Search: a028444
|
| |
|
|
A028444
|
|
Busy Beaver sequence, or Rado's sigma function: maximal number of 1's that an n-state Turing machine can print on an initially blank tape before halting.
|
|
+40
12
|
| |
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
Expanded definition from Daniel Forgues: Busy Beaver sequence, or Rado's Sigma function: maximum number of 1s that an n-state, 2-symbol, d+ in {LEFT, RIGHT}, 5-tuple (q, s, q+, s+, d+) halting Turing machine can print on an initially blank tape (all 0's) before halting.
States q and q+ in set Q_n of n distinct states (plus the Halt state), tape symbols s and s+ in set S = {0, 1}, shift direction d+ in {LEFT, RIGHT} (NONE is excluded here), + suffix meaning next and q+ = f(q, s), s+ = g(q, s), d+ = h(q, s).
The function Sigma(n) = Sigma(n, 2) (A028444) denotes the maximal number of tape marks (1's) which a halting Turing Machine H with n internal states, 2 symbols, and a two-way infinite tape can produce onto an initially blank tape (all 0's) and then halt. The function S(n) = S(n, 2) (A060843) denotes the maximal number of steps (thus shifts, since direction NONE is excluded) which a halting machine H can take (not necessarily the same Turing machine producing a maximum number of 1's and need not even produce many tape marks). For all n, S(n) >= Sigma(n).
Given that 5-state 2-symbol halting Turing machines can compute Collatz-like congruential functions (see references under A060843), it may be very hard to find the next term.
Rado's Sigma function grows faster than any computable function and is thus noncomputable.
H in H_(n, k) is a halting* Turing machine with n states and k symbols;
* (on a blank tape (all 0's) as input)
States q, q+ in set Q_n of n distinct states (plus the Halt state);
Symbols s, s+ in set S_k of k distinct symbols (0 as the blank symbol);
Shift direction d+ in {LEFT, RIGHT} (NONE is excluded here);
sigma(H) is the number of non-blank symbols left on the tape by H;
s(H) is the number of steps (or shifts in our case) taken by H;
Sigma(n, k) = max {sigma(H) : H is a halting Turing machine with n states and k symbols}
S(n, k) = max {s(H) : H is a halting Turing machine with n states and k symbols}
a(n) is Sigma(n) = Sigma(n, 2) since a 2-symbol BB-class Turing machine is assumed.
For all n, S(n, k) >= Sigma(n, k), k >= 2. (End)
|
|
|
REFERENCES
|
John Hopcroft, Turing Machines, Sci. Amer. vol. 250, #5, 86-98, May 1984, table on page 92 gives old lower bounds.
Shallit, Jeffrey. A second course in formal languages and automata theory. Cambridge University Press, 2008. See Fig. 6.2, p. 185.
|
|
|
LINKS
|
Bob Boonstra, Busy Beavers, Programmer's Challenge, MacTech Journal, Volume 16 (2000), Issue 9 - reports that in December 1984 George Uhing found a 5-state machine that produced 1915 1's before halting.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,hard,nice
|
|
|
AUTHOR
|
Scott Aaronson (sja8(AT)cornell.edu)
|
|
|
EXTENSIONS
|
The next two terms are at least 4098 and 10^^15.
|
|
|
STATUS
|
approved
|
| |
|
|
| |
| |
| |
|
|
|
OFFSET
|
0,2
|
|
|
REFERENCES
|
Dewdney, "The Armchair Universe," p. 164.
|
|
|
LINKS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A060843
|
|
Busy Beaver problem: a(n) = maximal number of steps that an n-state Turing machine can make on an initially blank tape before eventually halting.
|
|
+10
8
|
| |
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
"In 1965 Tibor Rado, together with Shen Lin, proved that a(3) is 21. (...) Next, in 1983, Allan Brady proved that a(4) is 107. (...) Then in 1989 Heiner Marxen and Juergen Buntrock discovered that a(5) is at least 47176870. (...) As for a(6), Marxen and Buntrock set another record in 1997 by proving that it is at least 8690333381690951." [Based on Aaronson's web page.]
The function Sigma(n) = Sigma(n, 2) (A028444) denotes the maximal number of tape marks (1's) that a Turing Machine with n internal states (plus the Halt state), 2 symbols, and a two-way infinite tape can write on an initially blank tape (all 0's) and then halt. The function a(n) (the present sequence) denotes the maximal number of steps S(n) = S(n, 2) (thus shifts, since direction NONE is excluded) that such a machine can make (not necessarily the same Turing machine producing a maximum number of 1s and need not even produce many tape marks).
Given that 5-state 2-symbol halting Turing machines can compute Collatz-like congruential functions (see references), it may be very hard to find the next term.
The sequence grows faster than any computable function of n and so is noncomputable.
A more precise definition might be as follows:
Busy Beaver Problem: a(n) is the maximal number of steps that an n-state, 2-symbol, d+ in {LEFT, RIGHT}, 5-tuple (q, s, q+, s+, d+) Turing machine can make on an initially blank tape and then halt.
Further comments:
H in H_(n, k) is a halting* Turing machine with n states and k symbols;
* (on a blank tape (all 0's) as input)
States q, q+ in set Q_n of n distinct states (plus the Halt state);
Symbols s, s+ in set S_k of k distinct symbols (0 as the blank symbol);
Shift direction d+ in {LEFT, RIGHT} (NONE is excluded here);
sigma(H) is the number of non-blank symbols left on the tape by H;
s(H) is the number of steps (or shifts in our case) taken by H;
Sigma(n, k) = max {sigma(H) : H is a halting Turing machine with n states and k symbols}
S(n, k) = max {s(H) : H is a halting Turing machine with n states and k symbols}
a(n) is S(n) = S(n, 2) since a 2-symbol BB-class Turing machine is assumed.
For all n, S(n, k) >= Sigma(n, k), k >= 2. (End)
a(6) > 10^^15, a tower of 10's of height 15 [Pavel Kropitz] - see Shawn Ligocki's blog. - N. J. A. Sloane, Jun 22 2022
|
|
|
REFERENCES
|
A. H. Brady, The busy beaver game and the meaning of life, in Herken, R. (Ed) The Universal Turing Machine: A Half-Century Survey, pp. 259-277, Oxford Univ Press 1988. Reprinted by Springer-Verlag, 1995 (see pages 237-254).
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010; see p. 33.
Yu. V. Rogozhin, Seven universal Turing machines (Russian), abstract, Fifth All-Union Conference on Math. Logic, Akad. Nauk. SSSR Sibirsk. Otdel., Inst. Mat., Novosibirsk, 1979, p. 127.
Yu. V. Rogozhin, Seven universal Turing machines (Russian), Systems and Theoretical Programming, Mat. Issled. no. 69, Akademiya Nauk Moldavskoi SSSR, Kishinev, 1982, pp. 76-90.
Jeffrey Shallit, A second course in formal languages and automata theory. Cambridge University Press, 2008. See Fig. 6.2, p. 185.
|
|
|
LINKS
|
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
hard,nice,nonn,more
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
Additional references from Bill Dubuque (wgd(AT)martigny.ai.mit.edu)
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A052200
|
|
Number of n-state, 2-symbol, d+ in {LEFT, RIGHT}, 5-tuple (q, s, q+, s+, d+) (halting or not) Turing machines.
|
|
+10
7
|
|
|
|
1, 64, 20736, 16777216, 25600000000, 63403380965376, 232218265089212416, 1180591620717411303424, 7958661109946400884391936, 68719476736000000000000000000, 739696442014594807059393047166976, 9711967541295580042210555933809967104, 152784834199652075368661148843397208866816
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
T in T_(n, k) is a Turing machine with n states and k symbols;
States q, q+ in set Q_n of n distinct states (plus the Halt state;)
Symbols s, s+ in set S_k of k distinct symbols (0 as the blank symbol;)
Shift direction d+ in {LEFT, RIGHT} (NONE is excluded here;)
+ suffix meaning next and q+ = f(q, s), s+ = g(q, s), d+ = h(q, s).
This sequence is computable. On the other hand, the busy beaver numbers are noncomputable, found only by examining each of the many n-state Turing machine programs' halting. - Michael Joseph Halm, Apr 29 2003
RE: Busy beaver halting Turing machines:
H in H_(n, k) is a halting* Turing machine with n states and k symbols;
* (on a blank tape, all 0's, as input)
sigma(H) is the number of non-blank symbols left on the tape by H;
s(H) is the number of steps (or shifts in our case) taken by H;
Sigma(n, k) = max {sigma(H) : H is a halting Turing machine with n states and k symbols}
S(n, k) = max {s(H) : H is a halting Turing machine with n states and k symbols}
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = (4*(n+1))^(2*n).
|
|
|
EXAMPLE
|
a(1) = 64 = (4*1+4)^(2*1) = 8^2.
|
|
|
MATHEMATICA
|
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A004147
|
|
Number of n-state Turing machines which halt.
(Formerly M5233)
|
|
+10
6
|
| |
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
This sequence is noncomputable, because it could be used to solve the halting problem. In fact, it is of the same degree of difficulty as the halting problem. - David Diamondstone (skeptical.scientist(AT)gmail.com), Dec 28 2007
|
|
|
REFERENCES
|
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
|
LINKS
|
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,nice,hard,bref
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
More terms from David Diamondstone (skeptical.scientist(AT)gmail.com), Dec 28 2007
a(4) from Jonathan Lee, who enumerated all 25.6 billion 4-state machines up to 107 steps, Mar 05 2016
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A353176
|
|
Maximum length of a finite snake in the Snake Number Problem with n-periodic instructions in an infinite square grid (see Comments).
|
|
+10
3
|
|
|
|
30, 79, 152, 450, 241, 257, 1098, 1448, 9520, 8804, 8338, 11348, 25316, 18823
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
5,1
|
|
|
COMMENTS
|
Given a list of n move instructions (up, right, down, left), the snake starts at the origin and moves according to the instructions, in order. If an instruction tells it to move to a square that has already been visited, the snake skips that instruction. After it has followed (or skipped) the last instruction in the list, it starts again with the first one. The snake is either finite (if it gets stuck at some point) or infinite (if it can go on forever). a(n) is the maximum number of squares visited by a finite snake. For n <= 4, all snakes are infinite. - Pontus von Brömssen, May 08 2022
|
|
|
LINKS
|
Ariel Futoransky, Snake Program, Snake Program to try the snakes, April 2022 (see bottom animation and label for different options).
Rodolfo Kurchan, Puzzle Fun, Snake Number Problem, March 2022.
|
|
|
EXAMPLE
|
a(5) = 30
URDDL: 30
-- -- 20 21 -- --
-- 18 19 22 23 --
16 17 02 03 24 --
15 14 01 04 25 26
12 13 06 05 30 27
11 10 07 -- 29 28
-- 09 08 -- -- --
.
a(6) = 79
UURDLL: 79
-- -- -- -- -- -- -- 74 75 -- --
-- -- -- -- -- 69 70 73 76 77 --
-- -- -- 64 65 68 71 72 79 78 --
-- -- -- 63 66 67 -- 27 28 -- --
-- -- 61 62 -- 22 23 26 29 30 --
56 57 60 17 18 21 24 25 32 31 --
55 58 59 16 19 20 03 04 33 34 --
54 53 52 15 14 13 02 05 06 35 36
-- -- 51 -- -- 12 01 08 07 38 37
-- -- 50 49 48 11 10 09 40 39 --
-- -- -- -- 47 -- 43 42 41 -- --
-- -- -- -- 46 45 44 -- -- -- --
.
n | Number of finite solutions | Maximum length | Instructions that give
| A352388(n) | a(n) | the maximum length
-------------------------------------------------------------------------
5 5 30 URDDL
6 21 79 UURDLL
7 127 152 URULLDD
8 618 450 URURUULD
9 2934 241 URRRRDLRR
10 13542 257 URRLDLRRUR
11 61803 1098 URUURUUULLD
12 276650 1448 URUULLDUDDDD
13 1219508 9520 URRRLLDLRRULL
14 5309179 8804 URRURRRLDLRULL
15 22868295 8338 UDDRULUUUULLULD
16 97663066 11348 URRURRRLDDLRUULL
17 414156142 25316 URRRDLULUUUUULURL
18 1746438478 18823 UDDDRULULLULLUULDU
Computer solutions a(5) to a(13) found by Giorgio Vecchi.
Computer solutions a(14) to a(18) found by Ariel Futoransky.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,more
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A333479
|
|
Busy Beaver for lambda calculus: the maximum normal form size of any closed lambda term of size n, or 0 if no closed term of size n exists.
|
|
+10
2
|
|
|
|
0, 0, 0, 4, 0, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 24, 26, 30, 42, 52, 44, 58, 223, 160, 267, 298, 1812, 327686, 38127987424941, 578960446186580977117854925043439539266349923328202820197287920039565648199686
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,4
|
|
|
COMMENTS
|
Sizes of terms are defined as in Binary Lambda Calculus (see Lambda encoding link) by size(lambda M)=2+size(M), size(M N)=2+size(M)+size(N), and size(V)=1+i for a variable V bound by the i-th enclosing lambda.
a(34), a(35) and a(36) correspond to Church numerals 2^2^2^2, 3^3^3, and 2^2^2^3, where numeral n has size 5*n+6.
a(38) > 10^19729, corresponding to Church numeral 2^2^2^2^2.
Only a finite number of entries can be known, as the function is uncomputable.
Quoting from Chaitin's paper below: "to information theorists it is clear that the correct measure is bits, not states [...] to deal with Sigma(number of bits) one would need a model of a binary computer as simple and compelling as the Turing machine model, and no obvious natural choice is at hand."
a(63) > Graham's number, as shown in the Googology Wiki article in the links. - John Tromp, Apr 11 2023
A universal form of this Busy Beaver, using the binary input feature of Binary Lambda Calculus, is given in sequence A361211. - John Tromp, May 24 2023
|
|
|
REFERENCES
|
Gregory Chaitin, Computing the Busy Beaver Function, in T. M. Cover and B. Gopinath, Open Problems in Communication and Computation, Springer, 1987, pages 108-112.
John Tromp, Binary Lambda Calculus and Combinatory Logic, in Randomness And Complexity, from Leibniz To Chaitin, ed. Cristian S. Calude, World Scientific Publishing Company, October 2008, pages 237-260.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
The smallest closed lambda term is lambda x.x with encoding 0010 of size 4, giving a(4)=4, as it is in normal form. There is no closed term of size 5, so a(5)=0. a(21)=22 because of term (lambda x. x x) (x (lambda y. x)).
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A081419
|
|
Largest value held in any register at the end of a halting computation by an n-instruction register Minski machine.
|
|
+10
1
|
| |
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
We start with initially empty registers and include exactly one Halt instruction. Analogous to the Busy Beaver function, Sigma, for Turing Machines.
n<=5 are proved. 5<n<=9 are good lower bounds.
|
|
|
REFERENCES
|
Tibor Rado, On Noncomputable Functions, Bell System Technical Journal, vol. 41, # 3, 877-884, May 1963.
|
|
|
LINKS
|
|
|
|
FORMULA
|
Noncomputable.
|
|
|
EXAMPLE
|
E.g. B(3) is of the form:
1: A+ -> 2
2: A+ -> 3
3: Halt
Halting with B(3)=2
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
hard,nice,nonn
|
|
|
AUTHOR
|
Rick J. Griffiths (rjg42(AT)cam.ac.uk), Apr 20 2003
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A164278
|
|
a(n) = (6*n + 1)^(2*n) - 1.
|
|
+10
1
|
|
|
|
0, 48, 28560, 47045880, 152587890624, 819628286980800, 6582952005840035280, 73885357344138503765448, 1104427674243920646305299200, 21209401372879911350250244140624, 508858109619679129936596364708525200
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
In Bátfai's notation, a(n) gives the total number of Turing machines with n states.
|
|
|
LINKS
|
|
|
|
MATHEMATICA
|
|
|
|
PROG
|
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A118874
|
|
A halting sequence: let f_n be the n-th recursive function, relative to the Godel numbering given in Cutland, then a(n) is f_n(n)+1 if the corresponding program halts on input n, 0 otherwise.
|
|
+10
0
|
|
|
|
1, 3, 1, 4, 2, 1, 1, 0, 1, 12, 2, 1, 1, 1, 1, 16, 0, 19, 1, 21, 3, 2, 2, 0, 1, 1, 1, 1, 1, 1, 1, 32, 1, 0, 0, 36, 2, 1, 1, 0, 2, 45, 3, 2, 2, 2, 2, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 64, 1, 67, 1, 0, 0, 0, 0, 0, 1, 76, 2, 1, 1, 1, 1, 81, 0, 84, 2, 86, 4, 3, 3, 0, 2, 2, 2, 2, 2, 2, 2, 1, 1, 0, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
The prototypical example of a noncomputable sequence.
|
|
|
REFERENCES
|
Nigel Cutland, "Computability: An introduction to recursive function theory". Cambridge University Press, 1980. p. 78.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
Using Cutland's Godel numbering, 80 corresponds to the URM program "Z(1) J(1,1,1) S(1)", which clearly loops forever on any input, so a(80)=0. On the other hand, 17 corresponds to the URM program "S(1) T(1,1)", which, on input 17, produces 18. So a(17)=18+1=19.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
Search completed in 0.013 seconds
|