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Mathieu Dutour Sikiric, Yoshiaki Itoh and Alexei Poyarkov- Simple random sequential packing of cubes
Transcript of Mathieu Dutour Sikiric, Yoshiaki Itoh and Alexei Poyarkov- Simple random sequential packing of cubes
Simple random sequentialpacking of cubes
Mathieu Dutour Sikiric
Rudjer Boskovic institute, Zagreb
Yoshiaki Itoh
Institut Statistical Mathematics,Tokyo
Alexei Poyarkov
Moscow State University, Moscow
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I. Packing
with
rigid boundaries
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�
-dim. random packing
Put sequentially at random intervals
�� � � �into
�� � � � untilone cannot do it any more.
step 0
step 1
step 2
step 3
Denote by
� � � the number of intervals put in
�� � � � .Renyi (1958) proved that
� � �� � �� � � � �� � ��� � �� �� � ��� � � �"! # $ %& ')( * '+
� �-, ./0 , , ,
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-dimensional random packing
Put sequentially at random cubes
�� � � � 1
into�� � � � 1
untilone cannot do it any more.
Denote by 1 � � � the number of cubes put in
�� � � � 1
.
Palasti conjectured that
� � �� � �� 2 � � � �� 2 exists and isequal to
� 1� .
Existence of the limit was proved by Penrose (2001) butthe second conjecture is probably false.
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A simplified model
We consider the cube
�� � / � 1
.
We put sequentially at random cubes 3 4 �� � � � 1with3 5 6 1
in it
0 1 2 30
1
2
3
until one cannot insert cubes any more.
Denote by 1 the number of cubes in the obtainednon-extendible packing.
We want to estimate the packing density
7 1 � �� 1 8 � 1 �
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Valueof 9
Computer simulations (Itoh and Ueda, (1983)) suggestthat 7 1 is asymptotically
'! :
with ; � �-, / / , , ,dim. <= >@? A < =CB
1
DFE GH I G J DFE G H I G2
DFE K J JL DFE KHM J J H DFEM NM G N J
3
DFEM JO K DFEM JO H HM DFE O G J KM O
4
DFE O I G J DFE O I J GM H DFE I GO M O K
5
DFE I NL K DFE I N D N K DFE I DO IL K
6
DFE IO D G DFE IO L NO K DFE H H G IO
7DFE IL JL DFE IL H JL H DFE L GL O H G
8
DFE H NO G DFE H N G G KH DFE L H O GM H
9
DFE H KM L DFE H K GM H N DFE J NM G N G
10
DFE HM H J DFE HM J I J D E JM IH K J
11
DFE H O JM DFE H IM O D J D E J H KL J K
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Extendedmodel
Consider the cube
�� � � P � 1
.
We put sequentially at random cubes 3 4 �� � P � 1with3 5 6 1
in it until one cannot insert cubes any more.
Denote by 1 � P �
the number of cubes in the obtainednon-extendible packing.
The main problem is to estimate the packing density:
7 1 � P � � �� 1 8 � 1 � P � �
Poyarkov, (2004) proved that 7 1 Q � � 4 �R � 1
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Lemma
Lemma First put the cube 3 4 �� � P � 1
in
�� � � P � 1and write
S � T �U V 3W � � or
P *Second put cube at random until one cannot do it anymore. Then:
The minimal number of cubes is
S 4 �
.The expectation of the number of cubes put isS 4 � 4 X � �R Y � �
S � � S � � S � �
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Expansionof 9
Denote by
8 � 1 � P � V S �
the expected number of cubesin a non-extendible random packing, while imposingthat the first cube 3 4 �� � P � 1
has 3 withS
coordinatesequal to
�
or
P
.
One has the expression
8 � 1 � P � � �1
Z\[ �� �
P 4 � � Z� P � �P 4 � � 1! Z S' 8 � 1 � P � V S �
So, one gets
8 � 1 � P � � � � R ! �R Y � � 1 8 � 1 � P � V � � 4 ' ]R Y � � R ! �R Y � � 1! � 8 � 1 � P � V � �
4 ' � ' � � � ]� R Y � �^ � R ! �R Y � � 1! ] 8 � 1 � P � V � � 4 X � �R Y � � _
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Expansionof 9
Clearly
8 � 1 � P � V � � � �
and8 � 1 � P � V � � � � 4 8 � 1! � � P � �
.
By above lemma,
8 � 1 � P � V � � � ` 4 X � �R Y � �
First, one has
8 � 1 � P � � � � 4 X � �R Y � �Second, one gets
8 � 1 � P � � � � � � ]R Y � � 1 4 ] 1R Y � � � � ]R Y � � 1! � � � 4 X � �R Y � � �
4 X � �� R Y � �^ �
� � � � ] 1R Y � 4 X � �� R Y � �^ � * 4 a 1R Y � 4 X � �� R Y � �^ �
� � 4 ] 1R Y � 4 X � �� R Y � �^ �
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Expansionof 9
Inserting this expression one gets
8 � 1 � P � � � � 4 � 'P 4 � 4 / ' � ' � � �
� P 4 � � ] 4 X � �P 4 � � _
One proves that for fixed
'
, there exists an asymptoticexpansion
8 � 1 � P � � � �Z [ �
b Zdc 1�
� P 4 � � Z
But finding the coefficients b Zc 1 for
S Q `
is less easy.
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II. Torus
cube tilings and
packings
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Torus cubetilings and packings
A cube tiling is a
/ 6 1
-periodic tiling of
e 1
by integraltranslates of the cube
�� � � � 1
.
There is only one cube tiling in dimension�
.
There are two cube tilings in dimension�
:
A cube packing is a packing of
e 1
by integral translatesof cubes
�� � � � 1, which is
/ 6 1-periodic.
If we cannot extend a cube packing by adding anothercube, then it is called non-extendible.
No non-extendible cube packings in dimension
�
and
�
.
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-dim. non-extendible cubepacking
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Resultsfor
In dimension
`
, there is a unique non-extendible cubepacking and there are
f
types of cube tilings.
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Resultsfor
In dimension
`
, there is a unique non-extendible cubepacking and there are
f
types of cube tilings.
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Resultsfor
In dimension
`
, there is a unique non-extendible cubepacking and there are
f
types of cube tilings.
In dimension
/
, the repartition is as follows:
g J L H I O M K G N J D J J JL J H J I JO JM
h i D D D D D D D H G M L I D K J D D D K I I
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Low densitycubepackings
Denote by
j � ' �
the smallest number of cubes ofnon-extendible cube packing.j � ` � � / and
j � / � �0 .
For any k � l 5 m
, the following inequality holds:
j � k 4 l � n j � k � j � l �,
The cube packing realizing this is constructed by“product” of two cube packings of
e o
and
e p
Conjecture:
j �q � � � �and
j �r � � � r
.
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III. Continuous
torus cube
packings
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Torus cubepackings
We consider the torus
6 1\s � P 6 1
and do sequentialrandom packing by cubes 3 4 �� � P � 1
with 3 5 6 1.
We denote 1 � P �
the number of cubes in the obtainedtorus cube packing and
; 1 � P � � 8 � 1 � P � �
As in the case of rigid boundaries, we are interested inthe limit
P t u.In the rigid boundary case, in the limit
P t u, one hasa single cube in the middle of another cube and nopossibility of adding any other cube.
But for torus case, the limit
P t u is more interesting.
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Continuouscubepackings
We consider the torus
e 1 s � 6 1
and do sequentialrandom packing by cubes 3 4 �� � � � 1
with 3 5 e 1.
Two cubes 3 4 �� � � � 1
and 3 v 4 �� � � � 1are non-overlapping
if and only if there is
� n U n '
with 3 vW w 3W 4 � � � x y � �
Fix a cube
z � 3 4 �� � � � 1
.
We want to insert a cube 3 v 4 �� � � � 1, which do not
overlap with
z
.The condition 3 vW � 3W 4 �
defines an hyperplane inthe torus
e 1\s � 6 1.
Those
'
hyperplanes have the same� ' � � �
-dimensional volume.In doing the sequential random packing, every oneof the
'hyperplanes is chosen with equal probability.
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Several cubes
One has several non-overlapping cubes 3 � 4 �� � � � 1, . . . ,3 { 4 �� � � � 1
. We want to add one more cube 3 4 �� � � � 1.
For every cube 3 | 4 �� � � � 1
, there should exist some� n U n '
such that 3 |W w 3W 4 � � � x y � �After enumerating all possible choices, one getsdifferent planes.
Their dimension might differ.
Only the one with maximal dimension have strictlypositive probability of being attained.
All planes of the highest dimension have the samevolume in the torus
e 1)s � 6 1
and so, the same probabilityof being attained.
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The two dimensional case
Put a cube 3 4 �� � � � ]
in
e ] s � 6 ]. 3 � � + � � + ] �In putting the next cube, two possibilities:
� + � 4 � � + _ � or� + _ � + ] 4 � �
. They correspond geometrically to:
}~���� ~�� �}~� ��� ~�� � }~���� ~�� �}~���� ~�� �� �
and they are equivalent.
Continuing the process, up to equivalence, one obtains:
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The -dimensional case
At first step, one puts the vector 3 � � � + � � + ] � + _ �At second step, up to equivalence, 3 ] � � + � 4 � � + a � +�� �
At third step, one generates six possibilities, all withequal probabilities:
� + � 4 � � + a 4 � � +�� � � + � � + ] 4 � � +�� � � + � � +�� � + _ 4 � �
� + � 4 � � + � � +�� 4 � � � + � � + ] 4 � � +�� 4 � � � + � � + a 4 � � + _ 4 � �
Up to equivalence, those possibilities split into
�
cases:� � + � � + ] � + _ � � � + � 4 � � + a � +�� � � � + � � +�� � + _ 4 � � *
withprobability
] _� � + � � + ] � + _ � � � + � 4 � � + a � +�� � � � +�� � + ] 4 � � + � 4 � � *
withprobability
� _
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The -dimensional case
Possible extensions of
� � + � � + ] � + _ � , � + � 4 � � + a � +�� �,� + � � + � � + _ 4 � � *
with probability
] _ are:� + � 4 � � +�� � + � 4 � �
with
�
parameter� + � 4 � � + a 4 � � + � �
with
�
parameter� + � � + ] 4 � � + _ � with
�
parameter� + � � +�� 4 � � + _ 4 � �
with
�
parameter
Cases with
�
parameters have probability
�
, so can beneglected.
So, up to equivalence, one obtains� � + � � + ] � + _ � �, , , � � + � � + ] 4 � � + � 4 � � � � + � � + � � + _ 4 � � � � + � 4� � + � � + � 4 � �with probability
� _� � + � � + ] � + _ � �, , , � � + � � + ] 4 � � + � 4 � � � � + � � + � � + _ 4 � � � � + � 4� � + a 4 � � +�� �with probability
� _
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The -dimensional case
Possible extensions of
� � + � � + ] � + _ � , � + � 4 � � + a � +�� �,� +�� � + ] 4 � � + � 4 � � *
with probability
� _ are:
� + � 4 � � + a 4 � � + _ 4 � � � + � 4 � � + ] � + � 4 � � � + � � + ] 4 � � + � �
� + � 4 � � + ] 4 � � +�� 4 � � � + � 4 � � + a 4 � � + � � � + � � + ] � + _ 4 � �
All those choices have
�
parameter.
Those possibilities are in two groups:� � + � � + ] � + _ � �, , , � � + � � + ] 4 � � + � 4 � � � � + � 4 � � + a 4 � � + _ 4 � � *
with probability���q
other cases with probability
� �� .
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The -dimensional case
At the end of the process, one obtains
.
parameters,probability
� _
.
parameters,probability
� _
r
parameters,probability
� ��
Also, with probability��� , one obtains the non-extendible
cube packing with/
cubes and
r
parameters.
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The -dimensional caseDoing the enumeration by computer, one obtains` �
non-extendible continuous cube packings` �
continuous cube tilings.
The number are lower than in the caseP � �
, since wedo the enumeration only of the one with strictly positiveprobability.
One of
` �
non-extendible continuous cube-packing hasr
cubes:
3 � 3 ] 3 _ 3 a 3� 3�
+ � + � 4 � +� + � � +� 4 � + � � 4 �
+ ] + � + ] 4 � + � 4 � + ] 4 � + ]+ _ + � + � 4 � + _ 4 � + _ + � 4 �
+ a + � +�� +�� 4 � +�� 4 � + a 4 �
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Packing density
Denote by ; 1 � u � the packing density for continuouscube packing and ; 1 � P �
the packing density in6 1s � P 6 1
.
One has
; � � u � � ; ] � u � � � � ; _ � u � � _� _� � �-, f . �,
and ; a � u � � � � ]� � � � �� _ _� � � � ] � � � ] � � � � , f / ., , ,
Thm. For any
' Q `, one has ; 1 � u ��� �
.
Thm. One has the limit �R �� ; 1 � P � � ; 1 � u �
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Non-extendible cubepackings
The number of cubes in non-extendible cube packingsis at least k 4 �
.
Problem find better lower bound on size ofnon-extendible cube packing.
In dimension
q
, we found a continuous non-extendiblecube packing with
0
cubes. But is it with strictly positiveprobability?
Problem Does there exist continuous non-extendiblecube packing of lower number of cubes than the one ofstrictly positive probability?
Problem Prove that there is no non-extendible cubepacking with
� o � �cubes with
� n `
.
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