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.
Related
This seems like a simple problem:
let slice = " some wacky text. ";
let trimmed = slice.trim();
// how do I get the index of the start and end within the original slice?
Attempt 1
Look for an alternative API. trim wraps trim_matches which deals with indices internally anyway: so lets copy this code! But this uses std::str::pattern::Pattern which is unstable, thus can't be used outside std in stable Rust.
Attempt 2
Just use trim and calculate the slice indices from the pointers. There's a nice as_ptr_range method, but its also unstable; luckily as the PR says there's an easy work-around.
let slice_ptr = slice.as_ptr();
let trimmed_ptr = trimmed.as_ptr();
// don't bother about the end (we can use trimmed.len())
Now that we've got some pointers, we need their difference. sub is not the right method for this. offset_from is, but it's unstable (as noted in the design, it's only valid use is to compare two pointers into the same slice, which is exactly what we want to do, unfortunately it's yet another thing delayed by the details).
Now, there are hackier ways of solving this problem. We could transmute the pointers to usize (we know the element size is 1 byte, so no need to multiply). But this is most likely the Undefined Behaviour type of unsafe, so lets not go there.
Attempt 3
Edit: the source problem is easy to solve directly, so probably the answer in this case is roll-my-own. Possibly I should just close this.
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'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.
I earlier asked a question about arrays in scheme (turns out they're called vectors but are basically otherwise the same as you'd expect).
Is there an easy way to do multidimensional arrays vectors in PLT Scheme though? For my purposes I'd like to have a procedure called make-multid-vector or something.
By the way if this doesn't already exist, I don't need a full code example of how to implement it. If I have to roll this myself I'd appreciate some general direction though. The way I'd probably do it is to just iterate through each element of the currently highest dimension of the vector to add another dimension, but I can see that being a bit ugly using scheme's recursive setup.
Also, this seems like something I should have been able to find myself so please know that I did actually google it and nothing came up.
The two common approaches are the same as in many languages, either use a vector of vectors, or (more efficiently) use a single vector of X*Y and compute the location of each reference. But there is a library that does that -- look in the docs for srfi/25, which you can get with (require srfi/25).