nimLog

This is problem 11 in Project Euler:

In the 20 x 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 x 20 grid?

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This is problem 10 on Project Euler:

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

Find the sum of all the primes below two million.

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This is Problem 9 on Project Euler:

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.

Pythagorean triplets can be generated by using the following formulas, where m and n are integers and m is greater than n:

m² – n²
2mn
m² + n²

So for the purposes of this problem we are looking for values for m and n where these three formulas add up to 1000, that is:

(m² – n²) + (2mn) + (m² + n²) = 1000
This can be simplified to 2m² + 2mn = 1000
Dividing by 2 and factoring by m we have m(m+n) = 500
Therefore m must be a factor of 500. Calculating m, m + n and n we get the following table:

Therefore the only solution is where m = 20 and n = 5. Plugging these values into our original expressions we have:

a = m² – n² = (20 * 20) – (5 * 5) = 375

b = 2mn = 2 * 20 * 5 = 200

c = m² + n² = (20 * 20) + (5 * 5) = 425

a + b + c = 1000 and ² + b² = c², so the answer to the problem is a * b * c.

We have not had to use a computer to solve this problem, but a brute-force approach in Delphi could be this:

procedure TfrmMain.Problem9;
var
  a, b, c : integer;
begin
  for a := 1 to 998 do begin
    for b := a + 1 to 999 do begin
      c := 1000 - (a + b);
      if (a * a) + (b * b) = (c * c) then begin
        ShowMessage(IntToStr(a * b * c));
        exit;
      end;
    end;
  end;
end;

This code just generates sets of numbers where a + b + c = 1000 and then checks the set for being a Pythagorean triple. If it is, the product is displayed. There are no optimizations in this code, which could be improved considerably.