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Project Euler #3 in Ruby solution times out
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Closed 9 years ago.
I'm trying to use a program to find the largest prime factor of 600851475143. This is for Project Euler here: http://projecteuler.net/problem=3
I first attempted this with this code:
#Ruby solution for http://projecteuler.net/problem=2
#Prepared by Richard Wilson (Senjai)
#We'll keep to our functional style of approaching these problems.
def gen_prime_factors(num) # generate the prime factors of num and return them in an array
result = []
2.upto(num-1) do |i| #ASSUMPTION: num > 3
#test if num is evenly divisable by i, if so add it to the result.
result.push i if num % i == 0
puts "Prime factor found: #{i}" # get some status updates so we know there wasn't a crash
end
result #Implicit return
end
#Print the largest prime factor of 600851475143. This will always be the last value in the array so:
puts gen_prime_factors(600851475143).last #this might take a while
This is great for small numbers, but for large numbers it would take a VERY long time (and a lot of memory).
Now I took university calculus a while ago, but I'm pretty rusty and haven't kept up on my math since.
I don't want a straight up answer, but I'd like to be pointed toward resources or told what I need to learn to implement some of the algorithms I've seen around in my program.
There's a couple problems with your solution. First of all, you never test that i is prime, so you're only finding the largest factor of the big number, not the largest prime factor. There's a Ruby library you can use, just require 'prime', and you can add an && i.prime? to your condition.
That'll fix inaccuracy in your program, but it'll still be slow and expensive (in fact, it'll now be even more expensive). One obvious thing you can do is just set result = i rather than doing result.push i since you ultimately only care about the last viable i you find, there's no reason to maintain a list of all the prime factors.
Even then, however, it's still very slow. The correct program should complete almost instantly. The key is to shrink the number you're testing up to, each time you find a prime factor. If you've found a prime factor p of your big number, then you don't need to test all the way up to the big number anymore. Your "new" big number that you want to test up to is what's left after dividing p out from the big number as many times as possible:
big_number = big_number/p**n
where n is the largest integer such that the right hand side is still a whole number. In practice, you don't need to explicitly find this n, just keep dividing by p until you stop getting a whole number.
Finally, as a SPOILER I'm including a solution below, but you can choose to ignore it if you still want to figure it out yourself.
require 'prime'
max = 600851475143; test = 3
while (max >= test) do
if (test.prime? && (max % test == 0))
best = test
max = max / test
else
test = test + 2
end
end
puts "Here's your number: #{best}"
Exercise: Prove that test.prime? can be eliminated from the if condition. [Hint: what can you say about the smallest (non-1) divisor of any number?]
Exercise: This algorithm is slow if we instead use max = 600851475145. How can it be improved to be fast for either value of max? [Hint: Find the prime factorization of 600851475145 by hand; it's easy to do and it'll make it clear why the current algorithm is slow for this number]
I am trying to run this code but it keeps crashing:
log10(x):=log(x)/log(10);
char(x):=floor(log10(x))+1;
mantissa(x):=x/10**char(x);
chop(x,d):=(10**char(x))*(floor(mantissa(x)*(10**d))/(10**d));
rnd(x,d):=chop(x+5*10**(char(x)-d-1),d);
d:5;
a:10;
Ibwd:[[30,rnd(integrate((x**60)/(1+10*x^2),x,0,1),d)]];
for n from 30 thru 1 step -1 do Ibwd:append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd);
Maxima crashes when it evaluates the last line. Any ideas why it may happen?
Thank you so much.
The problem is that the difference becomes negative and your rounding function dies horribly with a negative argument. To find this out, I changed your loop to:
for n from 30 thru 1 step -1 do
block([],
print (1/(2*n-1)-a*last(first(Ibwd))),
print (a*last(first(Ibwd))),
Ibwd: append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd),
print (Ibwd));
The last difference printed before everything fails miserably is -316539/6125000. So now try
rnd(-1,3)
and see the same problem. This all stems from the fact that you're taking the log of a negative number, which Maxima interprets as a complex number by analytic continuation. Maxima doesn't evaluate this until it absolutely has to and, somewhere in the evaluation code, something's dying horribly.
I don't know the "fix" for your specific example, since I'm not exactly sure what you're trying to do, but hopefully this gives you enough info to find it yourself.
If you want to deconstruct a floating point number, let's first make sure that it is a bigfloat.
say z: 34.1
You can access the parts of a bigfloat by using lisp, and you can also access the mantissa length in bits by ?fpprec.
Thus ?second(z)*2^(?third(z)-?fpprec) gives you :
4799148352916685/140737488355328
and bfloat(%) gives you :
3.41b1.
If you want the mantissa of z as an integer, look at ?second(z)
Now I am not sure what it is that you are trying to accomplish in base 10, but Maxima
does not do internal arithmetic in base 10.
If you want more bits or fewer, you can set fpprec,
which is linked to ?fpprec. fpprec is the "approximate base 10" precision.
Thus fpprec is initially 16
?fpprec is correspondingly 56.
You can easily change them both, e.g. fpprec:100
corresponds to ?fpprec of 335.
If you are diddling around with float representations, you might benefit from knowing
that you can look at any of the lisp by typing, for example,
?print(z)
which prints the internal form using the Lisp print function.
You can also trace any function, your own or system function, by trace.
For example you could consider doing this:
trace(append,rnd,integrate);
If you want to use machine floats, I suggest you use, for the last line,
for n from 30 thru 1 step -1 do :
Ibwd:append([[n-1,rnd(1/(2.0*n- 1.0)-a*last(first(Ibwd)),d)]],Ibwd);
Note the decimal points. But even that is not quite enough, because integration
inserts exact structures like atan(10). Trying to round these things, or compute log
of them is probably not what you want to do. I suspect that Maxima is unhappy because log is given some messy expression that turns out to be negative, even though it initially thought otherwise. It hands the number to the lisp log program which is perfectly happy to return an appropriate common-lisp complex number object. Unfortunately, most of Maxima was written BEFORE LISP HAD COMPLEX NUMBERS.
Thus the result (log -0.5)= #C(-0.6931472 3.1415927) is entirely unexpected to the rest of Maxima. Maxima has its own form for complex numbers, e.g. 3+4*%i.
In particular, the Maxima display program predates the common lisp complex number format and does not know what to do with it.
The error (stack overflow !!!) is from the display program trying to display a common lisp complex number.
How to fix all this? Well, you could try changing your program so it computes what you really want, in which case it probably won't trigger this error. Maxima's display program should be fixed, too. Also, I suspect there is something unfortunate in simplification of logs of numbers that are negative but not obviously so.
This is probably waaay too much information for the original poster, but maybe the paragraph above will help out and also possibly improve Maxima in one or more places.
It appears that your program triggers an error in Maxima's simplification (algebraic identities) code. We are investigating and I hope we have a bug fix soon.
In the meantime, here is an idea. Looks like the bug is triggered by rnd(x, d) when x < 0. I guess rnd is supposed to round x to d digits. To handle x < 0, try this:
rnd(x, d) := if x < 0 then -rnd1(-x, d) else rnd1(x, d);
rnd1(x, d) := (... put the present definition of rnd here ...);
When I do that, the loop runs to completion and Ibwd is a list of values, but I don't know what values to expect.
Firstly , For those of your who dont know - Anytime Algorithm is an algorithm that get as input the amount of time it can run and it should give the best solution it can on that time.
Weighted A* is the same as A* with one diffrence in the f function :
(where g is the path cost upto node , and h is the heuristic to the end of path until reaching a goal)
Original = f(node) = g(node) + h(node)
Weighted = f(node) = (1-w)g(node) +h(node)
My anytime algorithm runs Weighted A* with decaring weight from 1 to 0.5 until it reaches the time limit.
My problem is that most of the time , it takes alot time until this it reaches a solution , and if given somthing like 10 seconds it usaully doesnt find solution while other algorithms like anytime beam finds one in 0.0001 seconds.
Any ideas what to do?
If I were you I'd throw the unbounded heuristic away. Admissible heuristics are much better in that given a weight value for a solution you've found, you can say that it is at most 1/weight times the length of an optimal solution.
A big problem when implementing A* derivatives is the data structures. When I implemented a bidirectional search, just changing from array lists to a combination of hash augmented priority queues and array lists on demand, cut the runtime cost by three orders of magnitude - literally.
The main problem is that most of the papers only give pseudo-code for the algorithm using set logic - it's up to you to actually figure out how to represent the sets in your code. Don't be afraid of using multiple ADTs for a single list, i.e. your open list. I'm not 100% sure on Anytime Weighted A*, I've done other derivatives such as Anytime Dynamic A* and Anytime Repairing A*, not AWA* though.
Another issue is when you set the g-value too low, sometimes it can take far longer to find any solution that it would if it were a higher g-value. A common pitfall is forgetting to check your closed list for duplicate states, thus ending up in a (infinite if your g-value gets reduced to 0) loop. I'd try starting with something reasonably higher than 0 if you're getting quick results with a beam search.
Some pseudo-code would likely help here! Anyhow these are just my thoughts on the matter, you may have solved it already - if so good on you :)
Beam search is not complete since it prunes unfavorable states whereas A* search is complete. Depending on what problem you are solving, if incompleteness does not prevent you from finding a solution (usually many correct paths exist from origin to destination), then go for Beam search, otherwise, stay with AWA*. However, you can always run both in parallel if there are sufficient hardware resources.
I need to compare a query bit sequence with a database of up to a million bit sequences. All bit sequences are 100 bits long. I need the lookup to be as fast as possible. Are there any packages out there for fast determination of the similarity between two bit sequences? --Edit-- The bit sequences are position sensitive.
I have seen a possible algorithm on Bit Twiddling Hacks but if there is a ready made package that would be better.
If the database is rather static, you may want to build a tree data structure on it.
Search the tree recursively or in multiple threads and per search keep an actual difference variable. If the actual difference becomes greater than what you would consider 'similar', abort the search.
E.g. Suppose we have the following tree:
root
0 1
0 1 0 1
0 1 0 1 0 1 0 1
If you want to look for patterns similar to 011, and only want to allow 1 different bit at most, search like this (recursively or multi-threaded):
Start at the root
Take the left branch (0), this is similar, so difference is still 0
Take the left branch (0), this is different, so difference becomes 1, which is still acceptable
take the left branch (0), this is different, so difference becomes 2, which is too high. Abort looking in this branch.
take the right branch (1), this is equal, so difference remains 1, continue to search in this branch (not shown here)
Take the right branch (1), this is equal, so difference remains 0, go on
take the left branch (0), this is different, so difference becomes 1, which is still acceptable, go on.
This goes on until you have found your bit patterns.
If your bit patterns are more dynamic and being updated in your application, you will have to update the tree.
If memory is a problem, consider going to 64-bit.
If you want to look up the, let's say 50, most matching patterns, and we can assume that the input data set is rather static (or can be dynamically updated), you can repeat the initial phase of the previous comment, so:
For every bit pattern, count the bits.
Store the bit patterns in a multi_map (if you use STL, Java probably has something similar)
Then, use the following algorithm:
Make 2 collections: one for storing the found patterns, one for storing possibly good patterns (this second collection should probably be map, mapping 'distances' to patterns)
Take your own pattern and count the bits, assume this is N
Look in the multimap at index N, all these patterns will have the same sum, but not necessarily be completely identical
Compare all the patterns at index N. If they are equal store the result in the first collection. If they are not equal, store the result in the second collection/map, using the difference as key.
Look in the multimap at index N-1, all these patterns will have a distance of 1 or more
Compare all the patterns at index N-1. If they have a distance of 1, store them in the first collection. If they have a larger distance, store the result in the second collection/map, using the difference as key.
Repeat for index N+1
Now look in the second collection/map and see if there is something stored with distance 1. If it is, remove them from the second collection/map and store them in the first collection.
Repeat this for distance 2, distance 3, ... until you have enough patterns.
If the number of required patterns is not too big, and the average distance is also not too big, then the number of real compares between patterns is probably only a few %.
Unfortunately, since the patterns will be distributed using a Gaussian curve, there will still be quite some patterns to check. I didn't do a mathematical check on it, but in practice, if you don't want too many patterns out of the millions, and the average distance is not too far, you should be able to find the set of most-close patterns by checking only a few percent of the total bit patterns.
Please keep me updated of your results.
I came up with a second alternative.
For every bit pattern of the million ones count the number of bits and store the bit patterns in an STL multi_map (if you're writing in C++).
Then count the number of bits in your pattern. Suppose you have N bits set in your bit pattern.
If you now want to allow at most D differences, look up all the bit patterns in the multi_map having N-D, N-D+1, ..., N-1, N, N+1, ... N+D-1, N+D bits.
Unfortunately, the division of bit patterns in the multi_map will follow a Gaussian pattern, which means that in practice you will still have to compare quite some bit patterns.
(Originally I thought this could be solved by counting even 0's and uneven 1's but this isn't true.)
Assuming that you want to allow 1 difference, you have to look up 3 slots in the multi_map out of the 100 possible slots, leaving you with 3% of the actual bit patterns to do a full compare.
How does a stack overflow occur and what are the ways to make sure it doesn't happen, or ways to prevent one?
Stack
A stack, in this context, is the last in, first out buffer you place data while your program runs. Last in, first out (LIFO) means that the last thing you put in is always the first thing you get back out - if you push 2 items on the stack, 'A' and then 'B', then the first thing you pop off the stack will be 'B', and the next thing is 'A'.
When you call a function in your code, the next instruction after the function call is stored on the stack, and any storage space that might be overwritten by the function call. The function you call might use up more stack for its own local variables. When it's done, it frees up the local variable stack space it used, then returns to the previous function.
Stack overflow
A stack overflow is when you've used up more memory for the stack than your program was supposed to use. In embedded systems you might only have 256 bytes for the stack, and if each function takes up 32 bytes then you can only have function calls 8 deep - function 1 calls function 2 who calls function 3 who calls function 4 .... who calls function 8 who calls function 9, but function 9 overwrites memory outside the stack. This might overwrite memory, code, etc.
Many programmers make this mistake by calling function A that then calls function B, that then calls function C, that then calls function A. It might work most of the time, but just once the wrong input will cause it to go in that circle forever until the computer recognizes that the stack is overblown.
Recursive functions are also a cause for this, but if you're writing recursively (ie, your function calls itself) then you need to be aware of this and use static/global variables to prevent infinite recursion.
Generally, the OS and the programming language you're using manage the stack, and it's out of your hands. You should look at your call graph (a tree structure that shows from your main what each function calls) to see how deep your function calls go, and to detect cycles and recursion that are not intended. Intentional cycles and recursion need to be artificially checked to error out if they call each other too many times.
Beyond good programming practices, static and dynamic testing, there's not much you can do on these high level systems.
Embedded systems
In the embedded world, especially in high reliability code (automotive, aircraft, space) you do extensive code reviews and checking, but you also do the following:
Disallow recursion and cycles - enforced by policy and testing
Keep code and stack far apart (code in flash, stack in RAM, and never the twain shall meet)
Place guard bands around the stack - empty area of memory that you fill with a magic number (usually a software interrupt instruction, but there are many options here), and hundreds or thousands of times a second you look at the guard bands to make sure they haven't been overwritten.
Use memory protection (ie, no execute on the stack, no read or write just outside the stack)
Interrupts don't call secondary functions - they set flags, copy data, and let the application take care of processing it (otherwise you might get 8 deep in your function call tree, have an interrupt, and then go out another few functions inside the interrupt, causing the blowout). You have several call trees - one for the main processes, and one for each interrupt. If your interrupts can interrupt each other... well, there be dragons...
High-level languages and systems
But in high level languages run on operating systems:
Reduce your local variable storage (local variables are stored on the stack - although compilers are pretty smart about this and will sometimes put big locals on the heap if your call tree is shallow)
Avoid or strictly limit recursion
Don't break your programs up too far into smaller and smaller functions - even without counting local variables each function call consumes as much as 64 bytes on the stack (32 bit processor, saving half the CPU registers, flags, etc)
Keep your call tree shallow (similar to the above statement)
Web servers
It depends on the 'sandbox' you have whether you can control or even see the stack. Chances are good you can treat web servers as you would any other high level language and operating system - it's largely out of your hands, but check the language and server stack you're using. It is possible to blow the stack on your SQL server, for instance.
A stack overflow in real code occurs very rarely. Most situations in which it occurs are recursions where the termination has been forgotten. It might however rarely occur in highly nested structures, e.g. particularly large XML documents. The only real help here is to refactor the code to use an explicit stack object instead of the call stack.
Most people will tell you that a stack overflow occurs with recursion without an exit path - while mostly true, if you work with big enough data structures, even a proper recursion exit path won't help you.
Some options in this case:
Breadth-first search
Tail recursion, .Net-specific great blog post (sorry, 32-bit .Net)
Infinite recursion is a common way to get a stack overflow error. To prevent - always make sure there's an exit path that will be hit. :-)
Another way to get a stack overflow (in C/C++, at least) is to declare some enormous variable on the stack.
char hugeArray[100000000];
That'll do it.
Aside from the form of stack overflow that you get from a direct recursion (eg Fibonacci(1000000)), a more subtle form of it that I have experienced many times is an indirect recursion, where a function calls another function, which calls another, and then one of those functions calls the first one again.
This can commonly occur in functions that are called in response to events but which themselves may generate new events, for example:
void WindowSizeChanged(Size& newsize) {
// override window size to constrain width
newSize.width=200;
ResizeWindow(newSize);
}
In this case the call to ResizeWindow may cause the WindowSizeChanged() callback to be triggered again, which calls ResizeWindow again, until you run out of stack. In situations like these you often need to defer responding to the event until the stack frame has returned, eg by posting a message.
Usually a stack overflow is the result of an infinite recursive call (given the usual amount of memory in standard computers nowadays).
When you make a call to a method, function or procedure the "standard" way or making the call consists on:
Pushing the return direction for the call into the stack(that's the next sentence after the call)
Usually the space for the return value get reserved into the stack
Pushing each parameter into the stack (the order diverges and depends on each compiler, also some of them are sometimes stored on the CPU registers for performance improvements)
Making the actual call.
So, usually this takes a few bytes depeding on the number and type of the parameters as well as the machine architecture.
You'll see then that if you start making recursive calls the stack will begin to grow. Now, stack is usually reserved in memory in such a way that it grows in opposite direction to the heap so, given a big number of calls without "coming back" the stack begins to get full.
Now, on older times stack overflow could occur simply because you exausted all available memory, just like that. With the virtual memory model (up to 4GB on a X86 system) that was out of the scope so usually, if you get an stack overflow error, look for an infinite recursive call.
I have recreated the stack overflow issue while getting a most common Fibonacci number i.e. 1, 1, 2, 3, 5..... so calculation for fib(1) = 1 or fib(3) = 2.. fib(n) = ??.
for n, let say we will interested - what if n = 100,000 then what will be the corresponding Fibonacci number ??
The one loop approach is as below -
package com.company.dynamicProgramming;
import java.math.BigInteger;
public class FibonacciByBigDecimal {
public static void main(String ...args) {
int n = 100000;
BigInteger[] fibOfnS = new BigInteger[n + 1];
System.out.println("fibonacci of "+ n + " is : " + fibByLoop(n));
}
static BigInteger fibByLoop(int n){
if(n==1 || n==2 ){
return BigInteger.ONE;
}
BigInteger fib = BigInteger.ONE;
BigInteger fip = BigInteger.ONE;
for (int i = 3; i <= n; i++){
BigInteger p = fib;
fib = fib.add(fip);
fip = p;
}
return fib;
}
}
this quite straight forward and result is -
fibonacci of 100000 is : 25974069347221724166155034021275915414880485386517696584724770703952534543511273686265556772836716744754637587223074432111638399473875091030965697382188304493052287638531334921353026792789567010512765782716356080730505322002432331143839865161378272381247774537783372999162146340500546698603908627509966393664092118901252719601721050603003505868940285581036751176582513683774386849364134573388343651587754253719124105003321959913300622043630352137565254218239986908485563740801792517616293917549634585586163007628199160811098365263529954406942842065710460449038056471363460330005208522777075544467947237090309790190148604328468198579610159510018506082649192345873133991501339199323631023018641725364771362664750801339824312317034314529641817900511879573167668349799016820118499077566864568450662873924856039140476051995500662888263458771894106803700918793650017330117100283104739474562560914449328213748555738640805798130282666402703542944121049199958031318768058991865134251759599115205631553377039969410355182752749199598022575079020377981030899229849963044962558140455170002502997643221934621653662108418767454282982613982344783665815880408190033073829395000821320093747154851310272208173054322648669496309879147143629255542526240439996153269798768075106468190687921182991679644091782718685617029181022126792674013626504997849688436809752547001310045741864064482994858725517447466956518791269169932445648176733222571493149677633458466238303338202397024368594782876418757885729107101337003000942293335972927791914092128049015459762627910570552481588840517794181929052167695766087488155678601288183543542923073978101547857013284386127286201766539534449930019800629538936985500723286651317181135886613537472684585432548981137176605194616937916884425342594781263103889520479565943807153019112539648471126389007133628569101551453423329441284357220996286746119420951661002309740709965531900508158669911445442647882872642845017253320486483194578920399848938236367456182203750973485668474338872490493370316338265717607297788917989136673251906232471180372801739215723908227692280772924566627505383375006926077210593619421268920302567443565378008318306375933345023502569729065152853271943677560156660399164048825639676930792905029514886934137991251748566670747175149389790386533381395346848378086126737554383821108448976538368483182588363399173104558509056638462025014631311831087429077292622159430204291594740306101839816855066950261973761508571761199475875722129872053120607918649803615960923395941041186351688548839119185179061511562752936158490008721501922265117853150892510275280451512386037921846921215338292871369243215273327141574788295902601571954853164447945467502858402360002383447905203451080332820138038807089807348326201227952633606773669875783326254859449060219173688677862411205621098369850197290177157801120404586491539351157834995461006366357454485082418882790675313599505192062229760153765297973085881648731173082370598284894044874039320535929359764541655607954724778620299692329561389719894679422187273605123365595211331087787582288795975803204596084790245063851941743126163775104599211024868794963417068620929088930685252348056925998333775103901013166178123051145719327066291671254465121517468025481903583516889717075706778656188008220346836321018130262329960275994035799977740462449521145315883703579044832931500072461734173558055678321534543411700202585608091662941986374015145695722728369219632295111877625307534025947814482046574602884855000628069348113982760168555840795421620575435572915106415375929390228843561207926437055600623679865443824643739469724719459965557955058380348255978396827760847315302517889517186307227611036305093600742622617173630586132915440246954329046162586917746305785076749374879923291817501634840688134655343709975893536074051729094126976575932951568186247471276364688365517570183534172746626073065104511957628663499228486787805910851189856535554349587616640164475880286336297040462890970677362565843002353147494612339120686321466370878446992104275415694109122465685712047172411333784898167640969249816334211768571503116710400681753031921154156119580425706586931272762137106974722260296555246110537155545324997508432752001992143019105053629960070429632978051030666506387862681576587726837451289768507963663710593809112254288358391941211547737599813019216509521401333060709873137329265181692268450634439540567298120315463923249817937804691037934221694952291007930299492375072993250630509428139027930841344730614116433556147640931044259184813639305423693789765205264563476483182726333715121120306292338892864879492097378478618848682608046473195392008403983080088038690495574197562192939221108257663976813610444900247209483403267967688376213967440757138872928630798218493143438797780887379588968409461434159271317578365114578289355818599029235343888888465874521308381377794436361197628390368945957601203165022798579015453447473527069728514545998614229027372911314637820455162254475353567736227936485450357102086445412089842350389087702230398493802147348096874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Now another approach I have applied is through Divide and Concur via recursion
i.e. Fib(n) = fib(n-1) + Fib(n-2) and then further recursion for n-1 & n-2.....till 2 & 1. which is programmed as -
package com.company.dynamicProgramming;
import java.math.BigInteger;
public class FibonacciByBigDecimal {
public static void main(String ...args) {
int n = 100000;
BigInteger[] fibOfnS = new BigInteger[n + 1];
System.out.println("fibonacci of "+ n + " is : " + fibByDivCon(n, fibOfnS));
}
static BigInteger fibByDivCon(int n, BigInteger[] fibOfnS){
if(fibOfnS[n]!=null){
return fibOfnS[n];
}
if (n == 1 || n== 2){
fibOfnS[n] = BigInteger.ONE;
return BigInteger.ONE;
}
// creates 2 further entries in stack
BigInteger fibOfn = fibByDivCon(n-1, fibOfnS).add( fibByDivCon(n-2, fibOfnS)) ;
fibOfnS[n] = fibOfn;
return fibOfn;
}
}
When i ran the code for n = 100,000 the result is as below -
Exception in thread "main" java.lang.StackOverflowError
at com.company.dynamicProgramming.FibonacciByBigDecimal.fibByDivCon(FibonacciByBigDecimal.java:29)
at com.company.dynamicProgramming.FibonacciByBigDecimal.fibByDivCon(FibonacciByBigDecimal.java:29)
at com.company.dynamicProgramming.FibonacciByBigDecimal.fibByDivCon(FibonacciByBigDecimal.java:29)
Above you can see the StackOverflowError is created. Now the reason for this is too many recursion as -
// creates 2 further entries in stack
BigInteger fibOfn = fibByDivCon(n-1, fibOfnS).add( fibByDivCon(n-2, fibOfnS)) ;
So each entry in stack create 2 more entries and so on... which is represented as -
Eventually so many entries will be created that system is unable to handle in the stack and StackOverflowError thrown.
For Prevention :
For Above example perspective
Avoid using recursion approach or reduce/limit the recursion by again one level division like if n is too large then split the n so that system can handle with in its limit.
Use other approach, like the loop approach I have used in 1st code sample. (I am not at all intended to degrade Divide & Concur or Recursion as they are legendary approaches in many most famous algorithms.. my intention is to limit or stay away from recursion if I suspect stack overflow issues)
Considering this was tagged with "hacking", I suspect the "stack overflow" he's referring to is a call stack overflow, rather than a higher level stack overflow such as those referenced in most other answers here. It doesn't really apply to any managed or interpreted environments such as .NET, Java, Python, Perl, PHP, etc, which web apps are typically written in, so your only risk is the web server itself, which is probably written in C or C++.
Check out this thread:
https://stackoverflow.com/questions/7308/what-is-a-good-starting-point-for-learning-buffer-overflow
Stack overflow occurs when your program uses up the entire stack. The most common way this happens is when your program has a recursive function which calls itself forever. Every new call to the recursive function takes more stack until eventually your program uses up the entire stack.