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Moved ProjectEuler to its own repository

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Matthew Ellison
2018-09-27 13:39:26 -04:00
parent 7c04e9043c
commit de9b1a5162
19 changed files with 0 additions and 1085 deletions
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45
ProjectEuler/C++/Problem1.cpp
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//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 <iostream>
#include <vector>
#include <chrono>
int main(){
unsigned long fullSum = 0; //For the sum of all the numbers
std::vector<unsigned long> 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<std::chrono::nanoseconds>(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;
}

57
ProjectEuler/C++/Problem12.cpp
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//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 <iostream>
#include <chrono> //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::milliseconds>(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;
}

43
ProjectEuler/C++/Problem2.cpp
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//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 <iostream>
#include <vector>
#include <chrono>
int main(){
unsigned long fullSum = 2; //Holds the sum of all the numbers
std::vector<unsigned long> 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<std::chrono::nanoseconds>(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;
}

104
ProjectEuler/C++/Problem3.cpp
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//ProjectEuler/C++/Problem3.cpp
//Matthew Ellison
// Created: 9-28-18
//Modified: 9-28-18
//The largest prime factor of 600851475143
#include <iostream>
#include <cmath>
#include <vector>
#include <chrono>
std::vector<unsigned long long> generatePrimes(const unsigned long long number);
unsigned long long findLargest(const std::vector<unsigned long long> list);
int main(){
const unsigned long long goalNumber = 600851475143; //The number you are trying to find the factors of
std::vector<unsigned long long> primes; //Holds the list of prime numbers
std::vector<unsigned long long> 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<std::chrono::milliseconds>(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<unsigned long long> generatePrimes(const unsigned long long goalNumber){
std::vector<unsigned long long> 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<unsigned long long> list){
unsigned long long maxNum = 0;
for(unsigned long long num : list){
if(num > maxNum){
maxNum = num;
}
}
return maxNum;
}

30
ProjectEuler/Problem1.m
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%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;

6
ProjectEuler/Problem10.m
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%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))

152
ProjectEuler/Problem11.m
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%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;

51
ProjectEuler/Problem12.m
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@@ -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

52
ProjectEuler/Problem12_2.m
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@@ -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

214
ProjectEuler/Problem13.m
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@@ -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
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690
%}
%Variables
nums = [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,
86188088225875314529584099251203829009407770775672,
11306739708304724483816533873502340845647058077308,
82959174767140363198008187129011875491310547126581,
97623331044818386269515456334926366572897563400500,
42846280183517070527831839425882145521227251250327,
55121603546981200581762165212827652751691296897789,
32238195734329339946437501907836945765883352399886,
75506164965184775180738168837861091527357929701337,
62177842752192623401942399639168044983993173312731,
32924185707147349566916674687634660915035914677504,
99518671430235219628894890102423325116913619626622,
73267460800591547471830798392868535206946944540724,
76841822524674417161514036427982273348055556214818,
97142617910342598647204516893989422179826088076852,
87783646182799346313767754307809363333018982642090,
10848802521674670883215120185883543223812876952786,
71329612474782464538636993009049310363619763878039,
62184073572399794223406235393808339651327408011116,
66627891981488087797941876876144230030984490851411,
60661826293682836764744779239180335110989069790714,
85786944089552990653640447425576083659976645795096,
66024396409905389607120198219976047599490197230297,
64913982680032973156037120041377903785566085089252,
16730939319872750275468906903707539413042652315011,
94809377245048795150954100921645863754710598436791,
78639167021187492431995700641917969777599028300699,
15368713711936614952811305876380278410754449733078,
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;

27
ProjectEuler/Problem2.m
View File

@@ -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;

48
ProjectEuler/Problem3.m
View File

@@ -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;

50
ProjectEuler/Problem4.m
View File

@@ -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
%}

43
ProjectEuler/Problem5.m
View File

@@ -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;

21
ProjectEuler/Problem6.m
View File

@@ -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;

21
ProjectEuler/Problem7.m
View File

@@ -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;

63
ProjectEuler/Problem8.m
View File

@@ -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;

39
ProjectEuler/Problem9.m
View File

@@ -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;

19
ProjectEuler/Reverse.m
View File

@@ -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