We Keep Coding

Iterate through all list orders

So I have a set of 11 variables that I want to assign the numbers 1 to 11 using python. I want to then iterate through all combinations of these assignments and check they complete a series of tests. The 6 tests are such that 6 different combinations of the variables (including a separate 12th variable that retains a constant value (12)) are tested to see if they add to 26. For a permutation to be successful, all combinations must add to 26.
To simplify, say it was just 3 variables assigned the numbers 1-3 I want the program to output a=1,b=2,c=3 then to check this against the criteria before changing the order to a=1,b=3,c=2. Check then change: a=2,b=1,c=3 check and change etc until all combinations have been checked.
I first considered using a list to store the numbers 1 to 11 then just randomising the list order using the shuffle function. I’m not sure how else to iterate through the combinations. The random nature of shuffle would eventually do the job but it would be very slow and certainly not an elegant solution.
Thanks for any help in advance :)

I have since realised after looking through the answers and links provided that what I actually need here is the itertools.permutation function. Since this particular problem requires all permutations of the list of numbers 1-11 and not just combinations of these numbers in a list of a smaller size and the ordering matters.The following link was useful in realising the solution.
https://www.tutorialspoint.com/generate-all-permutation-of-a-set-in-python
import itertools
numbers = list(range(1,12))
permutations=list(itertools.permutations(numbers,11))
The result of this code provides all possible orders of the numbers 1-11 as a list of tuples which I can then iterate through to check if they pass the tests.

The itertools package includes a combinations function that may be helpful for you. For example:
import itertools
numbers = range(1, 12)
for (a,b,c) in itertools.combinations(numbers, 3):
# do something

How can I write a python code faster so that is passes the long given test cases?

def solution(worth,target):
limit=0
count=0
if target>sum(worth):
return -1
for buck in worth:
if limit<target:
limit+=buck
count+=1
return count
T=int(input())
worth=list(map(int,input().split()))
Q=int(input())
answer=[]
for i in range(Q):
target=int(input())
answer.append(solution(worth,target))
for i in answer:
print(i)
I want to omit something in this code so that it generates faster output.
T is the number of transactions made in a day
next line has T integers as input i.e It is worth of each transaction made
Next line has Q queries
Next q lines, the target set for each day

An examples would be nice but two things: Currently you go through all transactions for every query (twice actually since you use the sum command each time). Collect the queries first since you don't give outputs before having all of them anyway. Then sort the queries by size, smallest first. Then when summing up the transactions you can do the queries in one go by answering each when its value is reached. All remaining ones can be answered with -1 without needing the sum. (If you need the output in the right order simply safe the order.)
The second thing is that once the last query has been fulfilled, end the loop, no need to go through all remaining items. Whether that comes up depends on the queries of course.

The fastest solution would be to sort a cumulative sum version of the worth array and use bisection to directly find the number of transaction for each query. See accumulate in the itertools module and bisect_right in the bisect module.
A simpler (but not as efficient) improvement would be to compute the total worth once before the Q loop and pass it as a totalWorth parameter to the solution function. Using the totalWorth parameter instead of sum(worth) in the target>sum(worth) condition would save significant time if worth has many entries.

Prime numbers with python [duplicate]

This question already has answers here:
Sieve of Eratosthenes - Finding Primes Python
(22 answers)
Closed 7 years ago.
I'm fairly new to programming and I decided to do some exercises to improve my abilities. I'm stuck with an exercise: "Find the sum of all the primes below two million." My code is too slow.
Initially, I tried to solve as a normal prime problem and ended up with this:
sum = 2 + 3
for i in range (5, 2000000, 2):
for j in range (3, i, 2):
if i%j == 0:
break
else:
sum += i
print(sum)
In this way, all even numbers would be excluded from the loop. But it did not solve my problem. The magnitude here is really big.
So I tried to understand what was happening with this code. I have a loop inside a loop, and the loop inside the loop runs the index of the outside loop times (not quite that because the list doesn't start from 0), right? So, when I try to find prime numbers under 20, it runs the outside loop 8 times, but the inside loop 60 (I don't know if this math is correct, as I said, I'm quite knew to programming). But when I use it with 2,000,000, I'm running the inside loop something like 999,993,000,012 times in total, and that is madness.
My friend told me about the Sieve of Eratosthenes, and I tried to create a new code:
list = [2]
list.extend(range(3, 2000000, 2))
for i in list:
for j in list:
if j%i == 0 and j > i:
list.remove(j)
print(sum(list))
And that's what I achieved trying to simulate the sieve (Ignoring even numbers helped). it's a lot faster (with the other code, it would take a long time to find primes under 200,000, and with this new one I can do it) but it is not enough to compute 2,000,000,000 in a reasonable time. The code is running in the background since I started to write, and still nothing. I don't know how many times this thing is looping and I am too tired to think about it now.
I came here to ask for help. Why is it so slow? What should I learn/read/do to improve my code? Is there any other method more efficient than this sieve? Thank you for your time.

Because list.remove is a O(n) operation, and you're doing it a lot. And you're not performing a true sieve, just trial division in disguise; you're still doing all the remainder testing you did in the original code.
A Sieve of Eratosthenes typically is implemented with an array of flags; in the simplest form, each index corresponds to the same number, and the value is initially True for all indices but 0 and 1. You iterate along, and when you find a True value, you set all indices that are multiples of it to False. This means the work is sequential addition, not multiplication, not division (which are much more expensive.

Tuples birthday paradox

Tuples is n
The birthday problem equation is this:
Question:
For n = 200, write an algorithm (in Python) for enumerating the number of tuples in the sample space that satisfy the condition that at least two people have the same birthday. (Note that your algorithm will need to scan each tuple)
import itertools
print(list(itertools.permutations([0,0,0]))
I am wondering for this question how do I insert a n into this?

"how to get n in there":
n = 200
space = itertools.permutations(bday_pairs, n)
I've left out a couple parts of your code:
itertools returns a list; you don't need to coerce it.
Printing this result is likely not what you want with n = 200; that's a huge list.
Now, all you need to do is to build bday_pairs, the list of all possible pairs of birthdays. For convenience, I suggest that you use the integers 1-365. Since you haven't attacked that part of the problem at all, I'll leave that step up to you.
You still need to do the processing to count the sets with at least one matching birthday, another part of the problem you haven't attacked. However, I trust that the above code solves your stated problem?

Generating non-repeating random numbers in Python

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