Universal Turing machine
You are encouraged to solve this task according to the task description, using any language you may know.
One of the foundational mathematical constructs behind computer science is the universal Turing Machine. Indeed one way to definitively prove that a language is Turing complete is to implement a universal Turing machine in it.
The task
For this task you would simulate such a machine capable of taking the definition of any other Turing machine and executing it. You will not, of course, have an infinite tape, but you should emulate this as much as is possible. The three permissible actions on the tape are "left", "right" and "stay".
To test your universal Turing machine (and prove your programming language is Turing complete!), you should execute the following two Turing machines based on the following definitions.
Simple incrementer
- States: q0, qf
- Initial state: q0
- Terminating states: qf
- Permissible symbols: B, 1
- Blank symbol: B
- Rules:
- (q0, 1, 1, right, q0)
- (q0, B, 1, stay, qf)
The input for this machine should be a tape of 1 1 1
Three-state busy beaver
- States: a, b, c, halt
- Initial state: a
- Terminating states: halt
- Permissible symbols: 0, 1
- Blank symbol: 0
- Rules:
- (a, 0, 1, right, b)
- (a, 1, 1, left, c)
- (b, 0, 1, left, a)
- (b, 1, 1, right, b)
- (c, 0, 1, left, b)
- (c, 1, 1, stay, halt)
The input for this machine should be an empty tape.
Ruby
The universal machine
<lang ruby>class Turing
class Tape
def initialize(symbols, blank, starting_tape)
@symbols = symbols
@blank = blank
@tape = starting_tape
@index = 0
end
def read
retval = @tape[@index]
unless retval
retval = @tape[@index] = @blank
end
raise "invalid symbol '#{retval}' on tape" unless @tape.member?(retval)
return retval
end
def write(symbol)
@tape[@index] = symbol
end
def right
@index += 1
end
def left
if @index == 0
@tape.unshift @blank
else
@index -= 1
end
end
def stay
# nop
end
def get_tape
return @tape
end
end
def initialize(symbols, blank,
initial_state, halt_states, running_states,
rules, starting_tape = [])
@tape = Tape.new(symbols, blank, starting_tape)
@initial_state = initial_state
@halt_states = halt_states
@running_states = running_states
@rules = rules
@halted = false
end
def run
raise "machine already halted" if @halted
state = @initial_state
while (true)
break if @halt_states.member? state
raise "unknown state '#{state}'" unless @running_states.member? state
symbol = @tape.read
outsym, action, state = @rules[state][symbol]
@tape.write outsym
@tape.send action
end
@halted = true
return @tape.get_tape
end
end</lang>
The incrementer machine
<lang ruby>incrementer_rules = {
:q0 => { 1 => [1, :right, :q0],
:b => [1, :stay, :qf]}
} t = Turing.new([:b, 1], # permitted symbols
:b, # blank symbol
:q0, # starting state
[:qf], # terminating states
[:q0], # running states
incrementer_rules, # operating rules
[1, 1, 1]) # starting tape
print t.run, "\n"</lang>
The busy beaver machine
<lang ruby>busy_beaver_rules = {
:a => { 0 => [1, :right, :b],
1 => [1, :left, :c]},
:b => { 0 => [1, :left, :a],
1 => [1, :right, :b]},
:c => { 0 => [1, :left, :b],
1 => [1, :stay, :halt]}
} t = Turing.new([0, 1], # permitted symbols
0, # blank symbol
:a, # starting state
[:halt], # terminating states
[:a, :b, :c], # running states
busy_beaver_rules, # operating rules
[]) # starting tape
print t.run, "\n"</lang>
Mercury
The universal machine
Source for this example was lightly adapted from https://bitbucket.org/ttmrichter/turing. Of particular interest in this implementation is that because of the type parameterisation of the config type, the machine being simulated cannot be compiled if there is any mistake in the states, symbols and actions. Also, because of Mercury's determinism detection and enforcement, it's impossible to pass in a non-deterministic set of rules. At most one answer can come back from the rules interface.
<lang mercury>:- module turing.
- - interface.
- - import_module list.
- - import_module set.
- - type config(State, Symbol)
---> config(initial_state :: State,
halting_states :: set(State),
blank :: Symbol ).
- - type action ---> left ; stay ; right.
- - func turing(config(State, Symbol),
pred(State, Symbol, Symbol, action, State),
list(Symbol)) = list(Symbol).
- - mode turing(in,
pred(in, in, out, out, out) is semidet,
in) = out is det.
- - implementation.
- - import_module pair.
- - import_module require.
turing(Config@config(Start, _, _), Rules, Input) = Output :-
(Left-Right) = perform(Config, Rules, Start, ([]-Input)), Output = append(reverse(Left), Right).
- - func perform(config(State, Symbol),
pred(State, Symbol, Symbol, action, State),
State, pair(list(Symbol))) = pair(list(Symbol)).
- - mode perform(in, pred(in, in, out, out, out) is semidet,
in, in) = out is det.
perform(Config@config(_, Halts, Blank), Rules, State,
Input@(LeftInput-RightInput)) = Output :-
symbol(RightInput, Blank, RightNew, Symbol),
( set.member(State, Halts) ->
Output = Input
; Rules(State, Symbol, NewSymbol, Action, NewState) ->
NewLeft = pair(LeftInput, [NewSymbol|RightNew]),
NewRight = action(Action, Blank, NewLeft),
Output = perform(Config, Rules, NewState, NewRight)
;
error("an impossible state has apparently become possible") ).
- - pred symbol(list(Symbol), Symbol, list(Symbol), Symbol).
- - mode symbol(in, in, out, out) is det.
symbol([], Blank, [], Blank). symbol([Sym|Rem], _, Rem, Sym).
- - func action(action, State, pair(list(State))) = pair(list(State)).
action(left, Blank, ([]-Right)) = ([]-[Blank|Right]). action(left, _, ([Left|Lefts]-Rights)) = (Lefts-[Left|Rights]). action(stay, _, Tape) = Tape. action(right, Blank, (Left-[])) = ([Blank|Left]-[]). action(right, _, (Left-[Right|Rights])) = ([Right|Left]-Rights).</lang>
The incrementer machine
This machine has been stripped of the Mercury ceremony around modules, imports, etc.
<lang mercury>:- type incrementer_states ---> a ; halt.
- - type incrementer_symbols ---> b ; '1'.
- - func incrementer_config = config(incrementer_states, incrementer_symbols).
incrementer_config = config(a, % the initial state
set([halt]), % the set of halting states
b). % the blank symbol
- - pred incrementer(incrementer_states::in,
incrementer_symbols::in,
incrementer_symbols::out,
action::out,
incrementer_states::out) is semidet.
incrementer(a, '1', '1', right, a). incrementer(a, b, '1', stay, halt).
TapeOut = turing(incrementer_config, incrementer, [1, 1, 1]).</lang>
This will, on execution, fill TapeOut with [1, 1, 1, 1].
The busy beaver machine
This machine has been stripped of the Mercury ceremony around modules, imports, etc.
<lang mercury>:- type busy_beaver_states ---> a ; b ; c ; halt.
- - type busy_beaver_symbols ---> '0' ; '1'.
- - func busy_beaver_config = config(busy_beaver_states, busy_beaver_symbols).
busy_beaver_config = config(a, % initial state
set([halt]), % set of terminating states
'0'). % blank symbol
- - pred busy_beaver(busy_beaver_states::in,
busy_beaver_symbols::in,
busy_beaver_symbols::out,
action::out,
busy_beaver_states::out) is semidet.
busy_beaver(a, '0', '1', right, b). busy_beaver(a, '1', '1', left, c). busy_beaver(b, '0', '1', left, a). busy_beaver(b, '1', '1', right, b). busy_beaver(c, '0', '1', left, b). busy_beaver(c, '1', '1', stay, halt).
TapeOut = turing(busy_beaver_config, busy_beaver, []).</lang>
This will, on execution, fill TapeOut with [1, 1, 1, 1, 1, 1].
Prolog
The universal machine
Source for this example was lightly adapted from https://bitbucket.org/ttmrichter/turing. This machine, because of Prolog's dynamic nature, has to check its configuration and the rules' compliance to the same at run-time. This is the role of all but the first of the memberchk/2 predicates. In addition, calling the user-supplied rules has to be wrapped in a once/1 wrapper because there is no way to guarantee in advance that the rules provided are deterministic. (An alternative to doing this is to simply allow perform/5 to be non-deterministic or to check for multiple results and report an error on such.)
<lang prolog>turing(Config, Rules, TapeIn, TapeOut) :-
call(Config, IS, _, _, _, _),
perform(Config, Rules, IS, {[], TapeIn}, {Ls, Rs}),
reverse(Ls, Ls1),
append(Ls1, Rs, TapeOut).
perform(Config, Rules, State, TapeIn, TapeOut) :-
call(Config, _, FS, RS, B, Symbols),
( memberchk(State, FS) ->
TapeOut = TapeIn
; memberchk(State, RS) ->
{LeftIn, RightIn} = TapeIn,
symbol(RightIn, Symbol, RightRem, B),
memberchk(Symbol, Symbols),
once(call(Rules, State, Symbol, NewSymbol, Action, NewState)),
memberchk(NewSymbol, Symbols),
action(Action, {LeftIn, [NewSymbol|RightRem]}, {LeftOut, RightOut}, B),
perform(Config, Rules, NewState, {LeftOut, RightOut}, TapeOut) ).
symbol([], B, [], B). symbol([Sym|Rs], Sym, Rs, _).
action(left, {Lin, Rin}, {Lout, Rout}, B) :- left(Lin, Rin, Lout, Rout, B). action(stay, Tape, Tape, _). action(right, {Lin, Rin}, {Lout, Rout}, B) :- right(Lin, Rin, Lout, Rout, B).
left([], Rs, [], [B|Rs], B). left([L|Ls], Rs, Ls, [L|Rs], _).
right(L, [], [B|L], [], B). right(L, [S|Rs], [S|L], Rs, _).</lang>
The incrementer machine
<lang prolog>incrementer_config(IS, FS, RS, B, S) :-
IS = q0, % initial state FS = [qf], % halting states RS = [IS], % running states B = 0, % blank symbol S = [B, 1]. % valid symbols
incrementer(q0, 1, 1, right, q0). incrementer(q0, b, 1, stay, qf).
turing(incrementer_config, incrementer, [1, 1, 1], TapeOut).</lang>
This will, on execution, fill TapeOut with [1, 1, 1, 1].
The busy beaver machine
<lang prolog>busy_beaver_config(IS, FS, RS, B, S) :-
IS = 'A', % initial state FS = ['HALT'], % halting states RS = [IS, 'B', 'C'], % running states B = 0, % blank symbol S = [B, 1]. % valid symbols
busy_beaver('A', 0, 1, right, 'B'). busy_beaver('A', 1, 1, left, 'C'). busy_beaver('B', 0, 1, left, 'A'). busy_beaver('B', 1, 1, right, 'B'). busy_beaver('C', 0, 1, left, 'B'). busy_beaver('C', 1, 1, stay, 'HALT').
turing(busy_beaver_config, busy_beaver, [], TapeOut).</lang>
This will, on execution, fill TapeOut with [1, 1, 1, 1, 1, 1].