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I have read a lot about continuations and a very common definition I saw is, it returns the control state.
I am taking a functional programming course taught in SML.
Our professor defined continuations to be:
"What keeps track of what we still have to do"
; "Gives us control of the call stack"
A lot of his examples revolve around trees. Before this chapter, we did tail recursion. I understand that tail recursion lets go of the stack to hold the recursively called functions by having an additional argument to "build" up the answer. Reversing a list would be built in a new accumulator where we append to it accordingly. Also, he said something about functions are called(but not evaluated) except till we reach the end where we replace backwards. He said an improved version of tail recursion would be using CPS(Continuation Programming Style).
Could someone give a simplified explanation of what continuations are and why they are favoured over other programming styles?
I found this stackoverflow link that helped me, but still did not clarify the idea for me:
I just don't get continuations!
Continuations simply treat "what happens next" as first class objects that can be used once unconditionally, ignored in favour of something else, or used multiple times.
To address what Continuation Passing Style is, here is some expression written normally:
let h x = f (g x)
g is applied to x and f is applied to the result.
Notice that g does not have any control. Its result will be passed to f no matter what.
in CPS this is written
let h x next = (g x (fun result -> f result next))
g not only has x as an argument, but a continuation that takes the output of g and returns the final value. This function calls f in the same manner, and gives next as the continuation.
What happened? What changed that made this so much more useful than f (g x)? The difference is that now g is in control. It can decide whether to use what happens next or not. That is the essence of continuations.
An example of where continuations arise are imperative programming languages where you have control structures. Whiles, blocks, ordinary statements, breaks and continues are all generalized through continuations, because these control structures take what happens next and decide what to do with it, for example we can have
...
while(condition1) {
statement1;
if(condition2) break;
statement2;
if(condition3) continue;
statement3;
}
return statement3;
...
The while, the block, the statement, the break and the continue can all be described in a functional model through continuations. Each construct can be considered to be a function that accepts the
current environment containing
the enclosing scopes
optional functions accepting the current environment and returning a continuation to
break from the inner most loop
continue from the inner most loop
return from the current function.
all the blocks associated with it (if-blocks, while-block, etc)
a continuation to the next statement
and returns the new environment.
In the while loop, the condition is evaluated according to the current environment. If it is evaluated to true, then the block is evaluated and returns the new environment. The result of evaluating the while loop again with the new environment is returned. If it is evaluated to false, the result of evaluating the next statement is returned.
With the break statement, we lookup the break function in the environment. If there is no function found then we are not inside a loop and we give an error. Otherwise we give the current environment to the function and return the evaluated continuation, which would be the statement after the the while loop.
With the continue statement the same would happen, except the continuation would be the while loop.
With the return statement the continuation would be the statement following the call to the current function, but it would remove the current enclosing scope from the environment.
I just started learning Erlang and since I found out there is no for loop I tried recreating one with recursion:
display(Rooms, In) ->
Room = array:get(In, Rooms)
io:format("~w", [Room]),
if
In < 59 -> display(Rooms, In + 1);
true -> true
end.
With this code i need to display the content (false or true) of each array in Rooms till the number 59 is reached. However this creates a weird code which displays all of Rooms contents about 60 times (?). When I drop the if statement and only put in the recursive code it is working except for a exception error: Bad Argument.
So basically my question is how do I put a proper end to my "for loop".
Thanks in advance!
Hmm, this code is rewritten and not pasted. It is missing colon after Room = array:get(In, Rooms). The Bad argument error is probably this:
exception error: bad argument
in function array:get/2 (array.erl, line 633)
in call from your_module_name:display/2
This means, that you called array:get/2 with bad arguments: either Rooms is not an array or you used index out of range. The second one is more likely the cause. You are checking if:
In < 59
and then calling display again, so it will get to 58, evaluate to true and call:
display(Rooms, 59)
which is too much.
There is also couple of other things:
In io:format/2 it is usually better to use ~p instead of ~w. It does exactly the same, but with pretty printing, so it is easier to read.
In Erlang if is unnatural, because it evaluates guards and one of them has to match or you get error... It is just really weird.
case is much more readable:
case In < 59 of
false -> do_something();
true -> ok
end
In case you usually write something, that always matches:
case Something of
{One, Two} -> do_stuff(One, Two);
[Head, RestOfList] -> do_other_stuff(Head, RestOfList);
_ -> none_of_the_previous_matched()
end
The underscore is really useful in pattern matching.
In functional languages you should never worry about details like indexes! Array module has map function, which takes function and array as arguments and calls the given function on each array element.
So you can write your code this way:
display(Rooms) ->
DisplayRoom = fun(Index, Room) -> io:format("~p ~p~n", [Index, Room]) end,
array:map(DisplayRoom, Rooms).
This isn't perfect though, because apart from calling the io:format/2 and displaying the contents, it will also construct new array. io:format returns atom ok after completion, so you will get array of 58 ok atoms. There is also array:foldl/3, which doesn't have that problem.
If you don't have to have random access, it would be best to simply use lists.
Rooms = lists:duplicate(58, false),
DisplayRoom = fun(Room) -> io:format("~p~n", [Room]) end,
lists:foreach(DisplayRoom, Rooms)
If you are not comfortable with higher order functions. Lists allow you to easily write recursive algorithms with function clauses:
display([]) -> % always start with base case, where you don't need recursion
ok; % you have to return something
display([Room | RestRooms]) -> % pattern match on list splitting it to first element and tail
io:format("~p~n", [Room]), % do something with first element
display(RestRooms). % recursive call on rest (RestRooms is quite funny name :D)
To summarize - don't write forloops in Erlang :)
This is a general misunderstanding of recursive loop definitions. What you are trying to check for is called the "base condition" or "base case". This is easiest to deal with by matching:
display(0, _) ->
ok;
display(In, Rooms) ->
Room = array:get(In, Rooms)
io:format("~w~n", [Room]),
display(In - 1, Rooms).
This is, however, rather unidiomatic. Instead of using a hand-made recursive function, something like a fold or map is more common.
Going a step beyond that, though, most folks would probably have chosen to represent the rooms as a set or list, and iterated over it using list operations. When hand-written the "base case" would be an empty list instead of a 0:
display([]) ->
ok;
display([Room | Rooms]) ->
io:format("~w~n", [Room]),
display(Rooms).
Which would have been avoided in favor, once again, of a list operation like foreach:
display(Rooms) ->
lists:foreach(fun(Room) -> io:format("~w~n", [Room]) end, Rooms).
Some folks really dislike reading lambdas in-line this way. (In this case I find it readable, but the larger they get the more likely the are to become genuinely distracting.) An alternative representation of the exact same function:
display(Rooms) ->
Display = fun(Room) -> io:format("~w~n", [Room]) end,
lists:foreach(Display, Rooms).
Which might itself be passed up in favor of using a list comprehension as a shorthand for iteration:
_ = [io:format("~w~n", [Room]) | Room <- Rooms].
When only trying to get a side effect, though, I really think that lists:foreach/2 is the best choice for semantic reasons.
I think part of the difficulty you are experiencing is that you have chosen to use a rather unusual structure as your base data for your first Erlang program that does anything (arrays are not used very often, and are not very idiomatic in functional languages). Try working with lists a bit first -- its not scary -- and some of the idioms and other code examples and general discussions about list processing and functional programming will make more sense.
Wait! There's more...
I didn't deal with the case where you have an irregular room layout. The assumption was always that everything was laid out in a nice even grid -- which is never the case when you get into the really interesting stuff (either because the map is irregular or because the topology is interesting).
The main difference here is that instead of simply carrying a list of [Room] where each Room value is a single value representing the Room's state, you would wrap the state value of the room in a tuple which also contained some extra data about that state such as its location or coordinates, name, etc. (You know, "metadata" -- which is such an overloaded, buzz-laden term today that I hate saying it.)
Let's say we need to maintain coordinates in a three-dimensional space in which the rooms reside, and that each room has a list of occupants. In the case of the array we would have divided the array by the dimensions of the layout. A 10*10*10 space would have an array index from 0 to 999, and each location would be found by an operation similar to
locate({X, Y, Z}) -> (1 * X) + (10 * Y) + (100 * Z).
and the value of each Room would be [Occupant1, occupant2, ...].
It would be a real annoyance to define such an array and then mark arbitrarily large regions of it as "unusable" to give the impression of irregular layout, and then work around that trying to simulate a 3D universe.
Instead we could use a list (or something like a list) to represent the set of rooms, but the Room value would now be a tuple: Room = {{X, Y, Z}, [Occupants]}. You may have an additional element (or ten!), like the "name" of the room or some other status information or whatever, but the coordinates are the most certain real identity you're likely to get. To get the room status you would do the same as before, but mark what element you are looking at:
display(Rooms) ->
Display =
fun({ID, Occupants}) ->
io:format("ID ~p: Occupants ~p~n", [ID, Occupants])
end,
lists:foreach(Display, Rooms).
To do anything more interesting than printing sequentially, you could replace the internals of Display with a function that uses the coordinates to plot the room on a chart, check for empty or full lists of Occupants (use pattern matching, don't do it procedurally!), or whatever else you might dream up.
I've been writing (unsophisticated) code for a decent while, and I feel like I have a somewhat firm grasp on while and for loops and if/else statements. I should also say that I feel like I understand (at my level, at least) the concept of recursion. That is, I understand how a method keeps calling itself until the parameters of an iteration match a base case in the method, at which point the methods begin to terminate and pass control (along with values) to previous instances and eventually an overall value of the first call is determined. I may not have explained it very well, but I think I understand it, and I can follow/make traces of the structured examples I've seen. But my question is on creating recursive methods in the wild, ie, in unstructured circumstances.
Our professor wants us to write recursively at every opportunity, and has made the (technically inaccurate?) statement that all loops can be replaced with recursion. But, since many times recursive operations are contained within while or for loops, this means, to state the obvious, not every loop can be replaced with recursion. So...
For unstructured/non-classroom situations,
1) how can I recognize that a loop situation can/cannot be turned into a recursion, and
2) what is the overall idea/strategy to use when applying recursion to a situation? I mean, how should I approach the problem? What aspects of the problem will be used as recursive criteria, etc?
Thanks!
Edit 6/29:
While I appreciate the 2 answers, I think maybe the preamble to my question was too long because it seems to be getting all of the attention. What I'm really asking is for someone to share with me, a person who "thinks" in loops, an approach for implementing recursive solutions. (For purposes of the question, please assume I have a sufficient understanding of the solution, but just need to create recursive code.) In other words, to apply a recursive solution, what am I looking for in the problem/solution that I will then use for the recursion? Maybe some very general statements about applying recursion would be helpful too. (note: please, not definitions of recursion, since I think I pretty much understand the definition. It's just the process of applying them I am asking about.) Thanks!
Every loop CAN be turned into recursion fairly easily. (It's also true that every recursion can be turned into loops, but not always easily.)
But, I realize that saying "fairly easily" isn't actually very helpful if you don't see how, so here's the idea:
For this explanation, I'm going to assume a plain vanilla while loop--no nested loops or for loops, no breaking out of the middle of the loop, no returning from the middle of the loop, etc. Those other things can also be handled but would muddy up the explanation.
The plain vanilla while loop might look like this:
1. x = initial value;
2. while (some condition on x) {
3. do something with x;
4. x = next value;
5. }
6. final action;
Then the recursive version would be
A. def Recursive(x) {
B. if (some condition on x) {
C. do something with x;
D. Recursive(next value);
E. }
F. else { # base case = where the recursion stops
G. final action;
H. }
I.
J. Recursive(initial value);
So,
the initial value of x in line 1 became the orginial argument to Recursive on line J
the condition of the loop on line 2 became the condition of the if on line B
the first action inside the loop on line 3 became the first action inside the if on line C
the next value of x on line 4 became the next argument to Recursive on line D
the final action on line 6 became the action in the base case on line G
If more than one variable was being updated in the loop, then you would often have a corresponding number of arguments in the recursive function.
Again, this basic recipe can be modified to handle fancier situations than plain vanilla while loops.
Minor comment: In the recursive function, it would be more common to put the base case on the "then" side of the if instead of the "else" side. In that case, you would flip the condition of the if to its opposite. That is, the condition in the while loop tests when to keep going, whereas the condition in the recursive function tests when to stop.
I may not have explained it very well, but I think I understand it, and I can follow/make traces of the structured examples I've seen
That's cool, if I understood your explanation well, then how you think recursion works is correct at first glance.
Our professor wants us to write recursively at every opportunity, and has made the (technically inaccurate?) statement that all loops can be replaced with recursion
That's not inaccurate. That's the truth. And the inverse is also possible: every time a recursive function is used, that can be rewritten using iteration. It may be hard and unintuitive (like traversing a tree), but it's possible.
how can I recognize that a loop can/cannot be turned into a recursion
Simple:
what is the overall idea/strategy to use when doing the conversion?
There's no such thing, unfortunately. And by that I mean that there's no universal or general "work-it-all-out" method, you have to think specifically for considering each case when solving a particular problem. One thing may be helpful, however. When converting from an iterative algorithm to a recursive one, think about patterns. How long and where exactly is the part that keeps repeating itself with a small difference only?
Also, if you ever want to convert a recursive algorithm to an iterative one, think about that the overwhelmingly popular approach for implementing recursion at hardware level is by using a (call) stack. Except when solving trivially convertible algorithms, such as the beloved factorial or Fibonacci functions, you can always think about how it might look in assembler, and create an explicit stack. Dirty, but works.
for(int i = 0; i < 50; i++)
{
for(int j = 0; j < 60; j++)
{
}
}
Is equal to:
rec1(int i)
{
if(i < 50)
return;
rec2(0);
rec1(i+1);
}
rec2(int j)
{
if(j < 60)
return;
rec2(j + 1);
}
Every loop can be recursive. Trust your professor, he is right!
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.
Original question:
I know Mathematica has a built in map(f, x), but what does this function look like? I know you need to look at every element in the list.
Any help or suggestions?
Edit (by Jefromi, pieced together from Mike's comments):
I am working on a program what needs to move through a list like the Map, but I am not allowed to use it. I'm not allowed to use Table either; I need to move through the list without help of another function. I'm working on a recursive version, I have an empty list one down, but moving through a list with items in it is not working out. Here is my first case: newMap[#, {}] = {} (the map of an empty list is just an empty list)
I posted a recursive solution but then decided to delete it, since from the comments this sounds like a homework problem, and I'm normally a teach-to-fish person.
You're on the way to a recursive solution with your definition newMap[f_, {}] := {}.
Mathematica's pattern-matching is your friend. Consider how you might implement the definition for newMap[f_, {e_}], and from there, newMap[f_, {e_, rest___}].
One last hint: once you can define that last function, you don't actually need the case for {e_}.
UPDATE:
Based on your comments, maybe this example will help you see how to apply an arbitrary function:
func[a_, b_] := a[b]
In[4]:= func[Abs, x]
Out[4]= Abs[x]
SOLUTION
Since the OP caught a fish, so to speak, (congrats!) here are two recursive solutions, to satisfy the curiosity of any onlookers. This first one is probably what I would consider "idiomatic" Mathematica:
map1[f_, {}] := {}
map1[f_, {e_, rest___}] := {f[e], Sequence##map1[f,{rest}]}
Here is the approach that does not leverage pattern matching quite as much, which is basically what the OP ended up with:
map2[f_, {}] := {}
map2[f_, lis_] := {f[First[lis]], Sequence##map2[f, Rest[lis]]}
The {f[e], Sequence##map[f,{rest}]} part can be expressed in a variety of equivalent ways, for example:
Prepend[map[f, {rest}], f[e]]
Join[{f[e]}, map[f, {rest}] (#Mike used this method)
Flatten[{{f[e]}, map[f, {rest}]}, 1]
I'll leave it to the reader to think of any more, and to ponder the performance implications of most of those =)
Finally, for fun, here's a procedural version, even though writing it made me a little nauseous: ;-)
map3[f_, lis_] :=
(* copy lis since it is read-only *)
Module[{ret = lis, i},
For[i = 1, i <= Length[lis], i++,
ret[[i]] = f[lis[[i]]]
];
ret
]
To answer the question you posed in the comments, the first argument in Map is a function that accepts a single argument. This can be a pure function, or the name of a function that already only accepts a single argument like
In[1]:=f[x_]:= x + 2
Map[f, {1,2,3}]
Out[1]:={3,4,5}
As to how to replace Map with a recursive function of your own devising ... Following Jefromi's example, I'm not going to give to much away, as this is homework. But, you'll obviously need some way of operating on a piece of the list while keeping the rest of the list intact for the recursive part of you map function. As he said, Part is a good starting place, but I'd look at some of the other functions it references and see if they are more useful, like First and Rest. Also, I can see where Flatten would be useful. Finally, you'll need a way to end the recursion, so learning how to constrain patterns may be useful. Incidentally, this can be done in one or two lines depending on if you create a second definition for your map (the easier way), or not.
Hint: Now that you have your end condition, you need to answer three questions:
how do I extract a single element from my list,
how do I reference the remaining elements of the list, and
how do I put it back together?
It helps to think of a single step in the process, and what do you need to accomplish in that step.