isPrime Function for Python Language

Question

So I was able to solve this problem with a little bit of help from the internet and this is what I got  def isPrime n       for i in range 2 int n  0 5  1           if n i  0              return False              return True  But my question really is how to do it  but WHY  I understand that 1 is not considered a  quot prime quot  number even though it is  and I understand that if it divides by ANYTHING within the range it is automatically not a prime thus the return False statement  but my question is what role does the squar-rooting the  quot n quot  play here  P s  I am very inexperienced and have just been introduced to programming a month ago

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https   www youtube com watch v Vxw1b8f yts amp t 3384s    Avinash Jain   for i in range 2 5003       j   2     c   0     while j  lt  i          if i   j    0              c   1             j   j   1         else              j   j   1     if c    0          print str i      is a prime number       else          c   0

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The question was asked a bit ago  but I have a shorter solution for you  def isNotPrime Number       return 2 not in  Number 2  Number Number    The math operation will mostly return 2 if the number is a prime  instead of 2  But if 2 is the given number  it is appended to the list where we are looking into   Examples   2 5 32    32 5 2 2 7 128   128 7 2 2 11 2048 2048 11 2   Counter examples   2 341 341 2   but 341  11 31 2 561 561 2   but 561  3 11 17 2 645 645 2   but 645  3 5 43   isNotPrime   reliably returns True if Number is not prime though

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Srsly guys    Why so many lines of code for a simple method like this  Here s my solution   def isPrime a       div   a - 1     res   True     while div  gt  1           if a   div    0              res   False         div   div - 1     return res

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With n   5  you are not squaring n  but taking the square root   Consider the number 20  the integer factors are 1  2  4  5  10  and 20  When you divide 20 by 2 and get 10  you know that it is also divisible by 10  without having to check  When you divide it by 4 and get 5  you know it is divisible by both 4 and 5  without having to check for 5   After reaching this halfway point in the factors  you will have no more numbers to check which you haven t already recognized as factors earlier  Therefore  you only need to go halfway to see if something is prime  and this halfway point can be found by taking the number s square root   Also  the reason 1 isn t a prime number is because prime numbers are defined as having 2 factors  1 and itself  i e 2 is 1 2  3 is 1 3  5 is 1 5  But 1  1 1  only has 1 factor  itself  Therefore  it doesn t meet this definition

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def is prime n   if  n  2 or n  3   return True if n lt  1 or n 2  0 or n 3  0    return False for i in range 6 int  n  0 5     2 6       if n  i-1   0 or n  i 1   0           return False return True

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def is prime x         if x  lt  2            return False       for n in range 2   x  - 1             if x   n    0                return False       return True

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def isPrime num div 2       if num  div           return True     elif num   div    0           return False     else          return isPrime num div 1                                                   EDITED  def is prime num  div   2       if num    div  return True     elif num   div    0  return False     elif num    1  return False     else  return is prime num  div   1

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isPrime lambda x  all x   i    0 for i in range int x  0 5  1  2      and here goes how to use it  isPrime 2     False isPrime 5     True isPrime 7     True   To find all primes you might use    filter isPrime  range 4000  2     5    gt   2  3  5  7  11    Note that 5  in this case  denotes number of prime numbers to be found and 4000 max range of where primes will be looked for

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This is my np way    def is prime x       if x  lt  4          return True     if all   x  gt  2    x   2    0             return False     else          return np array   map lambda y    x   y     0  sum    np arange 1  x   1     sum      2   Here s the performance    timeit is prime 2   timeit is prime int 1e3    timeit is prime 5003   10000 loops  best of 3  31 1   s per loop 10000 loops  best of 3  33   s per loop 10 loops  best of 3  74 2 ms per loop

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I don t know if I am late but I will leave this here to help someone in future   We use the square root of  n  i e int n  0 5  to reduce the range of numbers your program will be forced to calculate   For example  we can do a trial division to test the primality of 100  Let s look at all the divisors of 100   2  4  5  10  20  25  50 Here we see that the largest factor is 100 2   50  This is true for all n  all divisors are less than or equal to n 2  If we take a closer look at the divisors  we will see that some of them are redundant  If we write the list differently   100   2    50   4    25   5    20   10    10   20    5   25    4   50    2 the redundancy becomes obvious  Once we reach 10  which is v100  the divisors just flip around and repeat  Therefore  we can further eliminate testing divisors greater than vn   Take another number like 16   Its divisors are  2 4 8  16   2   8  4   4  8   2   You can note that after reaching 4  which is the square root of 16  we repeated 8   2 which we had already done as 2 8  This pattern is true for all numbers   To avoid repeating ourselves  we thus test for primality up to the square root of a number n   So we convert the square root to int because we do not want a range with floating numbers   Read the primality test on wikipedia for more info

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Finding the square root of the number is for efficiency  eg  if I am trying to find the factors of 36  the highest number that can be multiplied by itself to form 36 is 6  7 7   49    therefore every factor of 36 has to be multiplied by 6 or a lesser number

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def is prime x         if x lt 2            return False       elif x    2            return True       else            for n in range 2  x                 if x n  0                    return False           return True

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I have a new solution which I think might be faster than any of the mentioned Function in Python  It s based on the idea that  N D   R for any arbitrary number N  the least possible number to divide N  if not prime  is D 2 and the corresponding result R is  N 2   highest    As D goes bigger the result R gets smaller ex  divide by D   3 results R    N 3  so when we are checking if N is divisible by D we are also checking if it s divisible by R  so as D goes bigger and R goes smaller till  D    R    square root N    then we only need to check numbers from 2 to sqrt N  another tip to save time  we only need to check the odd numbers as it the number is divisible by any even number it will also be divisible by 2   so the sequence will be 3 5 7 9        sqrt N    import math def IsPrime  n        if  n  lt   1 or n   2    0  return False     if n    2 return True     for i in range 3 int math sqrt n   1 2           if  n   i     0              return False     return True

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This method will be slower than the the recursive and enumerative methods here  but uses Wilson s theorem  and is just a single line   from math import factorial  def is prime x       return factorial x - 1     x    x - 1

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Here is mine   import math  def is prime num        if num   2    0 and num  gt  2          return False     for i in range 3  int math sqrt num     1  2           if num   i    0              return False     return True

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def is prime x       if x  lt  2          return False     elif x    2          return True       for n in range 2  x           if x   n   0              return False     return True

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The number 1 is a special case which is considered neither prime nor composite  For more info visit   http   mathworld wolfram com PrimeNumber html  And   n  0 5  --  This will give us the  square root  of  n   As it is  n raised to the power 0 5 or 1 2   And WHY do we do that  Take for example the number 400  We can represent it in the form a b  1 400   400 2 200   400 4 100   400 5 80   400 8 50   400 10 40   400 16 25   400 20 20   400 25 16   400 40 10   400 50 8   400 80 5   400 100 4   400 200 2   400 400 1   400   Square root of 400 is 20  and we can see that we only need to check the divisibility till 20 because  as  a  reaches 20  b  starts decreasing    So  ultimately we are checking divisibility with the numbers less than the square root

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Pretty simple   def prime x     if x    1      return False   else      for a in range 2 x         if x   a    0          return False   return True

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Every code you write should be efficient For a beginner like you the easiest way is to check the divisibility of the number  n  from 2 to  n-1   This takes a lot of time when you consider very big numbers  The square root method helps us make the code faster by less number of comparisons  Read about complexities in Design and Analysis of Algorithms

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Of many prime number tests floating around the Internet  consider the following Python function   def is prime n     if n    2 or n    3  return True   if n  lt  2 or n 2    0  return False   if n  lt  9  return True   if n 3    0  return False   r   int n  0 5      since all primes  gt  3 are of the form 6n    1     start with f 5  which is prime      and test f  f 2 for being prime     then loop by 6     f   5   while f  lt   r      print   t  f      if n   f    0  return False     if n    f 2     0  return False     f    6   return True       Since all primes   3 are of the form 6n    1  once we eliminate that n is     not 2 or 3  which are prime  and  not even  with n 2  and  not divisible by 3  with n 3  then we can test every 6th n    1    Consider the prime number 5003   print is prime 5003    Prints    5  11  17  23  29  35  41  47  53  59  65 True   The line r   int n  0 5  evaluates to 70  the square root of 5003 is 70 7318881411 and int   truncates this value   Consider the next odd number  since all even numbers other than 2 are not prime  of 5005  same thing prints    5 False   The limit is the square root since x y    y x The function only has to go 1 loop to find that 5005 is divisible by 5 and therefore not prime  Since 5 X 1001    1001 X 5  and both are 5005   we do not need to go all the way to 1001 in the loop to know what we know at 5     Now  let s look at the algorithm you have   def isPrime n       for i in range 2  int n  0 5  1           if n   i    0              return False      return True   There are two issues    It does not test if n is less than 2  and there are no primes less than 2  It tests every number between 2 and n  0 5 including all even and all odd numbers  Since every number greater than 2 that is divisible by 2 is not prime  we can speed it up a little by only testing odd numbers greater than 2    So     def isPrime2 n       if n  2 or n  3  return True     if n 2  0 or n lt 2  return False     for i in range 3  int n  0 5  1  2       only odd numbers         if n i  0              return False          return True   OK -- that speeds it up by about 30   I benchmarked it      The algorithm I used is prime is about 2x times faster still  since only every 6th integer is looping through the loop    Once again  I benchmarked it      Side note  x  0 5 is the square root    gt  gt  gt  import math  gt  gt  gt  math sqrt 100   100  0 5 True     Side note 2  primality testing is an interesting problem in computer science

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def is prime n       n abs n      if n lt 2      Numbers less than 2 are not prime numbers         return  False      elif n  2   2 is a prime number         return  True      else          for i in range 2 n     Highlights range numbers that can t be  a factor of prime number n               if n i  0                  return  False   if any of these numbers are factors of n  n is not a prime number     return  True    This is to affirm that n is indeed a prime number after passing all three tests

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Implemented a pseudocode  https   en wikipedia org wiki Primality test  in python  hope this help     original pseudocode https   en wikipedia org wiki Primality test def isPrime n         Corner Cases     if  n lt   1   return False     elif  n lt   3   return True     elif  n 2    0 or n 3    0   return False      i   5     while i i lt  n          if  n i  0 or n  i 2   0   return False         i    6      return True    timeit isPrime 800

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def fun N   prime test if N gt 1       for   in xrange 5           Num randint 1 N-1          if pow Num N-1 N   1              return False     return True return False   True if the number is prime otherwise false

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This is my method   import math  def isPrime n        Returns True if n is prime  False if n is not prime  Will not work if n is 0 or 1         Make sure n is a positive integer     n   abs int n          Case 1  the number is 2  prime      if n    2  return True        Case 2  the number is even  not prime      if n   2    0  return False        Case 3  the number is odd  could be prime or not         Check odd numbers less than the square root for possible factors      r   math sqrt n      x   3      while x  lt   r          if n   x    0  return False    A factor was found  so number is not prime         x    2   Increment to the next odd number        No factors found  so number is prime       return True    To answer the original question  n  0 5 is the same as the square of root of n  You can stop checking for factors after this number because a composite number will always have a factor less than or equal to its square root  This is faster than say just checking all of the factors between 2 and n for every n  because we check fewer numbers  which saves more time as n grows

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It was an exercise in codecademy and that s how i passed it below         def is prime x            If number x  is evenly divided by following dividers then number x  is not prime      divider    2  3  5  7         An empty list to be able to check whether number x  is evenly divided       remainder             exceptions for numbers 1 2 3 5 7      if x  lt  2          return False     if x in divider          return True     else          for nums in divider              remainder append x   nums          if 0 in remainder              return False         else              return True

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No floating point operations are done below  This is faster and will tolerate higher arguments  The reason you must go only to the square-root is that if a number has a factor larger than its square root  it also has a factor smaller than it   def is prime n           pre-condition  n is a nonnegative integer     post-condition  return True if n is prime and False otherwise         if n  lt  2            return False      if n   2    0                        return n    2    return False     k   3     while k k  lt   n           if n   k    0               return False          k    2     return True

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int n  0 5  is the floor value of sqrt n  which you confused with power 2 of n  n  2   If n is not prime  there must be two numbers 1  lt  i  lt   j  lt  n such that  i   j   n    Now  since sqrt n    sqrt n    n assuming one of i j is bigger than  or equals to  sqrt n  - it means that the other one has to be smaller than  or equals to  sqrt n    Since that is the case  it s good enough to iterate the integer numbers in the range   2  sqrt n    And that s exactly what the code that was posted is doing   If you want to come out as a real smartass  use the following one-liner function   import re def is prime n           return not re match r  1     11    1    n  1     An explanation for the  magic regex  can be found here

[python] isPrime Function for Python Language

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