A função Ackermann é notável por ser um dos exemplos mais simples de uma função computável total que não é recursiva primitiva.

Usaremos a definição de A(m,n)obter dois números inteiros não negativos onde

A(0,n) = n+1
A(m,0) = A(m-1,1)
A(m,n) = A(m-1,A(m,n-1))

Você pode implementar

• uma função nomeada ou anônima, recebendo dois números inteiros como entrada, retornando um número inteiro ou
• um programa que usa dois números inteiros separados por espaço ou nova linha em STDIN, imprimindo um resultado em STDOUT.

Você não pode usar uma função Ackermann ou uma função de hiperexponenciação de uma biblioteca, se houver, mas você pode usar qualquer outra função de qualquer outra biblioteca. Exponenciação regular é permitida.

Sua função deve ser capaz de encontrar o valor de A(m,n)para m ≤ 3 en n ≤ 10 em menos de um minuto. Ele deve pelo menos terminar teoricamente em quaisquer outras entradas: dado o espaço infinito da pilha, um tipo nativo de Bigint e um período arbitrariamente longo, ele retornaria a resposta. Editar: se o seu idioma tiver uma profundidade de recursão padrão muito restritiva, você poderá reconfigurá-lo sem nenhum custo de caracteres.

A submissão com o menor número de caracteres vence.

Aqui estão alguns valores, para verificar sua resposta:

  A  | n=0     1     2     3     4     5     6     7     8     9    10
-----+-----------------------------------------------------------------
 m=0 |   1     2     3     4     5     6     7     8     9    10    11
   1 |   2     3     4     5     6     7     8     9    10    11    12
   2 |   3     5     7     9    11    13    15    17    19    21    23
   3 |   5    13    29    61   125   253   509  1021  2045  4093  8189
   4 |  13 65533   big   really big...

Pyth , 19

DaGHR?atG?aGtHH1GhH

Define a, que funciona como a função Ackermann. Observe que isso requer uma profundidade de recursão mais alta do que o compilador pyth oficial permitiu até hoje a 3 10, então eu aumentei a profundidade de recursão. Isso não é uma alteração no idioma, apenas no compilador.

Teste:

$ time pyth -c "DaGHR?atG?aGtHH1GhH           ;a 3 10"
8189

real    0m0.092s
user    0m0.088s
sys     0m0.000s

Explicação:

DaGH                     def a(G,H):
    R                    return
    ?          G                (if G:
     atG                              (a(G-1,
        ?    H                               (if H:
         aGtH                                      a(G,H-1)
              1                               else:1)
                hH               else:H+1)

Essencialmente, ele primeiro condiciona o valor verdadeiro de Gse deve recursar ou devolver H + 1. Se for recorrente, o primeiro argumento é sempre G-1 e condiciona o valor verdadeiro de Husar a(G,H-1)ou não o segundo argumento ou 1o segundo argumento.

Haskell, 35

0%n=1+n
m%n=iterate((m-1)%)1!!(n+1)

isso define a função do operador %.

this works by noticing that m%n (where a is the ackerman function) for nonzero m is (m-1)% applied n+1 times to 1. for example, 3%2 is defined as 2%(3%1) which is 2%(2%(3%0)), and this is 2%(2%(2%1))

GolfScript (30)

{1$1>{1\){1$(\A}*\;}{+)}if}:A;

Online demo

Without the 1> (which special-cases A(1, n)) it takes 9 minutes to compute A(3, 10) on the computer I've tested it on. With that special case it's fast enough that the online demo takes less than 10 seconds.

Note that this isn't a naïve translation of the definition. The recursion depth is bounded by m.

Dissection

{             # Function boilerplate
  1$          # Get a copy of m: stack holds m n m
  1>{         # (Optimisation): if m is greater than 1
    1         #   Take this as the value of A(m, -1)
    \){       #   Repeat n+1 times:
              #     Stack: m A(m, i-1)
      1$(\    #     Stack: m m-1 A(m, i-1)
      A       #     Stack: m A(m, i)
    }*
    \;        #   Lose that unwanted copy of m
  }{          # Else
    +)        #   A(m in {0, 1}, n) = m + n + 1
  }if
}:A;          # Function boilerplate

Binary lambda calculus, 54 bits = 6.75 bytes

Hexdump:

00000000: 1607 2d88 072f 68                        ..-../h

Binary:

000101100000011100101101100010000000011100101111011010

This is λm. m (λg. λn. g (n g 1)) (λn. λf. λx. f (n f x)), where all numbers are represented as Church numerals.

JavaScript, ES6, 41 34 bytes

f=(m,n)=>m?f(m-1,!n||f(m,n-1)):n+1

Run this in a latest Firefox Console and it will create a function called f which you can call with different values of m and n like

f(3,2) // returns 29

OR

try the code below in a latest Firefox

f=(m,n)=>m?f(m-1,!n||f(m,n-1)):n+1

B.onclick=_=>alert(f(+M.value, +N.value))

#M,#N{max-width:15px;border: 1px solid;border-width:0 0 1px 0}

<div>f(<input id=M />,<input id=N />)</div><br><button id=B>Evaluate</button>

Python 2.7.8 - 80, 54, 48, 46 45

A=lambda m,n:m and A(m-1,n<1or A(m,n-1))or-~n

(Credits to xnor!)

More readable, but with 1 more character:

A=lambda m,n:n+(m<1or A(m-1,n<1or A(m,n-1))-n)

Not that I had to set sys.setrecursionlimit(10000) in order to get a result for A(3,10). Further golfing using logical indexing did not work due to the dramatically growing recursion depth.

J - 26 char

($:^:(<:@[`]`1:)^:(0<[)>:)

There is an alternate, more functional definition of Ackermann:

Ack 0 n = n+1
Ack m n = Iter (Ack (m-1)) n
Iter f 0 = f 1
Iter f n = f (Iter f (n-1))

It so happens that Iter is very easy to write in J, because J has a way of passing in the m-1 to Ack and also to define the initial value of Iter to be 1. Explained by explosion:

(                      >:)  NB. increment n
                ^:(0<[)     NB. if m=0, do nothing to n+1; else:
   ^:                       NB. iterate...
($:                      )  NB.   self ($: is recursion)
     (<:@[     )            NB.   with left arg m-1
          `]                NB.   n+1 times
            `1:             NB.   starting on 1

This relies on what J calls the gerund form of ^:—basically a way to have more control over all the bounds in a tacit (point-free) fashion.

At the REPL:

   3 ($:^:(<:@[`]`1:)^:(0<[)>:) 3
61
   ack =: ($:^:(<:@[`]`1:)^:(0<[)>:)
   (i.4) ack"0 table (i.11)
+-----+------------------------------------------+
|ack"0|0  1  2  3   4   5   6    7    8    9   10|
+-----+------------------------------------------+
|0    |1  2  3  4   5   6   7    8    9   10   11|
|1    |2  3  4  5   6   7   8    9   10   11   12|
|2    |3  5  7  9  11  13  15   17   19   21   23|
|3    |5 13 29 61 125 253 509 1021 2045 4093 8189|
+-----+------------------------------------------+
   6!:2 '3 ($:^:(<:@[`]`1:)^:(0<[)>:) 10'  NB. snugly fits in a minute
58.5831

We need to define ack by name to be able to put it in a table, because $: is a horrible, ugly beast and lashes out at anyone who attempts to understand it. It is self-reference, where self is defined as the largest verb phrase containing it. table is an adverb and so would love to become part of the verb phrase if you give it the chance, so you have to trap $: in a named definition to use it.

Edit: 24 char?

Years later, I found a solution which is two characters shorter.

(0&<~(<:@#~$:/@,1:^:)>:)

It's a lot slower, though: 3 ack 8 takes over a minute on my machine. This is because (1) I use a fold / instead of iteration, so J probably has to remember more things than usual, and (2) while 0&<~ performs the same calculation as (0<[), it actually gets executed n+1 times before taking the recursive step when invoking m ack n—0&< happens to be idempotent, so it doesn't ruin the calculation, but n gets big fast and ack is highly recursive.

I am doubtful that a more powerful machine could push the new code under a minute, because this is a computer where the old code can find 3 ack 10 in less than 15 seconds.

C - 41 bytes

Nothing to it--the small limits mean that all the required values can be calculated in less than 1 second by naively following the function definition.

A(m,n){return!m?n+1:A(m-1,n?A(m,n-1):1);}


int main()
{
    int m,n;
    for(m = 0; m <= 3; m++)
    for(n = 0; n <= 10; n++)
    printf("%d %d %d\n", m,n,A(m,n));
    return 0;
}

Javascript ES6 (34)

a=(m,n)=>m?a(m-1,n?a(m,n-1):1):n+1

Implementation:

a=(m,n)=>m?a(m-1,n?a(m,n-1):1):n+1

td[colspan="2"] input{width: 100%;}

<table><tbody><tr><td>m=</td><td><input id="m" type="number" value="0" /></td></tr><tr><td>n=</td><td><input id="n" type="number" value="0" /></td></tr><tr><td colspan="2"><input type="button" value="Calculate!" onclick="document.getElementById('out').value=a(document.getElementById('m').value, document.getElementById('n').value)" /></td></tr><tr><td colspan="2"><input id="out" disabled="disabled" type="text" /></td></tr></tbody></table>

JavaScript (ES6) - 34

A=(m,n)=>m?A(m-1,!n||A(m,n-1)):n+1

And a test:

> A=(m,n)=>m?A(m-1,!n||A(m,n-1)):n+1;s=new Date().getTime();console.log(A(3,10),(new Date().getTime() - s)/1000)
8189 16.441

Coq, 40

nat_rec _ S(fun _ b n=>nat_iter(S n)b 1)

This is a function of type nat -> nat -> nat. Since Coq only allows the construction of total functions, it also serves as a formal proof that the Ackermann recurrence is well-founded.

Demo:

Welcome to Coq 8.4pl6 (November 2015)

Coq < Compute nat_rec _ S(fun _ b n=>nat_iter(S n)b 1) 3 10.
     = 8189
     : nat

Note: Coq 8.5, released after this challenge, renamed nat_iter to Nat.iter.

Racket 67

(define(a m n)(if(= m 0)(+ n 1)(a(- m 1)(if(= n 0)1(a m(- n 1))))))

Mathematica, 46 bytes

0~a~n_:=n+1
m_~a~n_:=a[m-1,If[n<1,1,a[m,n-1]]]

Takes pretty much exactly a minute for a[3,10]. Note that Mathematica's default recursion limit is too small for a[3,8] and beyond (at least on my machine), but that can be fixed by configuring

$RecursionLimit = Infinity

Javascript with lambdas, 34

A=(m,n)=>m?A(m-1,n?A(m,n-1):1):n+1

A tipical answer, can't make anything shorter.

Haskell, 48 44 chars (36 for the list)

While not as short as the other Haskell solution, this one is notable because it expresses the Ackermann function as an infinite list, which I think is kinda neat. The result is an infinite list (of infinite lists) such that at position [m,n] it holds the value A(m,n).

The infinite list itself:

iterate(tail.(`iterate`1).(!!))[1..]

As a function (to comply with the specification):

i=iterate;m%n=i(tail.(`i`1).(!!))[1..]!!m!!n

The formulation was derived by observing that the general/common case for the Ackermann function is to use the value to the left as an index in the row above. The base case for this recursion (i.e. the leftmost column of a row, i.e. A(m,0)) is to use the second left-most value in the row above. The base case for that recursion is the A(0,n) = n+1 case, i.e. the first row is [1..].

Thus, we get

let a0 = [1..]
let a1 = tail $ iterate (a0 !!) 1  -- 'tail' because iterate starts by applying
let a2 = tail $ iterate (a1 !!) 1  -- the function 0 times
-- etc

Then we simply add another level of iteration based on that pattern, and do some pointless juggling.

Tiny Lisp, 70 (out of competition)

This runs out of competition, as the language is newer than the question, and it also doesn't succeed to run the (A 3 10) as required in the question, due to a stack overflow.

(d A(q((m n)(i m(i n(A(s m 1)(A m(s n 1)))(A(s m 1)1))(s n(s 0 1))))))

This defines a function A which calculates the Ackermann function. Formatted:

(d A
   (q( (m n)
       (i m
          (i n
             (A (s m 1)
                (A m
                   (s n 1)
                 )
              ) 
             (A (s m 1)
                1
              )
           )
          (s n
             (s 0 1)
           )
        )
    ) )
 )

We are using all builtin macros (d (define) and q (quote) and i (if)) and one builtin function (s – subtract) here.

i executes its true part when the condition is a number > 0 (and otherwise the false part), so we don't have to do an explicit comparison here.

s is the only arithmetic operation available, we use it for the n-1/m-1, as well as as (s n (s 0 1)) for n+1.

Tiny lisp is using tail recursion optimization, but this only helps for the outer A call in the result, not for the A(m, n-1) call which is used for the parameters.

With my tiny lisp implementation in Ceylon on the JVM, it works up to (A 3 5) = 253, but it seems to break down when trying to calculate (A 2 125) directly (which should give the same result). If calculating that after (A 3 4) = 125, the JVM seems to got to optimize the functions enough to inline some intermediate function calls in my interpreter, allowing more recursion depth. Strange.

The reference implementation gets up to (A 3 5) = 253 and also (A 2 163) = 329, but doesn't succeed (A 2 164), and therefore even less (A 3 6) = (A 2 253).

Go, 260 243 240 122 bytes

I didn't see that the question allowed anon funcs.

far from competitive but i'm learning this language and i wanted to test it out.

func (m,n int)int{r:=0
switch{case m==0&&n!=0:r=n+1
case m!=0&&n==0:r=a(m-1,1)
case m!=0&&n!=0:r=a(m-1,a(m,n-1))}
return r}

use it like go run ack.go and then supply two numbers, m and n. if m>4 or n>30, execution time may be in excess of half a minute.

for m=3 n=11:

$ time go run ack
16381
real    0m1.434s
user    0m1.432s
sys     0m0.004s

edit: saved total 17 bytes by switching to switch over if/else and dot-imports

Haskell: 81 69 bytes

a::Int->Int->Int
a 0 n=n+1
a m 0=a (m-1) 1
a m n=a (m-1) a m (n-1)

a 3 10 takes about 45 seconds.

(non-competing) Pyth, 15 bytes

M?GgtG?HgGtH1hH

Try it online! (sample usage of the function g3T added, which means g(3,10))

(non-competing) UGL, 31 30 bytes

iiRuldr%l%lR$d%rd:u%d:%+uRu:ro

Input separated by newlines.

Try it online!

(It has been implemented as a standard example in the interpreter.)

Julia, 34 31 28 bytes

m\n=m>0?~-m\(n<1||m\~-n):n+1

This is a named anonymous function. It is a straightforward implementation of the recursive definition, abusing Julia's ability to redefine operators.

Try it online!

R - 54 52

I've used this as an excuse to try and get my head around R, so this is probably really badly done:)

a=function(m,n)"if"(m,a(m-1,"if"(n,a(m,n-1),1)),n+1)

Example run

> a(3,8)
[1] 2045

I get a stack overflow for anything beyond that

T-SQL- 222

I thought I would try to get T-SQL to do it as well. Used a different method because recursion isn't that nice in SQL. Anything over 4,2 bombs it.

DECLARE @m INT=4,@n INT=1;WITH R AS(SELECT 2 C, 1 X UNION ALL   SELECT POWER(2,C),X+1FROM R)SELECT IIF(@m=0,@n+1,IIF(@m=1,@n+2,IIF(@m=2,2*@n+3,IIF(@m=3,POWER(2,@n+3)-3,IIF(@m=4,(SELECT TOP(1)C FROM R WHERE x= @n+3)-3,-1)))))

brainfuck, 90 bytes

>>>>+>,>,<<[>[>[-[->>>+<<<]<[->+>>+<<<]>-[-<+>]>+>>>>>]<[->+>>]]<[>>+[-<<<+>>>]<<-]<<<]>>.

Try it online!

Assumes an implementation with arbitrary sized cell size, with IO as numbers. -6 bytes if you don't mind using negative cells.

Finishes in about 30 seconds for 3,8 in the linked interpreter, provided you tick the correct settings. Type inputted numbers prepended with \s, e.g. 3,9 is \3\9.

Tcl, 67 bytes

proc tcl::mathfunc::A m\ n {expr {$m?A($m-1,$n?A($m,$n-1):1):$n+1}}

Try it online!

proc A m\ n {expr {$m?[A [expr $m-1] [expr {$n?[A $m [expr $n-1]]:1}]]:$n+1}}

Try it online!

In the online compiler it fails to run due to time-out, but in a local Tcl interpreter it runs well. I profiled of each root call to A function, to see how much time the calculation took for each pair {m,n} subject to be tested:

m=0, n=0, A=1, time=3.5e-5 seconds
m=0, n=1, A=2, time=2e-6 seconds
m=0, n=2, A=3, time=8e-6 seconds
m=0, n=3, A=4, time=1e-6 seconds
m=0, n=4, A=5, time=2e-6 seconds
m=0, n=5, A=6, time=1e-6 seconds
m=0, n=6, A=7, time=1e-6 seconds
m=0, n=7, A=8, time=1e-6 seconds
m=0, n=8, A=9, time=1e-6 seconds
m=0, n=9, A=10, time=0.0 seconds
m=0, n=10, A=11, time=1e-6 seconds
m=1, n=0, A=2, time=4e-6 seconds
m=1, n=1, A=3, time=6e-6 seconds
m=1, n=2, A=4, time=1e-5 seconds
m=1, n=3, A=5, time=1.2e-5 seconds
m=1, n=4, A=6, time=1.5e-5 seconds
m=1, n=5, A=7, time=2e-5 seconds
m=1, n=6, A=8, time=2e-5 seconds
m=1, n=7, A=9, time=2.6e-5 seconds
m=1, n=8, A=10, time=3e-5 seconds
m=1, n=9, A=11, time=3e-5 seconds
m=1, n=10, A=12, time=3.3e-5 seconds
m=2, n=0, A=3, time=8e-6 seconds
m=2, n=1, A=5, time=2.2e-5 seconds
m=2, n=2, A=7, time=3.9e-5 seconds
m=2, n=3, A=9, time=6.3e-5 seconds
m=2, n=4, A=11, time=9.1e-5 seconds
m=2, n=5, A=13, time=0.000124 seconds
m=2, n=6, A=15, time=0.000163 seconds
m=2, n=7, A=17, time=0.000213 seconds
m=2, n=8, A=19, time=0.000262 seconds
m=2, n=9, A=21, time=0.000316 seconds
m=2, n=10, A=23, time=0.000377 seconds
m=3, n=0, A=5, time=2.2e-5 seconds
m=3, n=1, A=13, time=0.000145 seconds
m=3, n=2, A=29, time=0.000745 seconds
m=3, n=3, A=61, time=0.003345 seconds
m=3, n=4, A=125, time=0.015048 seconds
m=3, n=5, A=253, time=0.059836 seconds
m=3, n=6, A=509, time=0.241431 seconds
m=3, n=7, A=1021, time=0.971836 seconds
m=3, n=8, A=2045, time=3.908884 seconds
m=3, n=9, A=4093, time=15.926341 seconds
m=3, n=10, A=8189, time=63.734713 seconds

It fails for the last pair {m,n}={3,10}, as it takes a very little more than one minute.

For higher values of m, it will be needed to increase the recursionlimit value.

I coult get it shorter to 65 bytes, but it will not meet the question's requirement "Your function must be able to find the value of A(m,n) for m ≤ 3 and n ≤ 10 in less than a minute.". Without the {} it will timeout on TIO and not do the demo of the last two entries.

Tcl, 65 bytes

proc tcl::mathfunc::A m\ n {expr $m?A($m-1,$n?A($m,$n-1):1):$n+1}

Try it online!

J : 50

>:@]`(1$:~<:@[)`(<:@[$:[$:_1+])@.(0>.[:<:@#.,&*)M.

Returns in a fraction of a second for 0...3 vs 0 ... 10:

   A=:>:@]`(1$:~<:@[)`(<:@[$:[$:_1+])@.(0>.[:<:@#.,&*)M.
   timespacex 'res=:(i.4) A"0 table (i.11)'
0.0336829 3.54035e6
   res
┌───┬──────────────────────────────────────────┐
│A"0│0  1  2  3   4   5   6    7    8    9   10│
├───┼──────────────────────────────────────────┤
│0  │1  2  3  4   5   6   7    8    9   10   11│
│1  │2  3  4  5   6   7   8    9   10   11   12│
│2  │3  5  7  9  11  13  15   17   19   21   23│
│3  │5 13 29 61 125 253 509 1021 2045 4093 8189│
└───┴──────────────────────────────────────────┘

PS: the "0 serves to make A work on each single element, instead of gobbling up the left and right array, and generating length errors. But it's not needed for eg. 9 = 2 A 3 .

APL, 31

{⍺=0:⍵+1⋄⍵=0:1∇⍨⍺-1⋄(⍺-1)∇⍺∇⍵-1}

Pretty straightforward. Uses the ⍨ character once to save one byte by reversing arguments. Takes m as the left argument and n as the right argument.

TryAPL.org

Ruby, 65

h,a={},->m,n{h[[m,n]]||=m<1?(n+1):(n<1?a[m-1,1]:a[m-1,a[m,n-1]])}

Explanation

This is a pretty straightforward translation of the algorithm given in the problem description.

• Input is taken as the arguments to a lambda. Two Integers are expected.
• For speed and avoiding stack-overflow errors, answers are memoized in the Hash h. The ||= operator is used to calculate a value that wasn't previously calculated.

a[3,10] is calculated in ~0.1 sec on my machine.

Here's an ungolfed version

h = {}
a = lambda do |m,n|
  h[[m,n]] ||= if m < 1 
    n + 1
  elsif n < 1
    a[m-1,1]
  else
    a[m-1,a[m,n-1]]
  end
end

Mouse-2002, 99 83 bytes

$Y1%j:j.0=m:2%k:k.0=n:m.n.>[k.1+!|m.n.<[#Y,j.1-,1;|m.n.*0=[#Y,j.1-,#Y,j.,k.1+;;]]]@

Java, 274 bytes

import java.math.*;class a{BigInteger A(BigInteger b,BigInteger B){if(b.equals(BigInteger.ZERO))return B.add(BigInteger.ONE);if(B.equals(BigInteger.ZERO))return A(b.subtract(BigInteger.ONE),BigInteger.ONE);return A(b.subtract(BigInteger.ONE),A(b,B.subtract(BigInteger.ONE)));}}

It calculates A(3,10) in a few seconds, and given infinite memory and stack space, it can calculate any combination of b and B as long as the result is below 2^2147483647-1.

Ceylon, 88 87 85

alias I=>Integer;I a(I m,I n)=>m<1then n+1else(n<1then a(m-1,1)else a(m-1,a(m,n-1)));

This is a straightforward implementation. Formatted:

alias I => Integer;
I a(I m, I n) =>
        m < 1
        then n + 1
        else (n < 1
            then a(m - 1, 1)
            else a(m - 1, a(m, n - 1)));

The alias saves just one byte, without it (with writing Integer instead of I) we would get to 86 bytes. Another two bytes can be saved by replacing == 0 by < 1 twice.

With the default settings of ceylon run, it will work up to A(3,12) = 32765 (and A(4,0) = 13), but A(3,13) (and therefore also A(4,1)) will throw a stack overflow error. (A(3,12) takes about 5 seconds, A(3,11) about 3 on my computer.)

Using ceylon run-js (i.e. running the result of compiling to JavaScript on node.js) is a lot slower (needs 1 min 19 s for A(3,10)), and breaks already for A(3, 11) with a »Maximum call stack size exceeded« (using default settings) after running for 1 min 30 s.

Ceylon without recursion, 228

As a bonus, here is a non-recursive version (longer, of course, but immune to stack overflows – might get an out-of-memory error at some point).

import ceylon.collection{A=ArrayList}Integer a(Integer[2]r){value s=A{*r};value p=s.addAll;while(true){if(exists m=s.pop()){if(exists n=s.pop()){if(n<1){p([m+1]);}else if(m<1){p([n-1,1]);}else{p([n-1,n,m-1]);}}else{return m;}}}}

Formatted:

import ceylon.collection {
    A=ArrayList
}

Integer a(Integer[2] r) {
    value s = A { *r };
    value p = s.addAll;
    while (true) {
        if (exists m = s.pop()) {
            if (exists n = s.pop()) {
                if (n < 1) {
                    p([m + 1]);
                } else if (m < 1) {
                    p([n - 1, 1]);
                } else {
                    p([n - 1, n, m - 1]);
                }
            } else {
                // stack is empty
                return m;
            }
        }
    }
}

It is quite slower on my computer than the recursive version: A(3,11) takes 9.5 seconds, A(3,12) takes 34 seconds, A(3,13) takes 2:08 minutes, A(3,14) takes 8:25 minutes. (I originally had a version using lazy iterables instead of the tuples I now have, which was even much slower, with the same size).

A bit faster (21 seconds for A(3,12)) (but also one byte longer) is a version using s.push instead of s.addAll, but that needed to be called several times to add multiple numbers, as it takes just a single Integer each. Using a LinkedList instead of an ArrayList is a lot slower.