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I want to find a seed that creates a specific number sequence:
[115,91,45,76,78,93,35,5,29,8,99,88,98,70,40,116,11,39,102,41,124,98,120,57,36,67,57,23,52,34,75,32,117,66,12,19,86,67,62,121,60,5,54,37,65,18,5,56,66,115,32,99,73,70,115,73,123,74,31]
I wonder if I could find one of seed that give me this result with the function get() I created :
def get():
seed(x)
return [choice(range(128)) for _ in range(59)]
with x a constant equal to the number that, apply as seed, give me the right top above sequence.
This is a little program I made to expect find it, but right now I'm about 1.6 milions tested seed and still nothing.
from random import choice, seed
lc =[115,91,45,76,78,93,35,5,29,8,99,88,98,70,40,116,11,39,102,41,124,98,120,57,36,67,57,23,52,34,75,32,117,66,12,19,86,67,62,121,60,5,54,37,65,18,5,56,66,115,32,99,73,70,115,73,123,74,31]
sd, h = 0,0
while 1:
seed(sd)
for c, o in enumerate(lc):
if not choice(range(128)) == o:
if c > h :
print(f"[Seeed {sd}] {c} matchs")
h = c
sd += 1
break
Can someone help me to find one of the right seed ?
I hope it is not possible.
Technically, it is possible to code a quasi-random generator that allows restoring a seed by a short sequence of results. But normal quasi-random generator should disallow that.
E.g. for quite common Mersenne Twister the internal state is 624 ints. But your seed is just one int.
Even if you brute-force the seed that gives you’re the same short sequence, the whole internal state actual may be different and consequent generation will goes completely other way.
Any seeded PRNG will have a formula to generate the next number from the internal data it holds. With something as simple as a Linear Congruential PRNG then it is easy to back-calculate the internal data and the numbers used in the formula from the output. With a more complex PRNG, such as Mersenne Twister, then the back-calculation becomes very difficult.
One solution would be to copy the sequence of numbers you want and store them somewhere, pulling them from the store as needed. Alternatively read the documentation of the PRNG used to generate those numbers initially to see if a back-calculation is possible.
If the numbers came from a cryptographically secure PRNG then your task becomes orders of magnitude more difficult.
Using brute force and assuming that every seed represent an extraction from your set of number (128), with replacement, you have a probability of
1/(128)^59 = 1 / 2.1153791001287955166461289857048673274508949854856999 × 10^124
for every extraction to get your exact set of numbers (assuming uniform distribution for every number extraction of your random function). Which is a probability pretty near to zero.
So yes. You could hang (almost) forever for that brute force search
I wrote a C++ wrapper class to some functions in LAPACK. In order to test the class, I use the Python C Extension, where I call numpy, and do the same operations, and compare the results by taking the difference
For example, for the inverse of a matrix, I generate a random matrix in C++, then pass it as a string (with many, many digits, like 30 digits) to Python's terminal using PyRun_SimpleString, and assign the matrix as numpy.matrix(...,dtype=numpy.double) (or numpy.complex128). Then I use numpy.linalg.inv() to calculate the inverse of the same matrix. Finally, I take the difference between numpy's result and my result, and use numpy.isclose with a specific relative tolerance to see whether the results are close enough.
The problem: The problem is that when I use C++ floats, the relative precision I need to be able to compare is about 1e-2!!! And yet with this relative precision I get some statistical failures (with low probability).
Doubles are fine... I can do 1e-10 and it's statistically safe.
While I know that floats have intrinsic bit precision of about 1e-6, I'm wondering why I have to go so low to 1e-2 to be able to compare the results, and it still fails some times!
So, going so low down to 1e-2 got me wondering whether I'm thinking about this whole thing the wrong way. Is there something wrong with my approach?
Please ask for more details if you need it.
Update 1: Eric requested example of Python calls. Here is an example:
//create my matrices
Matrix<T> mat_d = RandomMatrix<T>(...);
auto mat_d_i = mat_d.getInverse();
//I store everything in the dict 'data'
PyRun_SimpleString(std::string("data={}").c_str());
//original matrix
//mat_d.asString(...) will return in the format [[1,2],[3,4]], where 32 is 32 digits per number
PyRun_SimpleString(std::string("data['a']=np.matrix(" + mat_d.asString(32,'[',']',',') + ",dtype=np.complex128)").c_str());
//pass the inverted matrix to Python
PyRun_SimpleString(std::string("data['b_c']=np.matrix(" + mat_d_i.asString(32,'[',']',',') + ",dtype=np.complex128)").c_str());
//inverse in numpy
PyRun_SimpleString(std::string("data['b_p']=np.linalg.inv(data['a'])").c_str());
//flatten the matrices to make comparing them easier (make them 1-dimensional)
PyRun_SimpleString("data['fb_p']=((data['b_p']).flatten().tolist())[0]");
PyRun_SimpleString("data['fb_c']=((data['b_c']).flatten().tolist())[0]");
//make the comparison. The function compare_floats(f1,f2,t) calls numpy.isclose(f1,f2,rtol=t)
//prec is an integer that takes its value from a template function, where I choose the precision I want based on type
PyRun_SimpleString(std::string("res=list(set([compare_floats(data['fb_p'][i],data['fb_c'][i],1e-"+ std::to_string(prec) +") for i in range(len(data['fb_p']))]))[0]").c_str());
//the set above eliminates repeated True and False. If all results are True, we expect that res=[True], otherwise, the test failed somewhere
PyRun_SimpleString(std::string("res = ((len(res) == 1) and res[0])").c_str());
//Now if res is True, then success
Comments in the code describe the procedure step-by-step.
I am trying to create a list of points from a road network. Here, I try to put their coordinates in a List of [x,y] whose items have a float format. As a new point from the network is picked, it should be checked with the existing points in the list. if it exists, then the same index will be given to the feature of network, otherwise a new point will be added to the list and the new index will be given to the feature.
I know that a float number will be saved differently form integers, but for exactly the same float numbers, I still cannot use:
If new_point in list_of_points:
#do something
and I should use:
for point in list_of_points:
if abs(point.x-new_point.x)<0.01 and abs(point.y-new_point.y)<0.01
#do something
the points are supposed to be exactly the same as I snap them using the ArcGIS software, and when I check the coordinates in the software they are exactly the same.
I asked this question for:
1- I think using "in" can make my code tidy and also faster while using for-loop is kind of clumsy way of coding for this situation.
2- I want to know: does that mean even exactly the same float numbers are stored differently?
It's never a good idea to check for equality between two floating point numbers. However, there are built in functions to do a comparison like that. From numpy you can use allclose. For example,
>>> np.allclose( (1.0,2.0), (1.00000001,2.0000001) )
True
This checks if the two array like inputs are element-wise equal within a certain tolerance. You can adjust the relative and absolute tolerances with keyword arguments.
Any given Python implenetation should always store a given floating point number in the same, deterministic, non-random way within itself. I do not believe you can take the same floating point number, input it twice, and have it stored in two different ways. But I'm also reluctant to believe that you're going to be getting exact duplicates of coordinates out of a geographic program like ArcGIS, especially if the resolution is very small. There are many ways that floating point math can mess with your expectations, so you shouldn't ever expect that you'll have identical floats. And between different machines and different versions, you get even more possibilities for error.
If you're worried about the elegance of your code, you can just create a function to abstract out the for loop.
def coord_in(coord, coord_list):
for other_coord in coord_list:
if abs(coord.x-other_coord.x)<0.00001 and abs(coord.y-other_coord.y)<0.00001:
return True
return False
For a large number of points, numpy will always be faster (and perhapd more elegant). If you have separated the x and y coords into (float) arrays arrx and arry:
numpy.sometrue((arrx-point.x)**2+(arry-point.y)**2<tol**2)
will return True if point is within distance tol of an existing point.
2: exactly the same literal (e.g., "2.3") will be stored as exactly the same float representation for for a given platform and data-type, but in general it depends on the bit-ness, endian-ness and perhaps the compiler used to make python.
To be certain when comparing numbers, you should at least round to the precision of the least precise number, or (better) do the kind of thing you are doing here.
>>> 1==1.00000000000000000000000000000000001
True
Old thread but helped me develop my own solution using list comprehension. Because of course it's not a good idea to compare two floats using ==. The following returns list of indices of all elements of the input list that are reasonably close to the value we're looking for.
def findFloats(listOfFloats, value):
return [i for i, number in enumerate(listOfFloats)
if abs(number-value) < 0.00001]
beside the correct language ID langid.py returns a certain value - "The value returned is a score for the language. It is not a probability esimate, as it is not normalized by the document probability since this is unnecessary for classification."
But what does the value mean??
I'm actually the author of langid.py. Unfortunately, I've only just spotted this question now, almost a year after it was asked. I've tidied up the handling of the normalization since this question was asked, so all the README examples have been updated to show actual probabilities.
The value that you see there (and that you can still get by turning normalization off) is the un-normalized log-probability of the document. Because log/exp are monotonic, we don't actually need to compute the probability to decide the most likely class. The actual value of this log-prob is not actually of any use to the user. I should probably have never included it, and I may remove its output in the future.
I think this is the important chunk of langid.py code:
def nb_classify(fv):
# compute the log-factorial of each element of the vector
logfv = logfac(fv).astype(float)
# compute the probability of the document given each class
pdc = np.dot(fv,nb_ptc) - logfv.sum()
# compute the probability of the document in each class
pd = pdc + nb_pc
# select the most likely class
cl = np.argmax(pd)
# turn the pd into a probability distribution
pd /= pd.sum()
return cl, pd[cl]
It looks to me that the author is calculating something like the multinomial log-posterior of the data for each of the possible languages. logfv calculates the logarithm of the denominator of the PMF (x_1!...x_k!). np.dot(fv,nb_ptc) calculates the
logarithm of the p_1^x_1...p_k^x_k term. So, pdc looks like the list of language conditional log-likelihoods (except that it's missing the n! term). nb_pc looks like the prior probabilities, so pd would be the log-posteriors. The normalization line, pd /= pd.sum() confuses me, since one usually normalizes probability-like values (not log-probability values); also, the examples in the documentation (('en', -55.106250761034801)) don't look like they've been normalized---maybe they were generated before the normalization line was added?
Anyway, the short answer is that this value, pd[cl] is a confidence score. My understanding based on the current code is that they should be values between 0 and 1/97 (since there are 97 languages), with a smaller value indicating higher confidence.
Looks like a value that tells you how certain the engine is that it guessed the correct language for the document. I think generally the closer to 0 the number, the more sure it is, but you should be able to test that by mixing languages together and passing them in to see what values you get out. It allows you to fine tune your program when using langid depending upon what you consider 'close enough' to count as a match.
Ok this is one of those trickier than it sounds questions so I'm turning to stack overflow because I can't think of a good answer. Here is what I want: I need Python to generate a simple a list of numbers from 0 to 1,000,000,000 in random order to be used for serial numbers (using a random number so that you can't tell how many have been assigned or do timing attacks as easily, i.e. guessing the next one that will come up). These numbers are stored in a database table (indexed) along with the information linked to them. The program generating them doesn't run forever so it can't rely on internal state.
No big deal right? Just generate a list of numbers, shove them into an array and use Python "random.shuffle(big_number_array)" and we're done. Problem is I'd like to avoid having to store a list of numbers (and thus read the file, pop one off the top, save the file and close it). I'd rather generate them on the fly. Problem is that the solutions I can think of have problems:
1) Generate a random number and then check if it has already been used. If it has been used generate a new number, check, repeat as needed until I find an unused one. Problem here is that I may get unlucky and generate a lot of used numbers before getting one that is unused. Possible fix: use a very large pool of numbers to reduce the chances of this (but then I end up with silly long numbers).
2) Generate a random number and then check if it has already been used. If it has been used add or subtract one from the number and check again, keep repeating until I hit an unused number. Problem is this is no longer a random number as I have introduced bias (eventually I will get clumps of numbers and you'd be able to predict the next number with a better chance of success).
3) Generate a random number and then check if it has already been used. If it has been used add or subtract another randomly generated random number and check again, problem is we're back to simply generating random numbers and checking as in solution 1.
4) Suck it up and generate the random list and save it, have a daemon put them into a Queue so there are numbers available (and avoid constantly opening and closing a file, batching it instead).
5) Generate much larger random numbers and hash them (i.e. using MD5) to get a smaller numeric value, we should rarely get collisions, but I end up with larger than needed numbers again.
6) Prepend or append time based information to the random number (i.e. unix timestamp) to reduce chances of a collision, again I get larger numbers than I need.
Anyone have any clever ideas that will reduce the chances of a "collision" (i.e. generating a random number that is already taken) but will also allow me to keep the number "small" (i.e. less than a billion (or a thousand million for your europeans =)).
Answer and why I accepted it:
So I will simply go with 1, and hope it's not an issue, however if it is I will go with the deterministic solution of generating all the numbers and storing them so that there is a guarentee of getting a new random number, and I can use "small" numbers (i.e. 9 digits instead of an MD5/etc.).
This is a neat problem, and I've been thinking about it for a while (with solutions similar to Sjoerd's), but in the end, here's what I think:
Use your point 1) and stop worrying.
Assuming real randomness, the probability that a random number has already been chosen before is the count of previously chosen numbers divided by the size of your pool, i.e. the maximal number.
If you say you only need a billion numbers, i.e. nine digits: Treat yourself to 3 more digits, so you have 12-digit serial numbers (that's three groups of four digits – nice and readable).
Even when you're close to having chosen a billion numbers previously, the probability that your new number is already taken is still only 0,1%.
Do step 1 and draw again. You can still check for an "infinite" loop, say don't try more than 1000 times or so, and then fallback to adding 1 (or something else).
You'll win the lottery before that fallback ever gets used.
You could use Format-Preserving Encryption to encrypt a counter. Your counter just goes from 0 upwards, and the encryption uses a key of your choice to turn it into a seemingly random value of whatever radix and width you want.
Block ciphers normally have a fixed block size of e.g. 64 or 128 bits. But Format-Preserving Encryption allows you to take a standard cipher like AES and make a smaller-width cipher, of whatever radix and width you want (e.g. radix 10, width 9 for the parameters of the question), with an algorithm which is still cryptographically robust.
It is guaranteed to never have collisions (because cryptographic algorithms create a 1:1 mapping). It is also reversible (a 2-way mapping), so you can take the resulting number and get back to the counter value you started with.
AES-FFX is one proposed standard method to achieve this.
I've experimented with some basic Python code for AES-FFX--see Python code here (but note that it doesn't fully comply with the AES-FFX specification). It can e.g. encrypt a counter to a random-looking 7-digit decimal number. E.g.:
0000000 0731134
0000001 6161064
0000002 8899846
0000003 9575678
0000004 3030773
0000005 2748859
0000006 5127539
0000007 1372978
0000008 3830458
0000009 7628602
0000010 6643859
0000011 2563651
0000012 9522955
0000013 9286113
0000014 5543492
0000015 3230955
... ...
For another example in Python, using another non-AES-FFX (I think) method, see this blog post "How to Generate an Account Number" which does FPE using a Feistel cipher. It generates numbers from 0 to 2^32-1.
With some modular arithmic and prime numbers, you can create all numbers between 0 and a big prime, out of order. If you choose your numbers carefully, the next number is hard to guess.
modulo = 87178291199 # prime
incrementor = 17180131327 # relative prime
current = 433494437 # some start value
for i in xrange(1, 100):
print current
current = (current + incrementor) % modulo
If they don't have to be random, but just not obviously linear (1, 2, 3, 4, ...), then here's a simple algorithm:
Pick two prime numbers. One of them will be the largest number you can generate, so it should be around one billion. The other should be fairly large.
max_value = 795028841
step = 360287471
previous_serial = 0
for i in xrange(0, max_value):
previous_serial += step
previous_serial %= max_value
print "Serial: %09i" % previous_serial
Just store the previous serial each time so you know where you left off. I can't prove mathmatically that this works (been too long since those particular classes), but it's demonstrably correct with smaller primes:
s = set()
with open("test.txt", "w+") as f:
previous_serial = 0
for i in xrange(0, 2711):
previous_serial += 1811
previous_serial %= 2711
assert previous_serial not in s
s.add(previous_serial)
You could also prove it empirically with 9-digit primes, it'd just take a bit more work (or a lot more memory).
This does mean that given a few serial numbers, it'd be possible to figure out what your values are--but with only nine digits, it's not likely that you're going for unguessable numbers anyway.
If you don't need something cryptographically secure, but just "sufficiently obfuscated"...
Galois Fields
You could try operations in Galois Fields, e.g. GF(2)32, to map a simple incrementing counter x to a seemingly random serial number y:
x = counter_value
y = some_galois_function(x)
Multiply by a constant
Inverse is to multiply by the reciprocal of the constant
Raise to a power: xn
Reciprocal x-1
Special case of raising to power n
It is its own inverse
Exponentiation of a primitive element: ax
Note that this doesn't have an easily-calculated inverse (discrete logarithm)
Ensure a is a primitive element, aka generator
Many of these operations have an inverse, which means, given your serial number, you can calculate the original counter value from which it was derived.
As for finding a library for Galois Field for Python... good question. If you don't need speed (which you wouldn't for this) then you could make your own. I haven't tried these:
NZMATH
Finite field Python package
Sage, although it's a whole environment for mathematical computing, much more than just a Python library
Matrix multiplication in GF(2)
Pick a suitable 32×32 invertible matrix in GF(2), and multiply a 32-bit input counter by it. This is conceptually related to LFSR, as described in S.Lott's answer.
CRC
A related possibility is to use a CRC calculation. Based on the remainder of long-division with an irreducible polynomial in GF(2). Python code is readily available for CRCs (crcmod, pycrc), although you might want to pick a different irreducible polynomial than is normally used, for your purposes. I'm a little fuzzy on the theory, but I think a 32-bit CRC should generate a unique value for every possible combination of 4-byte inputs. Check this. It's quite easy to experimentally check this, by feeding the output back into the input, and checking that it produces a complete cycle of length 232-1 (zero just maps to zero). You may need to get rid of any initial/final XORs in the CRC algorithm for this check to work.
I think you are overestimating the problems with approach 1). Unless you have hard-realtime requirements just checking by random choice terminates rather fast. The probability of needing more than a number of iterations decays exponentially. With 100M numbers outputted (10% fillfactor) you'll have one in billion chance of requiring more than 9 iterations. Even with 50% of numbers taken you'll on average need 2 iterations and have one in a billion chance of requiring more than 30 checks. Or even the extreme case where 99% of the numbers are already taken might still be reasonable - you'll average a 100 iterations and have 1 in a billion change of requiring 2062 iterations
The standard Linear Congruential random number generator's seed sequence CANNOT repeat until the full set of numbers from the starting seed value have been generated. Then it MUST repeat precisely.
The internal seed is often large (48 or 64 bits). The generated numbers are smaller (32 bits usually) because the entire set of bits are not random. If you follow the seed values they will form a distinct non-repeating sequence.
The question is essentially one of locating a good seed that generates "enough" numbers. You can pick a seed, and generate numbers until you get back to the starting seed. That's the length of the sequence. It may be millions or billions of numbers.
There are some guidelines in Knuth for picking suitable seeds that will generate very long sequences of unique numbers.
You can run 1) without running into the problem of too many wrong random numbers if you just decrease the random interval by one each time.
For this method to work, you will need to save the numbers already given (which you want to do anyway) and also save the quantity of numbers taken.
It is pretty obvious that, after having collected 10 numbers, your pool of possible random numbers will have been decreased by 10. Therefore, you must not choose a number between 1 and 1.000.000 but between 1 an 999.990. Of course this number is not the real number but only an index (unless the 10 numbers collected have been 999.991, 999.992, …); you’d have to count now from 1 omitting all the numbers already collected.
Of course, your algorithm should be smarter than just counting from 1 to 1.000.000 but I hope you understand the method.
I don’t like drawing random numbers until I get one which fits either. It just feels wrong.
My solution https://github.com/glushchenko/python-unique-id, i think you should extend matrix for 1,000,000,000 variations and have fun.
I'd rethink the problem itself... You don't seem to be doing anything sequential with the numbers... and you've got an index on the column which has them. Do they actually need to be numbers?
Consider a sha hash... you don't actually need the entire thing. Do what git or other url shortening services do, and take first 3/4/5 characters of the hash. Given that each character now has 36 possible values instead of 10, you have 2,176,782,336 combinations instead of 999,999 combinations (for six digits). Combine that with a quick check on whether the combination exists (a pure index query) and a seed like a timestamp + random number and it should do for almost any situation.
Do you need this to be cryptographically secure or just hard to guess? How bad are collisions? Because if it needs to be cryptographically strong and have zero collisions, it is, sadly, impossible.
I started trying to write an explanation of the approach used below, but just implementing it was easier and more accurate. This approach has the odd behavior that it gets faster the more numbers you've generated. But it works, and it doesn't require you to generate all the numbers in advance.
As a simple optimization, you could easily make this class use a probabilistic algorithm (generate a random number, and if it's not in the set of used numbers add it to the set and return it) at first, keep track of the collision rate, and switch over to the deterministic approach used here once the collision rate gets bad.
import random
class NonRepeatingRandom(object):
def __init__(self, maxvalue):
self.maxvalue = maxvalue
self.used = set()
def next(self):
if len(self.used) >= self.maxvalue:
raise StopIteration
r = random.randrange(0, self.maxvalue - len(self.used))
result = 0
for i in range(1, r+1):
result += 1
while result in self.used:
result += 1
self.used.add(result)
return result
def __iter__(self):
return self
def __getitem__(self):
raise NotImplemented
def get_all(self):
return [i for i in self]
>>> n = NonRepeatingRandom(20)
>>> n.get_all()
[12, 14, 13, 2, 20, 4, 15, 16, 19, 1, 8, 6, 7, 9, 5, 11, 10, 3, 18, 17]
If it is enough for you that a casual observer can't guess the next value, you can use things like a linear congruential generator or even a simple linear feedback shift register to generate the values and keep the state in the database in case you need more values. If you use these right, the values won't repeat until the end of the universe. You'll find more ideas in the list of random number generators.
If you think there might be someone who would have a serious interest to guess the next values, you can use a database sequence to count the values you generate and encrypt them with an encryption algorithm or another cryptographically strong perfect has function. However you need to take care that the encryption algorithm isn't easily breakable if one can get hold of a sequence of successive numbers you generated - a simple RSA, for instance, won't do it because of the Franklin-Reiter Related Message Attack.
Bit late answer, but I haven't seen this suggested anywhere.
Why not use the uuid module to create globally unique identifiers
To generate a list of totally random numbers within a defined threshold, as follows:
plist=list()
length_of_list=100
upbound=1000
lowbound=0
while len(pList)<(length_of_list):
pList.append(rnd.randint(lowbound,upbound))
pList=list(set(pList))
I bumped into the same problem and opened a question with a different title before getting to this one. My solution is a random sample generator of indexes (i.e. non-repeating numbers) in the interval [0,maximal), called itersample. Here are some usage examples:
import random
generator=itersample(maximal)
another_number=generator.next() # pick the next non-repeating random number
or
import random
generator=itersample(maximal)
for random_number in generator:
# do something with random_number
if some_condition: # exit loop when needed
break
itersample generates non-repeating random integers, storage need is limited to picked numbers, and the time needed to pick n numbers should be (as some tests confirm) O(n log(n)), regardelss of maximal.
Here is the code of itersample:
import random
def itersample(c): # c = upper bound of generated integers
sampled=[]
def fsb(a,b): # free spaces before middle of interval a,b
fsb.idx=a+(b+1-a)/2
fsb.last=sampled[fsb.idx]-fsb.idx if len(sampled)>0 else 0
return fsb.last
while len(sampled)<c:
sample_index=random.randrange(c-len(sampled))
a,b=0,len(sampled)-1
if fsb(a,a)>sample_index:
yielding=sample_index
sampled.insert(0,yielding)
yield yielding
elif fsb(b,b)<sample_index+1:
yielding=len(sampled)+sample_index
sampled.insert(len(sampled),yielding)
yield yielding
else: # sample_index falls inside sampled list
while a+1<b:
if fsb(a,b)<sample_index+1:
a=fsb.idx
else:
b=fsb.idx
yielding=a+1+sample_index
sampled.insert(a+1,yielding)
yield yielding
You are stating that you store the numbers in a database.
Wouldn't it then be easier to store all the numbers there, and ask the database for a random unused number?
Most databases support such a request.
Examples
MySQL:
SELECT column FROM table
ORDER BY RAND()
LIMIT 1
PostgreSQL:
SELECT column FROM table
ORDER BY RANDOM()
LIMIT 1