For the above automaton, the regular expression that has been given in my textbook is the following :
a*(a*ba*ba*ba*)*(a+a*ba*ba*ba*)
I am having trouble deriving this...the following is my attempt at it :
aa* + aa*(ba*ba*ba*)* + ba*ba*ba* + ba*ba*ba*(ba*ba*ba*)*
Either I am wrong or I am not being able to simplify it into the form given in the book. Can someone please guide me here, point out the mistake or explain it to me step-by-step?
I'd really really thankful and appreciate that.
Check this out. It presents three good, algorithmic methods for answering questions like these. Learn one of them, or all three of them if you have the time or inclination. State removal is fairly intuitive, although I like Kleene's transitive closure method.
http://krchowdhary.com/toc/dfa-to-reg-exp.pdf
EDIT: Your RE is equivalent to the one provided. here's the reduction of theirs to yours:
0. a*(a*ba*ba*ba*)*(a+a*ba*ba*ba*)
1. = a*(a*ba*ba*ba*)*a + a*(a*ba*ba*ba*)*a*ba*ba*ba*
2. = a*(ba*ba*ba*)*a + a*(ba*ba*ba*)*ba*ba*ba*
3. = a*a + a*(ba*ba*ba*)*a + a*(ba*ba*ba*)*ba*ba*ba*
4. = aa* + a*(ba*ba*ba*)*ba*ba*ba*a + a*(ba*ba*ba*)*ba*ba*ba*
5. = aa* + a*(ba*ba*ba*)*ba*ba*ba*
6. = aa* + aa*(ba*ba*ba*)*ba*ba*ba* + (ba*ba*ba*)*ba*ba*ba*
7. = aa* + aa*(ba*ba*ba*)* + (ba*ba*ba*)*ba*ba*ba*
8. = aa* + aa*(ba*ba*ba*)* + ba*ba*ba* + (ba*ba*ba*)*ba*ba*ba*
9. = aa* + aa*(ba*ba*ba*)* + ba*ba*ba* + ba*ba*ba*(ba*ba*ba*)*
Step 1 is right since r(s+t) = rs + rt.
Step 2 is right since r*(r*sr*)* = r*(sr*)*.
Step 3 is right since r = r + s if L(s) is a subset of L(r).
Step 4 is right since r*r = rr* and rs + rq*s = rs + rqq*s.
Step 5 is right since rs + r = r.
Step 6 is right since r*s = rr*s + s.
Step 7 is right since rs + rqq*s = rs + rq*s.
Step 8 is right since r = r + s if L(s) is a subset of L(r).
Step 9 is right since r*r = rr*.
Please feel free to ask any questions or point out any mistakes I may have made.
EDIT2: If you are interested in these kinds of questions, show some love for the new Computer Science StackExchange by going to this link and committing!!!
http://area51.stackexchange.com/proposals/35636/computer-science-non-programming?referrer=rpnXA1_2BNYzXN85c5ibxQ2
The textbook seems correct. Taking it step by step:
a*(a*
If this part of the regular expression is true (in other words you do actually read in an 'a'), you will move to state 3. Following the rest of the expression:
ba*
will have you in state 2,
ba*
in state 4 and
ba*
will have you back in state 3.
Now, suppose that you did not read in an 'a' during a*(a*, reading the next b will move you to state 2. You then end up in exactly the same situation as previously, and by following the rest a*ba*ba*) you end up back in state 3.
Since you are now back in state 3, the part (a*ba*ba*ba*)* can execute as many times as it wants, as this will simply be the same as our first scenario (where you read in an 'a' during a*(a*).
The second part simply explains the first scenario again, where you have to read at least one 'a', and then the rest is the same.
Hope this helps, let me know if it still does not make sense. Don't know if my explanation is too clear.
Related
I try to write a simple console application in C++ which can read any chemical formula and afterwards compute its molar mass, for example:
Na2CO3, or something like:
La0.6Sr0.4CoO3, or with brackets:
Fe(NO3)3
The problem is that I don't know in detail how I can deal with the input stream. I think that reading the input and storing it into a char vector may be in this case a better idea than utilizing a common string.
My very first idea was to check all elements (stored in a char vector), step by step: When there's no lowercase after a capital letter, then I have found e.g. an element like Carbon 'C' instead of "Co" (Cobalt) or "Cu" (Copper). Basically, I've tried with the methods isupper(...), islower(...) or isalpha(...).
// first idea, but it seems to be definitely the wrong way
// read input characters from char vector
// check if element contains only one or two letters
// ... and convert them to a string, store them into a new vector
// ... finally, compute the molar mass elsewhere
// but how to deal with the numbers... ?
for (unsigned int i = 0; i < char_vec.size()-1; i++)
{
if (islower(char_vec[i]))
{
char arr[] = { char_vec[i - 1], char_vec[i] };
string temp_arr(arr, sizeof(arr));
element.push_back(temp_arr);
}
else if (isupper(char_vec[i]) && !islower(char_vec[i+1]))
{
char arrSec[] = { char_vec[i] };
string temp_arrSec(arrSec, sizeof(arrSec));
element.push_back(temp_arrSec);
}
else if (!isalpha(char_vec[i]) || char_vec[i] == '.')
{
char arrNum[] = { char_vec[i] };
string temp_arrNum(arrNum, sizeof(arrNum));
stoechiometr_num.push_back(temp_arrNum);
}
}
I need a simple algorithm which can handle with letters and numbers. There also may be the possibility working with pointer, but currently I am not so familiar with this technique. Anyway I am open to that understanding in case someone would like to explain to me how I could use them here.
I would highly appreciate any support and of course some code snippets concerning this problem, since I am thinking for many days about it without progress… Please keep in mind that I am rather a beginner than an intermediate.
This problem is surely not for a beginner but I will try to give you some idea about how you can do that.
Assumption: I am not considering Isotopes case in which atomic mass can be different with same atomic number.
Model it to real world.
How will you solve that in real life?
Say, if I give you Chemical formula: Fe(NO3)3, What you will do is:
Convert this to something like this:
Total Mass => [1 of Fe] + [3 of NO3] => [1 of Fe] + [ 3 of [1 of N + 3 of O ] ]
=> 1 * Fe + 3 * (1 * N + 3 * O)
Then, you will search for individual masses of elements and then substitute them.
Total Mass => 1 * 56 + 3 * (1 * 14 + 3 * 16)
=> 242
Now, come to programming.
Trust me, you have to do the same in programming also.
Convert your chemical formula to the form discussed above i.e. Convert Fe(NO3)3 to Fe*1+(N*1+O*3)*3. I think this is the hardest part in this problem. But it can be done also by breaking down into steps.
Check if all the elements have number after it. If not, then add "1" after it. For example, in this case, O has a number after it which is 3. But Fe and N doesn't have it.
After this step, your formula should change to Fe1(N1O3)3.
Now, Convert each number, say num of above formula to:
*num+ If there is some element after current number.
*num If you encountered ')' or end of formula after it.
After this, your formula should change to Fe*1+(N*1+O*3)*3.
Now, your problem is to solve the above formula. There is a very easy algorithm for this. Please refer to: https://www.geeksforgeeks.org/expression-evaluation/. In your case, your operands can be either a number (say 2) or an element (say Fe). Your operators can be * and +. Parentheses can also be present.
For finding individual masses, you may maintain a std::map<std::string, int> containing element name as key and its mass as value.
Hope this helps a bit.
please help
I'm a beginner to python programming and my problem is this:
I have to make a program which first reads a text file like this one->
A a 1 2 (line one)
A b 3 5 (line two)
A c 9 1
B d 2 4
B e 9 2
C r 3 4
...
and find out: for each First Value (A, B, C, ...), which second value (a, b, c, ...) has max (third value)*(fourth value) (1*2, 3*5, ...) value.
that is, in this example the result should be b, e, r.
And I need to do it 1) without using dictionary class and saving each data
or 2) devise a class and object and do the same thing.
(actually I have to make this program twice by using either methods)
What I'am really confused about is... I made this program first by using dictionary, but I have no idea how to do it with any of those two certain methods mentioned above.
I did this by making dictionary[dictionary[value]] format and (saving each line's data), and found out which one has max value for first value.
How can I do this not on this particular way?
Especially is it even possible to do this on method 1)? (without using dictionary class and saving each data)
thank you for reading my question
I'm really just beginning to learn about this programming and if any of you could give me some advice it would be really appreciated
here is what I've done so far:
The below code works by storing the maximum values and doing comparisons with the values currently being read from the file. This code is not complete as it does not intentionally handle instances where two of the products are the same and it also does not handle an edge case that you should be able to find using your example inputs. I've left those for you to complete.
max_vals = []
with open('FILE.TXT', 'r') as f:
max_first_val = None
max_second_val = None
max_prod = 0
for line in f:
vals = line.strip('\n').split(' ')
curr_prod = int(vals[2]) * int(vals[3])
if vals[0] != max_first_val and max_first_val is not None:
max_vals.append(max_second_val)
max_first_val = vals[0]
max_prod = 0
if curr_prod > max_prod:
max_first_val = vals[0]
max_second_val = vals[1]
max_prod = curr_prod
I'm trying to implement this algorithm but I keep getting a syntax error on the 12th line but I cannot pinpoint what is causing it. I'm new to ocaml and any help would be greatly appreciated.
"To find all the prime numbers less than or equal to a given integer n by Eratosthenes' method:
Create a list of consecutive integers from 2 through n: (2, 3, 4, ..., n).
Initially, let p equal 2, the first prime number.
Starting from p, enumerate its multiples by counting to n in increments of p, and mark them in the list (these will be 2p, 3p, 4p, ... ; the p itself should not be marked).
Find the first number greater than p in the list that is not marked. If there was no such number, stop. Otherwise, let p now equal this new number (which is the next prime), and repeat from step 3."
let prime(n) =
let arr = Array.create n false in
let set_marks (arr , n , prime ) = Array.set arr (n*prime) true in
for i = 2 to n do
set_marks(arr,i,2) done
let findNextPrimeNumberThatIsNotMarked (arr, prime , index ) =
let nextPrime = Array.get arr index in
let findNextPrimeNumberThatIsNotMarkedHelper (arr, prime, index) =
if nextPrime > prime then nextPrime
else prime in
;;
Adding to Jeffrey's answer,
As I have already answered to you at " What exactly is the syntax error here? ",
What you absolutely need to do right now is to install and use a proper OCaml indentation tool, and auto-indent lines. Unexpected auto-indent results often indicate syntactic mistakes like forgetting ;. Without such tools, it is very hard even for talented OCaml programmers to write OCaml code without syntax errors.
There are bunch of auto indenters for OCaml available:
ocp-indent for Emacs and Vim https://github.com/OCamlPro/ocp-indent
Caml mode and Tuareg mode for Emacs
Vim should have some other indenters but I do not know...
OCaml has an expression let a = b in c. Your code ends with in, but where is c? It looks like maybe you should just remove the in at the end.
Looking more closely I see there are more problems than this, sorry.
A function in OCaml is going to look like this roughly:
let f x =
let a = b in
let c = d in
val
Your definition for prime looks exactly like this, except that it ends at the for loop, i.e., with the keyword done.
The rest of the code forms a second, independent, function definition. It has a form like this:
let f x =
let a = b in
let g x = expr in
The syntactic problem is that you're missing an expression after in.
However, your use of indentation suggests you aren't trying to define two different functions. If this is true, you need to rework your code somewhat.
One thing that may be useful (for imperative style programming) is that you can write expr1; expr2 to evaluate two expressions one after the other.
I have a linear optimization goal to Maximize EE+FF, where EE and FF each consist of some C and D.
With code I've written, I can get solver to find:
EE_quantity: 0, FF_quantity: 7
...but I know there to be another solution:
EE_quantity: 1, FF_quantity: 6
In order to validate user input for other valid solutions, I added a constraint for both EE and FF. So I added the EE_quantity == 0, FF_quantity == 7 in the code below, which is a runnable example:
SolverContext c2 = SolverContext.GetContext();
Model m2 = c2.CreateModel();
p.elements = elements_multilevel_productmix();
Decision C_quantity = new Decision(Domain.IntegerNonnegative, "C_quantity");
Decision D_quantity = new Decision(Domain.IntegerNonnegative, "D_quantity");
Decision EE_quantity = new Decision(Domain.IntegerNonnegative, "EE_quantity");
Decision FF_quantity = new Decision(Domain.IntegerNonnegative, "FF_quantity");
m2.AddDecisions(C_quantity, D_quantity, EE_quantity, FF_quantity);
m2.AddConstraints("production",
6 * C_quantity + 4 * D_quantity <= 100,
1 * C_quantity + 2 * D_quantity <= 200,
2 * EE_quantity + 1 * FF_quantity <= C_quantity,
1 * EE_quantity + 2 * FF_quantity <= D_quantity,
EE_quantity == 0,
FF_quantity == 7
);
m2.AddGoal("fixed_EE_FF", GoalKind.Maximize, "EE_quantity + FF_quantity");
Solution sol = c2.Solve(new SimplexDirective());
foreach (var item in sol.Decisions)
{
System.Diagnostics.Debug.WriteLine(
item.Name + ": " + item.GetDouble().ToString()
);
}
It seems that Solver Foundation really doesn't like this specific combination. Using EE_quantity == 1, FF_quantity == 6 is fine, as is using just EE_quantity == 0 or FF_quantity == 7. But using both, AND having one of them being zero, throws an exception:
Index was outside the bounds of the array.
What is going on under the hood, here? And how do I specify that I want to find "all" solutions for a specific problem?
(Note: no new releases of Solver Foundation are forthcoming - it's essentially been dropped by Microsoft.)
The stack trace indicates that this is a bug in the simplex solver's presolve routine. Unfortunately the SimplexDirective does not have a way to disable presolve (unlike InteriorPointDirective). Therefore the way to get around this problem is to specify the fixed variables differently.
Remove the last two constraints that set EE_quantity and FF_quantity, and instead set both the upper and lower bounds to be 0 and 7 respectively when you create the Decision objects. This is equivalent to what you wanted to express, but appears to avoid the MSF bug:
Decision EE_quantity = new Decision(Domain.IntegerRange(0, 0), "EE_quantity");
Decision FF_quantity = new Decision(Domain.IntegerRange(7, 7), "FF_quantity");
The MSF simplex solver, like many mixed integer solvers, only returns the optimal solution. If you want MSF to return all solutions, change to the constraint programming solver (ConstraintProgrammingDirective). If you review the documentation for Solution.GetNext() you should figure out how to do this.
Of course the CP solver is not guaranteed to produce the globally optimal solution immediately. But if you iterate through solutions long enough, you'll get there.
What would be the most efficient algorithm to solve a linear equation in one variable given as a string input to a function? For example, for input string:
"x + 9 – 2 - 4 + x = – x + 5 – 1 + 3 – x"
The output should be 1.
I am considering using a stack and pushing each string token onto it as I encounter spaces in the string. If the input was in polish notation then it would have been easier to pop numbers off the stack to get to a result, but I am not sure what approach to take here.
It is an interview question.
Solving the linear equation is (I hope) extremely easy for you once you've worked out the coefficients a and b in the equation a * x + b = 0.
So, the difficult part of the problem is parsing the expression and "evaluating" it to find the coefficients. Your example expression is extremely simple, it uses only the operators unary -, binary -, binary +. And =, which you could handle specially.
It is not clear from the question whether the solution should also handle expressions involving binary * and /, or parentheses. I'm wondering whether the interview question is intended:
to make you write some simple code, or
to make you ask what the real scope of the problem is before you write anything.
Both are important skills :-)
It could even be that the question is intended:
to separate those with lots of experience writing parsers (who will solve it as fast as they can write/type) from those with none (who might struggle to solve it at all within a few minutes, at least without some hints).
Anyway, to allow for future more complicated requirements, there are two common approaches to parsing arithmetic expressions: recursive descent or Dijkstra's shunting-yard algorithm. You can look these up, and if you only need the simple expressions in version 1.0 then you can use a simplified form of Dijkstra's algorithm. Then once you've parsed the expression, you need to evaluate it: use values that are linear expressions in x and interpret = as an operator with lowest possible precedence that means "subtract". The result is a linear expression in x that is equal to 0.
If you don't need complicated expressions then you can evaluate that simple example pretty much directly from left-to-right once you've tokenised it[*]:
x
x + 9
// set the "we've found minus sign" bit to negate the first thing that follows
x + 7 // and clear the negative bit
x + 3
2 * x + 3
// set the "we've found the equals sign" bit to negate everything that follows
3 * x + 3
3 * x - 2
3 * x - 1
3 * x - 4
4 * x - 4
Finally, solve a * x + b = 0 as x = - b/a.
[*] example tokenisation code, in Python:
acc = None
for idx, ch in enumerate(input):
if ch in '1234567890':
if acc is None: acc = 0
acc = 10 * acc + int(ch)
continue
if acc != None:
yield acc
acc = None
if ch in '+-=x':
yield ch
elif ch == ' ':
pass
else:
raise ValueError('illegal character "%s" at %d' % (ch, idx))
Alternative example tokenisation code, also in Python, assuming there will always be spaces between tokens as in the example. This leaves token validation to the parser:
return input.split()
ok some simple psuedo code that you could use to solve this problem
function(stinrgToParse){
arrayoftokens = stringToParse.match(RegexMatching);
foreach(arrayoftokens as token)
{
//now step through the tokens and determine what they are
//and store the neccesary information.
}
//Use the above information to do the arithmetic.
//count the number of times a variable appears positive and negative
//do the arithmetic.
//add up the numbers both positive and negative.
//return the result.
}
The first thing is to parse the string, to identify the various tokens (numbers, variables and operators), so that an expression tree can be formed by giving operator proper precedences.
Regular expressions can help, but that's not the only method (grammar parsers like boost::spirit are good too, and you can even run your own: its all a "find and recourse").
The tree can then be manipulated reducing the nodes executing those operation that deals with constants and by grouping variables related operations, executing them accordingly.
This goes on recursively until you remain with a variable related node and a constant node.
At the point the solution is calculated trivially.
They are basically the same principles that leads to the production of an interpreter or a compiler.
Consider:
from operator import add, sub
def ab(expr):
a, b, op = 0, 0, add
for t in expr.split():
if t == '+': op = add
elif t == '-': op = sub
elif t == 'x': a = op(a, 1)
else : b = op(b, int(t))
return a, b
Given an expression like 1 + x - 2 - x... this converts it to a canonical form ax+b and returns a pair of coefficients (a,b).
Now, let's obtain the coefficients from both parts of the equation:
le, ri = equation.split('=')
a1, b1 = ab(le)
a2, b2 = ab(ri)
and finally solve the trivial equation a1*x + b1 = a2*x + b2:
x = (b2 - b1) / (a1 - a2)
Of course, this only solves this particular example, without operator precedence or parentheses. To support the latter you'll need a parser, presumable a recursive descent one, which would be simper to code by hand.