What is the maximum recursion depth, and how to increase it?

Question

I have this tail recursive function here:

def recursive_function(n, sum):
    if n < 1:
        return sum
    else:
        return recursive_function(n-1, sum+n)

c = 998
print(recursive_function(c, 0))

It works up to n=997, then it just breaks and spits out a RecursionError: maximum recursion depth exceeded in comparison. Is this just a stack overflow? Is there a way to get around it?

Answer 1

If you often need to change the recursion limit (e.g. while solving programming puzzles) you can define a simple context manager like this:

import sys

class recursionlimit:
    def __init__(self, limit):
        self.limit = limit

    def __enter__(self):
        self.old_limit = sys.getrecursionlimit()
        sys.setrecursionlimit(self.limit)

    def __exit__(self, type, value, tb):
        sys.setrecursionlimit(self.old_limit)

Then to call a function with a custom limit you can do:

with recursionlimit(1500):
    print(fib(1000, 0))

On exit from the body of the with statement the recursion limit will be restored to the default value.

P.S. You may also want to increase the stack size of the Python process for big values of the recursion limit. That can be done via the ulimit shell builtin or limits.conf(5) file, for example.

Answer 2

Edit: 6 years later I realized my "Use generators" was flippant and didn't answer the question. My apologies.

I guess my first question would be: do you really need to change the recursion limit? If not, then perhaps my or any of the other answers that don't deal with changing the recursion limit will apply. Otherwise, as noted, override the recursion limit using sys.getrecursionlimit(n).

Use generators?

def fib():
    a, b = 0, 1
    while True:
        yield a
        a, b = b, a + b

fibs = fib() #seems to be the only way to get the following line to work is to
             #assign the infinite generator to a variable

f = [fibs.next() for x in xrange(1001)]

for num in f:
        print num

Above fib() function adapted from Introduction to Python Generators.

Answer 3

RecursionError: maximum recursion depth exceeded in comparison

Solution :

First it’s better to know when you execute a recursive function in Python on a large input ( > 10^4), you might encounter a “maximum recursion depth exceeded error”.

The sys module in Python have a function getrecursionlimit() can show the recursion limit in your Python version.

import sys
print("Python Recursive Limitation = ", sys.getrecursionlimit())

The default in some version of Python is 1000 and in some other it was 1500

You can change this limitation but it’s very important to know if you increase it very much you will have memory overflow error.

So be careful before increase it. You can use setrecursionlimit() to increase this limitation in Python.

import sys
sys.setrecursionlimit(3000)

Please follow this link for more information about somethings cause this issue :

https://elvand.com/quick-sort-binary-search/

Answer 4

I'm not sure I'm repeating someone but some time ago some good soul wrote Y-operator for recursively called function like:

def tail_recursive(func):
  y_operator = (lambda f: (lambda y: y(y))(lambda x: f(lambda *args: lambda: x(x)(*args))))(func)
  def wrap_func_tail(*args):
    out = y_operator(*args)
    while callable(out): out = out()
    return out
  return wrap_func_tail

and then recursive function needs form:

def my_recursive_func(g):
  def wrapped(some_arg, acc):
    if <condition>: return acc
    return g(some_arg, acc)
  return wrapped

# and finally you call it in code

(tail_recursive(my_recursive_func))(some_arg, acc)

for Fibonacci numbers your function looks like this:

def fib(g):
  def wrapped(n_1, n_2, n):
    if n == 0: return n_1
    return g(n_2, n_1 + n_2, n-1)
  return wrapped

print((tail_recursive(fib))(0, 1, 1000000))

output:

..684684301719893411568996526838242546875

(actually tones of digits)

Answer 5

We can do that using @lru_cache decorator and setrecursionlimit() method:

import sys
from functools import lru_cache

sys.setrecursionlimit(15000)


@lru_cache(128)
def fib(n: int) -> int:
    if n == 0:
        return 0
    if n == 1:
        return 1

    return fib(n - 2) + fib(n - 1)


print(fib(14000))

Output

3002468761178461090995494179715025648692747937490792943468375429502230242942284835863402333575216217865811638730389352239181342307756720414619391217798542575996541081060501905302157019002614964717310808809478675602711440361241500732699145834377856326394037071666274321657305320804055307021019793251762830816701587386994888032362232198219843549865275880699612359275125243457132496772854886508703396643365042454333009802006384286859581649296390803003232654898464561589234445139863242606285711591746222880807391057211912655818499798720987302540712067959840802106849776547522247429904618357394771725653253559346195282601285019169360207355179223814857106405285007997547692546378757062999581657867188420995770650565521377874333085963123444258953052751461206977615079511435862879678439081175536265576977106865074099512897235100538241196445815568291377846656352979228098911566675956525644182645608178603837172227838896725425605719942300037650526231486881066037397866942013838296769284745527778439272995067231492069369130289154753132313883294398593507873555667211005422003204156154859031529462152953119957597195735953686798871131148255050140450845034240095305094449911578598539658855704158240221809528010179414493499583473568873253067921639513996596738275817909624857593693291980841303291145613566466575233283651420134915764961372875933822262953420444548349180436583183291944875599477240814774580187144637965487250578134990402443365677985388481961492444981994523034245619781853365476552719460960795929666883665704293897310201276011658074359194189359660792496027472226428571547971602259808697441435358578480589837766911684200275636889192254762678512597000452676191374475932796663842865744658264924913771676415404179920096074751516422872997665425047457428327276230059296132722787915300105002019006293320082955378715908263653377755031155794063450515731009402407584683132870206376994025920790298591144213659942668622062191441346200098342943955169522532574271644954360217472458521489671859465232568419404182043966092211744372699797375966048010775453444600153524772238401414789562651410289808994960533132759532092895779406940925252906166612153699850759933762897947175972147868784008320247586210378556711332739463277940255289047962323306946068381887446046387745247925675240182981190836264964640612069909458682443392729946084099312047752966806439331403663934969942958022237945205992581178803606156982034385347182766573351768749665172549908638337611953199808161937885366709285043276595726484068138091188914698151703122773726725261370542355162118164302728812259192476428938730724109825922331973256105091200551566581350508061922762910078528219869913214146575557249199263634241165352226570749618907050553115468306669184485910269806225894530809823102279231750061652042560772530576713148647858705369649642907780603247428680176236527220826640665659902650188140474762163503557640566711903907798932853656216227739411210513756695569391593763704981001125

Source

functools lru_cache

Answer 6

Of course Fibonacci numbers can be computed in O(n) by applying the Binet formula:

from math import floor, sqrt

def fib(n):                                                     
    return int(floor(((1+sqrt(5))**n-(1-sqrt(5))**n)/(2**n*sqrt(5))+0.5))

As the commenters note it's not O(1) but O(n) because of 2**n. Also a difference is that you only get one value, while with recursion you get all values of Fibonacci(n) up to that value.

Answer 7

resource.setrlimit must also be used to increase the stack size and prevent segfault

The Linux kernel limits the stack of processes.

Python stores local variables on the stack of the interpreter, and so recursion takes up stack space of the interpreter.

If the Python interpreter tries to go over the stack limit, the Linux kernel makes it segmentation fault.

The stack limit size is controlled with the getrlimit and setrlimit system calls.

Python offers access to those system calls through the resource module.

sys.setrecursionlimit mentioned e.g. at https://stackoverflow.com/a/3323013/895245 only increases the limit that the Python interpreter self imposes on its own stack size, but it does not touch the limit imposed by the Linux kernel on the Python process.

Example program:

main.py

import resource
import sys

print resource.getrlimit(resource.RLIMIT_STACK)
print sys.getrecursionlimit()
print

# Will segfault without this line.
resource.setrlimit(resource.RLIMIT_STACK, [0x10000000, resource.RLIM_INFINITY])
sys.setrecursionlimit(0x100000)

def f(i):
    print i
    sys.stdout.flush()
    f(i + 1)
f(0)

Of course, if you keep increasing setrlimit, your RAM will eventually run out, which will either slow your computer to a halt due to swap madness, or kill Python via the OOM Killer.

From bash, you can see and set the stack limit (in kb) with:

ulimit -s
ulimit -s 10000

The default value for me is 8Mb.

See also:

• Setting stacksize in a python script
• What is the hard recursion limit for Linux, Mac and Windows?

Tested on Ubuntu 16.10, Python 2.7.12.

Answer 8

It is a guard against a stack overflow, yes. Python (or rather, the CPython implementation) doesn't optimize tail recursion, and unbridled recursion causes stack overflows. You can check the recursion limit with sys.getrecursionlimit:

import sys
print(sys.getrecursionlimit())

and change the recursion limit with sys.setrecursionlimit:

sys.setrecursionlimit(1500)

but doing so is dangerous -- the standard limit is a little conservative, but Python stackframes can be quite big.

Python isn't a functional language and tail recursion is not a particularly efficient technique. Rewriting the algorithm iteratively, if possible, is generally a better idea.

Answer 9

As @alex suggested, you could use a generator function to do this sequentially instead of recursively.

Here's the equivalent of the code in your question:

def fib(n):
    def fibseq(n):
        """ Iteratively return the first n Fibonacci numbers, starting from 0. """
        a, b = 0, 1
        for _ in xrange(n):
            yield a
            a, b = b, a + b

    return sum(v for v in fibseq(n))

print format(fib(100000), ',d')  # -> no recursion depth error

Answer 10

It's to avoid a stack overflow. The Python interpreter limits the depths of recursion to help you avoid infinite recursions, resulting in stack overflows. Try increasing the recursion limit (sys.setrecursionlimit) or re-writing your code without recursion.

From the Python documentation:

sys.getrecursionlimit()

Return the current value of the recursion limit, the maximum depth of the Python interpreter stack. This limit prevents infinite recursion from causing an overflow of the C stack and crashing Python. It can be set by setrecursionlimit().

Answer 11

Looks like you just need to set a higher recursion depth:

import sys
sys.setrecursionlimit(1500)

Answer 12

We could also use a variation of dynamic programming bottom up approach

def fib_bottom_up(n):

    bottom_up = [None] * (n+1)
    bottom_up[0] = 1
    bottom_up[1] = 1

    for i in range(2, n+1):
        bottom_up[i] = bottom_up[i-1] + bottom_up[i-2]

    return bottom_up[n]

print(fib_bottom_up(20000))

Answer 13

I had a similar issue with the error "Max recursion depth exceeded". I discovered the error was being triggered by a corrupt file in the directory I was looping over with os.walk. If you have trouble solving this issue and you are working with file paths, be sure to narrow it down, as it might be a corrupt file.

Answer 14

import sys
sys.setrecursionlimit(1500)

def fib(n, sum):
    if n < 1:
        return sum
    else:
        return fib(n-1, sum+n)

c = 998
print(fib(c, 0))

Answer 15

I wanted to give you an example for using memoization to compute Fibonacci as this will allow you to compute significantly larger numbers using recursion:

cache = {}
def fib_dp(n):
    if n in cache:
        return cache[n]
    if n == 0: return 0
    elif n == 1: return 1
    else:
        value = fib_dp(n-1) + fib_dp(n-2)
    cache[n] = value
    return value

print(fib_dp(998))

This is still recursive, but uses a simple hashtable that allows the reuse of previously calculated Fibonacci numbers instead of doing them again.

Answer 16

If you want to get only few Fibonacci numbers, you can use matrix method.

from numpy import matrix

def fib(n):
    return (matrix('0 1; 1 1', dtype='object') ** n).item(1)

It's fast as numpy uses fast exponentiation algorithm. You get answer in O(log n). And it's better than Binet's formula because it uses only integers. But if you want all Fibonacci numbers up to n, then it's better to do it by memorisation.

Answer 17

Many recommend that increasing recursion limit is a good solution however it is not because there will be always limit. Instead use an iterative solution.

def fib(n):
    a,b = 1,1
    for i in range(n-1):
        a,b = b,a+b
    return a
print fib(5)

Answer 18

I realize this is an old question but for those reading, I would recommend against using recursion for problems such as this - lists are much faster and avoid recursion entirely. I would implement this as:

def fibonacci(n):
    f = [0,1,1]
    for i in xrange(3,n):
        f.append(f[i-1] + f[i-2])
    return 'The %.0fth fibonacci number is: %.0f' % (n,f[-1])

(Use n+1 in xrange if you start counting your fibonacci sequence from 0 instead of 1.)

Answer 19

Use a language that guarantees tail-call optimisation. Or use iteration. Alternatively, get cute with decorators.