Ruskey_BIRS

8/6/2019 Ruskey_BIRS

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Generating Irreducible Polynomialsover Small Finite Fields

Frank Ruskey1

(Joe Sawada)2

(Gilbert Lee, Kevin Cattell)1

1Department of Computer ScienceUniversity of Victoria, CANADA

2Department of Computer ScienceGuelph University, CANADA

BIRS 2006November 19, 2006
http://find/http://goback/


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The plan

Generating Necklaces and Lyndon WordsGenerating Related Combinatorial ObjectsGenerating Irreducible PolynomialsOptimizing Bit-Wise AdditionApplications of the generator: Tables, Tables, Tables.


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The plan



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The plan



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The plan



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The plan



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Necklaces, Lyndon Words

0 1 0 1 1 1 1 10 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

1 0 1 00 1 1 0

1 1 0 0

1 0 0 11 0 0 0

0 1 0 0

0 0 1 0

Alphabet: k = {0, 1, . . . , k 1}.Rotationally equivalent: .

Necklace: N k (n)def

= { nk | for all }Example: N 2(4) = { 0000, 0001, 0011, 0101, 0111, 1111 }.

Lyndon word: aperiodic necklace.Example: L2(4) = { 0001, 0011, 0111 }


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Generating all Lyndon wordsPrenecklace: prex of some necklace.Example: P 2(4) = N 2(4) { 0010, 0110 }.

Length of longest Lyndon prex of is denoted lyn( ).Example: lyn(0010101100101011001) = |00101011| = 8.

TheoremLet = a1a2 an 1 be a string in P k (n 1) and let p = lyn( ).The string b is in P

k (n) if and only if a

n p b k 1.

Furthermore,

lyn(b ) = p if b = an p n if an p < b k 1.

Proof.

= a1 ap a1 ap a1 ap a1 a j , where j p


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TheoremLet = a1a2 an 1 be a string in P k (n 1) and let p = lyn( ).The string b is in P k (n) if and only if an p b k 1.Furthermore,

lyn(b ) = p if b = an p n if an p < b k 1.

procedure gen( n, p : N );local j : N;begin

if n > N then { PrintIt( n ); return ; }a[n] := a[n p ]; gen( n + 1 , p );for j := a[n p ] + 1 to k 1 do

{ a[n] := j ; gen( n + 1 , n ); }end ;

Recursive version of Fredricksen, Kessler, Maiorana and Duval .


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Sequence type PrintIt(p)Pre-necklaces (P k (N )) Println( a[1..N ] )Lyndon words (Lk (N )) if p = N then Println( a[1..N ] )

Necklaces (N k (N )) if N mod p = 0 then Println( a[1..N ] )De Bruijn sequence if N mod p = 0 then Print( a[1..p ] )

procedure gen( n, p : N );local j : N;begin

if n > N then { PrintIt( n ); return ; }a[n] := a[n p ]; gen( n + 1 , p );for j := a[n p ] + 1 to k 1 do

{ a[n] := j ; gen( n + 1 , n ); }end ;


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Extensions of the basic algorithmCAT = Constant Amortized TimeAlley CAT = experimentally CAT, particularly if the ratiowork(n)/objects(n) is decreasing.

Fixed density necklaces and Lyndon words (CAT).Bracelets (001101 001011).Unlabelled binary necklaces (CAT, described later).Necklaces with forbidden substrings (CAT for Lyndonforbidden substring?).Bases for the n-th homogenous component of the free Liealgebra (amortized O(n) per basis element).

Chord diagrams (Alley CAT).Plane rooted and free trees (CAT, next slide).Gray codes (CAT, [Weston & Vajnovski])


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An algorithm I wish Id discovered... (Sawada)The plane rooted trees with 5 nodes:01234, 01232, 01123, 01122, 01112,

01233, 01223, 01222, 01212, 01111

procedure genRP( t , p : N );local j : N;

beginif t = n then if p | n 1 then { Visit(); return ; }a[t ] := a[t p ]; genRP( t +1 , p );for j := a[t p ] + 1 to a[t 1] + 1 do

{ a[t ] := j ; genRP( t +1 , t ); }end ;


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TheoremLet = a1a2 an 1 be a string in P 2(n 1) and let c = com( ).The string b is in P 2(n) if and only if (i)b P 2(n) and (ii)an c = 0 or b = an c . Furthermore,

com(b ) = p if b = an c n if b = an c = 0 .

procedure genU( t , p , c : N ); local j : N;begin

if t = n then { Visit(); return ; }if a[n c ] = 0 then {

if a[n p ] = 0 then { a[t ] := 0; genU( t +1 , p , n ); }if a[n p ] = 1 then { a[t ] := 1; genU( t +1 , p , c ); }

else { a[t ] := 1; genU( t +1 , t , c ); }} else { a[t ] := 0; genU( t +1 , p , c ); }

end ;


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Open Problem #1

Find a CAT algorithm and/or nice characteriza-tion of unlabeled k -ary necklaces for k > 2.

Note that the (lex least) unlabeled k -ary strings arerestricted-growth strings and are counted by j k

nk .


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Enumeration

The number of Lyndon words

k n =d |n

dLk (d ) Lk (n) =1n

d |n

(d )k n / d

Same as the number of irreducible polynomials overFq (solong a q is a prime power).The number of binary unlabelled Lyndon words

U (n) =1

2nd |n , d odd

(d )2n / d


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Enumeration

The number of Lyndon words

k n =d |n

dLk (d ) Lk (n) =1n

d |n

(d )k n / d

Same as the number of irreducible polynomials overFq (solong a q is a prime power).The number of binary unlabelled Lyndon words

U (n) =1

2nd |n , d odd (d )2

n / d


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Necklace polynomial correspondence (Golomb)

x + x + 1

1 0 1 0

0 1 0 1

5

aa10

2

0 0 1 1

0 1 1 0

1 1 0 0

1 0 0 1

3

a 6a

a12

a9

234x x +x +1 x +x+1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

0

a

7

11a

a

a

a14

13

a

1

2a

4a

8a

x+1

a

15

1 1 1 10 0 0 0

434x +x +x +x+1

Let a be a root of the primitive polynomialx 4 + x + 1.Exponents are taken mod 15 = 24 1Degree 4 irreducible polys correspond to Lyndon words.Primitive polynomials result if gcd(e , 15) = 1.
http://goback/http://find/http://find/http://goback/


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x + x + 1

1 0 1 0

0 1 0 1

5

aa10

2

0 0 1 1

0 1 1 0

1 1 0 0

1 0 0 1

3

a 6a

a12

a9

234x x +x +1 x +x+1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

0

a

7

11a

a

a

a14

13

a

1

2a

4a

8a

x+1

a

15

1 1 1 10 0 0 0

434x +x +x +x+1


kl l l d ( l b)


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x + x + 1

1 0 1 0

0 1 0 1

5

aa10

2

0 0 1 1

0 1 1 0

1 1 0 0

1 0 0 1

3

a 6a

a12

a9

234x x +x +1 x +x+1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

0

a

7

11a

a

a

a14

13

a

1

2a

4a

8a

x+1

a

15

1 1 1 10 0 0 0

434x +x +x +x+1


N kl l i l d (G l b)


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x + x + 1

1 0 1 0

0 1 0 1

5

aa10

2

0 0 1 1

0 1 1 0

1 1 0 0

1 0 0 1

3

a 6a

a12

a9

234x x +x +1 x +x+1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

0

a

7

11a

a

a

a14

13

a

1

2a

4a

8a

x+1

a

15

1 1 1 10 0 0 0

434x +x +x +x+1


N kl l i l d (G l b)


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x + x + 1

1 0 1 0

0 1 0 1

5

aa10

2

0 0 1 1

0 1 1 0

1 1 0 0

1 0 0 1

3

a 6a

a12

a9

234x x +x +1 x +x+1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

0

a

7

11a

a

a

a14

13

a

1

2a

4a

8a

x+1

a

15

1 1 1 10 0 0 0

434x +x +x +x+1


A littl th


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A little theory

a is a primitive root, meaning that{a1, a2, a3, . . . , a15} = F16 \ { 0}.By nite eld theory, each irreducible polynomial (of degree 4)is of the form

(x r 1)(x r 2)(x r 4)(x r 8).

For example, with r = a7.

(x a7)(x a14)(x a13)(x a11 ) = x 4 + x 3 + 1

Main problem in generating irreducible polynomials fromLyndon words: How to compute the above product efficiently?

(x 2 (a7 + a14)x + a7a14)(x a13)(x a11 )

= ( x 2 a1x + a8)(x a13)(x a11 ) = etc.


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Since a is a root of x 4 + x + 1, we have a4 = a + 1 .Elements of F24 are uniquely expressible asc 3a3 + c 2a2 + c 1a + c 0 where c i {0, 1}.Elements of F24 are represented by c 3c 2c 1c 0.

a7 = a3 + a + 1 = 1011a14 = a3 + 1 = 1001a13 = a3 + a2 + 1 = 1101a11 = a3 + a2 + a = 1110

An example computation


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An example computation

Recall a4 = a + 1 so a4 is 0011.(a7)2 = a14; i.e., 10112 = 1001.Simulate naive multiplication

1011 10111011 0101 = 101100011

1011 1110 1011 0111 = 0101000011

1100101 1001Operations done on computer words.

From the code (p = 0011, top bit = 1000):if (t & top_bit) t = ( (t & ~top_bit)


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Open Problem #2

Find a way to make use of past information.Each successive Lyndon word is converted to thecorresponding polynomial from scratch.

Or maybe there is some better approach altogether?

Technical Detail


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Technical Detail

For each n, we need a primitive polynomial of degreen to getthe process started.There were no good tables of primitive polynomials forF4, F7,or F8.

From Jungnickel: Let f be a primitive polynomial of degreenk over Fq . Considered as a polynomial overFq k , the polynomialf splits into k distinct primitive polynomials of degreen.Allows us to write a short Maple program to get primitivepolynomials for smalln; e.g., we computed primitivepolynomials overF4 of degrees 1, 2, . . . , 1000. Previously, wecould only nd a list up ton = 14 (Grieg).

Part of the code for F3


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Part of the code for F3/* A list of primitive polynomials

poly_table[i] contains what x^i is equivalent to.E.g. given primitive polynomial x^3 + 2x + 1:

x^3 ~= x + 2 --> {01,11} = {1,3}*/Poly_GF3 poly_table[MAX+1] = {

/* 0 */ { 0 , 1 },/* 1 */ { 0 , 1 },/* 2 */ { 0 , 3 },/* 3 */ { 1 , 3 },/* 4 */ { 0 , 3 },/* 5 */ { 1 , 3 },/* 6 */ { 0 , 3 },

/* 7 */ { 1 , 5 },/* 8 */ { 0 , 9 },/* 9 */ { 1 ,15 },/* 10 */ { 0 ,11 },/* 11 * / { 1 , 5 } ,/* 12 */ { 2 ,31 },/* 13 * / { 1 , 3 } ,

What about GF(3)?


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What about GF(3)?

F3 = 0 1 2W 1 0 0 1W 2 0 1 0

F3 = 0 1 2W 1 0 0 1W 2 0 1 1

What about GF(3)?


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What about GF(3)?F3 = 0 1 2W 1 0 0 1W 2 0 1 0

Standard encoding

F3 = 0 1 2W 1 0 0 1W 2 0 1 1

Optimum encoding

Input: (X 1, X 2) and (Y 1, Y 2). Output: ( Z 1, Z 2). Temporary:(U 1, U 2), (V 1, V 2).

(U 1, U 2) = ( X 1 Y 1, X 2 Y 2)(V 1, V 2) = ( U 1 Y 2, Y 1 U 2)

(Z 1, Z 2) = ( V 1 U 2, V 2 U 1)

8 operations, depth 3. This circuit from Knuth(http://www-cs-faculty.stanford.edu/ ~knuth/news.html ),Exercise 60, pre-fasicle 7.1.2. Boolean Evaluat ion.

What about GF(3)?
http://www-cs-faculty.stanford.edu/~knuth/news.htmlhttp://www-cs-faculty.stanford.edu/~knuth/news.htmlhttp://www-cs-faculty.stanford.edu/~knuth/news.htmlhttp://www-cs-faculty.stanford.edu/~knuth/news.htmlhttp://find/http://goback/


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What about GF(3)?

F3 = 0 1 2

W 1 0 0 1W 2 0 1 0

Standard encoding

F3 = 0 1 2

W 1 0 0 1W 2 0 1 1

Optimum encoding

Input: (X 1, X 2) and (Y 1, Y 2). Output: ( Z 1, Z 2). Temporary:(T 1, T 2).

(T 1, T 2) = ( X 2 Y 1, X 1 Y 2)(Z 1, Z 2) = ( T 1 T 2, (X 1 T 1) (Y 1 T 2))

6 ops, depth 2. This circuit is optimal.

+ 00 01 10 11


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+ 0 1 20 0 1 21 1 2 0

2 2 0 1

00 01 10 1100 00 01 * 1101 01 11 * 0010 * * * *11 11 00 * 01

Problem Specication to Optimizer:Output 1: 00*101*0****10*0Output 2: 01*111*0****10*1Encode each a Fq in binary using lg q bits.Form a boolean function (with dont cares) with 2k inputsand k outputs.Try to optimize number of operations (gates).

Knuths boolean footprint method. A bottom-up heuristicusing known good circuits. Knuth has exhaustive lists of 4input and 5 input (wow!); both with 1 output.Decomposition approach of Miller and Tomczuk. Splits thecircuit based on some variable.

0 1 2+ 00 01 10 11


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+ 0 1 20 0 1 21 1 2 0

2 2 0 1

00 00 01 * 1101 01 11 * 0010 * * * *11 11 00 * 01



+ 0 1 2 +00 01 10 11


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+ 0 1 20 0 1 21 1 2 0

2 2 0 1

00 00 01 * 1101 01 11 * 0010 * * * *11 11 00 * 01



+ 0 1 2 +00 01 10 11


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+ 0 1 20 0 1 21 1 2 0

2 2 0 1

00 00 01 * 1101 01 11 * 0010 * * * *11 11 00 * 01



F3, F4, F5, F7, F8, Z4, Z6


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3, 4, 5, 7, 8, 4, 6Standard representation: 0 , 1, 2, . . . = 000 , 001, 010, . . . .

F3: 6 ops (standard, 2 nots), 6 ops (non-standard, but no

nots)F5: 25 ops (standard, Knuth), 17 ops (non-standard), best ineither case is unknown.F7: ?? ops (standard, Open Problem), 18 ops (non-standard),

best unknown.F8: 3 ops (standard), optimum?Z4: 3 ops (standard), optimum.Z6: 7 ops (7, = 1+6, representation not unique), optimum?

The operations of Knuths model:

< >

Best known circuit forF5


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Representation used: 0 2 6 3 7 = 000 010 110 011 1117: 02 058: 02 059: 01 0410: 01 0411: 03 0612: 03 0613: 08 0914: 08 0915: 10 1216: 11 1317: 10 11

18: 11 1219: 13 > 1520: 15 1821: 14 20 = OUT122: 07 > 16 = OUT223: 17 19 = OUT3

22 23 21

11 13 15 18 8

1420

10 12

197 16 17

913 10

Using unlabelled Lyndon word generation


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g y g

Let f be a degree n polynomial. Dene f (x ) = x n f (1/ x ).f (x ) is irreducible if and only if f (x ) is irreducible.f (x ) is irreducible if and only if f (x ) is irreducible.The Lyndon words corresponding to f (x ) and f (x ) are unlabelledmates.Consequence: Program can be sped up by a factor of 2.Open Problem: Find a similar time-saving auto-morphism in the cases k > 2.

Exhaustive Lists, Past Work


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Randolph Church, Tables of Irreducible Polynomials for the First Four Prime Moduli , Annals of Mathematics, 36 (1935)

198209. In L&N.J.D. Alanen and D.E. Knuth, Tables of Finite Fields , Sankhya ,The Indian Journal of Statistics, A26 (1964) 305328.Reprinted with new material in D.E. Knuth, Selected Papers on Discrete Mathematics , 2003, Addison-Wesley, pp. 277-304.In L&N.S. Mossige, Table of irreducible polynomials over GF[2] of degree 10 through 20 , Mathematics of Computation, 26(1972) 10071009.

F. Chabaud, Polynomials of Galois Fields , http://fchabaud.free.fr/English/default.php?COUNT=1\&FILE0=Poly .Lots of others.COS: http://theory.cs.uvic.ca/gen/poly.html uses

our algorithms.

Exhaustive Lists, Our Work
http://fchabaud.free.fr/English/default.php?COUNT=1\&FILE0=Polyhttp://fchabaud.free.fr/English/default.php?COUNT=1\&FILE0=Polyhttp://fchabaud.free.fr/English/default.php?COUNT=1\&FILE0=Polyhttp://theory.cs.uvic.ca/gen/poly.htmlhttp://theory.cs.uvic.ca/gen/poly.htmlhttp://fchabaud.free.fr/English/default.php?COUNT=1\&FILE0=Polyhttp://fchabaud.free.fr/English/default.php?COUNT=1\&FILE0=Polyhttp://find/http://goback/


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F2: For n = 1 , 2, . . . , 32 we generated lists of all irreduciblepolynomials. For n = 32 the list has 134, 215, 680 entries. Forn = 33 , 34, . . . , 64 we generated 1,000,000 primitivepolynomials.F3: For n = 1 , 2, . . . , 20 we generated lists of all irreduciblepolynomials. For n = 20 the list has 174, 342, 216 entries. Forn = 21 , 22, . . . , 64 we generated 1,000,000 primitive

polynomials.F4: For n = 1 , 2, . . . , 15 we generated lists of all irreduciblepolynomials. For n = 15 the list has 71, 582, 716 entries. Forn = 16 , 17, . . . , 64 we generated 1,000,000 primitivepolynomials.F5: For n = 1 , 2, . . . , 12 we generated lists of all irreduciblepolynomials. For n = 12 the list has ?? entries.F7: Similar to above, up to n = 9 (4,483,696).F8: Similar to above (3 ops).

F9: Requires 3 words, yet to be done (12 = 6+ 6 ops).

Tables, Statistics


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Over F2; coefficient of x k is 1.


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n \ k 1 2 3 4 5 6 7 82 13 1 14 2 1 25 3 4 4 36

5 4 3 4 57 9 9 9 9 9 98 16 16 18 16 18 16 169 28 26 29 29 29 29 26 28

10 51 51 48 47 49 47 48 5111 93 93 93 96 96 96 96 9312 170 165 170 167 160 169 160 16713 315 320 320 315 323 312 312 32314 585 576 585 569 588 585 621 585

15 1091 1091 1078 1091 1080 1083 1091 109116 2048 2048 2064 2048 2064 2044 2068 204817 3855 3840 3840 3872 3872 3824 3808 379218 7280 7280 7294 7236 7270 7265 7292 729919 13797 13797 13797 13824 13824 13920 13792 1382420 26214 26163 26214 26227 26198 26155 26163 2620321 49929 49978 49905 49960 49782 49906 49929 4985322 95325 95232 95325 95104 95277 95309 95488 9550123 182361 182361 182361 182361 182361 182489 182425 18265624 349520 349520 349690 349456 350074 349408 349408 34942025 671088 670924 670924 670950 670745 671206 670982 67142726 1290555 1290555 1290240 1290491 1290651 1290395 1290496 1290352 127 2485504 2485504 2485960 2486092 2484896 2486008 2486006 2486592 228 4793490 4792905 4793490 4793801 4792786 4792882 4793225 4793618 429 9256395 9256960 9256960 9255371 9256523 9255552 9256384 9255659 930 17895679 17894588 17894041 17894297 17896588 17894651 17895689 17894555 1731 34636833 34636833 34636833 34636833 34636833 34639137 34635553 34636833 34

32 67108864 67108864 67110912 67107328 67110912 67109568 67105984 67108544 67

Over F2; coefficient of x k is 1; known formulae.


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Column 1: If t = 0, then the number of polynomials for which thecoefficient of x n 1 is t is [Carlitz-52]

C I q (n; n 1, t ) =1

qnd |n

(d )q n / d [[p d ]]. (1)

Column 2: The number of trace 0, subtrace 1 irreduciblepolynomials overF2 is

1n

k :k + n 0,3(4) d |gcd( n ,k )

(d )n/ d k / d

[CMRSS] The number of irreducible polynomials over GF (2 ) with given trace and subtrace. JCMCC 47 (2003),3164.

Kuzmin, E. N. A class of irreducible polynomials over a nite eld. Dokl. Akad. Nauk SSSR 313 (1990), no. 3,552555;

Column 3:Yucas, Joseph L.; Mullen, Gary L. Irreducible polynomials over GF (2) with prescribed coefficients. Discrete Math.274 (2004), no. 1-3, 265279.

Fitzgerald, Robert W.; Yucas, Joseph L. Irreducible polynomials over GF (2) with three prescribed coefficients.

Finite Fields Appl. 9 (2003), no. 3, 286299.

Largest power of 2 that divides entries from previous table.n \ k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17


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2 03 0 04 1 0 15 0 2 2 06 0 2 0 2 0

7 0 0 0 0 0 08 4 4 1 4 1 4 49 2 1 0 0 0 0 1 2

10 0 0 4 0 0 0 4 0 011 0 0 0 5 5 5 5 0 0 012 1 0 1 0 5 0 5 0 1 0 113 0 6 6 0 0 3 3 0 0 6 6 014 0 6 0 0 2 0 0 0 2 0 0 6 015 0 0 1 0 3 0 0 0 0 3 0 1 0 016 11 11 4 11 4 2 2 11 2 2 4 11 4 11 1117 0 8 8 5 5 4 5 4 4 5 4 5 5 8 8 018 4 4 1 2 1 0 2 0 1 0 2 0 1 2 1 4 419 0 0 0 9 9 5 5 9 6 6 9 5 5 9 9 0 020 1 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 121 0 1 0 3 1 1 0 0 2 0 0 2 0 0 1 1 322 0 10 0 7 0 0 8 0 0 7 0 7 0 0 8 0 023 0 0 0 0 0 0 0 7 7 4 6 6 4 7 7 0 0

24 4 4 1 4 1 5 5 2 6 3 1 4 1 3 6 2 525 4 2 2 1 0 1 1 0 1 2 1 2 2 1 2 1 026 0 0 12 0 0 0 8 4 2 3 6 0 0 0 6 3 227 8 8 3 2 5 3 1 6 0 0 2 5 3 3 5 2 028 1 0 1 0 1 1 0 1 0 0 1 5 1 0 1 5 129 0 14 14 0 0 7 6 0 0 5 5 0 0 4 4 0 030 0 2 0 0 2 0 0 0 1 1 0 3 1 0 0 0 131 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 032 26 26 11 9 11 6 6 6 8 3 3 5 3 3 3 7 3

Over F2; coefficient of x k is 1.


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Largest power of 2 that divides entries.

ConjectureIf n = 2 m then max power is 2m m 1 and min power is m 2.The max is attained with k = 1 .Note: This is consistent with the known formula for the number of

irreducible polynomials with trace 1.ConjectureIf n = 2 2m 1 1 then all table values are zero.

ConjectureIf n = 2 2m 1 then all non-zero table values are odd.This conjecture is (very) weakly supported, but a counter-examplemay be hard to nd!

Hamming distance to an irreducible polynomial


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n \ d 0 1 2 31 12 1 13 2 24 3 4 1

5 6 8 26 9 16 77 18 32 148 30 63 34 19 56 128 72

10 99 255 157 111 186 510 326 212 335 1020 689 413 630 2032 1418 1614 1161 4048 2935 4815 2182 8109 6010 8316 4080 16216 12304 16817 7710 32434 25058 33418 14532 64731 51004 80519 27594 129597 103478 147520 52377 258718 209767 342621 99858 517424 424430 686422 190557 1034430 858019 1414623 364722 2067780 1732430 2937224 698870 4132038 3495434 6226625 1342176 8262934 7046432 12567426 2580795 16515320 14196421 26189627 4971008 33021972 28583424 53246028 9586395 66029987 57522469 107887729 18512790 132008983 115704938 220874530 35790267 263944002 232645189 449145431 69273666 527772375 467597246 909853732 134215680 1055126462 939526144 18615362

Distance to a degree npolynomial with constant term

1 (row sums = 2n 1

).Leftmost column is irreduciblepolynomials.Is there ever any entry in

column 4?Calculation takes much space;must use packed words. Basicalgorithm is a BFS.How close is this to beingrandom?Rightmost column ( n 17) in Berczes and Hajdu,

Comp ... on the Distances of Polynomials to

Irreducible Polynomials , Mathematics of

Computa tion, 66 (1997) 391398.

Hamming distance to an irreducible polynomial\ d 0 1 2 3


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n \ d 0 1 2 31 12 1 13 2 24 3 4 1

5 6 8 26 9 16 77 18 32 148 30 63 34 19 56 128 72

10 99 255 157 111 186 510 326 212 335 1020 689 413 630 2032 1418 1614 1161 4048 2935 4815 2182 8109 6010 8316 4080 16216 12304 16817 7710 32434 25058 33418 14532 64731 51004 80519 27594 129597 103478 147520 52377 258718 209767 342621 99858 517424 424430 686422 190557 1034430 858019 1414623 364722 2067780 1732430 2937224 698870 4132038 3495434 6226625 1342176 8262934 7046432 12567426 2580795 16515320 14196421 26189627 4971008 33021972 28583424 53246028 9586395 66029987 57522469 107887729 18512790 132008983 115704938 220874530 35790267 263944002 232645189 449145431 69273666 527772375 467597246 909853732 134215680 1055126462 939526144 18615362


1 (row sums = 2n 1





Computa tion, 66 (1997) 391398.

Hamming distance to an irreducible polynomial\ d 0 1 2 3


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n \ d 0 1 2 31 12 1 13 2 24 3 4 1

5 6 8 26 9 16 77 18 32 148 30 63 34 19 56 128 72

10 99 255 157 111 186 510 326 212 335 1020 689 413 630 2032 1418 1614 1161 4048 2935 4815 2182 8109 6010 8316 4080 16216 12304 16817 7710 32434 25058 33418 14532 64731 51004 80519 27594 129597 103478 147520 52377 258718 209767 342621 99858 517424 424430 686422 190557 1034430 858019 1414623 364722 2067780 1732430 2937224 698870 4132038 3495434 6226625 1342176 8262934 7046432 12567426 2580795 16515320 14196421 26189627 4971008 33021972 28583424 53246028 9586395 66029987 57522469 107887729 18512790 132008983 115704938 220874530 35790267 263944002 232645189 449145431 69273666 527772375 467597246 909853732 134215680 1055126462 939526144 18615362


1 (row sums = 2n 1





Computa tion, 66 (1997) 391398.

Hamming distance to an irreducible polynomialn \ d 0 1 2 3


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n \ d 0 1 2 31 12 1 13 2 24 3 4 15 6 8 26 9 16 77 18 32 148 30 63 34 19 56 128 72

10 99 255 157 111 186 510 326 212 335 1020 689 413 630 2032 1418 1614 1161 4048 2935 4815 2182 8109 6010 8316 4080 16216 12304 16817 7710 32434 25058 33418 14532 64731 51004 80519 27594 129597 103478 147520 52377 258718 209767 342621 99858 517424 424430 686422 190557 1034430 858019 1414623 364722 2067780 1732430 2937224 698870 4132038 3495434 6226625 1342176 8262934 7046432 12567426 2580795 16515320 14196421 26189627 4971008 33021972 28583424 53246028 9586395 66029987 57522469 107887729 18512790 132008983 115704938 220874530 35790267 263944002 232645189 449145431 69273666 527772375 467597246 909853732 134215680 1055126462 939526144 18615362


1 (row sums = 2n 1

).Leftmost column is irreduciblepolynomials.Is there ever any entry incolumn 4?

Calculation takes much space;must use packed words. Basicalgorithm is a BFS.How close is this to beingrandom?Rightmost column ( n 17) in Berczes and Hajdu,



Computa tion, 66 (1997) 391398.



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n \ d 0 1 2 31 12 1 13 2 24 3 4 15 6 8 26 9 16 77 18 32 148 30 63 34 19 56 128 72

10 99 255 157 111 186 510 326 212 335 1020 689 413 630 2032 1418 1614 1161 4048 2935 4815 2182 8109 6010 8316 4080 16216 12304 16817 7710 32434 25058 33418 14532 64731 51004 80519 27594 129597 103478 147520 52377 258718 209767 342621 99858 517424 424430 686422 190557 1034430 858019 1414623 364722 2067780 1732430 2937224 698870 4132038 3495434 6226625 1342176 8262934 7046432 12567426 2580795 16515320 14196421 26189627 4971008 33021972 28583424 53246028 9586395 66029987 57522469 107887729 18512790 132008983 115704938 220874530 35790267 263944002 232645189 449145431 69273666 527772375 467597246 909853732 134215680 1055126462 939526144 18615362


1 (row sums = 2n 1





Computa tion, 66 (1997) 391398.



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n \ d 0 1 2 31 12 1 13 2 24 3 4 15 6 8 26 9 16 77 18 32 148 30 63 34 19 56 128 72

10 99 255 157 111 186 510 326 212 335 1020 689 413 630 2032 1418 1614 1161 4048 2935 4815 2182 8109 6010 8316 4080 16216 12304 16817 7710 32434 25058 33418 14532 64731 51004 80519 27594 129597 103478 147520 52377 258718 209767 342621 99858 517424 424430 686422 190557 1034430 858019 1414623 364722 2067780 1732430 2937224 698870 4132038 3495434 6226625 1342176 8262934 7046432 12567426 2580795 16515320 14196421 26189627 4971008 33021972 28583424 53246028 9586395 66029987 57522469 107887729 18512790 132008983 115704938 220874530 35790267 263944002 232645189 449145431 69273666 527772375 467597246 909853732 134215680 1055126462 939526144 18615362


1 (row sums = 2n 1





Computa tion, 66 (1997) 391398.

Hamming distance with a random start setA th t th bit t i g f l gth N h


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Assume that there are m bitstrings of length N chosen u.a.r.The expected number at Hamming distance 1 is

2N 1 1 2N 1 N

m2N 1

m

= 2 N 1 1 2N 1 m

N 2N 1

N

If N = n 1 and m = 2 n / n, then the expected number atHamming distance 1 is asymptotically

2n 2(1 e 4).How good of a t is this to the irreducible polynomial data?

Hamming distance with a random start set, cont.


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red = (1 e 4),green = D 1(n),blue points = m = 2n / n ,blue line = m = L2(n).

red = D 1(n)/ (1 e 4),green = D 1(n)/ exact.

Hamming distance with a random start set, cont.


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red = (1 e 4),green = D 1(n),

blue points = m = 2n / n ,blue line = m = L2(n).

red = D 1(n)/ (1 e 4),green = D 1(n)/ exact.

Hamming distance to a primitive polynomial overF2n \ d 0 1 2 3 4

1 1


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1 12 1 13 2 24 2 4 25 6 8 2

6 6 15 10 17 18 32 148 16 55 48 99 48 124 80 4

10 60 235 196 2111 176 507 336 512 144 843 880 18113 630 2032 1418 1614 756 3799 3340 29715 1800 7957 6392 23516 2048 14254 14336 213017 7710 32434 25058 33418 7776 57966 57760 757019 27594 129597 103478 147520 24000 220359 238144 4178521 84672 509646 439616 1464222 120032 968242 928544 8033423 356960 2064449 1740192 3270324 276480 3311349 3917823 882955 125 1296000 8240656 7092608 14795226 1719900 15626202 15057316 115101427 4202496 32485903 29351936 106852928 4741632 57628193 62367232 948067129 18407808 131952230 115809920 2265498

x 24+ x 23+ x 22+ x 21+ x 20+ x 16+ x 14+ x 12+ x 10+ x 8+ x 4+ x 3+ x 2+ x +1 .

Monic irreducible overF3Degree n and f (1) = k


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Degree n, and f (1) = k .n \ k 2 3 4 5 6 7 8 9 10 11 12 13 14 1

1 1 12 1 0 1 13 0 0 3 4 0 14 0 0 5 8 0 3 2 05 0 0 5 13 0 17 11 0 2 06 0 0 8 20 0 31 31 0 16 9 0 17 0 0 7 21 0 75 80 0 63 55 0 11 08 0 0 11 25 0 111 185 0 210 175 0 66 259 0 0 7 37 0 208 329 0 512 517 0 324 193

10 0 0 12 42 0 313 571 0 1143 1320 0 1131 88111 0 0 14 61 0 471 867 0 2269 3178 0 3577 297112 0 0 10 34 0 509 1417 0 4390 6372 0 9349 937513 0 0 19 98 0 873 1962 0 7659 12586 0 22196 2460214 0 0 26 90 0 1153 2827 0 12744 22613 0 47785 5980715 0 0 22 109 0 1489 3929 0 20603 39596 0 98499 13272816 0 0 25 97 0 1795 5421 0 32145 65206 0 190442 28013417 0 0 19 116 0 2221 7127 0 48542 105406 0 352070 55424618 0 0 28 180 0 3179 9370 0 72076 166030 0 622368 104704719 0 0 26 138 0 3322 11961 0 102304 252586 0 1063155 1900091

n \ k 21 22 23 24 25 26 27 28 29 30 3110 011 0 0 112 0 63 14 0 013 0 727 272 0 13 2 014 0 6449 2889 0 341 75 0 1 115 0 38849 20844 0 4052 1398 0 104 20 0 116 0 189911 117909 0 32460 14230 0 1708 527 0 1617 0 790810 554529 0 203034 105352 0 19909 7423 0 61618 0 2884634 2245467 0 1048172 624332 0 164154 72755 0 947019 0 9463457 8100841 0 4648215 3100091 0 1065621 546067 0 103819

# 1\ # 2 0 1# 1\ # 2 0 1 2

1 0 1

# 1\ # 2 0 1 2 31 0 1 1


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1 21 12 1 0

1 0 12 1 1 03 0 0 0

2 3 0 03 0 3 0 0

4 0 0 0# 1 \ # 2 0 1 2 3 4

1 0 3 02 3 0 2 03 0 4 3 0 04 2 0 0 0 05 1 0 0 0 0

# 1 \ # 2 0 1 2 3 4 51 0 1 4 02 3 0 9 2 03 0 10 9 0 0 04 2 0 2 0 0 05 2 4 0 0 0 06 0 0 0 0 0 0

# 1 \ # 2 0 1 2 3 4 5 61 0 2 3 0 2 12 4 0 10 8 0 03 0 13 21 0 7 0 04 4 0 20 8 0 0 05 5 6 0 0 0 0 06 0 1 0 0 0 0 0

7 1 0 0 0 0 0 0

# 1 \ # 2 0 1 2 3 4 5 6 71 0 1 6 0 6 1 02 3 0 27 16 0 03 0 14 43 0 35 10 0 04 4 0 36 47 0 0 0 05 6 22 0 14 0 0 0 06 0 17 0 0 0 0 07 4 0 0 0 0 0 0 08 0 0 0 0 0 0 0

Number of irreducible polynomials overF3 of degree1, 2, 3, 4, 5, 6, 7 classied according to number of coefficients thatare 1s and 2s, denoted I 3(n; k 1, k 2).

I 3(12; k 1 , k 2) and I 3(13; k 1 , k 2)# \ # 0 1 2 3 4 5 6 7 8 9 10 11 12


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# 1 \ # 2 0 1 2 3 4 5 6 7 8 9 10 11 121 0 2 5 0 46 95 0 84 23 0 12 2 0 151 291 0 644 528 0 79 28 0 03 0 15 156 0 1325 1743 0 1021 488 0 13 0 04 8 0 639 1731 0 3471 2621 0 355 35 0 0 05 17 280 0 2913 4385 0 2224 578 0 0 0 0 06 0 525 1824 0 3954 2602 0 146 0 0 0 0 07 68 0 1779 2621 0 922 196 0 0 0 0 0 08 102 527 0 1187 694 0 0 0 0 0 0 0 09 0 297 472 0 97 0 0 0 0 0 0 0 0

10 17 0 117 29 0 0 0 0 0 0 0 0 011 12 33 0 0 0 0 0 0 0 0 0 0 012 0 2 0 0 0 0 0 0 0 0 0 0 013 0 0 0 0 0 0 0 0 0 0 0 0

# 1 \ # 2 0 1 2 3 4 5 6 7 8 9 10 11 121 0 3 17 0 116 135 0 111 116 0 10 32 9 0 182 393 0 1092 1150 0 393 103 0 23 0 41 268 0 2031 3315 0 3172 1725 0 119 10 04 10 0 827 2811 0 7721 7148 0 1840 439 0 0 05 54 436 0 5605 9537 0 8009 3829 0 143 0 0 06 0 807 3326 0 10624 9537 0 1666 185 0 0 0 07 152 0 3942 7348 0 5275 1866 0 0 0 0 0 08 146 1059 0 4595 3754 0 249 0 0 0 0 0 09 0 842 1727 0 1017 229 0 0 0 0 0 0 0

10 70 0 555 467 0 0 0 0 0 0 0 0 011 50 130 0 66 0 0 0 0 0 0 0 0 012 0 15 12 0 0 0 0 0 0 0 0 0 013 4 0 0 0 0 0 0 0 0 0 0 0 014 0 0 0 0 0 0 0 0 0 0 0 0

Conjectures for F3, the I numbers.Since there are no self-reciprocal polynomials of odd degree the


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Since there are no self-reciprocal polynomials of odd degree, thenumber of monic irreducible polynomials overF3 of odd degreewhose coefficients are all 1s is even; i.e.,I 3(2m + 1; k 1, 0) is evenfor all values of m and k 1.

ConjectureThe number I 3(n; 4, 0) is even. If n 4 then I 3(n; 4, 0) 2.For n = 1 , 2, . . . , 19 the sequence of I

3(n; 4, 0) numbers is

0, 0, 0, 2, 2, 4, 4, 6, 4, 10, 8, 8, 10, 22, 16, 18, 16, 24, 20.

We can certainly extend this sequence!

ConjectureThe sequence of non-zero values in any row is unimodal.

ConjectureThe sequence of non-zero values in any column is unimodal.

C I 3(13; s , t ) and C I 3(14; s , t )


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t \ s 12 11 10 9 8 7 6 50 40880 40992 40880 41042 40992 40992 40992 40992 41 40880 40824 40880 40774 40824 40986 40824 40824 42 40880 40824 40880 40824 40824 40662 40824 40824 4

Number of irreducible polynomials overF3 of degree 13 classiedaccording to number that have coefficient s equal to t .

t \ p 13 12 11 10 9 8 70 113828 113724 113828 113744 114044 113346 113714 11 113828 113882 113828 113838 113720 113670 113885 12 113828 113878 113828 113902 113720 114468 113885 1

Number of irreducible polynomials overF3 of degree 14 classiedaccording to number that have coefficient s equal to t .

Conjectures for F3, the C I numbers, when equal?If n and k have different parity, then C I 3(n; k , 1) = C

I 3 (n; k , 2).


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If 1 k n 1, then C I 3 (n; k , 0) = C I 3 (n; n k , 0).ConjectureIf k is odd and 1 k n 1, then C I 3 (n; k , 1) = C

I 3 (n; n k , 1)

and C I 3 (n; k , 2) = C I 3 (n; n k , 2).

The lemma and conjectures below are concerned with those n andk for which

C I 3 (n; k , 0) = C I 3(n; k , 1) = C I 3 (n; k , 2).If the values of n and k are not covered by the lemma or theconjectures then we believe that one of the equalities above fails,except for some spurious cases whenk = n 3.

Lemma: If n 1, 2 mod 3, thenC I 3 (n; n 1, 0) = C

I 3 (n; n 1, 1) = C

I 3(n; n 1, 2).

ConjectureIf n 2, 4 mod 6, then C I 3 (n; 1, 0) = C

I 3 (n; 1, 1) = C

I 3 (n; 1, 2).

If n 5 mod 6 then CI 3 (n; 2 0) = C

I 3 (n; 2 1) = C

I 3 (n; 2 2)

Primitive polynomials:P (12; k 1 , k 2) and P (13; k 1 , k 2)# 1 \ # 2 0 1 2 3 4 5 6 7 8 9 10 11 12


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# 1 \ # 2 0 1 2 3 4 5 6 7 8 9 10 11 121 0 2 0 26 60 0 50 14 0 2 0 80 132 0 327 289 0 34 15 0 03 0 2 80 0 544 771 0 515 255 0 7 0 04 0 132 544 0 1391 1093 0 142 19 0 0 05 26 0 771 1391 0 841 245 0 0 0 0 06 0 60 327 0 1093 841 0 60 0 0 0 0 07 0 289 515 0 245 60 0 0 0 0 0 08 50 0 255 142 0 0 0 0 0 0 0 09 0 14 34 0 19 0 0 0 0 0 0 0 0

10 0 15 7 0 0 0 0 0 0 0 0 011 0 0 0 0 0 0 0 0 0 0 012 0 0 0 0 0 0 0 0 0 0 0 013 0 0 0 0 0 0 0 0 0 0 0 0

# 1 \ # 2 0 1 2 3 4 5 6 7 8 9 10 11 12 131 0 0 0 0 02 6 0 56 66 0 134 164 0 52 8 0 03 0 24 142 0 682 902 0 718 330 0 18 2 0 04 10 0 500 1462 0 2938 2388 0 464 112 0 0 0 05 54 320 0 3192 4754 0 3236 1436 0 52 0 0 0 06 0 672 2368 0 5864 4764 0 698 62 0 0 0 0 07 152 0 2956 4894 0 2882 898 0 0 0 0 0 0 08 146 948 0 3200 2378 0 126 0 0 0 0 0 0 09 0 726 1386 0 690 138 0 0 0 0 0 0 0 0

10 70 0 460 366 0 0 0 0 0 0 0 0 0 011 50 120 0 58 0 0 0 0 0 0 0 0 0 012 0 12 10 0 0 0 0 0 0 0 0 0 0 013 4 0 0 0 0 0 0 0 0 0 0 0 0 014 0 0 0 0 0 0 0 0 0 0 0 0 0

Conjectures concerning P (n ; k 1 , k 2)


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Some properties of the tables are explained by the facts that every

even degree polynomial has constant term 2, and every odd degreepolynomial has constant term 1 (norm must be primitive). Also,there are no self-reciprocal primitive polynomials.

Thus, for all odd n 3 the number P 3(n; k 1, k 2) is even.

Also, if n is even, then I 3(n; k 1, k 2) = I 3(n; k 2, k 1).It would be nice to have some conjectures other than the twobelow...

ConjectureThe sequence of non-zero values in any row is unimodal.The sequence of non-zero values in any column is unimodal.

The End


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Thanks for coming.

Please let me know if youve seen any of this before or can solvethe conjectures!

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