# Computing Hypercrossed Complex Pairings in Digital Images

Authors: Simge Öztunç; Ali Mutlu; Necdet Bildik
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We consider an additive group structure in digital images and introduce the commutator in digital images. Then we calculate the hypercrossed complex pairings which generates a normal subgroup in dimension 2 and in dimension 3 by using 8-adjacency and 26-adjacency.


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## Body

## 1. Introduction

In this paper we denote the set of integers byℤ. Then ℤn represents the set of lattice points in Euclidean n-dimensional spaces. A finite subset of  ℤn  with an adjacency relation is called a digital image.Definition 1 (see [1, 2]).
Consider the following.(1)
Two pointsp and q in ℤ are 2-adjacent if |p-q|=1.(2)
Two pointsp and q in ℤ2 are 8-adjacent if they are distinct and differ by at most 1 in each coordinate.(3)
Two pointsp and q in ℤ2 are 4-adjacent if they are 8-adjacent and differ by exactly one coordinate.(4)
Two pointsp and q in ℤ2 are 26-adjacent if they are distinct and differ by at most 1 in each coordinate.(5)
Two pointsp and q in ℤ2 are 18-adjacent if they are 26-adjacent and differ in at most two coordinates.(6)
Two pointsp and q in ℤ2 are 6-adjacent if they are 18-adjacent and differ by exactly one coordinate.Definition 2.
LetG be a subset of a digital image. A simplicial group G in digital images consists of a sequence of groups G and collections of group homomorphisms di:Gn→Gn-1 and si:Gn→Gn+1, 0≤i≤n, that satisfies the following axioms:
(1)didj=dj-1di,i<j,disj=sj-1di,i<j,djsj=dj+1sj=id,i=jori=j+1,disj=sjdi-1,i>j+1,sisj=sj+1si,i≤j.Definition 3.
Given a simplicial groupG with κ-adjacency, the Moore complex (NG,∂) of G is the chain complex defined by
(2)NGn=⋂i=0n-1Kerdi,
with ∂:NGn→NGn-1 induced from dn by restriction.
Thenth homology group of the Moore complex of G is
(3)Hn(NG,∂)=⋂i=0nKerdidn+1(⋂i=0nKerdi).


## 2. Hypercrossed Complex Pairings in Digital Images

First of all we adapt ideas from Carrasco and Cegarra [3–5] to get the construction in digital images. We define a set P(n) consisting of pairs of elements (α,β) from S(n) with α∩β=⌀ and β<α, with respect to lexicographic ordering in S(n) where α=(il,…,i1) and β=(jm,…,j1)∈S(n).Consider the following diagram:(4)
where
(5)sα=sil⋯si1:NGn-#α⟶Gn,sβ=sjm⋯sj1:NGn-#β⟶Gn,
and  define  p:Gn→NGn and p(x)=pn-1⋯p0(x) as pj(z)=z-sidiz and j=0,…,n-1. Since a digital image has the additive group structure, define the commutator as
(6)[x,y]=xy-yx.
Thus
(7)μ:Gn×Gn⟶Gn,Fα,β(xα,yβ)=pμ(sα×sβ)(xα,yβ)=p[sαxα,sβyβ].
The normal subgroup NGn of Gn is generated by the elements of the form
(8)Fα,β(xα,yβ),
where xα∈NGn-#α and yβ∈NGn-#β.Theorem 4.
2-dimensional normal subgroupN2 with 8-adjacency is generated by the elements of the form
(9)[s0x1-s1x1,s1y1].Proof.
Letα=(1) and β=(0) for n=2. For x1 and y1∈NG1=Kerd0,
(10)F(0),(1)(x1,y1)=p1p0[s0x1,s1y1]=p1{[s0x1,s1y1]-s0d0[s0x1,s1y1]}=[s0x1-s1x1,s1y1].
Thus F(0),(1)(x1,y1)=[s0x1-s1x1,s1y1] and this is the element generating N2 normal subgroups.Proposition 5.
3-dimensional normal subgroupN3 with 26-adjacency is generated by the elements of the following forms:(i)
[s1s0x1-s0s1x1,s2y2],(ii)
[s2s0x1-s2s1x1,s1y2-s2y2],(iii)
[s0x2-s1x2+s2x2,s1s1y1],(iv)
[s1x2-s2x2,s2y2],(v)
[s0x2,s2y2],(vi)
[s0x2-s1x2,s1y2]+[s2x2,s2y2].Proof.
Forn=3 the possible pairings are the following:(i)
F(1,0)(2),(ii)
F(2,0)(1),(iii)
F(0)(2,1),(iv)
F(1)(2),(v)
F(0)(2),(vi)
F(0)(1).For all x1∈NG1 and y2∈NG2 the corresponding generators of N3 are the following with Fα,β:NG1×NG2→NG3 and for n=3, p(x)=p2p1p0(x):
(i)(11)F(1,0)(2)(x1,y2)=p[s1s0x1,s2y2]=p2p1p0[s1s0x1,s2y2]=[s1s0x1-s2s1x1,s2y2],
(ii)(12)F(2,0)(1)(x1,y2)=p[s2s0x1,s1y2]=p2p1p0[s2s0x1,s1y2]=[s2s0x1-s2s1x1,s1y2-s2y2].
For all x2∈NG2 and y1∈NG1 and considering the map Fα,β:NG2×NG1→NG3, the corresponding generator of  N3 is
(iii)(13)F(0)(2,1)(x2,y1)=p[s0x2,s2s1y1]=p2p1p0[s0x2,s2s1y1]=[s0x2-s1x2+s2x2,s2s1y1].
For all x2, y2∈NG2 and for Fα,β:NG2×NG2→NG3 the corresponding generators of N3 are
(iv)(14)F(1)(2)(x2,y2)=p[s1x2,s2y2]=p2p1p0[s1x2,s2y2]=[s1x2-s2x2,s2y2],
(v)
(15)F(0)(2)(x2,y2)=p[s0x2,s2y2]=p2p1p0[s0x2,s2y2]=[s0x2,s2y2],
(vı)(16)F(0)(1)(x2,y2)=p[s0x2,s1y2]=p2p1p0[s0x2,s1y2]=[s0x2-s1x2,s1y2]+[s2x2,s2y2].Theorem 6.
LetNG2 be a 2-dimensional Moore complex of a simplicial group G. Then ∂2(NG2)=[Kerd0,Kerd1] where ∂2 is induced from d2 by restriction.Proof.
Forn=2, assume that α=(0), β=(1), and x1, y1∈NG1=Kerd0. Now calculate dnFα,β.
SinceF(0),(1)(x1,y1)=[s0x1-s1x1,s1y1], from Proposition 5(17)d2F(0),(1)(x1,y1)=[d2s0x1-d2s1︸idx1,d2s1︸idy1]=[s0d1x1-x1,y1].
At first we investigate whethers0d1x1-x1 is in Kerd0 or not. 							(18)d0s0︸idd1x1-d0x1︸=0=d1x1;
therefore s0d1x1-x1∉Kerd0.
Secondly we examine whethers0d1x1-x1 is in Kerd1 or not.
Since(19)d1s0︸idd1x1-d1x1︸=0=d1x1-d1x1=0,s0d1x1-x1∈Kerd1.
From the assumptiony1∈Kerd0, we get
(20)F(0),(1)(x1,y1)∈[Kerd1,Kerd0].Theorem 7.
LetNG3 be a 3-dimensional Moore complex of a simplicial group G with 26-adjacency. Then
(21)∂3(NG3)⊆[Kerd2,Kerd0∩Kerd1]+[Kerd1,Kerd0∩Kerd2]+[Kerd1∩Kerd2,Kerd0]+[Kerd1,Kerd0]+[Kerd0∩Kerd2,Kerd0∩Kerd1]+[Kerd2,Kerd0∩Kerd1],
where ∂3 is induced from d3 by restriction.Proof.
Forn=3 investigate dnFα,β where x1∈NG1 and y2∈NG2=Kerd0∩Kerd1.
From Proposition5 we have F(1,0)(2)(x1,y2)=[s1s0x1-s2s0x1,s2y2]. Then applying d3 to F(1,0)(2)(x1,y2), we get the following:
(22)d3F(1,0)(2)(x1,y2)=[d3s1s0x1-d3s2︸ids0x1,d3s2︸idy2]=[s1d2s0x1-s0x1,y2]=[s1s0d1x1-s0x1,y2].
Firstly, examine whether s1s0d1x1-s0x1 is  in Kerd0 or not:
(23)d0(s1s0d1x1-s0x1)=d0s1s0d1x1-d0s0x1=s0d0s0︸idd1x1-d0s0︸idx1=s0d1x1-x1.
So s1s0d1x1-s0x1∉Kerd0.
Secondly  we investigate whethers1s0d1x1-s0x1 is in Kerd1 or not:
(24)d1(s1s0d1x1-s0x1)=d1s1s0d1x1-d1s0x1=d1s1︸ids0d1x1-d1s0︸idx1=s0d1x1-x1.
Therefore s1s0d1x1-s0x1∉Kerd1.
Finally we check whethers1s0d1x1-s0x1 is in Kerd2 or not.
Since(25)d2(s1s0d1x1-s0x1)=d2s1︸ids0d1x1-d2s0x1=s0d1x1-s0d1x1=0,
therefore s1s0d1x1-s0x1∈Kerd2.
We get(26)F(0,1),(2)(x1,y1)∈[Kerd2,Kerd0∩Kerd1],
since y2∈Kerd0∩Kerd1.
If(27)F(2,0)(1)(x1,y2)=[s2s0x1-s2s1x1,s1y2-s2y2],
then
(28)d3F(2,0)(1)(x1,y2)=d3([s2s0x1-s2s1x1,s1y2-s2y2])=[d3s2︸ids0x1-d3s2︸ids1x1,d3s1y2-d3s2︸idy2]=[s0x1-s1x1,s1d2y2-y2].
At first  we check whethers0x1-s1x1 is in Kerd0, Kerd1, and Kerd2 or not. 							(29)d0(s0x1-s1x1)=d0s0︸idx1-d0s1x1=x1-d0s1x1.
Thus s0x1-s1x1∉Kerd0.
Next, since(30)d1(s0x1-s1x1)=d1s0︸idx1-d1s1︸idx1=x1-x1=0,s0x1-s1x1∈Kerd1,
and, finally,
(31)d2(s0x1-s1x1)=d2s0x1-d2s1︸idx1=s0d1x1-x1∉Kerd2.
Now examine whethers1d2y2-y2 is in Kerd0, Kerd1, and Kerd2 or not:
(32)d0(s1d2y2-y2)=d0s1d2y2-d0y2=s0d0d2y2-d0y2=s0d1d0y2︸=0-d0y2︸=0=0.
Therefore s1d2y2-y2∈Kerd0. We have the following:
(33)d1(s1d2y2-y2)=d1s1︸idd2y2-d1y2︸=0=d2y2∉Kerd1;d2(s1d2y2-y2)=d2s1︸idd2y2-d2y2=d2y2-d2y2=0⟹s1d2y2-y2∈Kerd2.
So F(2,0)(1)(x1,y2)=[Kerd1,Kerd0∩Kerd2].
For allx2∈NG2 and y1∈NG1 if
(34)F(0)(2,1)(x2,y1)=[s0x2-s1x2+s2x2,s1s1y1],
then
(35)d3F(0)(2,1)(x2,y1)=d3([s0x2-s1x2+s2x2,s2s1y1])=[d3s0x2-d3s1x2+d3s2︸idx2,d3s2︸ids1y1]=[s0d2x2-s1d2x2+x2,s1y1].
Firstly investigate whether s0d2x2-s1d2x2+x2 is in Kerd0, Kerd1, and Kerd2 or not:
(36)d0(s0d2x2-s1d2x2+x2)=d0s0︸idd2x2-d0s1d2x2+d0x2︸=0=d2x2-s0d1d0x2︸=0=d2x2.
Thereby s0d2x2-s1d2x2+x2∉Kerd0. We have
(37)d1(s0d2x2-s1d2x2+x2)=d1s0︸idd2x2-d1s1︸idd2x2+d1x2︸=0=d2x2-d2x2=0.
For this reason s0d2x2-s1d2x2+x2∈Kerd1. We also have
(38)d2(s0d2x2-s1d2x2+x2)=d2s0d2x2-d2s1︸idd2x2+d2x2=s0d1d2x2-d2x2+d2x2=s0d1d1x2︸=0=0.
Hence s0d2x2-s1d2x2+x2∈Kerd2.
Later on we research whethers2s1y1 is in Kerd0, Kerd1, and Kerd2 or not.
Sinced0(s1y1)=s0d0y1︸=0=0, s1y1∈Kerd0.
Sinced1(s1y1)=d1s1︸idy1=y1, s1y1∉Kerd1.
Sinced2(s1y1)=d2s1︸idy1=y1, s1y1∉Kerd2.
Thusd3F(0)(2,1)(x2,y1)∈[Kerd1∩Kerd2,Kerd0].
For allx2, y2∈NG2 since F(0)(2)(x2,y2)=[s0x2,s2y2],
(39)d3F(0)(2)(x2,y2)=d3([s0x2,s2y2])=[d3s0x2,d3s2︸idy2]=[s0d2x2,y2].
By using properties of the commutator we have
(40)[s0d2x2-s1d2x2+x2,y2]=[s0d2x2+(x2-s1d2x2),y2]=[s0d2x2,y2]+[x2-s1d2x2,y2],[s0d2x2-s1d2x2+x2,y2]+[y2,x2-s1d2x2]=[s0d2x2,y2]=d3F(0)(2)(x2,y2).
Thus
(41)d3F(0)(2)(x2,y2)∈[Kerd1∩Kerd2,Kerd1∩Kerd0]+[Kerd0∩Kerd1,Kerd0∩Kerd2].
If F(1)(2)(x2,y2)=[s1x2-s2x2,s2y2], then
(42)d3F(1)(2)(x2,y2)=d3([s1x2-s2x2,s2y2])=[d3s1x2-d3s2︸idx2,d3s2︸idy2]=[s1d2x2-x2,y2].
Firstly we check whether s1d2x2-x2 is in  Kerd0, Kerd1, and Kerd2 or not:
(43)d0(s1d2x2-x2)=d0s1d2x2-d0x2︸=0=s0d0d2x2=s0d1d0x2︸=0=0.
Therefore s1d2x2-x2∈Kerd0. Since
(44)d1(s1d2x2-x2)=d1s1︸idd2x2-d1x2︸=0=d2x2,s1d2x2-x2∉Kerd1. We have
(45)d2(s1d2x2-x2)=d2s1︸idd2x2-d2x2=d2x2-d2x2=0.
Hence s1d2x2-x2∈Kerd2.
Because of the casey2∈Kerd0∩Kerd1,
(46)d3F(1)(2)(x2,y2)∈[Kerd0∩Kerd2,Kerd0∩Kerd1].
If F(0)(1)(x2,y2)=[s0x2-s1x2,s1y2]+[s2x2,s2y2], then
(47)d3F(0)(1)(x2,y2)=[d3s0x2-d3s1x2,s1y2]+[d3s2︸idx2,d3s2︸idy2]=[s0d2x2-s1d2x2,s1d2y2]+[x2,y2].
Consider the following commutator:
(48)[s0d2x2-s1d2x2+x2,s1d2y2-y2],
and code the terms of this commutator such as
(49)a=s0d2x2,b=s1d2y2,c=s1d2x2,d=x2,e=y2,
in order to simplify the algebraic operations. Thus, by using the properties and definition of the commutator we obtain the following:
(50)[a-c+d,b-e]=[a-c,b]+[d,e],(a-c+d)(b-e)-(b-e)(a-c+d)=ab-cb+db-ae+ce-de-{ba-bc+bd-ea+ec-ed}.
Consider the following cases:
(51)ab-cb-ba+bc=(a-c)b-b(a-c)=[a-c,b],ce-de-ec+ed=(c-d)e-e(c-d)=[c-d,e].
And from the remaining terms we get
(52)db-bd-[d,e]=db-bd-de+ed=d(b-e)-(b-e)d=[d,b-e].
Consequently for n=3 we have
(53)∂3(NG3)⊆[Kerd2,Kerd0∩Kerd1]+[Kerd1,Kerd0∩Kerd2]+[Kerd1∩Kerd2,Kerd0]+[Kerd1,Kerd0]+[Kerd0∩Kerd2,Kerd0∩Kerd1]+[Kerd2,Kerd0∩Kerd1].Corollary 8.
LetNG3 be a 3-dimensional Moore complex of a simplicial group G with 26-adjacency. Then
(54)∂3(NG3)⊆[Kerd2,Kerd0∩Kerd1]+[Kerd1,Kerd0∩Kerd2]+[Kerd1∩Kerd2,Kerd0]+[Kerd1,Kerd0]+[Kerd0∩Kerd2,Kerd0∩Kerd1]+[Kerd2,Kerd0∩Kerd1].Proof.
Otherwise inclusion for the previous theorem is obtained from [4, 5]. Therefore
(55)∂3(NG3)=[Kerd2,Kerd0∩Kerd1]+[Kerd1,Kerd0∩Kerd2]+[Kerd1∩Kerd2,Kerd0]+[Kerd1,Kerd0]+[Kerd0∩Kerd2,Kerd0∩Kerd1]+[Kerd2,Kerd0∩Kerd1].


## 3. Conclusion

In this paper for dimension 2 and dimension 3, we obtained the Moore complex of simplicial groups generated by hypercrossed complex pairings in digital images.



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