work in progress journées de numération – tu graz – 20th ... · then the amortized cost of...
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ON THE AMORTIZED COST OF AN ODOMETER
V. Berthé (LIRMM/CNRS), C. Frougny (LIAFA/Univ. Paris 8),M. Rigo (Univ. Liège), J. Sakarovitch (ENST/CNRS)
Work in progress...Journées de Numération – TU Graz – 20th April 2007
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So far, different aspects of odometers have been studied :
Combinatorial, Metrical, Topological, Dynamics,Sequential properties, . . .
G. Barat, T. Downarowicz, C. Frougny, P. Grabner, P. Liardet,R. Tichy, A. M. Vershik, . . .
OUR MAIN QUESTION
What is the cost / complexity in average for computingthe odometer (i.e., successor map) on finite words,e.g. on integer representations ?
n −→ rep(n) ∈ Σ∗
↓ ↓n + 1 −→ rep(n + 1) ∈ Σ∗
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WHERE DOES IT COME FROM?
WORDS’05
E. Barcucci, R. Pinzani, M. Poneti, Exhaustive generation ofsome regular languages by using numeration systems.
For numeration systems built on some linear recurrentsequences of order 2, the “amortized cost” for computingrep(n + 1) from rep(n) is bounded by a constant (CAT).
J. SAKAROVITCH , ELTS. DE THÉORIE DES AUTOMATES’03
For any rational set R of A∗, the odometer on R is asynchronized function.
i.e., letter-to-letter (left or right) finite transducer with a terminalfunction appending values of the form (u, ε) or (ε, v)
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More than synchronized functions, we will often assume thatwe have a (right) sequential transducer to do the computation.
A transducer T is sequential if◮ T has a unique initial state,◮ the underlying input automaton is deterministic.
1/0 1/0
1/10/0
0/10/1
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More than synchronized functions, we will often assume thatwe have a (right) sequential transducer to do the computation.
A transducer T is sequential if◮ T has a unique initial state,◮ the underlying input automaton is deterministic.
1 1
00
01
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Usual binary system
A (TRIVIAL ) SEQUENTIAL FUNCTION
1/0 1/0
1/10/0
0/10/1
1010011110101000
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DEFINITION (COST)
We define the cost for computing rep(n + 1) from rep(n) as◮ the position up to where the carry propagates, or◮ the length of the path lying in the “transient part”,◮ for an integer base system, the number of changed digits.
1000100110101011
111 110 101 100 11 10 1
1/0 1/0
0/10/1
1/10/0
10101010
01111000
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ALTERNATIVE DEFINITION (COST)
Another interpretation for the cost in the lexicographic tree :◮ half of the distance between rep(n) and rep(n + 1)
◮ distance to the common ancestor of rep(n) and rep(n + 1)
10000 10001 10010 10100 10101
1000 1001 1010
100 101
1
10
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ALTERNATIVE DEFINITION (COST)
Another interpretation for the cost in the lexicographic tree :◮ half of the distance between rep(n) and rep(n + 1)
◮ distance to the common ancestor of rep(n) and rep(n + 1)
1000 1001 1010
100 101
1
10
10000 10001 10010 10100 10101
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So, cost can be expressed mainly on words
uav −→ ubv ′, a 6= b, |v | = |v ′|
cost = |av |
Let us introduce a different notion (computational aspects)
DEFINITION (COMPLEXITY)
The (algorithmic) complexity for computing rep(n + 1) fromrep(n) is the minimum number of operations required toperform this computation (in the sense of a Turing machine).
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REMARK
Consider a numeration system such that the odometer can berealized by a letter-to-letter (right) sequential transducer.
In that case, the cost is equal to the (algorithmic) complexity.
Indeed, it is not possible to do less computations,the Turing machine at least has to read the digits up to wherethe carry propagates
0 # ## 1 0 01
“cost ≤ complexity”
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COST 6= COMPLEXITY
X 2 − 3X + 1, β = 3+√
52 , dβ(1) = 21ω, (Un)n≥0 = 1, 3, 8, 21, . . .
rep(N) = {ε, 1, 2, 10, 11, 12, 20, 21, 100, 101, 102, . . .}
forbidden factors : 2 1∗ 2
100111111 → 100111112 but 102111111 → 110000000
10
0
0 1 2
1
1
1
1
1
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COST 6= COMPLEXITY
X 2 − 3X + 1, β = 3+√
52 , dβ(1) = 21ω, (Un)n≥0 = 1, 3, 8, 21, . . .
rep(N) = {ε, 1, 2, 10, 11, 12, 20, 21, 100, 101, 102, . . .}
forbidden factors : 2 1∗ 2
100111111 → 100111112 but 102111111 → 110000000
10
0
0 1 2
1
1
1
1
1
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COST 6= COMPLEXITY
X 2 − 3X + 1, β = 3+√
52 , dβ(1) = 21ω, (Un)n≥0 = 1, 3, 8, 21, . . .
rep(N) = {ε, 1, 2, 10, 11, 12, 20, 21, 100, 101, 102, . . .}
forbidden factors : 2 1∗ 2
100111111 → 100111112 but 102111111 → 110000000
10
0
0 1 2
1
1
1
1
1# 1 0 0 1 1 1 1 #1 1
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All this work on cost/complexity can be done in a general setting
DEFINITION
An abstract numeration system is a triple S = (L, A, <) whereL is a infinite (rational) language over a totally ordered alphabet(A, <).
The representation of n ∈ N is the (n + 1)-st word in thegenealogically (i.e., radix) ordered language L.
EXAMPLE
L = {(ab), (ac)}∗, a < b < c
0 1 2 3 4 5 6 7 · · ·
ε ab ac abab abac acab acac ababab · · ·
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EXAMPLE CONTINUES... AMORTIZED COST / COMPLEXITY
ε → 1 → 10 → 11 → 100 → 101 → 110 → 111 →1 2 1 3 1 2 1 4
In base k , kn words from ε to
n︷ ︸︸ ︷
(k − 1) · · · (k − 1),
kn + kn−1 + · · · + 1kn =
k − k−n
k − 1→
kk − 1
, as n → ∞
DEFINITION (AMORTIZED COST)
limn→∞
∑
w∈L,|w |≤n
cost(w)
/
#{w ∈ L : |w | ≤ n}
Same for amortized complexity
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EXAMPLE CONTINUES... AMORTIZED COST / COMPLEXITY
ε → 1 → 10 → 11 → 100 → 101 → 110 → 111 →1 2 1 3 1 2 1 4
In base k , kn words from ε to
n︷ ︸︸ ︷
(k − 1) · · · (k − 1),
kn + kn−1 + · · · + 1kn =
k − k−n
k − 1→
kk − 1
, as n → ∞
DEFINITION (AMORTIZED COST)
limn→∞
∑
w∈L,|w |≤n
cost(w)
/
#{w ∈ L : |w | ≤ n}
Same for amortized complexity
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FIRST EXERCISE...
For Fibonacci system. . .
ε → 1 → 10 → 100 → 101 → 1000 → 1010 →1 2 3 1 4 2 5
amortized cost = amortized complexity →τ
τ − 1≃ 2.618
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FIRST EXERCISE...
For Fibonacci system. . .
ε → 1 → 10 → 100 → 101 → 1000 → 1010 →1 2 3 1 4 2 5
amortized cost = amortized complexity →τ
τ − 1≃ 2.618
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THEOREM
Let L be a rational language having M as trim minimalautomaton.
If the adjacency matrix M of M is primitive with β > 1 asdominating Perron eigenvalue and if all states of M are final,then the amortized cost of the odometer on L is β
β−1 .
REMARK
◮ If the corresponding transducer is right sequential, thenthis is exactly the amortized (algorithmic) complexity.
◮ Otherwise, we get information on the average position upto where some change can occur. (More ?)
REMARK
All states final means L is prefix closed.
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PERRON THEORY
Let M be a d × d primitive matrix having β > 1 as dominatingeigenvalue. The following holds
∀i , j ∈ {0, . . . , d − 1}, ∃cij > 0 : (Mn)ij = cij βn + o(βn).
If x (resp. y) is a left 1 × d (resp. right d × 1) eigenvector of Mof eigenvalue β such that x.y = 1 then ∀0 ≤ i , j < d ,
cij = yi xj , i .e., limn→∞
Mn
βn = y.x.
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If w = pas is such that◮ q0.p = qj ,◮ a 6= max Aqj
◮ s ∈ max(Lq0.pa)
Fix qj ∈ Qpas −→ pbt , |s| = |t |
p a sj
n−k k
n−1∑
k=0
(Mn−k )0j (deg+(qj) − 1) k .
Then sum over Q. . .
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APPLICATIONS
RESULT
Let β > 1 be a Parry number. The amortized cost of theodometer for the canonical linear numeration systemassociated with β is β/(β − 1).
Same remark : cost = complexity when assuming that theodometer is realized with a right sequential transducer.
C. FROUGNY ’97
For such β-numeration systems (β being a Parry number), wehave
◮ a right sequential transducer in the finite type,◮ but NOT in the sofic case.
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simple Parry number
t t
0,...,t −11
1 2
0,...,t −12
0,...,t −1
q1 q2 q
m
tm−1m
non-simple case
t t
t
0,...,tN+p
0,...,t −11
1 2 tN−1
0,...,t −12
0,...,t −1N
N+p
N+1
tN+1
−1
0,...,t −1
q1 q2 Nq
qN+1
qN+p
N+p−1t
Nt
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APPLICATIONS
RESULT
Let S = (L, A, <) be an abstract numeration system built on arational language whose trim minimal automaton M is primitiveand has only final states. If β is the dominating eigenvalue ofM then the amortized cost of the odometer for S is β/(β − 1).
Same remark : cost = complexity when assuming that theodometer is realized with a right sequential transducer.
NEXT STEP, EASY TO HANDLE
Consider several primitive strongly connected components. . .
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Let’s have a look at the lexicographic tree
FIBONACCI WORDS OF LENGTH5
10000 10001 10010 10100 10101
1000 1001 1010
100 101
1
10
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Let’s have a look at the lexicographic tree
FIBONACCI WORDS OF LENGTH5
10000 10001 10010 10100 10101
1000 1001 1010
100 101
1
10
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Let’s have a look at the lexicographic tree
FIBONACCI WORDS OF LENGTH5
10000 10001 10010 10100 10101
1000 1001 1010
100 101
1
10
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Let’s have a look at the lexicographic tree
FIBONACCI WORDS OF LENGTH5
10000 10001 10010 10100 10101
1000 1001 1010
100 101
1
10
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Let’s have a look at the lexicographic tree
FIBONACCI WORDS OF LENGTH5
10000 10001 10010 10100 10101
1000 1001 1010
100 101
1
10
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Let’s have a look at the lexicographic tree
FIBONACCI WORDS OF LENGTH5
10000 10001 10010 10100 10101
1000 1001 1010
100 101
1
10
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Consequently, the total cost for all words of length n is
Cn := #{edges in Tn} + 1 = #{leaves in Tn} = #(L ∩ Σ≤n)
“Nice” hypothesis :◮ L is a prefix closed language (uv ∈ L ⇒ u ∈ L)◮ Any branch in the tree is infinite
REMARK
If uL : n 7→ #(L ∩ Σn) has a “nice asymptotic behavior”,then the amortized cost can be computed. . .
limn→∞
∑ni=0 Ci
#(L ∩ Σ≤n)= lim
n→∞
∑ni=0
∑ik=0 uL(k)
∑ni=0 uL(i)
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QUESTION
Can we compute the amortized complexity if there is nosequential transducer behind ?
If L is rational, when can the odometer be computedwith a right sequential transducer ? (local automaton)
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p/q-BASE (S. AKIYAMA , C. FROUGNY, J. SAKAROVITCH )
p > q ≥ 1 coprime integers,
N =k∑
i=0
ai
q
(pq
)i
, 0 ≤ ai < p
21011 21200 21202 21221
212221202101
212210
21
2
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p/q-BASE (S. AKIYAMA , C. FROUGNY, J. SAKAROVITCH )
p > q ≥ 1 coprime integers,
N =k∑
i=0
ai
q
(pq
)i
, 0 ≤ ai < p
21011 21200 21202 21221
21222101
212210
21
2
3 4
212065 7
9 10 118
2
1
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The language of numeration is “highly” non-rational :any two sub-trees of the lexicographic tree are non-isomorphicbut it is easy to build a digit-to-digit right sequential transducerthat realizes the odometer
PROPOSITION
For the base p/q system, p > q ≥ 1, the amortized cost (resp.complexity) is
pq
pq − 1
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EXAMPLE OF LANGUAGE WITH ZERO ENTROPY
a∗b∗ is a rational, prefix closed language (and any branch in thelexicographic tree is infinite)
u(n) = #(a∗b∗ ∩ {a, b}n) = n + 1, therefore
limn→∞
∑ni=0 #(L ∩ Σ≤i)
#(L ∩ Σ≤n)= lim
n→∞
16(n + 1)(n + 2)(n + 3)
12(n + 1)(n + 2)
= +∞.