Construction I

Input

{1, . . . , k} with propertyF

Output

The output of this construction is a sequencey1y2. . . ymwithm=β(bn/2c) (β

is defined below) over{1, . . . , k2+ 2}with propertyFφ. We will now define this functionφ(the idea behind this definition is given in the informal description below). In this definition we need the following function. Let β : _N → N be

defined recursively by

β(0) = 0

β(b+ 1) = β(b) + (b+ 3)·kb+2−b.

We now defineφas follows.

φ(i) =

β−1_{(i) + 1 ifβ}−1_(i+β−1_{(i) + 1) =β}−1_(i)

β−1_{(i) + 3 otherwise}

Example

If the input is 112222 (k= 2) then the output will be 1 1 0 0 S 1 1 0 1 S 1 1 1 0 S 1 1 1 1 T 2 2 0 0 S 1 2 2 0 0 1 S 1 2 2 0 1 0 S 1 2 2 0 1 1 S 1 2 2 1 0 0 S 1 2 2 1 0 1 S 1 2 2 1 1 0 S 1 2 2 1 1 1 T 2 2 0 0 S 2 2 2 2 0 0 0 1 S 2 2 2 2 0 0 1 0 S 2 2 2 2 0 0 1 1 S 2 2 2 2 0 1 0 0 S 2 2 2 2 0 1 0 1 S 2 2 2 2 0 1 1 0 S 2 2 2 2 0 1 1 1 S 2 2 2 2 1 0 0 0 S 2 2 2 2 1 0 0 1 S 2 2 2 2 1 0 1 0 S 2 2 2 2 1 0 1 1 S 2 2 2 2 1 1 0 0 S 2 2 2 2 1 1 0 1 S 2 2 2 2 1 1 1 0 S 2 2 2 2 1 1 1 1 T where a

c stand forN((a, c+ 1)),S stands fork

2_{+ 1 andT stands fork}2_{+ 2.}

Informal description

The output in the example consists of three blocks because the input 112222 has three windows, namely 11, 122 and 2222. The end of each block is marked by aT (k2+ 2). The ith block consists of repetitions of windowiof the input (xi. . . x2i) on the first line and these repetitions are counted in basek on the

second line. So theith block containski+1 repetitions since the length of win- dowi of the input isi+ 1. These repetitions are separated by an S (k2+ 1). The first repetition in each block is different. These repetitions only consist of the last two elements of the corresponding window of the input. This is because these two elements are new in the sense that the other elements in this window are also contained in the previous window. The functionβ maps a numberito the position ofithT. The functionφis defined in such a way that if a window of the output is contained in blockithen this window contains exactly oneS and a cyclic permutation of the corresponding window from the input. If a window

of the output starts in blockiand ends in blocki+ 1 then this window contains exactly oneS and oneT and a cyclic permutation of the windowxi+1. . . x2i+2

from the input.

Formal description

We will still useS=k2_{+ 1 andT =k}2_{+ 2 here. Given the inputx}

1. . . xn over {1, . . . k}with propertyFwe will define the outputy1. . . ymover{1, . . . , k2+2}

and prove that this sequence has propertyFφ. Letm=β(bn/2c). To improve

readability we will denoteβ−1_{(i) bybin the next definition.}

yi=            k2+ 2 ifβ(b) =i k2_{+ 1 ifβ(b)6=iand} i−β(b−1) +b−1 is a multiple ofb+ 2 N((xb+j−1, c+ 1))·k if∃q(i−β(b−1) +b−1 =q·(b+ 2) +j and

1≤j≤b+ 1 and 0≤c < kand bq/(kb+1−j_{)c ≡c (modk))}

One can check that for everyiexactly one of the three conditions is the case and in case of the third condition the values ofj,qandc are uniquely determined.

Lemma 5.2.1. Let 0 ≤ i0 ≤ i, 0 ≤ j0 ≤ j and let u1. . . ui and v1. . . vj be

sequences over{1, . . . , l}and letw > l. Ifu_i0

+1. . . uiwu1. . . ui0 is a subsequence

ofv_j0

+1. . . vjwv1. . . vj0 thenu1. . . ui is a subsequence ofv1. . . vj.

Proof. Sincew6∈ {1, . . . , l}it follows that it must be the case thatu_i0₊₁. . . uiis

a subsequence ofv_j0₊₁. . . vj andu₁. . . u_i0 is a subsequence ofv₁. . . v_j0. Hence u1. . . ui is a subsequence ofv1. . . vj.

Lemma 5.2.2. If u1. . . ui is a subsequence of v1. . . vj then for any function

hthe sequence h(ui). . . h(u_i0)is a subsequence of h(v₁). . . h(vj) with the same

embedding. Proof. Clear.

Lemma 5.2.3. If i < j, j+φ(j)≤ m, φ(i) = β−1_{(i) + 1, φ(j) = β}−1_{(j) +}

1, β−1(i) < β−1(j) and yi. . . yi+φ(i) is a subsequence of yj. . . yj+φ(j) then

xβ−1_(i). . . x_2β−1_(i) is a subsequence ofx_β−1_(j). . . x_2β−1_(j).

Proof. From the definition ofyiit follows that bothyi. . . yi+φ(i)andyj. . . yj+φ(j)

contain exactly one element that is bigger thank2 _{and since the one is a sub-}

sequence of the other these elements have to be equal. Lethequal the identity on numbers> k2_andp

1 on numbers≤k2. There exist i

0

,j0 such that

h(yi). . . h(yi+φ(i)) =xi0+1. . . x2β−1_(i)wx_β−1_(i). . . x_i0

and

h(yj). . . h(yj+φ(j)) =xj0+1. . . x2β−1_(j)wx_β−1_(j). . . x_j0

wherew > k2_{. By lemmas 5.2.2 and 5.2.1 we conclude that x}

β−1_(i). . . x_2β−1_(i)

Lemma 5.2.4. If i < j, j+φ(j)≤m, φ(i) =β−1(i) + 1,φ(j) =β−1(j) + 3

and yi. . . yi+φ(i) is a subsequence of yj. . . yj+φ(j) then xβ−1_(i). . . x_2β−1_(i) is a subsequence ofxβ−1_(j)+1. . . x_2β−1_(j)+2.

Proof. Lethbe defined as in the proof of the previous lemma. There existi0, j0

such thatβ−1(i)−1≤i0 ≤2β−1(i),β−1(j)≤j0 ≤2β−1(j) and

h(yi). . . h(yi+φ(i)) =xi0+1. . . x2β−1_(i)wx_β−1_(i). . . x_i0

wherew=S orw=T and

h(yj). . . h(yj+φ(j)) =xj0+1. . . x2β−1_(j)T x_2β−1_(j)+1x_2β−1_(j)+2Sx_β−1_(j)+1. . . x_j0.

By lemma 5.2.2 the first sequence is a subsequence of the second. In case

w =S this will still be the case if we delete the T from the second sequence and then by lemma 5.2.1 we conclude thatxβ−1_(i). . . x_2β−1_(i) is a subsequence

of xβ−1_(j)+1. . . x_2β−1_(j)+2. In case w = T the first sequence will still be a

subsequence of the second if we delete the S from the second sequence and then by lemma 5.2.1 we conclude that xβ−1_(i). . . x_2β−1_(i) is a subsequence of

xβ−1_(j)+1. . . x_2β−1_(j)+2.

Lemma 5.2.5. If i < j,j+φ(j)≤m,φ(i) =β−1(i) + 3, φ(j) =β−1(j) + 3,

β−1_{(i) + 1< β}−1_{(j) + 1 andy}

i. . . yi+φ(i) is a subsequence ofyj. . . yj+φ(j)then

xβ−1_(i)+1. . . x_2β−1_(i)+2 is a subsequence ofx_β−1_(j)+1. . . x_2β−1_(j)+2.

Proof. Lethbe as defined in the proof of the previous lemma. There existi0, j0

such thatβ−1_(i)≤i0 _≤2β−1_(i),β−1_(j)≤j0 _≤2β−1_{(j) and}

h(yi). . . h(yi+φ(i)) =xi0+1. . . x2β−1_(i)T x_2β−1_(i)+1x_2β−1_(i)+2Sx_β−1_(i)+1. . . x_i0.

and

h(yj). . . h(yj+φ(j)) =xj0+1. . . x2β−1_(j)T x_2β−1_(j)+1x_2β−1_(j)+2Sx_β−1_(j)+1. . . x_j0.

By lemma 5.2.2 the first sequence is a subsequence of the second. Two appli- cations of lemma 5.2.1 now yield the result.

Lemma 5.2.6. Ifi < j,j+φ(j)≤m,φ(i) =β−1(i) + 3andφ(j) =β−1(j) + 1

then it cannot be the case thatyi. . . yi+φ(i)is a subsequence of yj. . . yj+φ(j).

Proof. The sequence yi. . . yi+φ(i) contains an S and a T while the sequence

yj. . . yj+φ(j) does not contain anS or does not contain aT.

Lemma 5.2.7. Ifi < j,j+φ(j)≤m,φ(i) =β−1_{(i) + 1,φ(j) =β}−1_{(j) + 1and}

β−1_{(i) =β}−1_{(j)then it cannot be the case thaty}

i. . . yi+φ(i)is a subsequence of

Proof. Since the sequencesyi. . . yi+φ(i)andyj. . . yj+φ(j)have the same length,

all we have to do is show that they are not equal. Lethequal the identity on numbers> k2andp2 on numbers≤k2. If these sequences are equal then there

exists a numbercsuch that 0≤c≤β−1(i) + 1 and

h(yi). . . h(yi+φ(i)) =h(yj). . . h(yj+φ(j)) =ac+1. . . aβ−1_(i)+1wa1. . . ac

where w > k2_{. Let d be the number q that is used in the definition of y}

i if yi≤k2and letdbe one less than theqused in the definition ofyi+1 otherwise.

Soq=dis used in the definition of yi. . . yi+β−1_(i)−c and q=d+ 1 is used in

the definition ofyi+β−1_(i)−c+1. . . y_i+φ(i). The numberdis uniquely determined

bya1. . . aβ−1_(i)+1. This implies that i = j which contradicts the assumption

i < j.

Lemma 5.2.8. If i < j, j+φ(j)≤ m, φ(i) = β−1_{(i) + 3, φ(j) = β}−1_{(j) +}

3, β−1_{(i) + 1 = β}−1_{(j) + 1 then it cannot be the case that yi. . . y}

i+φ(i) is a

subsequence ofyj. . . yj+φ(j).

Proof. Since the length ofyi. . . yi+φ(i)is the same as the length ofyj. . . yj+φ(j)

it suffices to show that they are not equal. This is clear since both sequences must contain exactly oneT. This is theT at positionβ(β−1_{(i)) (=β(β}−1_(j)))

and sincei < j there cannot exist ac such thatyi+c=yj+c=T.

Lemma 5.2.9. There are no i < j such thatyi. . . yi+φ(i) is a subsequence of

yj. . . yj+φ(j).

Proof. For every i < j the conditions of one of the lemmas 5.2.3 - 5.2.8 are satisfied and it either follows directly thatyi. . . yi+φ(i) is not a subsequence of

yj. . . yj+φ(j)or the assumption thatyi. . . yi+φ(i)is a subsequence ofyj. . . yj+φ(j)

implies that there arei0 < j0 ≤ bn/2csuch that x_i0. . . x_2i0 is a subsequence of x_j0. . . x_2j0 which contradicts the assumption that the sequence x₁. . . xn has

propertyF.