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?
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.
sys and the resource modules: https://mcmap.net/q/42115/-setting-stacksize-in-a-python-script –
Sciolism sys.setrecursionlimit? What exactly would happen if I only set sys.setrecusionlimit, but not touch resource? –
Austreng PYTHONRECURSIONLIMIT=9999 python file.py –
Spongin sys and resource as mentioned by @ThomasAhle –
Veracity fetch_instances() and a configuration key max_crawl_depths. I used max(config["max_crawl_depths"], sys.getrecursionlimit() - 50) to make sure it won't exceed maximum minus a few for just being safe. –
Roseboro Looks like you just need to set a higher recursion depth:
import sys
sys.setrecursionlimit(1500)
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.
resource. Without it, you'll get a Segmentation Fault and the whole Python process will crash if you setrecursionlimit too high and try to use the new limit (about 8 megabytes of stack frames, which translates to ~30,000 stack frames with the simple function above, on my laptop). –
Lactalbumin 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().
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://mcmap.net/q/42074/-what-is-the-maximum-recursion-depth-and-how-to-increase-it 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:
Tested on Ubuntu 16.10, Python 2.7.12.
rlimit_stack after Stack Clash remediations may result in failure or related problems. Also see Red Hat Issue 1463241 –
Tritanopia Use a language that guarantees tail-call optimisation. Or use iteration. Alternatively, get cute with decorators.
ulimit -s of stack frames, yes it is stackoverflow.com/a/50120316 –
Lactalbumin 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.)
xrange is called simply range, in Python 3. –
Increscent 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.
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.
n because of floating point imprecision - the difference between (1+sqrt(5))**n and (1+sqrt(5))**(n+1) becomes less than 1 ulp, so you start getting incorrect results. –
Thermotherapy (1+sqrt(5))**n and ((1+sqrt(5))**n)+1 that becomes less than 1 ulp! (small typo) Also, {@}rwst That's not O(1)! Calculating 2**n takes at least O(n) time. –
Honduras 2**n is effectively O(log(n)) using Exponentiattion by squaring. –
Recovery fib(n) is O(n) digits long. –
Honduras 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.
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))
3002468761178461090995494179715025648692747937490792943468375429502230242942284835863402333575216217865811638730389352239181342307756720414619391217798542575996541081060501905302157019002614964717310808809478675602711440361241500732699145834377856326394037071666274321657305320804055307021019793251762830816701587386994888032362232198219843549865275880699612359275125243457132496772854886508703396643365042454333009802006384286859581649296390803003232654898464561589234445139863242606285711591746222880807391057211912655818499798720987302540712067959840802106849776547522247429904618357394771725653253559346195282601285019169360207355179223814857106405285007997547692546378757062999581657867188420995770650565521377874333085963123444258953052751461206977615079511435862879678439081175536265576977106865074099512897235100538241196445815568291377846656352979228098911566675956525644182645608178603837172227838896725425605719942300037650526231486881066037397866942013838296769284745527778439272995067231492069369130289154753132313883294398593507873555667211005422003204156154859031529462152953119957597195735953686798871131148255050140450845034240095305094449911578598539658855704158240221809528010179414493499583473568873253067921639513996596738275817909624857593693291980841303291145613566466575233283651420134915764961372875933822262953420444548349180436583183291944875599477240814774580187144637965487250578134990402443365677985388481961492444981994523034245619781853365476552719460960795929666883665704293897310201276011658074359194189359660792496027472226428571547971602259808697441435358578480589837766911684200275636889192254762678512597000452676191374475932796663842865744658264924913771676415404179920096074751516422872997665425047457428327276230059296132722787915300105002019006293320082955378715908263653377755031155794063450515731009402407584683132870206376994025920790298591144213659942668622062191441346200098342943955169522532574271644954360217472458521489671859465232568419404182043966092211744372699797375966048010775453444600153524772238401414789562651410289808994960533132759532092895779406940925252906166612153699850759933762897947175972147868784008320247586210378556711332739463277940255289047962323306946068381887446046387745247925675240182981190836264964640612069909458682443392729946084099312047752966806439331403663934969942958022237945205992581178803606156982034385347182766573351768749665172549908638337611953199808161937885366709285043276595726484068138091188914698151703122773726725261370542355162118164302728812259192476428938730724109825922331973256105091200551566581350508061922762910078528219869913214146575557249199263634241165352226570749618907050553115468306669184485910269806225894530809823102279231750061652042560772530576713148647858705369649642907780603247428680176236527220826640665659902650188140474762163503557640566711903907798932853656216227739411210513756695569391593763704981001125
sys.setrecursionlimit(15000) is almost certain to crash the Python interpreter on most systems if you actually used that much stack space. Instead of this workaround, write the algorithm iteratively. –
Wary 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 :
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.
[fibs().next() for ...] would make a new generator each time. –
Ursal islice docs.python.org/3/library/itertools.html#itertools.islice to take an element from your generator. –
Alkalosis islice would need to look like this (for 1001th number): value = next(islice(fib(), 1000, 1001)). –
Alkane 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
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)
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.
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))
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)
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))
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997then? Is it because Python interpreter occupies the first 3 levels of the stack? – Casperline <n>, in <module>in stack traces) and this code takes 2 stack frames forn=1(because the base case isn < 1, so forn=1it still recurses). And I guess the recursion limit is not inclusive, as in it's "error when you hit 1000" not "error if you exceed 1000 (1001)".997 + 2is less than 1000 so it works998 + 2doesn't because it hits the limit. – Lactalbumin997stack frames are occupied byrecursive_functionand the last2are occupied by the intepreter itself (builtin allocation)? Did I got it right? – Casperrecursive_function(997)works, it breaks at998. When you callrecursive_function(998)it uses 999 stack frames and 1 frame is added by the interpreter (because your code is always run as if it's part of top level module), which makes it hit the 1000 limit. – Lactalbuminn+1frames to calculate the result forn. – Lactalbuminrecursion limit = 1000,recursive_function(995, 0)works whilerecursive_function(996, 0)doesn't... So I guess there must be additional stack frames being used (I am using Python 3.6). – Caspermain()function or importing a file where you're defining the function? If you're running it with IPython, I wouldn't be surprised if that adds a few frames as well. – Lactalbuminpythonand then write and execute the code on the interpreter's>>>prompt. – Casperpythonand notpython3it's probably 2 – Lactalbuminpythonon the CLI:Python 3.6.5 (default, Mar 30 2018, 06:42:10) [GCC 4.2.1 Compatible Apple LLVM 9.0.0 (clang-900.0.39.2)] on darwin Type "help", "copyright", "credits" or "license" for more information. >>>– Casper