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We know that the problem “Does this Turing machine take at least this finite number of steps on that input?” is decidable, because it will always answer yes or no, where it will say yes if the machine reaches the given number of steps and no if it halts before that.
Now here is my doubt: if it halts before reaching those many steps — i.e. the input either (1) got accepted or (2) got rejected or maybe (3)if it doesn’t halt but rather goes into an infinite loop — then, when we are in case (3), how can we be sure that it will always be in that loop?
What I mean to say is that if it doesn't run forever but comes out of the loop at some point of time then it might cross the asked number of steps and the decision can be made now which was earlier not possible. If so, then how can we conclude that it's decidable when we know that being stuck in a loop we won’t be able to say anything about the outcome?
(I already more or less answered your question when I edited it.)
The thing is, the decision system (a Turing machine, an algorithm or any other equivalent formalism) that takes as inputs a Turing machine M, a number N and a value X, and returns yes or no, has total control over how it executes M on X. It simulates it step by step. So it can run one step of M(X), increment an instruction counter, compare it to N and, as soon as the given number of steps is reached, it stops and returns yes. At that point, there is no need that the simulated machine M be in a final state, and actually the full computation M(X) could very well diverge. We don’t care, because we only run the first N steps.
Most likely the "conditional structures where not being debuged/developed enough so that multiple conditions often conflicted each other..the error reporting where not as definitive, so it where used semi abstract notions as "decidable" and "undecidable"
as a semi example i writen years ago in vbs a "64 bit rom memory" simulator, as i tried to manage the memory cells, where i/o read/write locations where atributed , using manny formulas and conditions to set conversions from decimal to binary and all the operations, indexing, etc.
I had allso run into bugs becouse that the conditons where not perfect.Why? becouse the program had some unresolved somewhat arbitrary results that could had ended up in :
print.debug "decidable"
On Error Resume h
h:
print.debug "undecidable"
this was a example with a clear scope and with a debatable result.
to resume to your question : > "so how do we conclude that it's decidable??"
wikipedia :
The Turing machine was invented in 1936 by Alan Turing, who called it an "a-machine" (automatic machine). With this model, Turing was able to answer two questions in the negative:
Does a machine exist that can determine whether any arbitrary machine on its tape is "circular" (e.g., freezes, or fails to continue its computational task)?
Does a machine exist that can determine whether any arbitrary machine on its tape ever prints a given symbol?
Thus by providing a mathematical description of a very simple device capable of arbitrary computations, he was able to prove properties of computation in general—and in particular, the uncomputability of the ('decision problem').
I am trying to develop the following lock free code (c++11):
int val_max;
std::array<std::atomic<int>, 255> vector;
if (vector[i] > val_max) {
val_max =vector[i];
}
The problem is that when there are many threads (128 threads), the result is not correct, because if for example val_max=1, and three threads ( with vector[i] = 5, 15 and 20) will execute the code in the next line, and it will be a data race..
So, I don't know the best way to solve the problem with the available functions in c++11 (I cannot use mutex, locks), protecting the whole code.
Any suggestions? Thanks in advance!
You need to describe the bigger problem you need to solve, and why you want multiple threads for this.
If you have a lot of data and want to find the maximum, and you want to split that problem, you're doing it wrong. If all threads try to access a shared maximum, not only is it hard to get right, but by the time you have it right, you have fully serialized accesses, thus making the entire thing an exercise in adding complexity and thread overhead to a serial program.
The correct way to make it parallel is to give every thread a chunk of the array to work on (and the array members are not atomics), for which the thread calculates a local maximum, then when all threads are done have one thread find the maximum of the individual results.
Do an atomic fetch of val_max.
If the value fetched is greater than or equal to vector[i], stop, you are done.
Do an atomic compare exchange -- compare val_max to the value you read in step 1 and exchange it for the value of vector[i] if it compares.
If the compare succeeded, stop, you are done.
Go to step 1, you raced with another thread that made forward progress.
How would you determine a loop is a infinite loop and will break out of it.
Does anyone has the algorithm or can assist me on this one.
Thanks
There is no general case algorithm that can determine if a program is in an infinite loop or not for every turing complete language, this is basically the Halting Problem.
The idea of proving it is simple:
Assume you had such an algorithm A.
Build a program B that invokes A on itself [on B].
if A answers "the program will halt" - do an infinite loop
else [A answers B doesn't halt] - halt immidiately
Now, assume you invoke A on B - the answer will be definetly wrong, thus A doesn't exist.
Note: the above is NOT a formal proof, just a sketch of it.
As written by others, it cannot be determined.
However, if you want to have some checking, you can use the WatchDog design pattern.
This is a separate thread that checks if a task still is active. Your own thread should give a signal regularly to say it is alive. Make sure this signal is not set inside your (infinite) loop.
If there was no signal, the program is inside an infinite loop or has stopped and the watchdog can act on it.
I am required to design and build an 8 bit Pseudo Random Number Generator. I have looked at possible methods; using background noise, user input etc. I was wondering if anyone could give me some advice on where to start as this would be of great help to me.
random.org is perhaps the best place to start your investigation.
Below should get you started with the basics
howstuffworks.com
Construct your own random number generator
For a simple 8 bit PRNG you could ry something like a Linear Feedback Shift Register. This is very simple to implement in either software or hardware.
My plan is to use a temperature sensor. When the temps are being processed in the ADC, I am going to amplify the noise generated. This will then give me the random 8 bit number I require which will be used as the 'seed' for the PRNG in stdlib (C programming).
What do you's think?
I've found that the following works very well. This is implemented in MSP430 assembly, but would be easy enough to port to another processor. I've used this to generate 'white' noise for a synthesizer project, and there were no audible patterns in the output. Depending on what your requirements are, this might be sufficient. It uses two state variables, the previous output (8 bits), and a 16-bit state register. I found this online, http://www.avrfreaks.net/index.php?name=PNphpBB2&file=viewtopic&t=95614&highlight=radbrad, where it's listed in AVR assembly, and ported it to MSP.
Because it uses shifts and shifts the top bit out of one register into the bottom of another, it doesn't really lend itself to efficient implementation in C. Hence the assembly. I hope you find this as useful as I did.
mov.b &rand_out, r13
mov.b r13,r12
and.b #66, r13
jz ClearCarry
cmp.b #66, r13
xor.w #1, sr ; invert carry flag
jmp SkipClearCarry
ClearCarry:
clrc
SkipClearCarry:
rlc.w &rand_state
rlc.b r12
mov.b r12,&rand_out
ret
I have written a simple brainfuck interpreter in MATLAB script language. It is fed random bf programs to execute (as part of a genetic algorithm project). The problem I face is, the program turns out to have an infinite loop in a sizeable number of cases, and hence the GA gets stuck at the point.
So, I need a mechanism to detect infinite loops and avoid executing that code in bf.
One obvious (trivial) case is when I have
[]
I can detect this and refuse to run that program.
For the non-trivial cases, I figured out that the basic idea is: to determine how one iteration of the loop changes the current cell. If the change is negative, we're eventually going to reach 0, so it's a finite loop. Otherwise, if the change is non-negative, it's an infinite loop.
Implementing this is easy for the case of a single loop, but with nested loops it becomes very complicated. For example, (in what follows (1) refers to contents of cell 1, etc. )
++++ Put 4 in 1st cell (1)
>+++ Put 3 in (2)
<[ While( (1) is non zero)
-- Decrease (1) by 2
>[ While( (2) is non zero)
- Decrement (2)
<+ Increment (1)
>]
(2) would be 0 at this point
+++ Increase (2) by 3 making (2) = 3
<] (1) was decreased by 2 and then increased by 3, so net effect is increment
and hence the code runs on and on. A naive check of the number of +'s and -'s done on cell 1, however, would say the number of -'s is more, so would not detect the infinite loop.
Can anyone think of a good algorithm to detect infinite loops, given arbitrary nesting of arbitrary number of loops in bf?
EDIT: I do know that the halting problem is unsolvable in general, but I was not sure whether there did not exist special case exceptions. Like, maybe Matlab might function as a Super Turing machine able to determine the halting of the bf program. I might be horribly wrong, but if so, I would like to know exactly how and why.
SECOND EDIT: I have written what I purport to be infinite loop detector. It probably misses some edge cases (or less probably, somehow escapes Mr. Turing's clutches), but seems to work for me as of now.
In pseudocode form, here it goes:
subroutine bfexec(bfprogram)
begin
Looping through the bfprogram,
If(current character is '[')
Find the corresponding ']'
Store the code between the two brackets in, say, 'subprog'
Save the value of the current cell in oldval
Call bfexec recursively with subprog
Save the value of the current cell in newval
If(newval >= oldval)
Raise an 'infinite loop' error and exit
EndIf
/* Do other character's processings */
EndIf
EndLoop
end
Alan Turing would like to have a word with you.
http://en.wikipedia.org/wiki/Halting_problem
When I used linear genetic programming, I just used an upper bound for the number of instructions a single program was allowed to do in its lifetime. I think that this is sensible in two ways: I cannot really solve the halting problem anyway, and programs that take too long to compute are not worthy of getting more time anyway.
Let's say you did write a program that could detect whether this program would run in an infinite loop. Let's say for the sake of simplicity that this program was written in brainfuck to analyze brainfuck programs (though this is not a precondition of the following proof, because any language can emulate brainfuck and brainfuck can emulate any language).
Now let's say you extend the checker program to make a new program. This new program exits immediately when its input loops indefinitely, and loops forever when its input exits at some point.
If you input this new program into itself, what will the results be?
If this program loops forever when run, then by its own definition it should exit immediately when run with itself as input. And vice versa. The checker program cannot possibly exist, because its very existence implies a contradiction.
As has been mentioned before, you are essentially restating the famous halting problem:
http://en.wikipedia.org/wiki/Halting_problem
Ed. I want to make clear that the above disproof is not my own, but is essentially the famous disproof Alan Turing gave back in 1936.
State in bf is a single array of chars.
If I were you, I'd take a hash of the bf interpreter state on every "]" (or once in rand(1, 100) "]"s*) and assert that the set of hashes is unique.
The second (or more) time I see a certain hash, I save the whole state aside.
The third (or more) time I see a certain hash, I compare the whole state to the saved one(s) and if there's a match, I quit.
On every input command ('.', IIRC) I reset my saved states and list of hashes.
An optimization is to only hash the part of state that was touched.
I haven't solved the halting problem - I'm detecting infinite loops while running the program.
*The rand is to make the check independent of loop period
Infinite loop cannot be detected, but you can detect if the program is taking too much time.
Implement a timeout by incrementing a counter every time you run a command (e.g. <, >, +, -). When the counter reaches some large number, which you set by observation, you can say that it takes very long time to execute your program. For your purpose, "very long" and infinite is a good-enough approximation.
As already mentioned this is the Halting Problem.
But in your case there might be a solution: The Halting Problem is considering is about the Turing machine, which has unlimited memory.
In case you know that you have a upper limit of memory (e.g. you know you dont use more than 10 memory cells), you can execute your programm and stop it. The idea is that the computation space bounds computation time (as you cant write more than one cell at one step). After you executed as much steps as you can have different memory configurations, you can break. E.g. if you have 3 cells, with 256 conditions, you can have at most 3^256 different states, and so you can stop after executing that many steps. But be careful, there are implicit cells, like the instruction pointer and the registers. You do it even shorter, if you save every state configuration and as soon as you detect one, which you already had, you have an infite loop. This approach is definitly much better in the run time, but therefor needs much more space (here it might be suitable to hash the configurations).
This is not the halting problem, however, it is still not reasonable to try to detect halting even in such a limited machine as a 1000 cell BF machine.
Consider this program:
+[->[>]+<[-<]+]
This program will not repeat until it has filled up the entire of memory which for just 1000 cells will take about 10^300 years.
If I remember correctly, the halting problem proof was only true for some extreme case that involved self reference. However it's still trivial to show a practical example of why you can't make an infinite loop detector.
Consider Fermat's Last Theorem. It's easy to create a program that iterates through every number (or in this case 3 numbers), and detects if it's a counterexample to the theorem. If so it halts, otherwise it continues.
So if you have an infinite loop detector, it should be able to prove this theorem, and many many others (perhaps all others, if they can be reduced to searching for counterexamples.)
In general, any program that involves iterating through numbers and only stopping under some condition, would require a general theorem prover to prove if that condition can ever be met. And that's the simplest case of looping there is.
Off the top of my head (and I could be wrong), I would think it would be a little bit difficult to detect whether or not a program has an infinite loop without actually executing the program itself.
As the conditional execution of portions of the program depends on the execution state of the program, it will be difficult to know the particular state of the program without actually executing the program.
If you don't require that a program with an infinite loop be executed, you could try having an "instructions executed" counter, and only execute a finite number of instructions. This way, if a program does have an infinite loop, the interpreter can terminate the program which is stuck in an infinite loop.