posted 6 years ago

I've done the first 50 or so of the problems at Project Euler in Java, but thought I'd have a go at the same problems in Scala, for practice. I am a novice, but it would be fun if other people would like to join in and post their solutions, and/or suggestions for improving other peoples'. There must be a ton of different ways to solve these. Below are the first three. Later problems will be trickier.

Let the games begin!

Problem 1

Problem 2

Problem 3

Let the games begin!

Problem 1

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.

Problem 2

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.

Problem 3

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

posted 6 years ago

Cool, I didn't know about the sum method, and the anonymous function is more concise. When you were trying out functional style, how did you know if you were doing it right? I'm trying to do the same. I've read a little bit about functional programming and the things that have stuck are:

- use recursion or folds instead of loops

- use tail recursion where possible

- don't use objects / state in your functions, so that a function with the same arguments always yields the same results

Here are the next 3 if anyone wants to take a stab:

Problem 4

Problem 5

Problem 6

- use recursion or folds instead of loops

- use tail recursion where possible

- don't use objects / state in your functions, so that a function with the same arguments always yields the same results

Here are the next 3 if anyone wants to take a stab:

Problem 4

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.

Problem 5

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?

Problem 6

The sum of the squares of the first ten natural numbers is,

1^2 + 2^2 + ... + 10^2 = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)^2 = 55^2 = 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.

posted 6 years ago

My solutions:

Problem 5 was kind of interesting. I'm aware that there are much better ways to do this than brute force (you can do it in a couple of minutes with pen and paper), but found that using "for" or "forall" can make your code run literally 50 times slower (see Stack Overflow topic). Hence 2 versions : the first is more elegant but takes 40 seconds, and the second takes less than 1 second (which seems quite fast given it needs to loop over 100 million times).

Problem 6 wouldn't compile without a semicolon at the end of line 4.

Problem 5 was kind of interesting. I'm aware that there are much better ways to do this than brute force (you can do it in a couple of minutes with pen and paper), but found that using "for" or "forall" can make your code run literally 50 times slower (see Stack Overflow topic). Hence 2 versions : the first is more elegant but takes 40 seconds, and the second takes less than 1 second (which seems quite fast given it needs to loop over 100 million times).

Problem 6 wouldn't compile without a semicolon at the end of line 4.

posted 6 years ago

3 more... come on people, don't be shy

Problem 7

Problem 8

Problem 9

Problem 7

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 10001st prime number?

Problem 8

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

Problem 9

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.

posted 6 years ago

^ Just noticed P009 didn't format properly... should be

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a^2 + b^2 = c^2

For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.

Find the product abc.

Garrett Rowe

Ranch Hand

Posts: 1296

posted 6 years ago

OK, I'll chime in. I won't do number 10 since it's trivial if you use your Sieve function, but I will post my version of the sieve. I use the same algorithm as Luigi, but it's a few lines shorter.

Some problems are so complex that you have to be highly intelligent and well informed just to be undecided about them. - Laurence J. Peter

posted 6 years ago

Good stuff. That's pretty helpful and I've learnt a few things, because I couldn't think what a more idomatic / functional way to code the sieve algorithm would be, so had to cheat an use while loops. I'm still just picking up the language so I don't know about most of the methods available. It's not helped by these implicit defs on Arrays, which make it hard to find legal methods in the documentation.

Problem 10 is almost trivial, except for the fact you can't use the

Problem 10 is almost trivial, except for the fact you can't use the

`sum`function on the list, becuase it overflows the max value of Int. So I had to do Anyone know a better way of summing a list of Ints to get a Long?
posted 6 years ago

Problem 11 seems a bit harder. My Java solution wasn't that elegant, but then neither were any of my others.

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 any direction (up, down, left, right, or diagonally) in the 20×20 grid?

Garrett Rowe

Ranch Hand

Posts: 1296

posted 6 years ago
Some problems are so complex that you have to be highly intelligent and well informed just to be undecided about them. - Laurence J. Peter

We're getting into spoiler alert territory here. My solution would have been nice if I could have come up with a nicer algorithm for my transposeDiagonal function. But quick & dirty:

posted 6 years ago

Problem 12

Problem 13

Problem 14

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?

Problem 13

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

Problem 14

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

n → n/2 (n is even)

n → 3n + 1 (n is odd)

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

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 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.

Garrett Rowe

Ranch Hand

Posts: 1296

posted 6 years ago

^I did it the naive way first, but it took too long, so checked my Java solution and realised that that you only need to check factors up to the square root of the number and double it for reciprocals.

^easy one-liner. Learnt how to read a file and BigInt from String idiomAgain, pretty easy in Scala, the only complication being that the required answer was the start value, rather than the max number of steps. Don't know if reducing a list of tuples is really the way to go, but it seemed natural.

^easy one-liner. Learnt how to read a file and BigInt from String idiomAgain, pretty easy in Scala, the only complication being that the required answer was the start value, rather than the max number of steps. Don't know if reducing a list of tuples is really the way to go, but it seemed natural.

posted 6 years ago

My solution (spoilers!):

The number of paths to any point is just the sum of those above and to the left. So if you consider the number of routes to points on the grid in diagonalrows starting at the top left, you have

1

1 1

1 2 1

1 3 3 1

etc which happens to be Pascal's Triangle. Since each grid point only depends on routes from above and left, for a square we can ignore the fact that it is not an infinite grid.

Alternatively if you know about Pascal's Triangle, you can use the combination function to find out the value of the 20th element of row 40.

C(n,k) = n! / (k! (n - k)!)

So you can also do

The number of paths to any point is just the sum of those above and to the left. So if you consider the number of routes to points on the grid in diagonalrows starting at the top left, you have

1

1 1

1 2 1

1 3 3 1

etc which happens to be Pascal's Triangle. Since each grid point only depends on routes from above and left, for a square we can ignore the fact that it is not an infinite grid.

Alternatively if you know about Pascal's Triangle, you can use the combination function to find out the value of the 20th element of row 40.

C(n,k) = n! / (k! (n - k)!)

So you can also do

posted 6 years ago

Problem 16

Problem 17

Problem 18

215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number 21000?

Problem 17

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.

Problem 18

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75

95 64

17 47 82

18 35 87 10

20 04 82 47 65

19 01 23 75 03 34

88 02 77 73 07 63 67

99 65 04 28 06 16 70 92

41 41 26 56 83 40 80 70 33

41 48 72 33 47 32 37 16 94 29

53 71 44 65 25 43 91 52 97 51 14

70 11 33 28 77 73 17 78 39 68 17 57

91 71 52 38 17 14 91 43 58 50 27 29 48

63 66 04 68 89 53 67 30 73 16 69 87 40 31

04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

posted 6 years ago

Nice solution Stefan. Here are my comments if you're interested.

In your regex, if you put a "+" after your bracket expression, i.e. "[\n\t ]+", this will match 1 or more instances, so you don't need the trim or filter for length > 0. I thought you had to double-escape control characters but it looks like you don't have to with \n or \r or \t (but you do with \s).

You can probably see from the other solutions that map (_.toInt) can be used instead of the Java Integer.parseInt function.

It's normal in Scala code to make your indents 2 spaces, because you get more nesting with closures, and chaining functions means that lines tend to be longer. 8 is certainly overkill.

If you use curly braces instead of parentheses in your for-statement, you don't need the semicolons.

I was a bit confused by your maxProduct method because "max" is a function defined in List, until I realised it was a variable. From what I've read, it's better to avoid using vars, unless you have a good reason. Instead of updating a variable, you can put things in Lists, and use a "max" function. The "yield" keyword in a for-statements gives you a List of results. So you can change your methods to:

In your regex, if you put a "+" after your bracket expression, i.e. "[\n\t ]+", this will match 1 or more instances, so you don't need the trim or filter for length > 0. I thought you had to double-escape control characters but it looks like you don't have to with \n or \r or \t (but you do with \s).

You can probably see from the other solutions that map (_.toInt) can be used instead of the Java Integer.parseInt function.

It's normal in Scala code to make your indents 2 spaces, because you get more nesting with closures, and chaining functions means that lines tend to be longer. 8 is certainly overkill.

If you use curly braces instead of parentheses in your for-statement, you don't need the semicolons.

I was a bit confused by your maxProduct method because "max" is a function defined in List, until I realised it was a variable. From what I've read, it's better to avoid using vars, unless you have a good reason. Instead of updating a variable, you can put things in Lists, and use a "max" function. The "yield" keyword in a for-statements gives you a List of results. So you can change your methods to:

posted 6 years ago

Yes, thanks for your feedback.

I have to excuse myself by stating, that I solved Euler011 in Feb.2009, and I don't know, which methods where present then, and what was introduced later. Afaik, 'max' wasn't available then.

Well - Sting.split and [...]+ is from Java, and old enough. That simplifies much, while it isn't measurable for such a small grid. The method returns immediately on my old 2Ghz Laptop. And _.toInt is too more elegant.

Yes, I've given up my resistance to two-blank indent, and code to the Scala standards, when coding Scala now, but either I use 2 or 8 - nothing in between.

The problem with tabs is, that the REPL tries to autocomplete, if it finds a tab, so tabs aren't usable there. But I never got problems with 8-pos indentation either.

Yes. On the other hand - is there a problem with semicolons? They're more easy to type on a german keyboard than curly braces.

Well - I have a good reason, I want to find the max, and there is no problem, until you find one. Of course, the max-call you use is more elegant, but I think it wasn't available 2009, but a roll-your-own function to use in a fold would have been possible over then too.

Luigi Plinge wrote:Nice solution Stefan. Here are my comments if you're interested.

Yes, thanks for your feedback.

I have to excuse myself by stating, that I solved Euler011 in Feb.2009, and I don't know, which methods where present then, and what was introduced later. Afaik, 'max' wasn't available then.

In your regex, if you put a "+" after your bracket expression, i.e. "[\n\t ]+", this will match 1 or more instances, so you don't need the trim or filter for length > 0. I thought you had to double-escape control characters but it looks like you don't have to with \n or \r or \t (but you do with \s).

Well - Sting.split and [...]+ is from Java, and old enough. That simplifies much, while it isn't measurable for such a small grid. The method returns immediately on my old 2Ghz Laptop. And _.toInt is too more elegant.

It's normal in Scala code to make your indents 2 spaces, because you get more nesting with closures, and chaining functions means that lines tend to be longer. 8 is certainly overkill.

Yes, I've given up my resistance to two-blank indent, and code to the Scala standards, when coding Scala now, but either I use 2 or 8 - nothing in between.

The problem with tabs is, that the REPL tries to autocomplete, if it finds a tab, so tabs aren't usable there. But I never got problems with 8-pos indentation either.

If you use curly braces instead of parentheses in your for-statement, you don't need the semicolons.

Yes. On the other hand - is there a problem with semicolons? They're more easy to type on a german keyboard than curly braces.

I was a bit confused by your maxProduct method because "max" is a function defined in List, until I realised it was a variable. From what I've read, it's better to avoid using vars, unless you have a good reason.

Well - I have a good reason, I want to find the max, and there is no problem, until you find one. Of course, the max-call you use is more elegant, but I think it wasn't available 2009, but a roll-your-own function to use in a fold would have been possible over then too.

posted 6 years ago

If you use parentheses you need 6 extra semicolons, whereas replacing them with curly braces is no extra characters. Usually when people use parentheses + semicolons it for saving space by putting everything on one line (and maybe to make it look a bit more like a Java for-loop).

On a bit of a tangent,

That would work if you know you have at least 1 positive value; you could use Int.MinValue instead of 0, but better would be or equivalently For Int lists as in the case above, this could be simplified to This "max" method is the one on Int that has been around forever (I assume...), rather the one on lists that we're trying to emulate.

On even more of a tangent, I've done some research and found you can make a general version of this for anything that can be ordered (i.e. has the Ordered trait) using the Pimp My Library pattern: (Here

So you can now do things like this:

Yes. On the other hand - is there a problem with semicolons? They're more easy to type on a german keyboard than curly braces.

If you use parentheses you need 6 extra semicolons, whereas replacing them with curly braces is no extra characters. Usually when people use parentheses + semicolons it for saving space by putting everything on one line (and maybe to make it look a bit more like a Java for-loop).

On a bit of a tangent,

Well - I have a good reason, I want to find the max, and there is no problem, until you find one. Of course, the max-call you use is more elegant, but I think it wasn't available 2009, but a roll-your-own function to use in a fold would have been possible over then too.

That would work if you know you have at least 1 positive value; you could use Int.MinValue instead of 0, but better would be or equivalently For Int lists as in the case above, this could be simplified to This "max" method is the one on Int that has been around forever (I assume...), rather the one on lists that we're trying to emulate.

On even more of a tangent, I've done some research and found you can make a general version of this for anything that can be ordered (i.e. has the Ordered trait) using the Pimp My Library pattern: (Here

`<%`is a

*view bound*which translates into English as "can be seen as". Int is not an Ordered, but can be seen as a RichInt, which has the Ordered[Int] trait.)

So you can now do things like this: