Programming Challenge: Mini Calendar Display

# Print a calendar of this month and the two surrounding months.


# current date and weekday
y‿m‿d‿wd ← •BQN 1⊑ •SH ⟨"date", "+%Y‿%m‿%d‿%w"⟩
m‿d -↩ 1

# number of days in a month
Leap ← ≠´0=4‿100‿400|¨⊣
Days ← ⊢◶31‿(28+Leap)‿31‿30‿31‿30‿31‿31‿30‿31‿30‿31

# number of days and starting weekday of these three months
Offset ← {y‿m𝕊o: x←o+m+12×y ⋄ ⟨⌊x÷12,12|x⟩}
⟨prev,cur,·⟩ ← counts ← Days´¨ y‿m⊸Offset¨ ¯1‿0‿1
starts ← 7 | ⟨wd-d+prev, wd-d, cur+wd-d⟩

# format a single month as rank-2 character array
S ← ⥊⟜⟨"    "⟩
Chart ← { ∾˘ 6‿7⥊ (S𝕨)∾((¯4↑•Fmt)¨1+↕𝕩)∾(S 42) }


# output
•Out (32⥊" ")∾•Fmt⟨y,m+1,d+1⟩
•Out˘ (∾⟜"  │"⊸∾)˘˝> starts Chart¨ counts

import calendar
import datetime
import re
import sys


def _ansi(s, e):
    return lambda string: f"\033[{s}m{string}\033[{e}m"


_bold = _ansi(1, 0)
_dim = _ansi(2, 22)
_uline = _ansi(4, 24)
_inv = _ansi(7, 27)

if len(sys.argv) > 1:
    date = datetime.datetime.strptime(sys.argv[1], "%Y-%m-%d").date()
else:
    date = datetime.date.today()

curr = calendar.month(date.year, date.month, w=3)

# underline the current day in the header
day = calendar.day_abbr[date.weekday()]
curr = re.sub(rf"\b{day}\b", _uline(r"\g<0>"), curr)

# bold the current week
# be careful to place the bold tag _after_ any spaces at the start of the line.
# this makes it easier to do line merging later on.
curr = re.sub(rf'(?m)[^ \n].*?\b{date.day}\b.*$', _bold(r'\g<0>'), curr)

# highlight the current date
curr = re.sub(rf" ?\b{date.day}\b ?", _inv(r"\g<0>"), curr)

lines = curr.splitlines()

# merge last week of previous month
prev_month = date.replace(day=1) - datetime.timedelta(days=1)
prev_week = calendar.month(prev_month.year, prev_month.month, w=3).splitlines()[-1]
if lines[2].startswith("   "):
    # prev_week has no ansi codes yet. no problems.
    lines[2] = _dim(prev_week) + lines[2][len(prev_week) :]
else:
    lines.insert(2, _dim(prev_week))

# merge first week of next month
next_month = date.replace(day=1) + datetime.timedelta(days=32)
next_week = calendar.month(next_month.year, next_month.month, w=3).splitlines()[2]
if next_week.startswith("   "):
    # lines[-1] already has ansi codes. be careful.
    part = '   ' + next_week.lstrip(' ')
    lines[-1] += _dim(part)
else:
    lines.append(_dim(next_week))

print("\n".join(lines))

/* calendar.c
 * by Lily
 *
 * A small calendar display for the current month.
 *
 * For the "Mini Calendar Display" challenge on Tildes:
 * https://tildes.net/~comp/15l9/programming_challenge_mini_calendar_display
 */

#include <stdbool.h>
#include <stdio.h>
#include <time.h>

int main(void) {
    // get current date
    // (only the year needs extra calculation)
    time_t t = time(NULL);
    struct tm tm = *localtime(&t);
    int year = tm.tm_year + 1900;
    
    // display date
    printf("\x1b[1;4m%d-%02d-%02d                 \x1b[24m\n", year, tm.tm_mon,
           tm.tm_mday);

    // print weekdays, highlighting the current one
    for (int weekday = 0; weekday < 7; ++weekday) {
        char *name;
        switch (weekday) {
            case 0: name = "Sun"; break;
            case 1: name = "Mon"; break;
            case 2: name = "Tue"; break;
            case 3: name = "Wed"; break;
            case 4: name = "Thu"; break;
            case 5: name = "Fri"; break;
            case 6: name = "Sat"; break;
        }

        if (weekday == tm.tm_wday) {
            printf("\x1b[7m%s\x1b[27m ", name);
        } else {
            printf("%s ", name);
        }
    }
    puts("\x1b[0m");

    // hardcoding seems like the best option here
    int days_per_month[12] = {
        31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
    };

    // account for leap years
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0) {
        days_per_month[1] = 29;
    }

    // precalculate adjacent months
    int previous_month = (tm.tm_mon - 1) % 12;
    int next_month = (tm.tm_mon + 1) % 12;

    // start on the last Monday of the previous month
    int month = previous_month;
    int weekday = 0;
    
    // find what that Monday is (very hacky)
    int day = days_per_month[month]
              - (7 + (tm.tm_wday - (tm.tm_mday - 1)) % 7 - 1);

    // print days until we reach the first Friday in next month
    // days not in the current month are dimmed
    // today is highlighted
    fputs("\x1b[2m", stdout);
    while (month != next_month || weekday != 0) {
        if (day == tm.tm_mday && month == tm.tm_mon) {
            printf("\x1b[7m%3d\x1b[27m ", day);
        } else {
            printf("%3d ", day);
        }
        
        // move to next day, and month if needed
        ++day;
        if (day > days_per_month[month]) {
            ++month;
            day = 1;

            // set correct formatting for the new month
            if (month == tm.tm_mon) {
                fputs("\x1b[0m", stdout);
            } else {
                fputs("\x1b[2m", stdout);
            }
        }

        // move to next weekday
        ++weekday;
        if (weekday > 6) {
            weekday = 0;
            putchar('\n');
        }
    }

    // disable all formatting as to not mess up the shell
    fputs("\x1b[0m", stdout);
}