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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(2): 401-413

Published online June 30, 2020

### Periodic Solutions of a System of Piecewise Linear Difference Equations

Wirot Tikjha*, Evelina Lapierre

Faculty of Science and Technology, Pibulsongkram Rajabhat University, Phitsanu- lok, Thailand 65000, Centre of Excellence in Mathematics, PERDO, CHE, Thailand
e-mail : wirottik@psru.ac.th
Department of Mathematics, Johnson and Wales University, 8 Abbott Park Place, Providence, RI, USA
e-mail : Evelina.Lapierre@jwu.edu

Received: June 28, 2017; Revised: September 2, 2019; Accepted: August 27, 2019

### Abstract

In this article we consider the following system of piecewise linear difference equations: xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| − 1. We show that when the initial condition is an element of the closed second or fourth quadrant the solution to the system is either a prime period-3 solution or one of two prime period-4 solutions.

Keywords: difference equation, periodic solution, stability

### 1. Introduction

Nearly ten years ago we began studying the global behavior of the following system of piecewise linear difference equations

$(N) {xn+1=∣xn∣+ayn+b ,n=0,1,…yn+1=xn+c∣yn∣+d$

where the initial condition (x0, y0) ∈ R2 and the parameters a, b, c, and d ∈ {−1, 0, 1}. Since each parameter can be one of three values, there are 81 systems.

Each system is designated a number which is given by

$N=27(a+1)+9(b+1)+3(c+1)+(d+1)+1.$

Our purpose is to find patterns of behavior in order to better understand piecewise linear difference equations in general. We hope to develop methods to determine local asymptotic stability and global stability of such systems. The lack of these methods is evident by the fact that the global behavior of the Lozi equation and the Gingerbreadman Map (both can be expressed as a system of piecewise linear difference equations) are still not completely known. See [1, 2, 3, 5, 6].

After determining the global behavior of many of the 81 systems, we noticed a few trends. See [3, 4, 7, 8, 9]. Over half of the systems have exactly one equilibrium point, while some have two or three, and the remaining systems either have none or have infinitely many (which usually reside on a line). About a quarter of the systems have periodic solutions that are similar to the solutions of System(7). We were able to generalize a few systems; that is, we know their global behavior when some of the parameters are elements of R+, not just elements of {−1, 0, 1}. Within the next year we hope to complete a monograph that will share our results, detailed proofs, conjectures and open problems of all 81 systems.

Some of these systems are rather enigmatic. System(7) is one of them. It is the special case of System( ) where a = b = d = −1 and c = 1:

${xn+1=∣xn∣-yn-1 ,n=0,1,…yn+1=xn+∣yn∣-1 .$

After months of brute force calculations, we only had a partial result. Wirot Tikjha shared this partial result in the 2016 International Conference on Difference Equations and Applications. See [7]. At the time of that presentation we knew the behavior of the system for only a small set of initial conditions (a section of the x-axis). Since the presentation, with the aid of computer simulations using random initial values, we were able to extend the set of initial conditions to include the closed second and fourth quadrant.

In this paper we show that when the initial condition (x0, y0) is an element in the closed second or fourth quadrant System(7) has one prime period-3 solution and two prime period-4 solutions given by

$P31=(-13,-113,-13-13,-13),P41=(-1,-11,-11,1-1,1),and P42=(1,-33,3-1,5-5,3);$

where $(a1,a2b1,b2c1,c2)$ represents the consecutive solutions (a1, a2), (b1, b2), and (c1, c2) of the system. Please note that at the end of this article we share our conjecture for the global behavior of this system.

A solution ${(xn,yn)}n=0∞$ of a system of difference equations is called eventually periodic with prime period-p or eventually prime period-p solution if there exists an integer N > 0 and p is the smallest positive integer such that ${(xn,yn)}n=N∞$ periodic with period-p; that is,

$(xn+p,yn+p)=(xn,yn) for all n≥N.$

### 2. Main Results

Set ℒ1 = {(1, y)|y ≥ 0}, ℒ2 = {(1, y)|y ≤ 0}, ℒ3 = {(−1, y)|y ≥ 0}, ℒ4 = {(−1, y)|y ≤ 0}, ℒ5 = {(x,−1)|xR}, , .

### Theorem 2.1

Let ${(xn,yn)}n=0∞$be a solution of System(7) with . Then ${(xn,yn)}n=0∞$is eventually the prime period-3 solution $P31$or the prime period-4 solution $P41$or $P42$.

The proof of the theorem is a consequence of the following lemmas.

### Lemma 2.2

Suppose the initial condition (x0, y0) ∈ ℒ1 ∪ ℒ4. Then solution of System(7), ${(xn,yn)}n=2∞$is the prime period-4 solution $P41$.

Proof

Let (x0, y0) ∈ ℒ1. Then $(x2,y2)=(-1,-1)∈P41$. Let (x0, y0) ∈ ℒ4. Then

$x1=∣x0∣-y0-1=-y0≥0y1=x0+∣y0∣-1=-y0-2.$

If y1 = −y0 − 2 < 0, then

$x2=∣x1∣-y1-1=1y2=x1+∣y1∣-1=1.$

If y1 = −y0 − 2 ≥ 0, then

$x2=∣x1∣-y1-1=1y2=x1+∣y1∣-1=-2y0-3>0.$

Note that (x2, y2) ∈ ℒ1 and therefore $(x4,y4)∈P41$.

### Claim 2.3

Assume that there is a positive integer N such that yN = −xN −2 ≥ 0. Then ${(xn,yn)}n=N+1∞$is the prime period-4 solution $P42$.

Proof

Suppose (xN, yN) satisfies the hypothesis, then

$xN+1=∣xN∣-yN-1=-xN+xN+2-1=1yN+1=∣xN∣+∣yN∣-1=xN-xN-2-1=-3.$

Then the proof is complete.

### Lemma 2.4

Suppose the initial condition (x0, y0) ∈ ℒ2. Then ${(xn,yn)}n=0∞$is eventually the prime period-4 solution $P41$or $P42$.

Proof

Let (x0, y0) ∈ ℒ2. Then

$x1=∣x0∣-y0-1=-y0>0y1=x0+∣y0∣-1=-y0>0x2=∣x1∣-y1-1=-1y2=x1+∣y1∣-1=-2y0-1.$

Suppose y2 = −2y0 − 1 < 0, then (x2, y2) ∈ ℒ4. We now apply Lemma 2.2. and find that ${(xn,yn)}n=4∞$ is the prime period-4 solution $P41$.

Suppose y2 = −2y0 − 1 ≥ 0, that is $y0≤-12$, then

$x3=∣x2∣-y2-1=2y0+1≤0y3=x2+∣y2∣-1=-2y0-3.$

Suppose y3 = −2y0 − 3 ≥ 0, that is $y0≤-32$, then

$x4=∣x3∣-y3-1=1y4=x3+∣y3∣-1=-3,$

and so $(x4,y4)∈P42$, as required.

Suppose y3 = −2y0 −3 < 0, that is $-32, then we will progress using mathematical induction. For each integer n ≥ 0, let

$an=-22n+1-122n+1,bn=-22n+1+122n+1,cn=-22n+122n,and δn=22n-1.$

Observe that

$-32=a0b1>b2>…>-1 and limn→∞bn=-1,0=c0>c1>c2>…>-1 and limn→∞cn=-1.$

Furthermore for each integer n ≥ 1, let P(n) be the following set of statements. When y0 ∈ (an−1, bn−1], we have

$x4n=1y4n=22ny0+δn.$

When y0 ∈ [cn, bn−1], we have y4n ≥ 0, and so the solution is eventually the prime period-4 solution $P41$.

When y0 ∈ (an−1, cn), we have y4n < 0. Then

$x4n+1=-22ny0-δn>0y4n+1=-22ny0-δn>0x4n+2=-1y4n+2=-22n+1y0-(2δn+1).$

When y0 ∈ [bn, cn), we have y4n+2 ≤ 0, and so the solution is eventually the prime period-4 solution $P41$.

When y0 ∈ (an−1, bn), we have y4n+2> 0. Then

$x4n+3=22n+1y0+(2δn+1)<0y4n+3=-22n+1y0-(2δn+3).$

When y0 ∈ (an−1, an], we have y4n+3 ≥ 0, and so the solution is eventually the prime period-4 solution $P42$.

Finally, when y0 ∈ (an, bn], we have y4n+3< 0.

We shall now show that P(1) is true. For $y0∈(a1-1,b1-1]=(a0,b0]=(-32,-12]$, recall that x3 = 2y0 + 1 ≤ 0 and y3 = −2y0 − 3 < 0, then

$x4(1)=x4=∣x3∣-y3-1=1y4(1)=y4=x3+∣y3∣-1=4y0+3=22(1)y0+δ1.$

When $y0∈[c1,b1-1]=[c1,b0]=[-34,-12]$, then y4 = 4y0 + 3 ≥ 0. We apply Lemma 2.2. and find that ${(xn,yn)}n=6∞$ is the prime period-4 solution $P41$.

When $y0∈(a1-1,c1)=(a0,c1)=(-32,-34)$, then y4 = 4y0 + 3 < 0. Thus

$x4(1)+1=x5=-4y0-3=-22(1)y0-δ1>0y4(1)+1=y5=-4y0-3=-22(1)y0-δ1>0x4(1)+2=x6=-1y4(1)+2=y6=-8y0-7=-22(1)+1y0-(2δ1+1).$

When $y0∈[b1,c1)=[-78,-34)$, then y6 = −8y0 − 7 ≤ 0. We apply Lemma 2.2. and find that ${(xn,yn)}n=8∞$ is the prime period-4 solution $P41$.

When $y0∈(a1-1,b1)=(a0,b1)=(-32,-78)$, then y6 = −8y0 − 7 > 0. Thus

$x4(1)+3=x7=8y0+7=22(1)+1y0+2δ1+1<0y4(1)+3=y7=-8y0-9=-22(1)+1y0-(2δ1+3).$

When $y0∈(a1-1,a1]=(a0,a1]=(-32,-98]$, then y7 = −x7 −2 = −8y0 − 9 ≥ 0. We apply Claim 2.3. and find that ${(xn,yn)}n=8∞$ is the prime period-4 solution $P42$. When $y0∈(a1,b1]=(-98,-78]$, then y7 = −8y0 − 9 < 0. Hence P(1) is true.

Next, we assume that P(N) is true. We shall show that P(N +1) is true. Since P(N) is true, we know that when $y0∈(aN,bN]=(-22N+1-122N+1,-22N+1+122N+1]$, we have

$x4N+3=22N+1y0+(2δN+1)<0y4N+3=-22N+1y0-(2δN+3)<0.$

Then,

$x4(N+1)=x4N+4=1y4(N+1)=y4N+4=22(N+1)y0+4δN+3=22(N+1)y0+δN+1.$

Note that

$δN+1=22(N+1)-1=22N+2-1=22N+2-4+3=4δN+3.$

If $y0∈[cN+1,b(N+1)-1]=[cN+1,bN]=[-22N+2+122N+2,-22N+1+122N+1]$, then

$y4N+4=22(N+1)y0+δN+1=22N+2y0+22N+2-1≥0.$

Applying Lemma 2.2., we see that ${(xn,yn)}n=4N+6∞$ is the prime period-4 solution $P41$.

If $y0∈(a(N+1)-1,cN+1)=(aN,cN+1)=(-22N+1-122N+1,-22N+2+122N+2)$, then

$y4N+4=22(N+1)y0+δN+1=22N+2y0+22N+2-1<0.$

Thus,

$x4(N+1)+1=x4N+5=-22(N+1)y0-δN+1>0y4(N+1)+1=y4N+5=-22(N+1)y0-δN+1>0x4(N+1)+2=x4N+6=-1y4(N+1)+2=y4N+6=-22(N+1)+1y0-(2δN+1+1).$

If $y0∈[bN+1,cN+1)=[-22N+3+122N+3,-22N+2+122N+2)$, then

$y4N+6=-22(N+1)+1y0-2δN+1-1=-22N+3y0-22N+3+1≤0.$

Applying Lemma 2.2., we see that ${(xn,yn)}n=4N+8∞$ is the prime period-4 solution $P41$.

If $y0∈(a(N+1)-1,bN+1)=(aN,bN+1)=(-22N+1-122N+1,-22N+3+122N+3)$, then

$y4N+6=-22(N+1)+1y0-2δN+1-1=-22N+3y0-22N+3+1>0,$

thus

$x4(N+1)+3=x4N+7=22(N+1)+1y0+(2δN+1+1)<0y4(N+1)+3=y4N+7=-22(N+1)+1y0-(2δN+1+3).$

If $y0∈(a(N+1)-1,aN+1]=(aN,aN+1]=(-22N+1-122N+1,-22N+3-122N+3]$, then

$y4N+7=-22(N+1)+1y0-2δN+1+3=-22N+3y0-22N+3-1≥0.$

We note that y4N+7 = −x4N+7 − 2 ≥ 0. Applying Claim 2.3., we see that ${(xn,yn)}n=4N+8∞$ is the prime period-4 solution $P42$.

If $y0∈(a(N+1),bN+1]=(aN+1,bN+1]=(-22N+3-122N+3,-22N+3+122N+3]$, then

$y4N+7=-22(N+1)+1y0-(2δN+1+3)=-22N+3-22N+3-1<0.$

Hence, P(N + 1) is true. Therefore P(n) is true for all n ≥ 1.

Note that $limn→∞an=limn→∞bn=limn→∞cn=-1$ and $(1,-1)∈P41$.

### Lemma 2.5

Suppose the initial condition (x0, y0) ∈ ℒ3. Then ${(xn,yn)}n=0∞$is eventually the prime period-4 solution $P41$or $P42$.

Proof

Suppose (x0, y0) ∈ ℒ3. Then by direct computations we see that x2 = 1. We now apply Lemmas 2.2. and 2.4., and see that ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

### Claim 2.6

Assume that there is a positive integer N such that xN = yN ≥ 0. Then, ${(xn,yn)}n=N∞$is eventually the prime period-4 solution $P41$or $P42$.

Proof

Suppose that (xN, yN) satisfies the hypothesis, then xN+1 = −1. We apply Lemmas 2.2. and 2.5., and see that ${(xn,yn)}n=N∞$ is eventually the prime period-4 solution $P41$ or $P42$.

### Lemma 2.7

Suppose the initial condition (x0, y0) ∈ ℒ5. Then ${(xn,yn)}n=0∞$is eventually the prime period-3 solution $P31$or the prime period-4 solution $P41$or $P42$.

Proof

Suppose that (x0, y0) ∈ ℒ5 and suppose further that x0 ≥ 0. Then x1 = y1 = x0. We apply Claim 2.6., and find that ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

Now suppose that (x0, y0) ∈ ℒ5 but x0 ≤ 0. Then

$x1=∣x0∣-y0-1=-x0+1-1=-x0>0y1=x0+∣y0∣-1=x0+1-1=x0<0x2=∣x1∣-y1-1=-x0-x0-1=-2x0-1y2=x1+∣y1∣-1=-x0-x0-1=-2x0-1.$

If x2 = y2 = −2x0 − 1 ≥ 0, that is $x0≤-12$, then we can apply Claim 2.6. and see that ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

If x2 = y2 = −2x0 − 1 < 0, that is $-12. Then

$x3=∣x2∣-y2-1=2x0+1+2x0+1-1=4x0+1y3=x2+∣y2∣-1=-2x0-1+2x0+1-1=-1.$

If x3 = 4x0 + 1 ≥ 0, then since (x3, y3) ∈ ℒ5 by the above case, we see that ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

If x3 = 4x0 + 1 < 0, that is $-12, then we will progress by using mathematical induction.

For each integer n ≥ 0, let

$an=-22n+1-13×22n+1, bn=-22n+2+13×22n+2 and δn=22n-13.$

Observe that

$-12=a0b1>b2>…>-13 and limn→∞ bn=-13.$

Furthermore for each integer n ≥ 1, let P(n) be the following set of statements. When x0 ∈ (an−1, bn−1), we have

$x3n+1=-22nx0-δn>0y3n+1=22nx0+δn<0x3n+2=-22n+1x0-(2δn+1)y3n+2=-22n+1x0-(2δn+1).$

When x0 ∈ (an−1, an], we have x3n+2 = y3n+2 ≥ 0, and so by Claim 2.6. ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$. When x0 ∈ (an, bn−1), we have x3n+2 = y3n+2< 0, and so

$x3n+3=22n+2x0+4δn+1y3n+3=-1.$

When x0 ∈ [bn, bn−1), we have x3n+3 ≥ 0, and so by Claim 2.6. ${(xn,yn)}n=0∞$ eventually the prime period-4 solution $P41$ or $P42$. Finally, when x0 ∈ (an, bn), we have x3n+3< 0.

We shall show that P(1) is true.

For $x0∈(an-1,bn-1)=(a0,b0)=(-12,-14)$, recall that x3 = 4x0 + 1 < 0 and y3 = −1. Then,

$x3(1)+1=x4=-4x0-1=-22(1)x0-δ1>0y3(1)+1=y4=4x0+1=22(1)x0+δ1<0x3(1)+2=x5=-8x0-3=-22(1)+1x0-(2δ1+1)y3(1)+2=y5=-8x0-3=-22(1)+1x0-(2δ1+1).$

If $x0∈(a1-1,a1]=(a0,a1]=(-12,-38]$, then x5 = y5 = −8x0 − 3 ≥ 0, and so we apply Claim 2.6. and see that ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

If $x0∈(a1,b1-1)=(a1,b0)=(-38,-14)$, then x5 = y5 = −8x0 − 3 < 0, and so

$x3(1)+3=x6=16x0+5=22(1)+2x0+4δ1+1y3(1)+3=y6=-1.$

If $x0∈[b1,b1-1)=[b1,b0)=[-516,-14)$, then x6 = 16x0 + 5 ≥ 0, and (x6, y6) ∈ ℒ5. Applying earlier work in this proof we see that ${(xn,yn)}n=0∞$ eventually the prime period-4 solution $P41$ or $P42$.

If $x0∈(a1,b1)=(-38,-516)$, then x6 = 16x0 + 5 < 0. Hence P(1) is true.

Suppose that P(N) is true. We shall show that P(N + 1) is true.

Since P(N) is true, we know that $x0∈(a(N+1)-1,b(N+1)-1)=(aN,bN)=(-22N+1-13×22N+1,-22N+2+13×22N+2)$, and

$x3N+3=22N+2x0+4δN+1<0y3N+3=-1.$

Note that

$δN+1=22(N+1)-13=22N+2-43+33=4(22N-13)+1=4δN+1.$

Then

$x3(N+1)+1=x3N+4=-22N+2-4δN-1=-22(N+1)x0-δN+1>0y3(N+1)+1=y3N+4=22N+2+4δN+1=22(N+1)x0+δN+1<0x3(N+1)+2=x3N+5=-22(N+1)+1x0-(2δN+1+1)y3(N+1)+2=y3N+5=-22(N+1)+1x0-(2δN+1+1).$

If $x0∈(a(N+1)-1,aN+1]=(aN,aN+1]=(-22N+1-13×22N+1,-22N+3-13×22N+3]$, then$x3N+5=y3N+5=-22(N+1)+1x0-(2δN+1+1)=-22N+3x0+(-22N+3-13)≥0$

and so we apply Claim 2.6. and see that ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

If $x0∈(aN+1,b(N+1)-1)=(aN+1,bN)=(-22N+3-13×22N+3,-22N+2+13×22N+2)$, then$x3N+5=y3N+5=-22(N+1)+1x0-(2δN+1+1)=-22N+3x0+(-22N+3-13)<0,$

and so

$x3(N+1)+3=x3N+6=22(N+1)+2x0+4δN+1+1y3(N+1)+3=y3N+6=-1.$

If $x0∈[bN+1,b(N+1)-1)=[bN+1,bN)=[-22N+4+13×22N+4,-22N+2+13×22N+2)$, then

$x3N+6=22(N+1)+2x0+4δN+1+1=22N+4x0+22N+4-13≥0,$

and (x3N+6, y3N+6) ∈ ℒ5 so by previous work in this proof ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

If $x0∈(aN+1,bN+1)=(-22N+3-13×22N+3,-22N+4+13×22N+4)$, then

$x3N+6=22(N+1)+2x0+4δN+1+1=22N+4x0+22N+4-13<0.$

Hence, P(N + 1) is true. Therefore P(n) is true for all n ≥ 1. Please note that

$limn→∞an=limn→∞bn=-13$

and $(x0,y0)=(-13,-1)∈P31$.

### Lemma 2.8

Suppose the initial condition . Then ${(xn,yn)}n=0∞$is eventually the prime period-4 solution $P41$or $P42$.

Proof

Let (x0, y0) ∈ Q2. Then

$x1=∣x0∣-y0-1=-x0-y0-1y1=x0+∣y0∣-1=x0+y0-1.$
• Case 1: Suppose −x0 = y0, then $(x1,y1)=(-1,-1)∈P41$.

• Case 2: Suppose −x0> y0, then y1 = x0 + y0 − 1 < 0.

Suppose further that x1 = −x0y0 − 1 < 0, then

$x2=∣x1∣-y1-1=1y2=x1+∣y1∣-1=-2x0-2y0-1.$

We see that (x2, y2) ∈ ℒ1 ∪ ℒ2. Applying Lemmas 2.2. and 2.4., we see that ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

Now suppose that x1 = −x0y0 − 1 ≥ 0, then

$x2=∣x1∣-y1-1=-2x0-2y0-1>0y2=x1+∣y1∣-1=-2x0-2y0-1>0x3=∣x2∣-y2-1=-1y3=x2+∣y2∣-1=-4x0-4y0-3>0.$

We apply Lemma 2.5., and see that ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

• Case 3: Suppose −x0< y0, then

$x1=-x0-y0-1<0x2=∣x1∣-y1-1=1.$

We apply Lemmas 2.2. and 2.4., and the proof is complete.

### Lemma 2.9

Suppose the initial condition . Then ${(xn,yn)}n=0∞$is eventually the prime period-3 solution $P31$or the prime period-4 solution $P41$or $P42$.

Proof

Let (x0, y0) ∈ Q4. Then,

$x1=∣x0∣-y0-1=x0-y0-1y1=x0+∣y0∣-1=x0-y0-1.$

Suppose x0y0 − 1 ≥ 0, then we apply Claim 2.6. and see that ${(xn,yn)}n=0∞$ is eventually the prime period-4 solution $P41$ or $P42$.

Suppose that x0y0 − 1 < 0, then

$y2=x1+∣y1∣-1=-1.$

We see that (x2, y2) ∈ ℒ5. By Lemma 2.7. ${(xn,yn)}n=N∞$ is eventually the prime period-3 solution $P31$ or the prime period-4 solution $P41$ or $P42$.

### 3. Discussion and Conclusion

Returning our attention to the original family of System( ), number 7 of this group is one of the most interesting systems. Initially, when we only understood its behavior for a small set of initial conditions (a segments on x-axis), we were only able to prove that every solution was eventually prime period-4. See Ref. [7]. Now that we are able to include the closed second and fourth quadrant in the set of initial conditions we see that this is one of the few systems that exhibit solutions of varying periodicity. Although we have not yet proved the global behavior of System(7) we have a conjecture.

### Conjecture 3.1

Let ${(xn,yn)}n=0∞$be a solution of System(7) with (x0, y0) ∈ R2. Then ${(xn,yn)}n=0∞$is the unique equilibrium $(-15,-35)$, or eventually the prime period-3 solution $P31$or $P32$, or the prime period-4 solution $P41$or $P42$ where

$P31=(-13,-113,-13-13,-13),P32=(35,15-35,-15-15,-75),P41=(-1,-11,-11,1-1,1), and P42=(1,-33,3-1,5-5,3).$

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