I have written the following recursive function, but am incurring a runtime error due to maximum recursion depth. I was wondering is it possible to write an iterative function to overcome this:
def finaldistance(n):
if n%2 == 0:
return 1 + finaldistance(n//2)
elif n != 1:
a = finaldistance(n-1)+1
b = distance(n)
return min(a,b)
else:
return 0
What I have tried is this but it does not seem to be working,
def finaldistance(n, acc):
while n > 1:
if n%2 == 0:
(n, acc) = (n//2, acc+1)
else:
a = finaldistance(n-1, acc) + 1
b = distance(n)
if a < b:
(n, acc) = (n-1, acc+1)
else:
(n, acc) =(1, acc + distance(n))
return acc
Johnbot's solution shows you how to solve your specific problem. How in general can we remove this recursion? Let me show you how, by making a series of small, clearly correct, clearly safe refactorings.
First, here's a slightly rewritten version of your function. I hope you agree it is the same:
def f(n):
if n % 2 == 0:
return 1 + f(n // 2)
elif n != 1:
a = f(n - 1) + 1
b = d(n)
return min(a, b)
else:
return 0
I want the base case to be first. This function is logically the same:
def f(n):
if n == 1:
return 0
if n % 2 == 0:
return 1 + f(n // 2)
a = f(n - 1) + 1
b = d(n)
return min(a, b)
I want the code that comes after each recursive call to be a method call and nothing else. These functions are logically the same:
def add_one(n, x):
return 1 + x
def min_distance(n, x):
a = x + 1
b = d(n)
return min(a, b)
def f(n):
if n == 1:
return 0
if n % 2 == 0:
return add_one(n, f(n // 2))
return min_distance(n, f(n - 1))
Similarly, we add helper functions that compute the recursive argument:
def half(n):
return n // 2
def less_one(n):
return n - 1
def f(n):
if n == 1:
return 0
if n % 2 == 0:
return add_one(n, f(half(n))
return min_distance(n, f(less_one(n))
Again, make sure you agree that this program is logically the same. Now I'm going to simplify the computation of the argument:
def get_argument(n):
return half if n % 2 == 0 else less_one
def f(n):
if n == 1:
return 0
argument = get_argument(n) # argument is a function!
if n % 2 == 0:
return add_one(n, f(argument(n)))
return min_distance(n, f(argument(n)))
Now I'm going to do the same thing to the code after the recursion, and we'll get down to a single recursion:
def get_after(n):
return add_one if n % 2 == 0 else min_distance
def f(n):
if n == 1:
return 0
argument = get_argument(n)
after = get_after(n) # this is also a function!
return after(n, f(argument(n)))
Now I'm noticing that we're passing n to get_after, and then passing it right along to "after" again. I'm going to curry these functions to eliminate that problem. This step is tricky. Make sure you understand it!
def add_one(n):
return lambda x: x + 1
def min_distance(n):
def nested(x):
a = x + 1
b = d(n)
return min(a, b)
return nested
These functions did take two arguments. Now they take one argument, and return a function that takes one argument! So we refactor the use site:
def get_after(n):
return add_one(n) if n % 2 == 0 else min_distance(n)
and here:
def f(n):
if n == 1:
return 0
argument = get_argument(n)
after = get_after(n) # now this is a function of one argument, not two
return after(f(argument(n)))
Similarly we notice that we are calling get_argument(n)(n) to get the argument. Let's simplify that:
def get_argument(n):
return half(n) if n % 2 == 0 else less_one(n)
And let's make it just slightly more general:
base_case_value = 0
def is_base_case(n):
return n == 1
def f(n):
if is_base_case(n):
return base_case_value
argument = get_argument(n)
after = get_after(n)
return after(f(argument))
OK, we now have our program in an extremely compact form. The logic has been spread out into multiple functions, and some of them are curried, to be sure. But now that the function is in this form we can easily remove the recursion. This is the bit that is really tricky is turning the whole thing into an explicit stack:
def f(n):
# Let's make a stack of afters.
afters = [ ]
while not is_base_case(n) :
argument = get_argument(n)
after = get_after(n)
afters.append(after)
n = argument
# Now we have a stack of afters:
x = base_case_value
while len(afters) != 0:
after = afters.pop()
x = after(x)
return x
Study this implementation very carefully. You will learn a lot from it. Remember, when you do a recursive call:
after(f(something))
you are saying that after is the continuation -- the thing that comes next -- of the call to f. We typically implement continuations by putting information about the location in the callers code onto the "call stack". What we're doing in this removal of recursion is simply moving continuation information off of the call stack and onto a stack data structure. But the information is exactly the same.
The important thing to realize here is that we typically think of the call stack as "what is the thing that happened in the past that got me here?". That is exactly backwards. The call stack tells you what you have to do after this call is finished! So that's the information that we encode in the explicit stack. Nowhere do we encode what we did before each step as we "unwind the stack", because we don't need that information.
As I said in my initial comment: there is always a way to turn a recursive algorithm into an iterative one but it is not always easy. I've shown you here how to do it: carefully refactor the recursive method until it is extremely simple. Get it down to a single recursion by refactoring it. Then, and only then, apply this transformation to get it into an explicit stack form. Practice that until you are comfortable with this program transformation. You can then move on to more advanced techniques for removing recursions.
Note that of course this is almost certainly not the "pythonic" way to solve this problem; you could likely build a much more compact, understandable method using lazily evaluated list comprehensions. This answer was intended to answer the specific question that was asked: how in general do we turn recursive methods into iterative methods?
I mentioned in a comment that a standard technique for removing a recursion is to build an explicit list as a stack. This shows that technique. There are other techniques: tail recursion, continuation passing style and trampolines. This answer is already too long, so I'll cover those in a follow-up answer.
Read this answer after you read my first answer.
Again, we are answering the question in general of "how do you turn a recursive algorithm into an iterative algorithm", in this case in Python. As noted previously, this is about exploring the general idea of transforming a program; this is not the "pythonic" way to solve the specific problem.
In my first answer I started by rewriting the program into this form:
def f(n):
if is_base_case(n):
return base_case_value
argument = get_argument(n)
after = get_after(n)
return after(f(argument))
And then transformed it into this form:
def f(n):
# Let's make a stack of afters.
afters = [ ]
while not is_base_case(n) :
argument = get_argument(n)
after = get_after(n)
afters.append(after)
n = argument
# Now we have a stack of afters:
x = base_case_value
while len(afters) != 0:
after = afters.pop()
x = after(x)
return x
The technique here is to construct an explicit stack of "after" calls for a particular input, and then once we have it, run down the whole stack. We are essentially simulating what the runtime already does: constructs a stack of "continuations" that say what to do next.
A different technique is to let the function itself decide what to do with its continuation; this is called "continuation passing style". Let's explore it.
This time, we're going to add a parameter c to the recursive method f. c is a function that takes what would normally be the return value of f, and does whatever was suppose to happen after the call to f. That is, it is explicitly the continuation of f. The method f then becomes "void returning".
The base case is easy. What do we do if we're in the base case? We call the continuation with the value we would have returned:
def f(n, c):
if is_base_case(n):
c(base_case_value)
return
Easy peasy. What about the non-base case? Well, what were we going to do in the original program? We were going to (1) get the arguments, (2) get the "after" -- the continuation of the recursive call, (3) do the recursive call, (4) call "after", its continuation, and (5) return the computed value to whatever the continuation of f is.
We're going to do all the same things, except that when we do step (3) now we need to pass in a continuation that does steps 4 and 5:
argument = get_argument(n)
after = get_after(n)
f(argument, lambda x: c(after(x)))
Hey, that is so easy! What do we do after the recursive call? Well, we call after with the value returned by the recursive call. But now that value is going to be passed to the recursive call's continuation function, so it just goes into x. What happens after that? Well, whatever was going to happen next, and that's in c, so it needs to be called, and we're done.
Let's try it out. Previously we would have said
print(f(100))
but now we have to pass in what happens after f(100). Well, what happens is, the value gets printed!
f(100, print)
and we're done.
So... big deal. The function is still recursive. Why is this interesting? Because the function is now tail recursive! That is, the last thing it does in the non-base case is call itself. Consider a silly case:
def tailcall(x, sum):
if x <= 0:
return sum
return tailcall(x - 1, sum + x)
If we call tailcall(10, 0) it calls tailcall(9, 10), which calls (8, 19), and so on. But any tail-recursive method we can rewrite into a loop very, very easily:
def tailcall(x, sum):
while True:
if x <= 0:
return sum
x = x - 1
sum = sum + x
So can we do the same thing with our general case?
# This is wrong!
def f(n, c):
while True:
if is_base_case(n):
c(base_case_value)
return
argument = get_argument(n)
after = get_after(n)
n = argument
c = lambda x: c(after(x))
Do you see what is wrong? the lambda is closed over c and after, which means that every lambda will use the current value of c and after, not the value it had when the lambda was created. So this is broken, but we can fix it easily by creating a scope which introduces new variables every time it is invoked:
def continuation_factory(c, after)
return lambda x: c(after(x))
def f(n, c):
while True:
if is_base_case(n):
c(base_case_value)
return
argument = get_argument(n)
after = get_after(n)
n = argument
c = continuation_factory(c, after)
And we're done! We've turned this recursive algorithm into an iterative algorithm.
Or... have we?
Think about this really carefully before you read on. Your spider sense should be telling you that something is wrong here.
The problem we started with was that a recursive algorithm is blowing the stack. We've turned this into an iterative algorithm -- there's no recursive call at all here! We just sit in a loop updating local variables.
The question though is -- what happens when the final continuation is called, in the base case? What does that continuation do? Well, it calls its after, and then it calls its continuation. What does that continuation do? Same thing.
All we've done here is moved the recursive control flow into a collection of function objects that we've built up iteratively, and calling that thing is still going to blow the stack. So we haven't actually solved the problem.
Or... have we?
What we can do here is add one more level of indirection, and that will solve the problem. (This solves every problem in computer programming except one problem; do you know what that problem is?)
What we'll do is we'll change the contract of f so that it is no longer "I am void-returning and will call my continuation when I'm done". We will change it to "I will return a function that, when it is called, calls my continuation. And furthermore, my continuation will do the same."
That sounds a little tricky but really its not. Again, let's reason it through. What does the base case have to do? It has to return a function which, when called, calls my continuation. But my continuation already meets that requirement:
def f(n, c):
if is_base_case(n):
return c(base_case_value)
What about the recursive case? We need to return a function, which when called, executes the recursion. The continuation of that call needs to be a function that takes a value and returns a function that when called executes the continuation on that value. We know how to do that:
argument = get_argument(n)
after = get_after(n)
return lambda : f(argument, lambda x: lambda: c(after(x)))
OK, so how does this help? We can now move the loop into a helper function:
def trampoline(f, n, c):
t = f(n, c)
while t != None:
t = t()
And call it:
trampoline(f, 3, print)
And holy goodness it works.
Follow along what happens here. Here's the call sequence with indentation showing stack depth:
trampoline(f, 3, print)
f(3, print)
What does this call return? It effectively returns lambda : f(2, lambda x: lambda : print(min_distance(x)), so that's the new value of t.
That's not None, so we call t(), which calls:
f(2, lambda x: lambda : print(min_distance(x))
What does that thing do? It immediately returns
lambda : f(1,
lambda x:
lambda:
(lambda x: lambda : print(min_distance(x)))(add_one(x))
So that's the new value of t. It's not None, so we invoke it. That calls:
f(1,
lambda x:
lambda:
(lambda x: lambda : print(min_distance(x)))(add_one(x))
Now we're in the base case, so we *call the continuation, substituting 0 for x. It returns:
lambda: (lambda x: lambda : print(min_distance(x)))(add_one(0))
So that's the new value of t. It's not None, so we invoke it.
That calls add_one(0) and gets 1. It then passes 1 for x in the middle lambda. That thing returns:
lambda : print(min_distance(1))
So that's the new value of t. It's not None, so we invoke it. And that calls
print(min_distance(1))
Which prints out the correct answer, print returns None, and the loop stops.
Notice what happened there. The stack never got more than two deep because every call returned a function that said what to do next to the loop, rather than calling the function.
If this sounds familiar, it should. Basically what we're doing here is making a very simple work queue. Every time we "enqueue" a job, it is immediately dequeued, and the only thing the job does is enqueues the next job by returning a lambda to the trampoline, which sticks it in its "queue", the variable t.
We break the problem up into little pieces, and make each piece responsible for saying what the next piece is.
Now, you'll notice that we end up with arbitrarily deep nested lambdas, just as we ended up in the previous technique with an arbitrarily deep queue. Essentially what we've done here is moved the workflow description from an explicit list into a network of nested lambdas, but unlike before, this time we've done a little trick to avoid those lambdas ever calling each other in a manner that increases the stack depth.
Once you see this pattern of "break it up into pieces and describe a workflow that coordinates execution of the pieces", you start to see it everywhere. This is how Windows works; each window has a queue of messages, and messages can represent portions of a workflow. When a portion of a workflow wishes to say what the next portion is, it posts a message to the queue, and it runs later. This is how async await works -- again, we break up the workflow into pieces, and each await is the boundary of a piece. It's how generators work, where each yield is the boundary, and so on. Of course they don't actually use trampolines like this, but they could.
The key thing to understand here is the notion of continuation. Once you realize that you can treat continuations as objects that can be manipulated by the program, any control flow becomes possible. Want to implement your own try-catch? try-catch is just a workflow where every step has two continuations: the normal continuation and the exceptional continuation. When there's an exception, you branch to the exceptional continuation instead of the regular continuation. And so on.
The question here was again, how do we eliminate an out-of-stack caused by a deep recursion in general. I've shown that any recursive method of the form
def f(n):
if is_base_case(n):
return base_case_value
argument = get_argument(n)
after = get_after(n)
return after(f(argument))
...
print(f(10))
can be rewritten as:
def f(n, c):
if is_base_case(n):
return c(base_case_value)
argument = get_argument(n)
after = get_after(n)
return lambda : f(argument, lambda x: lambda: c(after(x)))
...
trampoline(f, 10, print)
and that the "recursive" method will now use only a very small, fixed amount of stack.
First you need to find all the values of n, luckily your sequence is strictly descending and only depends on the next distance:
values = []
while n > 1:
values.append(n)
n = n // 2 if n % 2 == 0 else n - 1
Next you need to calculate the distance at each value. To do that we need to start from the buttom:
values.reverse()
And now we can easily keep track of the previous distance if we need it to calculate the next distance.
distance_so_far = 0
for v in values:
if v % 2 == 0:
distance_so_far += 1
else:
distance_so_far = min(distance(v), distance_so_far + 1)
return distance_so_far
Stick it all together:
def finaldistance(n):
values = []
while n > 1:
values.append(n)
n = n // 2 if n % 2 == 0 else n - 1
values.reverse()
distance_so_far = 0
for v in values:
if v % 2 == 0:
distance_so_far += 1
else:
distance_so_far = min(distance(v), distance_so_far + 1)
return distance_so_far
And now you're using memory instead of stack.
(I don't program in Python so this is probably not be idiomatic Python)
Related
I'm practicing memoization to improve recursive functions, so I wrote this memoized Fibonacci generator:
memo = {}
def memo_fibo(n):
if n not in memo:
if n < 2:
memo[n] = n
else:
memo[n] = memo_fibo(n - 2) + memo_fibo(n - 1)
return memo[n]
And this actually works well! Next I wanted to generalize the idea of memoization, I wanted to write a function that basically adds memoization to some other function, so I wrote this:
def memoize(f):
cache = {}
def memoized(x):
if x not in cache:
cache[x] = f(x)
return cache[x]
return memoized
def fibo(n):
if n < 2:
return n
return fibo(n - 2) + fibo(n - 1)
mf = memoize(fibo)
But this just doesn't work for some reason, even though I'm just putting one function inside of another. What's even weirder to me is that if you do:
#memoize
def fibo(n):
if n < 2:
return n
return fibo(n - 2) + fibo(n - 1)
Then it works fine. But why? Why is simple function composition not working properly and giving a different result? Is there a syntax error in the way I compose those 2 functions? Or maybe the memoize(f) function is just not built correctly for composition?
I'm asking because I still don't understand decorators very well and if I knew how to make those 2 versions equivalent that'd help me a lot with them.
Your memoize function decorator returns a new function (that you called memoized) that "wraps" the function you give in.
The issue is that doing mf = memoize(fibo) you save the new decorated function in the mf function, but since it is recursive, when it does fibo(n - 2) + fibo(n - 1), it still calls the non-decorated original function.
To make it work as expected, you have to overwrite the original function writing fibo = memoize(fibo)
I've found a similar question as your with other explanations, hope that helps: https://stackoverflow.com/a/10757994/19288094
Suppose I have a recursive function like fibonacci:
fun fibonacci(n: Int): BigInteger =
if (n < 2)
n.toBigInteger()
else
fibonacci(n-1) + fibonacci(n-2)
This is slow because I'm recalculating known values a bunch of times. I can fix this by adding a "memo":
val memo = ConcurrentSkipListMap<Int, BigInteger>()
fun mFibonacci(n: Int): BigInteger =
memo.computeIfAbsent(n) {
if (n < 2)
it.toBigInteger()
else
mFibonacci(n-1) + mFibonacci(n-2)
}
Works like a charm, but can I do this without touching the function? My first thought was to use a wrapper class:
class Cached<in T, out R>(private val f: (T) -> R) : (T) -> R {
private val cache = ConcurrentSkipListMap<T, R>()
override fun invoke(x: T): R = cache.computeIfAbsent(x, f)
}
cFibonacci = Cached(::fibonacci)
... but the problem is, this only memoizes the outer-most call. If I call cFibonacci with a "big" number like 42, it takes a long time and then puts the correct value in the memo; subsequent calls with the 42 will be fast, but 41 will be slow again. Compare this to mFibonacci, which runs fast the first time, and populates the memo with values from 0 up to 42.
In Python, I can write a "decorator" which does this.
def memoized(f):
def helper(t):
if x in helper.memo:
return helper.memo[t]
else:
r = f(t)
helper.memo[t] = r
return r
helper.memo = {}
return helper
#memoized
def fib(n):
if n < 2:
return n
else:
return fib(n-1) + fib(n-2)
This works just like mFibonacci above. I can also call it as fib = memoized(fib) if I imported fib from somewhere else and don't have access to the definition. Interestingly, c_fib = memoized(fib) works like Cached/cFibonacci above, hinting that maybe mutability of function reference is necessary.
The question is: (how) can I wrap/"decorate" a recursive function in a way that affects the inner calls in Kotlin the way I can in Python?
I'll suggest a workaround in the absence of a solution. This pattern requires access to the definition of the function (i.e. it can't be an import):
object fibonacci: (Int) -> BigInteger {
private val memo = ConcurrentSkipListMap<Int, BigInteger>()
override fun invoke(n: Int): BigInteger = fibonacci(n)
private fun fibonacci(n: Int): BigInteger = memo.computeIfAbsent(n) {
if (n < 2)
n.toBigInteger()
else
fibonacci(n-1) + fibonacci(n-2)
}
}
There are few decisions here that may need justification:
I'm using a camelCase name instead of PascalCase. Despite the fact that object is a class, it's being called as a function and so I feel the naming convention of a function is better. With this, you can call the fibonacci function exactly as you normally would.
I've renamed invoke to fibonacci. Without this, the recursive calls use invoke which seems less readable to me. With this, you can read and write the fibonacci function (almost*) exactly as you normally would.
In general, the idea is to be the least intrusive as possible while still adding the desired functionality. I'm open to suggestions on how to improve it though!
*Something to note is that the function is defined using lambda syntax, so there is no return. If you have a single return at the end of the function, you just remove the return keyword. If you have multiple returns, you'll have to use the less-than-beautiful return#computeIfAbsent for the short-circuits.
As requested, this is a variation of #AlexJones's workaround that doesn't wrap the function in an object with unconventional naming. I did not test this--it's based on the assumption that the other solution works. The following code would be at the top level of a .kt file.
private val memo = ConcurrentSkipListMap<Int, BigInteger>()
fun fibonacci(n: Int): BigInteger = fibonacciImpl(n)
private fun fibonacciImpl(n: Int): BigInteger = memo.computeIfAbsent(n) {
if (n < 2)
n.toBigInteger()
else
fibonacci(n-1) + fibonacci(n-2)
}
Trying to understand recursive functions and so created this program but the output is incorrect. Would like to understand what am I doing wrong here
class Recursive:
def __init__(self, arr):
self.arr = arr
self.sum = 0
def sumRecursive(self):
if len(self.arr) == 0:
return self.sum
self.sum = self.arr.pop(0)
return self.sum + self.sumRecursive()
def main():
recur = Recursive([1,2,3])
print(recur.sumRecursive())
main()
output: 9
There are two types of recursion to consider: tail recursion, where the return value of a single recursive call is returned as-is, and "regular" recursion, where you do something with the return value(s) of the recursive call(s) before returning yourself.
You are combining the two. You either add a value from the list to the recursive sum, using no accumulator:
def non_tail_recursive(self):
if len(self.arr) == 0:
return 0
return self.arr.pop(0) + self.non_tail_recursive()
or you use an accumulator:
def tail_recursive(self):
if len(self.arr) == 0:
return self.sum
self.sum += self.arr.pop(0)
return self.tail_recursive()
You don't usually use an object with state to implement recursion. If you're keeping state, then you often don't need a recursive solution at all.
Here's how to do a "stateless" recursive sum.
def sumRecursive(arr):
if not arr:
return 0
return arr[0] + sumRecursive(arr[1:])
def main():
print(sumRecursive([1,2,3]))
main()
Your self.sum attribute is redundant. The information being processed by a recursive algorithm rarely needs members to pass information along.
class Recursive:
def __init__(self, arr):
self.arr = arr
def sumRecursive(self):
if not len(self.arr):
return 0
return self.arr.pop(0) + self.sumRecursive()
def main():
recur = Recursive([1,2,3])
print(recur.sumRecursive())
main()
Output: 6
To answer your question without rewriting your code (since obviously there are better ways to sum arrays), the answer is that the last leg of your recursion calls self.sum + self.sumRecursive which hits your sumRecursive function one last time which returns (from your if statement) self.sum which is the last element in your list (which you already summed).
Instead when the array is empty, return 0.
class Recursive:
def __init__(self, arr):
self.arr = arr
self.sum = 0
def sumRecursive(self):
if len(self.arr) == 0:
return 0
self.sum = self.arr.pop(0)
return self.sum + self.sumRecursive()
def main():
recur = Recursive([1,2,3,4])
print(recur.sumRecursive())
main()
Optionally move your if to the bottom where I personally think it makes more sense:
def sumRecursive(self):
self.sum = self.arr.pop(0)
if len(self.arr) == 0:
return self.sum
else:
return self.sum + self.sumRecursive()
fundamentals
You are making this more challenging for yourself because you are tangling your sum function with the class. The sum function simply needs to work on a list but the context of your object, self, is making it hard for you to focus.
Recursion is a functional heritage and this means writing your code in a functional and modular way. It's easier to read/write small, single-purpose functions and it promotes reuse within other areas of your program. Next time you need to sum a list in another class, do you want to rewrite sumRecursive again?
Write your summing function once and then import it where it's needed -
# mymath.py
def mysum(t):
if not t:
return 0
else:
return t.pop() + mysum(t)
See how mysum has no concern for context, self? Why should it? All it does it sum the elements of a list.
Now write your recursive module -
# recursive.py
from mymath import mysum
class Recursive:
def __init__(self, arr): self.arr = arr
def sum(self): return mysum(self.arr)
See how Recursive.sum just hands off self.arr to mysum? It doesn't have to be more complicated than that. mysum will work on every list, not just lists within your Recursive module. We don't have to know how mysum works, that is the concern of the mymath module.
Now we write the main module. Each module represents a barrier of abstraction. This means that the details of a module shouldn't spill over into other modules. We don't know how Recursive actually sums the input, and from the caller's point of view, we don't care. That is the concern of Recursive module.
# main.py
from recursive import Recursive
recur = Recursive([1,2,3])
print(recur.sum())
6
going functional
Above we wrote mysum using the .pop technique in your question. I did this because that seems to be how you are understanding the problem right now. But watch what happens when we do this -
x = [1,2,3]
print(mysum(x)) # 6
print(mysum(x)) # 0
Why does the mysum return a different answer the second time? Because as it is written now, mysum uses t.pop(), which mutates t. When mysum is finished running, t is completely emptied!
By why would we write our function like this? What if 5 + x returned a different result each time we called it?
x = 3
print(5 + x) # 8
print(5 + x) # 8
How annoying it would be if we could not depend on values not to change. The sum of the input, [1,2,3], is 6. But as it is written, the sum of the input is to return 6 and empty the input. This second part of emptying (changing) the input is known as a side effect. Ie, the desired effect is to sum and emptying of the input list is unintended but a consequence of us using .pop to calculate the result. This is not the functional way. Functional style means avoiding mutations, variable reassignments, and other side effects.
def mysum(t):
if not t:
return 0
else:
return t[0] + mysum(t[1:])
When written in this way, t, is not changed. This allows us to use equational reasoning whereby we can substitute any function call for its return value and always get the correct answer
x = [1,2,3]
mysum(x)
== 1 + mysum([2,3])
== 1 + 2 + mysum([3])
== 1 + 2 + 3 + mysum([])
== 1 + 2 + 3 + 0
== 1 + 2 + 3
== 1 + 5
== 6
And x was not changed as a result of running mysum -
print(x)
# [1,2,3]
Note, the Recursive module does not need to make any change to receive the benefit of rewriting mysum in this way. Before the change, we would've seen this behavior -
# main.py
from recursive import Recursive
recur = Recursive([1,2,3])
print(recur.sum())
print(recur.sum())
6
0
Because the first call to sum passes self.arr to mysum which empties self.arr as a side effect. A second call to recur.sum() will sum an empty self.arr! After fixing mysum we get the intended behaviour -
# main.py
from recursive import Recursive
recur = Recursive([1,2,3])
print(recur.sum())
print(recur.sum())
6
6
additional reading
I've written extensively about the techniques used in this answer. Follow the links to see them used in other contexts with additional explanation provided -
I want to reverse the stack but i dont know how to use recursion for reversing this… How can i reverse the stack without using Recursion
Finding all maze solutions with Python
Return middle node of linked list with recursion
How do i recursively find a size of subtree based on any given node? (BST)
Deleting node in BST Python
This is what I encounter when trying to do a LeetCode question. I wrote a function to call another recursive function.
This is the way that the output is correct.
def generateParenthesis(self, n: int) -> List[str]:
right = n
left = n
item = ''
result = []
self.helper(item, left, right, result)
return result
def helper(self, item, l, r, result):
if l==0 and r==0:
return result.append(item)
else:
if l > r:
return
else:
if l > 0:
self.helper(item+'(', l-1,r, result)
if r > 0:
self.helper(item+')', l, r-1, result)
However, when I change the first function to this, I will get a empty list [].
def generateParenthesis(self, n: int) -> List[str]:
right = n
left = n
item = ''
result = []
return self.helper(item, left, right, result)
Cannot quite figure out how the return works in the function, need some help from you guys.
The helper function as shown in your source code uses tail recursion. This is evident from the fact that the helper function essentially stores information from the previous recursive call in its argument parameter; hence, the arguments are updated in each recursive call.
But even if this concept is foreign to you, as pointed out by #63677, we can commence from the observation that the helper function does not return anything. The return result.append(item) might seem like it is returning something, but the append function in Python does not return anything (it returns None). Therefore, all that helper does is to update the argument parameter until it hits the base case.
Calling self.helper(item, left, right, result) within the generateParenthesis function is going to alter the contents of result according to the operations defined in helper. The generateParenthesis then simply returns that altered result.
Your modified code doesn't work since helper does not return anything as it is defined (to be more technical, it is tail recursive instead of just recursive). Therefore, the modified code will not work since calling self.helper will simply return None.
I'm new to programming.
def start():
x = 4
def addition():
n = 3
def exponential():
z = 2
def multiplication():
l = 2
print(x + n ** z * l)
return multiplication
equals = start()
equals()
why am I getting a "Nonetype" object is not callable error?
You're confusing a bunch of programming concepts:
Don't declare a function whenever you only need a statement
You're confusing function declaration with function call (invocation), and also the nesting is pointless. Declaring nested fn2 inside of fn1 doesn't magically also call fn2 and also transmit its return-value back to fn1. You still have to use an explicit return-statement from each fn.(If you forget that, you're implicitly returning None, which is almost surely not what you want)
For now, just don't ever nest functions at all.
Functions with no arguments are essentially useless, they can't take inputs and compute a result. Figure out what their arguments should be.
Specifically for the code you posted, addition(), multiplication() don't have any return value at all, i.e. None. exponential() returns multiplication, i.e. a function which only returns None. But then, both addition() and start() ignore that anyway, since they don't have a return-statement either, hence they implicitly return None.
Calling start() just gives you None, so you're just assigning equals = None. Not the result of some mathematical expression like you intended.
So:
reduce every unnecessary function to just a statement
declare each of your functions separately (non-nested)
each fn must have args (in this case at least two args, to make any sense)
each fn must have a return statement returning some value
only declaring a function and never calling it means it never gets run.
put an empty line in between function declarations (Then it's obvious if you forgot the return-statement)
Credits goes to #BrenBarn for being first to answer this. But I wanna post the code to make it more clear, and point out to some ways to make it better.
def start():
x = 4
def addition():
n = 3
def exponential():
z = 2
def multiplication():
l = 2
print (x + n ** z * l)
return multiplication()
return exponential()
return addition()
equals = start()
print equals #Output: 22
However, this is not the best way to list different methods. You should learn how to use a class in your python code.
I am going to define a class called "mathOperations". I will define three methods (functions): addition,exponential, multiplication. These functions are reusable.
class mathOperations():
def addition(self,x,y):
return x+y
def exponential(self,x,y):
return x**y
def multiplication(self,x,y):
return x*y
m= mathOperations()
z=2
l=2
x=4
n=3
result= m.addition(x,m.multiplication(m.exponential(n,z),l))
print result #Output:22
You should learn how to make your code reusable, try to google "procedural programming"; "Oriented Object Programming", or check "Learn Python the hard way" book. These are first and most used approach to make your code reusable. Think of it like a generic mathematical function to solve problems.