Re: I need help, again!
So while failing at an affine cipher, I decided to take a break. Sometimes it helps, sometimes it doesn't. In short, I coded something else:
Note, replace 2, 3, and 4 greater than signs with spaces. Then go to every function and edit the top of it, up till the loop starts. I'm sorry, but there isn't an easier way for me to do this, the forum trims my spaces. Anyways, here you go.
def factor(xs, factornum):
x1 = 1
x2 = 1
isfactored = False
while isfactored==False:
z = x1*x2
if z==factornum:
w = x1+x2
if w==xs:
print( str(x1)+" "+str(x2))
>>>>return
if x2>factornum:
>>>x2=1
x1+=1
if x1>factornum:
print("Unfactorable _expression_.")
return
x2+=1
def negfactor(xs, factornum):
x1 = -1
x2 = -1
isfactored=False
while isfactored==False:
z = x1*x2
if z==factornum:
w = x1+x2
if w==xs:
print(str(x1)+" "+str(x2))
return
if x2<factornum-factornum*2:
x2=-1
x1-=1
if x1<factornum-factornum*2:
print("Unfactorable _expression_.")
return
x2-=1
def hybrid(xs, factornum):
x1=-1
x2=1
isfactored=False
while isfactored==False:
z = x1*x2
if z==factornum:
w = x1+x2
if w==xs:
print(str(x1)+" "+str(x2))
return
if x2>factornum+factornum*-2:
x2=1
x1-=1
if x1<factornum:
print( "Unfactorable _expression_.")
return
x2+=1
def begin(xs, factornum):
if xs >0 and factornum>0:
factor(xs, factornum)
elif xs <0 and factornum >0:
negfactor(xs, factornum)
elif xs >0 and factornum <0:
hybrid(xs, factornum)
else:
print("It appears that you gave an invalid number. Try again.")Completely useless, I know, but something fun on the side. Also, let me know if you see any potential issues. I know you can crash the script by giving it letters or too big of a number, I'm looking into try statements to fix that problem as I write this.
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