Saying Goodbye to 2014!

It’s been a crazy journey, and I don’t know about you, but for me 2014 went by so fast, I still have troubles believing it’s already the end of December.

With the end of the year fast approaching (only a few hours left) I would like to wish you all a better 2015, filled with love, laughter, joy, accomplishments and beautiful moments spent next to your loved ones.

Before I go, I want to share a few pet safety tips, for tonight. While you enjoy the fireworks, please keep your pets safe inside. Loud noises terrifies them and can affect their sensitive hearing. Also, make sure you keep all alcohol, floral arrangements and party decorations (especially confetti) away from your pets. We want them to have a happy new year as well, right? 🙂

Also, I would like to share with you a few pictures with the New Year Fireworks, from December 2008, in Bucharest, Romania.

Thank you all for reading, and see you in 2015! Happy New Year!!! ❤

Remembering Why … Circle of Concern vs Circle of Control

I’m pretty sure not many are aware of this, but All Right Choices isn’t the first website I’ve administrated. Some of you might not even be aware of the fact that I’ve been an animal and environmental activist, for pretty much all my life.

When I was only 16 years old, I was quite active on ASPCA online community, signing petitions, getting involved in online adoptions, fighting against animal cruelty in any way possible.

Going fast from creating only a group against animal cruelty, to opening my own online community for animal lovers in Romania, a full time job started, which lasted for more than 2 years. That’s the moment when Romanian SPCA Online Community was born (September 10, 2008).

With almost 500 animal lovers actively involved, we really got it going. We became the voice of the voiceless  and did what we could to get animals adopted worldwide, send petitions all around the world to gather signatures, create petitions and forward them to the responsible authorities, write articles, inform people, start competitions, games and share our love for animals.

That community became my online family, bringing into my life a lot of wonderful people, which are up to this day, part of my family.

What went wrong?

The online platform which used to host the community went from free to … expensive. For a high school student, with no income, that translates into the end. I had no other option but to shut it down, and it was probably the hardest thing I ever had to do, up to that day.

This year is spring I managed to find the ruins of my community, on web archive, and I can’t even begin to describe what went through my heart when I saw the page. I went from happy to sad… That community meant the world to me at one point, and to take just another quick look at it, was a complete miracle to me.

Take Two

As I was working on becoming a programmer, I decided to bring it back to life, just as an exercise, to see if I can do it. Not being provided with the infrastructure of the online platform which used to host the original community, I had to start from scratch, with only a mental image of how it should look like.

Soon enough, I realized I don’t really have the knowledge to finish the job. A lot more learning was needed. So, I took it slow: learning how to create the database, then making the database interact with the website, building the homepage, working on the sign up form.

Little by little, it all started to fall into place, and made me wonder if I should finish it and put it back live again.

The Decision 

Even if the community is long gone, I never actually let go. You can’t really give in to animals being abused. Sometimes, however, it becomes a bit too much to handle. I have problems understanding why people like to project so much hurt around them. And even more, why would you hurt a defenseless soul?

I’ve decided to do a bit of reading, on the page of an animal protection group, and see how I feel about doing it all over again. In just a few seconds I came across an article about a guy who decided to kill a dog by splitting the animal’s head into half, with an ax. The answer: I can’t.

It hurts way to much to put myself in the middle of all of this, once again. I can’t go on a larger scale anymore. It’s bad for my sanity. Instead, keeping in mind the circle of concern and the circle of control, I’ve decided to help at a smaller scale, one animal at a time.

circle-concern-control

I can’t help all the animals in the world, but I can completely change the world of a few animals, so I better focus on that. The alternative is way to painful, and I just don’t have anymore what it takes to keep it up.

That being said, project RomanianSPCA Online Community will remain at project level. Good exercise for my coding skills, and a wonderful way to bring back great memories.

Through this post I want to thank the absolutely incredible people who came into my life, through the community. I am very happy to call you part of my family, and I hope this post will bring back some good memories for you as well!

Our First 100 Followers!

Since I started All Right Choices, on September 3rd, 2014, I had to honor to meet incredible writers, photographers, and make a lot of new friends, from pretty much all around the world.

When I first started writing, I only wanted to keep track of the chances I made in my life. I was asked before about what’s the theme of the blog, and at first I couldn’t put it into words. After a few days on WordPress, I’ve realized, I’m writing about the journey to happiness, and All Right Choices needed in my life, in order to get there.

Since the beginning of the blog, a few people told me I’ve inspired them as well, to make the right choices in their lives. All I can say is: my blog reached its goal! If what I write changes the life of at least one person, for me it’s enough to keep me going, as I’m definitely doing something right.

Today the blog reached its first 100 followers and I’d like to take this opportunity to thank everyone for your support, likes, shares, comments and time invested into scrolling though my posts.

Thank you everyone, and to many more posts to come! Lots of hugs to all of you! x

100 followers kisses

Project Euler – Challenge Accepted – Back to the basics

Someone told me we need more real programmers around, who actually understand what’s going on under the hood, and not the type who have no idea what they are building. The IT world doesn’t need more programmers who write a code patched around from different ‘Googleing missions’, only to break down when a website receives more traffic, and to put the business owners in the situation when they have to call a really skilled person to fix the bugs.

I don’t want to be part of the problem. I don’t want to just scratch the surface, I want to know it all. The way I went, so far, about teaching myself to code made me the painter, in the whole process of building a house. Yes, you do get the applause at the end of the job, but you have no idea about how to build the foundation. Actually, your work would be nothing without the guy who does the foundation.

Looking for a way to train my brain to understand how a computer works behinds the scene, I had to go back to the basics. Yes, no more attempts to build fancy website, only boring numbers until I get so good at it, that I can do it with my eyes closed.

That being said, I took the Project Euler Challenge and I’ve decided to share my Knowledge with everyone, while I complete a few problems.

Back to the basics, using Javascript to solve the following:

Problem 1 – Multiples of 3 and 5

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.

var sum = 0;
for (i=0; i<1000; i++) {
          if (i%3 === 0 || i%5 === 0) {
                             sum += i;
           }
}
console.log(sum);

Correct Answer: 233168

Problem 2 – Even Fibonacci numbers

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

var sum = 0;
var x = 0;
var y = 1;

while (y<4000000) {
              if (y%2 === 0) {
                      sum = sum + y;
              }
z = x + y;
x = y;
y = z;
}
console.log(sum);

Correct Answer: 4613732

Problem 3 – Largest prime factor

The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ?

function isaprime(p) {
         var k = p / 2;
         for (i=2; i<k; i++) {
                        if (p % i === 0) {
                                      return false;
                        }
          }
          return true;
}

n = 13195;

for (i=3; i<=n; i++) {
                if (n % i === 0) {
                            if (isaprime(i)) {
                                     console.log(i + ” ” + n);
                           }
               if ( n != i ) {
                        n = n / i;
                        i = 3;
                     }
            }
}

Correct Answer: 6857

Problem 4 – Largest palindrome product

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

Find the largest palindrome made from the product of two 3-digit numbers.

function reverse (n) {
             var r = ‘ ‘;
             while (n.length>0) {
                           r += n.substring(n.length-1,n.length);
                           n = n.substring(0, n.length-1);
             }
             return r;
}

function ispalindrome(n) {
                    return reverse(n) == n;
}

m = 0;

for (i=100; i<=999; i++) {
                   for (j=100; j<=999; j++) {
                                    n = i * j;
                                    if (ispalindrome(n.toString())) {
                                                           if ( n > m ) {
                                                                         m = n;
                                                          }
                                       }
                     }
}
console.log(m);

Correct Answer: 906609

Problem 5 – Smallest multiple

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

var i = 20;
while ( i % 2 !== 0 || i % 3 !== 0 || i % 4 !== 0 || i % 5 !== 0 || i % 6 !== 0 || i % 7 !== 0 || i % 8 !== 0 || i % 9 !== 0 || i % 10 !== 0 || i % 11 !== 0 || i % 12 !== 0 || i % 13 !== 0 || i % 14 !== 0 || i % 15 !== 0 || i % 16 !== 0 || i % 17 !== 0 || i % 18 !== 0 || i % 19 !== 0 || i % 20 !== 0 ) {
       i += 20;
}
console.log(i);

Correct Answer: 232792560

Problem 6 – Sum square difference

The sum of the squares of the first ten natural numbers is, 12 + 22 + … + 102 = 385

The square of the sum of the first ten natural numbers is, (1 + 2 + … + 10)2 = 552 = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

function sumsq(n){
               sum = 0;
               for (i=1; i<n; i++) {
                            sum += Math.pow(i,2);
                } return sum;
}

function sqsum(n) {
                sum = 0;
                for (i=1; i<n; i++) {
                           sum += Math.pow(sum,2);
                }
}
console.log(sumsq(100)-sqsum(100));

Correct Answer: 25164150

Problem 7 – 10001st prime

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10 001st prime number?

function isaprime(p) {
              if ( n<2 ) {
                     return false;
             }

             var n = Math.sqrt(p);

             for (i=2; i<=n; i++) {
                       if (p % i === 0) {
                              return false;
                     }
            }
           return true;
}

var a = 0;
var primes = 0;

while (primes <= 10001) {
          a++;
          if (isaprime(a)) {
                          primes++;
          }
}
console.log(a);

Correct Answer: 104743

Problem 8 – Largest product in a series

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

var n = “7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450”;

m = 0;

for (j=0; j<=n.length – 13; j++) {
           var p = n[j] * n[j+1] * n[j+2] * n[j+3] * n[j+4] * n[j+5] * n[j+6] * n[j+7] * n[j+8] * n[j+9] * n[j+10] * n[j+11] * n[j+12];

            if ( p>m ) {
                        m = p;
             }
}
console.log(Math.max(m));

Correct Answer: 23514624000

Problem 9 – Special Pythagorean triplet

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a2 + b2 = c2

For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

var a = 0;
var b = 0;
var c = 0;
var sum = 1000;

for (a=1; a<1000; a++) {
             for (b=a+1; b<1000; b++) {
                           c = 1000 – a – b;
                           if (a*a + b*b == c*c) {
                                             console.log(a, b , c);
                                             console.log(a*b*c);
                           }
              }
}

Correct Answer: 31875000

Problem 10 – Summation of primes

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

function isaprime(p) {
               if ( n<2 ) {
                        return false;
              }
              var n = Math.sqrt(p);
              for (i=2; i<=n; i++) {
                             if (p % i === 0) {
                                            return false;
                            }
              }
              return true;
}

var sum = 0;
var x = 2000000;

for (j=2; j<x; j++) {
             if (isaprime(j) == true) {
                            sum = sum + j;
            }
}
console.log(sum);

Correct Answer: 142913828922

Problem 11 – Largest product in a grid

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

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
The product of these numbers is 26 × 63 × 78 × 14 = 1788696.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

var num = [
[8,2,22,97,38,15,0,40,0,75,4,5,7,78,52,12,50,77,91,8],
[49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,4,56,62,0],
[81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,3,49,13,36,65],
[52,70,95,23,4,60,11,42,69,24,68,56,1,32,56,71,37,2,36,91],
[22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],
[24,47,32,60,99,3,45,2,44,75,33,53,78,36,84,20,35,17,12,50],
[32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],
[67,26,20,68,2,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21],
[24,55,58,5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],
[21,36,23,9,75,0,76,44,20,45,35,14,0,61,33,97,34,31,33,95],
[78,17,53,28,22,75,31,67,15,94,3,80,4,62,16,14,9,53,56,92],
[16,39,5,42,96,35,31,47,55,58,88,24,0,17,54,24,36,29,85,57],
[86,56,0,48,35,71,89,7,5,44,44,37,44,60,21,58,51,54,17,58],
[19,80,81,68,5,94,47,69,28,73,92,13,86,52,17,77,4,89,55,40],
[4,52,8,83,97,35,99,16,7,97,57,32,16,26,26,79,33,27,98,66],
[88,36,68,87,57,62,20,72,3,46,33,67,46,55,12,32,63,93,53,69],
[4,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36],
[20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,4,36,16],
[20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,5,54],
[1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48]
];
var m = 0;

for (i = 0; i < num.length; i++) {
      for (j = 0; j < num[i].length – 3; j++){
              m = Math.max(num[i][j] * num[i][j+1] * num[i][j+2] * num[i][j+3],m);
      }
}
for (i = 0; i < num.length – 3; i++) {
           for (j = 0; j < num[i].length; j++){
                      m = Math.max(num[i][j] * num[i+1][j] * num[i+2][j] * num[i+3][j],m);
           }
          for (j = 0; j < num[i].length – 3; j++){
                   m = Math.max( num[i][j] * num[i+1][j+1] * num[i+2][j+2] * num[i+3][j+3],m);
          }
          for (j = 3; j < num.length; j++) {
                 m = Math.max(num[i][j] * num[i+1][j-1] * num[i+2][j-2] * num[i+3][j-3],m);
         }
}
console.log(m);

Correct Answer: 70600674

 


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Mother Nature’s Winter Diamonds

I woke up this morning only to find out that even Mother Nature got ready for Christmas, wearing her most gorgeous ice diamonds.