I can't seem to get my function to work. When I type in 3 for a, 2 for b, and 3.61 for c. That works. However, when I try those values in a different order (Ex: 3.61 for a, 3 for b and 2 for c) It returns as false. I can't figure out what the problem is. Thanks in advance!
a = input("Enter a ")
b = input("Enter b ")
c = input("Enter c ")
def isright_angled():
if abs((a**2+b**2)-(c**2)) < 0.1 or abs((c**2-a**2)-(b**2)) < 0.1 or abs((c**2-b**2)-(a**2)) < 0.1:
return True
else:
return False
print isright_angled()
The hypotenuse, if the triangle is right-angled, will be the largest of a, b and c. You can use that to avoid duplicating the test 3 times (this is the "don't repeat yourself" principle). A second thing to avoid is that if something: return True else: return False. It's usually better expressed as simply return something. Thirdly, functions can take arguments rather than relying on global variables: this makes things easier to understand and there's then less chance of functions interfering with each other. I find a * a easier to understand than a ** 2 but that's personal taste. Putting all that together:
def is_approximately_right_angled(a, b, c):
a, b, c = sorted([a, b, c])
return abs(a * a + b * b - c * c) < 0.1
a = input('enter a ')
b = input('enter b ')
c = input('enter c ')
print is_approximately_right_angled(a, b, c)
If it's not working, you can speed up your development by adding some checks. If you were writing a big program you can write unit tests, but here just some asserts in the module will avoid you having to type a, b, c in each time to test.
I'd add something like this (before the a = input... line):
assert is_approximately_right_angled(3, 4, 5)
assert is_approximately_right_angled(3, 5, 4)
assert is_approximately_right_angled(3, 2, 3.61)
assert not is_approximately_right_angled(3, 5, 5)
With these lines in place, you can have some confidence in the code before you get to type numbers in. When you find cases where the code doesn't work you can add them as additional checks.
a = int(input("Enter the side length" ))
b = int(input("Enter the side length" ))
c = int(input("Enter the side length" ))
def is_right_triangle(a,b,c):
'''
is_right_triangle(a,b,c) -> bool
returns True if a,b,c is a right triangle with hypotenuse c
'''
a, b, c = sorted([a, b, c])
return a*a + b*b == c*c
print(is_right_triangle(a,b,c))
for more accuracy you can use
return abs(a * a + b * b - c * c) < 0.001
Related
I wrote a python code to find root of 2*x-4 using bisection method
def func(x):
return 2*x-4
def bisection(a,b):
if (func(a) * func(b) >= 0):
print("You have not assumed right a and b\n")
return
c = a
while ((b-a) >= 0.01):
c = (a+b)/2
if (func(c) == 0.0):
break
if (func(c)*func(a) < 0):
b = c
else:
a = c
print("The value of root is : ","%.0f"%c)
a =-2
b = 4
bisection(a, b)
Now i want that the function input should be given by the user in the form of mx+n where m and n are integers. Can anyone help how can i do that ?
m, n = list(map(int, input("Please enter the value of [m] [n] for f(x) = mx +n: ").split()))
def Input():
a, b = list(map(int, input("Enter values of [a] [b]: ").split()))
if f(a)*f(b)<0:
Bisection(a, b)
else:
print("Oops! The root of function doesn't belong to the above domain\nPlease try to again:")
Input()
def f(x):
return m*x + n
def Bisection(a, b):
c = (a+b)/2
if f(c)*f(b)<0:
Bisection(c, b)
elif f(c)*f(a)<0:
Bisection(c, a)
elif f(c)==0:
print(c)
Input()
See we know that Bisection, Newton-Raphson, or most of all Numerical methods are the iteration processes so better we use function inside of function: f(f(f(f.....) Of course! by checking the conditions.
Here, I have used elif f(c)==0 this is something which we can't use for quadratic/cubic or higher-order polynomials because getting the exact root will not be possible for all the types of equations say like f(x) = mx^2 - n where m, n > 0 However, we can define the iterations to be performed.
By asking like Please enter the number of iterations to be performed:
pow(a,x,c) operator in python returns (a**x)%c . If I have values of a, c, and the result of this operation, how can I find the value of x?
Additionally, this is all the information I have
pow(a,x,c) = pow(d,e,c)
Where I know the value of a,c,d, and e.
These numbers are very large (a = 814779647738427315424653119, d = 3, e = 40137673778629769409284441239, c = 1223334444555556666667777777) so I can not just compute these values directly.
I'm aware of the Carmichael's lambda function that can be used to solve for a, but I am not sure if and/or how this applies to solve for x.
Any help will be appreciated.
As #user2357112 says in the comments, this is the discrete logarithm problem, which is computationally very difficult for large c, and no fast general solution is known.
However, for small c there are still some things you can do. Given that a and c are coprime, there is an exponent k < c such that a^k = 1 mod c, after which the powers repeat. Let b = a^x. So, if you brute force it by calculating all powers of a until you get b, you'll have to loop at most c times:
def do_log(a, b, c):
x = 1
p = a
while p != b and p != 1:
x += 1
p *= a
p %= c
if p == b:
return x
else:
return None # no such x
If you run this calculation multiple times with the same a, you can do even better.
# a, c constant
p_to_x = {1: 0}
x = 1
p = a
while p != 1:
p_to_x[p] = x
x += 1
p *= a
p %= c
def do_log_a_c(b):
return p_to_x[b]
Here a cache is made in a loop running at most c times and the cache is accessed in the log function.
Define a function that takes three numbers as arguments and returns the sum of the squares of the two larger numbers.
For example, given 6,7,8, the function that I defined should return 113
When I gave my code, it solves most of the problems but apparently there is some possibility that I haven't tried?? I think my code is flawed but not sure what other possibilities are there. Would really appreciate some help thank you so much!
def bigger_sum(a,b,c):
if(a+b>b+c):
return(a*a+b*b)
if(a+c>b+c):
return(a*a+c*c)
if(b+c>a+c):
return(b*b+c*c)
You can use min for this problem:
def big2_sqrsum(a,b,c):
x = min(a,b,c)
return (a*a + b*b + c*c) - (x*x)
print(big2_sqrsum(6,7,8))
Output:
113
Alternate solution with if-else
def big2_sqrsum2(a,b,c):
if a < b and a <c:
return b*b + c*c
elif b < a and b < c:
return a*a + c*c
elif c < a and c < b:
return a*a + b*b
Just check for the smallest number. That known, assign the values to two new variables that will hold the largest and second largest value and sum their squares.
Something like this :
big1 = 0
big2 = 0
if ([a is smallest]):
big1 = b
big2 = c
elif ([b is smallest]):
big1 = a
big2 = c
elif ([c is smallest]):
big1 = a
big2 = b
allows you to have only one place to calculate your formula :
return big1 * big1 + big2 * big2
Let's take a look at why your code is flawed. Given a comparison like if a + b > b + c:, the implication that both a and b are both greater than c is false. b can be the smallest number. All you know is that a > c, since you can subtract b from both sides of the inequality.
You need to find and discard the smallest number. The simplest way is to compute the minimum with min and subtract it off, as #Sociopath's answer suggests.
If you want to keep your if-elsestructure, you have to compare numbers individually:
if a > b:
n1= a
n2 = b if b > c else c
elif a > c:
n1, n2 = a, b
else:
n1, n2 = b, c
You can Simply Define Function With Using min()
def two_bigger_sum(num1,num2,num3):
min_num = min(num1,num2,num3) # it returns minimum number
return ((num1**2 + num2**2 + num3**2)-(min_num**2)) # num**2 = square of num
print(two_bigger_sum(6,7,8))
Output = 113
Sociopath's answer works, but is inefficient since it requires two extra floating point multiplies. If you're doing this for a large number of items, it will take twice as long! Instead, you can find the two largest numbers directly. Basically, we're sorting the list and taking the two largest, this can be directly as follows:
def sumsquare(a,b,c):
# Strategy: swap, and make sure c is the smallest by the end
if c > b:
b, c = c, b
if c > a:
a, c = c, a
return a**2 + b**2
# Test:
print(sumsquare(3,1,2))
print(sumsquare(3,2,1))
print(sumsquare(1,2,3))
print(sumsquare(1,3,2))
print(sumsquare(2,1,3))
print(sumsquare(2,3,2))
I have tried to use list comprehension & list slicing with sorting method.
def b2(l):
return sum([x**2 for x in sorted(l)[1:]])
print(b2([1,2,3]))
OP:-
13
I am supposed to write a program that prompts the user for the lengths of three sides of a triangle, determines that the three lengths can form a triangle and if so uses Heron's formula to compute the area to 4 digits of precision.This is what I have so far I don't know where or how to put in the math
import math
def main():
print()
print("Triangle Area Program")
print()
a, b, c = eval(input("Enter three lengths separated by commas: "))
print()
s = (a+b+c) / 2.0
area = sqrt(s*(s-a)*(s-b)*(s-c))
if a > b:
a, b = b, a
if a > c:
a, c = c, a
if b > c:
b, c = c, b
else:
a + b > c
print("A triangle cannot be formed.")
main()
Here's another possible version of your mathy problem:
import math
def heron(a, b, c):
return 0.25 * math.sqrt((a + (b + c)) * (c - (a - b)) * (c + (a - b)) * (a + (b - c)))
if __name__ == "__main__":
print()
print("Triangle Area Program")
print()
print()
try:
description = "Enter three lengths separated by commas: "
sides = sorted(map(float, input(description).split(',')))
if (sides[1] + sides[2]) < sides[0]:
print("A triangle cannot be formed.")
else:
a, b, c = sides
print("Area of triangle {0}-{1}-{2} is {3:.4f}".format(
sides[0], sides[1], sides[2], heron(a, b, c)))
except Exception as e:
print("Check your input!!!")
print("--> Error: {0}".format(e))
Few notes about this version:
It's parsing your floating-point input values and sorting at the same time, that way you can check directly whether a triangle can be formed or not
It's not using the naive heron formula, instead is using another one which is numerically stable
I've decided to give you another version because in the comments you'll find some good advices about yours
Here's a slightly modified version of your program that checks if the inputs are compatible in one compound conditional expression and replaces the use of eval:
import math
def main():
print("\nTriangle Area Program\n")
a, b, c = map(float, input("Enter three lengths separated by commas: ").split(','))
if a + b > c and a + c > b and b + c > a:
s = (a + b + c) / 2.0
area = math.sqrt(s*(s-a)*(s-b)*(s-c))
return round(area, 4) # round area to four decimal places
else:
raise ValueError("The inputs you entered cannot form a triangle")
if __name__ == '__main__':
print(main())
More on avoiding eval when you can Why should exec() and eval() be avoided?
So, our teacher gave us an assignment to find three integers a, b c. They are in all between 0 and 450 using Python.
a = c + 11 if b is even
a = 2c-129 if b is odd
b = ac mod 2377
c = (∑(b-7k) from k = 0 too a-1) +142 (Edited. I wrote it wrong. Was -149)
I tired my code that looks like this: (Still a newbie. I guess a lot of my code is wrong)
for a, b, c in range(0, 450):
if b % 2 == 0:
a = c + 11
else:
a = 2 * c - 129
b = (a * c) % 2377
c = sum(b - 7 * k for k in range(0, a - 1))
but I get the error:
for a, b, c in range(0, 450):
TypeError: 'int' object is not iterable
What am I doing wrong and how can I make it check every number between 0 and 450?
The answers by Nick T and Eric hopefully helped you solve your issue with iterating over values of a, b, and c. I would like to also point out that the way you're approaching this problem isn't going to work. What's the point of iterating over various values of a if you're going to re-assign a to something anyway at each iteration of the loop? And likewise for b and c. A better approach involves checking that any given triple (a, b, c) satisfies the conditions given in the assignment. For example:
from itertools import product, tee
def test(a, b, c):
flags = {'a': False,
'b': False,
'c': False}
if (b % 2 == 0 and a == c+11) or (b % 2 == 1 and a == 2*c-129):
flags['a'] = True
if b == (a * c) % 2377:
flags['b'] = True
if c == sum(b - 7*k for k in range(a-1)) - 149:
flags['c'] = True
return all(flags.values()) # True if zero flags are False
def run_tests():
# iterate over all combinations of a=0..450, b=0..450, c=0..450
for a, b, c in product(*tee(range(451), 3)):
if test(a, b, c):
return (a, b, c)
print(run_tests())
NOTE: This is a slow solution. One that does fewer loops, like in glglgl's answer, or Duncan's comment, is obviously favorable. This is really more for illustrative purposes than anything.
import itertools
for b, c in itertools.product(*[range(450)]*2):
if b % 2 == 0:
a = c + 11
else:
a = 2 * c - 129
derived_b = (a * c) % 2377
derived_c = sum(b - 7 * k for k in range(0, a - 1))
if derived_b == b and derived_c == c:
print a, b, c
You need to nest the loops to brute-force it like you are attempting:
for a in range(451): # range(450) excludes 450
for b in range(451):
for c in range(451):
...
It's very obviously O(n3), but if you want a quick and dirty answer, I guess it'll work—only 91 million loops, worst case.
The stuff with [0, 450] is just as a hint.
In fact, your variables are coupled together. You can immediately eliminate at least one loop directly:
for b in range(0, 451):
for c in range(0, 451):
if b % 2: # odd
a = 2 * c - 129
else:
a = c + 11
if b != (a * c) % 2377: continue # test failed
if c != sum(b - 7 * k for k in range(a)): continue # test failed as well
print a, b, c
should do the job.
I won't post full code (after all, it is homework), but you can eliminate two of the outer loops. This is easiest if you iterate over c.
You code should then look something like:
for c in range(451):
# calculate a assuming b is even
# calculate b
# if b is even and a and b are in range:
# calculate what c should be and compare against what it is
# calculate a assuming b is odd
# calculate b
# if b is odd and a and b are in range:
# calculate what c should be and compare against what it is
Extra credit for eliminating the duplication of the code to calculate c
a = c + 11 if b is even
a = 2c-129 if b is odd
b = ac mod 2377
c = (∑(b-7k) from k = 0 to a-1) +142
This gives you a strong relation between all 3 numbers
Given a value a, there are 2 values c (a-11 or (a+129)/2), which in turn give 2 values for b (ac mod 2377 for both values of c, conditioned on the oddity of the result for b), which in turn gets applied in the formula for validating c.
The overall complexity for this is o(n^2) because of the formula to compute c.
Here is an implementation example:
for a in xrange(451):
c_even = a - 11
b = (a*c_even) % 2377
if b % 2 == 0:
c = sum(b - 7 * k for k in range(a)) + 142
if c == c_even:
print (a, b, c)
break
c_odd = (a+129)/2
b = (a*c_odd) % 2377
if b % 2 == 1:
c = sum(b - 7 * k for k in range(a)) + 142
if c == c_odd:
print (a, b, c)
break