Geekality » Big Int 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 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 8 http://www.geekality.net/2009/09/24/project-euler-problem-8/ http://www.geekality.net/2009/09/24/project-euler-problem-8/#comments Thu, 24 Sep 2009 17:14:56 +0000 Torleif http://www.geekality.net/?p=505 Continue reading →]]>

Find the greatest product of five consecutive digits in the 1000-digit number.

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

Solution

This one wasn’t too complicated to do, and there isn’t really that much optimizations or super fancy math stuff that can help either.

Simply put the numbers in an array, and walk through it.

var d = ("73167176531330624919225119674426574742355349194934"
       + "96983520312774506326239578318016984801869478851843"
       + "85861560789112949495459501737958331952853208805511"
       + "12540698747158523863050715693290963295227443043557"
       + "66896648950445244523161731856403098711121722383113"
       + "62229893423380308135336276614282806444486645238749"
       + "30358907296290491560440772390713810515859307960866"
       + "70172427121883998797908792274921901699720888093776"
       + "65727333001053367881220235421809751254540594752243"
       + "52584907711670556013604839586446706324415722155397"
       + "53697817977846174064955149290862569321978468622482"
       + "83972241375657056057490261407972968652414535100474"
       + "82166370484403199890008895243450658541227588666881"
       + "16427171479924442928230863465674813919123162824586"
       + "17866458359124566529476545682848912883142607690042"
       + "24219022671055626321111109370544217506941658960408"
       + "07198403850962455444362981230987879927244284909188"
       + "84580156166097919133875499200524063689912560717606"
       + "05886116467109405077541002256983155200055935729725"
       + "71636269561882670428252483600823257530420752963450")
    .Select(x => Convert.ToUInt64(x.ToString()))
    .ToArray();

var maxProduct = 0UL;

for (int i = 0; i < d.Length - 4; i++)
{
    var product = d[i]
        * d[i + 1]
        * d[i + 2]
        * d[i + 3]
        * d[i + 4];

    if (product > maxProduct)
        maxProduct = product;
}

var answer = maxProduct;

Nothing fancy, but works nicely. Finishes in around 3 milliseconds.

How did you do it? Do you have a more clever or efficient way to do this? Please share :)

]]> http://www.geekality.net/2009/09/24/project-euler-problem-8/feed/ 4