Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. States are q0,q1,q2,q3,q4. Z means empty stack. And q4 is final state. I designed it that way but i am not sure if it is correct. Can you check the answer and help me to fix it?
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First we have to count number of a's and that number should be equal to number of c's. But we have to take care b's coming between 'a' and 'c'. So we suggest you to keep doing the practice with the steps mentioned in the explnation. Just see the given problem in another perspective. As add number of a's and b's, and that will equal to number of c's. First we have to count number of a's and b's that total number should be equal to number of c's. That we will achieve by pushing a's and b's in STACK and then we will pop a's and b's whenever "c" comes.
Clarification First we have to count number of a's and that number should be equal to number of b's When all a's are finished by b's Secondly count number of b's and that should be equal to number of c's So in the end of the strings if nothing is left in the STACK then we can say that language is accepted in the PDA.
As add number of b's and c's, and that will equal to number of a's. First we have to count number of a's and that number should be equal to number of b's and c's.
Toggle navigation Scanftree.Join Stack Overflow to learn, share knowledge, and build your career. Connect and share knowledge within a single location that is structured and easy to search. I thought that for every 'a' in string I will push 3 'a' into the stack and for every 'b' in the string, I will pop 2 'a' from the stack now for every 'c' in the string I will still have 1 'a' in the stack. As Ami Tavory points out, there is no PDA for this language because this language is not context-free.
It is easy to recognize this language if you use a queue instead of a stack, use two stacks, or use a Turing machine all equivalent. One reason you've not managed to construct a pushdown automaton for this language, is because there isn't any.
The Bar Hillel pumping lemma shows this. To outline the proof, suppose it can be done. Then, for some peach string larger than p can be partitioned to uvwxys. The first rule implies that vwx can't span the three regions, only at most two for large enough strings.
The second and third rules now imply that you can pump so that the un-spanned region is smaller than the at least one of the other regions. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Asked 4 years, 4 months ago. Active 4 years, 4 months ago.
Viewed 6k times. First approach:- I thought that for every 'a' in string I will push 3 'a' into the stack and for every 'b' in the string, I will pop 2 'a' from the stack now for every 'c' in the string I will still have 1 'a' in the stack.
JFLAP files available for both if needed. NeoR NeoR 1 1 gold badge 9 9 silver badges 25 25 bronze badges. Add a comment.
Active Oldest Votes. Queue machine: Enqueue a s as long as you see a s, until you see a b. Dequeue a s and enqueue b s as long as you see b s, until you see a c Dequeue b s as long as you see c s. Accept if you end this process with no additional input and an empty queue. Patrick87 Patrick87 Firstly again I would like to point out "Fool's errand" and second i know PDA does not exists so please read the question description fully. NeoR You might consider changing the title to reflect your actual question and putting your actual question near the top of the question body, with any explanation or motivating discussion below it.
NeoR I'd make it two questions, one for each approach. The first question is "what strings does this PDA accept"; the second one is "is this language context free". The first question might have a more interesting answer depending on what you consider interesting. I have uploaded a new question at stackoverflow.Input tape: The input tape is divided in many cells or symbols.Vulcanair p68 observer 2 oy-sps
The input head is read-only and may only move from left to right, one symbol at a time. Finite control: The finite control has some pointer which points the current symbol which is to be read. Stack: The stack is a structure in which we can push and remove the items from one end only. It has an infinite size. In PDA, the stack is used to store the items temporarily. ID is an informal notation of how a PDA computes an input string and make a decision that string is accepted or rejected.
Solution: In this language, n number of a's should be followed by 2n number of b's. Hence, we will apply a very simple logic, and that is if we read single 'a', we will push two a's onto the stack. As soon as we read 'b' then for every single 'b' only one 'a' should get popped from the stack. Now when we read b, we will change the state from q0 to q1 and start popping corresponding 'a'. Thus this process of popping 'b' will be repeated unless all the symbols are read.
Note that popping action occurs in state q1 only.
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After reading all b's, all the corresponding a's should get popped. Hence the move will be:. Solution: In this PDA, n number of 0's are followed by any number of 1's followed n number of 0's. Hence the logic for design of such PDA will be as follows:. Push all 0's onto the stack on encountering first 0's. Then if we read 1, just do nothing.
Then read 0, and on each read of 0, pop one 0 from the stack. JavaTpoint offers too many high quality services. Mail us on hr javatpoint. Please mail your requirement at hr javatpoint. Duration: 1 week to 2 week.The following are the six problems given to the contestants. How many can you solve? The country names inside the parentheses are the problem proposers. What in the world is the pigeonhole principle? France Let m and n be positive integers.
Let aI, a2, taken from n boxes pigeonholesthen at least two of the objects will be from the To show how to apply this principle, we give a few Problem 2. Suppose that Example 1. Suppose51 numbersare i M is the midpoint of BC and 0 is the chosenfrom 1, 2, 3, Let us consider the 50 pairs of iii E lies on the line AB and F lies on the consecutive numbers 1,23,4Since 51 numbers are chosen, distinct and collinear.
Acknowledgment: Thanks to Martha A. The editors welcome contributions from all students. With your submission, please include your name, address, school, email, telephone and fax numbers if available. The deadline for receiving material for the next issue is January 31, Send all correspondence to: Dr. Note that the two examples look alike, however the boxes fornled are quite different. By now, the readers must have observed that forming the right boxes is the key to success.
Often a certain amount of experience as well as clever thinking are required to solve such problems. The additional examples below will help beginnersbecomefarniliar with this useful principle. Example 3.The Round in which the fight ends will be used for settlement purposes.Pushdown Automata problems with clear explanation
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In Test and County Championship matches, the whole match counts. In drawn games a minimum of 200 overs must be bowled, otherwise bets void, unless settlement of bets is already determined. In Twenty20 matches the match must be scheduled for the full 20 overs and there must be an official result unless settlement of bets is already determined. In One Day matches where the number of overs has been reduced and the outcome has not already been determined then bets will be void.Will I Be (8) odds 5.
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