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This is a question popped into my mind while reading the halting problem, collatz conjecture and Kolmogorov complexity. I have tried to search for something similar but I was unable to find a particular topic maybe because it is not of great value or it could just be a trivial question.
For the sake of simplicity I will give three examples of programs/functions.
function one(s):
return s
function two(s):
while (True):
print s
function three(s):
for i from 0 to 10^10:
print(s)
So my questions is, if there is a way to formalize the length of a program (like the bits used to describe it) and also the internal memory used by the program, to determine the minimum/maximum number of time/steps needed to decide whether the program will terminate or run forever.
For example, in the first function the program doesn't alter its internal memory and halts after some time steps.
In the second example, the program runs forever but the program also doesn't alter its internal memory. For example, if we considered all the programs with the same length as with the program two that do not alter their state, couldn't we determine an upper bound of steps, which if surpassed we could conclude that this program will never terminate ? (If not why ?)
On the last example, the program alters its state (variable i). So, at each step the upper bound may change.
[In short]
Kolmogorov complexity suggests a way of finding the (descriptive) complexity of an object such as a piece of text. I would like to know, given a formal way of describing the memory-space used by a program (computed in runtime), if we could compute a maximum number of steps, which if surpassed would allow us to know whether this program will terminate or run forever.
Finally, I would like to suggest me any source that I might find useful and help me figure out what I am exactly looking for.
Thank you. (sorry for my English, not my native language. I hope I was clear)
If a deterministic Turing machine enters precisely the same configuration twice (which we can detect b keeping a trace of configurations seen so far), then we immediately know the TM will loop forever.
If it known in advance that a deterministic Turing machine cannot possibly use more than some fixed constant amount of its input tape, then the TM must explicitly halt or eventually enter some configuration it has already visited. Suppose the TM can use at most k tape cells, the tape alphabet is T and the set of states is Q. Then there are (|T|+1)^k * |Q| unique configurations (the number of strings over (T union blank) of length k times the number of states) and by the pigeonhole principle we know that a TM that takes that many steps must enter some configuration it has already been to before.
one: because we are given that this function does not use internal memory, we know that it either halts or loops forever.
two: because we are given that this function does not use internal memory, we know that it either halts or loops forever.
three: because we are given that this function only uses a fixed amount of internal memory (like 34 bits) we can tell in fewer than 2^34 iterations of the loop whether the TM will halt or not for any given input s, guaranteed.
Now, knowing how much tape a TM is going to use, or how much memory a program is going to use, is not a problem a TM can solve. But if you have an oracle (like a person who was able to do a proof) that tells you a correct fixed upper bound on memory, then the halting problem is solvable.
I am optimizing algorithmic strategies. In the process of choosing from a pool of many optimized strategies, I am in the phase of searching (evaluating) for robustness of the strategy.
Following the guidelines of Dr. Pardo's book "The Evaluation of Trading Strategies" in page 231 Dr. Pardo recomends, in the Numeral 3 to apply the following ratio to the optimized data:
" 3. The ratio of the total profit of all profitable simulations divided by the
total profit of all simulationsis significantly positive"
The Question: from the optimization results, I am not being able to properly understand what does Mr. Pardo means by stating "...all simulationsis significantly positive"; what does Mr. Pardo means by 'significantly positive?
a.) with 95% confidence level?
b.) with a certain p value?
c.) the relation of the average net profit of each simulation minus it' standard deviation
Even though the sentence might seem 'simple' I would REALLY like to understand what Mr. Pardo means by the statement and HOW to calculate it, in order to filter the most robust algorithmic strategies.
The aim of analyzing the optimization profile of an algorithmic simulation is to be able to filter robust strategies.
Therefore the ratio should help us to uncover if the simulation results are on the right track or not.
So, we would like to impose some 'penalties' to our results, so we can select the robust cases from those of doubtful (not robust) result.
I came to the following penalizing measures (found in the book of Mr. Pardo and other sources).
a.) we can use a market return (yearly value) as a benchmark, so all the simulations whose result are below such level, can be excluded from our analysis,
b.) some other benchmark to divide those 'robust' results from those more 'doubtful' (for example, deducing to each result one standard deviation)
From (a) and (b), we can create the ratio:
the total sum of all profitable simulations divided by the profitable results considered robust
The ratio should be greater or equal than 1.
If the ratio is equal to 1 then it means that our simulation result has given interesting results (we are analyzing the positive values in this ratio, but profitable results should always be compared to the negative results also).
If the ratio is greater from 1, then we have not reach the possible scenario, and the result should be compared with the other tests for optimizations.
While simulating trading algorithms, no result is absolute but partial and it's value is taken in relationship to what we expect from the algorithm.
If someone has a better explanation or idea or concept you might find interesting please share, I would gladly read it.
Best regards to all.
Remark on the subject
With all due respect to the subject ( published in 2008 ) the term robustness has its own meaning if-and-only-if the statement also clarifies in which particular respect is the robustness measured and against what phenomena is it to be exposed & tested the Model-under-review's response ( against what perturbances -- type and scale -- shall the Model-under-test hold its robust behaviour, measures of which were both defined and quantified a-priori the test ).
In any case, where such context of the robustness is not defined, the material, be it printed by any bold name, sounds -- and forgive me to speak in plain English -- just like a PR-story, an over-hyped e-zine headline or like a paid advertorial.
Serious quantitative model evaluations, the more if one strives to perform an optimisation ( with respect to some defined quantitative goal ), requires a more thorough insight into the subject than to axiomatically post a trivial "must-have" imperative of
large-average && small-HiLo-range && small StDev.
Any serious Quant-Modelling effort, if it were not to just spoil the consumed hundreds-of-thousands CPU core hours of deep parametric-spaces' scans, shall incorporate a serious parametrisation decision in either dimension of the main TruTrading Strategy sub-spaces --
{ aSelectPOLICY, aDetectPOLICY, anActPOLICY, anAllocatePOLICY, aTerminatePOLICY }
A failure to do so, either cripples the model or leads to a blind-belief, where it is hard to guess, whether the former or the latter is a greater of the both Quant-sins.
Remark on the cited hypothesis
The book states, without any effort to proof the construction, that:
The more robust trading strategywill have an optimization profile with a: 1. Largeaverageprofit 2. Small maximum-minimumrange3. Small standarddeviation
Is it correct?
Now kindly spend a few moments and review this 4D-animated view of a Model-under-test ( visualisation of which is reduced into just four dimensions for easier visual perception ), where none of the above stands true.
<aMouseRightCLICK>.openPictureOnAnotherTab to see full HiRes picture details
Based on contemporary state-of-art adaptive money-management practice, that fails to be correct, be it due to a poor parametrisation ( thus artificially leading the model into a rather "flat-profits" sub-space of aParamSetVectorSPACE )
or due to a principal mis-concept or a poor practice ( including the lack thereof ) of the implementation of the most powerful profit-booster ever -- the very money-management model sub-space.
Item 1 becomes insignificant at all.
Item 2 works right on the contrary to the stated postulate.
Item 3 cannot yield anything but the opposite due to 1 & 2 above.
I'm thinking more about how much system memory my programs will use nowadays. I'm currently doing A level Computing at college and I know that in most programs the difference will be negligible but I'm wondering if the following actually makes any difference, in any language.
Say I wanted to output "True" or "False" depending on whether a condition is true. Personally, I prefer to do something like this:
Dim result As String
If condition Then
Result = "True"
Else
Result = "False"
EndIf
Console.WriteLine(result)
However, I'm wondering if the following would consume less memory, etc.:
If condition Then
Console.WriteLine("True")
Else
Console.WriteLine("False")
EndIf
Obviously this is a very much simplified example and in most of my cases there is much more to be outputted, and I realise that in most commercial programs these kind of statements are rare, but hopefully you get the principle.
I'm focusing on VB.NET here because that is the language used for the course, but really I would be interested to know how this differs in different programming languages.
The main issue making if's fast or slow is predictability.
Modern CPU's (anything after 2000) use a mechanism called branch prediction.
Read the above link first, then read on below...
Which is faster?
The if statement constitutes a branch, because the CPU needs to decide whether to follow or skip the if part.
If it guesses the branch correctly the jump will execute in 0 or 1 cycle (1 nanosecond on a 1Ghz computer).
If it does not guess the branch correctly the jump will take 50 cycles (give or take) (1/200th of a microsecord).
Therefore to even feel these differences as a human, you'd need to execute the if statement many millions of times.
The two statements above are likely to execute in exactly the same amount of time, because:
assigning a value to a variable takes negligible time; on average less than a single cpu cycle on a multiscalar CPU*.
calling a function with a constant parameter requires the use of an invisible temporary variable; so in all likelihood code A compiles to almost the exact same object code as code B.
*) All current CPU's are multiscalar.
Which consumes less memory
As stated above, both versions need to put the boolean into a variable.
Version A uses an explicit one, declared by you; version B uses an implicit one declared by the compiler.
However version A is guaranteed to only have one call to the function WriteLine.
Whilst version B may (or may not) have two calls to the function WriteLine.
If the optimizer in the compiler is good, code B will be transformed into code A, if it's not it will remain with the redundant calls.
How bad is the waste
The call takes about 10 bytes for the assignment of the string (Unicode 2 bytes per char).
But so does the other version, so that's the same.
That leaves 5 bytes for a call. Plus maybe a few extra bytes to set up a stackframe.
So lets say due to your totally horrible coding you have now wasted 10 bytes.
Not much to worry about.
From a maintainability point of view
Computer code is written for humans, not machines.
So from that point of view code A is clearly superior.
Imagine not choosing between 2 options -true or false- but 20.
You only call the function once.
If you decide to change the WriteLine for another function you only have to change it in one place, not two or 20.
How to speed this up?
With 2 values it's pretty much impossible, but if you had 20 values you could use a lookup table.
Obviously that optimization is not worth it unless code gets executed many times.
If you need to know the precise amount of memory the instructions are going to take, you can use ildasm on your code, and see for yourself. However, the amount of memory consumed by your code is much less relevant today, when the memory is so cheap and abundant, and compilers are smart enough to see common patterns and reduce the amount of code that they generate.
A much greater concern is readability of your code: if a complex chain of conditions always leads to printing a conditionally set result, your first code block expresses this idea in a cleaner way than the second one does. Everything else being equal, you should prefer whatever form of code that you find the most readable, and let the compiler worry about optimization.
P.S. It goes without saying that Console.WriteLine(condition) would produce the same result, but that is of course not the point of your question.
I've been searching the web and I'm finding somewhat contradictory answers. Some sources assert that a language/machine/what-have-you is Turing complete if and only if it has both conditional and unconditional branching (which I guess is kind of redundant), some say that only unconditional is required, others that only conditional is required.
Reading about the German Z3 and ENIAC, Wikipedia says:
The German Z3 (shown working in May
1941) was designed by Konrad Zuse. It
was the first general-purpose digital
computer, but it was
electromechanical, rather than
electronic, as it used relays for all
functions. It computed logically using
binary math. It was programmable by
punched tape, but lacked the
conditional branch. While not designed
for Turing-completeness, it
accidentally was, as it was found out
in 1998 (but to exploit this
Turing-completeness, complex, clever
hacks were necessary).
What complex, clever hacks, exactly?
A 1998 paper Abstract by R. Rojas also states (Note that I haven't read this paper, it's just a snippet from IEEE.):
The computing machine Z3, built by
Konrad Zuse between 1938 and 1941,
could execute only fixed sequences of
floating point arithmetical operations
(addition, subtraction,
multiplication, division, and square
root) coded in a punched tape. An
interesting question to ask, from the
viewpoint of the history of computing,
is whether or not these operations are
sufficient for universal computation.
The paper shows that, in fact, a
single program loop containing these
arithmetical instructions can simulate
any Turing machine whose tape is of a
given finite size. This is done by
simulating conditional branching and
indirect addressing by purely
arithmetical means. Zuse's Z3 is
therefore, at least in principle, as
universal as today's computers that
have a bounded addressing space.
In short, SOers, what type of branching is exactly required for Turing-completeness? Assuming infinite memory, can a language with only a goto or jmp branching construct (no if or jnz constructs) be considered Turing-complete?
The original Rojas paper can be found here. The basic idea is that the Z3 only supports a unconditional single loop (by gluing the ends of the instruction tape together). You build conditional execution of it by putting all code sections one after another in the loop, and having a variable z that determines which section to execute. At the beginning of section j, you set
if (z==j) then t=0 else t=1
and then make each assignment a = b op c in this section read
a = a*t + (b op c)*(1-t)
(i.e. each assignment is a no-op, except in the active section). Now, this still includes a conditional assignment: how to compare z==j? He proposes to use the binary representation of z (z1..zm) along with the negated binary representation of j (c1..cm), and then compute
t = 1 - sqr((c1-z1)(c2-z2)...(cm-zm))
This product will be 1 only if c and z differ in all bits, which will happen only if z==j. An assignment to z (which essentially is an indirect jump) must also assign to z1..zm.
Rojas has also written Conditional Branching is not Necessary for Universal Computation in von Neumann Computers. There he proposes a machine with self-modifying code and relative addressing, so that you can read the Turing instructions from memory, and modify the program to jump accordingly. As an alternative, he proposes the above approach (for Z3), in a version that only uses LOAD(A), STORE(A), INC and DEC.
If you have only arithmetical expressions you can use some properties of arithmetical operations. E.g., is A is either 0 or 1 depending on some condition (which is previously computed), then A*B+(1-A)*C computes the expression if A then B else C.
If you can compute the address for your goto or jmp, you can simulate arbritary conditionals. I occasionally used this to simulate "ON x GOTO a,b,c" in ZX Basic.
If "true" has the numerical value 1 and "false" 0, then a construction like:
if A then goto B else goto C
is identical to:
goto C+(B-C)*A
So, yes, with a "computed goto" or the ability to self-modify, a goto or jmp can act as a conditional.
You need something that can branch based on (results from) input.
One way to simulate conditional branches is with self-modifying code -- you do a computation that deposits its result into the stream of instructions being executed. You could put the op-code for an unconditional jump into the instruction stream, and do math on an input to create the correct target for that jump, depending on some set of conditions for the input. For example, subtract x from y, shift right to 0-fill if it was positive, or 1-fill if it was negative, then add a base address, and store that result immediately following the jmp op-code. When you get to that jmp, you'll go to one address if x==y, and another if x!=y.
You don't need conditional branching to build a Turing-complete machine, but of course any Turing-complete machine will provide conditional branching as a core feature.
It was proved that systems as simple as the Rule 110 Cellular Automaton can be used to implement a Turing machine. You sure don't need conditional branching to pull such a system from the bit bucket. Actually one could just use a bunch of rocks.
The point is that a Turing machine will provide the conditional branching, so what you are doing anyway by proving Turing completeness is somewhat implementing conditional branching. You have to do it without conditional branching at some point, be it rocks or PN-junctions in semi-conductors.
The Z3 was only Turing complete from an abstract point of view. You can have an arbitrarily long program tape and just have it compute both sides of every conditional branch. In other words, for each branch, it would compute both answers and tell you which one to ignore. Obviously this creates exponentially larger programs for every conditional branch you would have, so you could never use this machine in a Turing-complete manner.
If a machine can branch, then yes it's considered Turing complete.
The reason is having conditional-branching automatically makes any computer Turing complete. However, there are also machines that can't jump branch or even IF but are still considered Turing complete.
Processing is just the process of identifying inputs in-order to select outputs.
Branching is one way to mentalize this process, the condition of the jump is what can classify inputs, the place you branch to stores the correct output for that input.
So finally, to clarify things:
If you have conditional branching your computer is necessarily computationally equivalent to a Turing machine. However, there are plenty of other ways for a computer to achieve Turing completeness (lambda, IF's, CL).
I found this on an "interview questions" site and have been pondering it for a couple of days. I will keep churning, but am interested what you guys think
"10 Gbytes of 32-bit numbers on a magnetic tape, all there from 0 to 10G in random order. You have 64 32 bit words of memory available: design an algorithm to check that each number from 0 to 10G occurs once and only once on the tape, with minimum passes of the tape by a read head connected to your algorithm."
32-bit numbers can take 4G = 2^32 different values. There are 2.5*2^32 numbers on tape total. So after 2^32 count one of numbers will repeat 100%. If there were <= 2^32 numbers on tape then it was possible that there are two different cases – when all numbers are different or when at least one repeats.
It's a trick question, as Michael Anderson and I have figured out. You can't store 10G 32b numbers on a 10G tape. The interviewer (a) is messing with you and (b) is trying to find out how much you think about a problem before you start solving it.
The utterly naive algorithm, which takes as many passes as there are numbers to check, would be to walk through and verify that the lowest number is there. Then do it again checking that the next lowest is there. And so on.
This requires one word of storage to keep track of where you are - you could cut down the number of passes by a factor of 64 by using all 64 words to keep track of where you're up to in several different locations in the search space - checking all of your current ones on each pass. Still O(n) passes, of course.
You could probably cut it down even more by using portions of the words - given that your search space for each segment is smaller, you won't need to keep track of the full 32-bit range.
Perform an in-place mergesort or quicksort, using tape for storage? Then iterate through the numbers in sequence, tracking to see that each number = previous+1.
Requires cleverly implemented sort, and is fairly slow, but achieves the goal I believe.
Edit: oh bugger, it's never specified you can write.
Here's a second approach: scan through trying to build up to 30-ish ranges of contiginous numbers. IE 1,2,3,4,5 would be one range, 8,9,10,11,12 would be another, etc. If ranges overlap with existing, then they are merged. I think you only need to make a limited number of passes to either get the complete range or prove there are gaps... much less than just scanning through in blocks of a couple thousand to see if all digits are present.
It'll take me a bit to prove or disprove the limits for this though.
Do 2 reduces on the numbers, a sum and a bitwise XOR.
The sum should be (10G + 1) * 10G / 2
The XOR should be ... something
It looks like there is a catch in the question that no one has talked about so far; the interviewer has only asked the interviewee to write a program that CHECKS
(i) if each number that makes up the 10G is present once and only once--- what should the interviewee do if the numbers in the given list are present multple times? should he assume that he should stop execting the programme and throw exception or should he assume that he should correct the mistake by removing the repeating number and replace it with another (this may actually be a costly excercise as this involves complete reshuffle of the number set)? correcting this is required to perform the second step in the question, i.e. to verify that the data is stored in the best possible way that it requires least possible passes.
(ii) When the interviewee was asked to only check if the 10G weight data set of numbers are stored in such a way that they require least paases to access any of those numbers;
what should the interviewee do? should he stop and throw exception the moment he finds an issue in the algorithm they were stored in, or correct the mistake and continue till all the elements are sorted in the order of least possible passes?
If the intension of the interviewer is to ask the interviewee to write an algorithm that finds the best combinaton of numbers that can be stored in 10GB, given 64 32 Bit registers; and also to write an algorithm to save these chosen set of numbers in the best possible way that require least number of passes to access each; he should have asked this directly, woudn't he?
I suppose the intension of the interviewer may be to only see how the interviewee is approaching the problem rather than to actually extract a working solution from the interviewee; wold any buy this notion?
Regards,
Samba