 Мы говорим о уровне амбигюитации три атомата. Это Джон работал с моим студентом Деронте Ферритом. Рестриктанное детерминисство может быть полезным. Эта амбигуя есть, если на каждом инкуте есть один, а не рано. Эта амбигуя нет. Но это амбигуя, потому что эта транзиция должна быть сделана только на последний раз. Эта амбигуя есть, если на каждом инкуте есть один, а не рано. Эта амбигуя есть, если на каждом инкуте есть один, а не рано. Это амбигуя, если на каждом инкуте есть один, а не рано. Это амбигуя, если на каждом инкуте есть один, а не рано. На файне три атомата есть амбигуя. На амбигуя и на файне три атомата есть амбигуя, у которых есть много амбигуя, у которых есть много амбигуя. Рестриктанное детерминисство может быть полезным. Эта амбигуя, если на каждом инкуте есть один, а не рано. Например, универсальтирная проблема, в которой язык атомата содержит все стримы. И для генерального детерминисства атомата эта проблема будет комплина. Но для амбигуя атомата это в полномерном времени. Рестриктанное детерминисство может быть полезным для генерального детерминиса. Когда мы работаем с атоматом, для амбигуя атомата генерального детерминиса может быть лучше. Например, если мы consider an automata on finite words, it is easy to complement an ambiguous automata than non-deterministic automata. If we consider an automata on omega words it is easy to complement even finally ambiguous automata in the non-deterministic automata. Это сериум про полиномиальный алгоритм для решения дегрифа амбигитии от атомата на файне 3 и файне 4. Также есть полиномиальный алгоритм для решения дегрифа амбигитии от атомата на омега 4. Эта сериум про полиномиальный алгоритм для решения дегрифа амбигитии от атомата на файне 3 и файне 4. Эта сериум про полиномиальный алгоритм для решения дегрифа амбигитии от атомата на файне 3 и файне 4. Эта сериум про полиномиальный алгоритм для решения дегрифа амбигитии от атомата на файне 3 и файне 4. Теперь я буду говорить о том, что я рекомендую о том, что есть атомата на файне 3. Тогда я буду приводить структуру для решения дегрифа амбигитии от атомата на файне 4 и файне 4. И из этого структура, это очень легко для решения дегрифа амбигитии от атомата на файне 4 и файне 4. Эта сериум про полиномиальный алгоритм для решения дегрифа амбигитии от атомата на файне 3 и файне 4. Мы говорим о бинаритрии. В структура я использую структуру для решения дегрифа амбигитии от атомата на файне 3 и файне 4. Так что это полиномиальный алгоритм для решения дегрифа амбигитии от атомата на файне 3 и файне 4. И в структура я использую структуру для решения дегрифа амбигитии от атомата на файне 3. И структура должна быть лаборатором для решения дегрифа амбигитии от атомата на файне 4 и файне 5. И если у нас есть структура для решения дегрифа амбигитии от атомата на файне 4 и файне 5, то это должна быть лаборатором для решения дегрифа амбигитии от атомата на файне 4 и файне 5. И в структура я использую структура для решения дегрифа амбигитии от атомата на файне 5 и файне 5. Я использую структура для решения дегрифа амбигитии от атомата на файне 5. Вот это партнер атомата. Вот алгарит, поленомелалгарит для дегрифа амбигитии, и он базирует структура. Так, здесь характеризация для обновленной амбегии. Так, бюхетрия автомата не обновленная на обновленные амбегии, если она не вылечит, то хотя бы одна из полных условий. Так, амбегия не обновленная на обновленные обновленные амбегии, или амбегий transition system. Итак, в небольшом уровне я объясняю эти условия, но давайте просто представить, что эта ситуация может быть проверена в полномерном времени. Поэтому, амбегия обновленная на обновленном времени. Итак, я сейчас буду объяснить эти два условия. Но в первом этапе я провел notion. Думаю, у меня есть компьютация на три, и у меня есть бронж. Тогда проекция от компьютации на бронж это просто секунда статистичного бронжа. В этом случае, q1, q3, q6. Так, это проекция. Так, здесь definition of bounded-ambiguity. Atomata is K branch ambiguous. If for every tree and every branch the number of different projection of accepting computation to this branch is at most K. It is bounded branch ambiguous. If it is K ambiguous for some K, it is finally branch ambiguous. If for every tree and branch the number of different projection of accepting computation on T to this branch is finite. And the following observation is trivial. If A is bounded ambiguous, then it is bounded branch ambiguous. If A is bounded ambiguous, then there is K such that on every tree we have at most K accepting computations and then for every branch we have at most K projection. And the following proposition is also reasonable but requires a proof. Branch ambiguity degree is polynomial time reducible to the degree of ambiguity of an automata on words. Which can be constructed from our automaton trees. And since on word ambiguity is in polynomial time we obtain that branch ambiguity is also in polynomial time. So now we will explain the second condition. Suppose I have this transition. So from state P1 I see letter A and I am going to the state. So this transition is right ambiguous. If there is a tree and two different computation from state Q1 on this tree. Second possibility if there is another transition with the same right part. Left part are different and there is a tree such that there is a computation on this tree both accepting computation on this tree both from state Q1 and from state Q2. So this is the right ambiguous transition and left ambiguous transition is the finite dollar. If we have such a transition then we can plug here this tree and we will have at least two computations. Now the second condition. A has K ambiguous transition pattern if there is accepting computation from state Q which has the following form along some branch. So on some branch Q is at the root Q is somewhere deep in tree and on the path from this Q to this Q we have ambiguous transition. If we have such a pattern then we can plug here tree and we will have two accepting computations which differ only inside this sub tree. Now look at this pattern it starts from Q and end of Q so we can pump it. So we obtain something like this. So in this case we will have two possibility for computation here two possibility for computation here so two times two for possible computation. But you can pump it a bit more number of times. So if I pump it n times I will have at least two to the n different computations. If it has ambiguous transition pattern then A is not bounded ambiguous. So let me summarize what we did for bounded ambiguity. So I promised to prove this theorem so I explained these conditions and we proved direction 2 to 1. Direction 1 to 2 is variation of cell deproved for automata over finite trees. We get characterization for bounded ambiguity. So let me give characterization for finite ambiguity. A buggy tree automata is not finite ambiguous if at least one of the following condition calls. A is not finite to branch ambiguous so we replace it bounded branch ambiguity by finite branch ambiguity and A has F ambiguous transition pattern for final state F. So again we have the same pattern F here, F here and here we have an ambiguous transition system but now F is the final state. So we can pump it even infinitely many times and it will result in uncountable many possible exception computations. So if A has F ambiguous transition pattern then it is uncountable ambiguous. So we proved this direction so now I am going to explain second direction. So suppose we have a tree and suppose we have infinitely many exception computations of these three and then infinitely many of these computations share the first transition. And either there are infinitely many computation of this subtree or infinitely many computation of this subtree from state Q3 or from state Q2. Suppose there are infinitely many computation here and again infinitely many of them share the same first transition and either there are infinitely many computation here or here. So in such a way we will construct a branch such that from every node there are infinitely many computation of subtree rooted in this node from the corresponding states. There is a transition. So transition according to this branch are good. So now finite ambiguity will imply that there are infinitely many projection to this branch and from this one can derive that infinitely many projection share exactly what we constructed along this branch. So on this branch there are infinitely many F states so we will take one of the F states so we know that there are infinitely many computation so there are at least two. So they differ somewhere here as we take closest closest node on the branch to the point where they differ and then we obtain that at this node we have ambiguous transition system and because on the branch there are infinitely many F so we obtain that there is ambiguous pattern. So we assume that infinitely many computation and automata is finally branch ambiguous and we derive that there is ambiguous transition pattern. So we prove the second direction. Characterization for table ambiguity is similar but require one additional condition. So buheteria automata is not countable ambiguous if at least one of the following condition holds. So we replace here finally branch ambiguity by countable branch ambiguity. We have here F ambiguous transition pattern which already implies table ambiguity computation and we have one more condition which I am going to explain now. So the difficult part of the paper is implication one, two which I am not going to explain in the talk. So I will just explain short condition and even I will not explain how it implies this direction. So the short condition deals with a tree and two designated nodes in a tree and it requires the following. So there is a tree and two designated nodes in the tree and there are four computation on this tree. So two computations start from final states from final state F and reach designated node like in the picture in the state. And two additional computations so we have here Q1 so this starts from Q1 and reach designated node in Q1 this starts from Q2 and reach designated node in Q2. It is easy to see that this condition is testable in polynomial time and one can show that it also implies uncountable ambiguity but I am not going to do it. Buhi tree automata are less expressive than parity tree automata and the problem of deciding degree of ambiguity of parity tree automata is going to be hard. So the problem of deciding whether automata is not ambiguous is not boundary ambiguous or is not found Finitely ambiguous is going to be complete but for countable ambiguity it is open whether we can show this in deciding compete type. The degree of ambiguity of language is defined in a natural way so it is the degree of the minimal degree of finite automata which accepts the language. So every regular language of words finite tree and omega words is unambiguous. For words we teach in undergraduate students undergraduate course is it every automata is equivalent to deterministic automata therefore it is unambiguous. For finite tree and omega words it is less straightforward but it is old and the long results. For infinite tree we have completely different situation so here we have a hierarchy. For every K there is a ambiguous language which is not K-1 ambiguous there are finitely countable and ambiguous languages and problem which is difficult is to decide the degree of ambiguity of a language not of automata.