wholegrain

I was just wondering if a DFA can have a transition function that can take one of two input character and if it could have a dead end and if a state can loop on itself.

Show that if L is a regular language, then LR = fwR : w 2 Lg is also a regular language. (Note: this means that the words in LR are all of the words of L written in reverse.) (Hint: start with a DFA and make an NFA). If I understand well, I just have to write a random DFA with a random number of states and alphabet of my choosing and do exactly the reverse, and finally convert it to a NFA?

actually i was referring to another problem where we have the condition m not equal to n, and i have no idea what to do. actually, i think i am in a dead end and i have to choose another string.

What happens if you have 2 conditions? Do you have to prove that both conditions doesn't hold after pumping once? say we have condition A and B and we have something like L { something such that condition A holds or condition B holds} do we have to prove that both of them doesn't hold for a particular w? Also, how do we prove that n = m? When you pump, you get n+x how can you prove that n+x = m? there are infinite number of cases...

so can you pick a particular u v and x or do you only have the luxury to choose a particular w?

my proof goes like this w = a^m b^4m |uv| = n |v| > 0 uvx = w a^(k+i(m-k)b^4m let k = 2 if i = 2 then 2m-2 != m i am pretty sure this is wrong, can you tell me of a systematic way to solve these kind of problems, or a way that works 99% of the time? i have no intuition at all for these sort of things also, do i have to prove for all u, v and x, or can i choose a particular, u, v and x? someone told me this and it confused the hell out of me: Mr. Pumping Lemma gives you a pumping constant p. You pick a string s of length at least p. Mr. Pumping Lemma divides s into three parts uvw, subject to the restrictions that |uv|≤p, |v|≥1. You now "pump" the v part by picking an integer i≠1 to select a word uviw. If uviw is not in L, you win. But if uviw is in L, you lose.

ummm....... yeah actually my question wasn't: is this a valid proof, because it sure isn't but whether i can just pump once for that one by choosing a string that will make it immediately wrong once i pump it once, and also whether i can choose any v, or it has to work with any v.

Yeah, well I wrote something like Assume w is in the regular language, wrote the conditions for uvw and then just pumped once. It seems a bit too easy in some case, am I doing something wrong?

we have L {a^m b^n: n = 4m} we choose a string w = a^m b^4m if we pump 1 a we'd get a^(m+1)b^4m, which doesn't meet n = 4m that's it? is this a valid proof?

umm yeah thanks oh yeah its 1.24/1.18 nevermind thanks LOL

http://www.westga.edu/~rbest/finc3511/sampleprobs/chapter9/ch9p6/ch9p6.html I get 81.53. I am off by like 3. I calculated PV0 7.5(1.24)/(0.18-0.24) (1-(1.18/1.24)^2) = which gave me 14.64 then I calculated PV2 and converted it to PV0b 11.53(1.05)/(0.18-0.05)/1.18^2 which gave me 66.88 I added them 66.88+14.64 which gave me 81.53 I used the formula for growing perpetuity and growing annuity

I thought it was (1+01)* but apparently it's not since L((0+10)*) defines all expressions where 1 is followed by 0, I thought that the complement was all expressions where 0 is followed by 1.

thanks for the reply i thought it was 0.01 for some reason. probably because i divided the effective rate by 4 again. i forgot what i was thinking. can we use the effective rate and substitute with the k we find in financial formulas?

ahhhhhhh thanks a lot so they used 0.04 because 0.04 is pretty much the same as 0.046 also, how do we know that the interest rate given is APR or EAR or EFR? Actually, my question was why it is 0.01 instead of 0.046 shouldn't k (k = 0.01) be 0.046 since we calculated k with the effective interest rate formula?