From de9b1a51620348e6f4a6b884508b2e6c1c7e69d2 Mon Sep 17 00:00:00 2001 From: Matthew Ellison Date: Thu, 27 Sep 2018 13:39:26 -0400 Subject: [PATCH] Moved ProjectEuler to its own repository --- ProjectEuler/C++/Problem1.cpp | 45 ------- ProjectEuler/C++/Problem12.cpp | 57 --------- ProjectEuler/C++/Problem2.cpp | 43 ------- ProjectEuler/C++/Problem3.cpp | 104 ---------------- ProjectEuler/Problem1.m | 30 ----- ProjectEuler/Problem10.m | 6 - ProjectEuler/Problem11.m | 152 ----------------------- ProjectEuler/Problem12.m | 51 -------- ProjectEuler/Problem12_2.m | 52 -------- ProjectEuler/Problem13.m | 214 --------------------------------- ProjectEuler/Problem2.m | 27 ----- ProjectEuler/Problem3.m | 48 -------- ProjectEuler/Problem4.m | 50 -------- ProjectEuler/Problem5.m | 43 ------- ProjectEuler/Problem6.m | 21 ---- ProjectEuler/Problem7.m | 21 ---- ProjectEuler/Problem8.m | 63 ---------- ProjectEuler/Problem9.m | 39 ------ ProjectEuler/Reverse.m | 19 --- 19 files changed, 1085 deletions(-) delete mode 100644 ProjectEuler/C++/Problem1.cpp delete mode 100644 ProjectEuler/C++/Problem12.cpp delete mode 100644 ProjectEuler/C++/Problem2.cpp delete mode 100644 ProjectEuler/C++/Problem3.cpp delete mode 100644 ProjectEuler/Problem1.m delete mode 100644 ProjectEuler/Problem10.m delete mode 100644 ProjectEuler/Problem11.m delete mode 100644 ProjectEuler/Problem12.m delete mode 100644 ProjectEuler/Problem12_2.m delete mode 100644 ProjectEuler/Problem13.m delete mode 100644 ProjectEuler/Problem2.m delete mode 100644 ProjectEuler/Problem3.m delete mode 100644 ProjectEuler/Problem4.m delete mode 100644 ProjectEuler/Problem5.m delete mode 100644 ProjectEuler/Problem6.m delete mode 100644 ProjectEuler/Problem7.m delete mode 100644 ProjectEuler/Problem8.m delete mode 100644 ProjectEuler/Problem9.m delete mode 100644 ProjectEuler/Reverse.m diff --git a/ProjectEuler/C++/Problem1.cpp b/ProjectEuler/C++/Problem1.cpp deleted file mode 100644 index 447ed80..0000000 --- a/ProjectEuler/C++/Problem1.cpp +++ /dev/null @@ -1,45 +0,0 @@ -//ProjectEuler/C++/Problem1.cpp -//Matthew Ellison -// Created: 9-28-18 -//Modified: 9-28-18 -//What is the sum of all the multiples of 3 or 5 that are less than 1000 - - -#include -#include -#include - - -int main(){ - unsigned long fullSum = 0; //For the sum of all the numbers - std::vector numbers; //Holds all the numbers - - std::chrono::high_resolution_clock::time_point startTime = std::chrono::high_resolution_clock::now(); - //Step through every number < 1000 and see if either 3 or 5 divide it evenly - for(int cnt = 0;cnt < 1000;++cnt){ - //If either divides it then add it to the vector - if((cnt % 3) == 0){ - numbers.push_back(cnt); - } - else if((cnt % 5) == 0){ - numbers.push_back(cnt); - } - } - - //Get the sum of all numbers - for(unsigned long num : numbers){ - fullSum += num; - } - - //Calculate the time needed to run the algorithm - std::chrono::high_resolution_clock::time_point endTime = std::chrono::high_resolution_clock::now(); - std::chrono::high_resolution_clock::duration dur = std::chrono::duration_cast(endTime - startTime); - - //Print the output - std::cout << "The sum of all the numbers < 1000 that are divisible by 3 or 5 is " << fullSum - << "\nIt took " << dur.count() << " nanoseconds to run this algorithm" << std::endl; - - std::cin.get(); - - return 0; -} diff --git a/ProjectEuler/C++/Problem12.cpp b/ProjectEuler/C++/Problem12.cpp deleted file mode 100644 index 30ab552..0000000 --- a/ProjectEuler/C++/Problem12.cpp +++ /dev/null @@ -1,57 +0,0 @@ -//ProjectEuler/C++/Problem12.cpp -//Matthew Ellison -// Created: 9-27-18 -//Modified: 9-28-18 -//This file contains the program to calculate the answer to Problem 12 on ProjectEuler.net - - -#include -#include //For the timer - - -//Counter how many divisors number has -unsigned long countDivisors(unsigned long number); - - -int main(){ - bool found = false; //To flag whether the number has been found - unsigned long sum = 1; //The sum of the numbers up to counter - unsigned long counter = 2; //The next number to be added to sum - const unsigned long goalDivisors = 500; //The number of divisors that is being sought - - std::chrono::high_resolution_clock::time_point startTime = std::chrono::high_resolution_clock::now(); - while(!found){ - //If the number of divisors is correct set the flag - if(countDivisors(sum) > goalDivisors){ - found = true; - } - //Otherwise add to the sum and increase the next numeber - else{ - sum += counter; - ++counter; - } - } - std::chrono::high_resolution_clock::time_point endTime = std::chrono::high_resolution_clock::now(); - - //Print the results - std::cout << "The triangular number " << sum << " is made with all number >= " << counter - 1 << " and has " << countDivisors << " divisors" << std::endl; - std::cout << "The problem took " << std::chrono::duration_cast(std::chrono::high_resolution_clock::duration(endTime - startTime)).count() << " milliseconds" << std::endl; - - std::cin.get(); - - return 0; -} - - -unsigned long countDivisors(unsigned long number){ - unsigned long numDivisors = 0; //Holds the number of divisors - //You only need to go to sqrt(number). cnt * cnt is faster than sqrt() - for(int cnt = 1;cnt * cnt < number;++cnt){ - //Check if the counter evenly divides the number - //If it does the counter and the other number are both divisors - if((number % cnt) == 0){ - numDivisors += 2; - } - } - return numDivisors; -} diff --git a/ProjectEuler/C++/Problem2.cpp b/ProjectEuler/C++/Problem2.cpp deleted file mode 100644 index e013484..0000000 --- a/ProjectEuler/C++/Problem2.cpp +++ /dev/null @@ -1,43 +0,0 @@ -//ProjectEuler/C++/Problem2.cpp -//Matthew Ellison -// Created: 9-28-18 -//Modified: 9-28-18 -//The sum of the even Fibonacci numbers less than 4,000,000 - - -#include -#include -#include - - -int main(){ - unsigned long fullSum = 2; //Holds the sum of all the numbers - std::vector fib = {1, 1}; //Holds the Fibonacci numbers - unsigned long nextFib = 2; //Holds the next Fibonacci number - - std::chrono::high_resolution_clock::time_point startTime = std::chrono::high_resolution_clock::now(); //Start the timer - while(nextFib < 4000000){ - //If it is an even number add it to the sum - if(nextFib % 2){ - fullSum += nextFib; - } - //Move all the fib numbers down and calculate the next one - fib.at(0) = fib.at(1); - fib.at(1) = nextFib; - nextFib = fib.at(1) + fib.at(0); - //You could do this keeping all the fib numbers and sum at the end, - //but this way will be faster because you are only handling every number twice - //and you don't have to expand the vector - } - //Calculate the time needed for the algorithm - std::chrono::high_resolution_clock::time_point endTime = std::chrono::high_resolution_clock::now(); //End the timer - std::chrono::high_resolution_clock::duration dur = std::chrono::duration_cast(endTime - startTime); - - //Print the resultss - std::cout << "The sum of the even Fibonacci numbers less than 4,000,000 is " << fullSum - << "\nIt took " << dur.count() << " nanoseconds to run this algorithm" << std::endl; - - //Pause before ending - std::cin.get(); - return 0; -} \ No newline at end of file diff --git a/ProjectEuler/C++/Problem3.cpp b/ProjectEuler/C++/Problem3.cpp deleted file mode 100644 index 16728f9..0000000 --- a/ProjectEuler/C++/Problem3.cpp +++ /dev/null @@ -1,104 +0,0 @@ -//ProjectEuler/C++/Problem3.cpp -//Matthew Ellison -// Created: 9-28-18 -//Modified: 9-28-18 -//The largest prime factor of 600851475143 - - -#include -#include -#include -#include - -std::vector generatePrimes(const unsigned long long number); -unsigned long long findLargest(const std::vector list); - - -int main(){ - const unsigned long long goalNumber = 600851475143; //The number you are trying to find the factors of - std::vector primes; //Holds the list of prime numbers - std::vector factors; //Holds the factors of goalNumber - unsigned long long number = goalNumber; //The number that is left after taking the prime numbers out - bool found = false; - - - std::chrono::high_resolution_clock::time_point startTime = std::chrono::high_resolution_clock::now(); - //Generate the primes - primes = generatePrimes(ceil(sqrt(goalNumber))); - //Check the primes against the number - //62113 numbers at this point - while(!found){ - //Loop through the list of primes - for(int cnt = 0;cnt < primes.size();){ - //See if after dividing a prime out it is left with a whole number - if((number % primes.at(cnt)) == 0){ - number /= primes.at(cnt); - factors.push_back(primes.at(cnt)); - } - //If you didn't find a prime then advance to the next possible number - //If you did find a prime then stay where you are in the list incase it occurs more than once - else{ - ++cnt; - } - } - //If the remaining number is 1 then you just added the last number to the factors - if(number == 1){ - found = true; - } - } - - //Look for the largest number in the vector - unsigned long long maxNum = findLargest(factors); - - //Calculate the amount of time it took to run the algorithm - std::chrono::high_resolution_clock::time_point endTime = std::chrono::high_resolution_clock::now(); - std::chrono::high_resolution_clock::duration dur = std::chrono::duration_cast(endTime - startTime); - - //Print the results - std::cout << "The largest factor of the number " << goalNumber << " is " << maxNum - << "\nIt took " << dur.count() << " milliseconds to run this algorith" << std::endl; - - //Pause before exiting - std::cin.get(); - return 0; -} - -std::vector generatePrimes(const unsigned long long goalNumber){ - std::vector primes; - bool foundFactor = false; - - //Start at 2 so we can skip 1 - if(goalNumber >= 2){ - primes.push_back(2); - } - //We can skip all of the even numbers - for(unsigned long long possiblePrime = 3;possiblePrime < goalNumber;possiblePrime += 2){ - //Step through every element in the current primes. If you don't find anything that divides it, it must be a prime itself - for(int cnt = 0;(cnt < primes.size()) && ((primes.at(cnt) * primes.at(cnt)) < goalNumber);++cnt){ - if(fmod(((double)possiblePrime / primes.at(cnt)), 1) == 0){ - foundFactor = true; - break; - } - } - //If you didn't find a factor then it must be prime - if(!foundFactor){ - primes.push_back(possiblePrime); - } - //If you did find a factor you need to reset the flag - else{ - foundFactor = false; - } - } - return primes; -} - -unsigned long long findLargest(const std::vector list){ - unsigned long long maxNum = 0; - for(unsigned long long num : list){ - if(num > maxNum){ - maxNum = num; - } - } - - return maxNum; -} diff --git a/ProjectEuler/Problem1.m b/ProjectEuler/Problem1.m deleted file mode 100644 index 5e999bb..0000000 --- a/ProjectEuler/Problem1.m +++ /dev/null @@ -1,30 +0,0 @@ -%ProjectEuler/Problem1.m -%This is a script to answer Problem 1 for Project Euler -%What is the sum of all the multiples of 3 or 5 that are less than 1000 - -%Setup your variables -fullSum = 0; %To hold the sum of all the numbers -numbers = 0; %To hold all of the numbers -counter = 0; %The number. It must stay below 1000 - -while(counter < 1000) - %See if the number is a multiple of 3 - if(mod(counter, 3) == 0) - numbers(end + 1) = counter; - %See if the number is a multiple of 5 - elseif(mod(counter, 5) == 0) - numbers(end + 1) = counter; - end - - %Increment the number - ++counter; -end -%When done this way it removes the possibility of duplicate numbers - -fullSum = sum(numbers); -ans = fullSum - -%Cleanup your variables -clear fullSum; -clear numbers; -clear counter; diff --git a/ProjectEuler/Problem10.m b/ProjectEuler/Problem10.m deleted file mode 100644 index 08286cb..0000000 --- a/ProjectEuler/Problem10.m +++ /dev/null @@ -1,6 +0,0 @@ -%ProjectEuler/Problem10.m -%This is a script to answer Problem 10 for Project Euler -%Find the sum of all the primes below two million. - -%Print the answer -sum(primes(2000000)) diff --git a/ProjectEuler/Problem11.m b/ProjectEuler/Problem11.m deleted file mode 100644 index 731d0a9..0000000 --- a/ProjectEuler/Problem11.m +++ /dev/null @@ -1,152 +0,0 @@ -%ProjectEuler/Problem11.m -%This is a script to answer Problem 11 for Project Euler -%What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid? -%{ -08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 -49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 -81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 -52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 -22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 -24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 -32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 -67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 -24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 -21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 -78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 -16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 -86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 -19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 -04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 -88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 -04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 -20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 -20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 -01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48 -%} - -%Create your variables -grid = [08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08; - 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00; - 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65; - 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91; - 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80; - 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50; - 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70; - 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21; - 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72; - 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95; - 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92; - 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57; - 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58; - 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40; - 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66; - 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69; - 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36; - 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16; - 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54; - 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48]; -currentLocation = [1, 1]; %The location you are checking now -currentProduct = [0 0 0 0]; %The product you are ucrrently looking at -greatestProduct = [0 0 0 0]; %The greatest product of values you have found -finished = false; - - -%Loop until you reach the last element -startTime = clock(); -while(~finished) - left = false; - right = false; - down = false; - - %Check which directions you can go - %When moving you will be moving 4 locations - if((currentLocation(2) - 3) >= 1) - left = true; - end - if((currentLocation(2) + 3) <= size(grid)(2)) - right = true; - end - if((currentLocation(1) + 3) <= size(grid)(1)) - down = true; - end - - %Check the possible directions and check against greatest - %Left - if(left) - currentProduct = grid(currentLocation(1),currentLocation(2):-1:(currentLocation(2) - 3)); - %If the current numbers' product is greater than the greatest product so far, replace it - if(prod(currentProduct) > prod(greatestProduct)) - greatestProduct = currentProduct; - end - end - %Right - if(right) - currentProduct = grid(currentLocation(1), currentLocation(2):(currentLocation(2) + 3)); - %If the current numbers' product is greater than the greatest product so far, replace it - if(prod(currentProduct) > prod(greatestProduct)) - greatestProduct = currentProduct; - end - end - %Down - if(down) - currentProduct = grid(currentLocation(1):(currentLocation(1) + 3), currentLocation(2)); - %If the current numbers' product is greater than the greatest product so far, replace it - if(prod(currentProduct) > prod(greatestProduct)) - greatestProduct = currentProduct; - end - end - %LeftDown - if(left && down) - currentProduct(1) = grid(currentLocation(1), currentLocation(2)); - currentProduct(2) = grid(currentLocation(1) + 1,currentLocation(2) - 1); - currentProduct(3) = grid(currentLocation(1) + 2,currentLocation(2) - 2); - currentProduct(4) = grid(currentLocation(1) + 3,currentLocation(2) - 3); - %If the current numbers' product is greater than the greatest product so far, replace it - if(prod(currentProduct) > prod(greatestProduct)) - greatestProduct = currentProduct; - end - end - %RightDown - if(right && down) - currentProduct(1) = grid(currentLocation(1), currentLocation(2)); - currentProduct(2) = grid(currentLocation(1) + 1,currentLocation(2) + 1); - currentProduct(3) = grid(currentLocation(1) + 2,currentLocation(2) + 2); - currentProduct(4) = grid(currentLocation(1) + 3,currentLocation(2) + 3); - %If the current numbers' product is greater than the greatest product so far, replace it - if(prod(currentProduct) > prod(greatestProduct)) - greatestProduct = currentProduct; - end - end - - %Move to the next column - ++currentLocation(2); - %If you have moved too far in the columns move back to the beginning and to the next row - if(currentLocation(2) > size(grid)(2)) - currentLocation(2) = 1; - ++currentLocation(1); - end - %If the row is currently greater than what is available you have traversed the list - if(currentLocation(1) > size(grid)(1)) - finished = true; - end -end -greatestProduct = reshape(greatestProduct, 1, 4); %For some reason it is coming out a 4X1 instead of 1X4 -endTime = clock(); - -%Print the result -totalTime = etime(endTime, startTime) -greatestProduct -prod(greatestProduct) - -%Cleanup the variables -clear down; -clear right; -clear left; -clear finished; -clear startTime; -clear endTime; -clear totalTime; -clear grid; -clear greatestProduct; -clear currentLocation; -clear currentProduct; diff --git a/ProjectEuler/Problem12.m b/ProjectEuler/Problem12.m deleted file mode 100644 index 978a3d7..0000000 --- a/ProjectEuler/Problem12.m +++ /dev/null @@ -1,51 +0,0 @@ -%ProjectEuler/Problem12.m -%This is a script to answer Problem 12 for Project Euler -%What is the value of the first triangle number to have over five hundred divisors? - -%Setup your variables -counter = 1; %To hold the current number you have counted up to -number = 0; %To hold the current triangle number -found = false; %To tell when the answer has been found -numDivisors = 0; -maxDivisors = 0; -maxCounter = 0; - - -startTime = clock(); -while(~found) - %Get your next triangle number - number = sym(sum([1:counter])); - - %See if it has 500 divisors - numDivisors = size(divisors(number))(2); - if(numDivisors > maxDivisors) - maxDivisors = numDivisors; - maxCounter = counter; - end - if(numDivisors > 500) - found = true; - else - counter = counter + 1; - end -end -endTime = clock(); - -%Print your result -totalTime = etime(endTime, startTime) -topNumber = counter -double(number) - -%Cleanup your variables -clear counter; -clear number; -clear found; -clear topNumber; -clear startTime; -clear endTime; -clear totalTime; -clear numDivisors; -clear maxDivisors; -clear maxCounter; - -%This will take 6-7 hours to run. I got the C++ answer in 2.8 seconds -%Try to find a more efficient way to run this diff --git a/ProjectEuler/Problem12_2.m b/ProjectEuler/Problem12_2.m deleted file mode 100644 index dd4fda1..0000000 --- a/ProjectEuler/Problem12_2.m +++ /dev/null @@ -1,52 +0,0 @@ -%ProjectEuler/Problem12.m -%This is a script to answer Problem 12 for Project Euler -%What is the value of the first triangle number to have over five hundred divisors? - -found = false; -numSum = 1; -counter = 2; -numDivisors = 0; -goalDivisors = 500; - -startTime = clock(); -while(~found) - %Count the number of divisors - numDivisors = 0; - divCounter = 0; - while((divCounter * divCounter) < numSum) - if(mod(numSum, divCounter) == 0) - numDivisors += 2; - end - ++divCounter; - end - - %Check if there are enough divisors - if(numDivisors > goalDivisors) - found = true; - else - numSum += counter; - ++counter; - end -end -endTime = clock(); - -%Print the result -number = numSum -highestNumber = counter - 1 -numDivisors = numDivisors -totalTime = etime(endTime, startTime) - -%Cleanup the variables -clear found; -clear numSum; -clear counter; -clear numDivisors; -clear goalDivisors; -clear startTime; -clear endTime; -clear totalTime; -clear numDivisors; -clear divCounter; -clear ans; - -%Returns result in 7.5 minutes diff --git a/ProjectEuler/Problem13.m b/ProjectEuler/Problem13.m deleted file mode 100644 index 116ab24..0000000 --- a/ProjectEuler/Problem13.m +++ /dev/null @@ -1,214 +0,0 @@ -%ProjectEuler/Problem13.m -%This is a script to answer Problem 13 for Project Euler -%Work out the first ten digits of the sum of the following one-hundred 50-digit numbers -%{ -37107287533902102798797998220837590246510135740250 -46376937677490009712648124896970078050417018260538 -74324986199524741059474233309513058123726617309629 -91942213363574161572522430563301811072406154908250 -23067588207539346171171980310421047513778063246676 -89261670696623633820136378418383684178734361726757 -28112879812849979408065481931592621691275889832738 -44274228917432520321923589422876796487670272189318 -47451445736001306439091167216856844588711603153276 -70386486105843025439939619828917593665686757934951 -62176457141856560629502157223196586755079324193331 -64906352462741904929101432445813822663347944758178 -92575867718337217661963751590579239728245598838407 -58203565325359399008402633568948830189458628227828 -80181199384826282014278194139940567587151170094390 -35398664372827112653829987240784473053190104293586 -86515506006295864861532075273371959191420517255829 -71693888707715466499115593487603532921714970056938 -54370070576826684624621495650076471787294438377604 -53282654108756828443191190634694037855217779295145 -36123272525000296071075082563815656710885258350721 -45876576172410976447339110607218265236877223636045 -17423706905851860660448207621209813287860733969412 -81142660418086830619328460811191061556940512689692 -51934325451728388641918047049293215058642563049483 -62467221648435076201727918039944693004732956340691 -15732444386908125794514089057706229429197107928209 -55037687525678773091862540744969844508330393682126 -18336384825330154686196124348767681297534375946515 -80386287592878490201521685554828717201219257766954 -78182833757993103614740356856449095527097864797581 -16726320100436897842553539920931837441497806860984 -48403098129077791799088218795327364475675590848030 -87086987551392711854517078544161852424320693150332 -59959406895756536782107074926966537676326235447210 -69793950679652694742597709739166693763042633987085 -41052684708299085211399427365734116182760315001271 -65378607361501080857009149939512557028198746004375 -35829035317434717326932123578154982629742552737307 -94953759765105305946966067683156574377167401875275 -88902802571733229619176668713819931811048770190271 -25267680276078003013678680992525463401061632866526 -36270218540497705585629946580636237993140746255962 -24074486908231174977792365466257246923322810917141 -91430288197103288597806669760892938638285025333403 -34413065578016127815921815005561868836468420090470 -23053081172816430487623791969842487255036638784583 -11487696932154902810424020138335124462181441773470 -63783299490636259666498587618221225225512486764533 -67720186971698544312419572409913959008952310058822 -95548255300263520781532296796249481641953868218774 -76085327132285723110424803456124867697064507995236 -37774242535411291684276865538926205024910326572967 -23701913275725675285653248258265463092207058596522 -29798860272258331913126375147341994889534765745501 -18495701454879288984856827726077713721403798879715 -38298203783031473527721580348144513491373226651381 -34829543829199918180278916522431027392251122869539 -40957953066405232632538044100059654939159879593635 -29746152185502371307642255121183693803580388584903 -41698116222072977186158236678424689157993532961922 -62467957194401269043877107275048102390895523597457 -23189706772547915061505504953922979530901129967519 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40789923115535562561142322423255033685442488917353, - 44889911501440648020369068063960672322193204149535, - 41503128880339536053299340368006977710650566631954, - 81234880673210146739058568557934581403627822703280, - 82616570773948327592232845941706525094512325230608, - 22918802058777319719839450180888072429661980811197, - 77158542502016545090413245809786882778948721859617, - 72107838435069186155435662884062257473692284509516, - 20849603980134001723930671666823555245252804609722, - 53503534226472524250874054075591789781264330331690]; - -format long; %You need to be able to see 10 digits -sum(nums) -format short; %Set it back to normal - -%Cleanup the variables -clear nums; diff --git a/ProjectEuler/Problem2.m b/ProjectEuler/Problem2.m deleted file mode 100644 index 2fd839c..0000000 --- a/ProjectEuler/Problem2.m +++ /dev/null @@ -1,27 +0,0 @@ -%ProjectEuler/Problem2.m -%This is a script to answer Problem 2 for Project Euler -%The sum of the even Fibonacci numbers less than 4,000,000 - -%Setup your Variables -fib = [1, 1, 2]; %Holds the Fibonacci numbers -currentFib = fib(end) + fib(end - 1); %The current Fibonacci number to be tested -evenFib = [2]; %A subset of the even Fibonacci numbers - -while(currentFib < 4000000) - %Add the number to the list - fib(end + 1) = currentFib; - %If the number is even add it to the even list as well - if(mod(currentFib, 2) == 0) - evenFib(end + 1) = currentFib; - end - - %Set the next Fibonacci - currentFib = fib(end) + fib(end - 1); -end - -sum(evenFib) - -%Cleanup your variables -clear fib; -clear currentFib; -clear evenFib; diff --git a/ProjectEuler/Problem3.m b/ProjectEuler/Problem3.m deleted file mode 100644 index 47c9baa..0000000 --- a/ProjectEuler/Problem3.m +++ /dev/null @@ -1,48 +0,0 @@ -%ProjectEuler/Problem3.m -%This is a script to answer Problem 3 for Project Euler -%The largest prime factor of 600851475143 - -%Setup your variables -number = 600851475143; %The number we are trying to find the greatest prime factor of -primeNums = []; %A list of prime numbers. Will include all prime numbers <= number -factors = []; %For the list of factors of number -tempNum = number; %Used to track the current value if all of the factors were taken out of number -%number = 16; %Used for a test case - -%Get the prime numbers up to sqrt(number). If it is not prime there must be a value <= sqrt -primeNums = primes(sqrt(number)); - -%Setup the loop -counter = 1; -%Start with the lowest number and work your way up. When you reach a number > size(primeNums) you have found all of the factors -while(counter <= size(primeNums)(2)) - - %Divide the number by the next prime number in the list - answer = (tempNum/primeNums(counter)); - - %If it is a whole number add it to the factors - if(mod(answer,1) == 0) - factors(end + 1) = primeNums(counter); - %Set tempNum so that it reflects number/factors - tempNum = tempNum / primeNums(counter); - %Keep the counter where it is in case a factor appears more than once - %Get the new set of prime numbers - primeNums = primes(sqrt(tempNum)); - else - %If it was not an integer increment the counter - ++counter; - end -end -%When the last number is not divisible by a prime number it must be a prime number -factors(end + 1) = tempNum; - -%Print the answer -max(factors) - -%Cleanup your variables -clear counter; -clear tempNum; -clear answer; -clear number; -clear primeNums; -clear factors; diff --git a/ProjectEuler/Problem4.m b/ProjectEuler/Problem4.m deleted file mode 100644 index b615116..0000000 --- a/ProjectEuler/Problem4.m +++ /dev/null @@ -1,50 +0,0 @@ -%ProjectEuler/Problem4.m -%This is a script to answer Problem 4 for Project Euler -%Find the largest palindrome made from the product of two 3-digit numbers - -%Make your variables -answer = 0; %For the product of the two numbers -numbers = [100:999]; %Create an array with a list of all 3 digit numbers -palindromes = []; %Holds all the numbers that are palindromes -%Create 2 counters for an inner loop and an outer loop -%This allows you to multiply 2 numbers from the same array -outerCounter = 1; -innerCounter = 1; - -startTime = clock(); %This is for timing purposes -while(outerCounter < size(numbers)(2)) - innerCounter = outerCounter; %Once you have multiplied 2 numbers there is no need to multiply them again, so skip what has already been done - while(innerCounter < size(numbers)(2)) - %Multiply the two numbers - answer = numbers(outerCounter) * numbers(innerCounter); - - %See if the number is a palindromes - %%WARNING - Ocatave does not have a Reverse function. I had to create one that reversed strings - if(num2str(answer) == Reverse(num2str(answer))) - %Add it to the palindromes list - palindromes(end + 1) = answer; - end - ++innerCounter; %Increment - end - ++outerCounter; %Increment -end -endTime = clock(); %This is for timing purposes -timeTaken = etime(endTime - startTime) %This is for timing purposes - -max(palindromes) - -%Cleanup your variables -clear outerCounter; -clear innerCounter; -clear answer; -clear numbers; -clear palindromes; -clear startTime; -clear endTime; -clear timeTaken; - - -%This way is slow. I would like to find a faster way -%{ -The palindrome can be written as: abccba Which then simpifies to: 100000a + 10000b + 1000c + 100c + 10b + a And then: 100001a + 10010b + 1100c Factoring out 11, you get: 11(9091a + 910b + 100c) So the palindrome must be divisible by 11. Seeing as 11 is prime, at least one of the numbers must be divisible by 11 -%} diff --git a/ProjectEuler/Problem5.m b/ProjectEuler/Problem5.m deleted file mode 100644 index f28bbf6..0000000 --- a/ProjectEuler/Problem5.m +++ /dev/null @@ -1,43 +0,0 @@ -%ProjectEuler/Problem5.m -%This is a script to answer Problem 5 for Project Euler -%What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? - -%Create your variables -nums = [1:20]; -factors = [1]; %The factors that are already in the number -list = []; %For a temperary list of the factors of the current number -counter = 1; - -%You need to find the factors of all the numbers from 1->20 -while(counter <= size(nums)(2)) - list = factor(nums(counter)); - - %Search factors and try to match all elements in list - listCnt = 1; - factorCnt = 1; - while(listCnt <= size(list)(2)) - if((factorCnt > size(factors)(2)) || (factors(factorCnt) > list(listCnt))) - %If it was not found add the factor to the list for the number and reset the counters - factors(end + 1) = list(listCnt); - factors = sort(factors); - factorCnt = 1; - listCnt = 1; - elseif(factors(factorCnt) == list(listCnt)) - ++listCnt; - ++factorCnt; - else - ++factorCnt; - end - end - ++counter; -end - -prod(factors) - -%Cleanup your variables -clear counter; -clear factorCnt; -clear listCnt; -clear list; -clear nums; -clear factors; diff --git a/ProjectEuler/Problem6.m b/ProjectEuler/Problem6.m deleted file mode 100644 index 9b168ab..0000000 --- a/ProjectEuler/Problem6.m +++ /dev/null @@ -1,21 +0,0 @@ -%ProjectEuler/Problem4.m -%This is a script to answer Problem 4 for Project Euler -%Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum. - -%Setup your variables -nums = [1:100]; -squares = nums.^2; %Square every number in the list nums -sumOfSquares = sum(squares); %Get the sum of all the elements in the list squares -squareOfSums = sum(nums)^2; %Get the sum of all the elements in the list nums and square the answer -value = abs(squareOfSums - sumOfSquares); %Get the difference of the 2 numbers -%This could all be done on one line, but this is less confusing - -%Print the value -value - -%Cleanup your variables -clear nums; -clear squares; -clear sumOfSquares; -clear squareOfSums; -clear value; diff --git a/ProjectEuler/Problem7.m b/ProjectEuler/Problem7.m deleted file mode 100644 index 50db288..0000000 --- a/ProjectEuler/Problem7.m +++ /dev/null @@ -1,21 +0,0 @@ -%ProjectEuler/Problem7.m -%This is a script to answer Problem 7 for Project Euler -%What is the 10001th prime number? - -%Setup the variables -counter = 1000; -primeList = []; - -%Cycle through the prime numbers until you get the correct number -while(size(primeList)(2) < 10001) - primeList = primes(counter); - counter += 1000; -end - -%Print the number -primeList(10001) - -%Cleanup the variables -clear ans; -clear counter; -clear primeList; diff --git a/ProjectEuler/Problem8.m b/ProjectEuler/Problem8.m deleted file mode 100644 index ff907d8..0000000 --- a/ProjectEuler/Problem8.m +++ /dev/null @@ -1,63 +0,0 @@ -%ProjectEuler/Problem8.m -%This is a script to answer Problem 8 for Project Euler -%Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product? -%{ -73167176531330624919225119674426574742355349194934 -96983520312774506326239578318016984801869478851843 -85861560789112949495459501737958331952853208805511 -12540698747158523863050715693290963295227443043557 -66896648950445244523161731856403098711121722383113 -62229893423380308135336276614282806444486645238749 -30358907296290491560440772390713810515859307960866 -70172427121883998797908792274921901699720888093776 -65727333001053367881220235421809751254540594752243 -52584907711670556013604839586446706324415722155397 -53697817977846174064955149290862569321978468622482 -83972241375657056057490261407972968652414535100474 -82166370484403199890008895243450658541227588666881 -16427171479924442928230863465674813919123162824586 -17866458359124566529476545682848912883142607690042 -24219022671055626321111109370544217506941658960408 -07198403850962455444362981230987879927244284909188 -84580156166097919133875499200524063689912560717606 -05886116467109405077541002256983155200055935729725 -71636269561882670428252483600823257530420752963450 -%} - -%Setup your variables -%The string of the number -number = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'; -counter = 1; %Location of the first digit in the series -productNumbers = ['']; %The numbers in the current product -greatestProduct = []; %The numbers in the greatest product - -%Loop through the string until every element has been tested -while((counter + 12) < size(number)(2)) - innerCounter = 0; - productNumbers = ['']; %Clear the variable - while(innerCounter < 13) - %Octave throws an error if you don't take this round about way of adding the characters to the array - tempChar = ''; %Throw away variable - tempChar = [number(counter + innerCounter), ' ']; %Add the next number to what you already have and add a space at the end - productNumbers = [productNumbers, tempChar]; - ++innerCounter; - end - productNumbers = str2num(productNumbers); %Convert the characters to numbers - %Check whether the current product is greater than the current greatest product - if(prod(productNumbers) > prod(greatestProduct)) - greatestProduct = productNumbers; - end - ++counter; -end - -%Print the result -greatestProduct -prod(greatestProduct) - -%Cleanup your variables -clear number; -clear counter; -clear productNumbers; -clear greatestProduct; -clear tempChar; -clear innerCounter; diff --git a/ProjectEuler/Problem9.m b/ProjectEuler/Problem9.m deleted file mode 100644 index 824323e..0000000 --- a/ProjectEuler/Problem9.m +++ /dev/null @@ -1,39 +0,0 @@ -%ProjectEuler/Problem9.m -%This is a script to answer Problem 9 for Project Euler -%There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc. - -%Create the variable -a = 1; -b = 0; -c = 0; -found = false; - -%Start with the smallest possible a -while((a < 1000) && ~found) - b = a + 1; %b must be > a - c = sqrt(a^2 + b^2); %c^2 = a^2 + b^2 - %Loop through all possible b's. When the sum of a, b, c is > 1000. You done have the number. Try the next a - while(((a + b + c) <= 1000) && ~found) - %If the sum == 1000 you found the numbers - if((a + b + c) == 1000) - found = true; - %Otherwise try the next b and recalculate c - else - ++b; - c = sqrt(a^2 + b^2); - end - end - %If you haven't found the numbers yet, increment a and try again - if(~found) - ++a; - end -end - -%print the result -a * b * c - -%Cleanup the variables -clear a; -clear b; -clear c; -clear found; diff --git a/ProjectEuler/Reverse.m b/ProjectEuler/Reverse.m deleted file mode 100644 index 9a02df0..0000000 --- a/ProjectEuler/Reverse.m +++ /dev/null @@ -1,19 +0,0 @@ -function [rString] = Reverse(str) -%Reverse(string) -%This function Reverse the order of the elements in an array -%It was specifically designed for a string, but should work on other 1xX arrays -% - - if(nargin ~= 1) - error('That is not a valid number of arguments') - return; - end - - counter = size(str)(2); %Set the counter to the last element in string - %Loop until the counter reaches 0 - while(counter > 0) - %Add the current element of string to rString - rString(end + 1) = str(counter); - --counter; - end -end