My assingment
At each level the complete triangle for the previous level is placed into an extra outer triangle. The user should be asked to input the two characters to be used and the width of the innermost triangle, which must be odd. In addition to the test for negative input the function should test whether the supplied number is odd and display an appropriate message if it is not.
I need to print 3 triangles but every one of them includes other. It needs to get printed with two different character(like *-) and the user have to specify the length of innermost triangle and which has to be an odd number. Example,
Example output for 5 value
Ok, let me explain my way,
Every triangle should be in dictionary.
tri1 = {1:"*****", 2:"***", 3:"*"}
tri2 = {1:"..........", ...}
And couldn't find how I can deal with user input?
If enter 5,
length - 5 unit, height 3 unit
length - 11 unit, height 6 unit
length - 23 unit, height 12 unit.
How can i know? What is the logic?
Ok lets say if I did. I understand I should put a tringle in another triangle with nested loop, can simply iterate it another dictionary but, I need to check second character's position.
Thanks in advance.
My code,
ch1, ch2 = input("Please enter the characters you want to use: ")
num = int(input("Please specify the length of innermost triangle(only odd number): "))
if (num % 2 == 0) or (num < 3):
print("Number can not be even, less then 3 and negative")
num2 = (2 * num) + 1
num3 = (2 * num2) +1
tri1 = {}
tri2 = {}
tri3 = {}
for i in range(3):
tri1[i] = ch1*num
num -= 2
check = 1
cont = 0
var = 1
for ii in range(6):
tri2[ii] = ch2*check
check += 2
if (ii >= 3):
tri2[ii] = ch2*var + tri1[cont] + ch2*var
cont += 1
var += 2
for i in tri1:
print('{:^5}'.format(tri1[i]))
for i in tri2:
print('{:^11}'.format(tri2[i]))
The dictionary can be created using a simple function:
def create_tri_dict(tri_chars, tri_height):
level_str = {0:tri_chars[0]}
for i in range(1,tri_height):
level_length = i *2 +1
tri_char = tri_chars[i%2]
level_str[i] = level_str[i-1] + '\n' + tri_char * level_length
return level_str
Then the main logic of your program could be:
tri_chars = input('Input triangle characters: ')
tri_length = int(input('Input triangle base length: '))
tri_height = (tri_length + 1)//2
if tri_length %2 == 0:
raise Exception('Triangle base length not odd')
tri_dict = create_tri_dict(tri_chars, tri_length)
Then to print the final 3(?) triangles:
print(tri_dict[tri_height-2])
print(tri_dict[tri_height-1])
print(tri_dict[tri_height])
Related
number = str(input('Please input a number: '))
def add_numbers(number):
sum = 0
for i in range (1,4):
number = number * i
sum += int(number)
return sum
print(add_numbers(number)
Output for range (1,2) in the loop = 5
Output for range(1,3) in the loop = 60
Output for range (1,4) in the loop = 555615
I tried it for cases where the loop iterates only once or twice which results in 5, and 5+55 = 60, respectively. But when I run it for 3 times it returns 555615. I'm guessing that it returns the correct sum which is 615 but is then added with the string 555 (which I don't understand why this does happen.
This should work:
number = str(input('Please input a number: '))
def add_numbers(number):
number = int(number)
sum = 0
for i in range(1,4):
number = number ** i
sum += int(number)
return sum
print(add_numbers(number))
Output:
Please input a number: 5
15655
Instead of number * i you need number to the power of i or number ** i. Also, since you are passing add_numbers a value from input(), number is a string and you need it to be an int which number = int(number) does.
If you want the equation to be n + nn + nnn instead of n * nn * nnn this works:
number = str(input('Please input a number: '))
def add_numbers(number):
number = int(number)
sum = 0
for i in range(1,4):
new = number ** i
sum += int(new)
return sum
print(add_numbers(number))
Output:
Please input a number: 5
155
This is because number is not reset back to 5 after each loop. (As per John Gordon's comment above).
With one small change, the answer you seek can be found.
number = input('Please input a number: ')
def add_numbers(number):
sum_ = 0
for i in range (1,4):
new = number * i
sum_ += int(new)
return sum_
print(add_numbers(number))
sum will be like that = number *1 + number *2 + number *3 ...
and u want = number + number * number + number * number * number
so u should do that :
sum += pow(number,i)
I'm writing a program which should produce an output of something like this:
`Enter an integer (or Q to quit): 1
Enter an integer (or Q to quit): 2
Enter an integer (or Q to quit): 3
Enter an integer (or Q to quit): Q
(1 x 1) + (2 x 2) + (3 x 3) = 14`
So far, I've gotten the display of the equation right, but I can't seem to figure out how to actually get the total of the equation. Currently, the total displays 18 instead of the expected 14.
Here's my code so far:
`int_list = [] # initiate list
while True:
user_input = input("Enter an integer (or Q to quit): ") # get user input
if user_input == "Q": # break loop if user enters Q
break
integer = int(user_input) # convert user_input to an integer to add to list
int_list.append(integer) # add the integers entered to the list
for i in range(0, len(int_list)):
template = ("({0} x {1})".format(int_list[i], int_list[i]))
if i == len(int_list)-1:
trailing = " = "
else:
trailing = " + "
print(template, end="")
print(trailing, end="")
for i in range(0, len(int_list)):
x = (int_list[i]*int_list[i])
add = (x+x)
print(add)`
Any help would be greatly appreciated :))
Your problem exists in the here:
for i in range(0, len(int_list)):
x = (int_list[i]*int_list[i])
add = (x+x)
print(add)
Let us walk through what the code does to get a better understanding of what is going wrong.
With a list of [1, 2, 3] The for loop will iterate three times
On the first iteration, x will be assigned the value 1 because 1 * 1 is 1.
On the second iteration, x will be assigned the value 4 because 2 * 2 is 4. Notice that rather than the two values being added x's value being overwritten
On the third iteration, x will be assigned the value 9 because 3 * 3 is 9. Again the value is being overwritten.
After the loop, the variable add is created with the value x + x, or in our case 9 + 9 which is 18
This is why with the list [1, 2, 3] the value displayed is 18
Now that we have walked through the problem. We can see that the problem is overriding the value in the for loop then doubling the value before displaying it.
To fix these problems we can first remove the doubling giving the following code
for i in range(0, len(int_list)):
x = (int_list[i]*int_list[i])
print(x)
But the program still has the overriding problem so the value of a list [1, 2, 3] would be 9.
To fix this rather than overriding the value of x, let's create a new variable total that will have the result of the loop added to it every iteration rather than being overwritten.
total = 0
for i in range(0, len(int_list)):
total = total + (int_list[i]*int_list[i])
print(total)
Now let's walk through what the code does now.
With a list of [1, 2, 3] the for loop will iterate three times
On the first iteration, total will be assigned the value 1 because 0 + 1 * 1 is 1.
On the second iteration, total will be assigned the value 5 because 1 + 2 * 2 is 5. Notice how the previous value of the loop iteration is being added to the loop
On the third iteration, total will be assigned the value 14 because 5 + 3 * 3 is 14. Notice again how the previous value of the loop iteration is being added to the loop.
This gives us the correct result of 14.
One nice feature of python is the addition assignment which condentes A = A + B to A += B so it is possible to simply the following code to:
total = 0
for i in range(0, len(int_list)):
total += (int_list[i]*int_list[i])
print(total)
A reason this problem may have been so difficult is that the for loop is more complicated than it needs to be. It is possible that rather than iterating through a list of indices generated by a list of numbers then assessing the numbers from the list by their index. It is possible to iterate through the numbers directly.
With the that simplification your loop would look like this:
total = 0
for number in int_list:
total += number * number
print(total)
These changes would make your whole programme look like this:
int_list = [] # initiate list
while True:
user_input = input("Enter an integer (or Q to quit): ") # get user input
if user_input == "Q": # break loop if user enters Q
break
integer = int(user_input) # convert user_input to an integer to add to list
int_list.append(integer) # add the integers entered to the list
for i in range(0, len(int_list)):
template = ("({0} x {1})".format(int_list[i], int_list[i]))
if i == len(int_list)-1:
trailing = " = "
else:
trailing = " + "
print(template, end="")
print(trailing, end="")
total = 0 # start the total at zero as no values have been calculated yet
for number in int_list: # iterate through all values in the list
total += number * number # add to the total the number squared
print(total)
You duplicate only the last product (2 x 3 x 3 = 18).
Because you reassign x in your loop (x = (int_list[i]*int_list[i])) and than duplicate x with add = (x+x).
But you have to build the sum.
int_list = [] # initiate list
while True:
user_input = input("Enter an integer (or Q to quit): ") # get user input
if user_input == "Q": # break loop if user enters Q
break
integer = int(user_input) # convert user_input to an integer to add to list
int_list.append(integer) # add the integers entered to the list
for i in range(0, len(int_list)):
template = ("({0} x {1})".format(int_list[i], int_list[i]))
if i == len(int_list) - 1:
trailing = " = "
else:
trailing = " + "
print(template, end="")
print(trailing, end="")
x = 0
for i in range(0, len(int_list)):
x += (int_list[i] * int_list[i])
print(x)
You can try this
state= True
combi= []
while state:
user_input = input("Enter an integer (or Q to quit): ").lower()
if "q" in user_input:
state= False
else:
combi.append(int(user_input))
else:
final= sum([x+x for x in combi])
print(final)
experts.
I'm trying to define a function (collatz) that:
Asks for a number. If it is even it prints number // 2, if it odd it prints 3 * number + 1. (OK)
The result, whatever it is, must enter a loop until the result is 1. (NOK)
So, i´m not figure out because the result is not used and is in an infinite loop. Any suggestion?
def collatz():
number = int(input('Enter the number: '))
x = number % 2
while number != 1:
if x == 0:
print(f'{number // 2}')
else:
print(f'{3 * number + 1}')
number = number
print(f'{collatz()}')
You need to actually assign the result back to number.
As well:
The divisibility check needs to be in the loop.
The outer print() is not needed.
The f-strings are redundant. print() converts its arguments to string automatically.
def collatz():
number = int(input('Enter the number: '))
while number != 1:
if number % 2 == 0:
number //= 2 # Short for "number = number // 2"
else:
number = 3*number + 1
print(number)
collatz()
Example run:
Enter the number: 3
10
5
16
8
4
2
1
How do you write a program to count up and down from a users integer input, using only one while loop and no for or if statements?
I have been successful in counting up but can't seem to figure out how to count down. It needs to look like a sideways triangle also, with spaces before the number (equal to the number being printed).
this is my code so far;
n = int(input("Enter an integer: "))
i = " "
lines = 0
while lines <= n-1:
print(i * lines + str(lines))
lines += 1
You can solve this by making good use of negative numbers and the absolute value: abs() which will allow you to "switch directions" as the initial number passes zero:
n = int(input("Enter an integer: "))
s = " "
i = n * -1
while i <= n:
num = n - abs(i)
print(s * num + str(num))
i += 1
This will produce:
Enter an integer: 3
0
1
2
3
2
1
0
Move the print line to after the while loop. Form a list of lists (or an array) in the while loop using the content from your current print statement. Then create a second statement in your single while loop to count down, which creates a second list of lists. Now you can sequentially print your two lists of lists.
The program asks the user for a number N.
The program is supposed to displays all numbers in range 0-N that are "super numbers".
Super number: is a number such that the sum of the factorials of its
digits equals the number.
Examples:
12 != 1! + 2! = 1 + 2 = 3 (it's not super)
145 = 1! + 4! + 5! = 1 + 24 + 120 (is super)
The part I seem to be stuck at is when the program displays all numbers in range 0-N that are "super numbers". I have concluded I need a loop in order to solve this, but I do not know how to go about it. So, for example, the program is supposed to read all the numbers from 0-50 and whenever the number is super it displays it. So it only displays 1 and 2 since they are considered super
enter integer: 50
2 is super
1 is super
I have written two functions; the first is a regular factorial program, and the second is a program that sums the factorials of the digits:
number = int(input ("enter integer: "))
def factorial (n):
result = 1
i = n * (n-1)
while n >= 1:
result = result * n
n = n-1
return result
#print(factorial(number))
def breakdown (n):
breakdown_num = 0
remainder = 0
if n < 10:
breakdown_num += factorial(n)
return breakdown_num
else:
while n > 10:
digit = n % 10
remainder = n // 10
breakdown_num += factorial(digit)
#print (str(digit))
#print(str(breakdown_num))
n = remainder
if n < 10 :
#print (str(remainder))
breakdown_num += factorial(remainder)
#print (str(breakdown_num))
return breakdown_num
#print(breakdown(number))
if (breakdown(number)) == number:
print(str(number)+ " is super")
Existing answers already show how to do the final loop to tie your functions together. Alternatively, you can also make use of more builtin functions and libraries, like sum, or math.factorial, and for getting the digits, you can just iterate the characters in the number's string representation.
This way, the problem can be solved in a single line of code (though it might be better to move the is-super check to a separate function).
def issuper(n):
return sum(math.factorial(int(d)) for d in str(n)) == n
N = 1000
res = [n for n in range(1, N+1) if issuper(n)]
# [1, 2, 145]
First I would slightly change how main code is executed, by moving main parts to if __name__ == '__main__', which will execute after running this .py as main file:
if __name__ == '__main__':
number = int(input ("enter integer: "))
if (breakdown(number)) == number:
print(str(number)+ " is super")
After that it seems much clearer what you should do to loop over numbers, so instead of above it would be:
if __name__ == '__main__':
number = int(input ("enter integer: "))
for i in range(number+1):
if (breakdown(i)) == i:
print(str(i)+ " is super")
Example input and output:
enter integer: 500
1 is super
2 is super
145 is super
Small advice - you don't need to call str() in print() - int will be shown the same way anyway.
I haven't done much Python in a long time but I tried my own attempt at solving this problem which I think is more readable. For what it's worth, I'm assuming when you say "displays all numbers in range 0-N" it's an exclusive upper-bound, but it's easy to make it an inclusive upper-bound if I'm wrong.
import math
def digits(n):
return (int(d) for d in str(n))
def is_super(n):
return sum(math.factorial(d) for d in digits(n)) == n
def supers_in_range(n):
return (x for x in range(n) if is_super(x))
print(list(supers_in_range(150))) # [1, 2, 145]
I would create a lookup function that tells you the factorial of a single digit number. Reason being - for 888888 you would recompute the factorial of 8 6 times - looking them up in a dict is much faster.
Add a second function that checks if a number isSuper() and then print all that are super:
# Lookup table for single digit "strings" as well as digit - no need to use a recursing
# computation for every single digit all the time - just precompute them:
faks = {0:1}
for i in range(10):
faks.setdefault(i,faks.get(i-1,1)*i) # add the "integer" digit as key
faks.setdefault(str(i), faks [i]) # add the "string" key as well
def fakN(n):
"""Returns the faktorial of a single digit number"""
if n in faks:
return faks[n]
raise ValueError("Not a single digit number")
def isSuper(number):
"Checks if the sum of each digits faktorial is the same as the whole number"
return sum(fakN(n) for n in str(number)) == number
for k in range(1000):
if isSuper(k):
print(k)
Output:
1
2
145
Use range.
for i in range(number): # This iterates over [0, N)
if (breakdown(number)) == number:
print(str(number)+ " is super")
If you want to include number N as well, write as range(number + 1).
Not quite sure about what you are asking for. From the two functions you write, it seems you have solid knowledge about Python programming. But from your question, you don't even know how to write a simple loop.
By only answering your question, what you need in your main function is:
for i in range(0,number+1):
if (breakdown(i)) == i:
print(str(i)+ " is super")
import math
def get(n):
for i in range(n):
l1 = list(str(i))
v = 0
for j in l1:
v += math.factorial(int(j))
if v == i:
print(i)
This will print all the super numbers under n.
>>> get(400000)
1
2
145
40585
I dont know how efficient the code is but it does produce the desired result :
def facto():
minr=int(input('enter the minimum range :')) #asking minimum range
maxr=int(input('enter the range maximum range :')) #asking maximum range
i=minr
while i <= maxr :
l2=[]
k=str(i)
k=list(k) #if i=[1,4,5]
for n in k: #taking each element
fact=1
while int(n) > 0: #finding factorial of each element
n=int(n)
fact=fact*n
n=n-1
l2.append(fact) #keeping factorial of each element eg : [1,24,120]
total=sum(l2) # taking the sum of l2 list eg 1+24+120 = 145
if total==i: #checking if sum is equal to the present value of i.145=145
print(total) # if sum = present value of i than print the number
i=int(i)
i=i+1
facto()
input : minr =0 , maxr=99999
output :
1
2
145
40585