05AB1E, 130 128 133 131 124 123 122 121 120 bytes
žežfžg)V0[Y`UD3‹©12*+>₂*T÷®Xα©т%D4÷®т÷©4÷®·()ćsO7%2@+Y`т‰0Kθ4ÖUD<i\28X+ë<7%É31α}‹iY¬>0ëY1¾ǝDÅsD12‹i>1ë\1Dǝ¤>2}}ǝVYI'.¡Q#
I'm out of my mind..
For the golfing language 05AB1E it doesn't matter at all whether the input is with . or -. However, 05AB1E doesn't have any builtins for Date objects or calculations. The only builtin regarding dates it has is today's year/month/day/hours/minutes/seconds/microseconds.
So because of that, almost all of the code you see are manual calculations to go to the next day, and calculating the day of the week.
+5 bytes due to a part I forgot in Zeller's formula (year-1 for months January and February)..
-1 byte thanks to @Grimmy, by changing \$\left\lfloor{\frac{13(m+1)}{5}}\right\rfloor\$ to \$\left\lfloor{\frac{26(m+1)}{10}}\right\rfloor\$ in the code, since 05AB1E has single-byte builtins for both 26 and 10 (being ₂ and T respectively), as opposed to the 3 bytes of 13 and 5.
Try it online or Try it online with an emulated self-specified date of 'today'.
Explanation:
Wall of text incoming.
In general, the code follows the following pseudo-code:
1 Date currentDate = today;
2 Integer counter = 0;
3 Start an infinite loop:
4* If(currentDate is NOT a Saturday and currentDate is NOT a Sunday):
5 Counter += 1;
6* currentDate += 1; // Set currentDate to the next day in line
7 If(currentDate == parsed input-string):
8 Stop the infinite loop, and output the counter
1) Date currentDate = today; is this part of the 05AB1E program:
že # Push today's day
žf # Push today's month
žg # Push today's year
) # Wrap them into a single list
V # Pop and store this list in variable `Y`
2) Integer counter = 0; and 3) Start an infinite loop: are straight-forward in the 05AB1E program:
0 # Push 0 to the stack
[ # Start an infinite loop
4) If(currentDate is NOT a Saturday and currentDate is NOT a Sunday): is the first hard part with manual calculations. Since 05AB1E has no Date builtins, we'll have to calculate the Day of the Week manually.
The general formula to do this is:
$${\displaystyle h=\left(q+\left\lfloor{\frac{13(m+1)}{5}}\right\rfloor+K+\left\lfloor{\frac{K}{4}}\right\rfloor+\left\lfloor{\frac{J}{4}}\right\rfloor-2J\right){\bmod{7}}}$$
Where for the months March through December:
- \$q\$ is the \$day\$ of the month (
[1, 31])
- \$m\$ is the 1-indexed \$month\$ (
[3, 12])
- \$K\$ is the year of the century (\$year \bmod 100\$)
- \$J\$ is the 0-indexed century (\$\left\lfloor {\frac {year}{100}}\right\rfloor\$)
And for the months January and February:
- \$q\$ is the \$day\$ of the month (
[1, 31])
- \$m\$ is the 1-indexed \$month + 12\$ (
[13, 14])
- \$K\$ is the year of the century for the previous year (\$(year - 1) \bmod 100\$)
- \$J\$ is the 0-indexed century for the previous year (\$\left\lfloor {\frac {year-1}{100}}\right\rfloor\$)
Resulting in in the day of the week \$h\$, where 0 = Saturday, 1 = Sunday, ..., 6 = Friday.
Source: Zeller's congruence
We can see this in this part of the 05AB1E program:
Y # Push variable `Y`
` # Push the day, month, and year to the stack
U # Pop and save the year in variable `X`
D # Duplicate the month
3‹ # Check if the month is below 3 (Jan. / Feb.),
# resulting in 1 or 0 for truthy/falsey respectively
© # Store this in variable `®` (without popping)
12* # Multiply it by 12 (either 0 or 12)
+ # And add it to the month
# This first part was to make Jan. / Feb. 13 and 14
> # Month + 1
₂* # Multiplied by 26
T÷ # Integer-divided by 10
® # Push month<3 from variable `®` again
Xα # Take the absolute difference with the year
© # Store this potentially modified year as new `®` (without popping)
т% # mYear modulo-100
D4÷ # mYear modulo-100, integer-divided by 4
®т÷©4÷ # mYear integer-divided by 100, and then integer-divided by 4
®·( # mYear integer-divided by 100, doubled, and then made negative
) # Wrap the entire stack into a list
ć # Extract the head (the counter variable that was also on the stack)
s # Swap so the calculated values above are as list at the top
O # Take the sum of this entire list
7% # And then take modulo-7 to complete the formula,
# resulting in 0 for Saturday, 1 for Sunday, and [2, 6] for [Monday, Friday]
2@ # Check if the day is greater than or equal to 2 (so a working day)
5) Counter += 1; is straight-forward again:
# The >=2 check with `2@` results in either 1 for truthy and 0 for falsey
+ # So just adding it to the counter variable is enough
6) currentDate += 1; // Set currentDate to the next day in line is again more complex, because we have to do it manually. So this will be expanded to the following pseudo-code:
a Integer isLeapYear = ...;
b Integer daysInCurrentMonth = currentDate.month == 2 ?
c 28 + isLeapYear
d :
e 31 - (currentDate.month - 1) % 7 % 2;
f If(currentDate.day < daysInCurrentMonth):
g nextDate.day += 1;
h Else:
i nextDate.day = 1;
j If(currentDate.month < 12):
k nextDate.month += 1;
l Else:
m nextDate.month = 1;
n nextDate.year += 1;
Sources:
Algorithm for determining if a year is a leap year. (EDIT: No longer relevant, since I use an alternative method to check leap years which saved 7 bytes.)
Algorithm for determining the number of days in a month.
6a) Integer isLeapYear = ...; is done like this in the 05AB1E program:
Y # Push variable `Y`
` # Push the days, month and year to the stack
т‰ # Divmod the year by 100
0K # Remove all items "00" (or 0 when the year is below 100)
θ # Pop the list, and leave the last item
4Ö # Check if this number is visible by 4
U # Pop and save the result in variable `X`
Also used in this 05AB1E answer of mine, so there some example years are added to illustrate the steps.
6b) currentDate.month == 2 ? and 6c) 28 + isLeapYear are done like this:
D # Duplicate the month that is now the top of the stack
< # Decrease it by 1
i # And if this is now 1 (thus February):
\ # Remove the duplicated month from the top of the stack
28X+ # Add 28 and variable `X` (the isLeapYear) together
6d) : and 6e) 31 - (currentDate.month - 1) % 7 % 2; are done like this:
ë # Else:
< # Month - 1
7% # Modulo-7
É # Is odd (shortcut for %2)
31 # Push 31
α # Absolute difference between both
} # Close the if-else
6f) If(currentDate.day < daysInCurrentMonth): is done like this:
‹ # Check if the day that is still on the stack is smaller than the value calculated
i # And if it is:
6g) nextDate.day += 1; is done like this:
Y # Push variable `Y`
¬ # Push its head, the days (without popping the list `Y`)
> # Day + 1
0 # Push index 0
# (This part is done after the if-else clauses to save bytes)
}} # Close the if-else clauses
ǝ # Insert the day + 1 at index 0 in the list `Y`
V # Pop and store the updated list in variable `Y` again
6h) Else: and 6i) nextDate.day = 1; are then done like this:
ë # Else:
Y # Push variable `Y`
1 # Push a 1
¾ # Push index 0
ǝ # Insert 1 at index 0 (days part) in the list `Y`
6j) If(currentDate.month < 12)::
D # Duplicate the list `Y`
Ås # Pop and push its middle (the month)
D12‹ # Check if the month is below 12
i # And if it is:
6k) nextDate.month += 1;:
> # Month + 1
1 # Push index 1
# (This part is done after the if-else clauses to save bytes)
}} # Close the if-else clauses
ǝ # Insert the month + 1 at index 1 in the list `Y`
V # Pop and store the updated list in variable `Y` again
6l) Else:, 6m) nextDate.month = 1; and 6n) nextDate.year += 1; are then done like this:
ë # Else:
\ # Delete the top item on the stack (the duplicated month)
1 # Push 1
D # Push index 1 (with a Duplicate)
ǝ # Insert 1 at index 1 (month part) in the list `Y`
¤ # Push its tail, the year (without popping the list `Y`)
> # Year + 1
2 # Index 2
# (This part is done after the if-else clauses to save bytes)
}} # Close the if-else clauses
ǝ # Insert the year + 1 at index 2 in the list `Y`
V # Pop and store the updated list in variable `Y` again
And finally at 7) If(currentDate == parsed input-string): and 8) Stop the infinite loop, and output the counter:
Y # Push variable `Y`
I # Push the input
'.¡ '# Split it on dots
Q # Check if the two lists are equal
# # And if they are equal: stop the infinite loop
# (And output the top of the stack (the counter) implicitly)
