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
.
.
.
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 2/25
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
.
.
.
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 2/25
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
.
.
.
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 2/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 3/25
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 - - -
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 4/25
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!
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 5/25
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)
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 6/25
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)
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 6/25
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)
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 6/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 7/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 7/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 7/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 8/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 8/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 9/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 9/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 9/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 9/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 9/25
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.
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 10/25
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.
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 10/25
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.
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 10/25
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 . . .
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 11/25
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 . . .
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 11/25
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 . . .
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 11/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 12/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 12/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 12/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 12/25
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)
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 13/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 14/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 15/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 15/25
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...
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 16/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 17/25
Length difference for 256-bit integers
0
50
100
150
200
250
300
350
400
450
-5 0 5 10 15 20 25
freq
uenc
y
distance to full-greedy
256-bit integers
depth 2depth 3depth 4
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 18/25
Length difference for 512-bit integers
0
20
40
60
80
100
120
140
160
180
200
-5 0 5 10 15 20 25
freq
uenc
y
distance to full-greedy
512-bit integers
depth 3depth 4
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 19/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 20/25
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)
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 21/25
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)
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 21/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 22/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 23/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 24/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 24/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 24/25
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
Laurent Imbert ARITH – LIRMM, CNRS, Univ. Montpellier 2 25/25
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