I'm having trouble understanding how to write recursive pattern matching functions in TXR. Below I try to define a recursive directive for recognizing file paths. I know in this case I can represent this grammar with a regular expression ([a-z]+\/)+[a-z]+, but I have a more complex rule set in mind for my real code that would benefit from it. What is causing this directive to fail when there is a forward slash?
#(define location)#\
# (cases)#\
#/[a-z]+/#\
# (or)#\
#/[a-z]+//#(location)#\
# (end)#\
#(end)
#(repeat)
#(cases)
#{location (location)}
# (output)
#location is a valid location.
# (end)
#(or)
#location
# (output)
#location is not a valid location.
# (end)
#(end)
#(end)
Example valid inputs:
this/is/valid
this/is/also/valid
this
a/b/c
(Of course, you're almost certainly aware that location is matching a regular language which we can just munge with a regex: /[a-z]+(\/[a-z]+)*/. I'm assuming this is just a "recursion hello world" warm-up for something more complicated.)
The thing about cases is that it uses top to bottom, short-circuited evaluation. The second case cannot match since the first case matches its prefix. It's not like a regex branch operator where the order of the subexpressions doesn't matter.
If I simply swap the two cases, the sample works for me.
What also works (in this particular case) is changing cases to some. The some directive doesn't stop at the first match.
Using some is not a cure-all for this sort of cases ordering problem because sometimes you need to short circuit around a case to terminate recursion (for instance, avoid left-recursing when some condition is hit) or to avoid degenerate performance (exponential time). I'm reminded of a university professor's joke: "you've heard of divide and conquer; this is multiply and surrender".
some also has the property that the later clauses "see" bindings from earlier successfully matching clauses. That might interfere with it being a solution for a cases-ordering problem. That is to say, the later clause might fail to match due to a variable clash. The :resolve feature of some might be helpful in that situation or might not.
Related
Though I have studied and able am able to understand some programs in recursion, I am still not able to intuitively obtain a solution using recursion as I do easily using Iteration. Is there any course or track available in order to build an intuition for recursion? How can one master the concept of recursion?
if you want to gain a thorough understanding of how recursion works, I highly recommend that you start with understanding mathematical induction, as the two are very closely related, if not arguably identical.
Recursion is a way of breaking down seemingly complicated problems into smaller bits. Consider the trivial example of the factorial function.
def factorial(n):
if n < 2:
return 1
return n * factorial(n - 1)
To calculate factorial(100), for example, all you need is to calculate factorial(99) and multiply 100. This follows from the familiar definition of the factorial.
Here are some tips for coming up with a recursive solution:
Assume you know the result returned by the immediately preceding recursive call (e.g. in calculating factorial(100), assume you already know the value of factorial(99). How do you go from there?)
Consider the base case (i.e. when should the recursion come to a halt?)
The first bullet point might seem rather abstract, but all it means is this: a large portion of the work has already been done. How do you go from there to complete the task? In the case of the factorial, factorial(99) constituted this large portion of work. In many cases, you will find that identifying this portion of work simply amounts to examining the argument to the function (e.g. n in factorial), and assuming that you already have the answer to func(n - 1).
Here's another example for concreteness. Let's say we want to reverse a string without using in-built functions. In using recursion, we might assume that string[:-1], or the substring until the very last character, has already been reversed. Then, all that is needed is to put the last remaining character in the front. Using this inspiration, we might come up with the following recursive solution:
def my_reverse(string):
if not string: # base case: empty string
return string # return empty string, nothing to reverse
return string[-1] + my_reverse(string[:-1])
With all of this said, recursion is built on mathematical induction, and these two are inseparable ideas. In fact, one can easily prove that recursive algorithms work using induction. I highly recommend that you checkout this lecture.
I'm trying to understand the semicolon functionality.
I have this code:
del(X,[X|Rest],Rest).
del(X,[Y|Tail],[Y|Rest]) :-
del(X,Tail,Rest).
permutation([],[]).
permutation(L,[X|P]) :- del(X,L,L1), permutation(L1,P).
It's the simple predicate to show all permutations of given list.
I used the built-in graphical debugger in SWI-Prolog because I wanted to understand how it works and I understand for the first case which returns the list given in argument. Here is the diagram which I made for better understanding.
But I don't get it for the another solution. When I press the semicolon it doesn't start in the place where it ended instead it's starting with some deep recursion where L=[] (like in step 9). I don't get it, didn't the recursion end earlier? It had to go out of the recursions to return the answer and after semicolon it's again deep in recursion.
Could someone clarify that to me? Thanks in advance.
One analogy that I find useful in demystifying Prolog is that Backtracking is like Nested Loops, and when the innermost loop's variables' values are all found, the looping is suspended, the vars' values are reported, and then the looping is resumed.
As an example, let's write down simple generate-and-test program to find all pairs of natural numbers above 0 that sum up to a prime number. Let's assume is_prime/1 is already given to us.
We write this in Prolog as
above(0, N), between(1, N, M), Sum is M+N, is_prime(Sum).
We write this in an imperative pseudocode as
for N from 1 step 1:
for M from 1 step 1 until N:
Sum := M+N
if is_prime(Sum):
report_to_user_and_ask(Sum)
Now when report_to_user_and_ask is called, it prints Sum out and asks the user whether to abort or to continue. The loops are not exited, on the contrary, they are just suspended. Thus all the loop variables values that got us this far -- and there may be more tests up the loops chain that sometimes succeed and sometimes fail -- are preserved, i.e. the computation state is preserved, and the computation is ready to be resumed from that point, if the user presses ;.
I first saw this in Peter Norvig's AI book's implementation of Prolog in Common Lisp. He used mapping (Common Lisp's mapcan which is concatMap in Haskell or flatMap in many other languages) as a looping construct though, and it took me years to see that nested loops is what it is really all about.
Goals conjunction is expressed as the nesting of the loops; goals disjunction is expressed as the alternatives to loop through.
Further twist is that the nested loops' structure isn't fixed from the outset. It is fluid, the nested loops of a given loop can be created depending on the current state of that loop, i.e. depending on the current alternative being explored there; the loops are written as we go. In (most of the) languages where such dynamic creation of nested loops is impossible, it can be encoded with nested recursion / function invocation / inside the loops. (Here's one example, with some pseudocode.)
If we keep all such loops (created for each of the alternatives) in memory even after they are finished with, what we get is the AND-OR tree (mentioned in the other answer) thus being created while the search space is being explored and the solutions are found.
(non-coincidentally this fluidity is also the essence of "monad"; nondeterminism is modeled by the list monad; and the essential operation of the list monad is the flatMap operation which we saw above. With fluid structure of loops it is "Monad"; with fixed structure it is "Applicative Functor"; simple loops with no structure (no nesting at all): simply "Functor" (the concepts used in Haskell and the like). Also helps to demystify those.)
So, the proper slogan could be Backtracking is like Nested Loops, either fixed, known from the outset, or dynamically-created as we go. It's a bit longer though. :)
Here's also a Prolog example, which "as if creates the code to be run first (N nested loops for a given value of N), and then runs it." (There's even a whole dedicated tag for it on SO, too, it turns out, recursive-backtracking.)
And here's one in Scheme ("creates nested loops with the solution being accessible in the innermost loop's body"), and a C++ example ("create n nested loops at run-time, in effect enumerating the binary encoding of 2n, and print the sums out from the innermost loop").
There is a big difference between recursion in functional/imperative programming languages and Prolog (and it really became clear to me only in the last 2 weeks or so):
In functional/imperative programming, you recurse down a call chain, then come back up, unwinding the stack, then output the result. It's over.
In Prolog, you recurse down an AND-OR tree (really, alternating AND and OR nodes), selecting a predicate to call on an OR node (the "choicepoint"), from left to right, and calling every predicate in turn on an AND node, also from left to right. An acceptable tree has exactly one predicate returning TRUE under each OR node, and all predicates returning TRUE under each AND node. Once an acceptable tree has been constructed, by the very search procedure, we are (i.e. the "search cursor" is) on a rightmost bottommost node .
Success in constructing an acceptable tree also means a solution to the query entered at the Prolog Toplevel (the REPL) has been found: The variable values are output, but the tree is kept (unless there are no choicepoints).
And this is also important: all variables are global in the sense that if a variable X as been passed all the way down the call chain from predicate to predicate to the rightmost bottommost node, then constrained at the last possible moment by unifying it with 2 for example, X = 2, then the Prolog Toplevel is aware of that without further ado: nothing needs to be passed up the call chain.
If you now press ;, search doesn't restart at the top of the tree, but at the bottom, i.e. at the current cursor position: the nearest parent OR node is asked for more solutions. This may result in much search until a new acceptable tree has been constructed, we are at a new rightmost bottommost node. The new variable values are output and you may again enter ;.
This process cycles until no acceptable tree can be constructed any longer, upon which false is output.
Note that having this AND-OR as an inspectable and modifiable data structure at runtime allows some magical tricks to be deployed.
There is bound to be a lot of power in debugging tools which record this tree to help the user who gets the dreaded sphynxian false from a Prolog program that is supposed to work. There are now Time Traveling Debuggers for functional and imperative languages, after all...
OK regex nerds!
I am using regex lookahead assertions for password validation that is similar to the pattern described here:
\A(?=\w{6,10}\z)(?=[^a-z]*[a-z])(?=(?:[^A-Z]*[A-Z]){3})(?=\D*\d)
However, we want to only require that any 3 of the 4 assertions be valid - not necessarily all of them. Any thoughts on how this could be done?
To shorten any kind of pattern, factorize:
\A(?:
(?=\w{6,10}\z) (?=.*[a-z]) (?: (?:.*[A-Z]){3} | .*\d )
|
(?=.*\d) (?=(?:.*[A-Z]){3}) (?: .*[a-z] | \w{6,10}\z )
)
Note that you don't need a lookahead to test the last condition.
demo
Other way, where each condition is optional and that uses a named group to count (.net only):
\A
(?<c>(?=\w{6,10}\z))?
(?<c>(?=[^a-z]*[a-z]))?
(?<c>(?=(?:[^A-Z]*[A-Z]){3}))?
(?<c>(?=\D*\d))?
(?<-c>){3} # decrement c 3 times
(?(c)|(?!$)) # conditional: force the pattern to fail if too few conditions succeed.
demo
There's no "easy" way to do this in a single regular expression. The only way would be to define all possible permutations of the "three out of four" assertions - e.g.
\A(?=\w{6,10}\z)(?=[^a-z]*[a-z])(?=(?:[^A-Z]*[A-Z]){3})| # Maybe no digit
\A(?=[^a-z]*[a-z])(?=(?:[^A-Z]*[A-Z]){3})(?=\D*\d)| # Maybe wrong length
\A(?=\w{6,10}\z)(?=(?:[^A-Z]*[A-Z]){3})(?=\D*\d)| # Maybe no lower
\A(?=\w{6,10}\z)(?=[^a-z]*[a-z])(?=\D*\d) # Maybe not enough uppers
However, this mind-melting regex is clearly not a good solution.
A better approach would be to perform the four checks separately (with regex or otherwise), and count that there is at least three passed conditions.
...However, let's take a step back here and ask: Why are you doing this?? You're implementing a password entropy check. Based on your fuzzy rules, the following passwords are valid:
AAAa1
password1
LETmein
And the following passwords are invalid:
reallylongsecurepassword8374235359232
HorseBatteryStapleCorrect
I would strongly advise against such a bizarrely restrictive policy.
Brief
The easiest method would be to have separate regular expressions and check whether 3/4 of them are successful in your code's language. The only way to do this in regex is to present all cases. That being said, this is probably the easiest method (in regex) to present all options as it allows you to edit the patterns in one location (where they are defined) rather than multiple times (more prone to bugs). The DEFINE constructs in regex are seldom supported, but PCRE regex does.
You can also have your code generate each regex permutation. See this question about generating all permutations of a list in python
I don't know why you want to do this for passwords, it's considered malpractice, but, since you're asking for it, I figured I'd give you the easiest solution possible in regex... You really should only check minimum length (and complexity if you want [based on algorithms] to show the user how secure your system finds their password to be).
Code
(?(DEFINE)
(?<w>(?=\w{6,10}\z))
(?<l>(?=[^a-z]*[a-z]))
(?<u>(?=(?:[^A-Z]*[A-Z]){3}))
(?<d>(?=\D*\d))
)
\A(?:
(?&w)(?&l)(?&u)|
(?&w)(?&l)(?&d)|
(?&w)(?&u)(?&d)|
(?&l)(?&u)(?&d)
)
Note: The regex above uses the x modifier (ignore whitespace) so that we can nicely organize the content.
How can i write a substring function in ocaml without using any assignments lists and iterations, only recursions? i can only use string.length.
i tried so far is
let substring s s2 start stop=
if(start < stop) then
substring s s2 (start+1) stop
else s2;;
but obviously it is wrong, problem is that how can i pass the string that is being built gradually with recursive calls?
This feels like a homework problem that is intended to teach you think think about recursion. For me it would be easier to think about the recursion part if you decide on the basic operations you're going to use. You can't use assignments, lists, or iterations, okay. You need to extract parts of your input string somehow, but you obviously can't use the built-in substring function to do this, that would defeat the purpose of the exercise. The only other operation I can think of is the one that extracts a single character from a string:
# "abcd".[2];;
- : char = 'c'
You also need a way to add a character to a string, giving a longer string. But you're not allowed to use assignment to do this. It seems to me you're going to have to use String.make to translate your character to a string:
# String.make 1 'a';;
- : string = "a"
Then you can concatenate two strings using the ^ operator:
# "abc" ^ "def"
- : string = "abcdef"
Are you allowed to use these three operations? If so, you can start thinking about the recursion part of the substring problem. If not, then I probably don't understand the problem well enough yet to give advice. (Or maybe whoever set up the restrictions didn't expect you to have to calculate substrings? Usually the restrictions are also a kind of hint as to how you should proceed.)
Moving on to your specific question. In beginning FP programming, you don't generally want to pass the answer down to recursive calls. You want to pass a smaller problem down to the recursive call, and get the answer back from it. For the substring problem, an example of a smaller problem is to ask for the substring that starts one character further along in the containing string, and that is one character shorter.
(Later on, you might want to pass partial answers down to your recursive calls in order to get tail-recursive behavior. I say don't worry about it for now.)
Now I can't give you the answer to this, Partly because it's your homework, and partly because it's been 3 years since I've touched OCaml syntax, but I could try to help you along.
Now the Basic principle behind recursion is to break a problem down into smaller versions of itself.
You don't pass the string that is slowly being built up, instead use your recursive function to generate a string that is almost built up except for a single character, and then you add that character to the end of the string.
am making a function that will send me a list of all possible elemnts .. in each iteration its giving me the last answer .. but after the recursion am only getting the last answer back .. how can i make it give back every single answer ..
thank you
the problem is that am trying to find all possible distributions for a list into other lists .. the code
addIn(_,[],Result,Result).
addIn(C,[Element|Rest],[F|R],Result):-
member( Members , [F|R]),
sumlist( Members, Sum),
sumlist([Element],ElementLength),
Cap is Sum + ElementLength,
(Cap =< Ca,
append([Element], Members,New)....
by calling test .. am getting back all the list of possible answers .. now if i tried to do something that will fail like
bp(3,11,[8,2,4,6,1,8,4],Answer).
it will just enter a while loop .. more over if i changed the
bp(NB,C,OL,A):-
addIn(C,OL,[[],[],[]],A);
bp(NB,C,_,A).
to and instead of Or .. i get error :
ERROR: is/2: Arguments are not
sufficiently instantiated
appreciate the help ..
Thanks alot #hardmath
It sounds like you are trying to write your own version of findall/3, perhaps limited to a special case of an underlying goal. Doing it generally (constructing a list of all solutions to a given goal) in a user-defined Prolog predicate is not possible without resorting to side-effects with assert/retract.
However a number of useful special cases can be implemented without such "tricks". So it would be helpful to know what predicate defines your "all possible elements". [It may also be helpful to state which Prolog implementation you are using, if only so that responses may include links to documentation for that version.]
One important special case is where the "universe" of potential candidates already exists as a list. In that case we are really asking to find the sublist of "all possible elements" that satisfy a particular goal.
findSublist([ ],_,[ ]).
findSublist([H|T],Goal,[H|S]) :-
Goal(H),
!,
findSublist(T,Goal,S).
findSublist([_|T],Goal,S) :-
findSublist(T,Goal,S).
Many Prologs will allow you to pass the name of a predicate Goal around as an "atom", but if you have a specific goal in mind, you can leave out the middle argument and just hardcode your particular condition into the middle clause of a similar implementation.
Added in response to code posted:
I think I have a glimmer of what you are trying to do. It's hard to grasp because you are not going about it in the right way. Your predicate bp/4 has a single recursive clause, variously attempted using either AND or OR syntax to relate a call to addIn/4 to a call to bp/4 itself.
Apparently you expect wrapping bp/4 around addIn/4 in this way will somehow cause addIn/4 to accumulate or iterate over its solutions. It won't. It might help you to see this if we analyze what happens to the arguments of bp/4.
You are calling the formal arguments bp(NB,C,OL,A) with simple integers bound to NB and C, with a list of integers bound to OL, and with A as an unbound "output" Answer. Note that nothing is ever done with the value NB, as it is not passed to addIn/4 and is passed unchanged to the recursive call to bp/4.
Based on the variable names used by addIn/4 and supporting predicate insert/4, my guess is that NB was intended to mean "number of bins". For one thing you set NB = 3 in your test/0 clause, and later you "hardcode" three empty lists in the third argument in calling addIn/4. Whatever Answer you get from bp/4 comes from what addIn/4 is able to do with its first two arguments passed in, C and OL, from bp/4. As we noted, C is an integer and OL a list of integers (at least in the way test/0 calls bp/4).
So let's try to state just what addIn/4 is supposed to do with those arguments. Superficially addIn/4 seems to be structured for self-recursion in a sensible way. Its first clause is a simple termination condition that when the second argument becomes an empty list, unify the third and fourth arguments and that gives "answer" A to its caller.
The second clause for addIn/4 seems to coordinate with that approach. As written it takes the "head" Element off the list in the second argument and tries to find a "bin" in the third argument that Element can be inserted into while keeping the sum of that bin under the "cap" given by C. If everything goes well, eventually all the numbers from OL get assigned to a bin, all the bins have totals under the cap C, and the answer A gets passed back to the caller. The way addIn/4 is written leaves a lot of room for improvement just in basic clarity, but it may be doing what you need it to do.
Which brings us back to the question of how you should collect the answers produced by addIn/4. Perhaps you are happy to print them out one at a time. Perhaps you meant to collect all the solutions produced by addIn/4 into a single list. To finish up the exercise I'll need you to clarify what you really want to do with the Answers from addIn/4.
Let's say you want to print them all out and then stop, with a special case being to print nothing if the arguments being passed in don't allow a solution. Then you'd probably want something of this nature:
newtest :-
addIn(12,[7, 3, 5, 4, 6, 4, 5, 2], Answer),
format("Answer = ~w\n",[Answer]),
fail.
newtest.
This is a standard way of getting predicate addIn/4 to try all possible solutions, and then stop with the "fall-through" success of the second clause of newtest/0.
(Added) Suggestions about coding addIn/4:
It will make the code more readable and maintainable if the variable names are clear. I'd suggest using Cap instead of C as the first argument to addIn/4 and BinSum when you take the sum of items assigned to a "bin". Likewise Bin would be better where you used Members. In the third argument to addIn/4 (in the head of the second clause) you don't need an explicit list structure [F|R] since you never refer to either part F or R by itself. So there I'd use Bins.
Some of your predicate calls don't accomplish much that you cannot do more easily. For example, your second call to sumlist/2 involves a list with one item. Thus the sum is just the same as that item, i.e. ElementLength is the same as Element. Here you could just replace both calls to sumlist/2 with one such call:
sumlist([Element|Bin],BinSum)
and then do your test comparing BinSum with Cap. Similarly your call to append/3 just adjoins the single item Element to the front of the list (I'm calling) Bin, so you could just replace what you have called New with [Element|Bin].
You have used an extra pair of parentheses around the last four subgoals (in the second clause for addIn/4). Since AND is implied for all the subgoals of this clause, using the extra pair of parentheses is unnecessary.
The code for insert/4 isn't shown now, but it could be a source of some unintended "backtracking" in special cases. The better approach would be to have the first call (currently to member/2) be your only point of indeterminacy, i.e. when you choose one of the bins, do it by replacing it with a free variable that gets unified with [Element|Bin] at the next to last step.