Project Euler – Problem 19

Posted on 19 January, 2012 by Chin

Problem

You are given the following information, but you may prefer to do some research for yourself.

1 Jan 1900 was a Monday.
Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

Solution

$count = 0;

for ($year = 1901; $year <= 2000; $year++) {
  for ($month = 1; $month <= 12; $month++) {
    $day = date('w', mktime(0, 0, 0, $month, 1, $year));

    if ($day == 0) {
      $count++;
    }
  }
}

echo $count;

Project Euler – Problem 18

Posted on 17 January, 2012 by Chin

Problem

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.

3
7 4
2 4 6
8 5 9 3

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)

Solution

$triangle = '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';

$arrTemp1 = explode("\n", $triangle);
$arrTemp2 = array();
$arrTemp3 = array();

foreach ($arrTemp1 as $arrTemp) {
  $arrTemp2[] = explode(" ", $arrTemp);
}

for ($i = count($arrTemp2) - 2; $i >= 0; $i--) {
  for ($j = 0; $j < count($arrTemp2[$i]); $j++) {
    $arrTemp2[$i][$j] = ($arrTemp2[$i + 1][$j] > $arrTemp2[$i + 1][$j + 1] ? $arrTemp2[$i][$j] + $arrTemp2[$i + 1][$j] : $arrTemp2[$i][$j] + $arrTemp2[$i + 1][$j + 1]);
    if ($i == 0 && $j == 0) {
      echo $arrTemp2[0][0];
    }
  }
  unset($arrTemp2[$i + 1]);
}

Project Euler – Problem 17

Posted on 17 January, 2012 by Chin

Problem

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.

Solution

$countChar = 0;

for ($i = 1; $i <= 1000; $i++) {
  $tempChar = strlen(str_replace(' ', '', readable($i)));
  $countChar += $tempChar;
}

echo $countChar;

function readable($num) {
  $dict = array(
    0 => 'Zero',
    1 => 'One',
    2 => 'Two',
    3 => 'Three',
    4 => 'Four',
    5 => 'Five',
    6 => 'Six',
    7 => 'Seven',
    8 => 'Eight',
    9 => 'Nine',
    10 => 'Ten',
    11 => 'Eleven',
    12 => 'Twelve',
    13 => 'Thirteen',
    14 => 'Fourteen',
    15 => 'Fifteen',
    16 => 'Sixteen',
    17 => 'Seventeen',
    18 => 'Eighteen',
    19 => 'Nineteen',
    20 => 'Twenty',
    30 => 'Thirty',
    40 => 'Forty',
    50 => 'Fifty',
    60 => 'Sixty',
    70 => 'Seventy',
    80 => 'Eighty',
    90 => 'Ninety',
    100 => 'Hundred',
  );

  $readable = array();

  if ($num == 1000) {
    $num = 0;
    $readble[] = 'One Thousand';
  }

  if ($num >= 100) {
    $tempHund = floor($num / 100);
    $num %= 100;
    $readble[] = $dict[$tempHund];
    $readble[] = $dict[100];
    if ($num > 0) {
      $readble[] = 'And';
    }
  }

  if ($num >= 20) {
    $tempTen = floor($num / 10) * 10;
    $num %= 10;
    $readble[] = $dict[$tempTen];
  }

  if ($num > 0 && $num < 20) {
    $readble[] = $dict[$num];
  }

  return implode(' ', $readble);
}

Project Euler – Problem 16

Posted on 17 January, 2012 by Chin

Problem

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?

Solution

echo problem16(1000);

function problem16($num) {
  $sum = 0;
  $temp = 1;

  for ($i = 0; $i < $num; $i++) {
    $temp = bcmul($temp, 2);
  }

  $arrDigit = str_split($temp);

  foreach ($arrDigit as $digit) {
    $sum += $digit;
  }

  return $sum;
}

Project Euler – Problem 15

Posted on 17 January, 2012 by Chin

Problem

Starting in the top left corner of a 2×2 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 20×20 grid?

Solution

$numOfGrid = 20;

echo problem15(2 * $numOfGrid) / problem15($numOfGrid) / problem15($numOfGrid);

function problem15($num) {
  $val = 1;

  for ($i = 2; $i <= $num; $i++) {
    $val *= $i;
  }

  return $val;
}