Puzzle #4: Working Days - Almer S. Tigelaar

Calculating the number of days between two arbitrary dates is doable for humans and trivial on modern computers. However, calculating the number of working days: Mondays through Fridays, is more complicated. In this puzzle you are asked to devise an algorithm for doing this. It is something you can do on the back of a napkin, so you don’t need anything besides pen, paper and the rational part of your brain.

Problem

Given two dates $d_{1}, d_{2}$ where $d_{2}>d_{1}$ , what is the number of workdays $w$ between these two dates? Assume that only Saturdays and Sundays are not workdays: holidays can be ignored.

You may use the following definitions augmented with any standard mathematical functions, like $div$ and $min$ .

$d$ is a date
Subscripts are used to indicate different dates: $d_{1},d_{2}\cdots d_{n}$
$d_{x}-d_{y}$ yields the number of days between two dates. This is negative if $d_{x}<d_{y}$ and zero if the dates are identical
$weekday\left(d\right)$ gives the day of the week for the given date as a number (1 = Monday, 2 = Tuesday, …, 7 = Sunday)
$weeknumber\left(d\right)$ gives the number of the week for the specified date (in the range 1 up to including 53)
$workdays=\left{ 1,2,3,4,5\right}$ is the set of numbers that represent working days (a subset of the domain of the weekday function)

Try to work out a solution first, then expand mine below.

Solution

There are really two cases to consider.

If both days are in the same week:
$weeknumber\left(d_{1}\right)=weeknumber\left(d_{2}\right)\Rightarrow d_{2}-d_{1}\leq7$ ,
the solution might seem simple: $w=d_{2}-d_{1}$ . This will work provided that both dates have a $weekday$ less than or equal to five. If this is not the case, we need a more complex function: $w=min\left(weekday\left(d_{2}\right),5\right)-min\left(weekday\left(d_{1}\right),5\right)+1$ . The $min$ function returns the lesser of its two arguments (or either one in case they are equal). The +1 is to count the end date as a whole day. Note that we exploit here the fact that the workdays form a continuous range (1–5).

If both days are not in the same week:
$weeknumber\left(d_{1}\right)\neq weeknumber\left(d_{2}\right)$ ,
then we must devise a way to find the number of weekends in between the two dates (each weekend subtracts two days from the interval $\left[d_{2},d_{1}\right]$ ). We can do this in three steps:

Determine if the starting day is in a workweek: $5-weekday\left(d_{1}\right)\geq0$ . If so, then set this as the initial count: $c=\left(5-weekday\left(d_{1}\right)\right)$ , otherwise set $c=0$ .
$c$ now covers the number of working days in the first week of the range. To find the number of remaining days we first subtract $7-weekday\left(d_{1}\right)$ days from the days between the dates $d_{2}-d_{1}$ . We wil define this as: $r=\left(d_{2}-d_{1}+1\right)-\left(7-weekday\left(d_{1}\right)\right)$ .
The number of weekends must be $div\left(r,7\right)$ , since each weekend consists of two days, we must subtract this twice, giving: $w=c+r-\left(2\cdot div\left(r,7\right)\right)$ . The $div$ function is an integer division yielding a whole number. It is equivalent to $\leftl\floor x/y\rightr\floor$ (the floor of a division, in this case between $r$ and 7).

These steps are the full solution for the case where the weeks are not equal. Below are some examples of the method presented. However, perhaps you’ve come up with a simpler or more elegant solution? Let me know in the comments below.

Examples

For the number of working days between $d_{1}=$ 16-02-2009 (weekday 1) and $d_{2}=$ 18-02-2009 (weekday 3): since these are within the same week (week 8), the number of working days is:

$w=min\left(3,5\right)-min\left(1,5\right)+1=3-1+1=3$

For the number of working days between $d_{1}=$ 16-02-2009 (weekday 1) and $d_{2}=$ 22-02-2009 (weekday 7): these are also within the same week (week 8), but the end date is in the weekend. Still this should work:

$w=min\left(7,5\right)-min\left(1,5right)+1=5-1+1=5$

For the number of working days between $d_{1}=$ 16-02-2009 (weekday 1) and $d_{2}=$ 25-02-2009 (weekday 3): these are not within the same week, thus we must use the three step approach:

$c=\left(5-1\right)=4$
$r=\left(25-16+1\right)-\left(7-1\right)=10-6=4$
$w=4+4-\left(2\cdot0\right)=8$

Here we see that the last part of the formula has no effect. This is always true for adjacent weeks, since $div\left(r,7\right)\geq1$ only if $r\geq7$ .