I am trying to make sense of the different distribution objects in c++11 and I am finding it overwhelming. I hope some of you can and will help.
This is why I am looking into all this:
I need a random number generator that I can adjust every time it is used so that it is more likely to produce the same number again. The second requirement I need to fill is that I need the random numbers generated to only be these numbers:
{1, 2, 4, 8, 16, ..., 128}
Third and last requirement is that on certain occasions I need to skip one or more numbers from the above set.
My problem is that I don't understand the descriptions of various distribution objects. I, thus, cannot determine what tools I need to use to meet my above needs.
Can somebody tell me what tools I need and how I need to use them? The more clear, concise and detailed the response the better.
Your range can be generated with a random number j in the range [0, 7], then you compute:
1 << j
to get your number. std::uniform_int_distribution<> would be handy for generating the value in [0, 7].
Additionally you could use a std::bernoulli_distribution (which returns a random bool) to decide if the next number is going to be the same as the last one, or if you should generate a new number. The std::bernoulli_distribution defaults to a 50/50 chance of true/false, but you can customize that distribution in the bernoulli_distribution constructor to anything you like (e.g. 80/20 or whatever).
If this isn't clear enough, just jump in with some code. Try coding it up, and if it isn't working, post what you have, and I'm sure somebody will help.
Oh, forgot about your 3rd requirement: For that just put your [0, 7] generation in a loop, and if you come up with a number you're supposed to skip, then iterate the loop, else break out of it.
For skipping numbers I completely agree with Howard that manual checking is probably the way to go, but there might be a better way altering the probability of a given number being generated.
Another way to do this would be to use a discrete_distribution object, which allows you to specify the probability of generating any given value, so for your example it would be something like
std::default_random_engine entropy;
std::array<double, 128> probs;
probs.fill(1.0);
std::discrete_distribution<int> choose(probs.begin(), probs.end());
then when you're in your loop, in addition to deciding whether or not to skip, you can increment one of those values by some amount to increase the odds of it coming up again, making sure to reinitialize the discrete distribution, like this:
int x;
double myValue = 0.2;//or whatever increment you want
for (something; something else; something else else)
{
x = choose(entropy);
if (skip(x))
continue;//alternately you could set probs.at(x) = 0
//only if you never want to generate it again
probs.at(x) += myValue;
choose = std::discrete_distribution<int>(probs.begin(), probs.end());
output(x);
}
where skip and output are your functions to decide if x should be skipped and do whatever you want with the generated value respectively
Related
Im working on a project for which I need to make calculations with vectors (orthogonalizing a matrix using gram schmidt method). The length of this vectors is unknown now, the program must be able to adapt to different lengths. One of such calculations is calculating a new vector (C) which is the result of adding A and B. Each element of the vectors is a number in fixed-point.
I want C(i)=A(i)+B(i). For all the elements of the vector (for i=0 to N, where N is the vector length).
I can find 2 solutions for this but both present some problems:
1- I can declare in the entity, vectors whose length changes according to a generic and then just create a for loop which goes through all the vector.
for I in 0 to N loop
C(I)<=A(I)+B(I);
end loop;
The problem with this solution is that the execution would be sequential, and therefore slow. Im not completly sure about this and I dont know how to check it but I guess that the compiler is not smart enough to notice that it can be processed in parallel. In this application speed is a key factor.
2- I can declare vectors which are as long as the maximum possible length for the actual data and fill them with zeroes. Then I could just assign:
C(0)<=A(0)+B(0);
C(1)<=A(1)+B(1);
C(2)<=A(2)+B(2);
...
C(Nmax)<=A(Nmax)+B(Nmax);
This is not an elegant solution and in this application N can be between 3 and 300 therefore it could be a complete waste and tedious to program.
3- I want to find a third solution which could be able to create a number (asigned by the generic) of combinational calculations following a template such as C(i)=A(i)+B(i). Is there any solution like this? It is actually creating a loop which would not be executed sequentially but instead all at the same time.
I know that similar stuff can be done using CUDA but this project is actually a comparison between GPUs and FPGAs, so changing the platform is not a suitable solution either.
Thank you in advance
Edit: I have tought of another unsatisfactory solution but I want to share it in case it is helpful for somebody else checking this in the future. Given that A and B have the same length, you can write them in a 1-D format, that is: A(normal)=[1001,1100,0011], A(1-D)=100111000011. The same would be done with B.
If you know before hand that the sum of any two possible numbers can be expressed with the same amount of bits, there will be no problems. So with 4 unsigned bits you should make sure that in any possible case the numbers in A or B are !>0111 (not higher than 0111). You could just write C(1-D)=A(1-D)+B(1-D) and then just asign C(0)=C(1-D)(3 downto 0), C(1)=C(1-D)(7 downto 4) etc.
If you cannot make sure that the numbers are not higher than 0111 (in the 4 bit case) it wont work.
You might be able to use the length attribute to create a loop depending on the size of your vector.
https://www.csee.umbc.edu/portal/help/VHDL/attribute.html
As mentioned in the comment to the question the loop should be unrolled as long as it is not synchronized to the clock.
I was going through Google Interview Questions. to implement the random number generation from 1 to 7.
I did write a simple code, I would like to understand if in the interview this question asked to me and if I write the below code is it Acceptable or not?
import time
def generate_rand():
ret = str(time.time()) # time in second like, 12345.1234
ret = int(ret[-1])
if ret == 0 or ret == 1:
return 1
elif ret > 7:
ret = ret - 7
return ret
return ret
while 1:
print(generate_rand())
time.sleep(1) # Just to see the output in the STDOUT
(Since the question seems to ask for analysis of issues in the code and not a solution, I am not providing one. )
The answer is unacceptable because:
You need to wait for a second for each random number. Many applications need a few hundred at a time. (If the sleep is just for convenience, note that even a microsecond granularity will not yield true random numbers as the last microsecond will be monotonically increasing until 10us are reached. You may get more than a few calls done in a span of 10us and there will be a set of monotonically increasing pseudo-random numbers).
Random numbers have uniform distribution. Each element should have the same probability in theory. In this case, you skew 1 more (twice the probability for 0, 1) and 7 more (thrice the probability for 7, 8, 9) compared to the others in the range 2-6.
Typically answers to this sort of a question will try to get a large range of numbers and distribute the ranges evenly from 1-7. For example, the above method would have worked fine if u had wanted randomness from 1-5 as 10 is evenly divisible by 5. Note that this will only solve (2) above.
For (1), there are other sources of randomness, such as /dev/random on a Linux OS.
You haven't really specified the constraints of the problem you're trying to solve, but if it's from a collection of interview questions it seems likely that it might be something like this.
In any case, the answer shown would not be acceptable for the following reasons:
The distribution of the results is not uniform, even if the samples you read from time.time() are uniform.
The results from time.time() will probably not be uniform. The result depends on the time at which you make the call, and if your calls are not uniformly distributed in time then the results will probably not be uniformly distributed either. In the worst case, if you're trying to randomise an array on a very fast processor then you might complete the entire operation before the time changes, so the whole array would be filled with the same value. Or at least large chunks of it would be.
The changes to the random value are highly predictable and can be inferred from the speed at which your program runs. In the very-fast-computer case you'll get a bunch of x followed by a bunch of x+1, but even if the computer is much slower or the clock is more precise, you're likely to get aliasing patterns which behave in a similarly predictable way.
Since you take the time value in decimal, it's likely that the least significant digit doesn't visit all possible values uniformly. It's most likely a conversion from binary to some arbitrary number of decimal digits, and the distribution of the least significant digit can be quite uneven when that happens.
The code should be much simpler. It's a complicated solution with many special cases, which reflects a piecemeal approach to the problem rather than an understanding of the relevant principles. An ideal solution would make the behaviour self-evident without having to consider each case individually.
The last one would probably end the interview, I'm afraid. Perhaps not if you could tell a good story about how you got there.
You need to understand the pigeonhole principle to begin to develop a solution. It looks like you're reducing the time to its least significant decimal digit for possible values 0 to 9. Legal results are 1 to 7. If you have seven pigeonholes and ten pigeons then you can start by putting your first seven pigeons into one hole each, but then you have three pigeons left. There's nowhere that you can put the remaining three pigeons (provided you only use whole pigeons) such that every hole has the same number of pigeons.
The problem is that if you pick a pigeon at random and ask what hole it's in, the answer is more likely to be a hole with two pigeons than a hole with one. This is what's called "non-uniform", and it causes all sorts of problems, depending on what you need your random numbers for.
You would either need to figure out how to ensure that all holes are filled equally, or you would have to come up with an explanation for why it doesn't matter.
Typically the "doesn't matter" answer is that each hole has either a million or a million and one pigeons in it, and for the scale of problem you're working with the bias would be undetectable.
Using the same general architecture you've created, I would do something like this:
import time
def generate_rand():
ret = str(time.time()) # time in second like, 12345.1234
ret = ret % 8 # will return pseudorandom numbers 0-7
if ret == 0:
return 1 # or you could also return the result of another call to generate_rand()
return ret
while 1:
print(generate_rand())
time.sleep(1)
I wonder if there's a way to begin reading from an arbitrary position in a array. E.g. if I have a array of size 10 and it begins reading from position 4. Then it should continue on reading from position 5, 6, 7, 8, 9, 0, 1, 2, 3
I was uncertain with tag, so if have picked wrong tag please do change it for me.
Yes, you can index using the modulo operation which is written as % in most languages:
x = list[i % list.length]
This will give you the desired effect of wrapping around when you reach the end of the list instead of attempting to index out of bounds.
This assumes 0-based indexing. If you use 1-based indexing you have to add one to the result of the modulo operation.
offset = 4;
for(i=0; i<n; i++)
cout << x[(i+offset)%n] << ' ';
It depends on what you mean by "list". Traditionally in computer science, "list" usually means "linked list". In this case, you have to traverse the list in order to get to a particular element.
If you need/want to be able to start reading from arbitrary positions efficiently, you probably want to avoid linked lists, but the exact alternatives you have (easily) available will vary with the programming language, libraries, etc., you're using.
Edit: for an array, it's generally trivial -- just specify the starting position directly, and take the remainder after dividing by the array size.
One option is to make the list circular (like they do it in the Linux kernel).
I want to generate unique random, N-valued key.
This key can contain numbers and latin characters, i.e. A-Za-z0-9.
The only solution I am thinking about is something like this (pseudocode):
key = "";
smb = "ABC…abc…0123456789"; // allowed symbols
for (i = 0; i < N; i++) {
key += smb[rnd(0, smb.length() - 1)]; // select symbol at random position
}
Is there any better solution? What can you suggest?
I would look into GUIDs. From the Wikipedia entry, "the primary purpose of the GUID is to have a totally unique number," which sounds exactly like what you are looking for. There are several implementations out there that generate GUIDs, so it's likely you will not have to reinvent the wheel.
Keeping in mind that the whole field of cryptography relies on, amongst other things, making random numbers. Therefore the NSA, the CIA, and some of the best mathematicians in the world are working on this so I guarantee you that there are better ideas.
Me? I'd just do what fbrereto suggests, and just get a guid. Or look into cryptographic key generators, or y'know, some lava lamps and a camera.
Oh, and as to the code you have; depending on the language, you may need to seed the RNG, or it'll generate the same key every time.
Whatever you do, if you wind up generating a key that uses all numbers and all letters, and if a person is ever going to see that key (which is likely if you are using numbers and letters), omit the characters l, I, 1, O, and 0. People get them confused.
Nothing in your post addresses the question of uniqueness. You're going to have to have some way of not generating the same key twice. Usually, when I need a unique key, I have some unique information to start with. I usually take a one-way hash like MD5, then there are ways to convert that to a key with varying degrees of readability:
Convert to hex
Base64 encode it
Use bits of of the key to index into a list of words.
Example: the unique string computed by hashing the part of this answer above the horizontal line is
abduction's brogue's melted bragger's
You could do a base64 encoding of some random data and remove the +, /, and = characters from the result? I don't know if this would make a predictable distribution. Also, it seems like more work that what you're doing now, which is a fine solution.
Assuming you're using a language/library without an utterly pathetic random number generator, what you've got looks pretty good. N symbols uniformly distributed over a reasonable alphabet works for me, and no amount of applying fancier code is likely to make it more random (just slower).
(For the record, pathetic would include ditching the high-order bits of the underlying random numbers when choosing a value from the given range. While ideally all RNGs would make every bit equally random, in practice that's not so; the higher-order bits tend to be more random. This means that the modulus operator is totally the wrong thing to use when clamping to a restricted range.)
I am a beginner trying to do some engineering experiments using fortran 77. I am using Force 2.0 compiler and editor. I have the following queries:
How can I generate a random number between a specified range, e.g. if I need to generate a single random number between 3.0 and 10.0, how can I do that?
How can I use the data from a text file to be called in calculations in my program. e.g I have temperature, pressure and humidity values (hourly values for a day, so total 24 values in each text file).
Do I also need to define in the program how many values are there in the text file?
Knuth has released into the public domain sources in both C and FORTRAN for the pseudo-random number generator described in section 3.6 of The Art of Computer Programming.
2nd question:
If your file, for example, looks like:
hour temperature pressure humidity
00 15 101325 60
01 15 101325 60
... 24 of them, for each hour one
this simple program will read it:
implicit none
integer hour, temp, hum
real p
character(80) junkline
open(unit=1, file='name_of_file.dat', status='old')
rewind(1)
read(1,*)junkline
do 10 i=1,24
read(1,*)hour,temp,p,hum
C do something here ...
10 end
close(1)
end
(the indent is a little screwed up, but I don't know how to set it right in this weird environment)
My advice: read up on data types (INTEGER, REAL, CHARACTER), arrays (DIMENSION), input/output (READ, WRITE, OPEN, CLOSE, REWIND), and loops (DO, FOR), and you'll be doing useful stuff in no time.
I never did anything with random numbers, so I cannot help you there, but I think there are some intrinsic functions in fortran for that. I'll check it out, and report tomorrow. As for the 3rd question, I'm not sure what you ment (you don't know how many lines of data you'll be having in a file ? or ?)
You'll want to check your compiler manual for the specific random number generator function, but chances are it generates random numbers between 0 and 1. This is easy to handle - you just scale the interval to be the proper width, then shift it to match the proper starting point: i.e. to map r in [0, 1] to s in [a, b], use s = r*(b-a) + a, where r is the value you got from your random number generator and s is a random value in the range you want.
Idigas's answer covers your second question well - read in data using formatted input, then use them as you would any other variable.
For your third question, you will need to define how many lines there are in the text file only if you want to do something with all of them - if you're looking at reading the line, processing it, then moving on, you can get by without knowing the number of lines ahead of time. However, if you are looking to store all the values in the file (e.g. having arrays of temperature, humidity, and pressure so you can compute vapor pressure statistics), you'll need to set up storage somehow. Typically in FORTRAN 77, this is done by pre-allocating an array of a size larger than you think you'll need, but this can quickly become problematic. Is there any chance of switching to Fortran 90? The updated version has much better facilities for dealing with standardized dynamic memory allocation, not to mention many other advantages. I would strongly recommend using F90 if at all possible - you will make your life much easier.
Another option, depending on the type of processing you're doing, would be to investigate algorithms that use only single passes through data, so you won't need to store everything to compute things like means and standard deviations, for example.
This subroutine generate a random number in fortran 77 between 0 and ifin
where i is the seed; some great number such as 746397923
subroutine rnd001(xi,i,ifin)
integer*4 i,ifin
real*8 xi
i=i*54891
xi=i*2.328306e-10+0.5D00
xi=xi*ifin
return
end
You may modifies in order to take a certain range.