Archive
Python: Powerful Digit Counts
Python: Powerful Digit Counts
Problem No.63 @ ProjectEuler
Completed on: Completed on Thu, 11 Jul 2019, 17:21
Just to make my post simple, i am quoting from ProjectEuler page
The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power.
How many n-digit positive integers exist which are also an nth power?
Then, we need to find the loop that will solve this, and we did..
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The Code:
# P63
# Power digit count
# Solved
# Completed on Thu, 11 Jul 2019, 17:21
c = 0
for x in range (1,50):
for p in range (1,50) :
if (len(str(x**p)) == p ):
c += 1
print(‘\n We have {} n-digit integers exist which are also an nth power.’.format(c))
Follow me on Twitter..Python: Pentagon Numbers
Python: Pentagon Numbers
Problem No.44 on ProjectEuler
Completed on: Thu, 11 Jul 2019, 18:37
This problem talking about the Pentagonal numbers and gives us a formula. Using that formula for a certain range of numbers, the generated sequence showing that P4 + P7 = 22 + 70 = 92, 92 is the P8, but if we subtracting (P7 – P4) = 70 – 22 = 48, 48 is not in the generated sequence of pentagonal numbers, so 48 is not pentagonal.
The task here is to find the pair of pentagonal Pj,Pk which their sum and difference are Pentagonal D = Pk – Pj is minimised.(we need to get the D).
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The Code:
# P44
# Pentagon Numbers
# Solved
#Completed on Thu, 11 Jul 2019, 18:37
def pn(n):
return int(n*(3*n-1)/2)
pn_list=[]
for n in range (1000,3000) : # I start increasing the range step by step.
pn_list.append(pn(n))
we_found_it = False
for x in range (0,len(pn_list)-1) :
px= pn_list[x]
for y in range (x+1,len(pn_list)-1) :
py= pn_list[y]
if (px+py) in pn_list:
if (py-px) in pn_list:
print(‘\n We found one ‘,px,py,’D = ‘,py-px )
we_found_it = True
if we_found_it : break
print(‘Done’)
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Python is_prime and time consuming
Python: is_prime and time consuming
Function enhancement
Once i start solving projectEuler problems i notes that i need the prime numbers in most of cases, so I wrote a function called ‘is_prime’ and it works fine. Some time just to get all the primes in a given range takes some seconds, seconds in computer time means waiting a lot. With some other problems that we need to get the prime in large numbers my function looks slow, since I am not deep in math I search the net for a better way to get the primes or to check if a given number is prime or not.
Definition A prime number (or a prime) is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers. wikipedia.org. So as I understand a prime number is not dividable by any other numbers, so my is_prime function is taking a number let’s say n= 13, and start a loop from 2 to n=13 if we fond a number that divide 13 then 13 is not a prime.
Here is the code:
def is_prime1(num):
for t in range (2, num):
if num % t == 0 :
return False
return True
The function is working fine and we can get the prime numbers, but as I mention above, if we have a large number or a wide range, this will take some time. After searching the web, I found some facts regarding the Prime Numbers:
1. The only even prime number is 2. (So any other even numbers are not prime)
2. If the sum of a number digits is a multiple of 3, that number can be divided by 3.
3. No prime number greater than 5 ends in/with 5.
OK, now I can first cut any range to half by not going through even numbers (if even false). Then, I will see if the number end with 5 or not (if end with 5 false),last I will do a summation of the digits in the number if the sum divide by 3 (if yes false), and if the number pass then i will take it in the loop from 5 to n, and if any number divide it we will return false.
Here is the code after the enhancement:def is_prime2(num):
if num %2==0 : # pass the even numbers.
return False
num_d= str(num) # if last digits is 5, then not prime
t= len(num_d)
if (num_d[t-1]) == 5 :
return False
tot = 0
for each in str(num):
tot = tot + int(each)
if tot % 3 == 0 : # if digits sum divide by 3, then not prime
return False
for t in range (3, num, 2):
if num % t == 0 :
return False
return True
I test both function on my laptop, for different number ranges, and use the time function to see the time delays with each one. Here is the results. If any one know better way to do this please drop it here. Or on My Twitter.
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Python: Largest Palindrome Product
Python: Largest Palindrome Product
Problem No.4 @ Projecteuler
Complete on: on Fri, 5 Jul 2019, 08:53
The task was to find the largest palindromic number that been generated from multiplying two of 3 digits number.
Definition: Palindromic numbers are numbers that remains the same when its digits are reversed. Like 16461, we may say they are “symmetrical”.wikipedia.org
To solve this I first wrote a function to check if we can read a number from both side or not, Then using while and for loop through numbers 100 to 999, and store largest palindromic, we select the range (100,999) because the task is about tow number each with 3 digits.
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The Code:
# Problem 4
# Largest palindrome product
# SOLVED
# Completed on Fri, 5 Jul 2019, 08:53
palin =0
def palindromic(n) :
n_list=[]
for each in str(n) :
n_list.append(each)
n_last = len(n_list)-1
n_first =0
x=0
while (n_first+x != n_last-x) :
if n_list[n_first+x] != n_list[n_last-x] :
return False
else :
x +=1
if (n_first +x > n_last -x):
return True
return True
for set1 in range (1,999):
for set2 in range (set1,999):
if palindromic(set1 * set2) :
if (set1 * set2) > palin :
palin =(set1*set2)
print(‘\n We found it:’,palin, ‘coming from {} * {}’.format(set1,set2))
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Python: Champernowne’s constant
Python: Champernowne’s constant
Problem No.40 @ ProjectEuler
Completed on: Mon, 1 Jul 2019, 18:01
In This task No.40, basically we need to get some digits from a large decimal fraction, then finding the multiplication of those digits.
ProjectEuler assume that the fraction is: 0.123456789101112131415161718192021222324 …. until 1000000, then we should fined the digits in positions 1, and 10, 100, 1000, 10000, 100000 and 1000000. Here is a copy of the problem screen
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So to solve this I create a string variable n_list then using for loop i store the numbers from 1 to 1000000 in it as [12345678910111213141516 … 1000000], and simply get the digits I want using the list index, and Finally I calculate the needed multiplication as required. .. And we solve it. ..
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Python: Combinatoric Selections
Python: Combinatoric Selections
Problem 53 @ projecteuler
Completed on: Mon, 1 Jul 2019, 11:32
This is one of the tasks that based on equations. To solve this, first let me Simplify it.
Simplifying the task:
if 1<= n <= 100 , and r <= n
using the equation of C(n,r) is
c(n,r) = n! / (r! *(n-r)!
How many values of c(n,r) greater than one-million
Since we need the factorial of n (n!) and (r!) then I will re-use our function get_factorial(num), and I will write another function to calculate the C(n,r) as in the equation above, I will call it combi(n,r), also I will use a while loop, and we need to set some variables to help us in the execution. In this part, I will use a list called values_over_1m to store all the values in it, although this is Not required in the Problem 53, but i will keep it so later we can print them all to the screen or just get the total values of it. And so we did .. and we get the right answer.
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The Code:
# Combinatoric selections
# Problem 53 @ projecteuler
# Completed on Mon, 1 Jul 2019, 11:32
”’
if 1<= n <= 100 , and r <= n
using the equation of C(n,r) is
c(n,r) = n! / (r! *(n-r)!
How many values of c(n,r) greater than one-million
'''
# Function to get the Factorials of a number.
def get_factorial(num):
if (num == 0) :
return 1
else:
return num * get_factorial(num-1)
def combi(n,r):
fn = get_factorial(n)
fr = get_factorial(r)
fnr = get_factorial(n-r)
return fn / (fr*fnr)
# To store the values more than 1 million.
values_over_1m =[]
nn = 1
rr = 1
start_value= combi(nn,rr)
stop_me = False
while stop_me == False :
nn +=1
rr =2
while rr < = nn :
start_value = combi(nn,rr)
if start_value >1000000 :
values_over_1m.append([nn,rr])
rr +=1
if nn > 99 :
stop_me = True
print(‘Ther is ‘,len(values_over_1m),’ Values meeting the equation.’)
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Python: Permuted Multiples
Python: Permuted Multiples
Problem No.52 @ Projecteuler
This task Completed on Mon, 1 Jul 2019, 06:08, the goal is to find a number X that if we get the X, X*2, X*3, X*4, X*5, X*6 all the numbers are same digits but in a different order.
I use a function called if_x_in_y (n1,n2) will return True if the two numbers has same digits, False if not. Then a while found_one is not True a loop will examine all numbers starting from 2 until we found_one.
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The Code:
# Projecteuler.net
# Permuted multiples
# Problem No.52
# Completed on Mon, 1 Jul 2019, 06:08am (GMT+3)
# if x, x*2, x*3, x*4, x*5, x*6 has the same digits.
def if_x_in_y (n1,n2) :
for x in str(n1) :
if x not in str(n2) :
return False
for x in str(n2):
if x not in str(n1):
return False
return True
x_num = 2
found_one= False
while not found_one:
if if_x_in_y(x_num,x_num*2):
if if_x_in_y(x_num,x_num*3):
if if_x_in_y(x_num,x_num*4):
if if_x_in_y(x_num,x_num*5):
if if_x_in_y(x_num,x_num*6):
print(‘\n Yes, we found one ..’)
found_one = True
print(‘ The Number is ‘,x_num)
x_num += 1
print(x_num,x_num*2,x_num*3,x_num*4,x_num*5,x_num*6)
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Python: 10001st Prime
Python: 10001st Prime
Problem No.7 @ ProjectEuler
This is one of shot projects you may find on ProjectEuler, the task is find the prime No 10001, I just run my is_prime function, if the number is prime will add 1 to the counter, until we reach 10001, that will be the answer.
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The Code:
# 10001st prime
# Problem 7
# Solved: Wed, 26 Jun 2019, 05:57am (GMT+3)
def is_prime(num):
result = True
for t in range (2,num):
if num%t==0 :
result= False
break
return result
p_count=0
num =0
while p_count <=10001:
num +=1
if is_prime (num):
p_count +=1
print(p_count,num)
print(num,p_count-1)
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Python: Square Digit Chain
Python: Square Digit Chain
ProjectEuler Problem No.92
Here we have a mathematical definition called Happy Number..
A Happy Number is defined by the following process:
Starting with any positive integer, replace the number by the sum of the squares of its digits in base-ten, and repeat the process until the number either:
equals 1 (where it will stay), or
it loops endlessly in a cycle that does not include 1.
(Wikipedia).
In ProjectEuler the task stated in this problem as ” How many starting numbers below ten million will arrive at 89?”
Enhancement: Here we will do something else, we will try to solve the task and post the answer to the projecteuler portal, BUT we are not talking about this here, we will use the concept of this task to generate chains of looped number and I will use it later in another post (project) and trying to represent this chains in a graphic way.
So to do this we need two functions, First one will read a number, get its digits, squaring each digit and get the summation. To keep our eyes on the numbers we need to store it, so we will use list called the_chain.
To check if we have reach a closed chain then we need to ask if the new number (sum of square digit) exists in the chain list or not. If exists we finish and will return the chain for more manipulating.
I will solve this on my way .. 🙂
In this code we will do the following:
1. We will ask the user to enter a number.
2. We will run the function on that number.
3. Outputs:
If we ends with 1 then we have a Happy Number.
If we have closed chain (current number exists in the chain) then we will have tow cases:
If the current number is the same as the start number, then we will call this “Perfect Chain“. Otherwise we will call it “Tail Chain”
The Code:
# Square digit chain.
# Pprojecteuler problem No 92
num = 1
the_chain=[]
def get_square_digit_chain(n):
tot=0
the_chain.append(n)
while n != 0:
tot=0
for each in str(n):
tot= tot + int(each)**2
n = tot
if n in the_chain:
return n
else:
the_chain.append(n)
#We ask the user to enter a number.
num =int(input(“Enter a number “))
chain_closed = get_square_digit_chain(num)
if chain_closed == 1:
print(“We have a Happy Number”)
print(the_chain,’This is Open Chain’)
else:
if chain_closed == num:
print(“We have a Perfect Chain”)
print(the_chain,’Closed on’,chain_closed)
else:
print(“We have a Tail Chain”)
print(the_chain,’Closed on’,chain_closed)
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Python : Curious Number
Python: Curious Number
Problem No.34 in ProjectEuler
Definition: A number is Curious Number if the factorial of their digits equal to the number itself.
Example: 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
Our Task: We will write two functions, first one will get (return) all digits in the number, then another function to get the factorial of each digits in that number then with If statement we will examine the result.Enhancement: We will ask the user to enter a number and we will check if it is a Curious Number.
We will reuse some of our functions that we wrote in previous posts.
The Code:
digs=[]
print(‘\nEnter a number to see if it is a Curious Number or not.’)
num=input (‘\nEnter a number: ‘)
num=input (‘Enter a number: ‘)
tot=0
# To get the digits In a number
def digits_in_num (num):
for each in str(num):
digs.append(each)
# To get the Factorial of a number
def Factorial_digit_sum(num):
if (num == 0) :
return 1
else:
return num * Factorial_digit_sum(num-1)
for each in digs:
print(‘factorial :’,each,’ is ‘,Factorial_digit_sum(int(each)))
tot = tot + Factorial_digit_sum(int(each))
print(‘\nTotal sum of the Factorial of each digits is: ‘,tot)
if int(num) == tot:
print(num ,’Is a Curious Number.’)
else:
print(num ,’Is NOT Curious Number.’)
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