How to calculate if a year has 53 weeks

Question

According to formula https://en.wikipedia.org/wiki/ISO_week_date in the Weeks per year section you should be able to find each year with 53 weeks. I have copied the formula into PHP but it seems to mess up around 2020 returning 52 weeks instead of 53.

function weeks($year){
    $w = 52;
    $p = ($year+($year/4)-($year/100)+($year/400))%7;
    if ($p == 4 || ($p-1) == 3){
        $w++;
    }
    return $w." ".$p;
}

for ($i = 2000; $i <= 2144; $i++) {
    echo $i." (".weeks($i).") | ";
}

From Wikipedia

The following 71 years in a 400-year cycle have 53 weeks (371 days); years not listed have 52 weeks (364 days); add 2000 for current years:

004, 009, 015, 020, 026, 032, 037, 043, 048, 054, 060, 065, 071, 076, 082, 088, 093, 099, 105, 111, 116, 122, 128, 133, 139, 144

I know I can grab the total amount of weeks for a given year using the date() function but I'm just wondering if anyone knows the math since the formula on wikipedia seems to be wrong.

EDIT WORKING CODE

function p($year){
    return ($year+floor($year/4)-floor($year/100)+floor($year/400))%7;
}

function weeks($year){
    $w = 52;
    if (p($year) == 4 || p($year-1) == 3){
        $w++;
    }
    return $w; // returns the number of weeks in that year
}

for ($i = 2000; $i <= 2144; $i++) {
    echo $i." (".weeks($i).") | ";
}

The number of weeks and the ISO week number may be different things. I think I would just look at the ISO week number of the last day of the year. — Ray Paseur, Jul 14 '17 at 18:40
LOL @hobbs - There are years with 52 and a partial, but no years with a complete 53 weeks. — Jay Blanchard, Jul 14 '17 at 18:41
@JayBlanchard There may not be 53 "full" weeks Monday - Sunday but I need to take the half or quarter week into account — Mike Lewis, Jul 14 '17 at 18:46
I know @MikeLewis, ISO weeks are different animals than Gregorian and Julian weeks. — Jay Blanchard, Jul 14 '17 at 18:55

score 2 · Accepted Answer · answered Jul 14 '17 at 18:45

Two problems that I can see:

You missed the floor notation in the formula for p. In PHP that should be + int($year / 4) - int($year / 100) etc. not + ($year / 4) - ($year / 100) etc.
p is a function, and the overall formula uses p(year) and p(year - 1). Your $p is equal to p(year), but $p - 1 is not equal to p(year - 1) (note that your second condition, ($p - 1) == 3 is just the same as your first condition, $p == 4 — this should make it clear that that's not what was intended). The easiest fix is to write it as a function in PHP also.

Thank you I thought the floor notation was a square bracket and I tried to make the p function without using a function so I kinda messed myself up there. — Mike Lewis, Jul 14 '17 at 19:11

score 0 · Answer 2 · answered Jul 14 '17 at 18:42

0

You can use strftime to get the week number of the last day of the year.

for ($yr = 2000; $yr < 2144; ++$yr) {
    $weeks = (int) strftime("%U", strtotime("$yr-12-31 11:59"));
    if ($weeks === 53) {
        print("$yr has 53 weeks in it\n");
    }
}

answered Jul 14 '17 at 18:42

amphetamachine

27,620
12
60
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Thanks for the quick reply. I know you can use built in functions or formats but I just want to know the math to figure out a 53 week year – Mike Lewis Jul 14 '17 at 18:48
@MikeLewis I'm not sure that pursuit is worth the time and effort. – amphetamachine Jul 14 '17 at 18:54

score 0 · Answer 3 · answered Jul 14 '17 at 18:50

Not sure what the application needs in terms of "half or quarter weeks" but this little script will help you see the way the ISO week numbers work. The first ISO week of the year starts on Monday. The output is not exactly intuitive. https://iconoun.com/demo/temp_mikelewis.php

<?php // demo/temp_mikelewis.php
/**
 * Dealing with ISO Week Numbers
 *
 * https://stackoverflow.com/questions/45109650/how-to-calculate-if-a-year-has-53-weeks
 */
error_reporting(E_ALL);
echo '<pre>';

for ($i = 2000; $i <= 2035; $i++)
{
    echo PHP_EOL . "YEAR $i: " . date('D W', strtotime("$i-12-31"));
}

How to calculate if a year has 53 weeks

3 Answers3