## How to create the most compact mapping n → isprime(n) up to a limit N?

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Naturally, for `bool isprime(number)` there would be a data structure I could query. I define the best algorithm, to be the algorithm that produces a data structure with lowest memory consumption for the range (1, N], where N is a constant. Just an example of what I am looking for: I could represent every odd number with one bit e.g. for the given range of numbers (1, 10], starts at 3: `1110`

The following dictionary can be squeezed more, right? I could eliminate multiples of five with some work, but numbers that end with 1, 3, 7 or 9 must be there in the array of bits.

How do I solve the problem?

There are many ways to do the primality test.

There isn't really a data structure for you to query. If you have lots of numbers to test, you should probably run a probabilistic test since those are faster, and then follow it up with a deterministic test to make sure the number is prime.

You should know that the math behind the fastest algorithms is not for the faint of heart.

How to create the most compact mapping n → isprime(n) up to a , How to create the most compact mapping n → isprime(n) up to a , Naturally, for bool isprime(number) there would be a data structure I could query. I define the  An important special case of a compact mapping is a finite-to-one mapping. Topological properties are stable most often with respect to perfect mappings, which are the most natural analogues of continuous mappings of compacta in the class of all Hausdorff spaces. A product of compact mappings is a compact mapping. References

The fastest algorithm for general prime testing is AKS. The Wikipedia article describes it at lengths and links to the original paper.

If you want to find big numbers, look into primes that have special forms like Mersenne primes.

The algorithm I usually implement (easy to understand and code) is as follows (in Python):

```def isprime(n):
"""Returns True if n is prime."""
if n == 2:
return True
if n == 3:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False

i = 5
w = 2

while i * i <= n:
if n % i == 0:
return False

i += w
w = 6 - w

return True
```

It's a variant of the classic `O(sqrt(N))` algorithm. It uses the fact that a prime (except 2 and 3) is of form `6k - 1` or `6k + 1` and looks only at divisors of this form.

Sometimes, If I really want speed and the range is limited, I implement a pseudo-prime test based on Fermat's little theorem. If I really want more speed (i.e. avoid O(sqrt(N)) algorithm altogether), I precompute the false positives (see Carmichael numbers) and do a binary search. This is by far the fastest test I've ever implemented, the only drawback is that the range is limited.

Compact space, In mathematics, more specifically in general topology, compactness is a property that For any subset A of Euclidean space Rn, A is compact if and only if it is closed to the above statements, the pre-image of a compact space under a proper map is compact. Not logged in; Talk · Contributions · Create account · Log in  The compacity W p m + k → W q k is then easy to prove going back to the definition of a compact mapping and the completude of these spaces (cf. p. 487). The Kondrakov theorem does not hold if U is not bounded. For instance H 1 (R n) is not compactly embedded in L 2 (R n) [for more on compact embeddings, cf. Problem VI 6, H s,δ spaces].

The best method, in my opinion, is to use what's gone before.

There are lists of the first `N` primes on the internet with `N` stretching up to at least fifty million. Download the files and use them, it's likely to be much faster than any other method you'll come up with.

If you want an actual algorithm for making your own primes, Wikipedia has all sorts of good stuff on primes here, including links to the various methods for doing it, and prime testing here, both probability-based and fast-deterministic methods.

There should be a concerted effort to find the first billion (or even more) primes and get them published on the net somewhere so people can stop doing this same job over and over and over and ... :-)

Permutation, In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into The number of permutations of n distinct objects is n factorial, usually written This form is more compact, and is common in elementary combinatorics and Furthermore, Foata's mapping takes an n-permutation with k​-weak  Theorem 2. If f(X) = Y is a compact mapping and Y is a locally compact Hausdorff space, then f is closed and point inverses are compact. That point inverses are compact is obvious. Let F be a closed subset of X, and suppose yEf(F) and is an ac-cumulation point of/(F). Since Y is locally compact, an open set G

```bool isPrime(int n)
{
// Corner cases
if (n <= 1)  return false;
if (n <= 3)  return true;

// This is checked so that we can skip
// middle five numbers in below loop
if (n%2 == 0 || n%3 == 0) return false;

for (int i=5; i*i<=n; i=i+6)
if (n%i == 0 || n%(i+2) == 0)
return false;

return true;
}
```

this is just c++ implementation of above AKS algorithm

US7577558B2, A memory mapping system for providing compact mapping between The memory mapping system can compactly map contents from one or more first appears to be created in the processing devices by calculating the outputs of the user as compactly mapping any suitable number of first memory systems 200A-​N with  Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

I compared the efficiency of the most popular suggestions to determine if a number is prime. I used `python 3.6` on `ubuntu 17.10`; I tested with numbers up to 100.000 (you can test with bigger numbers using my code below).

This first plot compares the functions (which are explained further down in my answer), showing that the last functions do not grow as fast as the first one when increasing the numbers. And in the second plot we can see that in case of prime numbers the time grows steadily, but non-prime numbers do not grow so fast in time (because most of them can be eliminated early on). Here are the functions I used:

1. this answer and this answer suggested a construct using `all()`:

```def is_prime_1(n):
return n > 1 and all(n % i for i in range(2, int(math.sqrt(n)) + 1))
```
2. This answer used some kind of while loop:

```def is_prime_2(n):
if n <= 1:
return False
if n == 2:
return True
if n == 3:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False

i = 5
w = 2
while i * i <= n:
if n % i == 0:
return False
i += w
w = 6 - w

return True
```
3. This answer included a version with a `for` loop:

```def is_prime_3(n):
if n <= 1:
return False

if n % 2 == 0 and n > 2:
return False

for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False

return True
```
4. And I mixed a few ideas from the other answers into a new one:

```def is_prime_4(n):
if n <= 1:          # negative numbers, 0 or 1
return False
if n <= 3:          # 2 and 3
return True
if n % 2 == 0 or n % 3 == 0:
return False

for i in range(5, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False

return True
```

Here is my script to compare the variants:

```import math
import pandas as pd
import seaborn as sns
import time
from matplotlib import pyplot as plt

def is_prime_1(n):
...
def is_prime_2(n):
...
def is_prime_3(n):
...
def is_prime_4(n):
...

default_func_list = (is_prime_1, is_prime_2, is_prime_3, is_prime_4)

def assert_equal_results(func_list=default_func_list, n):
for i in range(-2, n):
r_list = [f(i) for f in func_list]
if not all(r == r_list for r in r_list):
print(i, r_list)
raise ValueError
print('all functions return the same results for integers up to {}'.format(n))

def compare_functions(func_list=default_func_list, n):
result_list = []
n_measurements = 3

for f in func_list:
for i in range(1, n + 1):
ret_list = []
t_sum = 0
for _ in range(n_measurements):
t_start = time.perf_counter()
is_prime = f(i)
t_end = time.perf_counter()

ret_list.append(is_prime)
t_sum += (t_end - t_start)

is_prime = ret_list
assert all(ret == is_prime for ret in ret_list)
result_list.append((f.__name__, i, is_prime, t_sum / n_measurements))

df = pd.DataFrame(
data=result_list,
columns=['f', 'number', 'is_prime', 't_seconds'])
df['t_micro_seconds'] = df['t_seconds'].map(lambda x: round(x * 10**6, 2))
print('df.shape:', df.shape)

print()
print('', '-' * 41)
print('| {:11s} | {:11s} | {:11s} |'.format(
'is_prime', 'count', 'percent'))
df_sub1 = df[df['f'] == 'is_prime_1']
print('| {:11s} | {:11,d} | {:9.1f} % |'.format(
'all', df_sub1.shape, 100))
for (is_prime, count) in df_sub1['is_prime'].value_counts().iteritems():
print('| {:11s} | {:11,d} | {:9.1f} % |'.format(
str(is_prime), count, count * 100 / df_sub1.shape))
print('', '-' * 41)

print()
print('', '-' * 69)
print('| {:11s} | {:11s} | {:11s} | {:11s} | {:11s} |'.format(
'f', 'is_prime', 't min (us)', 't mean (us)', 't max (us)'))
for f, df_sub1 in df.groupby(['f', ]):
col = df_sub1['t_micro_seconds']
print('|{0}|{0}|{0}|{0}|{0}|'.format('-' * 13))
print('| {:11s} | {:11s} | {:11.2f} | {:11.2f} | {:11.2f} |'.format(
f, 'all', col.min(), col.mean(), col.max()))
for is_prime, df_sub2 in df_sub1.groupby(['is_prime', ]):
col = df_sub2['t_micro_seconds']
print('| {:11s} | {:11s} | {:11.2f} | {:11.2f} | {:11.2f} |'.format(
f, str(is_prime), col.min(), col.mean(), col.max()))
print('', '-' * 69)

return df
```

Running the function `compare_functions(n=10**5)` (numbers up to 100.000) I get this output:

```df.shape: (400000, 5)

-----------------------------------------
| is_prime    | count       | percent     |
| all         |     100,000 |     100.0 % |
| False       |      90,408 |      90.4 % |
| True        |       9,592 |       9.6 % |
-----------------------------------------

---------------------------------------------------------------------
| f           | is_prime    | t min (us)  | t mean (us) | t max (us)  |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_1  | all         |        0.57 |        2.50 |      154.35 |
| is_prime_1  | False       |        0.57 |        1.52 |      154.35 |
| is_prime_1  | True        |        0.89 |       11.66 |       55.54 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_2  | all         |        0.24 |        1.14 |      304.82 |
| is_prime_2  | False       |        0.24 |        0.56 |      304.82 |
| is_prime_2  | True        |        0.25 |        6.67 |       48.49 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_3  | all         |        0.20 |        0.95 |       50.99 |
| is_prime_3  | False       |        0.20 |        0.60 |       40.62 |
| is_prime_3  | True        |        0.58 |        4.22 |       50.99 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_4  | all         |        0.20 |        0.89 |       20.09 |
| is_prime_4  | False       |        0.21 |        0.53 |       14.63 |
| is_prime_4  | True        |        0.20 |        4.27 |       20.09 |
---------------------------------------------------------------------
```

Then, running the function `compare_functions(n=10**6)` (numbers up to 1.000.000) I get this output:

```df.shape: (4000000, 5)

-----------------------------------------
| is_prime    | count       | percent     |
| all         |   1,000,000 |     100.0 % |
| False       |     921,502 |      92.2 % |
| True        |      78,498 |       7.8 % |
-----------------------------------------

---------------------------------------------------------------------
| f           | is_prime    | t min (us)  | t mean (us) | t max (us)  |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_1  | all         |        0.51 |        5.39 |     1414.87 |
| is_prime_1  | False       |        0.51 |        2.19 |      413.42 |
| is_prime_1  | True        |        0.87 |       42.98 |     1414.87 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_2  | all         |        0.24 |        2.65 |      612.69 |
| is_prime_2  | False       |        0.24 |        0.89 |      322.81 |
| is_prime_2  | True        |        0.24 |       23.27 |      612.69 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_3  | all         |        0.20 |        1.93 |       67.40 |
| is_prime_3  | False       |        0.20 |        0.82 |       61.39 |
| is_prime_3  | True        |        0.59 |       14.97 |       67.40 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_4  | all         |        0.18 |        1.88 |      332.13 |
| is_prime_4  | False       |        0.20 |        0.74 |      311.94 |
| is_prime_4  | True        |        0.18 |       15.23 |      332.13 |
---------------------------------------------------------------------
```

I used the following script to plot the results:

```def plot_1(func_list=default_func_list, n):
df_orig = compare_functions(func_list=func_list, n=n)
df_filtered = df_orig[df_orig['t_micro_seconds'] <= 20]
sns.lmplot(
data=df_filtered, x='number', y='t_micro_seconds',
col='f',
# row='is_prime',
markers='.',
ci=None)

plt.ticklabel_format(style='sci', axis='x', scilimits=(3, 3))
plt.show()
```

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• Your request is a little vague. You give a signature that tests a single number but then ask for a data structure of (1,N]. Do you want an algorithm that generates a dictionary<int,bool> or just a one-shot function that checks if a single number is prime?
• @Michael Sorry, that is the best description I could comeup with. What I am looking is excactly as you are saying: a boolean dictionary. I would like to minimize the space of the dictionary. Thanks :)
• If that's what you're looking for it's been asked already: stackoverflow.com/questions/1032427/…
• You would need to Ask the NSA
• related: Miller–Rabin primality test in Python
• Miller-Rabin is a popular fast probabilistic test to start out with.
• Two questions: Can you explain better what the variables `i` and `w` are, and what is meant by the form `6k-1` and `6k+1`? Thank you for your insight and the code sample (which I'm trying to understand)
• @Freedom_Ben Here you go, quora.com/…
• Wouldn't it be better to calculate the `sqrt` of `n` once and comparing `i` to it, rather than calculating `i * i` every cycle of the loop?
• @Dschoni ... but you can't fit the fastest implementation in the comment fields here to share with us?
• It fails for number 1 :(
• @hamedbh: Interesting. Have you tried to download those files? It appears they don't exist.
• Not yet, I'm afraid: I was just looking quickly during my lunch break. I'll delete that link in case there's anything malicious about it. So sorry, I really should have checked it first.
• Such lists do exist. Ive seen them years ago but never cared to download them. The truth is, they take up a lot of space (relatively speaking), and shouldn't be included in programs one sells or distributes. Furthermore, they will always and forever be incomplete. It kind of does make more sense to test each number that comes up in practice during a programs use, as far fewer will be tested that way than the length of any list you might own. Also, I think pax doesnt realize the purpose of prime algorithms, most of the time, is to test efficiency/speed rather than actually finding primes.