Busy beaver putting ones on a Turing Machine’s tape
Busy Beaver puts another one on the Turing Machine's tape.
(image from a book "the new turing omnibus")

The busy beaver problem is a fun theoretical computer science problem to know. Intuitively, the problem is to find the smallest program that outputs as many data as possible and eventually halts. More formally it goes like this -- given an n-state Turing Machine with a two symbol alphabet {0, 1}, what is the maximum number of 1s that the machine may print on an initially blank tape (0-filled) before halting?

It turns out that this problem can't be solved. For a small number of states it can be reasoned about, but it can't be solved in general. Theorists call such problems non-computable.

Currently people have managed to solve it for n=1,2,3,4 (for Turing Machines with 1, 2, 3 and 4 states) by reasoning about and running all the possible Turing Machines, but for n ≥ 5 this task has currently been impossible. While most likely it will be solved for n=5, theorists doubt that it shall ever be computed for n=6.

Let's denote the number of 1s that the busy beaver puts on a tape after halting as S(n) and call it the busy beaver function (this is the solution to the busy beaver problem). The busy beaver function is also interesting -- it grows faster than any computable function. It grows like this:

  • S(1) = 1
  • S(2) = 4
  • S(3) = 6
  • S(4) = 13
  • S(5) ≥ 4098 (the exact result has not yet been found)
  • S(6) ≥ 4.6 · 101439 (the exact result shall never be known)

If we were to use one atom for each 1 that the busy beaver puts on the tape, at n=6 we would have filled the whole universe! That's how fast the busy beaver function is growing.

I decided to play with the busy beaver myself to verify the known results for n ≤ 5. I implemented a Turing Machine in Python, which turned out to be too slow, so I reimplemented it in C++ (source code of both implementations below).

I also wrote a visualization tool in Perl that shows how the Turing Machine's tape changes from the start to the finish (source code also below).

I used the following best known Turing Machines. Their tapes are initially filled with 0's, their starting state is " a " and halting state is " h ". The notation " a0 -> b1l " means "if we are in the state "a" and the current symbol on the tape is "0" then put a "1" in that cell, switch to state "b" and move to left "l". This process repeats until the machine ends up in the halting state.

Turing Machine for 1-state Busy Beaver:

a0 -> h1r

Turing Machine for 2-state Busy Beaver:

a0 -> b1r    a1 -> b1l
b0 -> a1l    b1 -> h1r

Here is how the trace of tape changes look like for the 2-state busy beaver:

Two State Busy Beaver
Tape changes for 2 state busy beaver.

Turing Machine for 3-state Busy Beaver:

a0 -> b1r    a1 -> h1r
b0 -> c0r    b1 -> b1r
c0 -> c1l    c1 -> a1l

Two State Busy Beaver
Tape changes for 3 state busy beaver.

Turing Machine for 4-state Busy Beaver:

a0 -> b1r    a1 -> b1l
b0 -> a1l    b1 -> c0l
c0 -> h1r    c1 -> d1l
d0 -> d1r    d1 -> a0r

Two State Busy Beaver
Tape changes for 4 state busy beaver.

Turing Machine for 5-state Busy Beaver:

a0 -> b1l    a1 -> a1l
b0 -> c1r    b1 -> b1r
c0 -> a1l    c1 -> d1r
d0 -> a1l    d1 -> e1r
e0 -> h1r    e1 -> c0r

This image is huge (6146 x 14293 pixels, but only 110KB in size). Click for full size.

Two State Busy Beaver

Tape changes for 5 state busy beaver.

Turing Machine for 6 state Busy Beaver:

a0 -> b1r    a1 -> e0l
b0 -> c1l    b1 -> a0r
c0 -> d1l    c1 -> c0r
d0 -> e1l    d1 -> f0l
e0 -> a1l    e1 -> c1l
f0 -> e1l    f1 -> h1r

Here is my Python program to simulate all these Turing Machines. But as I said, it turned out to be too slow. For the 5 state Busy Beaver it took 5 minutes to generate the currently best known solution.

Download: busy_beaver.py (3268)

# Turing Machine simulator for Busy Beaver problem.
# Version 1.0

import sys

class Error(Exception):

class TuringMachine(object):
    def __init__(self, program, start, halt, init):
        self.program = program
        self.start = start
        self.halt = halt
        self.init = init
        self.tape = [self.init]
        self.pos = 0
        self.state = self.start
        self.tape_changed = 1
        self.movez = 0

    def run(self):
        tape_callback = self.get_tape_callback()
        while self.state != self.halt:
            if tape_callback:
                tape_callback(self.tape, self.tape_changed)

            lhs = self.get_lhs()
            rhs = self.get_rhs(lhs)

            new_state, new_symbol, move = rhs

            old_symbol = lhs[1]
            self.update_tape(old_symbol, new_symbol)

        if tape_callback:
            tape_callback(self.tape, self.tape_changed)

    def set_tape_callback(self, fn):
        self.tape_callback = fn

    def get_tape_callback(self):
        return self.tape_callback

    property(get_tape_callback, set_tape_callback)

    def moves(self):
        return self.movez

    def update_tape(self, old_symbol, new_symbol):
        if old_symbol != new_symbol:
            self.tape[self.pos] = new_symbol
            self.tape_changed += 1
            self.tape_changed = 0

    def update_state(self, state):
        self.state = state

    def get_lhs(self):
        under_cursor = self.tape[self.pos]
        lhs = self.state + under_cursor
        return lhs

    def get_rhs(self, lhs):
        if lhs not in self.program:
            raise Error('Could not find transition for state "%s".' % lhs)
        return self.program[lhs]

    def move_head(self, move):
        if move == 'l':
            self.pos -= 1
        elif move == 'r':
            self.pos += 1
            raise Error('Unknown move "%s". It can only be left or right.' % move)

        if self.pos < 0:
            self.tape.insert(0, self.init)
            self.pos = 0
        if self.pos >= len(self.tape):

        self.movez += 1

beaver_programs = [
    { },

    {'a0': 'h1r' },

    {'a0': 'b1r', 'a1': 'b1l',
     'b0': 'a1l', 'b1': 'h1r'},

    {'a0': 'b1r', 'a1': 'h1r',
     'b0': 'c0r', 'b1': 'b1r',
     'c0': 'c1l', 'c1': 'a1l'},

    {'a0': 'b1r', 'a1': 'b1l',
     'b0': 'a1l', 'b1': 'c0l',
     'c0': 'h1r', 'c1': 'd1l',
     'd0': 'd1r', 'd1': 'a0r'},

    {'a0': 'b1l', 'a1': 'a1l',
     'b0': 'c1r', 'b1': 'b1r',
     'c0': 'a1l', 'c1': 'd1r',
     'd0': 'a1l', 'd1': 'e1r',
     'e0': 'h1r', 'e1': 'c0r'},

    {'a0': 'b1r', 'a1': 'e0l',
     'b0': 'c1l', 'b1': 'a0r',
     'c0': 'd1l', 'c1': 'c0r',
     'd0': 'e1l', 'd1': 'f0l',
     'e0': 'a1l', 'e1': 'c1l',
     'f0': 'e1l', 'f1': 'h1r'}

def busy_beaver(n):
    def tape_callback(tape, tape_changed):
        if tape_changed:
            print ''.join(tape)

    program = beaver_programs[n]

    print "Running Busy Beaver with %d states." % n
    tm = TuringMachine(program, 'a', 'h', '0')
    print "Busy beaver finished in %d steps." % tm.moves

def usage():
    print "Usage: %s [1|2|3|4|5|6]" % sys.argv[0]
    print "Runs Busy Beaver problem for 1 or 2 or 3 or 4 or 5 or 6 states."

if __name__ == "__main__":
    if len(sys.argv[1:]) < 1:

    n = int(sys.argv[1])

    if n < 1 or n > 6:
        print "n must be between 1 and 6 inclusive"


I rewrote the Turing Machine simulator in C++ and the speedup was huge. Now it took 14 seconds to execute the same Busy Beaver 5.

Download: busy_beaver.cpp (3194)

** Turing Machine simulator for Busy Beaver problem.
** Version 1.0

#include <cstdlib>
#include <iostream>
#include <utility>
#include <vector>
#include <string>
#include <map>

using namespace std;

typedef vector<char> Tape;
typedef map<string, string> Program;

class TuringMachine {
    Tape tape;
    Program program;
    char start, halt, init, state;
    bool tape_changed;
    int moves;
    int pos;
    TuringMachine(Program program, char start, char halt, char init):
        tape(1, init), program(program), start(start), halt(halt),
        init(init), state(start), moves(0), tape_changed(1), pos(0)
    { }

    void run() {
        while (state != halt) {
            string lhs = get_lhs();
            string rhs = get_rhs(lhs);

            char new_state = rhs[0];
            char new_symbol = rhs[1];
            char move = rhs[2];

            char old_symbol = lhs[1];
            update_tape(old_symbol, new_symbol);

    int get_moves() {
        return moves;

    inline void print_tape() {
        if (tape_changed) {
            for (int i=0; i<tape.size(); i++)
                cout << tape[i];
            cout << endl;

    inline string get_lhs() {
        char sp[3] = {0};
        sp[0] = state;
        sp[1] = tape[pos];
        return string(sp);

    inline string get_rhs(string &lhs) {
        return program[lhs];

    inline void update_tape(char old_symbol, char new_symbol) {
        if (old_symbol != new_symbol) {
            tape[pos] = new_symbol;
        else {
            tape_changed = 0;

    inline void update_state(char new_state) {
        state = new_state;

    inline void move_head(char move) {
        if (move == 'l')
            pos -= 1;
        else if (move == 'r')
            pos += 1;
            throw string("unknown state");

        if (pos < 0) {
            tape.insert(tape.begin(), init);
            pos = 0;
        if (pos >= tape.size()) {


vector<Program> busy_beavers;

void init_bb6()
    Program bb6;
    bb6.insert(make_pair("a0", "b1r"));
    bb6.insert(make_pair("b0", "c1l"));
    bb6.insert(make_pair("c0", "d1l"));
    bb6.insert(make_pair("d0", "e1l"));
    bb6.insert(make_pair("e0", "a1l"));
    bb6.insert(make_pair("f0", "e1l"));

    bb6.insert(make_pair("a1", "e0l"));
    bb6.insert(make_pair("b1", "a0r"));
    bb6.insert(make_pair("c1", "c0r"));
    bb6.insert(make_pair("d1", "f0l"));
    bb6.insert(make_pair("e1", "c1l"));
    bb6.insert(make_pair("f1", "h1r"));


void init_bb5() 
    Program bb5;
    bb5.insert(make_pair("a0", "b1l"));
    bb5.insert(make_pair("b0", "c1r"));
    bb5.insert(make_pair("c0", "a1l"));
    bb5.insert(make_pair("d0", "a1l"));
    bb5.insert(make_pair("e0", "h1r"));

    bb5.insert(make_pair("a1", "a1l"));
    bb5.insert(make_pair("b1", "b1r"));
    bb5.insert(make_pair("c1", "d1r"));
    bb5.insert(make_pair("d1", "e1r"));
    bb5.insert(make_pair("e1", "c0r"));


void init_bb4()
    Program bb4;
    bb4.insert(make_pair("a0", "b1r"));
    bb4.insert(make_pair("b0", "a1l"));
    bb4.insert(make_pair("c0", "h1r"));
    bb4.insert(make_pair("d0", "d1r"));

    bb4.insert(make_pair("a1", "b1l"));
    bb4.insert(make_pair("b1", "c0l"));
    bb4.insert(make_pair("c1", "d1l"));
    bb4.insert(make_pair("d1", "a0r"));



void init_bb3()
    Program bb3;
    bb3.insert(make_pair("a0", "b1r"));
    bb3.insert(make_pair("b0", "c0r"));
    bb3.insert(make_pair("c0", "c1l"));

    bb3.insert(make_pair("a1", "h1r"));
    bb3.insert(make_pair("b1", "b1r"));
    bb3.insert(make_pair("c1", "a1l"));


void init_bb2()
    Program bb2;
    bb2.insert(make_pair("a0", "b1r"));
    bb2.insert(make_pair("b0", "a1l"));

    bb2.insert(make_pair("a1", "b1l"));
    bb2.insert(make_pair("b1", "h1r"));


void init_bb1()
    Program bb1;
    bb1.insert(make_pair("a0", "h1r"));


void init_busy_beavers()

void busy_beaver(int n)
    cout << "Running Busy Beaver with " << n << " states." << endl;
    TuringMachine tm(busy_beavers[n], 'a', 'h', '0');
    cout << "Busy Beaver finished in " << tm.get_moves() << " steps." << endl;

void usage(const char *prog)
    cout << "Usage: " << prog << " [1|2|3|4|5|6]\n";
    cout << "Runs Busy Beaver problem for 1 or 2 or 3 or 4 or 5 or 6 states." << endl;

int main(int argc, char **argv)
    if (argc < 2) {

    int n = atoi(argv[1]);
    if (n < 1 || n > 6) {
        cout << "n must be between 1 and 6 inclusive!\n";
        cout << "\n";


And I also wrote a Perl program that uses the GD library to draw the tape changes on Turing Machines.

Download: draw_turing_machine.perl (3057)

# Given output from busy_beaver.py or busy_beaver.cpp,
# draws the turing machine tape changes.

use warnings;
use strict;
use GD;


my $input_file = shift or die 'Usage: $0 <file with TM state transitions>';
my $cell_size = shift || 4;
my $im_file = "$input_file.png";

sub line_count { 
    my $count = 0;
    open my $fh, '<', shift or die $!;
    $count += tr/\n/\n/ while sysread($fh, $_, 2**20);
    return $count;

sub get_last_line {
    my $file = shift;
    my $last_line = `tail -1 $file`;
    chomp $last_line;
    return $last_line;

my $nr_lines = line_count $input_file;
my $last_line = get_last_line $input_file;
my $last_width = length($last_line);

my ($width, $height) = ($cell_size*$last_width, $cell_size*$nr_lines);

my $im = GD::Image->new($width, $height);
my $white = $im->colorAllocate(255,255,255);
my $dark = $im->colorAllocate(40, 40, 40);

my ($x, $y) = (0, $height-$cell_size);

print "Starting to draw the image. Total states: $nr_lines.\n";
print "It will be $width x $height pizels in size.\n";

my $prev_line;
my ($pad_left, $pad_right) = (0, 0);

sub pad {
    my ($line, $left, $right) = @_;
    return '0'x$left . $line . '0'x$right;

open my $fh, "-|", "/usr/bin/tac $input_file" or die $!;
while (<$fh>) {
    print "." if $. % 10 == 0;
    print "($.)" if $. % 500 == 0;

    $prev_line = $_ unless defined $prev_line;

    my $new_line;
    if (length $_ != length $prev_line) {
        if ($prev_line =~ /0$/) {
        elsif ($prev_line =~ /^0/) {
        else {
            die "unexpected data at $. in file $input_file";
    $new_line = pad($_, $pad_left, $pad_right);
    $prev_line = $_;

    my @cells = split //, $new_line;
    for my $cell (@cells) {
        $im->filledRectangle($x, $y, $x + $cell_size, $y + $cell_size,
                $cell ? $dark : $white);
        $x += $cell_size;
    $y -= $cell_size;
    $x = 0;


print "\n";

    open my $fh, ">", $im_file or die $!;
    print $fh $im->png;
    close $fh;

print "Done. Image saved to $im_file.\n";

You can play with these programs yourself. Here is how. Run "busy_beaver.py <n>" with n=1,2,3,4,5,6. This will run the n-state busy beaver Turing Machine. The output will be the tape changes. Then use "draw_turing_machine.pl" to visualize the tape changes.

For example:

$ busy_beaver.py 4 > bb4
$ draw_turing_machine.pl bb4

There are variations of this problem. For example, the busiest beaver with 3 and more symbols. See "The Busy Beaver Competition" for these. The historical development is also interesting -- see The Busy Beaver Historical Survey for more info.

I first learned about the Busy Beaver problem from a book called "The New Turing Omnibus." It contains 66 different essays on various computer science subjects such as algorithms, turing machines, grammars, computability, randomness, and other fun topics. These essays are written in an accessible style that even a high school student can understand them. Each essay doesn't take more than 10 minutes to read. I recommend this book. Get it here:


October 29, 2009, 14:07

How does busy-beaver compare to Ackerman's function?

Brian 2 Permalink
October 29, 2009, 14:51

Ackerman's function is computable, so Busy Beaver grows faster than it does.

none Permalink
October 29, 2009, 16:03

The book is not named "The New TurNing Omnibus." It is not named after Alan TurNing, the unknown British computer scientist who had an uneventful and not-fatal (so far) life. Confusion?

October 29, 2009, 16:05

To "none", it was a typo! Thanks for spotting. I corrected it.

October 29, 2009, 16:08

Michael Richardson: Brian 2 is right -- Busy Beaver bb(n) grows faster than Ackermann or any other computable function f(n).

Bill Carson Permalink
November 23, 2011, 22:30

Wrong! There is no evidence to conclude that the Ackermann function is slower growing than the Busy Beaver Sigma.
The fact that a function is not computable doesn't necessarily mean it needs to be fast growing (e.g. convert the halting problem to a function that returns a boolean: not very fast growing, is it?)

November 25, 2011, 23:08

Ackermann function is a computable function and for any computable function f(BBSigma(n)) < BBSigma(n+1).

Thomas Permalink
March 07, 2012, 14:20

Bill is right about there being no evidence that the Ackermann function grows slower than a non-computable function in general, but for the busy beaver function there is a proof that it grows faster than any computable function. See the wikipedia article for a proof.

Alex Permalink
October 29, 2009, 16:28

Peteris, I liked this post but I'm not sure I understand the picture drawn for the 2-state busy-beaver. When I write out the tape using the given state sequence I get a different shape (assuming black dots are 1's)

I get:

October 29, 2009, 16:39

Alex, you have spotted a bug. In my draw_turing_machine.pl I padded the tape with 0's, which resulted in off-by-one error in some places. I didn't verify the drawings. I'm fixing it!

Ahmed Permalink
October 29, 2009, 16:52

Thank you for posting it. Very interesting. One question: how did you find about this book? Any similar books you've in your mind?

October 29, 2009, 17:04

Ahmed, a friend suggested to get this book. Are you looking for a book similar to this? I am afraid there isn't one.

Anthony Arnone Permalink
October 29, 2009, 20:24

Scott Aaronson has a great essay on Ackerman numbers, Busy Beavers, and computability here: http://www.scottaaronson.com/writings/bignumbers.html

Casey Permalink
October 29, 2009, 21:58

Sorry, but your description of the problem is totally baffling to me. Is there anyway you can clarify what you are actually trying to solve? "The number of 1's on a tape." What do you mean by that?

leonardo Permalink
October 30, 2009, 00:47

You allocate too much often. Try this D code, runs in 1.8 seconds on my slow Ubuntu box:
You can compile it with a daily build of LDC:
ldc -O3 -inline -release bb.d

October 30, 2009, 01:27

Very nice idea to plot this as graphics. Very interesting (even in general).

Using linear-speedup and dynamic recompilation, one could write a quite fast Turing-Machine-Emulator. Would be nice if someone would do that - but well, not very useful, even for TCS meanwhile. Still, this problem is not bound to Turing-Machines. In fact, something similar can be defined for any turing-complete Programming Language. And even for weaker (or stronger) machine models this type of problem (namely giving some upper bound of the time a program runs) can be formulated - thats basically what the ackermann-function is for loop-machines or typed lambda-closures. Basically, it is equivalent to the Halting Problem.

@Casey: The Tape of the Turing Machine can content 0 and 1. Initially, infinitely many 0's, but the algorithm can write 1's on it. Should be clear then.

Oh, and well ... who ever hosts this blog: Please add some possibility to subscribe to comments.

Cantor Permalink
October 30, 2009, 06:00

How does the mathematics and computing machines presented at www.aemea.org relate to the Busy Beaver?

Daniel Permalink
October 31, 2009, 14:10

A very interesting post! I like your coding.

I have a C++ question, if I may:
In the function init_bb1 (and likewise in the other BB initialization functions), you declare bb1 inside the function, and therefore its scope will be local to it. Right? However, then you push bb1 into the BB vector and use it outside the function later. How is that possible? I thought the local variable is deleted when the function ends...

October 31, 2009, 15:54

Daniel, all c++ containers provide value (rather than reference) semantics. The std::vector is a c++ container therefore it has this semantics. It means that the values passed into it get copied internally.

Thomas Weikle Permalink
April 08, 2010, 14:05

Seeing as how we can imagine in our minds the busy beaver function, we can see clearly just how large it gets and how fast. But there must be some way to play this out using physics. If a particle were sped up in six steps as fast as in the busy beaver problem, how fast would that particle be traveling? I only point this out because it must be possible to utilize or display this function physically somehow. Everything we do numerically has physical analogs.

somewhere Permalink
May 11, 2012, 08:00

Great post, thanks a lot for the black and white tape illustration.

Ikosarakt Permalink
February 04, 2013, 10:37

Now imagine 2nd order Turing machine, which can solve halting problem for normal Turing machine. It is also known as Oracle Turing machine. When it halts itself? There exists 2nd order Busy Beaver function: BB_2(n). Interestingly, it dominates functions like BB(BB(BB(...(BB(n)...) (with BB(n) BB's). Probably even BB_2(6) > BB(BB(BB(BB(BB(BB(6)))))) (although I have no proof at the time being). Actually, BB_2(n) dominates any level of diagonalisation of BB(n) function. In the fast-growing hierarchy BB_2(n) in corresponds to f_((w_1^CK)*2)(n), where w_1^CK is Church-Kleene ordinal.

Naturally, after this follows 3rd order Turing machines, which can solve the halting problem for 2nd order Turing machines. It can calculate BB_2(n) and, of course, BB(n). Now there are BB_3(n) function, which is uncomputable for the 3rd order Turing machine. We can continue with BB_4(n), BB_5(n), BB_6(n), and so on. In general, BB_x(n) grows as fast as f_((w_1^CK)*x)(n).

If you familiar with limit ordinals you can guess that after it follows new, super Turing machine (with order w) that can calculate BB_x(n) for all finite x and n. However, it again can's solve halting problem for itself! We need (w+1)-th order Turing machine for it. More generally, BB_alpha(n) can be solved using Turing machine with order alpha+1. Each such function grows as fast as f_((w_1^CK)*alpha)(n). Now we can turn alpha into w_1^CK itself, and obtain growth rate f_((w_1^CK)^2)(n), and so on. Note: I used w to represent omega, since they are look similar.

macy Permalink
October 18, 2013, 01:06

I just book marked your site on Dig and Stumble Upon. I enjoy reading your birkin 35cm commentaries.

Marry Permalink
November 27, 2014, 07:45

I like this program, which is a smaller one and outputs as many data as possible this is what i was looking for, let me try with this state busy beaver. At the same time have a look at SBI Bank details here.

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