Unambiguous Morphic Images of Strings

Transcription

1 Unambiguous Morphic Images of Strings Daniel Reidenbach, University of Kaiserslautern A joint work with: Dominik D. Freydenberger, University of Kaiserslautern Johannes C. Schneider, University of Kaiserslautern

2 task Task: Map a finite string α over an infinite alphabet onto a finite word w over a binary alphabet such that w reflects the structure of α "optimally".

3 task Task: Map a finite string α over an infinite alphabet onto a finite word w over a binary alphabet such that w reflects the structure of α "optimally". Standard solution: Choose an injective morphism (i.e. a code) σ.

4 task Task: Map a finite string α over an infinite alphabet onto a finite word w over a binary alphabet such that w reflects the structure of α "optimally". Standard solution: Choose an injective morphism (i.e. a code) σ. A mapping σ: ï * {a, b} * that satisfies σ(α β) = σ (α) σ (β) for every α, β ï *.

5 task Task: Map a finite string α over an infinite alphabet onto a finite word w over a binary alphabet such that w reflects the structure of α "optimally". Standard solution: Choose an injective morphism (i.e. a code) σ. A mapping σ: ï * {a, b} * that satisfies σ(α β) = σ (α) σ (β) for every α, β ï *. Example: Let α = Let σ (i) = ab i. σ (α) = ab ab 2 ab 3 ab 4 ab ab 4 ab 3 ab 2

6 Ambiguity of words Assumption: We are confronted with a variety of morphic images of a single (unknown) string α ( inductive inference of pattern languages, PCP).

7 Ambiguity of words Assumption: We are confronted with a variety of morphic images of a single (unknown) string α ( inductive inference of pattern languages, PCP). Example: Let α = ababbabbbabbbbababbbbabbbabb

8 Ambiguity of words Assumption: We are confronted with a variety of morphic images of a single (unknown) string α ( inductive inference of pattern languages, PCP). Example: Let α = ababbabbbabbbbababbbbabbbabb ababbabbbabbbbababbbbabbbabb

9 Ambiguity of words Assumption: We are confronted with a variety of morphic images of a single (unknown) string α ( inductive inference of pattern languages, PCP). Example: Let α = ababbabbbabbbbababbbbabbbabb ababbabbbabbbbababbbbabbbabb example code leads to an ambiguous image. It does not prove the existence of the symbol "2" in α.

10 Ambiguity of words Assumption: We are confronted with a variety of morphic images of a single (unknown) string α ( inductive inference of pattern languages, PCP). Example: Let α = ababbabbbabbbbababbbbabbbabb ababbabbbabbbbababbbbabbbabb example code leads to an ambiguous image. It does not prove the existence of the symbol "2" in α. Note: Ambiguity of morphic images can cause that, even in a weak sense, "decoding" is a non-computable problem [Reidenbach, 2002].

11 A simple combinatorial question on morphisms in free monoids Question: For which pattern α is there a morphism σ: ï + {a, b} * such that σ (α) is unambiguous, i.e. there is no morphism ρ: ï + {a, b} * with, for some variable i in α, ρ (i) σ(i) and ρ (α) = σ (α)?

12 A simple combinatorial question on morphisms in free monoids Question: A string in ï +. For which pattern α is there a morphism σ: ï + {a, b} * such that σ (α) is unambiguous, i.e. there is no morphism ρ: ï + {a, b} * with, for some variable i in α, ρ (i) σ(i) and ρ (α) = σ (α)?

13 A simple combinatorial question on morphisms in free monoids Question: A string in ï +. For which pattern α is there a morphism σ: ï + {a, b} * such that σ (α) is unambiguous, i.e. there is no morphism ρ: ï + {a, b} * with, for some variable i in α, ρ (i) σ(i) and ρ (α) = σ (α)? A symbol in ï.

14 A simple combinatorial question on morphisms in free monoids Question: A string in ï +. For which pattern α is there a morphism σ: ï + {a, b} * such that σ (α) is unambiguous, i.e. there is no morphism ρ: ï + {a, b} * with, for some variable i in α, ρ (i) σ(i) and ρ (α) = σ (α)? A symbol in ï. Remark: Question can easily be rephrased in terms of, e.g., pattern languages and the Post Correspondence Problem, i.e. equality sets.

15 A simple combinatorial question on morphisms in free monoids Question: A string in ï +. For which pattern α is there a morphism σ: ï + {a, b} * such that σ (α) is unambiguous, i.e. there is no morphism ρ: ï + {a, b} * with, for some variable i in α, ρ (i) σ(i) and ρ (α) = σ (α)? A symbol in ï. Remark: Question can easily be rephrased in terms of, e.g., pattern languages and the Post Correspondence Problem, i.e. equality sets. Additionally, we provide an answer on finite fixed points of nontrivial morphisms...

16 Is this a trivial problem? Examples suggest that this might be a nontrivial problem: Let α = We know that the image of the morphism σ (i) = ab i is ambiguous. aabbabba is unambiguous, but the corresponding morphism is not injective. By the way, abbaaabb is ambiguous...

17 Is this a trivial problem? Examples suggest that this might be a nontrivial problem: Let α = We know that the image of the morphism σ (i) = ab i is ambiguous. aabbabba is unambiguous, but the corresponding morphism is not injective. By the way, abbaaabb is ambiguous... Let α = 1 2. Evidently, there is no unambiguous morphic image: Let σ(1) ε. n, with ρ(1) = ε and ρ(2) = σ(1) σ(2), ρ σand ρ (α) = σ (α).

18 A first negative result and its consequences orem: re is no nonerasing morphism σ such that, for every pattern α, σ (α) is unambiguous.

19 A first negative result and its consequences orem: σ: ï * {a, b} + re is no nonerasing morphism σ such that, for every pattern α, σ (α) is unambiguous.

20 A first negative result and its consequences orem: σ: ï * {a, b} + re is no nonerasing morphism σ such that, for every pattern α, σ (α) is unambiguous. Remarks: Obviously, for each pattern α =i j(with i, j ï and i j) and for each nonerasing morphism σ, σ (α) is unambiguous. We should generalise the structure of α...

21 A crucial (and natural) partition of ï + Definition: We call any α ï + prolix iff there exists a decomposition α = β 0 γ 1 β 1 γ 2 β 2... β n-1 γ n β n with n 1, β k ï * and γ k ï +, k n, such that 1. for every k, 1 k n, γ k 2, 2. for every k, 1 k n, and for every k', 0 k' n, var (γ k ) var (β k' ) = and 3. for every k, 1 k n, there exists an i var (γ k ) such that γ k i = 1 and, for every k', 1 k' n, if i var (γ k' ) then γ k = γ k'. We call α ï + succinct iff it is not prolix.

22 A crucial (and natural) partition of ï + Definition: We call any α ï + prolix iff there exists a decomposition α = β 0 γ 1 β 1 γ 2 β 2... β n-1 γ n β n with n 1, β k ï * and γ k ï +, k n, such that 1. for every k, 1 k n, γ k 2, 2. for every k, 1 k n, and for every k', 0 k' n, var (γ k ) var (β k' ) = and 3. for every k, 1 k n, there exists an i var (γ k ) such that γ k i = 1 and, for every k', 1 k' n, if i var (γ k' ) then γ k = γ k'. We call α ï + succinct iff it is not prolix. Examples: is prolix.

23 A crucial (and natural) partition of ï + Definition: We call any α ï + prolix iff there exists a decomposition α = β 0 γ 1 β 1 γ 2 β 2... β n-1 γ n β n with n 1, β k ï * and γ k ï +, k n, such that 1. for every k, 1 k n, γ k 2, 2. for every k, 1 k n, and for every k', 0 k' n, var (γ k ) var (β k' ) = and 3. for every k, 1 k n, there exists an i var (γ k ) such that γ k i = 1 and, for every k', 1 k' n, if i var (γ k' ) then γ k = γ k'. We call α ï + succinct iff it is not prolix. Examples: is prolix.

24 A crucial (and natural) partition of ï + Definition: We call any α ï + prolix iff there exists a decomposition α = β 0 γ 1 β 1 γ 2 β 2... β n-1 γ n β n with n 1, β k ï * and γ k ï +, k n, such that 1. for every k, 1 k n, γ k 2, 2. for every k, 1 k n, and for every k', 0 k' n, var (γ k ) var (β k' ) = and 3. for every k, 1 k n, there exists an i var (γ k ) such that γ k i = 1 and, for every k', 1 k' n, if i var (γ k' ) then γ k = γ k'. We call α ï + succinct iff it is not prolix. Examples: is prolix.

25 A crucial (and natural) partition of ï + Definition: We call any α ï + prolix iff there exists a decomposition α = β 0 γ 1 β 1 γ 2 β 2... β n-1 γ n β n with n 1, β k ï * and γ k ï +, k n, such that 1. for every k, 1 k n, γ k 2, 2. for every k, 1 k n, and for every k', 0 k' n, var (γ k ) var (β k' ) = and 3. for every k, 1 k n, there exists an i var (γ k ) such that γ k i = 1 and, for every k', 1 k' n, if i var (γ k' ) then γ k = γ k'. We call α ï + succinct iff it is not prolix. Examples: is prolix is succinct.

26 Why is this a natural partition? A succinct pattern is a shortest generator of a terminal-free E-pattern language: orem [Reidenbach, 2002]: Apatternα is succinct if and only if, for every pattern α' with L (α') = L (α), α' α.

27 Why is this a natural partition? A succinct pattern is a shortest generator of a terminal-free E-pattern language: orem [Reidenbach, 2002]: Apatternα is succinct if and only if, for every pattern α' with L (α') = L (α), α' α. set of prolix patterns exactly corresponds to the set of finite fixed points of nontrivial morphisms [Head, 1981]: orem [Head, 1981]: Apatternα is prolix if and only if there exists a morphism ϕ: ï * ï * such that ϕ id and ϕ (α) = α.

28 Unambiguous words for prolix patterns? No... orem: For any prolix pattern α and for any nonerasing morphism σ, σ (α) is ambiguous.

29 Unambiguous words for prolix patterns? No... orem: For any prolix pattern α and for any nonerasing morphism σ, σ (α) is ambiguous. Proof: α is prolix there is a ϕ with ϕ (α) = α and ϕ (i) = ε for some i in var (α) for every σ define ρ = σ ϕ ρ (α) = σ (α), but ρ is not nonerasing

30 Unambiguous words for prolix patterns? No... orem: For any prolix pattern α and for any nonerasing morphism σ, σ (α) is ambiguous. Proof: α is prolix there is a ϕ with ϕ (α) = α and ϕ (i) = ε for some i in var (α) for every σ define ρ = σ ϕ ρ (α) = σ (α), but ρ is not nonerasing...and yes: If we omit the claim that σ must be nonerasing then some prolix patterns have an unambiguous image, some have not.

31 A first approach to succinct patterns Proposition: re is no nonerasing morphism σ such that, for every succinct pattern α, σ (α) is unambiguous.

32 A first approach to succinct patterns Proposition: re is no nonerasing morphism σ such that, for every succinct pattern α, σ (α) is unambiguous. Proof idea: Particular types of (sub-)patterns in α are required such as α =... j k j k' j' k j' k'..., with no other occurrences of these variables. For every nonerasing morphism σ we can find a pattern α such that σ (α) is not unambiguous.

33 main result orem: For every succinct pattern α, there is an injective morphism σ α such that σ α (α) is unambiguous.

34 main result orem: For every succinct pattern α, there is an injective morphism σ α such that σ α (α) is unambiguous. This leads to a third characterisation of succinctness: Corollary: A pattern α is succinct if and only if there exists an injective morphism σ such that σ (α) is unambiguous.

35 morphism some ideas [1 of 7] Recall the above structure: Let α = bab baa a b a a b abb aaabaa

36 morphism some ideas [1 of 7] Recall the above structure: Let α = bab baa a b a a b abb aaabaa First observation: In the given example, ambiguity is caused by the fact that the morphic images of "1" and "4" end with the same letter, namely a.

37 morphism some ideas [1 of 7] Recall the above structure: Let α = bab baa a b a a b abb aaabaa First observation: In the given example, ambiguity is caused by the fact that the morphic images of "1" and "4" end with the same letter, namely a. Second observation: For every occurrence of "1" and "4", the right neighbour reads either "2" or "3", and the latter variables have no other left neighbours in α.

38 morphism some ideas [2 of 7] A first idea... For every variable i in a pattern α, identify the set L i (resp. R i ) of all of its left (resp. right) neighbours. Choose σ α such that each L i (resp. R i ) contains variables that have morphic images with different last (resp. first) letters. In other words: Choose σ α such that each L i (resp. R i ) is (morphically) heterogeneous.

39 morphism some ideas [2 of 7] A first idea... For every variable i in a pattern α, identify the set L i (resp. R i ) of all of its left (resp. right) neighbours. Choose σ α such that each L i (resp. R i ) contains variables that have morphic images with different last (resp. first) letters. In other words: Choose σ α such that each L i (resp. R i ) is (morphically) heterogeneous....that does not work: Example: Let α = L 1 = {2, 3}, L 2 = {1, 3}, L 3 = {1, 2}

40 morphism some ideas [2 of 7] A first idea... For every variable i in a pattern α, identify the set L i (resp. R i ) of all of its left (resp. right) neighbours. Choose σ α such that each L i (resp. R i ) contains variables that have morphic images with different last (resp. first) letters. In other words: Choose σ α such that each L i (resp. R i ) is (morphically) heterogeneous....that does not work: Example: Let α = L 1 = {2, 3}, L 2 = {1, 3}, L 3 = {1, 2} σ α (2) =... a σ α (3) =... b

41 morphism some ideas [2 of 7] A first idea... For every variable i in a pattern α, identify the set L i (resp. R i ) of all of its left (resp. right) neighbours. Choose σ α such that each L i (resp. R i ) contains variables that have morphic images with different last (resp. first) letters. In other words: Choose σ α such that each L i (resp. R i ) is (morphically) heterogeneous....that does not work: Example: Let α = L 1 = {2, 3}, L 2 = {1, 3}, L 3 = {1, 2} σ α (2) =... a σ α (3) =... b σ α (1) =... a σ α (3) =... b

42 morphism some ideas [2 of 7] A first idea... For every variable i in a pattern α, identify the set L i (resp. R i ) of all of its left (resp. right) neighbours. Choose σ α such that each L i (resp. R i ) contains variables that have morphic images with different last (resp. first) letters. In other words: Choose σ α such that each L i (resp. R i ) is (morphically) heterogeneous....that does not work: Example: Let α = L 1 = {2, 3}, L 2 = {1, 3}, L 3 = {1, 2} σ α (2) =... a σ α (3) =... b σ α (1) =... a σ α (3) =... b σ α (1) =... a σ α (2) =... a

43 morphism some ideas [3 of 7] Question: Can we weaken the claim that each L i (resp. R i ) must be morphically heterogeneous?

44 morphism some ideas [3 of 7] Question: Can we weaken the claim that each L i (resp. R i ) must be morphically heterogeneous? Example: Let α = L 2 = {1, 3}, L 5 = {4, 3}, L 6 = {1, 4}... a σ α (2)... b σ α (2)... a σ α (5)... b σ α (5)... a σ α (6)... a σ α (6)

45 morphism some ideas [3 of 7] Question: Can we weaken the claim that each L i (resp. R i ) must be morphically heterogeneous? Example: Let α = L 2 = {1, 3}, L 5 = {4, 3}, L 6 = {1, 4}... a σ α (2)... b σ α (2)... a σ α (5)... b σ α (5)... a σ α (6)... a σ α (6) ρ (6) = a σ α (6) ρ (6)

46 morphism some ideas [3 of 7] Question: Can we weaken the claim that each L i (resp. R i ) must be morphically heterogeneous? Example: Let α = L 2 = {1, 3}, L 5 = {4, 3}, L 6 = {1, 4}... a σ α (2)... b σ α (2)... a σ α (5)... b σ α (5)... a σ α (6)... a σ α (6) ρ (6) = a σ α (6) ρ (5) = a σ α (5) ρ (5) ρ (6)

47 morphism some ideas [3 of 7] Question: Can we weaken the claim that each L i (resp. R i ) must be morphically heterogeneous? Example: Let α = L 2 = {1, 3}, L 5 = {4, 3}, L 6 = {1, 4}... a σ α (2)... b σ α (2)... a σ α (5)... b σ α (5)... a σ α (6)... a σ α (6) ρ (6) = a σ α (6) ρ (5) ρ (6) ρ (5) = a σ α (5) but then ρ (5) = b σ α (5). Thus, there is no morphism ρ.

48 morphism some ideas [4 of 7] Observation: Whenever, for some j, k var (α), L j L k then the heterogeneity of L j "protects" the images of the variables in L k and vice versa.

49 morphism some ideas [4 of 7] Observation: Whenever, for some j, k var (α), L j L k then the heterogeneity of L j "protects" the images of the variables in L k and vice versa. In fact, this property even is "transitive", i.e. if, for some j, k var (α), there exist arbitrary L k,1, L k,2,..., L k,p with L k,i L k,i+1 and L k,1 = L j and L k,p = L k then the heterogeneity of L j makes an alternative morphic image for k impossible.

50 morphism some ideas [5 of 7] Idea: We introduce an equivalence relation ~ L on var (α): For each pair j, j' var (α), j ~ L j' if and only if there exist L k,1,l k,2,..., L k,p with L k,i L k,i+1 and j L k,1 and j' L k,p. We name the resulting equivalence classes L 1, L 2,..., L m. Thus, L 1 L 2... L m = var (α), and the L i are pairwise disjoint. Remark: Analogously, we introduce a relation ~ R.

51 morphism some ideas [6 of 7] Example: Let α = L 1 = {3}, L 2 = {1}, L 3 = {1, 2}, L 4 = {3, 6}, L 5 = {4, 6}, L 6 = {4, 5} L 1 = {1, 2}, L 2 = {3, 4, 5, 6}

52 morphism some ideas [6 of 7] Example: Let α = L 1 = {3}, L 2 = {1}, L 3 = {1, 2}, L 4 = {3, 6}, L 5 = {4, 6}, L 6 = {4, 5} L 1 = {1, 2}, L 2 = {3, 4, 5, 6} : For our needs, it is sufficient that each L i (resp. R i ) is morphically heterogeneous. As these sets are pairwise disjoint, there are no conflicting assignments when composing σ α.

53 morphism some ideas [7 of 7] : For every succinct pattern α, the following morphism σ α generates an unambiguous word: σ α (k) = k var (α). ab 3k aab 3k+1 aab 3k+2 a, i: k = min L i j: k = min R j, ba 3k b ab 3k+1 aab 3k+2 a, i: k = min L i j: k = min R j, ab 3k aab 3k+1 a ba 3k+2 b, i: k = min L i j: k = min R j, ba 3k b ab 3k+1 a ba 3k+2 b, i: k = min L i j: k = min R j,

54

55 1. re is no nonerasing morphism which, for every pattern, generates an unambiguous word.

56 1. re is no nonerasing morphism which, for every pattern, generates an unambiguous word. 2. image of any prolix pattern under any nonerasing morphism is not unambiguous.

57 1. re is no nonerasing morphism which, for every pattern, generates an unambiguous word. 2. image of any prolix pattern under any nonerasing morphism is not unambiguous. 3. re is no nonerasing morphism which, for every succinct pattern, generates an unambiguous word.

58 1. re is no nonerasing morphism which, for every pattern, generates an unambiguous word. 2. image of any prolix pattern under any nonerasing morphism is not unambiguous. 3. re is no nonerasing morphism which, for every succinct pattern, generates an unambiguous word. 4. For every succinct pattern there is an injective morphism generating an unambiguous word.