Friday, November 8, 2013

The Six Most Common Species Of Code


Michael Mandrus said...

A CS 101 student would never write a recursive function.

saurabh singh said...

@Michael Depending on the teacher taking the CS101. I am pretty much sure every one in my batch would have written a recursive function over an iterative solution

Animesh said...
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Animesh said...

@Saurabh, Michael
Recursive function for Fibonacci will be very inefficient. Its running time will grow exponentially with X. :) Watch this video :

Abhishek said...

I hope someone noticedthe point of this post and which in my view is that the more you know things, the more constrained your view becomes, instead of thinking in a straightforward manner, we think of all the ways something could go wrong and more often than not, it holds us back from doing anything.

Rooney said...


Recursion is not used without memoization. With it, it's a linear algorithm.

RockNes said...

Math guy's program will go in an infinite loop if b is a non integer number :/

meenu iyyer said...

This is sooo true and really funny.. i have been there and done that for all the different roles [Excpet the cat ofcourse ;)] .... code written at a large comany is the best.. ROFL!


why u guys senti by looking at this ?
this if for fun only :) :)

thanbo said...

The Math PhD's closed-form answer is less efficient than the basic iterative/recursive solution. It requres 2n (or for the rounding version, n) multiplications to do the exponentiation, while the brute-force solution requires n additions, which on most processors are faster than multiplications.

virgincoder said...

LOL ! Funny ! I laughed so much when I saw the "Code Written At a Large Company" part ! LOL

Bobo Pisaach said...

My version of fibonacci number:
int fibonacci(int n)
return rand();

Subhrajit said...

so no one gives a sh!t about the computational complexity. All the recursive implementations have exponential complexity. And I seriously have no clue what the author tried to prove with the totally gibberish large company or math PhD code.

How about the following:
int fibonacci(int x){
if (x <= 2)
return 1;
else {
int sum = 1, oldsum = 1, tmpsum;
for (int a = 3; a <= x; a++) {
tmpsum = sum;
sum = sum + oldsum;
oldsum = tmpsum;
return sum;

It has linear complexity.

Unknown said...

A database specialist would write

SELECT Value FROM dbo.Fibonacci WHERE n = @n;

Herman Saksono said...

I would be surprised if a large company has a fibonacci method that runs on O(2^n) time.

Qingshuo said...
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littleeden said...

@Unknown -- hahaha my thoughts exactly upon reading this. Just look it up!

funny comic

Miguel Ángel Valentín Naranjo said...

Ey Subhrajit, I´ll buy you a can of humor.

rakras said...
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Dao Le said...

I think I should copy cat

Gabrielle said...

When I took my bachelor degree, I used "cat" species code for my homework. The code worked, but guess what? Got 0 because my teacher didn't understand any sh*t I wrote :))

rathe amadeus said...
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rathe amadeus said...
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Manu said...

ROFL. Had a nice read.

Now for the folks who have commented with suggestions on optimizing the algorithm:
Seriously?!! Don't you guys get humour at all? (facepalm)

Rahul Thakur said...
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Rahul Thakur said...
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Rahul Thakur said...

I get the humour, but for those suggesting improvements, here's a simpler one -

void fibonacci(int number_of_terms){
int T1=0, T2=1;
T1 = T1 + T2;
T2 = T1 - T2;
printf("%d\n", T2);
} while(number_of_terms > 0);

This is in C btw, and here's a compiled version -

Mohamed Ameer said...

the best code is written by your cat.

Zohaib said...
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Scheyst Scheyst said...

Replace all of the method and variable names with variations on 'asdfjkl' and change the comments to '//does stuff' and '//does the rest of stuff' and that's basically what everything I code looks like.

aMIT sHaKyA said...
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aMIT sHaKyA said...

Here we go, complete imagination of author went to a toss. And post has become dead ground for recursive algo complexity discussion. Screw you coding wannabes.

Too good post. Don't do CS graffiti here.

Le Cong Nga said...
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Le Cong Nga said...

I love your cat, amazing code :D

Unknown said...

It doesn't take a Ph.D. to know the direct formula for the general element in the Fibonacci sequence.

I do wonder why the one() instead of numbers.

Ketan Vijayvargiya said...

Hilariously true!!

Unknown said...

I suppose `one()` may return an object with various methods, e.g. `.sign()`, `.magnitude()`, `.negative()` and `.reciprocal()`. This list may be expanded in the future. Not so with the language builtin `1`.

HuzursuZ said...

i do as

f(n)=( (1+sqrt(5))^n - (1-sqrt(5))^n ) / (sqrt(5)*2^n)

so what i become ?

Adrian said...

There is a solution which runs in O(log(n)) to compute the n'th term. No caching.

Matthias said...

It's one() and add() instead of 1 and + because the mathmathic field won't make a difference. So it's way more general!

ConceptJunkie said...

There's a solution that runs with O(1) posted above.

A Pythonista could just do this:

import mpmath

print( fib( n ) )

Paul K said...

Had a good laugh :D

Having worked at 5 *very* different companies in 5 years, I can testify that there is a lot of truth to this!

(Except perhaps the one with the cat)

Rahul Ghose said...

Ah the varied species! Found some more in the comments! :D

Mads said...

And code written by a student, that paid attention during algoritms, knows how to google and did remember to trust only reliable sources of information...

ac said...

is missing the kernel guy code:

int fib(int n) {
if (n < 0) {
errno = EDOM;
return -1;

return n == 0
? 0
: (n == 1
? 1
: (n == 2
? 2
: fib(n - 2) * fib(n-1)

sarath chandra said...

Lol so true, code written at large company does look like that, (why? :()

Daniel Dwire said...
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tsndiffopera said...

Phew! Then who'd write a O(lg(n)) algorithm using matrix exponentiation ? Only me? :P

[{1 1},{1 0}]^n = [{F_(n+1) F_n},{F_n F_(n-1)}]

Also, x^n = x^(n/2)*x^(n/2)

which has O(lg(n)) ;)

joe random said...

Just to be pedantic for my CS/math bros:
The CS101 code doesn't need recursion or memoization, and that would occur to most students, since that's how people do it by hand: they take the last two numbers, add them together, and get a new number. Then they can forget the oldest number. A simple for loop takes care of that. Admittedly, this is explicit memoization.

But worse, the code by a "math phd" isn't any faster than that, and is inexact if there is rounding error, unless it uses an overcomplicated math framework that handles sqrts symbolically.

Still, if you change the (math phd?) exponentiation function to do successive squaring, you get the best running time so far, O(log n). A CS101 student could even work out how to do it without a heavyweight math library, since all of the intermediate computations are on numbers of the form (a+b*sqrt(5))/2^n where a,b, and n are integers. So you only need integer arithmetic.

There other O(log n) algorithms, such as ones exploiting the recurrences
F_(2n-1) = (F_n)^2 + (F_(n-2))^2
F_(2n) = (2F_(n-1) + F_n)F_n

a math phd candidate

Леха Чебара said...

cat style looks like perl code

Haskell said...

The Math PhD would use haskell and produce an infinite list of fibonacci results.

Siberiano said...

I think math phd should write that in Lisp.

A simple version would be, but you may expand to add other parameter forms.

(defun fib (x) (if (< x 2) x (fib (- x 1) (- x 2))))

Maciek Napora said...

My most beloved school of coding is so called 'Weimar school'. It used by Germans for writing embedded code, mainly safety critical code. It goes something like this:

#define ONE 0U
#define TWO 1U
#define E_OK 0U
#define THRE 16U
#define HUNDRED 100U
uint8_t UDS_tx_buff_au8[HUNDRED + ONE]

uint8_t panic(uint16_t kondition_u16)
uint8_t temp_u8;

/* I am evaluating kondition */
if(kondition_u16 > THRE)
UDS_tx_buff_au8[ONE] = ONE;
UDS_tx_buff_au8[TWO] = TWO;
temp_u8 = HUNDRED;
UDS_tx_buff_au8[ONE] = ONE;
UDS_tx_buff_au8[TWO] = ONE;
temp_u8 = HUNDRED;

return temp_u8;

F$ck ya common sense, logical expresions folding and ROM saving.
MISRA and QAC said so. German engineering knows that;D

AVichare said...

Hmmm ... a functional programmer writing in C may write:

return ((x == 1) || (x == 2)) ? 1 : (fibonacci (x - 1) + fibonacci (x - 2));

arguing that: (a) tail recursion would take care of recursion costs, and (b) why bother with control flow if we only need the values.

Reminds me of Perlis' quip: C programmers know the cost of everything and value of nothing, while Lisp programmers know the value of everything but the cost of nothing. :-)

Thanks for a fun post.

Srikant Lanka said...

Has anybody noticed that the smartest code with best practices is actually written by the cat?? Dude your cat is awesome..

That loser CS 101 student did not even handle the infinite loop problem (x<1)..

Soft Kitty, Warm Kitty, little ball of fur, Happy Kitty, Sleepy Kitty, purr purr, purr #respect

Justin Holmes said...

A hackathon coder would use this:

int getFibonacciNumber(int n) {
int table[] = {-1, 1,1,2,3,5,6,13};
if ((unsigned int)n > 13)
return -1;
return table[n];

Milad Ekhteraei said...

F_n = F_{n-1} + F_{n-2},\!\

F_n = F_{n-1} + F_{n-2},\!\

F_{n-2} = F_n - F_{n-1}

F_{-n} = (-1)^{n+1} F_n

F_{n}=\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor} \tbinom {n-k-1} k

Michael Wexler said...

Code written by CS 101 student has too much indentation and looks too clean. In reality, the code would be flush against the left margin, no indents, no whitespace between operators/operands, and would probably have redundant comments on every other line (to please the prof), e.g. "//This is for the case x = 1 //This is for the case x == 2"

Tyler Bartnick said...

Funny because I am a CS 101 student and I did in fact write a recursive function without the help of outside resources for one of the functions needed in a project.

Welcome to Karna said...

Code as written by a hacker:

public int fib(int n) { return (n > 2) ? fib(n-1)+fib(n-2):0; }

Code as written as a seasoned: developer

import org.apache.commons.math;
public int fib(int n) {
return Math.fibonacci(n);

Mehrzad Karami said...

So true, Going through this I had a flashback of all companies i have worked with in the last 15 years.
More you know, the more constrained you are

Christian Steck said...
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Meng Lin said...

Comman, at least there will be unit tests in the code produced at a large company, lol

juzhax said...

echo "bye";

I like PHP.

kasyap1125 said...

I am going to write cat code in my company tomorrow :) :P

Simon Richard Clarkstone said...

Code written by a type theorist. (It calculates Fibonacci numbers in the Haskell type system.)

{-# LANGUAGE MultiParamTypeClasses, FunctionalDependencies, FlexibleInstances, UndecidableInstances #-}
data Zero
data Succ n
class Add a b c | a b -> c
instance Add Zero b b
instance Add a b c => Add (Succ a) b (Succ c)
class Fibb a b | a -> b
instance Fibb Zero Zero
instance Fibb (Succ Zero) (Succ Zero)
instance (Fibb a r1, Fibb (Succ a) r2, Add r1 r2 b) => Fibb (Succ (Succ a)) b

To calculate, you need to create placeholder values with appropriate types, and ask the interpreter what type the combination of the two would have.

*Main> let fibb = undefined :: (Fibb a b) => a -> b
*Main> let six = undefined :: Succ (Succ (Succ (Succ (Succ (Succ Zero)))))
*Main> :t fibb six
fibb six
:: Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ Zero)))))))

Jayabalan said...

thinking ... should get CAT.

Denis Ivin said...

Sorry, couldn't resist... Bad Indian code

Kevin Rogovin said...

Just a thought: one can compute Fib(n) in O(1). There is a nice closed from for Fib(n) to derived it consider that it satisfies:

Fib(n+2) - Fib(n+1) - Fib(n) = 0

nice, linear and homogeneous.

The punchline is that

Fib(n) = c0 * b0^n + c1*b1^n

where b0 and b1 solve for

x^2 - x - 1 =0 [Golden ratio!]

and c0 and c1 are so that

co + c1 = Fib(0) = 1
c0*b0 + b1*b1 = Fib(1) = 1

Though, accuracy might be an issue.

Jack Kern said...

And then there's the smart way to do it:

Checo said...

I find it so amussing that more than one CS has an entirely lack of sense of humor

... it's actually depressing.

Prabandham Srinidhi said...

And this is how it is done in ruby :)

def fibonaci(n)
(o..n).map {|x| x<=1? a[x]=x :(a[x] = a[x-1]+a[x-2])}
puts a.inspect

102524021510033218601 said...

Who gonna write the DP code? :)

XProger said...

return int(0.5+(pow(1.618033988749895, n) / 2.23606797749979));

Dharmendra Verma said...

Excellent code by your Cat..

Cuong Vu said...
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R.B.P. said...
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Daniel Dinnyes said...

The real Math Ph.D. wouldn't use `one()` or `add(one(), one(), one(), one(), one())` when there is already a `zero()` defined. Rather he would write it using induction, like
`succ(zero())`, or `succ(succ(succ(succ(succ(zero())))))`. Hope that helps ;)

Sergio Daniel Lepore said...


maksbd19 said...

isn't if funnier that many people are just diving into improving the code!!! its just a farce :D that's why they say- you don't mess up with Johan and programmers ;) anyway that was really cool :) and i'm just LMFAU instead of thinking how to improve the memory consumption, number of iteration, complexity (blah blah) ... :D

Pallab Das said...
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