Exercise for Hedy and Andy

Check if it is prime We should come back to this one over and over again. It is a hard problem.



In [1]:

    
def prime_checker1(x):
    '''decide if an integer is a prime'''
    for i in range(2,x):
        if x%i==0: return(False)
    
    return(True)



In [2]:

    
def prime_checker2(x):
    '''decide if an integer is a prime'''
    a=True
    for i in range(2,x):
        if x%i==0:
            a=False
    return(a)



In [21]:

    
def prime_checker3(x):
    '''WRONG ALGORITHM--decide if an integer is a prime'''
    a=True
    for i in range(2,x):
        if x%i==0:
            a=False
        else:
            a=True
    return(a)



In [12]:

    
print prime_checker1(272)
print prime_checker2(272)
print prime_checker3(272)









    



False
False
True



In [38]:

    
#Tester:
a=True
b=1
for i in range(1,1000):
    if prime_checker1(i)<>prime_checker3(i):
        a=False
        b=i
if a:
    print "OK"
else:
    print "Error-----last one: ", b,prime_checker1(b),prime_checker3(b)









    



Error-----last one:  999 False True



In [3]:

    
all(prime_checker1(i)==prime_checker2(i) for i in range(1,10000))









    Out[3]:





True



In [13]:

    
from timeit import timeit
print timeit(lambda:prime_checker1(177372),number= 1000)
print timeit(lambda:prime_checker2(177372),number= 1000)









    



2.73137116432
19.0169608593



In [11]:

    
prime_checker1(17737)









    Out[11]:





True

using While loop



In [18]:

    
def prime_checker4(x):
    a=2
    while a< x:
        if x%a==0: return(False)
        a=a+1
    return True



In [23]:

    
all(prime_checker1(i)==prime_checker4(i) for i in range(1,10000))









    Out[23]:





True



In [28]:

    
print timeit(lambda:prime_checker1(17777),number= 10000)
print timeit(lambda:prime_checker2(17777),number= 10000)
print timeit(lambda:prime_checker4(17777),number= 10000)









    



1.87202000618
15.4128699303
0.0714960098267



In [ ]: