Quelques systèmes de numération exotiques (et applications)imbert/talks/laurent_cacao_08.pdfQuelques syst`emes de num´eration exotiques (et applications) Laurent Imbert ARITH –

Quelques systemes de numeration exotiques(et applications)

Laurent Imbert

ARITH – LIRMM, CNRS, Univ. Montpellier 2

Seminaire CACAO, Nancy, fevrier 2008

Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 1/25

The double-base number system

Every integer n > 0 can be written as

n =m∑

i=0

2ai 3bi , ai , bi ≥ 0

Partition of n with distinct partsof the form 2a3b (2, 3-integers)

3-partition of n

n = p0 + 3p1 + 32p2 + · · ·+ 3mpm

Example: n = 23832098195

1

1

11

111

20 21 22 23 24 25 26 . . .

30

31

32

33

34

35

.

.

.




n =m∑

i=0

2ai 3bi , ai , bi ≥ 0


3-partition of n

n = p0 + 3p1 + 32p2 + · · ·+ 3mpm

Example: n = 23832098195

1

1

11

111

20 21 22 23 24 25 26 . . .

30

31

32

33

34

35

.

.

.




n =m∑

i=0

2ai 3bi , ai , bi ≥ 0


3-partition of n

n = p0 + 3p1 + 32p2 + · · ·+ 3mpm

Example: n = 23832098195

1

1

11

111

20 21 22 23 24 25 26 . . .

30

31

32

33

34

35

.

.

.


Properties

Redundancy: # of representations of a given n

f (n) =

f (n − 1) + f (n/3) if n ≡ 0 (mod 3),

f (n − 1) otherwise.

f (3n) = number of partitions of 3n into powers of 3

1, 2, 3, 5, 7, 9, 12, 15, 18, 23, 28, 33, 40, 47, 54, 63, 72, 81, 93, . . .

Sloane’s on-line encyclopedia of integer sequences #A005704


http://www.research.att.com/~njas/sequences/?q=A005704

Properties

Sparseness: # of parts (or length m of the expansion)

m ∈ O(log n/ log log n)

Numerical experiments show that the constant ' 1

Smallest n > 0 requiring m terms:

m unsigned DBNS Binary log n/ log log n

2 5 3 3.383 23 5 2.744 431 9 3.365 18 431 15 4.296 3 448 733 22 5.557 1 441 896 119 31 6.918 - - -


Canonic representations

Representation of minimal length (smallest partitions)

Example: f (127) = 783 among which 6 are canonic

1

1

1

20 21 22

30

31

32

33 1

1

1

20 21 22 23 24

30

31

32

33

1

1

1

20 21 22 23 24 25

30

31

32

33

20 21 22 23

30

31

32

33

1

1

1 1

1

1

20 21 22 23 24 25 26

30

31

32

33

1

1

1

20 21 22 23 24 25 26

30

31

32

33

Canonic representations are extremely difficult to compute!


Conversion: a greedy approach

Input: A positive integer nOutput: The sequence (ai , bi ) s.t. n =

∑i 2

ai 3bi with ai , bi ≥ 01: while n 6= 0 do2: Compute the best default approximation of n of the form z = 2a3b

3: print (a, b)4: n← n − z

Does not produce canonic representations... (41 = 36 + 4 + 1 = 32 + 9)

but satisfies m ∈ O(log n/ log log n)




∑i 2


3: print (a, b)4: n← n − z






∑i 2


3: print (a, b)4: n← n − z




Best default approximation of the form 2a3b

The problem: find a, b ≥ 0 such that

2a3b = max2c3d ≤ n : (c , d) ∈ N2

Equivalently, find a, b ≥ 0 such that, for all c 6= a, d 6= b

c log 2 + d log 3 < a log 2 + b log 3 ≤ log n

c log3 2 + d < a log3 2 + b ≤ log3 n




2a3b = max2c3d ≤ n : (c , d) ∈ N2



c log3 2 + d < a log3 2 + b ≤ log3 n




2a3b = max2c3d ≤ n : (c , d) ∈ N2



c log3 2 + d < a log3 2 + b ≤ log3 n


Geometric interpretation

cα + d < aα + b ≤ log3 n (α = log3 2, = fractional part)

Solutions: points under the line of equation y = −αx + log3 n

x

y

Best approx: (a, b) such that δ(a) = minδ(x) = −αx + log3 n


Geometric interpretation

cα + d < aα + b ≤ log3 n (α = log3 2, = fractional part)

Solutions: points under the line of equation y = −αx + log3 n

x

y

Best approx: (a, b) such that δ(a) = minδ(x) = −αx + log3 n


Continued fractions

I α irrational ; simple infinite CF: α = a0 +1

a1 +1

a2 + · · ·I Partial quotients: a0 = bαc, ai ≥ 1

I Convergents: pn/qn = [a0, a1, a2, . . . , an] −→ α

I

∣∣∣∣α− pn

qn

∣∣∣∣ ≤ 1

q2n

Ip0

q0= 0,

p1

q1= 1, . . . ,

pn+1

qn+1=

an+1pn + pn−1

an+1qn + qn−1

I even convergents < α < odd convergents


Continued fractions


a1 +1



I

∣∣∣∣α− pn

qn

∣∣∣∣ ≤ 1

q2n

Ip0

q0= 0,

p1

q1= 1, . . . ,

pn+1

qn+1=

an+1pn + pn−1

an+1qn + qn−1



Continued fractions


a1 +1



I

∣∣∣∣α− pn

qn

∣∣∣∣ ≤ 1

q2n

Ip0

q0= 0,

p1

q1= 1, . . . ,

pn+1

qn+1=

an+1pn + pn−1

an+1qn + qn−1



Continued fractions


a1 +1



I

∣∣∣∣α− pn

qn

∣∣∣∣ ≤ 1

q2n

Ip0

q0= 0,

p1

q1= 1, . . . ,

pn+1

qn+1=

an+1pn + pn−1

an+1qn + qn−1



Continued fractions


a1 +1



I

∣∣∣∣α− pn

qn

∣∣∣∣ ≤ 1

q2n

Ip0

q0= 0,

p1

q1= 1, . . . ,

pn+1

qn+1=

an+1pn + pn−1

an+1qn + qn−1



Ostrowski’s number system for integers

Every integer N can be written uniquely in the form

N =m∑

k=1

dkqk−1,

where 0 ≤ dk ≤ ak (d1 ≤ a1 − 1) and dk = 0 if dk+1 = ak+1

Example 1: α = log3 2 481 = 5q7 + 3q5 + q3 + q1

log3 2 0 1 1 1 2 2 3 1 5 2 . . .

pn 0 1 1 2 5 12 41 53 306 665 . . .

qn 1 1 2 3 8 19 65 84 485 1054 . . .

Example 2: α =1 +√

5

2= [1, 1, 1, 1, . . . ] 481 = F13 + F10 + F6 + F2

(qn): Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, . . . ), Zeckendorf repr.




N =m∑

k=1

dkqk−1,



log3 2 0 1 1 1 2 2 3 1 5 2 . . .

pn 0 1 1 2 5 12 41 53 306 665 . . .

qn 1 1 2 3 8 19 65 84 485 1054 . . .


5

2= [1, 1, 1, 1, . . . ] 481 = F13 + F10 + F6 + F2





N =m∑

k=1

dkqk−1,



log3 2 0 1 1 1 2 2 3 1 5 2 . . .

pn 0 1 1 2 5 12 41 53 306 665 . . .

qn 1 1 2 3 8 19 65 84 485 1054 . . .


5

2= [1, 1, 1, 1, . . . ] 481 = F13 + F10 + F6 + F2



Ostrowski’s number system for reals

Every real number −α ≤ β < 1− α can be written uniquely in the form

β =+∞∑k=1

bkθk−1, (θn)n = (qnα− pn)n

where 0 ≤ bk ≤ ak , (b1 ≤ a1 − 1), bk = 0 if bk+1 = ak+1

and bk 6= ak for infinitely many even and odd integers

Example 3: n = 26831

i ai pi qi θi = qiα− pi

0 0 0 1 0.630931 1 1 1 −0.369072 1 1 2 0.261863 1 2 3 −0.107214 2 5 8 0.047445 2 12 19 −0.012346 3 41 65 0.01043

β = log3 n = 0.281994 . . .

β = θ2 + 0.020135 . . .

β = θ2 + 2θ6 − 0.00073 . . .




β =+∞∑k=1






0 0 0 1 0.630931 1 1 1 −0.369072 1 1 2 0.261863 1 2 3 −0.107214 2 5 8 0.047445 2 12 19 −0.012346 3 41 65 0.01043

β = log3 n = 0.281994 . . .

β = θ2 + 0.020135 . . .

β = θ2 + 2θ6 − 0.00073 . . .




β =+∞∑k=1






0 0 0 1 0.630931 1 1 1 −0.369072 1 1 2 0.261863 1 2 3 −0.107214 2 5 8 0.047445 2 12 19 −0.012346 3 41 65 0.01043

β = log3 n = 0.281994 . . .

β = θ2 + 0.020135 . . .

β = θ2 + 2θ6 − 0.00073 . . .


Back to our problem

Compute the best approximation of n of the form 2a3b

with 0 ≤ a < log2 n, 0 ≤ b < log3 n

I n = 26831, log3 n = 9.281994 . . .

I log3 n = θ2 + 2θ6 + · · ·

I log3 n = (q2 + 2q6 + · · · )α− (p2 + 2p6 + · · · ) + blog3 nc

I 2q23blog3 nc−p2 = 2238 = 26244, ε = 587

I q2 + 2q6 = 132 > blog2 nc = 14


Back to our problem



I n = 26831, log3 n = 9.281994 . . .

I log3 n = θ2 + 2θ6 + · · ·


I 2q23blog3 nc−p2 = 2238 = 26244, ε = 587

I q2 + 2q6 = 132 > blog2 nc = 14


Back to our problem



I n = 26831, log3 n = 9.281994 . . .

I log3 n = θ2 + 2θ6 + · · ·


I 2q23blog3 nc−p2 = 2238 = 26244, ε = 587

I q2 + 2q6 = 132 > blog2 nc = 14


Back to our problem



I n = 26831, log3 n = 9.281994 . . .

I log3 n = θ2 + 2θ6 + · · ·


I 2q23blog3 nc−p2 = 2238 = 26244, ε = 587

I q2 + 2q6 = 132 > blog2 nc = 14


An algorithm for the best left approximation

Consider the sequence (fn)n = |θn|n

Input: Two irrationals 0 < α < 1 and 0 < β ≤ 1Output: The infinite sequence (knα− ln)n≥0 < β1: (k0, l0) := (0, 0)2: while true do3: Compute ni , ci , ei such that β − (kiα− li ) = ci fni + fni+1 + ei

4: if ni is even then5: (ki+1, li+1) := (ki + qni , li + pni )6: else7: (ki+1, li+1) := (ki − ciqni + qni+1, li − cipni + pni+1)


Complexity analysis

For all i ≥ 0, 0 < kiα− li < ki+1α− li+1 < β

For all i ≥ 0, ki+1 ≥ ki + ciqni−1

(1+√

52

)m

√5

− 1

2<

m∑i=0

ciqni−1 < um < blog2 nc < um+1,

Thus, there exists a constant C > 0 such that

m < C log log n


Example

n = 23832098195, blog3 nc = 21, log3 n = 0.7495 . . .

i β − (kiα− li ) ni ci ei ki+1 li+1 ki+1α− li+1 2k321−l

0 0.7495 1 1 0.1186 1 0 0.6309 20920706406

1 0.1186 4 2 0.0114 9 5 0.6784 22039921152

2 0.0712 4 1 0.0114 17 10 0.7258 23219011584

3 0.0237 5 1 0.0009 63 39 0.7486 −

n = 217311 + 27314 + 2738 + 2238 + 2930 + 2231 + 2031


Example

n = 23832098195, blog3 nc = 21, log3 n = 0.7495 . . .

i β − (kiα− li ) ni ci ei ki+1 li+1 ki+1α− li+1 2k321−l

0 0.7495 1 1 0.1186 1 0 0.6309 20920706406

1 0.1186 4 2 0.0114 9 5 0.6784 22039921152

2 0.0712 4 1 0.0114 17 10 0.7258 23219011584

3 0.0237 5 1 0.0009 63 39 0.7486 −

n = 217311 + 27314 + 2738 + 2238 + 2930 + 2231 + 2031


Approximate algorithms

DBNS decompositions of n = 23832098195 at depths 1, 2, 3

1

1

1

111

11

1

1

110 1 2

0123...

1

11

1

1

111

0 1 2 3 4 . . .

0123...

1

1

11

111

0 1 2 3 4 . . .

0123...


Length difference for 128-bit integers

0

100

200

300

400

500

600

700

800

900

1000

-6 -4 -2 0 2 4 6 8 10 12 14

freq

uenc

y

distance to full-greedy

128-bit integers

depth 2depth 3depth 4



0

50

100

150

200

250

300

350

400

450

-5 0 5 10 15 20 25

freq

uenc

y


256-bit integers

depth 2depth 3depth 4



0

20

40

60

80

100

120

140

160

180

200

-5 0 5 10 15 20 25

freq

uenc

y


512-bit integers

depth 3depth 4


2, 3-partition chains

n =∑

i

2ai 3bi , (ai , bi )

Ω(19) = (18, 1), (16, 2, 1), (12, 6, 1), (12, 4, 2, 1)

1

1

20 21 22 23

30

31

32

111

20 21 22 23

30

31

32

11

1

20 21 22 23

30

31

32

1

111

20 21 22 23

30

31

32

k-ary partitions: Euler

Chain partitions: Erdos, Loxton 79


Computing all 2, 3-partition chains

Basic relations:

Ω(n) = Ω∗(n) + 1Ω∗(n − 1)

Ω∗(n) = 2Ω(n/2) ∪ 3Ω(n/3)

= 2(Ω(n/2) \ 3Ω(n/6)

)+ 3Ω(n/3)

Ω(0) = () ou Ω(1) = (1)

Better relations:

Ω(6n + 1) = 1Ω(6n)

Ω(6n − 1) = 12Ω(3n − 1)

Ω(6n + 2) = 2Ω(3n + 1)

Ω(3n) = 3Ω(n) + 1Ω(3n − 1)

Ω(6n + 3) = 12Ω(3n + 1) + 3Ω(2n + 1)

Ω(6n + 4) = 13Ω(2n + 1) + 2Ω(3n + 2)


Computing all 2, 3-partition chains

Basic relations:

Ω(n) = Ω∗(n) + 1Ω∗(n − 1)

Ω∗(n) = 2Ω(n/2) ∪ 3Ω(n/3)

= 2(Ω(n/2) \ 3Ω(n/6)

)+ 3Ω(n/3)

Ω(0) = () ou Ω(1) = (1)Better relations:

Ω(6n + 1) = 1Ω(6n)

Ω(6n − 1) = 12Ω(3n − 1)

Ω(6n + 2) = 2Ω(3n + 1)

Ω(3n) = 3Ω(n) + 1Ω(3n − 1)

Ω(6n + 3) = 12Ω(3n + 1) + 3Ω(2n + 1)

Ω(6n + 4) = 13Ω(2n + 1) + 2Ω(3n + 2)


Representation of Ω(n) with a binary tree

19

18

6

2

1

2

3

5

2

1

2

12

1

3

17

8

4

1

13

2

1

2

2

2

12

1

1


Counting 2, 3-partition chains

0

100

200

300

400

500

600

700

800

900

1000

0 50000 100000 150000 200000 250000 300000 350000

w(n)Max w(n)

1/n Si=1..n-1 w(i)x0.435


Generating 2, 3-partition chains at random

Transitions:2 + 1 = 3

2(2m − 1 + 2m+1) = 3(2m+1 − 1) + 1

b is fixed, a is maximal going down

1 1 −→1

1 −→ 1

1

b is fixed, a is minimal going up

1

−→ 1 1 1

1

−→ 1

Transition graph: symmetric & connected




2(2m − 1 + 2m+1) = 3(2m+1 − 1) + 1


1 1 −→1

1 −→ 1

1


1

−→ 1 1 1

1

−→ 1





2(2m − 1 + 2m+1) = 3(2m+1 − 1) + 1


1 1 −→1

1 −→ 1

1


1

−→ 1 1 1

1

−→ 1



ANTS VIII

ORGANIZERS

INVITED SPEAKERS

PROGRAM COMMITTEE

Banff Centre, Banff, Alberta, CanadaMay 17 – 22, 2008

www.ants.math.ucalgary.ca

Mark Bauer, University of Calgary • Josh Holden, Rose-Hulman Institute • Mike Jacobson, University of Calgary • Renate Scheidler, University of Calgary • Jon Sorenson, Butler University

Igor Shparlinski, Chair, Dan Bernstein, Nils Bruin, Ernie Croot, Andrej Dujella, Steven Galbraith,Florian Hess, Ming-Deh Huang, Jürgen Klüners, Kristin Lauter, Stéphane Louboutin, Florian Luca,

Daniele Micciancio, Victor Miller, Oded Regev, Francesco Sica, Andreas Stein, Arne Storjohann, Tsuyoshi Takagi, Edlyn Teske, Felipe Voloch

Johannes Buchmann, Technical University of Darmstadt • Andrew Granville, University of Montreal • Francois Morain, Ecole Polytechnique • Hugh Williams, University of Calgary

8th Algorithmic Number Theory Symposium


Documents

Quelques systèmes de numération exotiques (et applications)imbert/talks/laurent_cacao_08.pdfQuelques syst`emes de num´eration exotiques (et applications) Laurent Imbert ARITH –