Question

need your help with question 2

Your solutions must be submitted as JFLAP solution files for full credit on questions 1, 2 and 5. A 10 % penalty will be assessed for non-JFLAP solutions The alphabet for all of the following problems is the same: A, B, C, null All tapes read from left to right the same as in JFLAP. 1) single tape Turing machine: initial tape has (A+B+C)* if AAA substring then it is rewritten as BBB if CCC substring appears then all previous BBB substrings are rewritten as ABA 2) single tape Turing machine: initial tape has (A+B+C)* final tape has A*B*C* for questions 3 and 4: */R,L,S indicate Turing command read/pop/push for stack 3) single tape Turing + single push down stack: an additional character D is available for this problem initial tape is blank initial stack is (A+B+C)* final tape is A*B* final stack is C* 4) single tape Turing + single push down stack: an additional character D is available for this problem. initial tape is (A+B+C)* initial stack is empty final tape is empty final stack contains only the (just one) character that is most commonly found on the initial tape 5. Two tape Turing machine both tapes initially contain (A+B+C)* each tape is read sequentially at the same time, e.g. position 0 on both tapes, then position 1 on both tapes etc. if the characters match on both tapes at one specific position then nothing is changed swapped if the characters do not match then the characters are

Answer 1

Answer:

Question 2:

Here what I am doing in Turing Machine is that appending X to the string in start.

Then we changes all A's to Y in original string and for each A append A in the end by replacing blank symbol.

Now when all A's are shifted we know focus on B.

Now B are replaced by Y and for each B append B in the end by replacing blank symbol.

Now when all B's are shifted we know focus on C.

Now C are replaced by Y and for each C append C in the end by replacing blank symbol.

After whole original string symbol becomes Y, we change the Y to blank symbol and X to blank symbol and reach final state.

if we could not reach final state, string given is invalid.

If we could reach the final state, string is as per requirement in the tape as A*B*C*.

I could not attach jflap file but i can attach screenshots.

JFLAP v8.0 (Beta)(1jflap) File Input Help X Multiple Run Automaton Editor Table Text Size Input Outout Result A; A, L B; B O A, L q2 O; X, L A; Y, R 13 X; X, L FR3. B; Y, R BB,R ;B. L X: X,Lq5 q7 X; X, L 0,R Y; Y, L X; X L q9 C: Y. R 0; c, L qli q1 q12 O;,R X; X, L Load Inputs Run Inputs Enter AView Trace Clear 20:43 )ENG 27.11-2017 O Type here to search

Initial steps and final steps Execution trace for string ACBA is as follows:

You can dry run for some string if you want. It is doing the desired function only.

You could easily understand.

Traceback X BADOOOD BADOOD0 OOODOA nnnnnAC 0pggDAC BAODOOD DOOD ODOODACBA ql ODOODACR q2 00000ACBYXOOOO OOODOACBYXDDO0 O000OACBYKAOO0 q. D000DACBYXADDO 2040 OType here to search ENG 27-11-2017

Traceback X DOOODYYYYXAA BCO qs OoDOOY YYXAAB C D YYYXAABCO ПООД YYYYXAABCO oOooD YYXAABC T2 ODOO00 YYXAAB CD 412 YХААВСО q12 OOOOO0ODYXAABCD DOOD O0DXAABCO OOOO0D000DAABCD 2040 OType here to search ENG 27-11-2017
Alright Dude, If that worked for you... don't forget to give THUMBS UP.(that will work for me!)