Errors in the program to sort a triangle [closed]
want to improve this question? Update the question so it's on-topic for Stack Overflow.
Closed 8 months ago .
improve this questionI'm having some problems and don't know how to fix:
1) an equilateral triangle, three equal sides and angles = 60°, is read as an isosceles triangle (two sides equal).
2) other than printing wrong, I can't pull out this 'NONE'.
3) when the Triangle has one of the angles = 90, it is rectangle. But the program does not recognize.
4) the fourth and last problem is that I don't know how I can restart the program: 'Type S to restart or N to end the program'
Can anyone help me ???
import math
# lado a
a = int(input('Digite o valor de um dos lados do triângulo '))
# lado b
b = int(input('Digite o valor do outro lado do triângulo '))
# angulo entre a e b
m = int(input('Digite o ângulo entre estes dois lados '))
ab = (m*math.pi)/180
# calculo do lado c
def lado_c ():
global a,b,c,ab
c = math.sqrt(a**2 + b**2 - 2*a*b*math.cos(ab))
return round(c,2)
# angulo entre a e c
def angulo_ac ():
global a,b,c,ac
n = ((a**2+c**2)-(b**2))/(2*a*c)
ac = (math.acos(n)*180)/math.pi
return (round(ac,2))
# angulo b e c
def angulo_bc ():
global a,b,c,bc
p = ((c**2+b**2)-(a**2))/(2*b*c)
bc =(math.acos(p)*180)/math.pi
return (round(bc,2))
# perímetro
def perimetro ():
global P,a,b,c
P =(a+b+c)
return P
# área
def area ():
global a,b,c,P,A
p = (a+b+c)/2
A = float(math.sqrt(p*(p-a)*(p-b)*(p-c)))
return A
# altura
def altura ():
global A
h = (A*2)/c
return h
# classificação quanto aos lados
def lado_class ():
global a,b,c,ab
if a!=b and a!=c and b!=c:
print ('Escaleno')
elif ab == 60 and a == b and b == c:
print ('Equilátero')
elif a != b and b == c or b !=c and c == a or c !=a and a==b:
print ('Isósceles')
return
# classificação quanto aos angulos
def angulo_class ():
global ab, ac, bc
if ab == 90 or bc == 90 or ac == 90:
print ('Retângulo')
elif ab < 90 and bc < 90 and ac < 90:
print ('Acutângulo')
elif ab > 90 and bc < 90 and ac < 90 or bc > 90 and ac < 90 and ab < 90 or ac > 90 and ac < 90 and ab < 90 :
print ('Obtusângulo')
return
#-----------------------------------------------------------------------------
print ('Lado c = '+ str(round(lado_c (),2)))
print ('Ângulo entre a e c '+str(angulo_ac ()))
print ('Ângulo entre b e c '+str(angulo_bc ()))
print('Área = '+str(round(area (),2))+' unidades de área')
print('Perímetro = '+str(round(perimetro(),2))+' unidades de comprimento')
print ('Altura, em relação ao lado c = '+str(round(altura (),2))+' unidades de comprimento')
print(lado_class ())
print (angulo_class ())
> import math
# lado a
a = int(input('Digite o valor de um dos lados do triângulo '))
# lado b
b = int(input('Digite o valor do outro lado do triângulo '))
# angulo entre a e b
m = int(input('Digite o ângulo entre estes dois lados '))
ab = (m*math.pi)/180
# calculo do lado c
def lado_c ():
global a,b,c,ab
c = math.sqrt(a**2 + b**2 - 2*a*b*math.cos(ab))
return round(c,2)
# angulo entre a e c
def angulo_ac ():
global a,b,c,ac
n = ((a**2+c**2)-(b**2))/(2*a*c)
ac = (math.acos(n)*180)/math.pi
return (round(ac,2))
# angulo b e c
def angulo_bc ():
global a,b,c,bc
p = ((c**2+b**2)-(a**2))/(2*b*c)
bc =(math.acos(p)*180)/math.pi
return (round(bc,2))
# perímetro
def perimetro ():
global P,a,b,c
P =(a+b+c)
return P
# área
def area ():
global a,b,c,P,A
p = (a+b+c)/2
A = float(math.sqrt(p*(p-a)*(p-b)*(p-c)))
return A
# altura
def altura ():
global A
h = (A*2)/c
return h
# classificação quanto aos lados
def lado_class ():
global a,b,c,ab
if a!=b and a!=c and b!=c:
print ('Escaleno')
elif ab == 60 and a == b and b == c:
print ('Equilátero')
elif a != b and b == c or b !=c and c == a or c !=a and a==b:
print ('Isósceles')
return
# classificação quanto aos angulos
def angulo_class ():
global ab, ac, bc
if ab == 90 or bc == 90 or ac == 90:
print ('Retângulo')
elif ab < 90 and bc < 90 and ac < 90:
print ('Acutângulo')
elif ab > 90 and bc < 90 and ac < 90 or bc > 90 and ac < 90 and ab < 90 or ac > 90 and ac < 90 and ab < 90 :
print ('Obtusângulo')
return
#-----------------------------------------------------------------------------
print ('Lado c = '+ str(round(lado_c (),2)))
print ('Ângulo entre a e c '+str(angulo_ac ()))
print ('Ângulo entre b e c '+str(angulo_bc ()))
print('Área = '+str(round(area (),2))+' unidades de área')
print('Perímetro = '+str(round(perimetro(),2))+' unidades de comprimento')
print ('Altura, em relação ao lado c = '+str(round(altura (),2))+' unidades de comprimento')
print(lado_class ())
print (angulo_class ())
I also posted the code here (it's better for view): https://code.sololearn.com/cUh4dY3oul2A/#py
2 answers
The questions 1 and 3 can be answered by analyzing their variables and the returns of their functions. In an example where I set one side of the Triangle to 50, another side to 50, and the angle between them 60°, I'll get as an answer from your script:
a: 50 # Valor do lado a - int
b: 50 # Valor do lado b - int
c: 49.99999999999999 # Valor do lado c não arredondado
m: 60 # Ângulo ab (graus) - int
ab: 1.0471975511965976 # Ângulo ab (radianos) não arredondado
ac: 60.00000000000001 # Ângulo ac (graus) não arredondado
bc: 60.00000000000001 # Ângulo bc (graus) não arredondado
lado_c: 50.0 # Valor do lado c arredondado ---- Retorno de função
angulo_ac: 60.0 # Ângulo ac (graus) arredondado ---- Retorno de função
angulo_bc: 60.0 # Ângulo bc (graus) arredondado ---- Retorno de função
From this data you can see why your functions return with some errors. When parsing the function lado_class() we have:
def lado_class ():
global a,b,c,ab
if a != b and a != c and b != c:
""" Comparação a, b e c.
Sua variável c não está arredondada, o que vai influenciar na sua comparação.
"""
print ('Escaleno')
elif ab == 60 and a == b and b == c:
""" Comparação a, b e c + análise do ângulo ab.
Novamente sua variável c não está arredondada.
A variável ab está definida como o ângulo entre a e b em RADIANOS,
porém você compara ela com o valor em graus.
"""
print ('Equilátero')
elif a != b and b == c or b !=c and c == a or c != a and a == b:
""" Comparação a, b e c.
Novamente o mesmo problema com a variável c.
"""
print ('Isósceles')
return
The same problem will repeat in your function angulo_class(), but with other variables. I recommend always having knowledge of the return of all its variables, even if it is just for testing.
To fix this problem you have two options: adjust your comparisons or modify your functions.
# Opção 1. Ajuste dos condicionais.
if a != b and a != lado_c() and b != lado_c():
(...)
# Opção 2. Ajuste das funções.
def lado_c ():
""" Arredonde a variável c, e não apenas o retorno da função.
NOTA: Lembre-se que arredondar c irá influenciar
os próximos cálculos que o utilizarem.
"""
global a,b,c,ab
c = round(math.sqrt(a**2 + b**2 - 2*a*b*math.cos(ab)))
return c
Regarding Question 2: None is the empty return of the function that arises from the very way you wrote your functions. I will work with only one of them, but the reasoning is the same for both of them.
def lado_class():
global a,b,c,ab
if a != b and a != c and b != c:
print ('Escaleno')
elif ab == 60 and a == b and b == c:
print ('Equilátero')
elif a != b and b == c or b != c and c == a or c != a and a == b:
print ('Isósceles')
return # <----- Essa é a linha que está definindo o retorno vazio.
You have two options to fix this. The first is to change the return of your function to:
def lado_class():
global a,b,c,ab
if a != b and a != c and b != c:
return 'Escaleno' # Nesse caso o retorno será 'Escaleno'.
elif ab == 60 and a == b and b == c:
return 'Equilátero' # Nesse caso o retorno será 'Equilátero'.
elif a != b and b == c or b != c and c == a or c != a and a == b:
return 'Isósceles' # E finalmente aqui o retorno será 'Isósceles'.
Written in this way the function will not give the print when used, but will only return the indicated value. In this case, if you want to give a print on the return of the function you will write:
print(lado_class())
If you don't want to modify your function you can use it alone. Without the need for print:
angulo_class()
So you will not print the return None of the its function. As you have already programmed the print inside the function, there is no need to give a print on the return of it.
In short: when defining a function be aware of when to use return and print. (doc. return; doc. print )
Question 4: I would put everything inside a while, but I don't know if this is the best way to do it. While runs your script until you give the condition for it to stop (doc. while ).
Example with while:
import math
repetir = True
while repetir is True:
##### Seu script aqui #####
consulta_repet = input('Deseja repetir? (S/N)')
comando_ok = False
while comando_ok is False:
if consulta_repet.lower() == 's':
repetir = True
comando_ok = True
elif consulta_repet.lower() == 'n':
repetir = False
comando_ok = True
else:
consulta_repet = input('Comando não entendido. Deseja repetir? (S/N)')
You can use the code below. This code will receive the measurements of the three sides of the Triangle in the same line separated by only one space , then the program will check if the measurements of the sides actually form a triangle and, if they do form a triangle, it will return the type of the Triangle.
The code is as follows...
continuar = True
while continuar == True:
try:
l = list(map(float, input('Medida dos lados: ').split()))
l.sort(reverse=True)
print()
if l[0] >= (l[1] + l[2]):
print('NAO FORMA TRIANGULO')
else:
if (l[0] ** 2) == (l[1] ** 2 + l[2] ** 2):
print('TRIANGULO RETANGULO')
if (l[0] ** 2) > (l[1] ** 2 + l[2] ** 2):
print('TRIANGULO OBTUSANGULO')
if (l[0] ** 2) < (l[1] ** 2 + l[2] ** 2):
print('TRIANGULO ACUTANGULO')
if l[0] == l[1] == l[2]:
print('TRIANGULO EQUILATERO')
if l[0] == l[1] != l[2] or l[1] == l[2] != l[0] or l[0] == l[2] != l[1]:
print('TRIANGULO ISOSCELES')
resposta = input('Digite "S" para reiniciar ou "N" para encerrar: ')
while resposta not in 'SsNn':
print(f'Valor INVÁLIDO! Digite apenas uma das duas letras sugeridas!')
resposta = input('Digite "S" para reiniciar ou "N" para encerrar: ')
if resposta == 's' or resposta == 'S':
continuar = True
else:
break
except:
break
As you may notice, this algorithm does not need to know the angle measurements to identify them. He identifies them analyzing only the behavior of the largest side in relation to others.
For you to use this program correctly, simply run the program and type the measurements of the sides in the same line, separated by a space, as shown below...
Medida dos lados: 3 4 5
... then type "Enter" and the program will solve the situation by displaying the following result...
TRIANGULO RETANGULO
Digite "S" para reiniciar ou "N" para encerrar:
... and then, you can Type " S "or" N", according to the situation.