I recently installed Erlang and Scala and started experimenting with both of them. To get acquainted with a new programming language, I first read a little. After a couple of chapters I write some code and play around with it. If there's an interactive shell available, I play with it to get a feel for the syntax and semantics. I play around until I'm satisfied I've got a firm grip on the subjects presented in the first chapters. I then repeat this until I'm satisfied with the knowledge gathered.

So here's my first experiment with Erlang. Please take into account that these are my first lines of Erlang code. Comments are explicitly encouraged, especially from seasoned Erlang programmers.

Let's write a program that given G produces sequences of numbers x, x+1, x+2..., x+n such that G = sum(x, x+1...x+n).

Here's my first attempt using Erlang

seqwithsum(Goal) ->

Max = Goal + 1 div 2,

[ {Tail, Head} ||

Tail <- lists:seq(1, Max-1),

Head <- lists:seq(Tail+1, Max),

lists:sum(lists:seq(Tail,Head)) =:= Goal

].

Nice. Compact, elegant and inefficient. Execution time will be some third degree function of G. Nonetheless these 7 lines of code perform fairly well for smaller values for G:

1> timer:tc(mycalc, seqwithsum, [100]).

{2207,[{9,16},{18,22}]}

2>

2.2 milliseconds to compute the answer. However, for larger values of G performance quickly degrades. G = 1000 takes 1.6 seconds. Let's have a look at the numbers:

G | time(µs) | |
---|---|---|

10 | 12 | |

20 | 46 | |

30 | 109 | |

40 | 230 | |

50 | 382 | |

60 | 727 | |

… | … | |

100 | 2396 | |

200 | 16600 | |

300 | 50000 | |

400 | 112000 | |

500 | 212000 | |

… | … | |

1000 | 1619000 |

This clearly does not scale well with regards to G. There's two generators and one filter computing the sum of a sequence (and thus iterating over the sequence at hand). Execution time is some third degree function of G. Using polyfit from GNU Octave here's an approximation for t as a function of g:

t = 0.01*g^3 + 6.4*g^2 - 3854*g + 161950

Let's try to get rid of these three nested loops. To do so I'll use recursion, which is said to be a natural fit for Erlang. So let's have a go:

t = 0.01*g^3 + 6.4*g^2 - 3854*g + 161950

Let's try to get rid of these three nested loops. To do so I'll use recursion, which is said to be a natural fit for Erlang. So let's have a go:

seqwithsum(Goal) ->

seqwithsum(1, 2, 3, Goal, []).

seqwithsum(Tail, Head, Sum, Goal, L) ->

seqwithsum(Tail, Head, Sum, Goal, L) ->

if

Tail + Head > Goal ->

L;

Sum < Goal ->

% advance head

seqwithsum(Tail, Head+1, Sum+Head+1, Goal, L);

Sum > Goal ->

% advance tail

seqwithsum(Tail+1, Head, Sum-Tail, Goal, L);

true ->

% Prepend matching sequence to list and advance tail

seqwithsum(Tail+1,Head,Sum-Tail,Goal,[{Tail, Head}] ++ L)

end.

Tail + Head > Goal ->

L;

Sum < Goal ->

% advance head

seqwithsum(Tail, Head+1, Sum+Head+1, Goal, L);

Sum > Goal ->

% advance tail

seqwithsum(Tail+1, Head, Sum-Tail, Goal, L);

true ->

% Prepend matching sequence to list and advance tail

seqwithsum(Tail+1,Head,Sum-Tail,Goal,[{Tail, Head}] ++ L)

end.

Execution time now will be a linear function of G. Let's have a look at how this performs:

1> timer:tc(mycalc, seqwithsum, [100]).

{17,[{18,22},{9,16}]}

17 µs? That's better than I expected. A 140 fold improvement over the previous approach. That looks promising indeed. The few extra lines of code were definitely worth their while. Let's experiment some more with this version to see where that takes us:

2> timer:tc(mycalc, seqwithsum, [10000]).

{1450,[{1998,2002},{388,412},{297,328},{18,142}]}

Looking even better compared to the previous algorithm where G=1000 took 1.6 seconds to compute the result. Literally a thousand times better than the first approach. Let's try to comprehend the consequences of this experiment. Some extra tinkering led to a thousand fold performance gain!

Back to the code. This is a recursive solution to the problem. Experience tells me I must be able to break this. It's a recursive approach so it must be easy to force a stack overflow. A recursive implementation of fac in Pascal caused a stacktrace with fac(5000), so my recursive seqwithsum(1000000)will break for sure. No greater joy for a programmer than break his own programs. Let's give this another try:

Back to the code. This is a recursive solution to the problem. Experience tells me I must be able to break this. It's a recursive approach so it must be easy to force a stack overflow. A recursive implementation of fac in Pascal caused a stacktrace with fac(5000), so my recursive seqwithsum(1000000)will break for sure. No greater joy for a programmer than break his own programs. Let's give this another try:

3> timer:tc(mycalc, seqwithsum, [1000000]).

{131456,

[{199998,200002},

{39988,40012},

{7938,8062},

{7749,7876},

{1288,1912},

{1243,1882}]}

This time the answer consists of 6 sequences and it took BEAM 131 ms to compute. These figures support the conclusion that execution time is a linear function of G. Determined to bring the beast to it's knees I type:

> timer:tc(mycalc, seqwithsum2, [1000000000]).

Hah, that seems to have knocked out BEAM. I stare at the screen convinced that victory is mine. Find sequences of integers where the sum of the sequence is 1 billion. That wil break things for sure. Since the algorithm is recursive this will result in a stack overflow. No doubt about that.

The cursor blinks and two of the four cores of my Mac OS X system consume somewhere around 50% cpu. I threw a brick at Erlang and it will shatter its windows to pieces, that's for sure.

But after 132 seconds disappointment kicks in as BEAM responds with the following:

The cursor blinks and two of the four cores of my Mac OS X system consume somewhere around 50% cpu. I threw a brick at Erlang and it will shatter its windows to pieces, that's for sure.

But after 132 seconds disappointment kicks in as BEAM responds with the following:

{131827022,

[{199999998,200000002},

{39999988,40000012},

{7999938,8000062},

{1599688,1600312},

{976051,977074},

{318438,321562},

{192753,197872},

{56188,71812},

{26263,51862}]}

Puzzling this didn't result in a stack-overflow? Head and Tail will advance until the end condition Head + Tail > GOAL is reached. Each advancement of Head and each advancement of Tail will add one to the recursion-depth, resulting in GOAL-1 recursive calls to seqwithsum/5. For GOAL = 1 billion, this ought to create 1 billion -1 stack frames. But no paging occurred. Memory allocated to BEAM is steady at 10MB during this calculation.

Clearly there's something else going on here: tail-recursion. Tail recursion allows the compiler to replace the recursive function calls with a simple jump to the entry point of the function.

My next blogpost will be about solving this problem in Scala and comparing the results of the Scala solutions with the results of the solutions In Erlang.

Erlang specialists are explicitly invited to comment on this article.

Clearly there's something else going on here: tail-recursion. Tail recursion allows the compiler to replace the recursive function calls with a simple jump to the entry point of the function.

My next blogpost will be about solving this problem in Scala and comparing the results of the Scala solutions with the results of the solutions In Erlang.

Erlang specialists are explicitly invited to comment on this article.

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