Generalized Milewski sequences with perfect

Generalized Milewski sequenceswith

perfect autocorrelation/optimal crosscorrelation(and its relation with circular Florentine arrays)

ITA 2019, San Diego, California, USAMin Kyu Song, Gangsan Kim and Hong-Yeop Song

Yonsei University, Seoul, Korea

based on the recent submission to IEEE IT Trans in Jan 2019New framework for sequences with perfect autocorrelation and optimal crosscorrelation

Hong-Yeop Song

Contents of Talk

• Some preliminary concepts

▫ Circular Florentine arrays

▫ Perfect sequences▫ Interleaved sequences▫ Milewski construction – review

• Proposed generalization

• Concluding remarks

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Contents of Talk






A resurrection of a combinatorial structure after more than 30 years

Circular Florentine arrays

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A 𝑘𝑘 × 𝑁𝑁 circular Florentine array is equivalent to a set of 𝑘𝑘 distinct permutations 𝜋𝜋1,𝜋𝜋2, … ,𝜋𝜋𝑘𝑘 of the integers modulo 𝑁𝑁 such that

𝜋𝜋𝑖𝑖 𝑥𝑥 + 𝜏𝜏 = 𝜋𝜋𝑗𝑗(𝑥𝑥)has exactly one solution 𝑥𝑥for any two distinct permutations 𝜋𝜋𝑖𝑖, 𝜋𝜋𝑗𝑗 for any shift 𝜏𝜏.

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A 𝑘𝑘 × 𝑁𝑁 circular Florentine array is equivalent to a set of 𝑘𝑘 distinct permutations 𝜋𝜋1,𝜋𝜋2, … ,𝜋𝜋𝑘𝑘 of the integers modulo 𝑁𝑁 such that

𝜋𝜋𝑖𝑖 𝑥𝑥 + 𝜏𝜏 = 𝜋𝜋𝑗𝑗(𝑥𝑥)has exactly one solution 𝑥𝑥for any two distinct permutations 𝜋𝜋𝑖𝑖, 𝜋𝜋𝑗𝑗 for any shift 𝜏𝜏.

S. W. Golomb and H. Taylor,Tuscan squares – a new family of combinatorial designs, Ars Combinatoria, 1985.

more than 30 years ago!

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10×11 circular Florentine array

0000000000

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10×11 circular Tuscan-1 array

0000000000

• Each circular row is a permutation.

• For any symbol 𝒂𝒂 and distance 1, the symbols from 𝒂𝒂 in a circular distance 1 are all distinct

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0000000000


• In addition, for any symbol 𝒂𝒂 and distance 2, the symbols from 𝒂𝒂 in a circular distance 2 are all distinct

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0000000000


• In addition, for any symbol 𝒂𝒂 and distance 3, the symbols from 𝒂𝒂 in a circular distance 3 are all distinct

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…10×11 circular Tuscan-10 array

=10×11 circular Florentine array

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10×11 circular Florentine array

0000000000


• For any symbol 𝒂𝒂and each distance d=1,2,…,N-1, the symbols from 𝒂𝒂 in a circular distance dare all distinct

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A transform

0000000000

We may rotateeach row without

violating the property so that

a common symbol comes to the left-most column, and delete the column.

This gives …

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S. W. Golomb and H. Taylor,Tuscan squares – a new family of combinatorial designs,Ars Combinatoria, 1985


10×10 Florentine square

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Property of rows

Property of columns, in addition

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H.-Y. Song and J. H. Dinitz, "Tuscan Squares,"CRC Handbook of Combinatorial Designs, edited by C. J. Colbournand J. H. Dinitz, CRC Press, pp. 480-484, 1996.




http://www.emba.uvm.edu/%7Edinitz/contents.html

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• Let 𝐹𝐹𝑐𝑐 𝑁𝑁 be the largest integer such that an 𝐹𝐹𝑐𝑐 𝑁𝑁 × 𝑁𝑁circular Florentine array exists. Then,

𝑝𝑝min − 1 ≤ 𝐹𝐹𝑐𝑐 𝑁𝑁 ≤ 𝑁𝑁 − 1

.

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• It is known that ▫ if 𝑁𝑁 is prime, then 𝐹𝐹𝑐𝑐 𝑁𝑁 = 𝑁𝑁 − 1.

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• It is known that ▫ if 𝑁𝑁 is prime, then 𝐹𝐹𝑐𝑐 𝑁𝑁 = 𝑁𝑁 − 1.▫ Especially, if 𝑁𝑁 is even, then 𝐹𝐹𝑐𝑐 𝑁𝑁 = 1.

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• It is known that ▫ if 𝑁𝑁 is prime, then 𝐹𝐹𝑐𝑐 𝑁𝑁 = 𝑁𝑁 − 1.▫ Especially, if 𝑁𝑁 is even, then 𝐹𝐹𝑐𝑐 𝑁𝑁 = 1.▫ For all other odd 𝑁𝑁, the exact value 𝐹𝐹𝑐𝑐 𝑁𝑁 is widely

open.

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A 4 × 15 circular Florentine array𝜋𝜋1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

𝜋𝜋2 0 7 1 8 2 12 3 11 9 4 13 5 14 6 10

𝜋𝜋3 0 4 11 7 10 1 13 9 5 8 3 6 2 14 12

𝜋𝜋4 0 13 7 2 11 6 14 10 3 5 12 9 1 4 8

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A 4 × 15 circular Florentine array

• 15 = 3 � 5 and 𝑝𝑝𝑚𝑚𝑖𝑖𝑚𝑚 = 3. Therefore, 2 ≤ 𝐹𝐹𝑐𝑐 15 ≤ 14.• It turned out that

𝑭𝑭𝒄𝒄 𝟏𝟏𝟏𝟏 = 𝟒𝟒and the above example has 4 rows.

𝜋𝜋1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

𝜋𝜋2 0 7 1 8 2 12 3 11 9 4 13 5 14 6 10

𝜋𝜋3 0 4 11 7 10 1 13 9 5 8 3 6 2 14 12

𝜋𝜋4 0 13 7 2 11 6 14 10 3 5 12 9 1 4 8

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A 4 × 15 circular Florentine array

• 15 = 3 � 5 and 𝑝𝑝𝑚𝑚𝑖𝑖𝑚𝑚 = 3. Therefore, 2 ≤ 𝐹𝐹𝑐𝑐 15 ≤ 14.• It turned out that

𝑭𝑭𝒄𝒄 𝟏𝟏𝟏𝟏 = 𝟒𝟒and the above example has 4 rows.

𝜋𝜋1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

𝜋𝜋2 0 7 1 8 2 12 3 11 9 4 13 5 14 6 10

𝜋𝜋3 0 4 11 7 10 1 13 9 5 8 3 6 2 14 12

𝜋𝜋4 0 13 7 2 11 6 14 10 3 5 12 9 1 4 8

PHD Thesis, USC, by Hong-Yeop Song (1991)and also, later in

Computers & Mathematics with Applications (2000)

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Check𝝅𝝅𝟐𝟐 𝒙𝒙 + 𝝉𝝉 = 𝝅𝝅𝟏𝟏(𝒙𝒙)

has exactly one solution 𝑥𝑥 for any 𝜏𝜏

01

2

3

4

5

6

78

9

10

11

12

13

140

7

1

8

2

12

3

119

4

13

5

14

6

10

𝜋𝜋1 𝜋𝜋2

01

2

3

4

5

6

78

9

10

11

12

13

14

0 7

1

8

2

12

3

119

4

13

5

14

6

10

𝜏𝜏 = 0

01

2

3

4

5

6

78

9

10

11

12

13

14

71

8

2

12

3

11

94

13

5

14

6

10

0

𝜏𝜏 = 1

01

2

3

4

5

6

78

9

10

11

12

13

14

18

2

12

3

11

9

4135

14

6

10

07

𝜏𝜏 = 2

01

2

3

4

5

6

78

9

10

11

12

13

14

82

12

3

11

9

4

135

14

6

10

0

7

1

𝜏𝜏 = 3

01

2

3

4

5

6

78

9

10

11

12

13

14

212

3

11

9

4

13

514

6

10

0

7

1

8

𝜏𝜏 = 4

01

2

3

4

5

6

78

9

10

11

12

13

14

123

11

9

4

13

5

14610

0

7

1

8

2

𝜏𝜏 = 5

01

2

3

4

5

6

78

9

10

11

12

13

14

𝜏𝜏 = 63

11

9

4

13

5

14

6100

7

1

8

2

12

etc…

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• For complex-valued sequences 𝒙𝒙,𝒚𝒚 of length 𝐿𝐿, the periodic correlation of 𝒙𝒙 and 𝒚𝒚 at shift 𝝉𝝉 is

𝑪𝑪𝒙𝒙,𝒚𝒚 𝝉𝝉 = �𝒏𝒏=𝟎𝟎

𝑳𝑳−𝟏𝟏

𝒙𝒙 𝒏𝒏 + 𝝉𝝉 𝒚𝒚∗(𝒏𝒏)

Sequences and Correlation

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• For complex-valued sequences 𝒙𝒙,𝒚𝒚 of length 𝐿𝐿, the periodic correlation of 𝒙𝒙 and 𝒚𝒚 at shift 𝝉𝝉 is

𝑪𝑪𝒙𝒙,𝒚𝒚 𝝉𝝉 = �𝒏𝒏=𝟎𝟎

𝑳𝑳−𝟏𝟏

𝒙𝒙 𝒏𝒏 + 𝝉𝝉 𝒚𝒚∗(𝒏𝒏)

▫ If 𝒚𝒚 is a cyclic shift of 𝒙𝒙, it is called autocorrelation, and denoted by 𝑪𝑪𝒙𝒙 𝝉𝝉

▫ Otherwise, it is called crosscorrelation

Sequences and Correlation

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Perfect Sequences

• A sequence 𝒙𝒙 of length 𝑳𝑳 is called perfect if

C𝒙𝒙 𝜏𝜏 = �𝑬𝑬, 𝜏𝜏 ≡ 0 mod 𝐿𝐿0, 𝜏𝜏 ≢ 0 (mod 𝐿𝐿)

Here, 𝑬𝑬 is called the energy of 𝒙𝒙

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Perfect Sequences




• (Sarwate, 79) Crosscorrelation of any two perfect sequences of length 𝑳𝑳 with the same energy 𝑬𝑬 is lower bounded by 𝑬𝑬/ 𝑳𝑳.

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Perfect Sequences




• (Sarwate, 79) Crosscorrelation of any two perfect sequences of length 𝑳𝑳 with the same energy 𝑬𝑬 is lower bounded by 𝑬𝑬/ 𝑳𝑳.▫ An optimal pair of perfect sequences of length 𝑳𝑳▫ An optimal set of perfect sequences of length 𝑳𝑳

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Interleaved Sequence

• Consider two sequences 𝒔𝒔0 = 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 and 𝒔𝒔1 = 𝑑𝑑, 𝑒𝑒,𝑓𝑓 of length 3 each

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• Write each as a column of an array:

𝒔𝒔0, 𝒔𝒔1 =𝑎𝑎 𝑑𝑑𝑏𝑏 𝑒𝑒𝑐𝑐 𝑓𝑓

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• Write each as a column of an array:

𝒔𝒔0, 𝒔𝒔1 =𝑎𝑎 𝑑𝑑𝑏𝑏 𝑒𝑒𝑐𝑐 𝑓𝑓

• Read the array row-by-row and obtain a sequence of length 6:

𝒔𝒔 = 𝐼𝐼(𝒔𝒔𝟎𝟎, 𝒔𝒔𝟏𝟏) = 𝑎𝑎,𝑑𝑑, 𝑏𝑏, 𝑒𝑒, 𝑐𝑐, 𝑓𝑓is called an interleaved sequence of 𝒔𝒔0 and 𝒔𝒔1

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History of Perfect Polyphase Sequences

1960’s

1970’s

1980’s

1990’s

Frank, ZadoffIRE T-IT, 1962

(Heimiller 1961)

Kumar, Scholtz, WelchJ. Comb. Theory Series

A. 1985

P1 code

Mow’s two unified construction (dissertation 1993, ISSTA 1996)

Alphabet size : 𝑁𝑁Period : 𝑁𝑁2

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1960’s

1970’s

1980’s

1990’s


(Heimiller 1961)


A. 1985

P1 code



Chu(Frank,Zadoff)

IEEE T-IT, 1973

PopovicIEEE T-IT, 1992

P2 codeAlphabet size : 𝑁𝑁

Period : 𝑁𝑁

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1960’s

1970’s

1980’s

1990’s


(Heimiller 1961)


A. 1985

P1 code



Chu(Frank,Zadoff)

IEEE T-IT, 1973



Period : 𝑁𝑁

MilewskiIBM J. R&D, 1983

P3 codeAlphabet size : 𝑁𝑁𝑘𝑘+1

Period : 𝑁𝑁2𝑘𝑘+1

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1960’s

1970’s

1980’s

1990’s


(Heimiller 1961)


A. 1985

P1 code



Chu(Frank,Zadoff)

IEEE T-IT, 1973



Period : 𝑁𝑁

Chung, KumarIEEE T-IT, 1989

P4 codeAlphabet size : 𝑠𝑠𝑁𝑁

Period : 𝑠𝑠𝑁𝑁2

MilewskiIBM J. R&D, 1983

P3 codeAlphabet size : 𝑁𝑁𝑘𝑘+1

Period : 𝑁𝑁2𝑘𝑘+1

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The original Milewski constructionLength: 𝒎𝒎 → 𝒎𝒎 �𝒎𝒎𝟐𝟐𝟐𝟐

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MilewskiConstruction

perfect polyphase Sequence of length 𝑚𝑚

𝜷𝜷 = 𝛼𝛼 𝑛𝑛 𝑚𝑚=0𝑚𝑚−1

A positive integer

𝐾𝐾

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Output perfect polyphase sequence

𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎𝟐𝟐𝟐𝟐+𝟏𝟏−1



A positive integer

𝐾𝐾

where

𝒔𝒔 𝒏𝒏 = 𝜷𝜷 𝒒𝒒 𝝎𝝎𝒒𝒒𝒓𝒓

𝜔𝜔 = 𝑒𝑒−𝑗𝑗2𝜋𝜋

𝑚𝑚1+𝐾𝐾

Here, we use𝑛𝑛 = 𝒒𝒒𝒎𝒎𝟐𝟐 + 𝒓𝒓 ↔ (𝒒𝒒, 𝒓𝒓)

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𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎𝟐𝟐𝟐𝟐+𝟏𝟏−1



A positive integer

𝐾𝐾

where



𝑚𝑚1+𝐾𝐾


𝑚𝑚 � 𝑚𝑚𝐾𝐾 × 𝑚𝑚𝐾𝐾 array form of 𝒔𝒔

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𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎𝟐𝟐𝟐𝟐+𝟏𝟏−1



A positive integer

𝐾𝐾

where



𝑚𝑚1+𝐾𝐾



⋮ ⋮ ⋮

⋮ ⋮ ⋮

⋮ ⋮ ⋮

𝛽𝛽(0) 𝛽𝛽(0)⋯𝛽𝛽(1) 𝛽𝛽(1)⋯

𝛽𝛽(2) 𝛽𝛽(2)⋯

𝛽𝛽(𝑚𝑚− 1) 𝛽𝛽(𝑚𝑚− 1) 𝛽𝛽(𝑚𝑚− 1)⋯

𝛽𝛽(0) 𝛽𝛽(0)⋯

𝛽𝛽(1) 𝛽𝛽(1)⋯

𝛽𝛽(2) 𝛽𝛽(2)⋯

𝛽𝛽(𝑚𝑚− 1) 𝛽𝛽(𝑚𝑚− 1) 𝛽𝛽(𝑚𝑚− 1)⋯

Input sequence 𝜷𝜷of period 𝑚𝑚

is repeated 𝑚𝑚𝐾𝐾 times

⋮

⋮

⋮

𝛽𝛽(0)𝛽𝛽(1)

𝛽𝛽(2)

𝛽𝛽(𝑚𝑚− 1)

𝛽𝛽(0)

𝛽𝛽(1)

𝛽𝛽(2)

𝛽𝛽(𝑚𝑚− 1)

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𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎𝟐𝟐𝟐𝟐+𝟏𝟏−1



A positive integer

𝐾𝐾

where


𝝎𝝎 = 𝒆𝒆−𝒋𝒋𝟐𝟐𝝅𝝅

𝒎𝒎𝟏𝟏+𝟐𝟐



⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱

𝛽𝛽(0) 𝛽𝛽(0) 𝛽𝛽(0)𝟏𝟏× × ×⋯ 𝟏𝟏𝟏𝟏𝛽𝛽(1) 𝛽𝛽(1) 𝛽𝛽(1)𝝎𝝎× × ×⋯ 𝝎𝝎𝑵𝑵−𝟏𝟏 𝟏𝟏𝟏𝟏

𝛽𝛽(2) 𝛽𝛽(2) 𝛽𝛽(2)𝝎𝝎𝟐𝟐× × ×⋯ 𝝎𝝎𝑵𝑵−𝟏𝟏 𝟐𝟐𝟏𝟏

𝛽𝛽(𝑚𝑚− 1) 𝛽𝛽(𝑚𝑚− 1) 𝛽𝛽(𝑚𝑚− 1)𝝎𝝎𝒎𝒎−𝟏𝟏× × ×⋯ 𝝎𝝎𝑵𝑵−𝟏𝟏 𝒎𝒎−𝟏𝟏𝟏𝟏

𝛽𝛽(0) 𝛽𝛽(0) 𝛽𝛽(0)𝝎𝝎𝒎𝒎(𝑵𝑵−𝟏𝟏)× × ×⋯ 𝝎𝝎𝑵𝑵−𝟏𝟏 𝒎𝒎(𝑵𝑵−𝟏𝟏)𝟏𝟏

𝛽𝛽(1) 𝛽𝛽(1) 𝛽𝛽(1)𝝎𝝎𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟏𝟏× × ×⋯ 𝝎𝝎𝑵𝑵−𝟏𝟏 𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟏𝟏𝟏𝟏

𝛽𝛽(2) 𝛽𝛽(2) 𝛽𝛽(2)𝝎𝝎𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟐𝟐× × ×⋯ 𝝎𝝎𝑵𝑵−𝟏𝟏 𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟐𝟐𝟏𝟏

𝛽𝛽(𝑚𝑚− 1) 𝛽𝛽(𝑚𝑚− 1) 𝛽𝛽(𝑚𝑚− 1)𝝎𝝎𝒎𝒎𝑵𝑵−𝟏𝟏× × ×⋯ 𝝎𝝎𝑵𝑵−𝟏𝟏 𝒎𝒎𝑵𝑵−𝟏𝟏𝟏𝟏



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𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎𝟐𝟐𝟐𝟐+𝟏𝟏−1



A positive integer

𝐾𝐾

where



𝒎𝒎𝟏𝟏+𝟐𝟐



⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱


𝛽𝛽(2) 𝛽𝛽(2) 𝛽𝛽(2)𝝎𝝎𝟐𝟐× × ×⋯ 𝝎𝝎𝑵𝑵−𝟏𝟏 𝟐𝟐𝟏𝟏






Here, 𝑁𝑁 = 𝑚𝑚𝐾𝐾 and the exponent runs from 0 to 𝑚𝑚𝑁𝑁 − 1



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𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎𝟐𝟐𝟐𝟐+𝟏𝟏−1



A positive integer

𝐾𝐾

where



𝒎𝒎𝟏𝟏+𝟐𝟐



⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱


𝛽𝛽(2) 𝛽𝛽(2) 𝛽𝛽(2)𝝎𝝎𝟐𝟐× × ×⋯ 𝝎𝝎𝑵𝑵−𝟏𝟏 𝟐𝟐𝟏𝟏








the exponent runs from 0 to 𝑁𝑁 − 1

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Contents of Talk






A resurrection of a combinatorial structure after more than 30 years


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Our framework(A special type of interleaved sequences)

Interleaving technique

A positive integer𝑁𝑁

A polyphase sequenceof length 𝑁𝑁

𝝁𝝁 𝜋𝜋

A functionℤ𝑵𝑵 ⟶ ℤ𝒎𝒎𝑵𝑵

A collection of 𝑁𝑁 sequence of length 𝑚𝑚𝑩𝑩 = 𝜷𝜷0,𝜷𝜷1, … ,𝜷𝜷𝑁𝑁

not necessarily polyphasenot necessarily all distinct

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where

with 𝒏𝒏 = 𝒒𝒒𝑵𝑵 + 𝒓𝒓, and 𝝎𝝎 = 𝐞𝐞𝐞𝐞𝐞𝐞(−𝒋𝒋𝟐𝟐𝝅𝝅/𝒎𝒎𝑵𝑵).




𝝁𝝁 𝜋𝜋




Output sequence 𝑠𝑠 = 𝒔𝒔 𝒏𝒏 𝑚𝑚=0𝑚𝑚𝑁𝑁2−1

𝒔𝒔 𝒏𝒏 = 𝝁𝝁 𝒓𝒓 𝜷𝜷𝒓𝒓 𝒒𝒒 𝝎𝝎𝒒𝒒𝝅𝝅(𝒓𝒓)

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where

with 𝒏𝒏 = 𝒒𝒒𝑵𝑵 + 𝒓𝒓, and 𝝎𝝎 = 𝐞𝐞𝐞𝐞𝐞𝐞(−𝒋𝒋𝟐𝟐𝝅𝝅/𝒎𝒎𝑵𝑵).




𝝁𝝁 𝜋𝜋




Output sequence 𝑠𝑠 = 𝒔𝒔 𝒏𝒏 𝑚𝑚=0𝑚𝑚𝑁𝑁2−1

𝒔𝒔 𝒏𝒏 = 𝝁𝝁 𝒓𝒓 𝜷𝜷𝒓𝒓 𝒒𝒒 𝝎𝝎𝒒𝒒𝝅𝝅(𝒓𝒓)

Definition. We define A 𝑩𝑩,𝝅𝝅 be a family of interleaved sequences constructed by the above procedure using all possible polyphase sequences 𝝁𝝁.

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Array FormAssume that 𝝁𝝁 is the all-one sequence, ※ 𝜔𝜔 = 𝑒𝑒−𝑗𝑗

2𝜋𝜋𝑚𝑚𝑁𝑁

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Array Form

Row index 𝒒𝒒

=𝟎𝟎,𝟏𝟏,𝟐𝟐,…

,𝒎𝒎𝑵𝑵−𝟏𝟏

Column index 𝒓𝒓 = 𝟎𝟎,𝟏𝟏,𝟐𝟐, … ,𝑵𝑵− 𝟏𝟏Assume that 𝝁𝝁 is the all-one sequence, ※ 𝜔𝜔 = 𝑒𝑒−𝑗𝑗


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Array Form

Row index 𝒒𝒒

=𝟎𝟎,𝟏𝟏,𝟐𝟐,…




⋮ ⋮ ⋮

⋮ ⋮ ⋮

⋮ ⋮ ⋮

𝛽𝛽0(0) 𝛽𝛽1(0) 𝛽𝛽𝑁𝑁−1(0)⋯

𝛽𝛽0(1) 𝛽𝛽1(1) 𝛽𝛽𝑁𝑁−1(1)⋯

𝛽𝛽0(2) 𝛽𝛽1(2) 𝛽𝛽𝑁𝑁−1(2)⋯

𝛽𝛽0(𝑚𝑚− 1) 𝛽𝛽1(𝑚𝑚− 1) 𝛽𝛽𝑁𝑁−1(𝑚𝑚− 1)⋯

𝛽𝛽0(0) 𝛽𝛽1(0) 𝛽𝛽𝑁𝑁−1(0)⋯

𝛽𝛽0(1) 𝛽𝛽1(1) 𝛽𝛽𝑁𝑁−1(1)⋯

𝛽𝛽0(2) 𝛽𝛽1(2) 𝛽𝛽𝑁𝑁−1(2)⋯

𝛽𝛽0(𝑚𝑚− 1) 𝛽𝛽1(𝑚𝑚− 1) 𝛽𝛽𝑁𝑁−1(𝑚𝑚− 1)⋯

Input sequence 𝜷𝜷0of period 𝑚𝑚

Repeating 𝑁𝑁 times

Input sequence 𝜷𝜷1of period 𝑚𝑚


Input sequence 𝜷𝜷𝑁𝑁−1of period 𝑚𝑚


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Array Form

Row index 𝒒𝒒

=𝟎𝟎,𝟏𝟏,𝟐𝟐,…




⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱

𝛽𝛽0(0) 𝛽𝛽1(0) 𝛽𝛽𝑁𝑁−1(0)𝜔𝜔𝜋𝜋(1) 𝟎𝟎× × ×⋯ 𝜔𝜔𝜋𝜋(𝑁𝑁−1) 𝟎𝟎𝜔𝜔𝜋𝜋(0) 0

𝛽𝛽0(1) 𝛽𝛽1(1) 𝛽𝛽𝑁𝑁−1(1)𝜔𝜔𝜋𝜋(1) 𝟏𝟏× × ×⋯ 𝜔𝜔𝜋𝜋(𝑁𝑁−1) 𝟏𝟏𝜔𝜔𝜋𝜋(0) 𝟏𝟏

𝛽𝛽0(2) 𝛽𝛽1(2) 𝛽𝛽𝑁𝑁−1(2)𝜔𝜔𝜋𝜋(1) 𝟐𝟐× × ×⋯ 𝜔𝜔𝜋𝜋(𝑁𝑁−1) 𝟐𝟐𝜔𝜔𝜋𝜋(0) 𝟐𝟐

𝛽𝛽0(𝑚𝑚− 1) 𝛽𝛽1(𝑚𝑚− 1) 𝛽𝛽𝑁𝑁−1(𝑚𝑚− 1)𝜔𝜔𝜋𝜋(1) 𝒎𝒎−𝟏𝟏× × ×⋯ 𝜔𝜔𝜋𝜋(𝑁𝑁−1) 𝒎𝒎−𝟏𝟏𝜔𝜔𝜋𝜋(0) (𝒎𝒎−𝟏𝟏)

𝛽𝛽0(0) 𝛽𝛽1(0) 𝛽𝛽𝑁𝑁−1(0)𝜔𝜔𝜋𝜋(1) 𝒎𝒎(𝑵𝑵−𝟏𝟏)× × ×⋯ 𝜔𝜔𝜋𝜋(𝑁𝑁−1) 𝒎𝒎(𝑵𝑵−𝟏𝟏)𝜔𝜔𝜋𝜋(0) 𝒎𝒎(𝑵𝑵−𝟏𝟏)

𝛽𝛽0(1) 𝛽𝛽1(1) 𝛽𝛽𝑁𝑁−1(1)𝜔𝜔𝜋𝜋(1) 𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟏𝟏× × ×⋯ 𝜔𝜔𝜋𝜋(𝑁𝑁−1) 𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟏𝟏𝜔𝜔𝜋𝜋(0) 𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟏𝟏

𝛽𝛽0(2) 𝛽𝛽1(2) 𝛽𝛽𝑁𝑁−1(2)𝜔𝜔𝜋𝜋(1) 𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟐𝟐× × ×⋯ 𝜔𝜔𝜋𝜋(𝑁𝑁−1) 𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟐𝟐𝜔𝜔𝜋𝜋(0) 𝒎𝒎 𝑵𝑵−𝟏𝟏 +𝟐𝟐

𝛽𝛽0(𝑚𝑚− 1) 𝛽𝛽1(𝑚𝑚− 1) 𝛽𝛽𝑁𝑁−1(𝑚𝑚− 1)𝜔𝜔𝜋𝜋(1) 𝒎𝒎𝑵𝑵−𝟏𝟏× × ×⋯ 𝜔𝜔𝜋𝜋(𝑁𝑁−1) 𝒎𝒎𝑵𝑵−𝟏𝟏𝜔𝜔𝜋𝜋(0) 𝒎𝒎𝑵𝑵−𝟏𝟏

Input function 𝜋𝜋:ℤ𝑵𝑵 ⟶ ℤ𝒎𝒎𝑵𝑵

Hong-Yeop Song

Array Form

Row index 𝒒𝒒

=𝟎𝟎,𝟏𝟏,𝟐𝟐,…


Column index 𝒓𝒓 = 𝟎𝟎,𝟏𝟏,𝟐𝟐, … ,𝑵𝑵− 𝟏𝟏Assume that 𝝁𝝁 is the all-one sequence,

⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱









the exponent runs from 𝜋𝜋(0) to 𝜋𝜋(𝑁𝑁 − 1)

Hong-Yeop Song

Array Form

Row index 𝒒𝒒

=𝟎𝟎,𝟏𝟏,𝟐𝟐,…




⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱

⋮ ⋮ ⋮⋱









Here, the exponent of 𝝎𝝎𝝅𝝅(𝒓𝒓)

runs from 𝟎𝟎 to 𝒎𝒎𝑵𝑵− 𝟏𝟏

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Milewski Construction is a Special Case



𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎𝟐𝟐𝟐𝟐+𝟏𝟏−1

perfect polyphase sequence


A positive integer

𝐾𝐾

where𝑠𝑠 𝑛𝑛 = 𝛽𝛽 𝑞𝑞 𝜔𝜔𝑞𝑞𝒓𝒓


𝑁𝑁 𝝁𝝁 𝜋𝜋𝐵𝐵 = 𝜷𝜷0,𝜷𝜷1, … ,𝜷𝜷𝑁𝑁

Output sequence

=?

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𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎𝟐𝟐𝟐𝟐+𝟏𝟏−1



A positive integer

𝐾𝐾

where𝑠𝑠 𝑛𝑛 = 𝛽𝛽 𝑞𝑞 𝜔𝜔𝑞𝑞𝒓𝒓


𝑁𝑁 𝝁𝝁 𝜋𝜋𝐵𝐵 = 𝜷𝜷0,𝜷𝜷1, … ,𝜷𝜷𝑁𝑁

Output sequence

𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎 𝒎𝒎𝟐𝟐 𝟐𝟐

−1

perfect polyphase sequences

𝜷𝜷0 = 𝜷𝜷1 = ⋯ = 𝜷𝜷𝑁𝑁−1An integer𝑁𝑁 = 𝑚𝑚𝐾𝐾

all-one sequence

The identityfunction𝜋𝜋 𝑟𝑟 = 𝑟𝑟

=

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𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎𝟐𝟐𝟐𝟐+𝟏𝟏−1



A positive integer

𝐾𝐾

where𝑠𝑠 𝑛𝑛 = 𝜷𝜷 𝒒𝒒 𝝎𝝎𝒒𝒒𝒓𝒓


𝑁𝑁 𝝁𝝁

where𝑠𝑠 𝑛𝑛 = 𝛽𝛽𝒓𝒓 𝑞𝑞 𝜔𝜔𝑞𝑞𝝅𝝅(𝒓𝒓) = 𝜷𝜷 𝒒𝒒 𝝎𝝎𝒒𝒒𝒓𝒓

𝜋𝜋𝐵𝐵 = 𝜷𝜷0,𝜷𝜷1, … ,𝜷𝜷𝑁𝑁

Output sequence

𝒔𝒔 = 𝑠𝑠 𝑛𝑛 𝑚𝑚=0𝒎𝒎 𝒎𝒎𝟐𝟐 𝟐𝟐

−1

perfect polyphase sequences

𝜷𝜷0 = 𝜷𝜷1 = ⋯ = 𝜷𝜷𝑁𝑁−1An integer𝑁𝑁 = 𝑚𝑚𝐾𝐾

all-one sequence

The identityfunction𝜋𝜋 𝑟𝑟 = 𝑟𝑟

=

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Important Link

Definition. Let 𝝅𝝅,𝝈𝝈 be two functions from ℤ𝑵𝑵 to ℤ𝒎𝒎𝑵𝑵. We define

When 𝝅𝝅 = 𝝈𝝈, we use 𝜳𝜳𝝅𝝅 𝝉𝝉 simply.

𝜳𝜳𝝅𝝅,𝝈𝝈 𝝉𝝉 = 𝒙𝒙 ∈ ℤ𝑵𝑵 𝝅𝝅 𝒙𝒙 + 𝝉𝝉 ≡ 𝝈𝝈 𝒙𝒙 𝒎𝒎𝒎𝒎𝒎𝒎 𝑵𝑵 .

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Condition on perfectness(Main result 1)

Theorem. Any sequence inA 𝑩𝑩,𝝅𝝅 is perfect if and only if the following conditions are satisfied:

1) 𝚿𝚿𝝅𝝅(𝒓𝒓) = 𝟎𝟎 for 𝒓𝒓 = 𝟏𝟏,𝟐𝟐, … ,𝑵𝑵− 𝟏𝟏.That is, 𝝅𝝅 𝒓𝒓 (𝐦𝐦𝐦𝐦𝐦𝐦 𝑵𝑵) for 𝒓𝒓 = 𝟎𝟎,𝟏𝟏, … ,𝑵𝑵− 𝟏𝟏 is a permutation over ℤ𝑵𝑵.

2) 𝑩𝑩 is a collection of perfect sequences all of period 𝑚𝑚 with the same energy.




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Condition on perfectness(Main result 1)

Theorem. Any sequence inA 𝑩𝑩,𝝅𝝅 is perfect if and only if the following conditions are satisfied:

1) 𝚿𝚿𝝅𝝅(𝒓𝒓) = 𝟎𝟎 for 𝒓𝒓 = 𝟏𝟏,𝟐𝟐, … ,𝑵𝑵− 𝟏𝟏.That is, 𝝅𝝅 𝒓𝒓 (𝐦𝐦𝐦𝐦𝐦𝐦 𝑵𝑵) for 𝒓𝒓 = 𝟎𝟎,𝟏𝟏, … ,𝑵𝑵− 𝟏𝟏 is a permutation over ℤ𝑵𝑵.

2) 𝑩𝑩 is a collection of perfect sequences all of period 𝑚𝑚 with the same energy.

We now call themthe generalized Milewski sequences




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Examples

Generalized Milewski

Construction

𝑁𝑁 𝝁𝝁𝐵𝐵 = 𝜷𝜷0,𝜷𝜷1 𝜋𝜋

• 𝜷𝜷0 = 𝜷𝜷1 = 0,−1,1,0,1,1which is a perfect sequence of length 6,

• 𝑁𝑁 = 2,• 𝜋𝜋(𝑟𝑟) = 𝑟𝑟, and• 𝝁𝝁 is the all-one sequence.

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Examples

※ 𝜔𝜔 = 𝑒𝑒−𝑗𝑗2𝜋𝜋12


Construction


𝒔𝒔 = 0, 0,−1,−𝜔𝜔, 1,𝜔𝜔2, 0, 0, 1,𝜔𝜔4, 1,𝜔𝜔5, 0, 0,−1,−𝜔𝜔7, 1,𝜔𝜔8, 0, 0, 1,𝜔𝜔10, 1,𝜔𝜔11

is a perfect sequence of length 24.



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ExamplesConstellation of

𝒔𝒔

※ 𝜔𝜔 = 𝑒𝑒−𝑗𝑗2𝜋𝜋12


Construction


𝒔𝒔 = 0, 0,−1,−𝜔𝜔, 1,𝜔𝜔2, 0, 0, 1,𝜔𝜔4, 1,𝜔𝜔5, 0, 0,−1,−𝜔𝜔7, 1,𝜔𝜔8, 0, 0, 1,𝜔𝜔10, 1,𝜔𝜔11

is a perfect sequence of length 24.



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Examples

※ 𝜔𝜔 = 𝑒𝑒− 𝑗𝑗2𝜋𝜋12


Construction

𝑁𝑁 𝝁𝝁𝐵𝐵 = 𝜷𝜷0,𝜷𝜷1,𝜷𝜷2 𝜋𝜋

𝒔𝒔 is a perfect sequence of length 90.

• 𝜷𝜷0 = 𝜷𝜷1 = 𝜷𝜷2 =3,−2,3,−2,−2,3,−2,−7,−2,−2

which is a perfect sequence of period 10• 𝑁𝑁 = 3,• 𝜋𝜋(𝑟𝑟) = 𝑟𝑟, and• 𝝁𝝁 is the all-one sequence.

ASK constellation

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Examples

※ 𝜔𝜔 = 𝑒𝑒− 𝑗𝑗2𝜋𝜋12


Construction

𝑁𝑁 𝝁𝝁𝐵𝐵 = 𝜷𝜷0,𝜷𝜷1,𝜷𝜷2 𝜋𝜋

𝒔𝒔 is a perfect sequence of length 90.

• 𝜷𝜷0 = 𝜷𝜷1 = 𝜷𝜷2 =3,−2,3,−2,−2,3,−2,−7,−2,−2

which is a perfect sequence of period 10• 𝑁𝑁 = 3,• 𝜋𝜋(𝑟𝑟) = 𝑟𝑟, and• 𝝁𝝁 is the all-one sequence.

Constellation of 𝒔𝒔

APSK constellation

ASK constellation

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Direct vs Indirect (when 𝑁𝑁 is composite)

Perfect sequences of length 𝑚𝑚


sequences of length 𝑚𝑚𝑁𝑁12



Two-step synthesis

Direct synthesis

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Theorem. Assume that 𝑁𝑁 is a composite number.1) Any generalized Milewski sequence of length 𝑚𝑚𝑁𝑁2 from the two-step

method can be also obtained by the direct method.2) There exists a generalized Milewski sequence of length 𝑚𝑚𝑁𝑁2 from the direct

method which can not be obtained by the two-step method.

Perfect sequences of length 𝑚𝑚





Two-step synthesis

Direct synthesis

Direct vs Indirect (when 𝑁𝑁 is composite)

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Condition on optimal pair(Main result 2)

Theorem. Let 𝐵𝐵1 = 𝜷𝜷0,𝜷𝜷1, … ,𝜷𝜷𝑁𝑁−1 and 𝐵𝐵2 = 𝜸𝜸0,𝜸𝜸1, … ,𝜸𝜸𝑁𝑁−1 , all of length 𝑚𝑚 and the same energy 𝐸𝐸𝐵𝐵, and perfect.

Construct 𝒔𝒔 ∈ A 𝐵𝐵1,𝜋𝜋 and 𝒇𝒇 ∈ A(𝐵𝐵2,𝜎𝜎).

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Condition on optimal pair(Main result 2)

Theorem. Let 𝐵𝐵1 = 𝜷𝜷0,𝜷𝜷1, … ,𝜷𝜷𝑁𝑁−1 and 𝐵𝐵2 = 𝜸𝜸0,𝜸𝜸1, … ,𝜸𝜸𝑁𝑁−1 , all of length 𝑚𝑚 and the same energy 𝐸𝐸𝐵𝐵, and perfect.

Construct 𝒔𝒔 ∈ A 𝐵𝐵1,𝜋𝜋 and 𝒇𝒇 ∈ A(𝐵𝐵2,𝜎𝜎).

Then, 𝒔𝒔 and 𝒇𝒇 have optimal crosscorrelation if and only if the following conditions are satisfied for each 𝒓𝒓 = 𝟎𝟎,𝟏𝟏, … ,𝑵𝑵− 𝟏𝟏:1) Ψ𝜋𝜋,𝜎𝜎(𝑟𝑟) = 1, i.e., Ψ𝜋𝜋,𝜎𝜎 𝑟𝑟 = 𝑥𝑥 .2) For the unique 𝑥𝑥 ∈ Ψ𝜋𝜋,𝜎𝜎(𝑟𝑟), the pair of sequences

𝛽𝛽𝑥𝑥+𝑟𝑟 𝑡𝑡 𝜔𝜔𝑚𝑚𝜋𝜋 𝑥𝑥+𝑟𝑟 𝑡𝑡

𝑡𝑡=0

𝑚𝑚−1and 𝛾𝛾𝑥𝑥 𝑡𝑡 𝜔𝜔𝑚𝑚

𝜎𝜎 𝑥𝑥 𝑡𝑡𝑡𝑡=0

𝑚𝑚−1is optimal.

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Condition on optimal pair(Simple Special Case)

Corollary. Let 𝐵𝐵1 = 𝜷𝜷0,𝜷𝜷1, … ,𝜷𝜷𝑁𝑁−1 and 𝐵𝐵2 = 𝜸𝜸0,𝜸𝜸1, … ,𝜸𝜸𝑁𝑁−1 , all of length 𝑚𝑚 and the same energy 𝐸𝐸𝐵𝐵, and perfect.

Assume that 𝝅𝝅 and 𝝈𝝈 have the same range.Construct 𝒔𝒔 ∈ A 𝐵𝐵1,𝜋𝜋 and 𝒇𝒇 ∈ A(𝐵𝐵2,𝜎𝜎).

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Then, 𝒔𝒔 and 𝒇𝒇 have optimal crosscorrelation if and only if the following conditions are satisfied for each 𝒓𝒓 = 𝟎𝟎,𝟏𝟏, … ,𝑵𝑵− 𝟏𝟏:1) Ψ𝜋𝜋,𝜎𝜎(𝑟𝑟) = 1, i.e., Ψ𝜋𝜋,𝜎𝜎 𝑟𝑟 = 𝑥𝑥 .2) For the unique 𝑥𝑥 ∈ Ψ𝜋𝜋,𝜎𝜎(𝑟𝑟),

the pair of sequences 𝜷𝜷𝒙𝒙+𝒓𝒓 and 𝜸𝜸𝒙𝒙 is optimal.



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Condition on input pairs

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Condition on input pairs

Condition on input

permutations

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when 𝒎𝒎 = 𝟏𝟏• The all-one sequence of length 1 is a trivial perfect sequence.• Therefore,

“the all-one sequence and itself is a (trivial) optimal pair of perfect sequences of length 1”

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when 𝒎𝒎 = 𝟏𝟏• The all-one sequence of length 1 is a trivial perfect sequence.• Therefore,

“the all-one sequence and itself is a (trivial) optimal pair of perfect sequences of length 1”

an optimal 𝑘𝑘-set of generalized Milewski sequences of length 𝑵𝑵𝟐𝟐 exists

if and only if a 𝑘𝑘 × 𝑁𝑁 circular Florentine array exists

• Therefore, for 𝑚𝑚 = 1, we have:

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Example

• For a 4 × 15 circular Florentine array

we have optimal 𝟒𝟒-set of generalized Milewski sequences of length 𝑵𝑵𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟐𝟐 by picking up any single perfect sequence from each and every family

𝜋𝜋1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

𝜋𝜋2 0 7 1 8 2 12 3 11 9 4 13 5 14 6 10

𝜋𝜋3 0 4 11 7 10 1 13 9 5 8 3 6 2 14 12

𝜋𝜋4 0 13 7 2 11 6 14 10 3 5 12 9 1 4 8

A 1 ,𝜋𝜋1 , A 1 ,𝜋𝜋2 , A 1 ,𝜋𝜋3 , and

A 1 ,𝜋𝜋4 .

(Song, 91 and 00)

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Example

• For a 𝟒𝟒 × 𝟏𝟏𝟏𝟏 circular Florentine array

we have optimal 𝟒𝟒-set of generalized Milewski sequences of length 𝑵𝑵𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟐𝟐 by picking up any single perfect sequence from each and every family

• New, in the sense of size 4 (previously known only of size 2) for length 𝟏𝟏𝟏𝟏𝟐𝟐 or 𝟏𝟏𝟏𝟏

𝜋𝜋1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

𝜋𝜋2 0 7 1 8 2 12 3 11 9 4 13 5 14 6 10

𝜋𝜋3 0 4 11 7 10 1 13 9 5 8 3 6 2 14 12

𝜋𝜋4 0 13 7 2 11 6 14 10 3 5 12 9 1 4 8

A 1 ,𝜋𝜋1 , A 1 ,𝜋𝜋2 , A 1 ,𝜋𝜋3 , and

A 1 ,𝜋𝜋4

(Song, 91 and 00)

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when 𝒎𝒎 > 𝟏𝟏

𝝅𝝅𝟏𝟏 0 1 2 3 4

𝝅𝝅𝟐𝟐 0 2 4 1 3Assume we have an optimal pair 𝜷𝜷,𝜸𝜸and a 𝟐𝟐 × 𝟏𝟏 circular Florentine array:

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𝝅𝝅𝟏𝟏 0 1 2 3 4

𝝅𝝅𝟐𝟐 0 2 4 1 3

1) Ψ𝜋𝜋,𝜎𝜎(𝑟𝑟) = 1, i.e., Ψ𝜋𝜋,𝜎𝜎 𝑟𝑟 = 𝑥𝑥 .2) For the unique 𝑥𝑥 ∈ Ψ𝜋𝜋,𝜎𝜎(𝑟𝑟), the pair of sequences 𝜷𝜷𝒙𝒙+𝒓𝒓 and 𝜸𝜸𝒙𝒙 is optimal.

Assume we have an optimal pair 𝜷𝜷,𝜸𝜸and a 𝟐𝟐 × 𝟏𝟏 circular Florentine array:

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• Construct 𝑠𝑠 ∈ A 𝐵𝐵1,𝜋𝜋1 , 𝑓𝑓 ∈ A 𝐵𝐵2,𝜋𝜋2 with𝐵𝐵1 = 𝛽𝛽0,𝛽𝛽1, … ,𝛽𝛽𝑁𝑁−1 and 𝐵𝐵2 = 𝛾𝛾0, 𝛾𝛾1, … , 𝛾𝛾𝑁𝑁−1 , where

𝝅𝝅𝟏𝟏 0 1 2 3 4

𝝅𝝅𝟐𝟐 0 2 4 1 3

𝜷𝜷0𝜷𝜷1

𝜷𝜷4

⋮

= 𝜷𝜷= 𝜷𝜷

= 𝜷𝜷

𝜸𝜸0𝜸𝜸1

𝜸𝜸4

⋮

= 𝜸𝜸= 𝜸𝜸

= 𝜸𝜸



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𝝅𝝅𝟏𝟏 0 1 2 3 4

𝝅𝝅𝟐𝟐 0 2 4 1 3

𝜷𝜷0𝜷𝜷1

𝜷𝜷4

⋮


= 𝜷𝜷

𝜸𝜸0𝜸𝜸1

𝜸𝜸4

⋮


= 𝜸𝜸



Then, any 𝑠𝑠 ∈ A 𝐵𝐵1,𝜋𝜋1and 𝑓𝑓 ∈ A 𝐵𝐵2,𝜋𝜋2 is an

optimal pair

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𝝅𝝅𝟏𝟏 0 1 2 3 4

𝝅𝝅𝟐𝟐 0 2 4 1 3

𝜷𝜷0𝜷𝜷1

𝜷𝜷4

⋮


= 𝜷𝜷

𝜸𝜸0𝜸𝜸1

𝜸𝜸4

⋮


= 𝜸𝜸



𝜷𝜷0𝜷𝜷1

𝜷𝜷4

⋮

= 𝜸𝜸= 𝜷𝜷

= 𝜷𝜷

𝜸𝜸0𝜸𝜸1

𝜸𝜸4

⋮

= 𝜷𝜷= 𝜸𝜸

= 𝜸𝜸

Then, any 𝑠𝑠 ∈ A 𝐵𝐵1,𝜋𝜋1and 𝑓𝑓 ∈ A 𝐵𝐵2,𝜋𝜋2 is an

optimal pair

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𝜋𝜋1 0 1 2 3 4

𝜋𝜋2 0 2 4 1 3

Definition. Let 𝝅𝝅,𝝈𝝈 be two functions from ℤ𝑵𝑵 to ℤ𝒎𝒎𝑵𝑵. 𝜳𝜳𝝅𝝅,𝝈𝝈 𝒓𝒓 = 𝒙𝒙 ∈ ℤ𝑵𝑵 𝝅𝝅 𝒙𝒙 + 𝒓𝒓 ≡ 𝝈𝝈 𝒙𝒙 𝒎𝒎𝒎𝒎𝒎𝒎 𝑵𝑵 .

𝜳𝜳𝟏𝟏,𝟐𝟐 𝒓𝒓 = 𝒙𝒙 ∈ ℤ𝑵𝑵 𝝅𝝅𝟏𝟏 𝒙𝒙 + 𝒓𝒓 ≡ 𝝅𝝅𝟐𝟐 𝒙𝒙 𝒎𝒎𝒎𝒎𝒎𝒎 𝟏𝟏 ↔ 𝜷𝜷𝒙𝒙+𝒓𝒓 and 𝜸𝜸𝒙𝒙

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𝜋𝜋1 0 1 2 3 4

𝜋𝜋2 0 2 4 1 3


𝛹𝛹1,2 𝟎𝟎 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟎𝟎 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 0 ↔ 𝛽𝛽0+𝟎𝟎 = 𝛽𝛽𝟎𝟎 and 𝛾𝛾𝟎𝟎𝛹𝛹1,2 𝟏𝟏 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟏𝟏 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 2 ↔ 𝛽𝛽2+𝟏𝟏 = 𝛽𝛽𝟑𝟑 and 𝛾𝛾𝟐𝟐𝛹𝛹1,2 𝟐𝟐 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟐𝟐 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 4 ↔ 𝛽𝛽4+𝟐𝟐 = 𝛽𝛽𝟏𝟏 and 𝛾𝛾𝟒𝟒𝛹𝛹1,2 𝟑𝟑 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟑𝟑 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 1 ↔ 𝛽𝛽1+𝟑𝟑 = 𝛽𝛽𝟒𝟒 and 𝛾𝛾𝟏𝟏𝛹𝛹1,2 𝟒𝟒 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟒𝟒 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 3 ↔ 𝛽𝛽3+𝟒𝟒 = 𝛽𝛽𝟐𝟐 and 𝛾𝛾𝟑𝟑


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𝜋𝜋1 0 1 2 3 4

𝜋𝜋2 0 2 4 1 3


𝛹𝛹1,2 𝟎𝟎 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟎𝟎 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 0 ↔ 𝛽𝛽0+𝟎𝟎 = 𝛽𝛽𝟎𝟎 and 𝛾𝛾𝟎𝟎𝛹𝛹1,2 𝟏𝟏 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟏𝟏 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 2 ↔ 𝛽𝛽2+𝟏𝟏 = 𝛽𝛽𝟑𝟑 and 𝛾𝛾𝟐𝟐𝛹𝛹1,2 𝟐𝟐 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟐𝟐 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 4 ↔ 𝛽𝛽4+𝟐𝟐 = 𝛽𝛽𝟏𝟏 and 𝛾𝛾𝟒𝟒𝛹𝛹1,2 𝟑𝟑 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟑𝟑 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 1 ↔ 𝛽𝛽1+𝟑𝟑 = 𝛽𝛽𝟒𝟒 and 𝛾𝛾𝟏𝟏𝛹𝛹1,2 𝟒𝟒 = 𝑥𝑥 ∈ ℤ𝑁𝑁 𝜋𝜋1 𝑥𝑥 + 𝟒𝟒 ≡ 𝜋𝜋2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑑𝑑 5 = 3 ↔ 𝛽𝛽3+𝟒𝟒 = 𝛽𝛽𝟐𝟐 and 𝛾𝛾𝟑𝟑


(𝜷𝜷0,

(𝜷𝜷4,

𝜸𝜸0) =

𝜸𝜸1) =

(𝜷𝜷,𝜸𝜸) or (𝜸𝜸,𝜷𝜷)

(𝜷𝜷3, 𝜸𝜸2) =

(𝜷𝜷,𝜸𝜸) or (𝜸𝜸,𝜷𝜷)(𝜷𝜷,𝜸𝜸) or (𝜸𝜸,𝜷𝜷)(𝜷𝜷,𝜸𝜸) or (𝜸𝜸,𝜷𝜷)(𝜷𝜷,𝜸𝜸) or (𝜸𝜸,𝜷𝜷)

(𝜷𝜷2, 𝜸𝜸3) =(𝜷𝜷1, 𝜸𝜸4) =

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Maximum set size

• Let 𝑭𝑭𝒄𝒄 𝑵𝑵 be the maximum such that an 𝑭𝑭𝒄𝒄 𝑵𝑵 × 𝑵𝑵circular Florentine array exists.

• Let 𝑶𝑶𝑮𝑮(𝒎𝒎𝑵𝑵𝟐𝟐) be the maximum such that an optimal 𝑶𝑶𝑮𝑮(𝒎𝒎𝑵𝑵𝟐𝟐)-set of generalized Milewski sequences of length 𝒎𝒎𝑵𝑵𝟐𝟐 from perfect sequences of length 𝒎𝒎.

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Maximum set size



1) When 𝑚𝑚 = 1,𝑂𝑂𝐺𝐺 mN2 = 𝑂𝑂𝐺𝐺 N2 = 𝐹𝐹𝑐𝑐 𝑁𝑁 .

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Maximum set size



1) When 𝑚𝑚 = 1,𝑂𝑂𝐺𝐺 mN2 = 𝑂𝑂𝐺𝐺 N2 = 𝐹𝐹𝑐𝑐 𝑁𝑁 .

2) When 𝑚𝑚 ≥ 2, 𝑂𝑂𝐺𝐺 mN2 = min 𝑂𝑂𝑃𝑃 𝑚𝑚 ,𝐹𝐹𝑐𝑐 𝑁𝑁

where 𝑂𝑂𝑃𝑃 𝑚𝑚 is the maximum such that an optimal 𝑂𝑂𝑃𝑃 𝑚𝑚 -set of perfect sequences of period 𝑚𝑚.

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Concluding remarks

• To obtain an optimal 𝒌𝒌-set of generalized Milewski sequences of length 𝒎𝒎𝑵𝑵𝟐𝟐, we need both:▫ A 𝒌𝒌 × 𝑵𝑵 circular Florentine array, and▫ An optimal 𝒌𝒌-set of perfect sequences of length 𝒎𝒎.

(When 𝑚𝑚 = 1, a trivial example will work always)

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Concluding remarks

• To obtain an optimal 𝒌𝒌-set of generalized Milewski sequences of length 𝒎𝒎𝑵𝑵𝟐𝟐, we need both:▫ A 𝒌𝒌 × 𝑵𝑵 circular Florentine array, and▫ An optimal 𝒌𝒌-set of perfect sequences of length 𝒎𝒎.

(When 𝑚𝑚 = 1, a trivial example will work always)

Some open problems:• Find any other positive odd integer 𝑵𝑵 such that 𝑭𝑭𝒄𝒄(𝑵𝑵)

is greater than 𝒑𝒑𝒎𝒎𝒎𝒎𝒏𝒏 − 𝟏𝟏.• Determine the exact value of 𝑭𝑭𝒄𝒄(𝑵𝑵) for every odd 𝑵𝑵.

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Thanks !

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Generalized Milewski sequences with perfect