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# Palindrome dates

Palindrome dates
You are encouraged to solve this task according to the task description, using any language you may know.

Today   (2020-02-02,   at the time of this writing)   happens to be a palindrome,   without the hyphens,   not only for those countries which express their dates in the   yyyy-mm-dd   format but,   unusually,   also for countries which use the   dd-mm-yyyy   format.

Write a program which calculates and shows the next 15 palindromic dates for those countries which express their dates in the   yyyy-mm-dd   format.

## ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

Uses the Algol 68G local time routine which is non-standard.

`BEGIN # print future palindromic dates                                           #    # a palindromic date must be of the form demn-nm-ed                          #    # returns a string representation of n with at least 2 digits                #    PROC two digits = ( INT n )STRING:         BEGIN             STRING result := whole( ABS n, 0 );             IF ( UPB result - LWB result ) + 1 < 2 THEN "0" +=: result FI;             IF n < 0 THEN "-" +=: result FI;             result         END # two digits # ;    # possible years for a palindromic date                                      #    []INT mn     = (  1, 10, 11, 20, 21, 30, 40, 50, 60, 70, 80, 90 );    # months corresponding to the year for for a palindromic date                #    []INT nm     = ( 10,  1, 11,  2, 12,  3,  4,  5,  6,  7,  8,  9 );    # possible centuaries for a palindromic date                                 #    []INT de     = ( 10, 11, 12, 13, 20, 21, 22, 30, 31, 32, 40, 41, 42, 50                   , 51, 52, 60, 61, 62, 70, 71, 72, 80, 81, 82, 90, 91, 92                   );    # days corresponding to the centuary for a palindromic date                  #    []INT ed     = (  1, 11, 21, 31,  2, 12, 22,  3, 13, 23,  4, 14, 24,  5                   , 15, 25,  6, 16, 26,  7, 17, 27,  8, 18, 28,  9, 19, 29                   );    # max days per month ( february handled specifically in code )               #    []INT max dd = ( 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 );    # current date in local time (Algol 68G extension)                           #    []INT date   = local time;    INT   yy now = local time[ 1 ] MOD  100;    INT   cc now = local time[ 1 ] OVER 100;     INT dates to print := 15; # maximum number of dates to print                 #    FOR c pos FROM LWB de TO UPB de WHILE dates to print > 0 DO        INT cc = de[ c pos ];        INT dd = ed[ c pos ];        FOR y pos FROM LWB nm TO UPB nm WHILE dates to print > 0 DO            INT mm = nm[ y pos ];            INT yy = mn[ y pos ];            IF cc > cc now OR ( cc = cc now AND yy > yy now ) THEN                # have a possible future date                                    #                IF dd <= max dd[ mm ]                OR ( mm = 2 AND dd = 29 AND yy MOD 4 = 0 )                THEN                    # have a valid future date                                   #                    # no need to test yy = 0 as dd = 0 is impossible             #                    dates to print -:= 1;                    print( ( two digits( cc )                           , two digits( yy )                           , "-"                           , two digits( mm )                           , "-"                           , two digits( dd )                           , newline                           )                         )                FI            FI        OD    ODEND`
Output:
```2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
```

`with Ada.Text_IO;with Ada.Calendar.Formatting;with Ada.Calendar.Arithmetic; procedure Palindrome_Dates is   Desired_Count : constant := 15;   Start_Date    : constant String := "2020-01-01 00:00:00";    use Ada.Calendar;    function Is_Palindrome_Date (Date : Time) return Boolean is      Image : String renames Formatting.Image (Date);   begin      return        Image (1) = Image (10) and        Image (2) = Image (9)  and        Image (3) = Image (7)  and        Image (4) = Image (6);   end Is_Palindrome_Date;    Date  : Ada.Calendar.Time := Formatting.Value (Start_Date);   Count : Natural := 0;    use type Ada.Calendar.Arithmetic.Day_Count;begin   loop      if Is_Palindrome_Date (Date) then         Ada.Text_IO.Put_Line (Formatting.Image (Date) (1 .. 10));         Count := Count + 1;      end if;      exit when Count = Desired_Count;      Date := Date + 1;   end loop;end Palindrome_Dates;`

## AppleScript

### Procedural

`on palindromeDates(startYear, targetNumber)    script o        property output : {}    end script     set counter to 0    set y to startYear    repeat until ((counter = targetNumber) or (y > 9999))        -- Derive a month number from the last two digits of the current year number. It's valid if it's in the range 1 to 12.        set m to y mod 10 * 10 + y mod 100 div 10        if ((m > 0) and (m < 13)) then            -- Derive a day number from the first two digits of the year number.            set d to y div 100 mod 10 * 10 + y div 1000            -- It's valid if it's between 1 and 28. Otherwise, if it's between 29 and 31, check that it fits the month and year.            -- In fact though, it'll only ever be 2 or 12 in the period containing the 15 palindromic dates after 2020.            if ((d > 0) and ¬                ((d < 29) ¬                    or ((d < 31) and ((m is not 2) or ((d is 29) and (y mod 4 is 0) and ((y mod 100 > 0) or (y mod 400 is 0))))) ¬                    or ((d is 31) and (m is not in {2, 4, 9, 6, 11})))) then                -- If the figures represent a valid date, add a yyyy-mm-dd format text to the end of the output list.                tell ((100000000 + y * 10000 + m * 100 + d) as text) to ¬                    set end of o's output to text 2 thru 5 & ("-" & text 6 thru 7) & ("-" & text 8 thru 9)                set counter to counter + 1            end if        end if        set y to y + 1    end repeat     return o's outputend palindromeDates palindromeDates(2021, 15)`
Output:
`{"2021-12-02", "2030-03-02", "2040-04-02", "2050-05-02", "2060-06-02", "2070-07-02", "2080-08-02", "2090-09-02", "2101-10-12", "2110-01-12", "2111-11-12", "2120-02-12", "2121-12-12", "2130-03-12", "2140-04-12"}`

### Functional

`use AppleScript version "2.4"use framework "Foundation"use scripting additions  -- palinYearsInRange :: Int -> Int -> [String]on palinYearsInRange(fromYear, toYear)     concatMap(palinDay(iso8601Formatter()), ¬        enumFromTo(fromYear, toYear)) end palinYearsInRange  -- palinDay :: DateFormatter -> Int -> [String]on palinDay(formatter)    script        property fmtr : formatter        on |λ|(y)            -- Either an empty list or a list containing a valid            -- palindromic date for a year in the range [1000 .. 9999]            if 10000 > y and 999 < y then                set s to y as string                set {m, m1, d, d1} to reverse of characters of s                set mbDate to s & "-" & m & m1 & "-" & d & d1                 if missing value is not ¬                    (fmtr's dateFromString:(mbDate & ¬                        "T00:00:00+00:00")) then                    {mbDate}                else                    {}                end if            else                {}            end if        end |λ|    end scriptend palinDay  --------------------------- TEST ---------------------------on run    set xs to palinYearsInRange(2021, 9999)     unlines({¬        "Count of palindromic dates [2021..9999]: " & ¬        ((length of xs) as string), ¬        "", ¬        "First 15:", unlines(items 1 thru 15 of xs), "", ¬        "Last 15:", unlines(items -15 thru -1 of xs)})end run  -------------------- GENERIC FUNCTIONS --------------------- -- concatMap :: (a -> [b]) -> [a] -> [b]on concatMap(f, xs)    set lng to length of xs    set acc to {}    tell mReturn(f)        repeat with i from 1 to lng            set acc to acc & (|λ|(item i of xs, i, xs))        end repeat    end tell    return accend concatMap  -- enumFromTo :: Int -> Int -> [Int]on enumFromTo(m, n)    if m ≤ n then        set lst to {}        repeat with i from m to n            set end of lst to i        end repeat        lst    else        {}    end ifend enumFromTo  -- mReturn :: First-class m => (a -> b) -> m (a -> b)on mReturn(f)    -- 2nd class handler function lifted into 1st class script wrapper.     if script is class of f then        f    else        script            property |λ| : f        end script    end ifend mReturn  -- iso8601Formatter :: () -> NSISO8601DateFormatteron iso8601Formatter()    tell current application        set formatter to its NSISO8601DateFormatter's alloc's init()        set formatOptions of formatter to ¬            (its NSISO8601DateFormatWithInternetDateTime as integer)        return formatter    end tellend iso8601Formatter  -- unlines :: [String] -> Stringon unlines(xs)    -- A single string formed by the intercalation    -- of a list of strings with the newline character.    set {dlm, my text item delimiters} to ¬        {my text item delimiters, linefeed}    set str to xs as text    set my text item delimiters to dlm    strend unlines`
Output:
```Count of palindromic dates [2021..9999]: 284

First 15:
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12

Last 15:
9170-07-19
9180-08-19
9190-09-19
9201-10-29
9210-01-29
9211-11-29
9220-02-29
9221-12-29
9230-03-29
9240-04-29
9250-05-29
9260-06-29
9270-07-29
9280-08-29
9290-09-29```

## AWK

` # syntax: GAWK -f PALINDROME_DATES.AWKBEGIN {    show = 15    year_b = 2020    year_e = 9999    split("31,28,31,30,31,30,31,31,30,31,30,31",daynum_array,",") # days per month in non leap year    for (y=year_b; y<=year_e; y++) {      daynum_array[2] = (y % 400 == 0 || (y % 4 == 0 && y % 100)) ? 29 : 28      for (m=1; m<=12; m++) {        for (d=1; d<=daynum_array[m]; d++) {          ymd = sprintf("%04d%02d%02d",y,m,d)          if (substr(ymd,1,1) == substr(ymd,8,1)) { # speed up            if (ymd == reverse(ymd)) {              arr[++n] = ymd            }          }        }      }    }    printf("%04d0101-%04d1231=%d years, %d palindromes, showing first and last %d\n",year_b,year_e,year_e-year_b+1,n,show)    printf("YYYYMMDD YYYYMMDD\n")    for (i=1; i<=show; i++) {      printf("%s %s\n",arr[i],arr[n-show+i])    }    exit(0)}function reverse(str,  i,rts) {    for (i=length(str); i>=1; i--) {      rts = rts substr(str,i,1)    }    return(rts)} `
Output:
```20200101-99991231=7980 years, 285 palindromes, showing first and last 15
YYYYMMDD YYYYMMDD
20200202 91700719
20211202 91800819
20300302 91900919
20400402 92011029
20500502 92100129
20600602 92111129
20700702 92200229
20800802 92211229
20900902 92300329
21011012 92400429
21100112 92500529
21111112 92600629
21200212 92700729
21211212 92800829
21300312 92900929
```

## BBC BASIC

`      INSTALL @lib\$ + "DATELIB"      DIM B% 8      TestDate%=FN_today      REPEAT        \$B%=FN_date\$(TestDate%, "yyyyMMdd")        FOR I%=0 TO 3          IF ?(B% + I%) <> ?(B% + 7 - I%) EXIT FOR        NEXT        IF I%=4 PRINT FN_date\$(TestDate%, "yyyy-MM-dd")        TestDate%+=1      UNTIL VPOS=15      END`
Output:
```2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
```

## C

This only works if time_t is a 64-bit type.

`#include <stdbool.h>#include <stdio.h>#include <string.h>#include <time.h> bool is_palindrome(const char* str) {    size_t n = strlen(str);    for (size_t i = 0; i + 1 < n; ++i, --n) {        if (str[i] != str[n - 1])            return false;    }    return true;} int main() {    time_t timestamp = time(0);    const int seconds_per_day = 24*60*60;    int count = 15;    char str[32];    printf("Next %d palindrome dates:\n", count);    for (; count > 0; timestamp += seconds_per_day) {        struct tm* ptr = gmtime(&timestamp);        strftime(str, sizeof(str), "%Y%m%d", ptr);        if (is_palindrome(str)) {            strftime(str, sizeof(str), "%F", ptr);            printf("%s\n", str);            --count;        }    }    return 0;}`
Output:
```Next 15 palindrome dates:
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
```

## C#

`using System;using System.Linq;using System.Collections.Generic; public class Program{    static void Main()    {        foreach (var date in PalindromicDates(2021).Take(15)) WriteLine(date.ToString("yyyy-MM-dd"));    }     public static IEnumerable<DateTime> PalindromicDates(int startYear) {        for (int y = startYear; ; y++) {            int m = Reverse(y % 100);            int d = Reverse(y / 100);            if (IsValidDate(y, m, d, out var date)) yield return date;        }         int Reverse(int x) => x % 10 * 10 + x / 10;        bool IsValidDate(int y, int m, int d, out DateTime date) => DateTime.TryParse(\$"{y}-{m}-{d}", out date);    }}`
Output:
```2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12```

## C++

Library: Boost
`#include <iostream>#include <string>#include <boost/date_time/gregorian/gregorian.hpp> bool is_palindrome(const std::string& str) {    for (size_t i = 0, j = str.size(); i + 1 < j; ++i, --j) {        if (str[i] != str[j - 1])            return false;    }    return true;} int main() {    using boost::gregorian::date;    using boost::gregorian::day_clock;    using boost::gregorian::date_duration;     date today(day_clock::local_day());    date_duration day(1);    int count = 15;    std::cout << "Next " << count << " palindrome dates:\n";    for (; count > 0; today += day) {        if (is_palindrome(to_iso_string(today))) {            std::cout << to_iso_extended_string(today) << '\n';            --count;        }    }    return 0;}`
Output:
```Next 15 palindrome dates:
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
```

## Clojure

` (defn valid-date? [[y m d]]  (and (<= 1 m 12)       (<= 1 d 31))) (defn date-str [[y m d]]  (format "%4d-%02d-%02d" y m d)) (defn yr->date [y]  (let [[_ m d] (re-find #"(..)(..)" (apply str (reverse (str y))))]    [y (Long. m) (Long. d)])) (defn palindrome-dates [start-yr n]  (->> (iterate inc start-yr)       (map yr->date)       (filter valid-date?)       (map date-str)       (take n))) `
Output:
```("2021-12-02" "2030-03-02" "2040-04-02" "2050-05-02" "2060-06-02" "2070-07-02" "2080-08-02" "2090-09-02" "2101-10-12" "2110-01-12" "2111-11-12" "2120-02-12" "2121-12-12" "2130-03-12" "2140-04-12")
```

## F#

`// palindrome_dates.fsxopen System let is_palindrome_date =    let date_string (date: DateTime) = date.ToString "yyyyMMdd"    let is_palindrome s =        let rev_string = Seq.rev >> Seq.map string >> String.concat ""        s = rev_string s    date_string >> is_palindrome let palindrome_dates =    let rec loop date =        seq {            if is_palindrome_date date            then                yield date                yield! loop (date.AddDays 1.0)            else                yield! loop (date.AddDays 1.0)        }    loop DateTime.Now let print_date =    let iso_string (date: DateTime) = date.ToString "yyyy-MM-dd"    iso_string >> printfn "%s" palindrome_dates|> Seq.take 15|> Seq.iter print_date `
Output:
```> dotnet fsi palindrome_dates.fsx
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12```

## Factor

### Brute force

A simple brute force solution that repeatedly increments a timestamp's day by one and checks whether it's a palindrome:

Works with: Factor version 0.99 2020-01-23
`USING: calendar calendar.format io kernel lists lists.lazysequences sets ; : palindrome-dates ( -- list )    2020 2 2 <date> [ 1 days time+ ] lfrom-by    [ timestamp>ymd ] lmap-lazy    [ "-" without dup reverse = ] lfilter ; 15 palindrome-dates ltake [ print ] leach`
Output:
```2020-02-02
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12```

### Faster

A faster version that directly generates palindromic numbers such as 20200202 and keeps those which are valid dates:

Works with: Factor version 0.99 2020-01-23
`USING: calendar calendar.format continuations io kernel listslists.lazy math math.functions math.parser math.ranges sequences ; : create-palindrome ( n odd? -- m )    dupd [ 10 /i ] when swap [ over 0 > ]    [ 10 * [ 10 /mod ] [ + ] bi* ] while nip ; : palindromes ( -- list )    3 lfrom [        10 swap ^ dup 10 * [a,b)        [ [ t create-palindrome ] map ]        [ [ f create-palindrome ] map ] bi        [ sequence>list ] [email protected] lappend    ] lmap-lazy lconcat [ 20200202 >= ] lfilter ; : palindrome-dates ( -- list )    palindromes [        number>string 4 cut* 2 cut [ string>number ] [email protected]        [ <date> ] [ 4drop f ] recover    ] lmap-lazy [ f = not ] lfilter ; "10,000th palindrome date after 2020-02-02: " write10,000 palindrome-dates lnth timestamp>ymd print`
Output:
```10,000th palindrome date after 2020-02-02: 1250101-05-21
```

## Go

Simple brute force as speed is not an issue here.

`package main import (    "fmt"    "time") func reverse(s string) string {    chars := []rune(s)    for i, j := 0, len(chars)-1; i < j; i, j = i+1, j-1 {        chars[i], chars[j] = chars[j], chars[i]    }    return string(chars)} func main() {    const (        layout  = "20060102"        layout2 = "2006-01-02"    )    fmt.Println("The next 15 palindromic dates in yyyymmdd format after 20200202 are:")    date := time.Date(2020, 2, 2, 0, 0, 0, 0, time.UTC)    count := 0    for count < 15 {        date = date.AddDate(0, 0, 1)        s := date.Format(layout)        r := reverse(s)        if r == s {            fmt.Println(date.Format(layout2))            count++        }    }}`
Output:
```2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
```

Or, a more ambitious version.

`package main import (    "fmt"    "sort"    "time") func reverse(s string) string {    chars := []rune(s)    for i, j := 0, len(chars)-1; i < j; i, j = i+1, j-1 {        chars[i], chars[j] = chars[j], chars[i]    }    return string(chars)} func findIndex(sl []string, s string) int {    return sort.Search(len(sl), func(i int) bool {        return sl[i] > s    })} func main() {    const (        layout  = "20060102"        layout2 = "2006-01-02"    )    palins := []string{}    for i := 0; i < 10000; i++ {        y := fmt.Sprintf("%04d", i)        r := reverse(y)        if r[:2] > "12" || r[2:] > "31" {            continue        }        d := fmt.Sprintf("%s%s", y, r)        t, err := time.Parse(layout, d)        if err == nil {            palins = append(palins, t.Format(layout2))        }    }    le := len(palins)    i1 := findIndex(palins, "1001-01-01")    i2 := findIndex(palins, "2020-02-02")    fmt.Printf("There are %d palindromic dates after 0000-01-01 of which:\n", le)    fmt.Printf("          %d are after 1000-01-01\n", le-i1)    fmt.Printf("          %d are after 2020-02-02\n", le-i2)    fmt.Println("\nThe first 15 after 2020-02-02 are:")    for i := 0; i < 15; i++ {        if i != 0 && i%5 == 0 {            fmt.Println()        }        fmt.Printf("%s   ", palins[i+i2])    }    fmt.Println("\n\nThe last 15 before 9999-12-31 are:")    for i := 15; i >= 1; i-- {        if i != 15 && i%5 == 0 {            fmt.Println()        }        fmt.Printf("%s   ", palins[le-i])    }    fmt.Println()}`
Output:
```There are 366 palindromic dates after 0000-01-01 of which:
331 are after 1000-01-01
284 are after 2020-02-02

The first 15 after 2020-02-02 are:
2021-12-02   2030-03-02   2040-04-02   2050-05-02   2060-06-02
2070-07-02   2080-08-02   2090-09-02   2101-10-12   2110-01-12
2111-11-12   2120-02-12   2121-12-12   2130-03-12   2140-04-12

The last 15 before 9999-12-31 are:
9170-07-19   9180-08-19   9190-09-19   9201-10-29   9210-01-29
9211-11-29   9220-02-29   9221-12-29   9230-03-29   9240-04-29
9250-05-29   9260-06-29   9270-07-29   9280-08-29   9290-09-29
```

`import Data.Time.Calendar (Day, fromGregorianValid)import Data.List.Split (chunksOf)import Data.List (unfoldr)import Data.Tuple (swap)import Data.Bool (bool)import Data.Maybe (mapMaybe) palinDates :: [Day]palinDates = mapMaybe palinDay [2021 .. 9999] palinDay :: Integer -> Maybe DaypalinDay y = fromGregorianValid y m d  where    [m, d] = unDigits <\$> chunksOf 2 (reversedDecimalDigits (fromInteger y)) reversedDecimalDigits :: Int -> [Int]reversedDecimalDigits =  unfoldr ((flip bool Nothing . Just . swap . flip quotRem 10) <*> (0 ==)) unDigits :: [Int] -> IntunDigits = foldl ((+) . (10 *)) 0 main :: IO ()main = do  let n = length palinDates  putStrLn \$ "Count of palindromic dates [2021..9999]: " ++ show n  putStrLn "\nFirst 15:"  mapM_ print \$ take 15 palinDates  putStrLn "\nLast 15:"  mapM_ print \$ take 15 (drop (n - 15) palinDates)`
Output:
```Count of palindromic dates [2021..9999]: 284

First 15:
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12

Last 15:
9170-07-19
9180-08-19
9190-09-19
9201-10-29
9210-01-29
9211-11-29
9220-02-29
9221-12-29
9230-03-29
9240-04-29
9250-05-29
9260-06-29
9270-07-29
9280-08-29
9290-09-29```

## Java

` import java.time.LocalDate;import java.time.format.DateTimeFormatter; public class PalindromeDates {     public static void main(String[] args) {        LocalDate date = LocalDate.of(2020, 2, 3);        DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyyMMdd");        DateTimeFormatter formatterDash = DateTimeFormatter.ofPattern("yyyy-MM-dd");        System.out.printf("First 15 palindrome dates after 2020-02-02 are:%n");        for ( int count = 0 ; count < 15 ; date = date.plusDays(1) ) {            String dateFormatted = date.format(formatter);            if ( dateFormatted.compareTo(new StringBuilder(dateFormatted).reverse().toString()) == 0 ) {                count++;                System.out.printf("date = %s%n", date.format(formatterDash));            }        }    } } `
Output:
```First 15 palindrome dates after 2020-02-02 are:
date = 2021-12-02
date = 2030-03-02
date = 2040-04-02
date = 2050-05-02
date = 2060-06-02
date = 2070-07-02
date = 2080-08-02
date = 2090-09-02
date = 2101-10-12
date = 2110-01-12
date = 2111-11-12
date = 2120-02-12
date = 2121-12-12
date = 2130-03-12
date = 2140-04-12
```

## JavaScript

### Procedural

`/** * Adds zeros for 1 digit days/months * @param date: string */const addMissingZeros = date => (/^\d\$/.test(date) ? `0\${date}` : date); /** * Formats a Date to a string. If readable is false, * string is only numbers (used for comparison), else * is a human readable date. * @param date: Date * @param readable: boolean */const formatter = (date, readable) => {  const year = date.getFullYear();  const month = addMissingZeros(date.getMonth() + 1);  const day = addMissingZeros(date.getDate());   return readable ? `\${year}-\${month}-\${day}` : `\${year}\${month}\${day}`;}; /** * Returns n (palindromesToShow) palindrome dates * since start (or 2020-02-02) * @param start: Date * @param palindromesToShow: number */function getPalindromeDates(start, palindromesToShow = 15) {  let date = start || new Date(2020, 3, 2);   for (    let i = 0;    i < palindromesToShow;    date = new Date(date.setDate(date.getDate() + 1))  ) {    const formattedDate = formatter(date);    if (formattedDate === formattedDate.split("").reverse().join("")) {      i++;      console.log(formatter(date, true));    }  }} getPalindromeDates();`
Output:
```2021-12-02 ​
2030-03-02 ​
2040-04-02 ​
2050-05-02 ​
2060-06-02 ​
2070-07-02 ​
2080-08-02 ​
2090-09-02
2101-10-12 ​
2110-01-12 ​
2111-11-12 ​
2120-02-12​
2121-12-12 ​
2130-03-12 ​
2140-04-12 ​```

### Functional

`(() => {    'use strict';     // ----------------- PALINDROME DATES ------------------     // palindromeDate :: Int -> [String]    const palindromeDate = year => {        // Either an empty list, if no palindromic date        // can be derived from this year, or a list        // containing a palindromic IS0 8601 date.        const            s = year.toString(),            r = reverse(s),            iso = [                s,                r.slice(0, 2),                r.slice(2, 4)            ].join('-');        return isNaN(new Date(iso)) ? (            []        ) : [iso];    };     // ----------------------- TEST ------------------------    const main = () => {        const            xs = enumFromTo(2021)(9999).flatMap(                palindromeDate            );        return [            `Count of palindromic dates [2021..9999]: \${                xs.length            }`,            '',            `First 15: \${'\n' + xs.slice(0, 15).join('\n')}`,            '',            `Last 15: \${'\n' + xs.slice(-15).join('\n')}`        ].join('\n');    };     // ----------------- GENERIC FUNCTIONS -----------------     // enumFromTo :: Int -> Int -> [Int]    const enumFromTo = m =>        n => Array.from({            length: 1 + n - m        }, (_, i) => m + i);     // reverse :: [a] -> [a]    const reverse = xs =>        'string' !== typeof xs ? (            xs.slice(0).reverse()        ) : xs.split('').reverse().join('');     // MAIN ---    return main();})();`
Output:
```Count of palindromic dates [2021..9999]: 284

First 15:
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12

Last 15:
9170-07-19
9180-08-19
9190-09-19
9201-10-29
9210-01-29
9211-11-29
9220-02-29
9221-12-29
9230-03-29
9240-04-29
9250-05-29
9260-06-29
9270-07-29
9280-08-29
9290-09-29```

## Julia

Uses the built-in Dates package to check date validity but not for iteration.

`using Dates function datepalindromes(nextcount=20)    println("Date palindromes:")    count, d = 0, Date(1000, 1, 1)    for year in 2021:9200        try            dig = digits(year)            month = 10 * dig[1] + dig[2]            day = 10 * dig[3] + dig[4]            d = Date(year, month, day)        catch            continue        end        println(d)        count += 1        if count >= nextcount            break        end    endend datepalindromes() `
Output:
```Date palindromes:
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
2150-05-12
2160-06-12
2170-07-12
2180-08-12
2190-09-12
```

## Perl

### Date calculation

The more robust solution, using a date/time module.

`use Time::Piece;my \$d = Time::Piece->strptime("2020-02-02", "%Y-%m-%d"); for (my \$k = 1 ; \$k <= 15 ; \$d += Time::Piece::ONE_DAY) {    my \$s = \$d->strftime("%Y%m%d");    if (\$s eq reverse(\$s) and ++\$k) {        print \$d->strftime("%Y-%m-%d\n");    }}`
Output:
```2020-02-02
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
```

### String manipulation

Given the limited look-ahead required by the task, processing date-like strings can also work.

Library: ntheory
`use strict;use warnings;use feature 'say';use ntheory qw/forsetproduct/; my \$start = '2020-02-02' =~ s/-//gr;my(\$y) = substr(\$start,0,4); my(@dates,\$cnt);forsetproduct { push @dates, "@_" } [\$y..\$y+999],['01'..'12'],['01'..'31'];for (@dates) {    (my \$date = \$_) =~ s/ //g;    next unless \$date > \$start and \$date eq reverse \$date;    say s/ /-/gr;    last if 15 == ++\$cnt;}`
Output:
```2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12```

## Phix

Note that parse_date_string() copes with 1/2/4 digit years, but (reasonably enough) throws a wobbly given 5-digit years and beyond.

`include builtins\timedate.e sequence res = {}for d=2021 to 9999 do    string s = sprintf("%4d",d),           t = reverse(s)    s &= "-"&t[1..2]&"-"&t[3..4]    sequence td = parse_date_string(s, {"YYYY-MM-DD"})    if timedate(td) then res = append(res,s) end ifend forprintf(1,"Count of palindromic dates [2021..9999]: %d\n\n",length(res))printf(1,"first 15:\n%s\n",join_by(res[1..15],3,5))printf(1,"last 15:\n%s\n",join_by(res[-15..-1],3,5))`
Output:
```Count of palindromic dates [2021..9999]: 284

first 15:
2021-12-02   2050-05-02   2080-08-02   2110-01-12   2121-12-12
2030-03-02   2060-06-02   2090-09-02   2111-11-12   2130-03-12
2040-04-02   2070-07-02   2101-10-12   2120-02-12   2140-04-12

last 15:
9170-07-19   9201-10-29   9220-02-29   9240-04-29   9270-07-29
9180-08-19   9210-01-29   9221-12-29   9250-05-29   9280-08-29
9190-09-19   9211-11-29   9230-03-29   9260-06-29   9290-09-29
```

## Python

### Functional

Defined in terms of string reversal:

Works with: Python version 3.7
`'''Palindrome dates''' from datetime import datetimefrom itertools import chain  # palinDay :: Int -> [ISO Date]def palinDay(y):    '''A possibly empty list containing the palindromic       date for the given year, if such a date exists.    '''    s = str(y)    r = s[::-1]    iso = '-'.join([s, r[0:2], r[2:]])    try:        datetime.strptime(iso, '%Y-%m-%d')        return [iso]    except ValueError:        return []  # --------------------------TEST---------------------------# main :: IO ()def main():    '''Count and samples of palindromic dates [2021..9999]    '''    palinDates = list(chain.from_iterable(        map(palinDay, range(2021, 10000))    ))    for x in [            'Count of palindromic dates [2021..9999]:',            len(palinDates),            '\nFirst 15:',            '\n'.join(palinDates[0:15]),            '\nLast 15:',            '\n'.join(palinDates[-15:])    ]:        print(x)  # MAIN ---if __name__ == '__main__':    main()`
Output:
```Count of palindromic dates [2021..9999]:
284

First 15:
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12

Last 15:
9170-07-19
9180-08-19
9190-09-19
9201-10-29
9210-01-29
9211-11-29
9220-02-29
9221-12-29
9230-03-29
9240-04-29
9250-05-29
9260-06-29
9270-07-29
9280-08-29
9290-09-29```

Or, defined in terms of integer operations, rather than string reversals:

Works with: Python version 3.7
`'''Palindrome dates''' from functools import reducefrom itertools import chainfrom datetime import date  # palinDay :: Integer -> [ISO Date]def palinDay(y):    '''A possibly empty list containing the palindromic       date for the given year, if such a date exists.    '''    [m, d] = [undigits(pair) for pair in chunksOf(2)(        reversedDecimalDigits(y)    )]    return [] if (        1 > m or m > 12 or 31 < d    ) else validISODate((y, m, d))  # --------------------------TEST---------------------------# main :: IO ()def main():    '''Count and samples of palindromic dates [2021..9999]    '''    palinDates = list(chain.from_iterable(        map(palinDay, range(2021, 10000))    ))    for x in [            'Count of palindromic dates [2021..9999]:',            len(palinDates),            '\nFirst 15:',            '\n'.join(palinDates[0:15]),            '\nLast 15:',            '\n'.join(palinDates[-15:])    ]:        print(x)  # -------------------------GENERIC------------------------- # Just :: a -> Maybe adef Just(x):    '''Constructor for an inhabited Maybe (option type) value.       Wrapper containing the result of a computation.    '''    return {'type': 'Maybe', 'Nothing': False, 'Just': x}  # Nothing :: Maybe adef Nothing():    '''Constructor for an empty Maybe (option type) value.       Empty wrapper returned where a computation is not possible.    '''    return {'type': 'Maybe', 'Nothing': True}  # chunksOf :: Int -> [a] -> [[a]]def chunksOf(n):    '''A series of lists of length n, subdividing the       contents of xs. Where the length of xs is not evenly       divible, the final list will be shorter than n.    '''    return lambda xs: reduce(        lambda a, i: a + [xs[i:n + i]],        range(0, len(xs), n), []    ) if 0 < n else []  # reversedDecimalDigits :: Int -> [Int]def reversedDecimalDigits(n):    '''A list of the decimal digits of n,       in reversed sequence.    '''    return unfoldr(        lambda x: Nothing() if (            0 == x        ) else Just(divmod(x, 10))    )(n)  # unDigits :: [Int] -> Intdef undigits(xs):    '''An integer derived from a list of decimal digits    '''    return reduce(lambda a, x: a * 10 + x, xs, 0)  # unfoldr(lambda x: Just((x, x - 1)) if 0 != x else Nothing())(10)# -> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]# unfoldr :: (b -> Maybe (a, b)) -> b -> [a]def unfoldr(f):    '''Dual to reduce or foldr.       Where catamorphism reduces a list to a summary value,       the anamorphic unfoldr builds a list from a seed value.       As long as f returns Just(a, b), a is prepended to the list,       and the residual b is used as the argument for the next       application of f.       When f returns Nothing, the completed list is returned.    '''    def go(v):        xr = v, v        xs = []        while True:            mb = f(xr[0])            if mb.get('Nothing'):                return xs            else:                xr = mb.get('Just')                xs.append(xr[1])        return xs    return go  # validISODate :: (Int, Int, Int) -> [Date]def validISODate(ymd):    '''A possibly empty list containing the       ISO8601 string for a date, if that date exists.    '''    try:        return [date(*ymd).isoformat()]    except ValueError:        return []  # MAIN ---if __name__ == '__main__':    main()`
Output:
```Count of palindromic dates [2021..9999]:
284

First 15:
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12

Last 15:
9170-07-19
9180-08-19
9190-09-19
9201-10-29
9210-01-29
9211-11-29
9220-02-29
9221-12-29
9230-03-29
9240-04-29
9250-05-29
9260-06-29
9270-07-29
9280-08-29
9290-09-29```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2020.01

Pretty basic, but good enough. Could start earlier but 3/2/1 digit years require different handling that isn't necessary for this task. (And would be pretty pointless anyway assuming we need 2 digits for the month and two digits for the day. ISO:8601 anybody?)

`my \$start = '1000-01-01'; my @palindate = {     state \$year = \$start.substr(0,4);     ++\$year;     my \$m = \$year.substr(2, 2).flip;     my \$d = \$year.substr(0, 2).flip;     next if not try Date.new("\$year-\$m-\$d");     "\$year-\$m-\$d"} … *; my \$date-today = Date.today; # 2020-02-02 my \$k = @palindate.first: { Date.new(\$_) > \$date-today }, :k; say join "\n", @palindate[\$k - 1 .. \$k + 14]; say "\nTotal number of four digit year palindrome dates:\n" ~my \$four = @palindate.first( { .substr(5,1) eq '-' }, :k );say "between {@palindate[0]} and {@palindate[\$four - 1]}."; my \$five = @palindate.first: { .substr(6,1) eq '-' }, :k; say "\nTotal number of five digit year palindrome dates:\n" ~+@palindate[\$four .. \$five]`
Output:
```2020-02-02
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12

Total number of four digit year palindrome dates:
331
between 1001-10-01 and 9290-09-29.

Total number of five digit year palindrome dates:
3303```

## REXX

This REXX version works with   Regina REXX.

The   date   BIF   (with the   base   argument)   converts a date to the number of years since the beginning of
the Gregorian calendar,   the date is in the   ISO   format   (International Standards Organization   8601:2004).

`/*REXX program finds & displays the next  N  palindromic dates starting after 2020─02─02*//*                                                                      ─────           */parse arg n from .                               /*obtain optional argumets from the CL*/if    n=='' |    n==","  then    n= 15           /*Not specified?  Then use the default.*/if from=='' | from==","  then from= '2020-02-02' /* "      "         "   "   "     "    */#= 0                                             /*the count of palindromic dates so far*/     do j=date('Base', from, "ISO")+1 until #==n /*find palindromic dates 'til  N  found*/     aDate= date('ISO', j, "Base")               /*convert a "base" date to ISO format. */     \$= space( translate(aDate, , '-'),  0)      /*elide the dashes  (-)  in this date. */     if \$\==reverse(\$)  then iterate             /*Not palindromic?  Then skip this date*/     say 'a palindromic date: '        aDate     /*display a palindromic date ──► term. */     #= # + 1                                    /*bump the counter of palindromic dates*/     end   /*j*/                                 /*stick a fork in it,  we're all done. */`
output   when using the default inputs:
```a palindromic date:  2021-12-02
a palindromic date:  2030-03-02
a palindromic date:  2040-04-02
a palindromic date:  2050-05-02
a palindromic date:  2060-06-02
a palindromic date:  2070-07-02
a palindromic date:  2080-08-02
a palindromic date:  2090-09-02
a palindromic date:  2101-10-12
a palindromic date:  2110-01-12
a palindromic date:  2111-11-12
a palindromic date:  2120-02-12
a palindromic date:  2121-12-12
a palindromic date:  2130-03-12
a palindromic date:  2140-04-12
```

## Ring

` load "stdlib.ring" dt		= 0num		= 0limit	        = 15 ? "First 15 palindromic dates:" + nl while num < limit 	dt++	dateStr  = adddays(date(),dt)	newDate  = substr(dateStr,7,4) + substr(dateStr,4,2) + substr(dateStr,1,2)	newDate2 = substr(dateStr,7,4) + "-" + substr(dateStr,4,2) + "-" + substr(dateStr,1,2)	if ispalindrome(newDate)                num++		? newDate2 		ok	if num > limit		exit	ok end `
Output:
```First 15 palindromic dates:

2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
```

## Ruby

`require 'date' palindate = Enumerator.new do |yielder|  ("2020"..).each do |y|    m, d = y.reverse.scan(/../) # let the Y10K kids handle 5 digit years    strings = [y, m, d]    yielder << strings.join("-") if Date.valid_date?( *strings.map( &:to_i ) )  endend puts palindate.take(15)`
Output:
```2020-02-02
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
```

## Rust

`// [dependencies]// chrono = "0.4" fn is_palindrome(s: &str) -> bool {    s.chars().rev().eq(s.chars())} fn main() {    let mut date = chrono::Utc::today();    let mut count = 0;    while count < 15 {        if is_palindrome(&date.format("%Y%m%d").to_string()) {            println!("{}", date.format("%F"));            count += 1;        }        date = date.succ();    }}`
Output:

Output when program is run on 2020-07-13:

```2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
```

## Sidef

Translation of: Ruby
`var palindates = gather {    for y in (2020 .. 9999) {        var (m, d) = Str(y).flip.last(4).split(2)...        with ([y,m,d].join('-')) {|t|            take(t) if Date.valid(t, "%Y-%m-%d")        }    }} say "Count of palindromic dates [2020..9999]: #{palindates.len}" for a,b in ([    ["First 15:", palindates.head(15)],    ["Last 15:",  palindates.tail(15)]]) {    say ("\n#{a}\n", b.slices(5).map { .join("   ") }.join("\n"))}`
Output:
```Count of palindromic dates [2020..9999]: 285

First 15:
2020-02-02   2021-12-02   2030-03-02   2040-04-02   2050-05-02
2060-06-02   2070-07-02   2080-08-02   2090-09-02   2101-10-12
2110-01-12   2111-11-12   2120-02-12   2121-12-12   2130-03-12

Last 15:
9170-07-19   9180-08-19   9190-09-19   9201-10-29   9210-01-29
9211-11-29   9220-02-29   9221-12-29   9230-03-29   9240-04-29
9250-05-29   9260-06-29   9270-07-29   9280-08-29   9290-09-29
```

## Swift

`import Foundation func isPalindrome(_ string: String) -> Bool {    let chars = string.lazy    return chars.elementsEqual(chars.reversed())} let format = DateFormatter()format.dateFormat = "yyyyMMdd" let outputFormat = DateFormatter()outputFormat.dateFormat = "yyyy-MM-dd" var count = 0let limit = 15let calendar = Calendar.currentvar date = Date() while count < limit {    if isPalindrome(format.string(from: date)) {        print(outputFormat.string(from: date))        count += 1    }    date = calendar.date(byAdding: .day, value: 1, to: date)!}`
Output:

Output when executed on 2020-07-26:

```2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
```

## UNIX Shell

printf format, rev and date commands are the keys :

`is_palyndrom_date() { date -d "\$1" 1>/dev/null 2>&1 && echo "\$1" ; } for _H in {2..9}; do for _I in {0..9}; do  for _J in {0..99}; do   is_palyndrom_date \${_H}\${_I}`printf "%02d%s" \${_J}`-`printf "%02d%s" \${_J} | rev`-`printf "%02d%s" \${_H}\${_I} | rev`  done donedone `
Output:
```2001-10-02
2010-01-02
2011-11-02
2020-02-02
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
...
```

## Wren

Library: Wren-fmt
Library: Wren-date
`import "/fmt" for Fmtimport "/date" for Date var isPalDate = Fn.new { |date|    date = date.format(Date.rawDate)    return date == date[-1..0]} Date.default = Date.isoDateSystem.print("The next 15 palindromic dates in yyyy-mm-dd format after 2020-02-02 are:")var date = Date.new(2020, 2, 2)var count = 0while (count < 15) {    date = date.addDays(1)    if (isPalDate.call(date)) {        System.print(date)        count = count + 1    }}`
Output:
```The next 15 palindromic dates in yyyy-mm-dd format after 2020-02-02 are:
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
2140-04-12
```

## zkl

`TD,date,n := Time.Date, T(2020,02,02), 15;while(n){   ds:=TD.toYMDString(date.xplode()) - "-";   if(ds==ds.reverse()){ n-=1; println(TD.toYMDString(date.xplode())); }   date=TD.addYMD(date,0,0,1);}`
Output:
```2020-02-02
2021-12-02
2030-03-02
2040-04-02
2050-05-02
2060-06-02
2070-07-02
2080-08-02
2090-09-02
2101-10-12
2110-01-12
2111-11-12
2120-02-12
2121-12-12
2130-03-12
```