When executing the following Prolog program with YAP, the output is always the same, namely the integer 233.
:- use_module(library(random)).
x:- random(1,1000,X), writeln(X).
For instance, if I execute the following bash script, the output is always the same integer (233).
for k in `seq 0`
do
yap -l test.pl << %
x.
%
done
If I repeat this procedure using swipl, then the output is different each time, i.e random.
Can anyone explain this?
First things first!
For many good reasons (reproducibility of past results being the most important one) computer programs do not work with actual random numbers, but with pseudo random numbers.
PRNGs are fully deterministic functions that, given the same internal state (a.k.a. "seed") as initialization, produce exactly the same sequence of numbers (from now until eternity).
Quick fix: find yourself a suitable seed (date, time, phase of the moon, ...) and explicitly initialize the PRNG with that seed. Record the seed so you can later deterministically re-run past experiments.
usually, random generators require something like set_seed(SomeReallyRandomValue) call, in C was often used seed(time(0)).
So, I guess
datime(datime(_Year, _Month, _DayOfTheMonth, _Hour, Minute, Second)),
X is Minute * Second,Y=X,Z=X,
setrand(rand(X,Y,Z)),
could work
Related
When I repeatedly run this code,
srand 1;
my #x = (1..1000).pick: 100;
say sum #x;
I get different answers each time. If I'm resetting with srand why shouldn't it produce the same random numbers each time?
The error occurs in the REPL.
The error occurs in this file:
use v6.d;
srand 1;
my $x = rand;
say $x; # OUTPUT: 0.5511548437617427
srand 1;
$x = rand;
say $x; # OUTPUT: 0.308302962221659
say $*KERNEL; # OUTPUT: darwin
I'm using:
Welcome to Rakudo™ v2022.07.
Implementing the Raku® Programming Language v6.d.
Built on MoarVM version 2022.07.
It should produce the same numbers for a given piece of code all of the time. And I haven't been able to reproduce your observation in any way.
There may be something spooky going on under the hood, though:
$ raku -e 'srand 1; (my $x = (1..1000).pick(1)).say'
(761)
$ raku -e 'srand 1; (my #x = (1..1000).pick(1)).say'
[471]
On the surface, you'd say that these values should be the same, as each only generates a single value. But apparently a different number of random values is actually calculated under the hood, causing the visibly different values. Is that perhaps what is going on in your case?
(This answer is a paraphrase of jnthn's comment in the GitHub issue opened based on this question).
Setting srand 1 will cause the same sequence of random numbers to be generated -- that is, the nth random number will be the same. However, since Raku (really, Rakudo and/or MoarVM, assuming you're using those backends) uses random numbers internally, you won't always be in the same position in that sequence (i.e., your n might be different) and thus you might not get the same random number.
This is further complicated by Rakudo's optimizer. Naively, repeating the same code later in the program should consume the same number of random numbers from the sequence. However, the optimizer may well remove some of those random number uses from subsequent calls, which can result in different random numbers.
I'm unclear to what degree the current behavior is intended versus a bug in Rakudo/MoarVM's implementation; please see the previously linked issue for additional details.
I am making some software that need to work with integers.
Also I need to apply some formula to those integers, repeatedly over time (example, do x/=z several times in a row for a indefinite amount).
All tools, algorithms and formulas I could think or find, or don't work with integers at all, or work as approximations at best.
For example the x/=z several times in a row for example, you can theoretically calculate what x will be in the 10th time by doing x = x/(z^10), but that will be wrong if the result is fractional, you can use floor(x/(z^10)), but the result will STILL be wrong.
Plotting software that I found also don't have integers at all, or has "floor()/ceil()" functions support, at best, and still the result would fall in the problem of the previous paragraph.
So how I do it?
Here's something to get you going for the iteration of x/=z:
(that should have ended in "all three terms are 0 with regard to integer division")
Now if x or z are negative, you can try and see whether this still holds; I did not invest the time to make the necessary case distinctions, but they should be fairly analogous.
As Karoly Horvath mentions in a comment, without a clear specification of the kinds of functions for which you would like to find a shortcut to replace iterative evaluation, helping you out won't be possible since there are uncountably many functions over the integers, and the same approach won't work for all of them.
Its well known in theoretical computer science that the "Hello world tester" program is an undecidable problem.(Here is a link what i mean by hello world tester
My question is since given a program as input we can't say what the program will do,can we solve the reverse problem:
Given set of input and output,is there an algorithm for writing a program that writes a program to achieve a one to one mapping between the given input and output.
I know about metaprogramming but my question is more of theoretical interest. Something which can apply for a generic case.
With these kind of musings one has to be very careful. A lot of confusion arises from not clearly distinguishing about a program x for which proposition P(x) holds or any program x for which proposition P(x) hold. As long as the set of programs for which P(x) holds is finite there always is a proof, of their correctness (although this proof may not be known).
At this point you also have to distinguish between programs, which are and can be known and programs which can only be shown to exist by full enumeration of all posibilities. Let's make an example:
Take 10 Programs, which take no input and may or may not terminate and produce "hello World". Then there is a program which decides exactly which of these programs are correct, and which aren't. Lets call these programs (x_1,...,x_10). Then take the programs (X_0,...,X_{2^10}) where X_i output true for program x_j if the j-th bit in the binary representation of i is set. One of these programs has to be the one which decides correctly for all ten x_i, there just might not be any way to ever figure out which one of these 100 X_js is the correct one (a meta-problem at this point).
This goes to show that considering finite sets of programs and input/output pairs one can always resolve to full enumeration and all halting-problem type of paradoxies instantly disappear. In your case the set of generated programs for each input is of size one and the set of input/output pairs is of finite size (because you have to supply it to the meta-program). Hence full enumeration solves your problem very simple and you can also easily proof both the correctness of the corrected program as well as the correctness of the meta-program.
Note: Since the set of generated programs is infinite, this is one of the few cases where you can proof P(x) for a infinite set of programs (actually you can proof P(x,input,output) for this set). This shows that the set being infinite is only a necessary, not a sufficient condition for this type of paradoxes to appear.
Your question is ambiguously phrased.
How would you specify "what a program should do"?
Any precise, complete, and machine-readable specification of a program's functionality is already a program.
Thus, the answer to your question is, a compiler.
Now, you're asking how to find a function based on a sample of its input and output.
That is a question about statistical analysis that I cannot answer.
Sounds like you want to generate a state machine that learns by being given an input sequence and then updates itself to produce the appropriate output sequence. Assuming your output sequences are always the same for the same input sequence it should be simple enough to write. If your output is not deterministic, such as changing the output depending on the time of day, then you cannot automatically generate a state machine.
Depends on what you mean by "one to one mapping". (And also, I suppose, "input" and "output".)
My guess is that you're asking whether, given an example of inputs and outputs for a given arbitrary program, can one devise an algorithm to write an equivalent program? If so, the answer is no. Eg, you could have a program with the inputs/outputs of 1/1, 2/2, 3/3, 4/4, and yet, if the next input value was 5, the output would be 3782. There's no way to know, from a given set of results, what the next result might be.
The question is underspecified since you did not say how the input and output are presented. For finite lists, the answer is "yes", as in this Python code:
def f(input,output):
print "def g():"
print " x = {" + ",".join(repr(x) + ":" + repr(y) for x,y in zip(input,output)) + "}"
print " print x[raw_input()]"
>>> f(['2','3','4'],['a','b','x'])
def g():
x = {'2':'a','3':'b','4':'x'}
print x[raw_input()]
>>> def g():
... x = {'2':'a','3':'b','4':'x'}
... print x[raw_input()]
...
>>> g()
3
b
for infinite sets how are you going to present them? If you show only a small sample of input this does not specify the whole algorithm. Guessing the best fit is undecidable. If you have a "magic blackbox" then there are continuum many mappings but only a countable number of programs, so it's impossible.
I think I agree with SLaks, but taking things from a different angle, what does a compiler do?
(EDIT: I see SLaks edited his original answer, which was essentially 'you're describing the identity function').
It takes a program in one source language that describes the intended behaviour of a program, and "writes" another program in a target language that exhibits that behaviour.
We could also think of this in terms of things like process refinement --- given an abstract specification, we can construct a refinement mapping to some "more concrete" (read: less non-deterministic, usually) implementation.
But based on your question, it's really very difficult to tell which of these you meant, if any.
I'm trying to seed a random number generator with the output of a hash. Currently I'm computing a SHA-1 hash, converting it to a giant integer, and feeding it to srand to initialize the RNG. This is so that I can get a predictable set of random numbers for an set of infinite cartesian coordinates (I'm hashing the coordinates).
I'm wondering whether Kernel::srand actually has a maximum value that it'll take, after which it truncates it in some way. The docs don't really make this obvious - they just say "a number".
I'll try to figure it out myself, but I'm assuming somebody out there has run into this already.
Knowing what programmers are like, it probably just calls libc's srand(). Either way, it's probably limited to 2^32-1, 2^31-1, 2^16-1, or 2^15-1.
There's also a danger that the value is clipped when cast from a biginteger to a C int/long, instead of only taking the low-order bits.
An easy test is to seed with 1 and take the first output. Then, seed with 2i+1 for i in [1..64] or so, take the first output of each, and compare. If you get a match for some i=n and all greater is, then it's probably doing arithmetic modulo 2n.
Note that the random number generator is almost certainly limited to 32 or 48 bits of entropy anyway, so there's little point seeding it with a huge value, and an attacker can reasonably easily predict future outputs given past outputs (and an "attacker" could simply be a player on a public nethack server).
EDIT: So I was wrong.
According to the docs for Kernel::rand(),
Ruby currently uses a modified Mersenne Twister with a period of 2**19937-1.
This means it's not just a call to libc's rand(). The Mersenne Twister is statistically superior (but not cryptographically secure). But anyway.
Testing using Kernel::srand(0); Kernel::sprintf("%x",Kernel::rand(2**32)) for various output sizes (2*16, 2*32, 2*36, 2*60, 2*64, 2*32+1, 2*35, 2*34+1), a few things are evident:
It figures out how many bits it needs (number of bits in max-1).
It generates output in groups of 32 bits, most-significant-bits-first, and drops the top bits (i.e. 0x[r0][r1][r2][r3][r4] with the top bits masked off).
If it's not less than max, it does some sort of retry. It's not obvious what this is from the output.
If it is less than max, it outputs the result.
I'm not sure why 2*32+1 and 2*64+1 are special (they produce the same output from Kernel::rand(2**1024) so probably have the exact same state) — I haven't found another collision.
The good news is that it doesn't simply clip to some arbitrary maximum (i.e. passing in huge numbers isn't equivalent to passing in 2**31-1), which is the most obvious thing that can go wrong. Kernel::srand() also returns the previous seed, which appears to be 128-bit, so it seems likely to be safe to pass in something large.
EDIT 2: Of course, there's no guarantee that the output will be reproducible between different Ruby versions (the docs merely say what it "currently uses"; apparently this was initially committed in 2002). Java has several portable deterministic PRNGs (SecureRandom.getInstance("SHA1PRNG","SUN"), albeit slow); I'm not aware of something similar for Ruby.
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.