A reddit thread brought up an apparently interesting question:
Tail recursive functions can trivially be converted into iterative functions. Other ones, can be transformed by using an explicit stack. Can every recursion be transformed into iteration?
The (counter?)example in the post is the pair:
(define (num-ways x y)
(case ((= x 0) 1)
((= y 0) 1)
(num-ways2 x y) ))
(define (num-ways2 x y)
(+ (num-ways (- x 1) y)
(num-ways x (- y 1))
Can you always turn a recursive function into an iterative one? Yes, absolutely, and the Church-Turing thesis proves it if memory serves. In lay terms, it states that what is computable by recursive functions is computable by an iterative model (such as the Turing machine) and vice versa. The thesis does not tell you precisely how to do the conversion, but it does say that it's definitely possible.
In many cases, converting a recursive function is easy. Knuth offers several techniques in "The Art of Computer Programming". And often, a thing computed recursively can be computed by a completely different approach in less time and space. The classic example of this is Fibonacci numbers or sequences thereof. You've surely met this problem in your degree plan.
On the flip side of this coin, we can certainly imagine a programming system so advanced as to treat a recursive definition of a formula as an invitation to memoize prior results, thus offering the speed benefit without the hassle of telling the computer exactly which steps to follow in the computation of a formula with a recursive definition. Dijkstra almost certainly did imagine such a system. He spent a long time trying to separate the implementation from the semantics of a programming language. Then again, his non-deterministic and multiprocessing programming languages are in a league above the practicing professional programmer.
In the final analysis, many functions are just plain easier to understand, read, and write in recursive form. Unless there's a compelling reason, you probably shouldn't (manually) convert these functions to an explicitly iterative algorithm. Your computer will handle that job correctly.
I can see one compelling reason. Suppose you've a prototype system in a super-high level language like [donning asbestos underwear] Scheme, Lisp, Haskell, OCaml, Perl, or Pascal. Suppose conditions are such that you need an implementation in C or Java. (Perhaps it's politics.) Then you could certainly have some functions written recursively but which, translated literally, would explode your runtime system. For example, infinite tail recursion is possible in Scheme, but the same idiom causes a problem for existing C environments. Another example is the use of lexically nested functions and static scope, which Pascal supports but C doesn't.
In these circumstances, you might try to overcome political resistance to the original language. You might find yourself reimplementing Lisp badly, as in Greenspun's (tongue-in-cheek) tenth law. Or you might just find a completely different approach to solution. But in any event, there is surely a way.
Is it always possible to write a non-recursive form for every recursive function?
Yes. A simple formal proof is to show that both µ recursion and a non-recursive calculus such as GOTO are both Turing complete. Since all Turing complete calculi are strictly equivalent in their expressive power, all recursive functions can be implemented by the non-recursive Turing-complete calculus.
Unfortunately, I’m unable to find a good, formal definition of GOTO online so here’s one:
A GOTO program is a sequence of commands P executed on a register machine such that P is one of the following:
HALT, which halts executionr = r + 1 where r is any registerr = r – 1 where r is any registerGOTO x where x is a labelIF r ≠ 0 GOTO x where r is any register and x is a labelHowever, the conversions between recursive and non-recursive functions isn’t always trivial (except by mindless manual re-implementation of the call stack).
For further information see this answer.
Recursion is implemented as stacks or similar constructs in the actual interpreters or compilers. So you certainly can convert a recursive function to an iterative counterpart because that's how it's always done (if automatically). You'll just be duplicating the compiler's work in an ad-hoc and probably in a very ugly and inefficient manner.
Basically yes, in essence what you end up having to do is replace method calls (which implicitly push state onto the stack) into explicit stack pushes to remember where the 'previous call' had gotten up to, and then execute the 'called method' instead.
I'd imagine that the combination of a loop, a stack and a state-machine could be used for all scenarios by basically simulating the method calls. Whether or not this is going to be 'better' (either faster, or more efficient in some sense) is not really possible to say in general.
Recursive function execution flow can be represented as a tree.
The same logic can be done by a loop, which uses a data-structure to traverse that tree.
Depth-first traversal can be done using a stack, breadth-first traversal can be done using a queue.
So, the answer is: yes. Why: https://mcmap.net/q/82465/-which-recursive-functions-cannot-be-rewritten-using-loops-duplicate.
Can any recursion be done in a single loop? Yes, because
a Turing machine does everything it does by executing a single loop:
- fetch an instruction,
- evaluate it,
- goto 1.
Yes, using explicitly a stack (but recursion is far more pleasant to read, IMHO).
Yes, it's always possible to write a non-recursive version. The trivial solution is to use a stack data structure and simulate the recursive execution.
In principle it is always possible to remove recursion and replace it with iteration in a language that has infinite state both for data structures and for the call stack. This is a basic consequence of the Church-Turing thesis.
Given an actual programming language, the answer is not as obvious. The problem is that it is quite possible to have a language where the amount of memory that can be allocated in the program is limited but where the amount of call stack that can be used is unbounded (32-bit C where the address of stack variables is not accessible). In this case, recursion is more powerful simply because it has more memory it can use; there is not enough explicitly allocatable memory to emulate the call stack. For a detailed discussion on this, see this discussion.
All computable functions can be computed by Turing Machines and hence the recursive systems and Turing machines (iterative systems) are equivalent.
Sometimes replacing recursion is much easier than that. Recursion used to be the fashionable thing taught in CS in the 1990's, and so a lot of average developers from that time figured if you solved something with recursion, it was a better solution. So they would use recursion instead of looping backwards to reverse order, or silly things like that. So sometimes removing recursion is a simple "duh, that was obvious" type of exercise.
This is less of a problem now, as the fashion has shifted towards other technologies.
Recursion is nothing just calling the same function on the stack and once function dies out it is removed from the stack. So one can always use an explicit stack to manage this calling of the same operation using iteration. So, yes all-recursive code can be converted to iteration.
Removing recursion is a complex problem and is feasible under well defined circumstances.
The below cases are among the easy:
Appart from the explicit stack, another pattern for converting recursion into iteration is with the use of a trampoline.
Here, the functions either return the final result, or a closure of the function call that it would otherwise have performed. Then, the initiating (trampolining) function keep invoking the closures returned until the final result is reached.
This approach works for mutually recursive functions, but I'm afraid it only works for tail-calls.
I'd say yes - a function call is nothing but a goto and a stack operation (roughly speaking). All you need to do is imitate the stack that's built while invoking functions and do something similar as a goto (you may imitate gotos with languages that don't explicitly have this keyword too).
Have a look at the following entries on wikipedia, you can use them as a starting point to find a complete answer to your question.
Follows a paragraph that may give you some hint on where to start:
Solving a recurrence relation means obtaining a closed-form solution: a non-recursive function of n.
Also have a look at the last paragraph of this entry.
It is possible to convert any recursive algorithm to a non-recursive one, but often the logic is much more complex and doing so requires the use of a stack. In fact, recursion itself uses a stack: the function stack.
More Details: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Functions
tazzego, recursion means that a function will call itself whether you like it or not. When people are talking about whether or not things can be done without recursion, they mean this and you cannot say "no, that is not true, because I do not agree with the definition of recursion" as a valid statement.
With that in mind, just about everything else you say is nonsense. The only other thing that you say that is not nonsense is the idea that you cannot imagine programming without a callstack. That is something that had been done for decades until using a callstack became popular. Old versions of FORTRAN lacked a callstack and they worked just fine.
By the way, there exist Turing-complete languages that only implement recursion (e.g. SML) as a means of looping. There also exist Turing-complete languages that only implement iteration as a means of looping (e.g. FORTRAN IV). The Church-Turing thesis proves that anything possible in a recursion-only languages can be done in a non-recursive language and vica-versa by the fact that they both have the property of turing-completeness.
Here is an iterative algorithm:
def howmany(x,y)
a = {}
for n in (0..x+y)
for m in (0..n)
a[[m,n-m]] = if m==0 or n-m==0 then 1 else a[[m-1,n-m]] + a[[m,n-m-1]] end
end
end
return a[[x,y]]
end
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choosex = (x+y)!/(x!y!), which doesn't need recursion. – Refund