Pandigital products : Problem 32 : Project Euler : Python Code

Pandigital products : Problem 32 : Project Euler

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.
The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

 

 

First note that if m X n = P, by concatenating m,n and P it will be a 9-digit pandigital number if both in m & n none of the digits are repeated. I have use the function checkOneDigit(numb) to check whether in m or n frequncy of any digit is greater than 1. checkPandigital(numb) simply checks whether a number is n-digit pandigital or not. Next task is to determine the upper limit of m and n. Observe that if m or n is a 5-digit number, after concatenation it will never form a 9-digit pandigital number. Since P will be of atleast 5-digits. So after concatenation the number will consist of 11 digits (even if m is of 1 digit number) Ex. 12345 X m = abcde, after concatenation it results in 12345mabcde, (at-least 11 digits) by Pigeon Hole Principle at least on of the digits 1,2,3,....,9 has to be repeated. Hence Max(m,n) <= 4999, within this range m and n are filtered using checkOneDigit(numb)

Python Code 

  

def pandigitalProducts():

    def checkPandigital(numb):
        flag = True
        strNumber = str(numb)
        length = len(strNumber)
        if '0' in strNumber:
            return False
        nList = [0 for i in range (length)]
        for i in range (length):
            nList[i] = int(strNumber[i])
            if int(strNumber[i]) > length:
                flag = False
        if 0  in nList:
            return False       
        if flag == False:
            return False
        else:
            for i in range (length):
                if nList.count(int(strNumber[i])) > 1 :
                    flag = False
                    break
            return flag
          
    def checkOneDigit(numb):
        flag = True
        strNumber = str(numb)
        length = len(strNumber)
        if '0' in strNumber:
            return False
        nList = [0 for i in range (length)]
        for i in range (length):
            nList[i] = int(strNumber[i])
        for i in range (length):
            if nList.count(int(strNumber[i])) > 1 :
                    flag = False
                    break
        return flag
              
    sumPandigital = 0
    upperLimit = 9999
    n = 1
    pandigitalList = ()
    """ pandigitalList contains the all possible values of m and n """
    while n <= upperLimit:
        if checkOneDigit(n) == True:
            pandigitalList = pandigitalList + (n,)
        n = n+1  
    """print pandigitalList    """
    panProduct = ()
    sumP = 0  
    for d in pandigitalList:
        m = d
        for x in pandigitalList:
            product = 1
            n = x
            product = m*n
            strNumb = ""
            strNumb = strNumb + (str(m)+str(n)+str(product))
            if len(strNumb) == 9:
                if checkPandigital(int(strNumb)) == True:
                    if product not in panProduct:
                        panProduct = panProduct + (product,)
                        sumP = sumP + product
                        print m,"X",n,"=",product,"Partial Sum--------------",sumP  
    print "Required Sum is:",sumP  
pandigitalProducts()        

Pandigital prime : Problem 41 : Project Euler : Python Code

Pandigital prime : Problem 41 : Project Euler

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
What is the largest n-digit pandigital prime that exists?

 Given that 2143 is a 4-digit pandigital prime. A 5-digit or 6-digit pandigital number cannot be prime because sum of there digits is
1+2+3+4+5 = 15 which is divisible by 3 and also
1+2+3+4+5+6 = 21 which is also divisible by 3 hence the numbers will be divisible by 3 and cannot be prime so we can set the lower limit to be 654321 and upper limit to be 7654321, since a 8-digit or 9-digit pandigital number cannot be a prime by similar arguments.

Python Code

def PandigitalPrime():
    upperLimit = 7654321
    def checkPrime(numb):
        prime = True
        if numb == 1:
            return False
        else:   
            for i in range(int(numb/2)):
                i = i+2
                if numb%i == 0:
                    prime = False
                    break
            if numb == 2:
                return True
            else:           
                return prime
               
    def checkPandigital(numb):
        flag = True
        strNumber = str(numb)
        length = len(strNumber)
        nList = [0 for i in range (length)]
        for i in range (length):
            nList[i] = int(strNumber[i])
            if int(strNumber[i]) > length:
                flag = False
        if 0  in nList:
            return False        
        if flag == False:
            return False
        else:
            for i in range (length):
                if nList.count(int(strNumber[i])) > 1 :
                    flag = False
                    break
            return flag                        
                   
    n =  654321
    largestPandigitalPrime = 0
    while n <= upperLimit:
        if checkPandigital(n) == False:
            n = n + 2
            continue
        if checkPrime(n) == True:
            largestPandigitalPrime = n
            print len(str(n)),"th digit pandigital prime:",n
        n = n + 2
    print "largest Pandigital Prime is:",largestPandigitalPrime   
PandigitalPrime()                       

Coded triangle numbers : Problem 42 : Project Euler : JAVA Code

Coded triangle numbers : Problem 42 : Project Euler

The n^th term of the sequence of triangle numbers is given by, t_n = ½n(n+1); so the first ten triangle numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t₁₀. If the word value is a triangle number then we shall call the word a triangle word.
Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words?

Note that the file words.txt has been modified, an extra ',' is added at the end of the file for easy extraction of the names. Download it and then edit it to add a ',' after "YOUTH" (the last word of the file)

JAVA Code

import java.io.*;
import java.util.*;
public class CodedTriangleNumbers {
    public static void main(String[] args) throws IOException {
        String fileName = "words.txt";
        String allNames = "";
       
        FileReader fr = new FileReader(fileName);   
        BufferedReader br = new BufferedReader(fr);
        allNames=br.readLine();
        br.close();
        fr.close();
       
        NameExtractionScores nms = new NameExtractionScores();
        System.out.println("Total number of names: "+nms.namesCount(allNames));
        System.out.println("LIST of the names:");
        nms.namesExtraction(allNames);
        System.out.println("Number of triangle words:"+nms.namesTraingular());
        }
    }
class NameExtractionScores {
    public int namesCount(String allNames) {
        for (int i = 0; i < allNames.length(); i++) {
            if (allNames.charAt(i) == ',')
                nameCount++;
            }
        return nameCount;   
        }
   
    public void namesExtraction (String allNames) {
        names = new String[nameCount];
        String temp = "";
        int k = 0;
        for (int i = 0; i < allNames.length(); i++) {
            temp = "";
           
            while ( allNames.charAt(i) != ',' ) {
                if (allNames.charAt(i) != '"')
                    temp += allNames.charAt(i++);
                else   
                    i++;
                }
               
            names[k] = "";
            names[k] +=temp;
            System.out.println(names[k++]);   
            }
        }   
    public int namesTraingular() {
        for (int i = 0; i < nameCount; i++) {
            int cSum = 0;
            boolean tFlag = false;
            for (int j = 0; j < names[i].length(); j++)
                cSum += (names[i].charAt(j)-64);
            int n = 1;
            while (n <= cSum) {
                if (n*(n+1) == 2*cSum) {
                    tFlag = true;
                    break;
                    }
                n++;   
                }   
            if (tFlag == true)
                nameTraingularTotal++;   
            }
        return nameTraingularTotal;   
        }   
    private int nameTraingularTotal = 0;   
    private int nameCount = 0;
    private String[] names;
    }       

Digit factorial chains : Problem 74 : Project Euler Python Code

Digit factorial chains : Problem 74 : Project Euler Python Code

The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:

1! + 4! + 5! = 1 + 24 + 120 = 145

Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that exist:

169 → 363601 → 1454 → 169
871 → 45361 → 871
872 → 45362 → 872

It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,

69 → 363600 → 1454 → 169 → 363601 (→ 1454)
78 → 45360 → 871 → 45361 (→ 871)
540 → 145 (→ 145)

Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.
How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?

Python Code 

 

def digitFactorialChains():
    def fact(digit):
        l=1
        for i in range(digit):
            l=l*digit
            digit=digit-1
        return l
    number = 1
    digitfactsum = 0
    counter = 0
    while number < 10**6:
            chain = ()
            chain = chain + (number,)
        strnumber = str(number) 
        length = 1
            while True:
                digitfactsum = 0
            for i in range(len(strnumber)):
                    digitfactsum = digitfactsum + fact(int(strnumber[i]))
                if digitfactsum in chain:
                    break
                else:
                    chain = chain + (digitfactsum,)  
                    length = length + 1
                    strnumber = str(digitfactsum)
        """print chain,length  prints the sequnce with length"""         
        if length == 60:
                 counter = counter + 1  
             """You wont get bored if you print the loop iteration!"""  
             print number       
             number = number + 1
    print "chains, with a starting number below one million, contain exactly sixty non-repeating terms:",counter
digitFactorialChains() 

 

Bouncy numbers : Problem 112 : Project Euler : Python Code

Bouncy numbers : Problem 112 : Project Euler

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.

Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.
Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.
Find the least number for which the proportion of bouncy numbers is exactly 99%.

Python Code

def bouncyNumbers():
    def increasing(numb):
        iflag = True
        strNumb = str(numb)
        for i in range(len(strNumb)-1):
            if int(strNumb[i]) > int(strNumb[i+1]):
                iflag = False
                break
        return iflag
    def decreasing(numb):
        dflag = True
        strNumb = str(numb)
        for i in range(len(strNumb)-1):
            if int(strNumb[i]) < int(strNumb[i+1]):
                dflag = False
                break
        return dflag
    bouncyCount = 0
    percentagebCount = 0.0
    n = 99
    while int(percentagebCount) != 99:   
        n = n + 1
        if increasing(n) == False and decreasing(n) == False:
            bouncyCount = bouncyCount + 1
        percentagebCount = (bouncyCount*100.0)/n
        print percentagebCount,n
    print "least number for which the proportion of bouncy numbers is exactly 99% is :",n   
bouncyNumbers()               

Names scores : Problem 22 : Project Euler JAVA Code

Names scores : Problem 22 : Project Euler

Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 × 53 = 49714.

What is the total of all the name scores in the file?

 JAVA CODE

import java.io.*;
import java.util.*;
public class NamesScore {
    public static void main(String[] args) throws IOException {
        String fileName = "names.txt";
        String allNames = "";
       
        FileReader fr = new FileReader(fileName);   
        BufferedReader br = new BufferedReader(fr);
        allNames=br.readLine();
        br.close();
        fr.close();
       
        NameExtractionScores nms = new NameExtractionScores();
        System.out.println("Total number of names: "+nms.namesCount(allNames));
        System.out.println("LIST of the names (unsorted):");
        nms.namesExtraction(allNames);
        System.out.println("Score is: "+nms.namesScore());
        }
    }
class NameExtractionScores {
    public int namesCount(String allNames) {
        for (int i = 0; i < allNames.length(); i++) {
            if (allNames.charAt(i) == ',')
                nameCount++;
            }
        return nameCount;   
        }
   
    public void namesExtraction (String allNames) {
        names = new String[nameCount];
        String temp = "";
        int k = 0;
        for (int i = 0; i < allNames.length(); i++) {
            temp = "";
           
            while ( allNames.charAt(i) != ',' ) {
                if (allNames.charAt(i) != '"')
                    temp += allNames.charAt(i++);
                else   
                    i++;
                }
               
            names[k] = "";
            names[k] +=temp;
            System.out.println(names[k++]);   
            }
        }   
    public int namesScore() {
        Arrays.sort(names);
        for (int i = 0; i < nameCount; i++) {
            int pSum = 0;
            System.out.println(names[i]);   
            for (int j = 0; j < names[i].length(); j++)
                pSum += (names[i].charAt(j)-64);
            nameScoreTotal += (pSum*(i+1));   
            }
        return nameScoreTotal;   
        }    
    private int nameScoreTotal = 0;   
    private int nameCount = 0;
    private String[] names;
    }       

Square digit chains : Problem 92 : Project Euler

Square digit chains : Problem 92 : Project Euler

A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.
For example,

44 → 32 → 13 → 10 → 1 → 1
85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89

Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.
How many starting numbers below ten million will arrive at 89?

Python Code

def squareDigitChains():
    squareDigitCount = 0
    for i in range(1,10000000):
        strnumber = ""
        strnumber = str(i)
        sqnumber = 0
        temp = strnumber
        while True:
            sqnumber = 0
            for j in range(len(strnumber)):
                sqnumber = sqnumber + int(strnumber[j])**2
            strnumber = str(sqnumber)   
            if sqnumber == 1 or sqnumber == 89:
                break
        if sqnumber == 89:
            print temp
            squareDigitCount = squareDigitCount + 1
    print "Required Counter is:",squareDigitCount               
squareDigitChains()