def show_diff(n):
p=1
num_trues=0
num_falses=0
while p <= n:
num=generateprimes.genprime(p)
num1=generateprimes.genprime(p+1)
#num_log=math.log(num)
#num_log=int(abs(num_log))
num1_log=math.log(num1)
num1_log=int(round(num1_log))
diff=num1-num
isit=False
#num1_log_p=num1/num1_log
if diff <= num1_log:
isit=True
num_trues+=1
else:
isit=False
num_falses+=1
print num1, "-", num, "=", diff, "-is diff <= log(", num1, ")", "...", isit
p+=1
num_falses=float(num_falses)
num_trues=float(num_trues)
total=num_falses+num_trues
percent_true=num_trues/total*100
percent_false=num_falses/total*100
print "Total 'True's: ", num_trues, " ", percent_true,"%"
print "Total 'False's: ", num_falses, " ", percent_false,"%"
def f(k,j):
'''
This function 'f' takes a give number 'k' and produces the Prime-Factorization in the form of a list, excluding the number one and the given number.
'j' here is a counter for the 'j'th prime number.
If you would like this in a pretty print format I have made a for loop inside "numlist_prime.py" which turns the list into "a * b * c" etc..
'''
global listofprimefactors
if k==0:
temp=0
print temp #I've never encountered this, but just in case ;)
elif k==1:
temp=0 #basically do nothing, this still has to be here so it doesn't do the "else" statement an extra time
elif k==2:
listofprimefactors.append(2) #Adds 2 as a factor if k is equal 2
elif k==3:
listofprimefactors.append(3) #Adds 3 as a factor if k is equal to 3
elif k % 2 == 0:
#If divisble by add 2 as a factor to k
listofprimefactors.append(2)
f(k/2,j) #call to recursion, but the new value of k is half of k, because we havae just accounted for k being divisble by 2
elif k % generateprimes.genprime(j)==0:
#This is a bit copmlicated, but basically what this does is: if the value of k is evenly divisble by the 'j'th prime number then add that 'j'th prime nubmer as a factor. The first time this elif statement runs the "generateprimes.genprime(j)" will equal 3
primenum=generateprimes.genprime(j)
listofprimefactors.append(primenum)
b=k/primenum #since we have just accounted for the "prime factor" we need to move on, so we divide k by that prime number and send that new value of k (b) to the call of recursion
j=1 #j is always set to 1 here because the first time a number (k) is divisble by a prime number, the next time it goes to the call of recursion the value of j will be 2.
f(b,j+1) #j will be equal to 2, the first time this elif statement runs through.
else:
f(k,j+1) #if k was not divisble by anthing above, increase the value of j. This allows the 'j'th prime number to increase by one. So the first time this runs if k is not divisble by the (j=2) 2nd prime number (which is 3), then increase j to 3 (j=3). This way the second time through it will check to see if k is evenly divisble by the 3rd prime number (which is 5). and so on...
return listofprimefactors
def newfunc(m):
m = int(m)
for eachnum in range(2, m):
k = 3 / 2
compareme = eachnum ** k
print "here is the 'compare-me' to the power of 3/2: ", compareme
comparemeagain = compareme / 2
comparemeagain = int(comparemeagain)
generatedprime = generateprimes.genprime(eachnum)
if comparemeagain < generatedprime:
print eachnum, " is TRUE for this case. Here is the prime ", generatedprime, "here is the 'compare-me': ", comparemeagain
else:
print eachnum, " is FALSE."
eachnum += 1