Factoring Polynomials

[collin.day.0]

I am trying to write a simple application to factor polynomials. I
wrote (simple) raw_input lines to collect the a, b, and c values from
the user, but I dont know how to implement the quadratic equation

x = (-b +or- (b^2 - 4ac)^1/2) / 2a

into python. Any ideas?

 

[eric]

I am trying to write a simple application to factor polynomials. I
wrote (simple) raw_input lines to collect the a, b, and c values from
the user, but I dont know how to implement the quadratic equation

x = (-b +or- (b^2 - 4ac)^1/2) / 2a

into python. Any ideas?

with numpy:
from numpy import *

s=[1,-1]
x = -b+s*sqrt( b**2-4*a*c )/(2*a)

Eric
http://codeslash.blogspot.com

 

[Collin D]

I am trying to write a simple application to factor polynomials. I
wrote (simple) raw_input lines to collect the a, b, and c values from
the user, but I dont know how to implement the quadratic equation

x = (-b +or- (b^2 - 4ac)^1/2) / 2a

into python. Any ideas?

with numpy:
from numpy import *

s=[1,-1]
x = -b+s*sqrt( b**2-4*a*c )/(2*a)

Erichttp://codeslash.blogspot.com

Ahh. Great.. thank you. I didnt know about the sqrt function.. saves
me from doing "^1/2". Thanks again.
-CD

 

[Mark Dickinson]

     else: # a single result (discriminant is zero)
         return (-b / (2 * a),)

Maybe make that (-b / (2. * a)) to avoid getting funny results
when a and b are integers. (Or do a from __future__ import
division, or use Python 3.0, or ....)

And to make the function more bullet-proof, you might want to
do something like (untested):

from math import copysign

[rest of example as in Scott's post]

if discriminant: # two results
root1 = (-b - copysign(discriminant, b))/(2*a)
root2 = c/(a*root1)
return (root1, root2)

to avoid numerical problems when b*b is much larger
than abs(a*c). Compare with the results of the usual
formula when a = c = 1, b = 10**9, for example. But
that still doesn't help you when the computation
of the discriminant underflows or overflows...

Isn't floating-point a wonderful thing!

Mark

 

[eric]

with numpy:
from numpy import *

s=[1,-1]
x = -b+s*sqrt( b**2-4*a*c )/(2*a)

Eric

Without the Nump.

def polynomial(a, b, c):
    N = ((b**2 - 4*a*c)**.5) / 2*a
    return (-b + N, -b - N)

there is a little mistake in your formula :

def roots_of(a, b, c):
N = ((b**2 - 4*a*c)**.5)
return ((-b + N)/(2*a), (-b - N)/(2*a))

but I stick with numpy, numpy is heavy, but not as much as using a
computer to solve this problem ! And numpy should be standard module
for computer science student !

Eric
http://codeslash.blogspot.com

 

[Collin D]

     else: # a single result (discriminant is zero)
         return (-b / (2 * a),)

Maybe make that (-b / (2. * a)) to avoid getting funny results
when a and b are integers.  (Or do a from __future__ import
division, or use Python 3.0, or ....)

And to make the function more bullet-proof, you might want to
do something like (untested):

    from math import copysign

    [rest of example as in Scott's post]

    if discriminant: # two results
        root1 = (-b - copysign(discriminant, b))/(2*a)
        root2 = c/(a*root1)
        return (root1, root2)

to avoid numerical problems when b*b is much larger
than abs(a*c). Compare with the results of the usual
formula when a = c = 1, b = 10**9, for example.  But
that still doesn't help you when the computation
of the discriminant underflows or overflows...

Isn't floating-point a wonderful thing!  

Mark

Thanks for all your help! Its good to know how to do it w/ without
numpy.
And yes, floating point is the best thing since sliced bread. ^^

-CD

 

[Gabriel Genellina]

I am trying to write a simple application to factor polynomials. I
wrote (simple) raw_input lines to collect the a, b, and c values from
the user, but I dont know how to implement the quadratic equation

x = (-b +or- (b^2 - 4ac)^1/2) / 2a

into python. Any ideas?

Read the section on "Numeric Types" (under "Built-in Types") in the
Library Reference: http://docs.python.org/library/index.html
Then you should be able to write two expressions to evaluate your formula
above.

 

[James Mills]

Why is this so hard ? This is simple simple
expression. Reading through the Python
tutorial and reading up on how to define
functions is all you need!

Here goes:
... x = (-1 * b) + ((b**2 - (4 * a * c)) / (2 * a))
... return (-1 * x), x
...
cheers
James

 

[Collin D]

Why is this so hard ? This is simple simple
expression. Reading through the Python
tutorial and reading up on how to define
functions is all you need!

Here goes:


...     x = (-1 * b) + ((b**2 - (4 * a * c)) / (2 * a))
...     return (-1 * x), x
...

(6, -6)

cheers
James

Hiya James!

Just one small problem with your equation above...
The quadratic formula is:
x = -b +or- (b**2 - (4 * a * c))^1/2 / 2a

You just forgot the square root which makes quadratic a bit more
complicated.
You would have to download and import sqrt() from numpy or **.5

Also.. I need to build in functionality so the user does not have to
directly call the function like:
f(a,b,c)

Instead.. they should be able to just raw_input their values.
Also.. as with some of the examples above its a good idea to analyze
the discriminant to make sure we have a real solution.
Of course.. thats all pretty simple to build in. Thanks a lot!

 

[Steven D'Aprano]

I am trying to write a simple application to factor polynomials. I wrote
(simple) raw_input lines to collect the a, b, and c values from the
user, but I dont know how to implement the quadratic equation

x = (-b +or- (b^2 - 4ac)^1/2) / 2a

into python. Any ideas?

def quadratic_solution(a, b, c):
sol1 = (-b + (b**2 - 4*a*c)**0.5)/2*a
sol2 = (-b - (b**2 - 4*a*c)**0.5)/2*a
return (sol1, sol2)


Because this looks like homework, I've deliberately left in two errors in
the above. One of them is duplicated in the two lines above the return,
and you must fix it or you'll get radically wrong answers.

The second is more subtle, and quite frankly if this is homework you
could probably leave it in and probably not even lose marks. You will
need to do significant research into numerical methods to learn what it
is, but you will then get significantly more accurate results.

 

[James Mills]

Hi Collin,

Here you go:

jmills@atomant:~/tmp$ cat polycalc.py
#!/usr/bin/env python

from math import sqrt

def f(a, b, c):
if (b**2 - (4 * a * c)) < 0:
return None, None # Can't solve
x = (-1 * b) + (((b**2 - (4 * a * c)) ** 0.5) / (2 * a))
return (-1 * x), x

print "Polynomial Solver..."
print

while True:
a = float(raw_input("a: "))
b = float(raw_input("b: "))
c = float(raw_input("c: "))

x = f(a, b, c)
if None in x:
print "Can't solve!"
else:
print "x = -%0.2f" % x[0]
print "x = +%0.2f" % x[1]
jmills@atomant:~/tmp$

cheers
James

 

[James Mills]

UPDATE:

jmills@atomant:~/tmp$ cat polycalc.py
#!/usr/bin/env python

from math import sqrt

def f(a, b, c):
if (b**2 - (4 * a * c)) < 0:
return None, None # Can't solve
x1 = -b - (sqrt(b**2 - (4 * a * c)) / (2 * a))
x2 = -b + (sqrt(b**2 - (4 * a * c)) / (2 * a))
return x1, x2

print "Polynomial Solver..."
print

while True:
a = float(raw_input("a: "))
b = float(raw_input("b: "))
c = float(raw_input("c: "))

x = f(a, b, c)
if None in x:
print "Can't solve!"
else:
print "x = (%0.2f, %0.2f)" % x
jmills@atomant:~/tmp$ ./polycalc.py
Polynomial Solver...

a: 1
b: 8
c: 5
x = (-11.32, -4.68)

 

[Collin D]

UPDATE:

jmills@atomant:~/tmp$ cat polycalc.py
#!/usr/bin/env python

from math import sqrt

def f(a, b, c):
    if (b**2 - (4 * a * c)) < 0:
        return None, None # Can't solve
    x1 = -b - (sqrt(b**2 - (4 * a * c)) / (2 * a))
    x2 = -b + (sqrt(b**2 - (4 * a * c)) / (2 * a))
    return x1, x2

print "Polynomial Solver..."
print

while True:
    a = float(raw_input("a: "))
    b = float(raw_input("b: "))
    c = float(raw_input("c: "))

    x = f(a, b, c)
    if None in x:
        print "Can't solve!"
    else:
        print "x = (%0.2f, %0.2f)" % x
jmills@atomant:~/tmp$ ./polycalc.py
Polynomial Solver...

a: 1
b: 8
c: 5
x = (-11.32, -4.68)

Ahh. Great.. that answers a lot of questions.
Originally I was using just a = raw_input('a: ')
And was getting errors because you cant perform mathmatical operations
on strings. >.<
Thanks again!

 

[Collin D]

def quadratic_solution(a, b, c):
    sol1 = (-b + (b**2 - 4*a*c)**0.5)/2*a
    sol2 = (-b - (b**2 - 4*a*c)**0.5)/2*a
    return (sol1, sol2)

Because this looks like homework, I've deliberately left in two errors in
the above. One of them is duplicated in the two lines above the return,
and you must fix it or you'll get radically wrong answers.

The second is more subtle, and quite frankly if this is homework you
could probably leave it in and probably not even lose marks. You will
need to do significant research into numerical methods to learn what it
is, but you will then get significantly more accurate results.

The corrected function is:
def quadratic_solution(a,b,c)
sol1 = -1*b + ((b**2 - 4*a*c)**.5)/2*a
sol1 = -1*b - ((b**2 - 4*a*c)**.5)/2*a
return (sol1, sol2)

Squaring the -b would give you some strange solutions....

-CD

 

[James Mills]

Ahh. Great.. that answers a lot of questions.
Originally I was using just a = raw_input('a: ')
And was getting errors because you cant perform mathmatical operations
on strings. >.<
Thanks again!

No worries. Please take an hour or two to
go through the Python Tutorial at
http://docs.python.org/

cheers
James

 

[Steven D'Aprano]

Because this looks like homework...

Homework or not, of course others have answered it completely, more or
less error-free.

 

[Steven D'Aprano]

The corrected function is:
def quadratic_solution(a,b,c)
sol1 = -1*b + ((b**2 - 4*a*c)**.5)/2*a
sol1 = -1*b - ((b**2 - 4*a*c)**.5)/2*a
return (sol1, sol2)

Squaring the -b would give you some strange solutions....

So it would, but I didn't do that. Try again.

You have a typo in the above: you've written "sol1" twice and there is no
sol2.

There is no difference between -1*b and -b (except that -b is probably a
smidgen faster).

In the quadratic formula, the -b must be divided by 2a as well as the
discriminant. You're only dividing the discriminant.

The *second* error has already been given away by Mark Dickson, who
rightly points out that the quadratic equation is subject to round-off
errors. The first error is still there

(Hint: think about the precedence of operators and the need for brackets.)

 

[Collin D]

I am trying to write a simple application to factor polynomials. I
wrote (simple) raw_input lines to collect the a, b, and c values from
the user, but I dont know how to implement the quadratic equation

x = (-b +or- (b^2 - 4ac)^1/2) / 2a

into python. Any ideas?

I completed the code:

#import
from math import sqrt

# collect data
a = float(raw_input('Type a value: '))
b = float(raw_input('Type b value: '))
c = float(raw_input('Type c value: '))

# create solver
def solver(a,b,c):
if b**2 - 4*a*c < 0:
return 'No real solution.'
else:
sol1 = -1 * b + (sqrt(b**2 - 4*a*c)) / 2*a
sol2 = -1 * b - (sqrt(b**2 - 4*a*c)) / 2*a
return (sol1, sol2)

# execute
print solver(a,b,c)

Thanks to everyone who helped...
This really expanded my knowledge on some of the mathematical
functions in Python.

 

[Collin D]

I completed the code:

#import
from math import sqrt

# collect data
a = float(raw_input('Type a value: '))
b = float(raw_input('Type b value: '))
c = float(raw_input('Type c value: '))

# create solver
def solver(a,b,c):
    if b**2 - 4*a*c < 0:
        return 'No real solution.'
    else:
        sol1 = -1 * b + (sqrt(b**2 - 4*a*c)) / 2*a
        sol2 = -1 * b - (sqrt(b**2 - 4*a*c)) / 2*a
        return (sol1, sol2)

# execute
print solver(a,b,c)

Thanks to everyone who helped...
This really expanded my knowledge on some of the mathematical
functions in Python.

UPDATE:
'

#import
from math import sqrt

# collect data
a = float(raw_input('Type a value: '))
b = float(raw_input('Type b value: '))
c = float(raw_input('Type c value: '))

# create solver
def solver(a,b,c):
if b**2 - 4*a*c < 0:
return 'No real solution.'
else:
sol1 = (-1 * b + (sqrt(b**2 - 4*a*c))) / 2*a
sol2 = (-1 * b - (sqrt(b**2 - 4*a*c))) / 2*a
return (sol1, sol2)

# execute
print solver(a,b,c)

 

[Russ P.]

UPDATE:
'

#import
from math import sqrt

# collect data
a = float(raw_input('Type a value: '))
b = float(raw_input('Type b value: '))
c = float(raw_input('Type c value: '))

# create solver
def solver(a,b,c):
    if b**2 - 4*a*c < 0:
        return 'No real solution.'
    else:
        sol1 = (-1 * b + (sqrt(b**2 - 4*a*c))) / 2*a
        sol2 = (-1 * b - (sqrt(b**2 - 4*a*c))) / 2*a
        return (sol1, sol2)

# execute
print solver(a,b,c)

You need to put your denominator, 2*a, in parens. The way it stands,
you are dividing by 2, then multiplying by a. That's not what you
want.

Also, for better style, I suggest you compute the discriminanat once
and store it for reuse rather than repeating the expression three
times.