I'm working on a Caesar Cypher with Python 3 where s is the string input and k is the amount that you shift the letter. I'm currently just trying to work through getting a letter like 'z' to wrap around to equal 'B'(I know the case is wrong, I'll fix it later). However when I run caesarCipher using the the following inputs: s = 'z' and k = 2, the line: s[n] = chr((122-ord(s[n]) + 64 + k)) causes s[n] to equal 'D'. If i adjust it down two(logically on the unicode scale this would equal 'B'), it makes s[n] = #. What am I doing wrong on that line that's causing 'B' not to be the output?
def caesarCipher(s, k):
# Write your code here
n = 0
s = list(s)
while n < len(s):
if s[n].isalpha() == True:
if (ord(s[n].lower())+k) < 123:
s[n] = (chr(ord(s[n])+k))
n += 1
else:
s[n] = chr((122-ord(s[n]) + 64 + k))
else:
n += 1
s = ''.join(s)
return s
You forgot to add 1 to n in the test of (ord(s[n].lower())+k) < 123 so that it would count s[n] twice or more.
Change it to
else:
s[n] = chr((122 - ord(s[n]) + 64 + k))
n += 1
and if you input "z" and 2, you'll get "B"
print(caesarCipher("z", 2))
# B
and if you adjust it down two, you'll get "#", which is the previous previous character of B in ASCII.
...
else:
s[n] = chr((122 - ord(s[n]) + 62 + k))
n += 1
...
print(caesarCipher("z", 2))
# #
Related
I'm new to python and want to make a program that generates Pi with the given decimal numbers. Problem is that I don't know how to check if the user has inputted a positive number.
This is the function that generates Pi (I don't know for sure how it works)
def PiBerekening(limiet):
q = 1
r = 0
t = 1
k = 1
n = 3
l = 3
decimaal = limiet
teller = 0
while teller != decimaal + 1:
if 4 * q + r - t < n * t:
# yield digit
yield n
# insert period after first digit
if teller == 0:
yield '.'
# end
if decimaal == teller:
print('')
break
teller += 1
nr = 10 * (r - n * t)
n = ((10 * (3 * q + r)) // t) - 10 * n
q *= 10
r = nr
else:
nr = (2 * q + r) * l
nn = (q * (7 * k) + 2 + (r * l)) // (t * l)
q *= k
t *= l
l += 2
k += 1
n = nn
r = nr
And this is how I ask the user how many decimals he wants to see
while not verlaatloop:
try:
pi_cijfers = PiBerekening(int(input("With how many decimals would you like to calculate Pi?")))
assert pi_cijfer > 0 # This is one of the things I've tried but I get the "NameError: name 'pi_cijfer' is not defined" error and I don't know what to do to check if the inputted number is negative
except ValueError:
print("This isn't a valid number, try again")
except AssertionError:
print("This number is negative, try again")
else:
verlaatloop = True
This is how I show the calculated Pi
for pi_cijfer in pi_cijfers:
print(pi_cijfer, end='')
You can first validate the input and then pass it to the PiBerekening function. Something like this:
while not verlaatloop:
try:
no_decimals = int(input("With how many decimals would you like to calculate Pi?"))
if no_decimals > 0:
pi_cijfers = PiBerekening(no_decimals)
#assert pi_cijfer > 0 # This is one of the things I've tried but I get the "NameError: name 'pi_cijfer' is not defined" error and I don't know what to do to check if the inputted number is negative
except ValueError:
print("This isn't a valid number, try again")
except AssertionError:
print("This number is negative, try again")
else:
verlaatloop = True
I have this list
a = [1,2,3,4,5,6,7,8,9]
I want to find out that how many co-prime pair elements of the list add up to sum=9
Ex, (1+8) = 9 , (2+7) = 9 , (3+6)=9 , (4+5)=9, (5+4)=9 , (6+3)=9, (7+2)=9 , (8+1)=9
Note that i don't want (3+6) as they are prime numbers. And i also don't want (7+2)=9 as it has already occurred (means 2,7 has been already taken in account)
I tried this But it takes repeated values too.
a = [1,2,3,4,5,6,7,8,9]
count=0
for m in a:
for n in a:
total=m+n
if(total==9):
s=str(m) + '+'+ str(n) + "="
print(s , m+n)
count=count+1
print("Count =" ,count)
The result should have count=3
Your mistake is in the way of doing the loops, so you repeat values.
Try this:
#from math import gcd as bltin_gcd
a = [1,2,3,4,5,6,7,8,9]
count = 0
def __gcd(a, b):
# Everything divides 0
if (a == 0 or b == 0): return 0
# base case
if (a == b): return a
# a is greater
if (a > b):
return __gcd(a - b, b)
return __gcd(a, b - a)
# Only python 3
# def coprime(a, b):
# return bltin_gcd(a, b) == 1
for i in range(0,9):
for j in range(i+1,9):
if __gcd(a[i], a[j]) == 1 and a[i] + a[j] == 9:
count += 1
print str(a[i]) + ' ' + str(a[j])
print 'Count = ' + str(count)
In number theory, two integers a and b are said to be relatively prime, mutually prime, or coprime if the only positive integer that divides both of them is 1. Consequently, any prime number that divides one does not divide the other. This is equivalent to their greatest common divisor being 1.
for m in a:
for n in a:
You are not selecting pairs by using this loops, ie. you are picking the first element in both the outer and inner loop during your first iteration.
if(total==9):
You are not checking the condition if the selected pair of numbers are coprime. You are only verifying the sum.
A pythonic solution may be obtained with a one-liner:
from math import gcd
a = [1,2,3,4,5,6,7,8,9]
pairs = [(m,n) for m in a for n in a if n > m and m+n == 9 and gcd(m,n) == 1]
Result :
pairs --> [(1, 8), (2, 7), (4, 5)]
If you are sure to never, never need the pairs but only the number of pairs (as written in the OP), the most efficient solution may be:
count = len([1 for m in a for n in a if n > m and m+n == 9 and gcd(m,n) == 1])
EDIT : I inversed the three conditions in the if statement for improved benefit from lazy boolean evaluation
You can solve this if you have something that calculates your prime factorization in python:
from functools import lru_cache
# cached function results for pime factorization of identical nr
#lru_cache(maxsize=100)
def factors(nr):
# adapted from https://stackoverflow.com/a/43129243/7505395
i = 2
factors = []
while i <= nr:
if (nr % i) == 0:
factors.append(i)
nr = nr / i
else:
i = i + 1
return factors
start_at = 1
end_at = 9
total = 9
r = range(start_at, end_at+1)
# create the tuples we look for, smaller number first - set so no duplicates
tupls = set( (a,b) if a<b else (b,a) for a in r for b in r if a+b == total)
for n in tupls:
a,b = n
f_a = set(factors(a))
f_b = set(factors(b))
# if either set contains the same value, the f_a & f_b will be truthy
# so not coprime - hence skip it
if f_a & f_b:
continue
print(n)
Output:
(2, 7)
(1, 8)
(4, 5)
I want to write a program that receives 10 entries and at the end it should print the number that has the greatest number of prime factors along with the number of its prime factors in the output. And, if some of the inputs have the same condition, the program will print the biggest one.
The code that I wrote executed all conditions except the final condition.
How should I write a code for a case where two numbers have the same numbers of prime factors (when I want the bigger number to be my output)?
For example both 678 and 84 have 3 prime factors . The output of my code is 84, while I want the output to be 678 (the bigger number).
input:
123
43
54
12
76
84
98
678
543
231
Correct output:
678 3
but my output:
84 3
a = [0,0,0,0,0,0,0,0,0,0]
#a = [123,43,54,12,76,84,98,678,543,231]
b = [0,0,0,0,0,0,0,0,0,0]
def is_first (number):
Prime_number= 0
for m in range(1, (number //2)+1, 1):
if number % m == 0:
Prime_number += 1
if Prime_number > 1:
is_prime = 0
else:
is_prime = 1
return is_prime
for i in range(0,10,1):
a[i] = input()
for j in a:
numbers= 0
for k in range(2, int(j)//2, 1):
if int(j) % k == 0:
if is_first (k) == 1:
numbers += 1
b[a.index(j)] = numbers
index_of_same = [i for i, e in enumerate(b) if e == max(b)]
n = []
for t in index_of_same:
n.append(a[t])
print(str(max(n))+ ' ' + str(max(b)))
When you append the numbers that have the highest number of prime factors to the list n, you are storing them as strings and not integers. Hence when you do the max() function on n, which holds 84,678 and 231, 84 is returned. I would suggest you to convert the numbers using int(), when storing into n and then perform the max()function on it. Just tweak your code as under:
for t in index_of_same:
n.append(int(a[t]))
Your code is cumbersome IMO, so I'd like to suggest a simpler version:
def getNumOfFactors(n):
num = 0
factor = 1
while n > 1:
factor += 1
if n % factor == 0:
num += 1
n /= factor
while n % factor == 0:
n /= factor
return num
def getMaxValue(values):
factors = [getNumOfFactors(n) for n in values]
tuples = zip(factors,values)
sorted_tuples = sorted([tuple for tuple in tuples],key=lambda x: x[0])
max_tuples = [tuple for tuple in sorted_tuples if tuple[0] == sorted_tuples[-1][0]]
sorted_max_tuples = sorted([tuple for tuple in max_tuples],key=lambda x: x[1])
return sorted_max_tuples[-1]
maxValue = getMaxValue([123,43,54,12,76,84,98,678,543,231])
print(maxValue)
I try to find out the index of a certain Fibonacci number. However my program returned to me this result "Your program took too long to execute. Check your code for infinite loops, or extra input requests" after typing in 1134903171.
num = 1
num_prev = 1
n = int(input())
i = 1
if n < 2:
print(1, 2)
else:
while i <= n + 2:
num_prev, num = num, num + num_prev
i += 1
if n == num:
print(i + 1)
break
elif i == n + 3:
print(-1)
#break`
Thank you guys. The problem of last code is that: if the number isn't a Fibonacci number and meanwhile it is too large, it will took to many loops for the calculation. As I used a web compiler to calculate, they do not allow such "infinity" loop to operate. Then I used a math methode to limite the loop.
import math
N=int(input())
root1=math.sqrt(5*N*N+4)
root2=math.sqrt(5*N*N-4)
i=1
num, num_prev = 1, 1
if root1%1==0 or root2%1==0:
while i <= N+2:
num_prev,num = num,(num+num_prev)
i+=1
if N==num:
print(i+1)
break
else:
print(-1)
But the best answer could be:
prev, next = 1, 1
index = 2
possible_fib = int(input())
while possible_fib > next:
prev, next = next, prev + next
index += 1
if possible_fib == next:
print(index)
else:
print(-1)
This code is to convert decimals to binary.
What I'm trying to do is to chop off the decimal part after diving by 2.
binary = []
n = 25
while n != 0:
binary.append(n % 2)
n = n / 2
int(n) #this part
print(binary)
print(n)
choose = input("continue?[Y/N]")
if choose == 'y':
continue
else:
break
print(list(reversed(binary)))
Is this what you want?
binary = []
n = 25
while n != 0:
binary.append(n % 2)
n = n / 2
n = int(n) #assign result to n
print(binary)
print(n)
choose = input("continue?[Y/N]")
if choose == 'y':
continue
else:
break
print(list(reversed(binary)))