Geekality » Project Euler http://www.geekality.net With a hint of Social Ineptitude Tue, 27 Jul 2010 16:15:56 +0000 en hourly 1 http://wordpress.org/?v=3.0.1 Project Euler: Problem 25 http://www.geekality.net/2009/11/06/project-euler-problem-25/ http://www.geekality.net/2009/11/06/project-euler-problem-25/#comments Fri, 06 Nov 2009 22:51:56 +0000 Torleif http://www.geekality.net/?p=751 Continue reading →]]>

The Fibonacci sequence is defined by the recurrence relation:
F_n = F_{n-1} + F_{n-2}, where F_1 = 1 and F_2 = 1.

Hence the first 12 terms will be:

  • F_1=1
  • F_2=1
  • …
  • F_{11}=89
  • F_{12}=144

The 12th term, F_{12}, is the first term to contain three digits.

What is the first term in the Fibonacci sequence to contain 1000 digits?

Solution

We touched on the Fibonacci sequence a while ago in the second euler problem. This time, however, we are in a whole different league. That time we were supposed to sum up all even Fibonacci numbers below 4 million, which might sound like a lot, but not when we compare it to the number we are asked for this time! The last Fibonacci number below 4 million is 3524578. That would be 7 digits. The maximum value of the datatype I used in the generator I made for that (ulong) is 18446744073709551615. That would be 20 digits. Quite a distance left to a thousand!

So, once again we turn to our big integer class, IntX. And since we have that, a solution to this problem is actually pretty straight forward. We simply have to make a new generator that uses the IntX data type instead of tiny ulong (it’s all relative…).

public class LargeFibonacciSequence : IEnumerable<IntX>
{
    public IEnumerator<IntX> GetEnumerator()
    {
        var a = new IntX(0);
        var b = new IntX(1);

        while (true)
        {
            yield return a;

            var c = a + b;
            a = b;
            b = c;
        }
    }

    IEnumerator IEnumerable.GetEnumerator()
    {
        return GetEnumerator();
    }
}

The straight forward solution should then be fairly obvious.

var n = 0;
var sequence = new LargeFibonacciSequence();

using (var e = sequence.GetEnumerator())
{
    while (e.MoveNext() && e.Current.ToString().Length < 1000)
        n++;
}

var answer = n;

That pretty much just moves through one Fibonacci number after the other, increasing n as it goes, until it reaches the first one that has 1000 or more digits.

Very brute-force, not very fast (takes around 500 ms) and not very interesting, but that’s pretty much it… ooor is it?

Take two!

Alright, while reading up on Fibonacci numbers I found some interesting mathematical formulas and such. I’m not going to come with an in-depth explanation or lots proofs here, but I will share the formulas that matters and go through how we use them. If you want the elaborate explanations you can find them where smarter people than I reign. For example at Wikipedia, or at this site. Anyways, here we go. Take cover if you fear math…

The Fibonacci sequence

The nth number in the Fibonacci sequence, F_n, is defined by the recurrence relation

F_n = F_{n-1} + F_{n-2}

with the seed values F_0=0 and F_1 = 1.

This recurrence, since it is linear, can be expressed by a closed-form solution known as the Binet’s formula:

Fib(n)=\dfrac{\varphi^n-(1-\varphi)^n}{\sqrt{5}}=\dfrac{\varphi^n-(\frac{-1}{\varphi})^n}{\sqrt{5}}
\varphi=\dfrac{1+\sqrt{5}}{2}\approx1.6180339887\ldots — (The Golden ratio)

Now, why 1-\varphi and \frac{-1}{\varphi} is considered the same, or even why that formula works or looks like that, I will leave as an exercise for the reader (Cause I have no idea :P Please share if you do…). But anyways, as an example, if you put 20 into that formula, you will get 6765 which is the 20th number in the Fibonacci sequence. Fantastic. We could now use that formula to make another brute-force solution to find the term we are after. But that wouldn’t make much sense and would be much much slower. So, we must move on further into this crazy land.

Before we move on to the next part, since we will be working with quite large values of n, we can simplify that expression a bit. That is because \frac{-1}{\varphi}^n will move pretty fast towards 0 as n increases. In other words, we can pretty much just skip that part and use the following formula instead:

Fib(n)=\dfrac{\varphi^n}{\sqrt{5}}

The length of a number

How do you calculate the length of a number? I actually had to do this in an earlier post and then I kind of cheated. I just converted the number into a string and just count the characters (In C#, string.Length). But, I recently discovered that there actually is a way you can do this mathematically! I had no idea… The key lays in the logarithm to base 10.

\log_{10}{n} (often written as just \log{n}) for any 1-digit number will give you 0.something. For any 2-digit number it gives you 1.something, and so on. So a formula for getting the length of any number:

Len(n)=\lfloor\log{n}\rfloor+1

In case you were wondering, those weird square brackets are called the floor function.

The length of a Fibonacci number

Now we are able to calculate the length of a certain Fibonacci number by using the following formula, which I for no good reason will call G:

G(n)=Len(Fib(n))=\lfloor\log{\dfrac{\varphi^n}{\sqrt{5}}}\rfloor+1

There is a problem here though. And the problem is \varphi^n. Why is it a problem? Because it get’s seriously large very quick. A value of around 500 actually makes my calculator overflow and just give me an error.

But fear not! There are a couple of formulas having to do with roots and logarithms that we can use to our advantage:

\sqrt{c}=\sqrt[2]{c}

\sqrt[b]{c}=c^{\frac{1}{b}}

\log_a{\dfrac{b}{c}}=\log_a{b}-\log_a{c}

\log_a{b^c}=c\log_a{b}

With those formulas we can handily rewrite our formula into one that is easier to handle:

G(n)=\lfloor\log{\dfrac{\varphi^n}{\sqrt{5}}}\rfloor+1

G(n)=\lfloor\log{\varphi^n}-\log{5^{\frac{1}{2}}}\rfloor+1

G(n)=\lfloor n\log{\varphi}-\dfrac{\log{5}}{2}\rfloor+1

Tadaa! Now we can push in 20 and get 4. And as we calculated earlier, the 20th Fibonacci number is 6769, which in fact is 4 digits long! Amazing…

The first Fibonacci number with 1000 digits

Now we are almost ready to solve our problem. The formula we have now works perfectly. The only problem is that it’s backwards! We are not looking for the length of a certain Fibonacci number, but rather what Fibonacci number has a certain length. In other words, we want n, not G(n). Luckily, using pretty basic algebra and some guessing, we can make a new formula that finds exactly what we are looking for. (My way of skipping the floor stuff and introducing a ceiling function in the end is purely based on guessing and observation. But it gives the correct answer :P Now, if you know the proper way of handling it, please let me know! )

G(n)=\lfloor n\log{\varphi}-\dfrac{\log{5}}{2}\rfloor+1

n\log{\varphi}=G(n)+\dfrac{\log{5}}{2}-1

n=\lceil\dfrac{G(n)+\dfrac{\log{5}}{2}-1}{\log\varphi}\rceil

Now we can just substitute G(n) with 1000, and calculate n, which should be the answer to this problem!

As usual, I will not put the actual answer here though ;)

This has been the longest blog post yet about these problems, I think, but I must say it was pretty fun to solve this one. I didn’t really get much at first, just saw that the formulas worked, but now that I have written about it and gone through it all step by step, it actually makes sense. Most of it anyways. If you know how to properly handle that floor and ceiling stuff, let me know :P

Anyways, that was it for now! Stay tuned for more good stuff like this :P

]]> http://www.geekality.net/2009/11/06/project-euler-problem-25/feed/ 1 Project Euler: Problem 16 http://www.geekality.net/2009/11/05/project-euler-problem-16/ http://www.geekality.net/2009/11/05/project-euler-problem-16/#comments Thu, 05 Nov 2009 16:46:16 +0000 Torleif http://www.geekality.net/?p=740 Continue reading →]]>

2^{15} = 32768 and the sum of its digits is
3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number 2^{1000}?

Solution

With the big integer type IntX (which I added to my project to solve an earlier problem) this one is pretty much as easy as you’d think.

var answer = IntX
        .Pow(2, 1000)
        .ToString()
        .Select(x => x - '0')
        .Sum();

It first does the calculation and then gets the string representation of the result. And in case you were wondering, that would be:

10715086071862673209484250490600018105614048117
05533607443750388370351051124936122493198378815
69585812759467291755314682518714528569231404359
84577574698574803934567774824230985421074605062
37114187795418215304647498358194126739876755916
55439460770629145711964776865421676604298316526
24386837205668069376

And as a comparison, here is the largest number that can be represented using built-in .net types. decimal.MaxValue.

79228162514264337593543950335

As you can see, the first number is kind of bigger than the second :P Aaaanyways, the next bit might be a bit cryptic. But fear not, it is actually quite simple:

  • A string is an array of characters.
  • All characters have a character code (a numeric value).
  • The character codes for the numeric characters (’0′, ’1′, and so on) comes after each other, which means that for example ’1′ has a character code value which is 1 larger than the character code of ’0′.
  • If we take a character code and subtracts it from itself, we get zero. (Thank you captain obvious…)
  • This means that if we subtract ’0′ from ’0′, we get 0. And if we subtract ’0′ from ’5′, we get 5. (Oh, right, clever!)

So, the last two lines in my code simply converts all the characters in the string into actual numbers and then sums them together. Which would get us the answer! Tadaa ^_^

How did you do? Have you found a more interesting solution? Please do share :)

]]> http://www.geekality.net/2009/11/05/project-euler-problem-16/feed/ 0 The Sieve of Eratosthenes in C# http://www.geekality.net/2009/10/19/the-sieve-of-eratosthenes-in-c/ http://www.geekality.net/2009/10/19/the-sieve-of-eratosthenes-in-c/#comments Mon, 19 Oct 2009 18:37:47 +0000 Torleif http://www.geekality.net/?p=626 Continue reading →]]> In some of the Project Euler problems we have needed a source of primes. One algorithm for finding primes is called the Sieve of Eratosthenes. This algorithm is both pretty simple to understand and to implement. It is also fairly fast and usable, at least for the lower primes.

My implementation is based upon the algorithm described on the Wikipedia page and some helpful optimizations I found in an article at Black Wasp. The differences from the original algorithm and the solution at Black Wasp is that it finds the primes incrementally and that it only looks for a new prime when asked.

My code should be pretty self explanatory, but the basic idea is that we have a list of known primes and for each odd number we check it against the these. If a number is not evenly divisible by any of the known primes, then we have found a new one and can add it to our list.

  public class Eratosthenes : IEnumerable<ulong>
  {
      private readonly List<ulong> knownPrimes;

      public Eratosthenes()
      {
          knownPrimes = new List<ulong> {2, 3};
      }

      public IEnumerator<ulong> GetEnumerator()
      {
          // Return the ones we know first
          foreach (var prime in knownPrimes)
              yield return prime;

          // Then find new ones
          var possible = primes.Last();
          while (true)
              if (IsPrime(possible += 2))
              {
                  yield return possible;
                  knownPrimes.Add(possible);
              }
      }

      IEnumerator IEnumerable.GetEnumerator()
      {
          return GetEnumerator();
      }
     
      private bool IsPrime(ulong value)
      {
          var sqrt = (ulong) Math.Sqrt(value);
          return !knownPrimes
              .TakeWhile(x => x <= sqrt)
              .Any(x => value%x == 0);
      }
  }

How do you use it? Simple! For example, to print the first 500 primes you can do this:

var primes = new Eratosthenes().Take(500);

foreach(var prime in primes)
    Console.WriteLine(prime);

And that’s all there is to it! What do you think? Let me know if you have any questions about it or any suggestions to how it can be improved. I want to know ;)

]]> http://www.geekality.net/2009/10/19/the-sieve-of-eratosthenes-in-c/feed/ 0 Project Euler: Problem 14 http://www.geekality.net/2009/10/19/project-euler-problem-14/ http://www.geekality.net/2009/10/19/project-euler-problem-14/#comments Mon, 19 Oct 2009 17:47:38 +0000 Torleif http://www.geekality.net/?p=616 Continue reading →]]>

The following iterative sequence is defined for the set of positive integers:

n \to n/2 (n is even)
n \to 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

Solution

This problem has to do with the Collatz conjecture. In short, from Wikipedia:

The Collatz conjecture is an unsolved conjecture in mathematics. It is named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, as the Ulam conjecture (after Stanislaw Ulam), or as the Kakutani’s Problem, or as the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers, or as wondrous numbers.

We take any whole number n greater than 0. If n is even, we halve it (n/2), else we do “triple plus one” and get 3n+1. The conjecture is that for all numbers this process converges to 1. Hence it has been called “Half Or Triple Plus One”, sometimes called HOTPO.

Paul Erdős said about the Collatz conjecture: “Mathematics is not yet ready for such confusing, troubling, and hard problems.” He offered $500 for its solution. (Lagarias 1985)

Interesting stuff. Ok, maybe not that interesting, but it is at least a problem that we can solve (not the Collatz Problem, but the Euler problem ;) ). And solving it is pretty simple. Just go through all the numbers below one million, and follow the mentioned algorithm for each number and see which one results in the longest chain.

var startOfLongest = 0U;
var longestChain = 0U;

for (var start = 1000000U; start > 0; start--)
{
    var s = start;
    var length = 1U;

    while (s != 1)
    {
        if (s % 2 == 0)
            s /= 2;
        else
            s = 3 * s + 1;
        length++;
    }

    if (length < longestChain)
        continue;

    startOfLongest = start;
    longestChain = length;
}

var answer = startOfLongest;

I don’t know if there really is much to say about that code. It is pretty straight forward. For each number, do the chain stuff until you reach 1. If the chain was longer than what we have found so far, store it. Then move on to the next one.

It runs in around 480 milliseconds so not among the fastest solutions I have had so far, but fast enough. I tried to come up with a more fancy solution, but they just took longer or didn’t work at all, so I’ll spare you the details of those :P

How did you do? Do you see any obvious inefficient parts in my code? Can it be optimized somehow? Please let me know in the comments below :)

]]> http://www.geekality.net/2009/10/19/project-euler-problem-14/feed/ 0 Project Euler: Problem 13 http://www.geekality.net/2009/10/04/project-euler-problem-13/ http://www.geekality.net/2009/10/04/project-euler-problem-13/#comments Sun, 04 Oct 2009 20:58:59 +0000 Torleif http://www.geekality.net/?p=612 Continue reading →]]>

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

Solution

This would’ve been pretty straight forward if it wasn’t for the fact that the numbers are too long. The UInt32 datatype can hold numbers up to 4294967295 (10 digits) and the UInt64 up to 18446744073709551615 (20 digits). That is at least 30 digits too little for those numbers we have to sum up.

We have mainly two options here. We could implement our own addition algorithm, for example by adding the numbers kind of like you would do by hand. Or we can find something that handles this large numbers already.

I decided to go with the last option. Later problems will probably become easier with such a class too, so might as well go hunting for one now.

After some looking around I found one on CodePlex called IntX. When that was added to my project, solving the problem was easy. (The code looks a bit long, but it’s only because of the 50 numbers :P )

var Numbers = new[]
    {
        new IntX("37107287533902102798797998220837590246510135740250"),
        new IntX("46376937677490009712648124896970078050417018260538"),
        new IntX("74324986199524741059474233309513058123726617309629"),
        new IntX("91942213363574161572522430563301811072406154908250"),
        new IntX("23067588207539346171171980310421047513778063246676"),
        new IntX("89261670696623633820136378418383684178734361726757"),
        new IntX("28112879812849979408065481931592621691275889832738"),
        new IntX("44274228917432520321923589422876796487670272189318"),
        new IntX("47451445736001306439091167216856844588711603153276"),
        new IntX("70386486105843025439939619828917593665686757934951"),
        new IntX("62176457141856560629502157223196586755079324193331"),
        new IntX("64906352462741904929101432445813822663347944758178"),
        new IntX("92575867718337217661963751590579239728245598838407"),
        new IntX("58203565325359399008402633568948830189458628227828"),
        new IntX("80181199384826282014278194139940567587151170094390"),
        new IntX("35398664372827112653829987240784473053190104293586"),
        new IntX("86515506006295864861532075273371959191420517255829"),
        new IntX("71693888707715466499115593487603532921714970056938"),
        new IntX("54370070576826684624621495650076471787294438377604"),
        new IntX("53282654108756828443191190634694037855217779295145"),
        new IntX("36123272525000296071075082563815656710885258350721"),
        new IntX("45876576172410976447339110607218265236877223636045"),
        new IntX("17423706905851860660448207621209813287860733969412"),
        new IntX("81142660418086830619328460811191061556940512689692"),
        new IntX("51934325451728388641918047049293215058642563049483"),
        new IntX("62467221648435076201727918039944693004732956340691"),
        new IntX("15732444386908125794514089057706229429197107928209"),
        new IntX("55037687525678773091862540744969844508330393682126"),
        new IntX("18336384825330154686196124348767681297534375946515"),
        new IntX("80386287592878490201521685554828717201219257766954"),
        new IntX("78182833757993103614740356856449095527097864797581"),
        new IntX("16726320100436897842553539920931837441497806860984"),
        new IntX("48403098129077791799088218795327364475675590848030"),
        new IntX("87086987551392711854517078544161852424320693150332"),
        new IntX("59959406895756536782107074926966537676326235447210"),
        new IntX("69793950679652694742597709739166693763042633987085"),
        new IntX("41052684708299085211399427365734116182760315001271"),
        new IntX("65378607361501080857009149939512557028198746004375"),
        new IntX("35829035317434717326932123578154982629742552737307"),
        new IntX("94953759765105305946966067683156574377167401875275"),
        new IntX("88902802571733229619176668713819931811048770190271"),
        new IntX("25267680276078003013678680992525463401061632866526"),
        new IntX("36270218540497705585629946580636237993140746255962"),
        new IntX("24074486908231174977792365466257246923322810917141"),
        new IntX("91430288197103288597806669760892938638285025333403"),
        new IntX("34413065578016127815921815005561868836468420090470"),
        new IntX("23053081172816430487623791969842487255036638784583"),
        new IntX("11487696932154902810424020138335124462181441773470"),
        new IntX("63783299490636259666498587618221225225512486764533"),
        new IntX("67720186971698544312419572409913959008952310058822"),
        new IntX("95548255300263520781532296796249481641953868218774"),
        new IntX("76085327132285723110424803456124867697064507995236"),
        new IntX("37774242535411291684276865538926205024910326572967"),
        new IntX("23701913275725675285653248258265463092207058596522"),
        new IntX("29798860272258331913126375147341994889534765745501"),
        new IntX("18495701454879288984856827726077713721403798879715"),
        new IntX("38298203783031473527721580348144513491373226651381"),
        new IntX("34829543829199918180278916522431027392251122869539"),
        new IntX("40957953066405232632538044100059654939159879593635"),
        new IntX("29746152185502371307642255121183693803580388584903"),
        new IntX("41698116222072977186158236678424689157993532961922"),
        new IntX("62467957194401269043877107275048102390895523597457"),
        new IntX("23189706772547915061505504953922979530901129967519"),
        new IntX("86188088225875314529584099251203829009407770775672"),
        new IntX("11306739708304724483816533873502340845647058077308"),
        new IntX("82959174767140363198008187129011875491310547126581"),
        new IntX("97623331044818386269515456334926366572897563400500"),
        new IntX("42846280183517070527831839425882145521227251250327"),
        new IntX("55121603546981200581762165212827652751691296897789"),
        new IntX("32238195734329339946437501907836945765883352399886"),
        new IntX("75506164965184775180738168837861091527357929701337"),
        new IntX("62177842752192623401942399639168044983993173312731"),
        new IntX("32924185707147349566916674687634660915035914677504"),
        new IntX("99518671430235219628894890102423325116913619626622"),
        new IntX("73267460800591547471830798392868535206946944540724"),
        new IntX("76841822524674417161514036427982273348055556214818"),
        new IntX("97142617910342598647204516893989422179826088076852"),
        new IntX("87783646182799346313767754307809363333018982642090"),
        new IntX("10848802521674670883215120185883543223812876952786"),
        new IntX("71329612474782464538636993009049310363619763878039"),
        new IntX("62184073572399794223406235393808339651327408011116"),
        new IntX("66627891981488087797941876876144230030984490851411"),
        new IntX("60661826293682836764744779239180335110989069790714"),
        new IntX("85786944089552990653640447425576083659976645795096"),
        new IntX("66024396409905389607120198219976047599490197230297"),
        new IntX("64913982680032973156037120041377903785566085089252"),
        new IntX("16730939319872750275468906903707539413042652315011"),
        new IntX("94809377245048795150954100921645863754710598436791"),
        new IntX("78639167021187492431995700641917969777599028300699"),
        new IntX("15368713711936614952811305876380278410754449733078"),
        new IntX("40789923115535562561142322423255033685442488917353"),
        new IntX("44889911501440648020369068063960672322193204149535"),
        new IntX("41503128880339536053299340368006977710650566631954"),
        new IntX("81234880673210146739058568557934581403627822703280"),
        new IntX("82616570773948327592232845941706525094512325230608"),
        new IntX("22918802058777319719839450180888072429661980811197"),
        new IntX("77158542502016545090413245809786882778948721859617"),
        new IntX("72107838435069186155435662884062257473692284509516"),
        new IntX("20849603980134001723930671666823555245252804609722"),
        new IntX("53503534226472524250874054075591789781264330331690")
    };

var sum = Numbers.Aggregate(new IntX(0), (s, x) => s + x);
var answer = Convert.ToUInt64(sum.ToString().Truncate(10));

Runs in less than a millisecond, around 335 ticks. Not much more to say about this one really… how did you do it?

]]> http://www.geekality.net/2009/10/04/project-euler-problem-13/feed/ 2 Project Euler: Problem 12 http://www.geekality.net/2009/10/04/project-euler-problem-12/ http://www.geekality.net/2009/10/04/project-euler-problem-12/#comments Sat, 03 Oct 2009 23:43:54 +0000 Torleif http://www.geekality.net/?p=588 Continue reading →]]>

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

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

Let us list the factors of the first seven triangle numbers:

 1: 1
 3: 1,3
 6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

At first it seemed like a piece of cake, however it wasn’t as easy as I thought it was…

Solution

Well, what we first need here, irregardless of how you do it (pretty much), is a source of triangle numbers.

public class TriangleSequence : IEnumerable<ulong>
{
    public IEnumerator<ulong> GetEnumerator()
    {
        ulong sum = 0;
        for (ulong n = 1;; n++)
        {
            sum += n;
            yield return sum;
        }
    }
    IEnumerator IEnumerable.GetEnumerator()
    {
        return GetEnumerator();
    }
}

Pretty simple and straight forward. We can then move on to the next issue.

Take one

Like mentioned, I thought this one would be a piece of cake. I mean, having the sequence of triangle numbers in place, how hard could it be? My first solution, probably the most obvious one, is to go through each triangle number, find the factors of it, see if there are more than 500 of them, and move to the next if it isn’t.

I created a basic factorization method using the rather naive trial-division method and wrote a simple Linq statement to get the answer.

public static IEnumerable<ulong> GetDivisors(ulong number)
{
    if (number == 0)
        yield break;

    var limit = number / 2;
    for (ulong i = 1; i <= limit; i++)
        if (number % i == 0)
            yield return i;

    yield return number;
}


var answer = new TriangleSequence()
    .Select(x => new
        {
            Number = x,
            DivisorCount = Factorization.GetDivisors(x).Count(),
        })
    .First(x => x.DivisorCount > 500)
    .Number;

And then I ran it. A few seconds went by and no answer. Minutes went by and still no answer. Considering the fact that most of my other problems was solved in less than a few seconds, this was just taking too long. A smarter solution is needed.

Take two

The way to speed this up is “of course” to find a faster factorization algorithm, so I went hunting for a more effective one. But, while I was on that hunt, I realized that I was perhaps on the wrong path here. I mean, what do I even need those factors for? What I really need is just the count of the factors. I wasn’t sure if such a method existed though, but while roaming around on StackOverflow (like I tend to do sometimes) I stumbled over some Python code. Later I also found some formulas. So, time for some math.

Any integer can be expressed as

N = p_1^{a_1} \cdot p_2^{a_2} \cdot p_3^{a_3} \cdot \ldots

where p_n is a distinct prime number and a_n is its exponent. The count of divisors D(N) can then be found by the formula

D(N) = (a_1 + 1) \cdot (a_2 + 1) \cdot (a_3 + 1) \cdot \ldots

For example:

N = 36 = 3^2 \cdot 2^2 = 9 \cdot 4

D(32) = (2 + 1) \cdot (2 + 1) = 9

With this knowledge and an example implementation in python, I was able to throw together the following code.

public class ProbablePrimeSequence : IEnumerable<ulong>
{
    public IEnumerator<ulong> GetEnumerator()
    {
        yield return 2;
        yield return 3;
        ulong i = 5;
        while (true)
        {
            yield return i;
            if (i % 6 == 1)
                i += 2;
            i += 2;
        }
    }
    IEnumerator IEnumerable.GetEnumerator()
    {
        return GetEnumerator();
    }
}

public static ulong GetCountOfDivisors(ulong number)
{
    if (number == 0)
        return 0;

    var divisors = 1UL;

    foreach (var prime in new ProbablePrimeSequence())
    {
        var exponent = 0UL;
        while (number % prime == 0)
        {
            exponent += 1;
            number /= prime;
        }

        if (exponent > 0)
            divisors *= exponent + 1;

        if (number == 1)
            break;
    }
    return divisors;
}

Now, why this works by just using probable primes and not pure primes, I am not sure to be honest. But it sure is a lot faster. My tests doesn’t fail and I get the right answer so I’ll just move on :P (However, leave a comment if you know, cause I would like to know too :) )

Anyways, now we can finally find our answer using the following straight forward statement.

var answer = new TriangleSequence()
        .First(x =>GetCountOfDivisors(x) > 500);

The running time of that lays around 320 milliseconds, which I think is pretty acceptable although not blazingly fast.

What do you think of my solution? Have I overlooked something obvious? How would you solve this one? I am curious and would like to know, so please leave a comment :D

]]> http://www.geekality.net/2009/10/04/project-euler-problem-12/feed/ 0 Project Euler: Problem 11 http://www.geekality.net/2009/10/03/project-euler-problem-11/ http://www.geekality.net/2009/10/03/project-euler-problem-11/#comments Sat, 03 Oct 2009 17:44:58 +0000 Torleif http://www.geekality.net/?p=569 Continue reading →]]>

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 x 63 x 78 x 14 = 1788696.

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

Solution

Not really much interesting you can do with this one really. Going through the whole grid while checking every single possibility is really the only way you can do it. We have take every number in the grid and multiply it with the three numbers to the left of it, below it, diagonally up to the right, and so on, like shown below.

EulerProblem11-Example-1

However, this way we would actually be checking all the possibilities twice because a product doesn’t care in what order its factors are. This means we can for example just check the ones to the right and not the ones to the left. We then end up with only four of the original eight “arms” of factors to multiply.

EulerProblem11-Example-2

I put the grid in a two dimensional array and then just looped through all of them looking for the highest product. It is very simple to make. Only thing that can be a bit tricky is to make sure you do try out all the possibilities without going out the side the array bounds. Anyways, you can see my result below.

var grid = new[,]
    {
        {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},
    };

var rows = grid.GetLength(0);
var columns = grid.GetLength(1);

var greatest = 0;

for (var r = 0; r < rows; r++)
    for (var c = 0; c < columns; c++)
    {
        if (c < columns - 3)
        {
            // Right and "Left"
            if (greatest < grid[r,c] * grid[r,c+1] * grid[r,c+2] * grid[r,c+3])
                greatest = grid[r,c] * grid[r,c+1] * grid[r,c+2] * grid[r,c+3];
        }

        if (r < rows - 3)
        {
            // Down and "Up"
            if (greatest < grid[r,c] * grid[r+1,c] * grid[r+2,c] * grid[r+3,c])
                greatest = grid[r,c] * grid[r+1,c] * grid[r+2,c] * grid[r+3,c];

            // Diagonally, down to the right
            if (c < columns - 3)
                if (greatest < Grid[r,c] * Grid[r+1,c+1] * Grid[r+2,c+2] * grid[r+3,c+3])
                    greatest = Grid[r,c] * Grid[r+1,c+1] * Grid[r+2,c+2] * grid[r+3,c+3];

            // Diagonally, down to the left
            if (c > 3)
                if (greatest < grid[r,c] * grid[r+1,c-1] * grid[r+2,c-2] * grid[r+3,c-3])
                    greatest = grid[r,c] * grid[r+1,c-1] * grid[r+2,c-2] * grid[r+3,c-3];
        }
    }

var answer = greatest;

And that’s pretty much it! Takes around 150 ticks, and is not really interesting to be honest :P Next Euler problem however is a bit more tricky… will write about that one next, so stay tuned :)

]]> http://www.geekality.net/2009/10/03/project-euler-problem-11/feed/ 0