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A lower bound for the Graver complexity of the incidence matrix of a complete bipartite graph

Taisei Kudo  and  Akimichi Takemura11footnotemark: 1 Department of Mathematical Informatics, Graduate School of Information Science and Technology, University of Tokyo.JST, CREST
(February 2011)
Abstract

We give an exponential lower bound for the Graver complexity of the incidence matrix of a complete bipartite graph of arbitrary size. Our result is a generalization of the result by Berstein and Onn [2] for the complete bipartite graph K3,rK_{3,r}, r3r\geq 3.

Keywords and phrases: algebraic statistics, contingency table, three-way transportation program.

1 Introduction and the main result

The Graver complexity of an integer matrix is currently actively investigated for its importance to integer programming, algebraic statistics and other applications ([2], [4], [5], [1]). In particular, from the universality of the three-way transportation program to general integer programs (De Loera and Onn [3]), the Graver complexity of the incidence matrix of the complete bipartite graph K3,rK_{3,r} is particularly important. Berstein and Onn [2] proved that the Graver complexity g(r)g(r) for the incidence matrix of K3,rK_{3,r}, r3r\geq 3, is bounded below as g(r)=Ω(2r)g(r)=\Omega(2^{r}), where g(r)172r37g(r)\geq 17\cdot 2^{r-3}-7. It is a natural question to generalize this result to the complete bipartite graph Kt,rK_{t,r} of arbitrary size t,rt,r. We prove that the Graver complexity for Kt,rK_{t,r} is Ω((t1)r)\Omega((t-1)^{r}), where t4t\geq 4 is fixed and rr diverges to infinity. For proving our result, we employ double induction on rr and tt starting from the result of [2].

Let At,rA_{t,r} denote the incidence matrix of the complete bipartite graph Kt,rK_{t,r} and let g(At,r)g(A_{t,r}) denote its Graver complexity. Here we state our main theorem. Relevant notations and definitions will be given in the next section.

Theorem 1.1.

The Graver complexity of At,rA_{t,r} for any 4tr4\leq t\leq r is bounded from below as

g(At,r)(t1)rt(bt+1t2)1t2,g(A_{t,r})\geq(t-1)^{r-t}(b_{t}+\frac{1}{t-2})-\frac{1}{t-2},

where

bt=(t2)!(15+i=1t4i+4(i+2)!).b_{t}=(t-2)!\left(15+\sum_{i=1}^{t-4}\frac{i+4}{(i+2)!}\right).

We give a proof of this theorem in Section 3 after giving necessary definitions and reviewing relevant known results in Section 2. We conclude the paper with some discussion in Section 4.

2 Preliminaries

In this section we summarize our notation and review relevant known results on the Graver complexity following Berstein and Onn [2].

The integer kernel of an s×ts\times t integer matrix AA is denoted by ker(A)={xtAx=0}\ker_{\mathbb{Z}}(A)=\{x\in{\mathbb{Z}}^{t}\mid Ax=0\}. Define a partial order \sqsubseteq on t{\mathbb{Z}}^{t}, which extends the coordinate-wise order \leq on +t{\mathbb{Z}}_{+}^{t}, as follows: For two vectors u,vtu,v\in{\mathbb{Z}}^{t}, uvu\sqsubseteq v if |ui||vi||u_{i}|\leq|v_{i}| and uivi0u_{i}v_{i}\geq 0 for i=1,,ti=1,\dots,t. The Graver basis 𝒢(A){\mathcal{G}}(A) of AA is the finite set of \sqsubseteq-minimal elements in the set ker(A){0}\ker_{\mathbb{Z}}(A)\setminus\{0\}.

For any fixed positive integer hh, write an htht-dimensional integer vector xhtx\in{\mathbb{Z}}^{ht} as x=(x1,,xh)x=(x^{1},\dots,x^{h}) with each block xix^{i} belonging to t{\mathbb{Z}}^{t}. The type of x=(x1,,xh)x=(x^{1},\dots,x^{h}) is the number type(x):=#({ixi0}){\rm type}(x):=\#(\{i\mid x^{i}\neq 0\}) of nonzero blocks of xx. The hh-th Lawrence lifting of an s×ts\times t matrix AA is the following (t+hs)×ht(t+hs)\times ht matrix, with ItI_{t} denoting the t×tt\times t identity matrix:

A(h):=(A0000A00000AItItItIt).A^{(h)}:=\left(\begin{array}[]{ccccc}A&0&0&\dots&0\\ 0&A&0&\dots&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&A\\ I_{t}&I_{t}&I_{t}&\dots&I_{t}\\ \end{array}\right). (1)

The Graver complexity of AA is defined as

g(A)=sup({0}{type(x)|xh1𝒢(A(h))}).{g}(A)=\sup\left(\{0\}\cup\Set{{\rm type}(x)}{x\in\bigcup_{h\geq 1}{\mathcal{G}}\left(A^{(h)}\right)}\right). (2)

Let 𝒢(𝒢(A)){\mathcal{G}}({\mathcal{G}}(A)) denote the Graver basis of a matrix whose columns are the elements of 𝒢(A){\mathcal{G}}(A) ordered arbitrarily. The following result shows that the Graver complexity of AA is determined by 𝒢(𝒢(A)){\mathcal{G}}({\mathcal{G}}(A)).

Proposition 2.1.

[6]  The Graver complexity of AA satisfies

g(A)=max{x1:x𝒢(𝒢(A))},{g}(A)={\rm max}\{\|x\|_{1}:x\in{\mathcal{G}}({\mathcal{G}}(A))\},

where 1\|\cdot\|_{1} denotes the 1-norm of a vector.

A circuit of an integer matrix AA is a nonzero integer vector xker(A)x\in\ker_{\mathbb{Z}}(A), that has inclusion-minimal support with respect to ker(A)\ker_{\mathbb{Z}}(A), and whose nonzero entries are relatively prime. Let 𝒞(A){\mathcal{C}}(A) denote the set of circuits of a matrix AA. Then 𝒞(A)𝒢(A){\mathcal{C}}(A)\subseteq{\mathcal{G}}(A) (cf. [7]). An integer relation h=(h1,,hk)h=(h_{1},\dots,h_{k}) on integer vectors v1,,vktv^{1},\dots,v^{k}\in{\mathbb{Z}}^{t}

0=h1v1++hkvk0=h_{1}v^{1}+\dots+h_{k}v^{k}

is primitive if h1,,hkh_{1},\dots,h_{k} are relatively prime positive integers and no k1k-1 of the {vi}i=1k\{v^{i}\}_{i=1}^{k} satisfy any nontrivial linear relation. By 𝒞(A)𝒢(A){\mathcal{C}}(A)\subseteq{\mathcal{G}}(A) and Proposition 2.1 we have the following result.

Proposition 2.2.

[2]  Suppose that hh is a primitive relation on some set of circuits {xi}i=1k\{x^{i}\}_{i=1}^{k} of an integer matrix AA. Then the Graver complexity of AA satisfies g(A)i=1khig(A)\geq\sum_{i=1}^{k}h_{i}.

In this paper we consider the Graver complexity of the incidence matrix At,rA_{t,r} for the complete bipartite graph Kt,rK_{t,r}. Let 𝟏t=(1,1,,1){\bf 1}_{t}=(1,1,\dots,1) denote the 1×t1\times t matrix consisting of 11’s. Then the rr-th Lawrence lifting At,r=𝟏t(r)A_{t,r}={\bf 1}_{t}^{(r)} of 𝟏t{\bf 1}_{t} is the incidence matrix of Kt,rK_{t,r}. In algebraic statistics, At,rA_{t,r} is the design matrix specifying the row sums and the column sums of a two-way contingency table. Another Lawrence lifting (𝟏t(r))(h)({\bf 1}_{t}^{(r)})^{(h)} of 𝟏t(r){\bf 1}_{t}^{(r)} is the design matrix for no-three-factor interaction model for t×r×ht\times r\times h three-way contingency tables ([1], [5]). It is also the coefficient matrix for the three-way transportation program. The Graver complexity g(At,r)=g(𝟏t(r))g(A_{t,r})=g({\bf 1}_{t}^{(r)}) gives the bound of complexity of the Graver basis for the toric ideal associated with the no-three-factor interaction model for t×r×ht\times r\times h three-way contingency tables as hh\rightarrow\infty.

We employ below the following notation, where t,rt,r are positive integers. Let

V:={v1,,vt},U:={u1,,ur}.V:=\{v_{1},\dots,v_{t}\},\ U:=\{u_{1},\dots,u_{r}\}. (3)

Then VUV\oplus U and V×UV\times U denote the set of vertices and the set of edges of the complete bipartite graph Kt,rK_{t,r}, respectively. They index the rows and the columns of the incidence matrix At,rA_{t,r} of Kt,rK_{t,r}. Here we explain interpretations of a circuit of At,rA_{t,r} referring to [2]. We interpret each vector xV×Ux\in{\mathbb{Z}}^{V\times U} as:

  1. 1.

    an integer valued function on the set of edges V×UV\times U;

  2. 2.

    a t×rt\times r matrix with its rows and columns indexed by V and U.

With these interpretations, xx is in 𝒞(At,r){\mathcal{C}}(A_{t,r}) if and only if:

  1. 1.

    as a function on V×UV\times U with the following properties: its support is a circuit of Kt,rK_{t,r}, along which its values ±1\pm 1 alternate. It can be expressed by the sequence (vi1,ui1,vi2,ui2,,vil,uil)(v_{i_{1}},u_{i_{1}},v_{i_{2}},u_{i_{2}},\dots,v_{i_{l}},u_{i_{l}}) of vertices of the circuit of Kt,rK_{t,r} on which it is supported, with the convention that its value is +1+1 on the first edge (vi1,ui1)(v_{i_{1}},u_{i_{1}}).

  2. 2.

    as a nonzero matrix with the following properties: its elements are 0,±10,\pm 1, its row sums and columns sums are zeros, and it has an inclusion-minimal support with respect to these properties.

The following example is the base case for our inductive argument for the lower bound of the Graver complexity.

Table 1: Circuits for A3,4A_{3,4} in a 3×43\times 4 matrix form
u1u_{1} u2u_{2} u3u_{3} u4u_{4}
0 0 1-1 11 v1v_{1}
x1=(v1,u4,v3,u2,v2,u3)=x^{1}=(v_{1},u_{4},v_{3},u_{2},v_{2},u_{3})= 0 1-1 11 0 v2v_{2}
0 11 0 1-1 v3v_{3}
1-1 11 0 0 v1v_{1}
x2=(v1,u2,v3,u3,v2,u1)=x^{2}=(v_{1},u_{2},v_{3},u_{3},v_{2},u_{1})= 11 0 1-1 0 v2v_{2}
0 1-1 11 0 v3v_{3}
0 1-1 0 11 v1v_{1}
x3=(v1,u4,v2,u1,v3,u2)=x^{3}=(v_{1},u_{4},v_{2},u_{1},v_{3},u_{2})= 11 0 0 1-1 v2v_{2}
1-1 11 0 0 v3v_{3}
1-1 0 0 11 v1v_{1}
x4=(v1,u4,v2,u2,v3,u1)=x^{4}=(v_{1},u_{4},v_{2},u_{2},v_{3},u_{1})= 0 11 0 1-1 v2v_{2}
11 1-1 0 0 v3v_{3}
11 0 1-1 0 v1v_{1}
x5=(v1,u1,v2,u2,v3,u3)=x^{5}=(v_{1},u_{1},v_{2},u_{2},v_{3},u_{3})= 1-1 11 0 0 v2v_{2}
0 1-1 11 0 v3v_{3}
0 1-1 11 0 v1v_{1}
x6=(v1,u3,v2,u4,v3,u2)=x^{6}=(v_{1},u_{3},v_{2},u_{4},v_{3},u_{2})= 0 0 1-1 11 v2v_{2}
0 11 0 1-1 v3v_{3}
0 11 0 1-1 v1v_{1}
x7=(v1,u2,v2,u3,v3,u4)=x^{7}=(v_{1},u_{2},v_{2},u_{3},v_{3},u_{4})= 0 1-1 11 0 v2v_{2}
0 0 1-1 11 v3v_{3}
Example 2.1.

[2]  Let t=3t=3 and r=4r=4. Consider seven circuits {xi}i=17\{x^{i}\}_{i=1}^{7} in Table 1 (written in a 3×43\times 4 matrix form) of A3,4=(1,1,1)(4)A_{3,4}=(1,1,1)^{(4)}. They satisfy a primitive relation

x1+2x2+3x3+3x4+5x5+6x6+7x7=0.x^{1}+2x^{2}+3x^{3}+3x^{4}+5x^{5}+6x^{6}+7x^{7}=0.

Therefore from Proposition 2.2

g(A3,4)1+2+3+3+5+6+7=27.{g}(A_{3,4})\geq 1+2+3+3+5+6+7=27.

3 Proof of the main theorem

In this section we give a proof of our main theorem. Our proof is based on recursive construction of primitive relations for circuits of At,rA_{t,r}. We need recursions for tt and rr, separately. In Lemma 3.1 we give a recursion for tt and in Lemma 3.2 we give a recursion for rr.

Lemma 3.1.

Let t4t\geq 4. Suppose that there are circuits {xi}i=1k\{x^{i}\}_{i=1}^{k} of At,t+1=𝟏t(t+1)A_{t,t+1}={\bf 1}_{t}^{(t+1)} admitting a primitive relation hh, where the kk-th circuit and the kk-th coefficient are

xk\displaystyle x^{k} =(v1,u2,v2,u3,,vt,ut+1),\displaystyle=(v_{1},u_{2},v_{2},u_{3},\dots,v_{t},u_{t+1}),
hk\displaystyle h_{k} =1.\displaystyle=1.

Then there are circuits {x¯i}i=1k+t\{\bar{x}^{i}\}_{i=1}^{k+t} of At+1,t+1=𝟏t+1(t+1)A_{t+1,t+1}={\bf 1}_{t+1}^{(t+1)} admitting a primitive relation h¯\bar{h}, where the (k+t)(k+t)-th circuit and the (k+t)(k+t)-th coefficient are

x¯k+t\displaystyle\bar{x}^{k+t} =(v1,u1,v2,u2,,vt+1,ut+1),\displaystyle=(v_{1},u_{1},v_{2},u_{2},\dots,v_{t+1},u_{t+1}),
h¯k+t\displaystyle\bar{h}_{k+t} =1.\displaystyle=1.
Proof.

Using the natural embedding of Kt,t+1K_{t,t+1} into Kt+1,t+1K_{t+1,t+1}, we can interpret circuits of the former also as circuits of the latter. Put

yi=xi,i=1,,k1,y^{i}=x^{i},\ \forall i=1,\dots,k-1,

and define

yk+j1\displaystyle y^{k+j-1} =(v1,uj,vt+1,uj+1),j=1,,t,\displaystyle=(v_{1},u_{j},v_{t+1},u_{j+1}),\ \forall j=1,\dots,t,
yk+t\displaystyle y^{k+t} =(v1,u2,v2,u3,,vt,ut+1,vt+1,u1).\displaystyle=(v_{1},u_{2},v_{2},u_{3},\dots,v_{t},u_{t+1},v_{t+1},u_{1}).

Table 2 displays {yk+j1}j=1t+1\{y^{k+j-1}\}_{j=1}^{t+1} as matrices, where

p=(1,0,,0,1)t+1.p=(1,0,\dots,0,-1)^{\top}\in{\mathbb{Z}}^{t+1}.

Blank entries are zeros. Note that these circuits satisfy j=1t+1yk+j1=xk\sum_{j=1}^{t+1}y^{k+j-1}=x^{k}.

Table 2: Circuits for recursion on tt
u1u_{1} u2u_{2} u3u_{3} \dots utu_{t} ut+1u_{t+1}
yk=(v1,u1,vt+1,u2)=y^{k}=(v_{1},u_{1},v_{t+1},u_{2})= pp p-p
yk+1=(v1,u2,vt+1,u3)=y^{k+1}=(v_{1},u_{2},v_{t+1},u_{3})= pp p-p
\vdots \ddots \ddots
yk+t1=(v1,ut,vt+1,ut+1)=y^{k+t-1}=(v_{1},u_{t},v_{t+1},u_{t+1})= pp p-p
yk+t1=(v1,ut,vt+1,ut+1)=y^{k+t-1}=(v_{1},u_{t},v_{t+1},u_{t+1})= pp p-p
1-1 11
1-1 11
yk+t=(v1,u2,v2,u3,,vt,ut+1,vt+1,u1)=y^{k+t}=(v_{1},u_{2},v_{2},u_{3},\dots,v_{t},u_{t+1},v_{t+1},u_{1})= \ddots \ddots
1-1 11
1-1 11
11 1-1

Suppose that h¯k+t\bar{h}\in{\mathbb{Z}}^{k+t} satisfies

h¯i\displaystyle\bar{h}_{i} =hi,i=1,,k1,\displaystyle=h_{i},\ \forall i=1,\dots,k-1,
h¯k+j1\displaystyle\bar{h}_{k+j-1} =hk,j=1,,t+1.\displaystyle=h_{k},\ \forall j=1,\dots,t+1.

Then

i=1k+th¯iyi=i=1k1hiyi+j=1t+1hkyk+j1=i=1k1hixi+hkxk=0.\sum_{i=1}^{k+t}\bar{h}_{i}y^{i}=\sum_{i=1}^{k-1}h_{i}y^{i}+\sum_{j=1}^{t+1}h_{k}y^{k+j-1}=\sum_{i=1}^{k-1}h_{i}x^{i}+h_{k}x^{k}=0.

Therefore h¯\bar{h} is an integer relation of circuits {yi}i=1k+t\{y^{i}\}_{i=1}^{k+t}.

Next, we show that h¯\bar{h} is primitive. Suppose that hk+th^{\prime}\in{\mathbb{Z}}^{k+t} is a nontrivial relation on the {yi}i=1k+t\{y^{i}\}_{i=1}^{k+t}. Without loss of generality we may assume that the {hi}i=1k+t\{h^{\prime}_{i}\}_{i=1}^{k+t} are relatively prime integers, at least one of which is positive. We look at the row of vt+1v_{t+1}. Then it follows that

hk=hk+1==hk+t.h^{\prime}_{k}=h^{\prime}_{k+1}=\dots=h^{\prime}_{k+t}.

Therefore

0=i=1k+thiyi=i=1k1hiyi+hkj=1t+1yk+j1=i=1k1hixi+hkxk.0=\sum_{i=1}^{k+t}h^{\prime}_{i}y^{i}=\sum_{i=1}^{k-1}h^{\prime}_{i}y^{i}+h^{\prime}_{k}\sum_{j=1}^{t+1}y^{k+j-1}=\sum_{i=1}^{k-1}h^{\prime}_{i}x^{i}+h^{\prime}_{k}x^{k}.

This is an integer relation on {xi}i=1k\{x^{i}\}_{i=1}^{k}, and because hh is primitive,

hi=hi,i=1,,k.h^{\prime}_{i}=h_{i},\ \forall i=1,\dots,k.

Therefore h=h¯h^{\prime}=\bar{h} and h¯\bar{h} is primitive.

Now apply to {yi}i=1k+t\{y^{i}\}_{i=1}^{k+t} a permutation of columns so that yk+ty^{k+t} becomes (v1,u1,v2,u2,,vt+1,ut+1)(v_{1},u_{1},v_{2},u_{2},\dots,v_{t+1},u_{t+1}). For i=1,,k+ti=1,\dots,k+t, let x¯i\bar{x}^{i} be the circuit of At+1,t+1A_{t+1,t+1} which is the image of yiy^{i} under this permutation. Then {x¯i}i=1k+t\{\bar{x}^{i}\}_{i=1}^{k+t} also satisfy the primitive relation i=1k+th¯ix¯i=0\sum_{i=1}^{k+t}\bar{h}_{i}\bar{x}^{i}=0 with the same coefficients h¯\bar{h}. This completes the proof. ∎

Lemma 3.2.

Let rt4r\geq t\geq 4. Suppose that there are circuits {xi}i=1k\{x^{i}\}_{i=1}^{k} of At,r=𝟏t(r)A_{t,r}={\bf 1}_{t}^{(r)} admitting a primitive relation hh, where the kk-th circuit and the kk-th coefficient are

xk\displaystyle x^{k} =(v1,urt+1,v2,urt+2,,vt,ur),\displaystyle=(v_{1},u_{r-t+1},v_{2},u_{r-t+2},\dots,v_{t},u_{r}),
hk\displaystyle h_{k} =1.\displaystyle=1.

Then there are circuits {x¯i}i=1k+t1\{\bar{x}^{i}\}_{i=1}^{k+t-1} of At,r+1=𝟏t(r+1)A_{t,r+1}={\bf 1}_{t}^{(r+1)} admitting primitive relation h¯\bar{h}, where the (k+t1)(k+t-1)-th circuit is

x¯k+t1=(v1,urt+2,v2,urt+3,,vt,ur+1)\bar{x}^{k+t-1}=(v_{1},u_{r-t+2},v_{2},u_{r-t+3},\dots,v_{t},u_{r+1})

and the elements of h¯\bar{h} are

h¯i=(t1)hi,i=1,,k1,h¯k=h¯k+1==h¯k+t1=hk=1.\bar{h}_{i}=(t-1)h_{i},\ \forall i=1,\dots,k-1,\\ \ \bar{h}_{k}=\bar{h}_{k+1}=\dots=\bar{h}_{k+t-1}=h_{k}=1.
Proof.

Using the natural embedding of Kt,rK_{t,r} into Kt,r+1K_{t,r+1}, we can interpret circuits of the former also as circuits of the latter. Put

yi=xi,i=1,,k1,y^{i}=x^{i},\ \forall i=1,\dots,k-1,

and for all j=1,,tj=1,\dots,t, let yk+j1y^{k+j-1} denote vectors obtained by changing vertex urj+1u_{r-j+1} of xkx^{k} to ur+1u_{r+1}. Table 3 displays these circuits as matrices. Here for each i=1,,ti=1,\dots,t, qitq^{i}\in{\mathbb{Z}}^{t} denotes a vector satisfying

qii=1,qi+1i=1,q^{i}_{i}=1,\ q^{i}_{i+1}=-1,

and the rest are zeros. Here we identify t+1t+1 with 11.

Table 3: Circuits for recursion on rr
\ \dots urt+1u_{r-t+1} urt+2u_{r-t+2} urt+3u_{r-t+3} \dots uru_{r} ur+1u_{r+1}
yk=(v1,ur+1,v2,urt+2,,vt1,ur1,vt,ur)=y^{k}=(v_{1},u_{r+1},v_{2},u_{r-t+2},\dots,v_{t-1},u_{r-1},v_{t},u_{r})= \dots 0 q2q^{2} q3q^{3} \dots qtq^{t} q1q^{1}
yk+1=(v1,urt+1,v2,ur+1,,vt1,ur1,vt,ur)=y^{k+1}=(v_{1},u_{r-t+1},v_{2},u_{r+1},\dots,v_{t-1},u_{r-1},v_{t},u_{r})= \dots q1q^{1} 0 q3q^{3} \dots qtq^{t} q2q^{2}
\vdots \vdots
yk+t2=(v1,urt+1,v2,urt+2,,vt1,ur+1,vt,ur)=y^{k+t-2}=(v_{1},u_{r-t+1},v_{2},u_{r-t+2},\dots,v_{t-1},u_{r+1},v_{t},u_{r})= \dots q1q^{1} q2q^{2} q3q^{3} \dots qtq^{t} qt1q^{t-1}
yk+t1=(v1,urt+1,v2,urt+2,,vt1,ur1,vt,ur+1)=y^{k+t-1}=(v_{1},u_{r-t+1},v_{2},u_{r-t+2},\dots,v_{t-1},u_{r-1},v_{t},u_{r+1})= \dots q1q^{1} q2q^{2} q3q^{3} \dots 0 qtq^{t}

Notice that

j=1tyk+j1=(t1)xk.\sum_{j=1}^{t}y^{k+j-1}=(t-1)x^{k}.

Define

h¯i\displaystyle\bar{h}_{i} =(t1)hi,i=1,,k1,\displaystyle=(t-1)h_{i},\ \forall i=1,\dots,k-1,
h¯k+j1\displaystyle\bar{h}_{k+j-1} =hk=1,j=1,,t+1.\displaystyle=h_{k}=1,\ \forall j=1,\dots,t+1.

Then

i=1k+t1h¯iyi=i=1k1(t1)hiyi+j=1thkyk+j1=i=1k1hixi+hkxk=0.\sum_{i=1}^{k+t-1}\bar{h}_{i}y^{i}=\sum_{i=1}^{k-1}(t-1)h_{i}y^{i}+\sum_{j=1}^{t}h_{k}y^{k+j-1}=\sum_{i=1}^{k-1}h_{i}x^{i}+h_{k}x^{k}=0.

Therefore h¯\bar{h} is an integer relation on circuits {yi}i=1k+t1\{y^{i}\}_{i=1}^{k+t-1}.

Next we show that h¯\bar{h} is primitive. Suppose that hk+t1h^{\prime}\in{\mathbb{Z}}^{k+t-1} is a nontrivial relation on the {yi}i=1k+t1\{y^{i}\}_{i=1}^{k+t-1}. Without loss of generality we may assume that {hi}i=1k+t1\{h^{\prime}_{i}\}_{i=1}^{k+t-1} are relatively prime integers, at least one of which is positive. Consider the column of ur+1u_{r+1}. Then

hk=hk+1==hk+t1.h^{\prime}_{k}=h^{\prime}_{k+1}=\dots=h^{\prime}_{k+t-1}.

Therefore

0=i=1k+t1hiyi=i=1k1hiyi+hkj=1tyk+j1=i=1k1hixi+(t1)hkxk.0=\sum_{i=1}^{k+t-1}h^{\prime}_{i}y^{i}=\sum_{i=1}^{k-1}h^{\prime}_{i}y^{i}+h^{\prime}_{k}\sum_{j=1}^{t}y^{k+j-1}=\sum_{i=1}^{k-1}h^{\prime}_{i}x^{i}+(t-1)h^{\prime}_{k}x^{k}.

This is an integer relation on {xi}i=1k\{x^{i}\}_{i=1}^{k}. Therefore there exists α\alpha\in{\mathbb{Z}} such that

hi\displaystyle{h}^{\prime}_{i} =αhi,i=1,,k1,\displaystyle=\alpha h_{i},\ \forall i=1,\dots,k-1, (4)
(t1)hk\displaystyle(t-1)h^{\prime}_{k} =αhk=α.\displaystyle=\alpha h_{k}=\alpha. (5)

Since hi>0h_{i}>0 for all ii and there is an ii such that hi>0h^{\prime}_{i}>0, equations (4) and (5) imply α>0\alpha>0. Therefore (4) and (5) imply that hi>0h^{\prime}_{i}>0 for all ii. Hence h¯\bar{h} is primitive.

Now apply to {yi}i=1k+t1\{y^{i}\}_{i=1}^{k+t-1} a permutation of columns so that yk+t1y^{k+t-1} becomes (v1,urt+2,v2,urt+3,,vt,ur+1)(v_{1},u_{r-t+2},v_{2},u_{r-t+3},\allowbreak\dots,v_{t},u_{r+1}). For i=1,,k+t1i=1,\dots,k+t-1, let x¯i\bar{x}^{i} be the circuit of At,r+1A_{t,r+1} which is the image of yiy^{i} under this permutation. Then {x¯i}i=1k+t1\{\bar{x}^{i}\}_{i=1}^{k+t-1} also satisfy the primitive relation i=1k+t1h¯ix¯i=0\sum_{i=1}^{k+t-1}\bar{h}_{i}\bar{x}^{i}=0 with the same coefficients h¯\bar{h}. This completes the proof. ∎

We are now ready to prove Theorem 1.1. In the proof we use the following notation. Let 𝒜({xi}i=1k)={x¯i}i=1k+t{\mathscr{A}}(\{x^{i}\}_{i=1}^{k})=\{\bar{x}^{i}\}_{i=1}^{k+t} and (h)=h¯=(h¯1,,h¯k+t1,1){\mathscr{B}}(h)=\bar{h}=(\bar{h}_{1},\dots,\bar{h}_{k+t-1},1) denote circuits of At+1,t+1A_{t+1,t+1} and the primitive relation which are obtained by the operation of Lemma 3.1 to circuits {xi}i=1k\{x^{i}\}_{i=1}^{k} of At,t+1A_{t,t+1} and the primitive relation hh. Note that (h)1=h1+t\|{\mathscr{B}}(h)\|_{1}=\|h\|_{1}+t. Furthermore let 𝒜({xi}i=1k)={x¯i}i=1k+t1{\mathscr{A}}^{\prime}(\{x^{i}\}_{i=1}^{k})=\{\bar{x}^{i}\}_{i=1}^{k+t-1} and (h)=h¯=(h¯1,,h¯k+t2,1){\mathscr{B}}^{\prime}(h)=\bar{h}=(\bar{h}_{1},\dots,\bar{h}_{k+t-2},1) denote circuits of At,r+1A_{t,r+1} and the primitive relation which are obtained by the operation of Lemma 3.2 to circuits {xi}i=1k\{x^{i}\}_{i=1}^{k} of At,rA_{t,r} and the primitive relation hh. Note that (h)1=(t1)(h11)+t\|{\mathscr{B}}^{\prime}(h)\|_{1}=(t-1)(\|h\|_{1}-1)+t.

Our proof uses induction on t,rt,r. We will construct a primitive relation h(t×r)h^{(t\times r)} on circuits 𝒳(t×r){\mathscr{X}}_{(t\times r)} of At,rA_{t,r} by induction. Therefore we obtain g(At,r)h(t×r)1g(A_{t,r})\geq\|h^{(t\times r)}\|_{1}. Our induction is illustrated in Figure 1. There, a down arrow corresponds to the operation of Lemma 3.1, and a right arrow corresponds to the operation of Lemma 3.2.

h(3×4)1\|h^{(3\times 4)}\|_{1} \rightarrow h(3×5)1\|h^{(3\times 5)}\|_{1} \rightarrow h(3×6)1\|h^{(3\times 6)}\|_{1} \rightarrow \cdots
\downarrow
h(4×4)1\|h^{(4\times 4)}\|_{1} \rightarrow h(4×5)1\|h^{(4\times 5)}\|_{1} \rightarrow h(4×6)1\|h^{(4\times 6)}\|_{1} \rightarrow \cdots
\downarrow
h(5×5)1\|h^{(5\times 5)}\|_{1} \rightarrow h(5×6)1\|h^{(5\times 6)}\|_{1} \rightarrow \cdots
\downarrow
\vdots
Figure 1: Induction on t,rt,r
Proof of Theorem 1.1.

By induction on tt we will prove that for all t4t\geq 4 there exist k(t)=t22t+2k(t)=t^{2}-2t+2 circuits 𝒳(t×t)={x(t×t)i}i=1k(t)𝒞(At,t){\mathscr{X}}_{(t\times t)}=\{x_{(t\times t)}^{i}\}_{i=1}^{k(t)}\subset{\mathcal{C}}(A_{t,t}) and the primitive relation h(t×t)h^{(t\times t)} such that

x(t×t)k(t)\displaystyle x_{(t\times t)}^{k(t)} =(v1,u1,v2,u2,,vt,ut),\displaystyle=(v_{1},u_{1},v_{2},u_{2},\dots,v_{t},u_{t}),
i=1k(t)hi(t×t)xi\displaystyle\sum_{i=1}^{k(t)}h^{(t\times t)}_{i}x^{i} =0,\displaystyle=0,
hk(t)(t×t)\displaystyle h^{(t\times t)}_{k(t)} =1,\displaystyle=1,
h(t×t)1\displaystyle\|h^{(t\times t)}\|_{1} =(t2)!(15+i=1t4i+4(i+2)!).\displaystyle=(t-2)!\left(15+\sum_{i=1}^{t-4}\frac{i+4}{(i+2)!}\right).

Exchange x1x^{1} and x7x^{7} of circuits of Example 2.1 and apply to the circuits a permutation of vertices so that

x7=(v1,u2,v2,u3,v3,u4).x^{7}=(v_{1},u_{2},v_{2},u_{3},v_{3},u_{4}).

Let 𝒳(3×4)={x(3×4)i}i=17{\mathscr{X}}_{(3\times 4)}=\{x_{(3\times 4)}^{i}\}_{i=1}^{7} be the image of {xi}i=17\{x^{i}\}_{i=1}^{7} under this permutation. The primitive relation h(3×4)h^{(3\times 4)} on these circuits satisfy

h(3×4)=(7,2,3,3,5,6,1).h^{(3\times 4)}=(7,2,3,3,5,6,1).

Notice that h7(3×4)=1h^{(3\times 4)}_{7}=1 holds.

Let 𝒳(4×4)=𝒜(𝒳(3×4)){\mathscr{X}}_{(4\times 4)}={\mathscr{A}}({\mathscr{X}}_{(3\times 4)}) and h(4×4)=(h(3×4))h^{(4\times 4)}={\mathscr{B}}(h^{(3\times 4)}) denote the image of 𝒳(3×4){\mathscr{X}}_{(3\times 4)} and h(3×4)h^{(3\times 4)} under the operation of Lemma 3.1. Then we have h(4×4)=(7,2,3,3,5,6,1,1,1,1)10h^{(4\times 4)}=(7,2,3,3,5,6,1,1,1,1)\in{\mathbb{Z}}^{10} and

x(4×4)10\displaystyle x_{(4\times 4)}^{10} =(v1,u1,v2,u2,v3,u3,v4,u4),\displaystyle=(v_{1},u_{1},v_{2},u_{2},v_{3},u_{3},v_{4},u_{4}),
h10(4×4)\displaystyle h^{(4\times 4)}_{10} =1,\displaystyle=1,
h(4×4)1\displaystyle\|h^{(4\times 4)}\|_{1} =h(3×4)1+3=30.\displaystyle=\|h^{(3\times 4)}\|_{1}+3=30.

Therefore we have verified the initial condition at t=4t=4 for the induction.

Suppose now that the result holds for t4t\geq 4. Let 𝒳(t×(t+1))=𝒜(𝒳(t×t)){\mathscr{X}}_{(t\times(t+1))}={\mathscr{A}}^{\prime}({\mathscr{X}}_{(t\times t)}) and h(t×(t+1))=(h(t×t))h^{(t\times(t+1))}={\mathscr{B}}^{\prime}(h^{(t\times t)}) denote the image of 𝒳(t×t){\mathscr{X}}_{(t\times t)} and h(t×t)k(t)h^{(t\times t)}\in{\mathbb{Z}}^{k(t)} under the operation of Lemma 3.2.

x(t×(t+1))k(t)+t1\displaystyle x_{(t\times(t+1))}^{k(t)+t-1} =(v1,u2,v2,u3,,vt,ut+1),\displaystyle=(v_{1},u_{2},v_{2},u_{3},\dots,v_{t},u_{t+1}),
hk(t)+t1(t×(t+1))\displaystyle h^{(t\times(t+1))}_{k(t)+t-1} =1,\displaystyle=1,
h(t×(t+1))1\displaystyle\|h^{(t\times(t+1))}\|_{1} =(t1)(h(t×t)11)+t\displaystyle=(t-1)(\|h^{(t\times t)}\|_{1}-1)+t

follows from Lemma 3.2. Now let 𝒳((t+1)×(t+1))=𝒜(𝒳(t×(t+1))){\mathscr{X}}_{((t+1)\times(t+1))}={\mathscr{A}}({\mathscr{X}}_{(t\times(t+1))}) and h((t+1)×(t+1))=(h(t×(t+1)))h^{((t+1)\times(t+1))}={\mathscr{B}}(h^{(t\times(t+1))}) denote the image of 𝒳(t×(t+1)){\mathscr{X}}_{(t\times(t+1))} and h(t×(t+1))h^{(t\times(t+1))} under the operation of Lemma 3.1. Then

x((t+1)×(t+1))k(t)+2t1\displaystyle x_{((t+1)\times(t+1))}^{k(t)+2t-1} =(v1,u1,v2,u2,,vt+1,ut+1),\displaystyle=(v_{1},u_{1},v_{2},u_{2},\dots,v_{t+1},u_{t+1}),
hk(t)+2t1((t+1)×(t+1))\displaystyle h^{((t+1)\times(t+1))}_{k(t)+2t-1} =1,\displaystyle=1,
h((t+1)×(t+1))1\displaystyle\|h^{((t+1)\times(t+1))}\|_{1} =(t1)(h(t×t)11)+2t\displaystyle=(t-1)(\|h^{(t\times t)}\|_{1}-1)+2t
=(t1)!(15+i=1t4i+4(i+2)!)+t+1\displaystyle=(t-1)!\left(15+\sum_{i=1}^{t-4}\frac{i+4}{(i+2)!}\right)+t+1
=((t+1)2)!(15+i=1(t+1)4i+4(i+2)!)\displaystyle=((t+1)-2)!\left(15+\sum_{i=1}^{(t+1)-4}\frac{i+4}{(i+2)!}\right)

follows from Lemma 3.1. Here k(t+1)=k(t)+2t1k(t+1)=k(t)+2t-1 and k(4)=10k(4)=10 imply k(t)=t22t+2k(t)=t^{2}-2t+2. Therefore the result holds for t+1t+1. Henceforth, let bt=h(t×t)1b_{t}=\|h^{(t\times t)}\|_{1}.

We fix t4t\geq 4 arbitrarily. We prove by induction on rr that, for all rtr\geq t, there are circuits 𝒳(t×r)={x(t×r)i}i=1k(t)𝒞(At,r){\mathscr{X}}_{(t\times r)}=\{x_{(t\times r)}^{i}\}_{i=1}^{k(t)}\subset{\mathcal{C}}(A_{t,r}) and the primitive relation h(t×r)h^{(t\times r)} such that

x(t×r)k(t)\displaystyle x_{(t\times r)}^{k(t)} =(v1,urt+1,v2,urt+2,,vt,ur),\displaystyle=(v_{1},u_{r-t+1},v_{2},u_{r-t+2},\dots,v_{t},u_{r}),
i=1k(t)hi(t×r)x(t×r)i\displaystyle\sum_{i=1}^{k(t)}h^{(t\times r)}_{i}x_{(t\times r)}^{i} =0,\displaystyle=0,
hk(t)(t×r)\displaystyle h^{(t\times r)}_{k(t)} =1,\displaystyle=1,
h(t×r)1\displaystyle\|h^{(t\times r)}\|_{1} =(t1)rt(bt+1t2)1t2.\displaystyle=(t-1)^{r-t}\left(b_{t}+\frac{1}{t-2}\right)-\frac{1}{t-2}.

The initial condition of the induction, at r=tr=t, follows from h(t×t)1=bt\|h^{(t\times t)}\|_{1}=b_{t}.

Suppose now that the result holds for some rtr\geq t. Let 𝒳(t×(r+1))=𝒜(𝒳(t×r)){\mathscr{X}}_{(t\times(r+1))}={\mathscr{A}}^{\prime}({\mathscr{X}}_{(t\times r)}) and h(t×(r+1))=(h(t×r))h^{(t\times(r+1))}={\mathscr{B}}^{\prime}(h^{(t\times r)}) denote the image of 𝒳(t×r){\mathscr{X}}_{(t\times r)} and h(t×r)h^{(t\times r)} under the operation of Lemma 3.2. Then

x(t×(r+1))k(t)+t1\displaystyle x_{(t\times(r+1))}^{k(t)+t-1} =(v1,urt+2,v2,urt+3,,vt,ur+1),\displaystyle=(v_{1},u_{r-t+2},v_{2},u_{r-t+3},\dots,v_{t},u_{r+1}),
hk(t)+t1(t×(r+1))\displaystyle h^{(t\times(r+1))}_{k(t)+t-1} =1,\displaystyle=1,
h(t×(r+1))1\displaystyle\|h^{(t\times(r+1))}\|_{1} =(t1)(h(t×r)11)+t\displaystyle=(t-1)(\|h^{(t\times r)}\|_{1}-1)+t
=(t1)((t1)rt(bt+1t2)1t21)+t\displaystyle=(t-1)\left((t-1)^{r-t}\left(b_{t}+\frac{1}{t-2}\right)-\frac{1}{t-2}-1\right)+t
=(t1)r+1t(bt+1t2)1t2\displaystyle=(t-1)^{r+1-t}\left(b_{t}+\frac{1}{t-2}\right)-\frac{1}{t-2}

follows from Lemma 3.2. Therefore the result holds for r+1r+1 and

g(At,r)(t1)rt(bt+1t2)1t2g(A_{t,r})\geq(t-1)^{r-t}(b_{t}+\frac{1}{t-2})-\frac{1}{t-2}

follows from Lemma 2.2. ∎

4 Discussion

In this paper we provided a lower bound in Theorem 1.1 by the induction on t,rt,r. Here we discuss some ideas for improving our lower bound.

Look at Figure 1 again. On the step h(3×4)1h(4×4)1\|h^{(3\times 4)}\|_{1}\rightarrow\|h^{(4\times 4)}\|_{1}, we can construct a larger primitive relation than the relation constructed in the proof.

Example 4.1.

Let {xi}i=17\{x^{i}\}_{i=1}^{7} denote the circuits in Example 2.1. Using the natural embedding of K3,4K_{3,4} into K4,4K_{4,4}, let

x¯i=xi,i=1,,6.\bar{x}^{i}=x^{i},\ \forall i=1,\dots,6.

and for i=7,,10i=7,\dots,10, we define x¯i\bar{x}^{i} as shown in Table 4.

Table 4: Circuits of A4,4A_{4,4}
u1u_{1} u2u_{2} u3u_{3} u4u_{4}
0 11 0 1-1 v1v_{1}
x¯7=(v1,u2,v4,u1,v3,u4)=\bar{x}^{7}=(v_{1},u_{2},v_{4},u_{1},v_{3},u_{4})= 0 0 0 0 v2v_{2}
1-1 0 0 11 v3v_{3}
11 1-1 0 0 v4v_{4}
0 0 0 0 v1v_{1}
x¯8=(v2,u3,v4,u2)=\bar{x}^{8}=(v_{2},u_{3},v_{4},u_{2})= 0 1-1 11 0 v2v_{2}
0 0 0 0 v3v_{3}
0 11 1-1 0 v4v_{4}
0 0 0 0 v1v_{1}
x¯9=(v3,u4,v4,u3)=\bar{x}^{9}=(v_{3},u_{4},v_{4},u_{3})= 0 0 0 0 v2v_{2}
0 0 1-1 11 v3v_{3}
0 0 11 1-1 v4v_{4}
0 11 0 1-1 v1v_{1}
x¯10=(v1,u2,v2,u3,v3,u1,v4,u4)=\bar{x}^{10}=(v_{1},u_{2},v_{2},u_{3},v_{3},u_{1},v_{4},u_{4})= 0 1-1 11 0 v2v_{2}
11 0 1-1 0 v3v_{3}
1-1 0 0 11 v4v_{4}

Then {x¯i}i=110\{\bar{x}^{i}\}_{i=1}^{10} are circuits of ker(𝟏4(4)){\rm ker}_{\mathbb{Z}}({\bf 1}_{4}^{(4)}) and h¯=(2,4,6,6,10,12,7,7,7,7)\bar{h}=(2,4,6,6,10,12,7,7,7,7) is its primitive relation. Then, by Proposition 2.2,

g(𝟏4(4))2+4+6+6+10+12+7+7+7+7=68.g({\bf 1}_{4}^{(4)})\geq 2+4+6+6+10+12+7+7+7+7=68. (6)

Equation (6) is sharper than the evaluation in Theorem 1.1.

We could start induction from circuits and its primitive relation in Example 4.1. Then we obtain a sharper evaluation for some small t,rt,r. However, unfortunately it turns out that, if we start from Example 4.1 then on the step h(8×r)1h(8×(r+1))1\|h^{(8\times r)}\|_{1}\rightarrow\|h^{(8\times(r+1))}\|_{1}, we can not obtain an exponential lower bound. Therefore we did not use Example 4.1 in the proof of Theorem 1.1. However this example suggests that there may be some other better initial set of circuits for our induction.

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