v
u
r(1+
r
(−
−x5+ 1
u
√5
√5
√5
4 (−1+√5)+1)2
u
u
1
1
2 )( 5 −
5
u
5−
u
√5
√5
√5
√5
(1+ 1
u
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
8
8
2 )( 5 −
5
5
u1−
u
r
√5
+x5+ 1
r(1+
√5
√5
√5
8
8 )(x1+r 8 − 8 )+x4+r 8 + 8 )2+(
4 (1−√5)−1)2 u
t(r(1+
u
5
1
1
2 )( 5 −
5
u
√5
√5
√5
√5
(1+ 1
8− 8
4 (1−√5)+ 4 (−1+√5)− 4
8
8 )(1+√5))(x1+r 8 − 8 )
u
5
5
2 )( 5 −
u
r
(r(1+ √5
+x5+ 1
√5
8
8 )(x1+r 8 − 8 )+x4+r 8 + 8 )2+(
4 (1−√5)−1)2
t
5
8− 8
√
+2
5−2√5
1
Wöden Kusner TU Graz
10 / 35
((
Pentagon Packing
t
Objective
(1+ 1
4 (1
p5)+ 1
4 ( 1+p5)
)(1+p5))(x1+r 5
8
1
4
r(1+
r
5
8
2 )( 5
p5
8
p5
p5
5
8 )(1+p5))(x1+r 8
8 )
1
p5
8 )
2 )( 5
x2+x5+r(1+ p5
8
8
p5
2 )( 5
x2+x5+r(1+ p5
8
p5
1
8 )+ 4 (1
p5))2
p
cos2( ⇡
5
x3)+sin2( ⇡
5
v
u
u
u
u
u
u
u
u
x3)u
u1
u
u
u
u
t
p5
+
1
8 )+ 4 (1
((
p5))
(1+ 1
4 (1
cos( ⇡
5
2 )( 5
x3)+( x1+r(1+ p5
8
p5)+ 1
4 ( 1+p5)
2 )( 5
(( x1+r(1+ p5
8
p5
1
4
r(1+
5
8 )(x1+r 8
r
5
8
2 )( 5
p5
8
p5
p5
5
8 )(x1+r 8
p5
p5
5
⇡
8 )+x4+r 8 + 8 ) sin( 5
p5
5
8 )(1+p5))(x1+r 8
x3))
+2
p5
8 )
p5
2 )( 5
x2+x5+r(1+ p5
8
8
p5
(1+ 1
4 (1
5
8 )+x4+r 8 + 8 )2+(
p5
p5)+ 1
4 ( 1+p5)
1
4
r(1+
r
5
8
1
8 )+ 4 (1
2 )( 5
p5
8
p5
8
p5
p5))
cos( ⇡
5
2 )( 5
x3)+( x1+r(1+ p5
8
5
8 )(1+p5))(x1+r 8
p5
5
8 )(x1+r 8
p5
8 )
2 )( 5
x2+x5+r(1+ p5
8
p5
p5
1
8 )+ 4 (1
p5
5
⇡
8 )+x4+r 8 + 8 ) sin( 5
p5))2)(cos2( ⇡
5
x3))2
x3)+sin2( ⇡
5
x3))
p
2((
(1+ 1
4 (1
p5)+ 1
4 ( 1+p5)
5 2p5
v
u
u r(1+
u
u
2 )( 5
u
p5
8
t(
p5
5
8 )(r 8
p5
8
x1) x1+x7
r
p5
(1+ 1
4 (1
5+
8
8 )2+(
p5)+ 1
4 ( 1+p5)
1
4
r(1+
r
5
8
2 )( 5
p5
8
p5
8
p5
5
8 )(1+p5))(r 8
p5
8
x1)
1
4
r(1+
r
5
8
2 )( 5
p5
8
p5
p5
5
8 )(1+p5))(r 8
p5
8
8
2 )( 5
x2+x8+r(1+ p5
8
p5
1
8 )+ 4 (1
p5))2
p
1
x1)
2 )( 5
x2+x8+r(1+ p5
8
v
u
u
u
u
u
u
u
u
⇡ u1
cos2(x3+ ⇡
u
5 )+sin2(x3+ 5 )u
u
u
u
t
p5
1
8 )+ 4 (1
((
((
p5))
(1+ 1
4 (1
r(1+
cos(x3+ ⇡
5) (
r(1+
p5)+ 1
4 ( 1+p5)
2 )( 5
p5
8
p5
5
8 )(r 8
p5
8
1
4
p5
2 )( 5
p5
8
r(1+
r
5
8
5
8 )(r 8
2 )( 5
p5
8
p5
p5
p5
8
x1) x1+x7
5
8 )(1+p5))(r 8
p5
8
r
+
p5
5+
⇡
8
8 ) sin(x3+ 5 ))
+2
x1)
r
p5
(1+ 1
4 (1
5+
8
8 )2+(
p5
2 )( 5
x2+x8+r(1+ p5
8
8
x1) x1+x7
p5)+ 1
4 ( 1+p5)
1
4
r(1+
r
5
8
1
8 )+ 4 (1
2 )( 5
p5
8
p5
p5))
p5
cos(x3+ ⇡
5) (
5
8 )(1+p5))(r 8
p5
8
x1)
r(1+
2 )( 5
p5
8
p5
5
8 )(r 8
2 )( 5
x2+x8+r(1+ p5
8
p5
8
p5
x1) x1+x7
1
8 )+ 4 (1
r
p
2((
(1+ 1
4 (1
p5)+ 1
4 ( 1+p5)
1
4
r(1+
5 2p5
p5
5+
⇡
8
8 ) sin(x3+ 5 ))2
v
u
u r(1+
u
u
2 )( 5
u
p5
8
t(
p5))2)(cos2(x3+ ⇡
⇡
5 )+sin2(x3+ 5 ))
8
1
2(
1
4
r
5(3+p5) 2
p5 x
1
5
p5
5
8 )(x1+r 8
1
4
q
p5
8 ) x4
3+p5
p5 x2
1
5
r
+ 2q 5
p5
5+
8
8 )2+(
2p5 2
x1
(1+ 1
4 (1
1
2
q
1
2 (3
p5)+ 1
4 ( 1+p5)
+
p5)x
2 x1
1
4
r(1+
r
5
8
2 )( 5
p5
8
p5
p5
5
8 )(1+p5))(x1+r 8
r
5
8
2 )( 5
p5
8
p5
8 )
p5
5
8 )(1+p5))(x1+r 8
1
p5
8 )
⇡
x5+ 1
4 ( 1+p5)+1)(sin( 10
8
p5
x5+ 1
4 ( 1+p5)+1)2
p
⇡
cos2( 10
⇡
x6)+(sin( 10
1
x6)+ 1
4 ( 1+p5)+ 4 (1
8
+ x5 x1 +
1
2
q
1
2 (1
+
2
p5
)( 58
p5
8
)(3 +
p5)x
1
1
8
q
5
2 (3
+
p5)x
1
+
3
8
q
1
2 (3
+
p5)x
1
v
u
u
u
u
u
u
u
p5))2 u
u1
u
u
u
u
u
t
p5x
4
1
1
x6)+ 1
4 ( 1+p5)+ 4 (1
((
((
+
2 )( 5
p5
8
1
2
q
p5))
(
p5)+ 1
4 ( 1+p5)
(1+ 1
4 (1
r(1+
x1
4
p5
+
p5)x
r(1+
1
4
2 )( 5
p5
8
r(1+
r
p5
5
8 )(x1+r 8
1
2 (5
5
8
8 ) x4
2
1
8
2 )( 5
p5
8
p5
p5
5
8 )(x1+r 8
p5
8 ) x4
p5
5
8 )(1+p5))(x1+r 8
r
+
p5
5+
⇡
8
8 ) cos( 10
p5
5+
8
8 )2+(
p5)(3
(1+ 1
4 (1
+
8 )
+2
⇡
x5+ 1
4 ( 1+p5)+1)(sin( 10
p5)+ 1
4 ( 1+p5)
p5)x
2
x6))
p5
8
r
q
(5
x2 x4 +
1
4
r(1+
r
5
8
2 )( 5
p5
8
p5
q
(1 +
p5))
1
x6)+ 1
4 ( 1+p5)+ 4 (1
p5
5
8 )(1+p5))(x1+r 8
p5
8
(
r(1+
2 )( 5
p5
8
p5
5
8 )(x1+r 8
p5
8 ) x4
r
p5
5+
⇡
8
8 ) cos( 10
)x4 + x4 +
8 )
1
2
⇡
x5+ 1
4 ( 1+p5)+1)2)(cos2( 10
q
1
2 (1
+
2
p5
)( 58
p5
8
⇡
x6)+(sin( 10
)(5 +
p5)
+
1
x6)+ 1
4 ( 1+p5)+ 4 (1
1
2
q
1
2 (5
+
p5)
p
2((
+
v
u
u
u
u
u
t(x1
p5))2)
1
8
q
(1 +
(1+ 1
4 (1
p5)+ 1
4 ( 1+p5)
1
4
r(1+
r
5 2p5
x6))2
p5
8
2
p5
)( 58
2
p5
)(5
p5)( 5
8
r(1+
2 )( 5
p5
8
p5
8
)(3 +
p5
5
8 )(x1+r 8
p5)
+
1
8
p5
q
8 ) x4
(5
r
p5
5+
8
8 )2+(
p5)(3
+
(1+ 1
4 (1
p5))
+ 12 (
p5)+ 1
4 ( 1+p5)
1
4
r
1
4
r(1+
5(3+p5) 2
p5 x
1
5
r
5
8
2 )( 5
p5
8
p5
1
4
p5
5
8 )(1+p5))(x1+r 8
p5
8 )
+x2
x5+ 1
4 ( 1+p5)
8
q
3+p5
p5 x2
1
5
+ 2q 5
2p5 2
x1
+
1
2
q
1
2 (3
+
5
8
r(1+
p5)x
2 x1
2 )( 5
p5
8
p5
p5
5
8 )(1+p5))(x1+r 8
1
p5
8 )
+x2
x5+ 1
4 ( 1+p5)
8
2 )( 5
p5
8
p5
x8 x1
8 ))2
p
1
2
⇡
cos2( 10
q
1
2 (1
+
⇡
x6)+(sin( 10
2
p5
)( 58
r(1+
1
x6)+ 1
4 ( 1+p5)+ 4 (1
p5
8
)(3 +
p5)x
1
+
2 )( 5
p5
8
p5
⇡
8 ))(sin( 10
v
u
u
u
u
u
u
u
p5))2 u
u1
u
u
u
u
u
t
1
8
q
5
2 (3
((
((x1
+
1
x6)+ 1
4 ( 1+p5)+ 4 (1
(1+ 1
4 (1
r(1+
p5)x
p5)+ 1
4 ( 1+p5)
2 )( 5
p5
8
1
3
8
p5
5
8 )(x1+r 8
q
1
2 (3
+
p5))
1
4
r
5
8
+
2 )( 5
p5
8
p5
p5
5
8 )(x1+r 8
p5
2 )( 5
p5
8
p5
8 ) x4
1
r(1+
(x1
r(1+
p5
p5)x
5
8 )(1+p5))(x1+r 8
8 ) x4
r
+
p5
5+
⇡
8
8 ) cos( 10
r
4
p5
5+
8
8 )2+(
1
x1
4
(1+ 1
4 (1
1
2
q
8 )
+x2
x5+ 1
4 ( 1+p5)
p5)+ 1
4 ( 1+p5)
1
2 (5
+
x6))
+2
p5
8
p5x
p5)x
2
1
4
r(1+
1
8
q
(5
r
5
8
r(1+
2 )( 5
p5
8
p5
2 )( 5
p5
8
p5
⇡
8 ))(sin( 10
p5
5
8 )(1+p5))(x1+r 8
1
x6)+ 1
4 ( 1+p5)+ 4 (1
p5)(3
+
p5)x
2
p5))
p5
8 )
8
+ x2 x7
+x2
x5+ 1
4 ( 1+p5)
q
(1 +
2
p5
)( 58
(x1
r(1+
r(1+
2 )( 5
p5
8
p5
8
)x7
2 )( 5
p5
8
p5
5
8 )(x1+r 8
p5
⇡
8 ))2)(cos2( 10
x7 +
1
2
q
p5
8 ) x4
⇡
x6)+(sin( 10
1
2 (1
+
2
p5
)( 58
r
p5
5+
⇡
8
8 ) cos( 10
p5
8
)(5 +
p
2 )( 5
2((r(1+ p5
8
p5)
v
u
u r(1+
u
u
2 )( 5
u
p5
8
t(
p5))2)
+
1
2
q
1
2 (5
p5
5
8 )(r 8
p5
8
p5
⇡
x1) x7+r 5
8 + 8 ) cos(x9+ 10 )+(
(1+ 1
4 (1
p5)+ 1
4 ( 1+p5)
5 2p5
x6))2
1
x6)+ 1
4 ( 1+p5)+ 4 (1
+
p5)
+
1
8
q
(1 +
p5
5
8 )(r 8
2
p5
)(5
p5
p5
x1) x7+r 5
8 + 8 )2+(
8
p5)( 5
8
p5
8
)(3 +
p5)
+
(1+ 1
4 (1
1
8
q
(5
p5)+ 1
4 ( 1+p5)
p5)(3
+
1
4
r(1+
r
p5))
5
8
2 )( 5
p5
8
p5
8
p5
5
8 )(1+p5))(r 8
p5
8
x1)
x8+ 1
4 ( 1+p5)+1)2
p
⇡ )+(sin(x9+ ⇡ )+ 1 ( 1+p5)+ 1 (1
cos2(x9+ 10
10
4
4
1
4
r(1+
r
v
u
u
u
u
u
u
u
p5))2 u
u1
u
u
u
u
u
t
5
8
1
2 )( 5
p5
8
p5
p5
p5
5
8 )(1+p5))(r 8
8
+
x1)
1 ( 1+p5)+1)(sin(x9+ ⇡ )+ 1 ( 1+p5)+ 1 (1
x8+ 4
10
4
4
p5)))
8
2 )( 5
((r(1+ p5
8
2 )( 5
((r(1+ p5
8
p5
p5
5
8 )(r 8
5
8 )(r 8
p5
8
p5
8
p5
⇡
x1) x7+r 5
8 + 8 ) cos(x9+ 10 )+(
p5
x1) x7+r 5
8 + 8 )2+(
(1+ 1
4 (1
(1+ 1
4 (1
p5)+ 1
4 ( 1+p5)
p5)+ 1
4 ( 1+p5)
1
4
r(1+
r
5
8
2 )( 5
p5
8
p5
8
1
4
r(1+
r
p5
5
8
2 )( 5
p5
8
p5
p5
5
8 )(1+p5))(r 8
8
5
8 )(1+p5))(r 8
p5
8
x1)
p5
8
+2
x1)
⇡
1
1
x8+ 1
4 ( 1+p5)+1)(sin(x9+ 10 )+ 4 ( 1+p5)+ 4 (1
p5)))2
⇡
⇡
1
1
x8+ 1
4 ( 1+p5)+1)2)(cos2(x9+ 10 )+(sin(x9+ 10 )+ 4 ( 1+p5)+ 4 (1
p5))2)
p
2 )( 5
2((r(1+ p5
8
p5
5
8 )(r 8
p5
8
x1)+x1
p5
⇡
x7+r 5
8 + 8 ) cos(x9+ 10 )+(
(1+ 1
4 (1
p5)+ 1
4 ( 1+p5)
1
4
r(1+
5 2p5
v
u
u r(1+
u
u
2 )( 5
u
p5
8
t(
p5
5
8 )(r 8
p5
8
x1)+x1
p5
x7+r 5
8 + 8 )2+(
(1+ 1
4 (1
p5)+ 1
4 ( 1+p5)
1
4
r(1+
r
5
8
2 )( 5
p5
8
p5
8
p5
5
8 )(1+p5))(r 8
p5
8
x1)
+x2
x8+ 1
4 ( 1+p5)
r(1+
2 )( 5
p5
8
p5
8 ))2
p
⇡ )+(sin(x9+ ⇡ )+ 1 ( 1+p5)+ 1 (1
cos2(x9+ 10
10
4
4
r
1
2 )( 5
p5
8
p5
5
8
8
v
u
u
u
u
u
u
u
p5))2 u
u1
u
u
u
u
u
t
p5
5
8 )(1+p5))(r 8
2 )( 5
((r(1+ p5
8
2 )( 5
((r(1+ p5
8
p5
p5
8
x1)
+x2
p5
5
8 )(r 8
5
8 )(r 8
p5
8
p5
8
x8+ 1
4 ( 1+p5)
x1)+x1
x1)+x1
r(1+
2 )( 5
p5
8
p5
⇡
1
1
8 ))(sin(x9+ 10 )+ 4 ( 1+p5)+ 4 (1
p5
⇡
x7+r 5
8 + 8 ) cos(x9+ 10 )+(
p5
x7+r 5
8 + 8 )2+(
(1+ 1
4 (1
(1+ 1
4 (1
p5)+ 1
4 ( 1+p5)
p5)+ 1
4 ( 1+p5)
1
4
r(1+
r
1
Wöden Kusner TU Graz
11 / 35
5
8
2 )( 5
p5
8
p5
8
p5)))
1
4
r(1+
r
p5
5
8
2 )( 5
p5
8
p5
p5
5
8 )(1+p5))(r 8
8
5
8 )(1+p5))(r 8
p5
8
x1)
+x2
Pentagon Packing
t
Objective
+
+2
p
2((
(1+ 1
4 (1
p
p
5)+ 1
4 ( 1+ 5)
1
4
r
p
5 2 5
p
5 + 5 ) sin(x3+ ⇡ ))2
8
8
5
v
u
u r
u
u
u
(1+ p2 )( 5
t(
5 8
⇡
s2(x3+ ⇡
5 )+sin2(x3+ 5 ))
1
2(
1
4
r
p
5(3+ 5) 2
p x
1
5
5
r
p
5
5
8 )(x1+ 8
1
4
q
p
5
8 ) x4
p
3+p5 2
x
5
5 1
+2
r
q
5
p
5 + 5 )2+(
8
8
2p
x2
5 1
(1+ 1
4 (1
1
2
q
1
2 (3
p
p
5)+ 1
4 ( 1+ 5)
+
p
1
4
r
(1+ p2 )( 5
5 8
r
p
5
5
8
8
5)x2 x1 + x5 x1 +
1
2
p
p
5
8 )(1+ 5))(x1+
q
1
2 (1
+
r
p2 )( 5
5 8
5
8
p
5
8 )
p
x5+ 1
4 ( 1+
5
8 )(3
+
p
2 )( 5
(1+ p
5 8
r
p
5
5
8
8
p
5)+1)2
5)x1
p
1
8
p
p
5
8 )(1+ 5))(x1+
⇡
cos2( 10
q
5
2 (3
+
r
⇡
x6)+(sin( 10
p
5)x1 +
5
8
p
5
8 )
1
x5+ 1
4 ( 1+
x6)+ 1
4 ( 1+
3
8
q
1
2 (3
p
+
p
5)+ 1
4 (1
p
5)x1
⇡
5)+1)(sin( 10
v
u
u
u
u
u
u
u
u
p
5))2 u
u1
u
u
u
u
t
p
5x1
4
x6)+ 1
4 ( 1+
((
((
+
r
x1
4
p
p
5)+ 1
5)) (
4 (1
(1+ 1
4 (1
p
p
5)+ 1
4 ( 1+ 5)
(1+ p2 )( 5
5 8
r
p
5
5
8 )(x1+ 8
1
2
q
1
2 (5
+
p
5)x2
r
1
4
(1+ p2 )( 5
5 8
r
(1+ p2 )( 5
5 8
r
p
5
5
8
8
p
5
8 ) x4
r
q
1
(5
8
p
5
8 )(x1+
r
5
8
p
5
8 ) x4
p
p
5
8 )(1+ 5))(x1+
p
5 + 5 )2+(
8
8
p
(1+ 1
4 (1
5)(3 +
p
r
p
5
8
r
+
p
5 + 5 ) cos( ⇡
8
8
10
5
8 )
5)x2
x6))
+2
p
x5+ 1
4 ( 1+
5)+ 1
4 ( 1+
p
5)
1
4
r
p
⇡
5)+1)(sin( 10
(1+ p2 )( 5
5 8
r
p
5
5
8
8
q
x2 x4 + (1 +
x6)+ 1
4 ( 1+
p
p2 )( 5
5 8
p
5
8 )x4
p
5)+ 1
4 (1
r
p
p
5
5
8 )(1+ 5))(x1+ 8
5)) (
r
(1+ p2 )( 5
5 8
p
+ x4 +
5
8 )
1
2
x5+ 1
4 ( 1+
q
1
2 (1
+
p
5
8 )(x1+
p
⇡
5)+1)2)(cos2( 10
p2 )( 5
5 8
p
5
8 )(5
r
5
8
p
5
8 ) x4
⇡
x6)+(sin( 10
+
p
5) +
r
p
5 + 5 ) cos( ⇡
8
8
10
x6)+ 1
4 ( 1+
1
2
q
1
2 (5
+
p
p
q
5) + 18 (1 +
(1+ 1
4 (1
2((
p
p
5)+ 1
4 ( 1+ 5)
1
4
p
5 2 5
v
u
u
u
u
u
t(x1
x6))2
p
p
5)+ 1
5))2)
4 (1
p2 )(5
5
p
5)( 58
r
2 )( 5
(1+ p
5 8
p
5
8 )(3
+
p
5
8 )(x1+
p
r
5) +
5
8
1
8
p
5
8 ) x4
q
(5
p
r
p
5 + 5 )2+(
8
8
5)(3 +
p
(1+ 1
4 (1
5)) + 12 (
p
p
5)+ 1
4 ( 1+ 5)
1
4
r
1
4
r
p
5(3+ 5) 2
p x
1
5
5
(1+ p2 )( 5
5 8
r
p
5
5
8
8
1
4
q
p
p
5
8 )(1+ 5))(x1+
p
3+p5 2
x
5
5 1
+2
q
5
r
5
8
2p
p
5
8 )
x2
5 1
+x2
+
1
2
x5+ 1
4 ( 1+
q
1
2 (3
+
p
p
5)
r
2 )( 5
(1+ p
5 8
r
p
5
5
8
8
r
(1+ p2 )( 5
5 8
5)x2 x1
p
p
5
8 )(1+ 5))(x1+
p
5
8 ))2
x8 x1
p
1
2
r
⇡
cos2( 10
q
1
2 (1
+
5
8
1
p
5
8 )
+x2
⇡
x6)+(sin( 10
p2 )( 5
5 8
x5+ 1
4 ( 1+
p
x6)+ 1
4 ( 1+
p
5
8 )(3
+
5)
p
p
r
(1+ p2 )( 5
5 8
5)+ 1
4 (1
5)x1 +
p
5
⇡
8 ))(sin( 10
v
u
u
u
u
u
u
u
u
p
5))2 u
u1
u
u
u
u
t
1
8
q
5
2 (3
+
x6)+ 1
4 ( 1+
(1+ 1
4 (1
((
((x1
r
p
3
8
r
p
5
5
8 )(x1+ 8
q
r
p
p
5)+ 1
5)) (x1
4 (1
p
p
5)+ 1
4 ( 1+ 5)
(1+ p2 )( 5
5 8
5)x1
1
2 (3
+
p
1
4
r
(1+ p2 )( 5
5 8
(1+ p2 )( 5
5 8
r
p
5
5
8
8
p
5
8 ) x4
5)x1 +
r
p
r
p
5
5
8 )(x1+ 8
p
p
5
8 )(1+ 5))(x1+
p
5 + 5 )2+(
8
8
5x1
4
x1
4
r
(1+ 1
4 (1
1
2
q
5
8
p
1
2 (5
p
5
8 ) x4
r
+
p
5 + 5 ) cos( ⇡
8
8
10
5
8 )
+
p
x6))
+2
p
+x2
5)+ 1
4 ( 1+
x5+ 1
4 ( 1+
p
5)
5)x2
p
1
4
r
1
8
q
(5
5)
r
(1+ p2 )( 5
5 8
r
p
5
5
8
8
(1+ p2 )( 5
5 8
p
p
5
⇡
8 ))(sin( 10
r
p
p
5
5
8 )(1+ 5))(x1+ 8
5)(3 +
p
x6)+ 1
4 ( 1+
p
5
8 )
5)x2 + x2 x7
+x2
p
p
5)+ 1
5)) (x1
4 (1
x5+ 1
4 ( 1+
q
(1 +
p
5)
p2 )( 5
5 8
r
r
2 )( 5
(1+ p
5 8
(1+ p2 )( 5
5 8
p
5
8 )x7
p
5
8 )(x1+
p
5
⇡
8 ))2)(cos2( 10
x7 +
1
2
q
r
5
8
p
5
8 ) x4
⇡
x6)+(sin( 10
1
2 (1
+
p2 )( 5
5 8
r
p
5 + 5 ) cos( ⇡
8
8
10
x6)+ 1
4 ( 1+
p
5
8 )(5
p
p
5 2 5
v
u
u r
u
u
5
u
p2
t( (1+ 5 )( 8
x6))2
p
p
5)+ 1
5))2)
4 (1
+
p
5) +
1
2
q
1
2 (5
+
p
5) +
1
8
q
(1 +
p
5
8 )(
p2 )(5
5
1
Wöden Kusner TU Graz
12 / 35
r
5
8
p
p
5
8
5)( 58
x1) x7
p
5
8
+2
+
Pentagon Packing
t
Objective
+
p
(1+ 1
4 (1
2((
p
p
5)+ 1
4 ( 1+ 5)
1
4
p
5 2 5
v
u
u
u
u
u(x1
t
p2 )(5
5
p
5)( 58
r
2 )( 5
(1+ p
5 8
p
5
8 )(3
+
p
5
8 )(x1 +
p
r
5) +
5
8
1
8
p
5
8 ) x4
q
(5
p
r
p
5 + 5 )2+(
8
8
5)(3 +
p
p
p
(1+ 1
5)+ 1
4 (1
4 ( 1+ 5)
5)) + 12 (
1
4
r
1
4
r
p
5(3+ 5) 2
p x
1
5
5
(1+ p2 )( 5
5 8
r
p
5
5
8
8
1
4
q
p
p
5
8 )(1+ 5))(x1 +
p
3+p5 2
x
5
5 1
+2
q
5
r
5
8
2p
p
5
8 )
x2
5 1
+x2
+
1
2
x5 + 1
4 ( 1+
q
1
2 (3
+
p
p
5)
r
2 )( 5
(1+ p
5 8
r
p
5
5
8
8
r
(1+ p2 )( 5
5 8
5)x2 x1
p
p
5
8 )(1+ 5))(x1 +
p
5
8 ))2
x8 x1
p
1
2
r
⇡
cos2( 10
q
1
2 (1
+
5
8
1
p
5
8 )
+x2
⇡
x6 )+(sin( 10
p2 )( 5
5 8
x5 + 1
4 ( 1+
p
5)
r
(1+ p2 )( 5
5 8
p
5
⇡
8 ))(sin( 10
v
u
u
u
u
u
u
u
u
p
p
1
x6 )+ 1
5))2 u
u1
4 ( 1+ 5)+ 4 (1
u
u
u
u
t
p
5
8 )(3
+
p
5)x1 +
1
8
q
5
2 (3
((x1
+
x6 )+ 1
4 ( 1+
(1+ 1
4 (1
((
r
p
(1+ p2 )( 5
5 8
5)x1
3
8
r
p
5
5
8 )(x1 + 8
q
r
p
p
5)+ 1
5)) (x1
4 (1
p
p
5)+ 1
4 ( 1+ 5)
1
2 (3
+
p
1
4
r
(1+ p2 )( 5
5 8
(1+ p2 )( 5
5 8
r
p
5
5
8
8
p
5
8 ) x4
5)x1 +
r
p
p
5
8 )(1+ 5))(x1 +
p
5 + 5 )2+(
8
8
5x1
4
r
p
5
5
8 )(x1 + 8
p
x1
4
r
(1+ 1
4 (1
1
2
q
5
8
p
1
2 (5
p
5
8 ) x4
r
p
5 + 5 ) cos( ⇡
8
8
10
p
5
8 )
+x2
5)+ 1
4 ( 1+
+
p
x5 + 1
4 ( 1+
p
5)
5)x2
1
4
r
p
5)
r
(1+ p2 )( 5
5 8
r
p
5
5
8
8
q
1
(5
8
x6 ))
(1+ p2 )( 5
5 8
p
p
5
⇡
8 ))(sin( 10
r
p
p
5
5
8 )(1+ 5))(x1 + 8
5)(3 +
p
5)x2 + x2
1
Wöden Kusner TU Graz
p
13 / 35
5
8
Pentagon Packing
t
Objective
(1+ 1
4 (1
2((
p
p
5
8 )(1+ 5))(x1 +
p
+p5 2
x
5 1
+2
q
5
r
5
8
2p
p
p
5)+ 1
4 ( 1+ 5)
p
5
8 )
+x2
x2 +
5 1
1
2
p
x5 + 1
4 ( 1+ 5)
q
Wöden Kusner TU Graz
1
2 (3
+
p
1
4
r
2 )( 5
(1+ p
5 8
r
p
5
5
8
8
r
(1+ p2 )( 5
5 8
5)x2 x1
p
p
5
8 )(1+ 5))(x1 +
p
5 2
8 ))
x8 x1
p
1
2
r
⇡
cos2 ( 10
q
1
2 (1
+
5
8
p
5
8 )
+x2
⇡
x6 )+(sin( 10
p2 )( 5
5 8
x5 + 1
4 ( 1+
p
5)
r
(1+ p2 )( 5
5 8
p
5
8
v
u
u
u
u
u
u
u
u
p
p
1
1
2
x6 )+ 4 ( 1+ 5)+ 4 (1
5)) u
u1
u
u
u
u
t
p
5
8 )(3
+
p
5)x1 +
1
8
q
5
2 (3
14 / 35
(
Pentagon Packing
t
Constraints
The constraint functions for the nonlinear program are conditions
forcing non-intersection of the pentagons. Locally, it is sufficient to
require that no vertex of a pentagon be in the interior of another. Let
pij be the vertex of Pi that is in contact with Pj . The constraint that a
vertex of Pk does not lie in Pj may be considered as an angle
constraint, written as
Angle{Ci − pij , pki − pij } −
3π
≥0
10
for appropriate choices of i, j, k in {1, 2, 3, 4}.
Wöden Kusner TU Graz
15 / 35
Pentagon Packing
t
Constraints
Violation of an angle constraint.
Wöden Kusner TU Graz
16 / 35
Slicing Programming Problems
t
Numerical Check
Trying to solve the nonlinear program destroyed a computer. A linear
approximation of the program can be found and easily solved
numerically, indicating a (possibly degenerate) critical point at 0 .
A restricted bordered Hessian test indicates 0 is a local max. There
are some directions where the geometry forces the objective function
be constant or linear.
Careful error analysis should show 0 is a local maximum.
Wöden Kusner TU Graz
17 / 35
Slicing Programming Problems
t
Numerical Check
Trying to solve the nonlinear program destroyed a computer. A linear
approximation of the program can be found and easily solved
numerically, indicating a (possibly degenerate) critical point at 0 .
A restricted bordered Hessian test indicates 0 is a local max. There
are some directions where the geometry forces the objective function
be constant or linear.
Careful error analysis should show 0 is a local maximum.
Wöden Kusner TU Graz
17 / 35
Slicing Programming Problems
t
Numerical Check
Trying to solve the nonlinear program destroyed a computer. A linear
approximation of the program can be found and easily solved
numerically, indicating a (possibly degenerate) critical point at 0 .
A restricted bordered Hessian test indicates 0 is a local max. There
are some directions where the geometry forces the objective function
be constant or linear.
Careful error analysis should show 0 is a local maximum.
Wöden Kusner TU Graz
17 / 35
Slicing Programming Problems
t
Conical Programs
The nonlinear programming problem for pentagons satisfies certain
assumptions that allow it to be sliced and analyzed. In general,
Proposition
a nonlinear program
maxn f (x) subject to gr (x) ≥ 0, r ∈ I
x∈R
satisfying the following assumptions, has an isolated local maximum
at 0 with f(0) = 0.
Wöden Kusner TU Graz
18 / 35
Slicing Programming Problems
t
Conical Programs
Definitions
Let e1 be the standard unit vector {1, 0, . . . , 0} in Rn .
Let F (t) = ∇f (te1 ).
Let Gr (t) = ∇gr (te1 ).
Let E = {te1 : t ∈ R}.
Let H be the orthogonal complement of E so that Rn = E ⊕ H.
Assumptions
The index set I is a finite set.
For r in I, the objective and constraint functions f and gr are
analytic functions on a neighborhood of 0.
Wöden Kusner TU Graz
19 / 35
Slicing Programming Problems
t
Conical Programs
Assumptions
Assume f (0) = gr (0) = 0 for all r in I.
There is a bounded solution at 0 to the linear program
maxn F (0) · x subject to Gr (0) · x ≥ 0, r ∈ I
x∈R
E is the set of solutions in Rn to the program
F (0) · x = 0 subject to Gr (0) · x ≥ 0, r ∈ I.
There is an > 0 so the functions gr (te1 ) = 0 for all t ∈ (−, ),
for all r in I.
Assume
∂
∂t f (te1 )|t=0
Wöden Kusner TU Graz
= 0,
∂2
f (te1 )|t=0
∂t 2
< 0.
20 / 35
Slicing Programming Problems
t
Sketch of Proof
Lemma
Given the assumptions, the linear program
max F (0) · x subject to Gr (0) · x ≥ 0, r ∈ I
x∈H
has a unique maximum at x = 0
Wöden Kusner TU Graz
21 / 35
Slicing Programming Problems
t
Sketch of Proof
Proof.
From the assumptions, the linear program
maxn F (0) · x subject to Gr (0) · x ≥ 0, r ∈ I
x∈R
is maximized exactly on E. The restricted program feasible set
{x : Gr (0) · x ≥ 0, r ∈ I and x ∈ H} is a subset of the feasible set
{x : Gr (0) · x ≥ 0, r ∈ I} for the full program. Thus, the program
max F (0) · x subject to Gr (0) · x ≥ 0, r ∈ I
x∈H
attains its maximum exactly on the (non-empty) intersection
E ∩ {x : Gr (0) · x ≥ 0, r ∈ I} ∩ H = 0.
Wöden Kusner TU Graz
22 / 35
Slicing Programming Problems
t
Sketch of Proof
Definition
A finitely generated cone is a subset of Rn which is the
non-negative span of a finite set of non-zero vectors {v1 , . . . vm } in
Rn , which are called the generators of the cone.
Definition
A conical linear program is a linear program with a constraint set
that is a finitely generated cone.
The linear programs described throughout this section are always
constrained to be on the intersection of half-spaces with 0 on the
boundary. These are conical programs.
Definition
For a cone C, the set C p := {x ∈ Rn : v · x ≤ 0 for all v ∈ C} is the
polar cone of C.
Wöden Kusner TU Graz
23 / 35
Slicing Programming Problems
t
Sketch of Proof
Lemma
A conical linear program with F 6= 0 given by
maxn F · x subject to Gr · x ≥ 0, r ∈ I
x∈R
(a) has a unique maximum at x = 0 iff F is in the interior of the polar
cone C p of C = {x : Gr · x ≥ 0, r ∈ I} (b) has a bounded solution iff F
is in the polar cone C p of C = {x : Gr · x ≥ 0, r ∈ I} and attains its
maximum exactly on the non-negative span of the generators vi
satisfying F · vi = 0.
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Proof.
If F is in the interior of the polar cone C p , then F · vi < 0 for all
generators vi . Therefore F · x is uniquely maximized in C at the
vertex. If F is on the boundary of the polar cone, then F · x is
maximized in C exactly on the span of the generators vi for which
F · vi = 0 as F · vj < 0 otherwise. If F is outside the polar cone, then
F · vi > 0 for some generator vi . Then F · x is unbounded in C.
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Lemma
Given the assumptions, there exists > 0 such that for all t in (−, ),
the linear program
max F (t) · yt
yt ∈H
subject to
Gr (t) · yt ≥ 0, r ∈ I
has a unique maximum at yt = 0.
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Proof.
The program for t ∈ (−, ), for yt in H, for each t in (−, ), for some
> 0, is a conical program on all of Rn with a cone Ct in Rn of
co-dimension ≥ 1 with constraints e1 · yt ≥ 0 and −e1 · yt ≥ 0. By
previous lemmas, F (0) is in the polar cone of
C0 = {y0 : Gr (0) · y0 ≥ 0, e1 · y0 ≥ 0, −e1 · y0 ≥ 0}.
As f , gr ∈ C ω , the condition of F (t) being in the interior of the polar
cone Ctp is open and the feasible set
Ct = {yt : Gr (t) · yt ≥ 0, e1 · yt ≥ 0, −e1 · yt ≥ 0}
is conical. By a previous lemma, the program has a unique maximum
at yt = 0 for each t in (−, ) for some > 0.
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Lemma
Given the assumptions and as in the previous lemmas, for all
t ∈ (−, ) there exists δ(t) > 0 and a cube Q(t) ⊂ Rn of side length
2δ(t) such that
{(F (t) + Q(t)) ∩ (∂(Ctp ) + Q(t))} = ∅.
Proof.
This follows from a previous lemma, which shows F (t) is in the
interior of the polar cone Ctp . Then F (t) and the boundary of Ctp can
be separated and the existence of Q is trivial.
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Corollary
Given the assumptions and as in previous lemmas, for all t ∈ (−, ),
(F (t) + ∆) · yt ≤ 0
whenever yt satisfies
(Gr (t) + ∆r ) · yt ≥ 0, r ∈ I and e1 · yt ≥ 0, −e1 · yt ≥ 0
where ∆ and ∆r are any points in the 2δ(t)-cube Q(t) and yt is in H.
Proof.
By a previous lemma, F (t) + ∆ is in the interior of the polar cone
p
Ct,∆
, where
Ct,∆ = {yt : (Gr (t) + ∆r ) · yt ≥ 0, e1 · yt ≥ 0, −e1 · yt ≥ 0, r ∈ I}.
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Lemma
Given the assumptions and as in the previous lemmas, for all
t ∈ (−, ), let yt = x − te1 ∈ H. Choose ∆ = ∆(yt ) and ∆r = ∆r (yt )
in the 2δ(t)-cube Q(t) to be the corner given by the sign of
x − te1 = yt . Then there is an t for which
(F (t) + ∆(yt )) · yt ≤ 0 =⇒ f (x) − f (te1 ) ≤ 0
and
(Gr (t) + ∆r (yt )) · yt ≤ 0 =⇒ gr (x) − gr (te1 ) = gr (x) ≤ 0
for all yt satisfying kyt k ≤ t .
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Proof.
This follows from the local expansions of the nonlinear program. By
this choice of ∆(yt ) and ∆r (yt ),
f (x) − f (te1 ) = F (t) · (x − te1 ) + O(t 2 ) = F (t) · yt + O(t 2 )
≤ F (t) · yt + δ(t)kyt k1 = (F (t) + ∆(yt )) · yt
and
gr (x) = gr (x) − gr (te1 ) = Gr (t) · (x − te1 ) + O(t 2 ) = Gr (t) · yt + O(t 2 )
≤ Gr (t) · yt + δ(t)kyt k1 = (Gr (t) + ∆r (yt )) · yt .
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By the previous lemmas, for t in (−, ), the program
max(F (t) + ∆) · yt subject to (Gr + ∆r ) · yt
yt ∈H
is uniquely maximized at yt = 0 for any choice of ∆, ∆r in the 2δ(t)
cube Q(t). Then there is an t neighborhood of 0 where f (yt + te1 ) is
less than f (te1 ) on
∪∆r ∈Q(t) {yt : (Gr + ∆r ) · yt ≥ 0, r ∈ I, yt ∈ H},
which contains the feasible set {yt : gr (yt + te1 ) ≥ 0, r ∈ I, yt ∈ H}.
Therefore the nonlinear programs f (yt + te1 ) subject to
gr (yt + te1 ) ≥ 0, yt ∈ H, parameterized by t in (−, ), have local
maxima at yt = 0. This gives the following:
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Theorem
Given the assumptions, a fixed t in (−, ) and choosing ∆ and ∆r as
in the previous lemmas, for x satisfying gr (x) ≥ 0 for all r in I and
yt = x − te1 in H, there exist linear programs
max(F (t) + ∆(yt )) · yt subject to (Gr (t) + ∆r (yt )) · yt ≥ 0
yt ∈H
that give solutions to the nonlinear programs
max f (x) subject to gr (x) ≥ 0
x∈H+te1
in an t neighborhood of te1 in H + te1 .
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By choice of a sufficiently small and a minimal non-zero t , the
previous theorem gives an open neighborhood of 0 in which the
maximum value of the original nonlinear program occurs on E. The
assumptions for the first and second t-derivatives at 0 shows 0 to be
a local maximum for the nonlinear program
maxn f (x) subject to gr (x) ≥ 0.
x∈R
Theorem
A nonlinear program satisfying the assumptions has an isolated local
maximum at 0 with f(0) = 0.
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