python algorithm - find prime numbers less than 100,000

1, define the calculation using prime number

# - * - Coding: UTF-. 8 - * - 
# Version: python3.7 

'' ' 
   @ File: prime_number 
   @ author: zhangyangyang 
   @ Create: 2020/3/22 
   @ the remark: ' 
'' 

# version1 for 

Import datetime     # Import module computational efficiency
 Start = datetime.datetime.now () 
COUNT = 0
 for X in range (2,100000):      # find a prime number within the specified range 
    for I in range (2, X):       # divided by itself other than 1 and number 
        IF X% I == 0:
             BREAK 
    the else :
         #Print (X) 
        COUNT =. 1 + 
Delta = (datetime.datetime.now () - Start) .total_seconds ()     # total_seconds () Total number of seconds 
Print ( ' COUNT = ' , COUNT, ' Delta = ' , Delta) # Wall time on
 
the results: 
COUNT = = 148.146291 # 9592 Delta extremely inefficient

 

2, optimization 1: is calculated, the critical value square value

# Version2: Optimization 

Import datetime     # import modules computational efficiency 
Start = datetime.datetime.now () 
COUNT = 0
 for X in Range (2,100000 ):
     for I in Range (2, int (X + 0.5 **. 1)) :    # optimization 1, tested: critical value square value 
        IF X% I == 0:
             BREAK 
    the else :
         # Print (X) 
        COUNT = 1 + 
Delta = (datetime.datetime.now () - Start) .total_seconds ( )     # total_seconds () total number of seconds 
Print ( ' COUNT = ', COUNT, ' Delta = ' , Delta) 

execution results: 
COUNT = 9592 = 1.084154 # Delta efficiency greatly improved

 

3, optimization 2: the even number greater than two full engagement

# Version3: Optimization + 

Import datetime     # import modules computational efficiency 
Start = datetime.datetime.now () 
COUNT =. 1
 # Print (2) # 3 from the beginning, their printing 2 
for X in Range (3,100000,2):                    # optimization 2: 3 from the beginning of the odd 
    # for i in range (3, int (x ** 0.5 +. 1)): optimization # 3: odd and not modulo 2 
    for I in Range (3, int (X ** 0.5 ) + 2):    # optimization: i.e., they do not and the even modulus 
        IF X% I == 0:
             BREAK 
    the else :
         # Print (X) 
        COUNT =. 1 + 
Delta= (Datetime.datetime.now () - Start) .total_seconds ()     # total_seconds () Total number of seconds 
Print ( ' COUNT = ' , COUNT, ' Delta = ' , Delta)     # Wall time 

execution result of: 
COUNT = 9592 delta = 0.553471 # to further improve performance

 

4, Optimization 3: 5 are all multiple composite number, excluding multiples of 5

# Version4: Optimization ++ 

Import datetime     # import modules computational efficiency 
Start = datetime.datetime.now () 
COUNT =. 1
 # Print (2) # 3 from the beginning, their printing 2 
for X in Range (3,100000,2):                    # optimization 2: 3 starting from the odd 
    IF X> 10 and X 5% == 0:
         Continue                               # optimization: 5 excluding multiples of 
    # for I in Range (3, int (X + 0.5 **. 1)): # optimization 3: odd and not modulo 2 
    for I in Range (3, int (X ** 0.5) + 1,2):    # optimization: i.e., they do not and the even modulus 
        IF X% I == 0:
             BREAK
    the else :
         # Print (X) 
        COUNT =. 1 + 
Delta = (datetime.datetime.now () - Start) .total_seconds ()     # total_seconds () Total number of seconds 
Print ( ' COUNT = ' , COUNT, ' Delta = ' , Delta )     # time on the wall of 

the results: 
COUNT = 9592 = 0.493866 Delta

 

5, the number of quality applications:

　　Application in the field of cryptography, have to use a large prime number