I need some help with designing Turing machine that accepts language L= {a^n+1 b^2n c^3n: n>=0}
There are a lot of correct ways to do this. I will just walk through one of them I hope illustrates a useful way to attack these problems.
First, we notice the commonality n between the three segments. We will cross off symbols from each section, one at a time, to make sure they have the right relationships. First, we can verify that the relationship between a and b is right. Then, we can check the relationship between a and c. If those are both right, we are done.
First, let's get rid of the pesky "+1" from the a. This means we have at least one a regardless of what n is. So, we can begin by changing a to X. Now, the remaining input should have n instances of a, 2n instances of b and 3n instances of c. If we don't have the one a, we can halt-reject immediately; we can't have n+1 instances of a if we don't have at least one.
Now, if there are more instances of a, cross it off by writing A in its place, and go cross off two instances of b by writing B in their places. Then, go back and look for more instances of a, bouncing back and forth until there are no more instances of a. Then, verify there are no more instances of b; if so, there were twice as many instances of b as there were of a, and the first two sections have the right relationship. If at any point you didn't have enough b to cross off, or if after you ran out of a you still had b, then you can safely halt-reject at this point.
Next, you can do the same thing for instances of A and c, just cross off three instances of c instead of two. If we find the A get exhausted at the same time as the c do, we are done and halt-accept.
The transitions might look like this:
Q T Q' T' d comment
q0 a q1 X right account for +1
q1 a q2 A right n>0 case, continue
q1 # hA # same n=0 case, accept
q2 a q2 a right skip uncrossed a
q2 B q2 B right skip crossed b
q2 b q3 B right find first uncrossed b, cross it
q3 b q4 B left find next b, cross it
q4 B q4 B left find last uncrossed a
q4 a q2 A right cross it
q4 A q4 A left skip crossed a, if any
q4 X q5 X right ran out of a to cross
q5 A q5 A right skip crossed a
q5 B q5 B right skip crossed b
q5 c q6 c left verify existence of some c
q6 C q6 C left skip crossed c
q6 B q6 B left skip crossed b
q6 A q7 a right find last crossed a, uncross
q6 X q10 X right ran out of crossed a
q7 a q7 a right skip uncrossed a
q7 B q7 B right skip crossed b
q7 C q7 C right skip crossed c
q7 c q8 C right find first uncrossed c, cross
q8 c q9 C right cross 2nd uncrossed c
q9 c q6 C left cross 3rd uncrossed c
q10 a q10 a right skip uncrossed a
q10 B q10 B right skip crossed b
q10 C q10 C right skip crossed c
q10 # hA # same accept if no leftover symbols until end