Cheryl's birthday

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Task
Cheryl's birthday
You are encouraged to solve this task according to the task description, using any language you may know.

Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is.

Cheryl gave them a list of ten possible dates:

     May 15,     May 16,     May 19
     June 17,    June 18
     July 14,    July 16
     August 14,  August 15,  August 17

Cheryl then tells Albert the   month   of birth,   and Bernard the   day   (of the month)   of birth.

 1)  Albert:   I don't know when Cheryl's birthday is, but I know that Bernard does not know too.
 2)  Bernard:  At first I don't know when Cheryl's birthday is, but I know now.
 3)  Albert:   Then I also know when Cheryl's birthday is.


Task

Write a computer program to deduce, by successive elimination, Cheryl's birthday.


Related task


References



11l

Translation of: Nim

<lang 11l>T Date

  String month
  Int day
  F (month, day)
     .month = month
     .day = day

V dates = [Date(‘May’, 15), Date(‘May’, 16), Date(‘May’, 19), Date(‘June’, 17), Date(‘June’, 18),

          Date(‘July’, 14), Date(‘July’, 16), Date(‘August’, 14), Date(‘August’, 15), Date(‘August’, 17)]

DefaultDict[Int, Set[String]] monthTable L(date) dates

  monthTable[date.day].add(date.month)

DefaultDict[String, Set[Int]] dayTable L(date) dates

  dayTable[date.month].add(date.day)

Set[String] possibleMonths Set[Int] possibleDays

L(month, days) dayTable

  I days.len > 1
     possibleMonths.add(month)

L(month, days) dayTable

  L(day) days
     I monthTable[day].len == 1
        possibleMonths.remove(month)

print(‘After first Albert's sentence, possible months are ’Array(possibleMonths).join(‘, ’)‘.’)

L(day, months) monthTable

  I months.len > 1
     possibleDays.add(day)

Set[Int] impossibleDays L(day) possibleDays

  I monthTable[day].intersection(possibleMonths).len > 1
     impossibleDays.add(day)

L(day) impossibleDays

  possibleDays.remove(day)

print(‘After Bernard's sentence, possible days are ’Array(possibleDays).join(‘, ’)‘.’)

Set[String] impossibleMonths L(month) possibleMonths

  I dayTable[month].intersection(possibleDays).len > 1
     impossibleMonths.add(month)

L(month) impossibleMonths

  possibleMonths.remove(month)

assert(possibleMonths.len == 1) V month = possibleMonths.pop() print(‘After second Albert's sentence, remaining month is ’month‘...’)

possibleDays = possibleDays.intersection(dayTable[month]) assert(possibleDays.len == 1) V day = possibleDays.pop() print(‘and thus remaining day is ’day‘.’)

print() print(‘So birthday date is ’month‘ ’day‘.’)</lang>

Output:
After first Albert's sentence, possible months are August, July.
After Bernard's sentence, possible days are 15, 16, 17.
After second Albert's sentence, remaining month is July...
and thus remaining day is 16.

So birthday date is July 16.

AppleScript

<lang applescript>use AppleScript version "2.4" use framework "Foundation" use scripting additions

property M : 1 -- Month property D : 2 -- Day

on run

   -- The MONTH with only one remaining day
   -- among the DAYs with unique months,
   -- EXCLUDING months with unique days,
   -- in Cheryl's list:
   
   showList(uniquePairing(M, ¬
       uniquePairing(D, ¬
           monthsWithUniqueDays(false, ¬
               map(composeList({tupleFromList, |words|, toLower}), ¬
                   splitOn(", ", ¬
                       "May 15, May 16, May 19, June 17, June 18, " & ¬
                       "July 14, July 16, Aug 14, Aug 15, Aug 17"))))))
   
   --> "[('july', '16')]"

end run


-- QUERY FUNCTIONS ----------------------------------------

-- monthsWithUniqueDays :: Bool -> [(Month, Day)] -> [(Month, Day)] on monthsWithUniqueDays(blnInclude, xs)

   set _months to map(my fst, uniquePairing(D, xs))
   script uniqueDay
       on |λ|(md)
           set bln to elem(fst(md), _months)
           if blnInclude then
               bln
           else
               not bln
           end if
       end |λ|
   end script
   filter(uniqueDay, xs)

end monthsWithUniqueDays


-- uniquePairing :: DatePart -> [(M, D)] -> [(M, D)] on uniquePairing(dp, xs)

   script go
       property f : my mReturn(item dp of {my fst, my snd})
       on |λ|(md)
           
           set dct to f's |λ|(md)
           script unique
               on |λ|(k)
                   set mb to lookupDict(k, dct)
                   if Nothing of mb then
                       false
                   else
                       1 = length of (Just of mb)
                   end if
               end |λ|
           end script
           set uniques to filter(unique, keys(dct))
           
           script found
               on |λ|(tpl)
                   elem(f's |λ|(tpl), uniques)
               end |λ|
           end script
           filter(found, xs)
       end |λ|
   end script
   bindPairs(xs, go)

end uniquePairing


-- bindPairs :: [(M, D)] -> ((Dict Text [Text], Dict Text [Text]) -- -> [(M, D)]) -> [(M, D)] on bindPairs(xs, f)

   tell mReturn(f)
       |λ|(Tuple(dictFromPairs(xs), ¬
           dictFromPairs(map(my swap, xs))))
   end tell

end bindPairs

-- dictFromPairs :: [(M, D)] -> Dict Text [Text] on dictFromPairs(mds)

   set gps to groupBy(|on|(my eq, my fst), ¬
       sortBy(comparing(my fst), mds))
   script kv
       on |λ|(gp)
           Tuple(fst(item 1 of gp), map(my snd, gp))
       end |λ|
   end script
   mapFromList(map(kv, gps))

end dictFromPairs


-- LIBRARY GENERICS ---------------------------------------

-- comparing :: (a -> b) -> (a -> a -> Ordering) on comparing(f)

   script
       on |λ|(a, b)
           tell mReturn(f)
               set fa to |λ|(a)
               set fb to |λ|(b)
               if fa < fb then
                   -1
               else if fa > fb then
                   1
               else
                   0
               end if
           end tell
       end |λ|
   end script

end comparing

-- composeList :: [(a -> a)] -> (a -> a) on composeList(fs)

   script
       on |λ|(x)
           script
               on |λ|(f, a)
                   mReturn(f)'s |λ|(a)
               end |λ|
           end script
           
           foldr(result, x, fs)
       end |λ|
   end script

end composeList

-- drop :: Int -> [a] -> [a] -- drop :: Int -> String -> String on drop(n, xs)

   set c to class of xs
   if c is not script then
       if c is not string then
           if n < length of xs then
               items (1 + n) thru -1 of xs
           else
               {}
           end if
       else
           if n < length of xs then
               text (1 + n) thru -1 of xs
           else
               ""
           end if
       end if
   else
       take(n, xs) -- consumed
       return xs
   end if

end drop

-- dropAround :: (a -> Bool) -> [a] -> [a] -- dropAround :: (Char -> Bool) -> String -> String on dropAround(p, xs)

   dropWhile(p, dropWhileEnd(p, xs))

end dropAround

-- dropWhile :: (a -> Bool) -> [a] -> [a] -- dropWhile :: (Char -> Bool) -> String -> String on dropWhile(p, xs)

   set lng to length of xs
   set i to 1
   tell mReturn(p)
       repeat while i ≤ lng and |λ|(item i of xs)
           set i to i + 1
       end repeat
   end tell
   drop(i - 1, xs)

end dropWhile

-- dropWhileEnd :: (a -> Bool) -> [a] -> [a] -- dropWhileEnd :: (Char -> Bool) -> String -> String on dropWhileEnd(p, xs)

   set i to length of xs
   tell mReturn(p)
       repeat while i > 0 and |λ|(item i of xs)
           set i to i - 1
       end repeat
   end tell
   take(i, xs)

end dropWhileEnd

-- elem :: Eq a => a -> [a] -> Bool on elem(x, xs)

   considering case
       xs contains x
   end considering

end elem

-- enumFromToInt :: Int -> Int -> [Int] on enumFromToInt(M, n)

   if M ≤ n then
       set lst to {}
       repeat with i from M to n
           set end of lst to i
       end repeat
       return lst
   else
       return {}
   end if

end enumFromToInt

-- eq (==) :: Eq a => a -> a -> Bool on eq(a, b)

   a = b

end eq

-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)

   tell mReturn(f)
       set lst to {}
       set lng to length of xs
       repeat with i from 1 to lng
           set v to item i of xs
           if |λ|(v, i, xs) then set end of lst to v
       end repeat
       return lst
   end tell

end filter

-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)

   tell mReturn(f)
       set v to startValue
       set lng to length of xs
       repeat with i from 1 to lng
           set v to |λ|(v, item i of xs, i, xs)
       end repeat
       return v
   end tell

end foldl

-- foldr :: (a -> b -> b) -> b -> [a] -> b on foldr(f, startValue, xs)

   tell mReturn(f)
       set v to startValue
       set lng to length of xs
       repeat with i from lng to 1 by -1
           set v to |λ|(item i of xs, v, i, xs)
       end repeat
       return v
   end tell

end foldr

-- fst :: (a, b) -> a on fst(tpl)

   if class of tpl is record then
       |1| of tpl
   else
       item 1 of tpl
   end if

end fst

-- Typical usage: groupBy(on(eq, f), xs) -- groupBy :: (a -> a -> Bool) -> [a] -> a on groupBy(f, xs)

   set mf to mReturn(f)
   
   script enGroup
       on |λ|(a, x)
           if length of (active of a) > 0 then
               set h to item 1 of active of a
           else
               set h to missing value
           end if
           
           if h is not missing value and mf's |λ|(h, x) then
               {active:(active of a) & {x}, sofar:sofar of a}
           else
               {active:{x}, sofar:(sofar of a) & {active of a}}
           end if
       end |λ|
   end script
   
   if length of xs > 0 then
       set dct to foldl(enGroup, {active:{item 1 of xs}, sofar:{}}, rest of xs)
       if length of (active of dct) > 0 then
           sofar of dct & {active of dct}
       else
           sofar of dct
       end if
   else
       {}
   end if

end groupBy

-- insertMap :: Dict -> String -> a -> Dict on insertMap(rec, k, v)

   tell (current application's NSMutableDictionary's ¬
       dictionaryWithDictionary:rec)
       its setValue:v forKey:(k as string)
       return it as record
   end tell

end insertMap

-- intercalateS :: String -> [String] -> String on intercalateS(sep, xs)

   set {dlm, my text item delimiters} to {my text item delimiters, sep}
   set s to xs as text
   set my text item delimiters to dlm
   return s

end intercalateS

-- Just :: a -> Maybe a on Just(x)

   {type:"Maybe", Nothing:false, Just:x}

end Just

-- keys :: Dict -> [String] on keys(rec)

   (current application's NSDictionary's dictionaryWithDictionary:rec)'s allKeys() as list

end keys

-- lookupDict :: a -> Dict -> Maybe b on lookupDict(k, dct)

   set ca to current application
   set v to (ca's NSDictionary's dictionaryWithDictionary:dct)'s objectForKey:k
   if v ≠ missing value then
       Just(item 1 of ((ca's NSArray's arrayWithObject:v) as list))
   else
       Nothing()
   end if

end lookupDict

-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: First-class m => (a -> b) -> m (a -> b) on mReturn(f)

   if class of f is script then
       f
   else
       script
           property |λ| : f
       end script
   end if

end mReturn

-- map :: (a -> b) -> [a] -> [b] on map(f, xs)

   tell mReturn(f)
       set lng to length of xs
       set lst to {}
       repeat with i from 1 to lng
           set end of lst to |λ|(item i of xs, i, xs)
       end repeat
       return lst
   end tell

end map

-- mapFromList :: [(k, v)] -> Dict on mapFromList(kvs)

   set tpl to unzip(kvs)
   script
       on |λ|(x)
           x as string
       end |λ|
   end script
   (current application's NSDictionary's ¬
       dictionaryWithObjects:(|2| of tpl) ¬
           forKeys:map(result, |1| of tpl)) as record

end mapFromList

-- min :: Ord a => a -> a -> a on min(x, y)

   if y < x then
       y
   else
       x
   end if

end min

-- Nothing :: Maybe a on Nothing()

   {type:"Maybe", Nothing:true}

end Nothing

-- e.g. sortBy(|on|(compare, |length|), ["epsilon", "mu", "gamma", "beta"]) -- on :: (b -> b -> c) -> (a -> b) -> a -> a -> c on |on|(f, g)

   script
       on |λ|(a, b)
           tell mReturn(g) to set {va, vb} to {|λ|(a), |λ|(b)}
           tell mReturn(f) to |λ|(va, vb)
       end |λ|
   end script

end |on|

-- partition :: predicate -> List -> (Matches, nonMatches) -- partition :: (a -> Bool) -> [a] -> ([a], [a]) on partition(f, xs)

   tell mReturn(f)
       set ys to {}
       set zs to {}
       repeat with x in xs
           set v to contents of x
           if |λ|(v) then
               set end of ys to v
           else
               set end of zs to v
           end if
       end repeat
   end tell
   Tuple(ys, zs)

end partition

-- show :: a -> String on show(e)

   set c to class of e
   if c = list then
       showList(e)
   else if c = record then
       set mb to lookupDict("type", e)
       if Nothing of mb then
           showDict(e)
       else
           script
               on |λ|(t)
                   if "Either" = t then
                       set f to my showLR
                   else if "Maybe" = t then
                       set f to my showMaybe
                   else if "Ordering" = t then
                       set f to my showOrdering
                   else if "Ratio" = t then
                       set f to my showRatio
                   else if class of t is text and t begins with "Tuple" then
                       set f to my showTuple
                   else
                       set f to my showDict
                   end if
                   tell mReturn(f) to |λ|(e)
               end |λ|
           end script
           tell result to |λ|(Just of mb)
       end if
   else if c = date then
       "\"" & showDate(e) & "\""
   else if c = text then
       "'" & e & "'"
   else if (c = integer or c = real) then
       e as text
   else if c = class then
       "null"
   else
       try
           e as text
       on error
           ("«" & c as text) & "»"
       end try
   end if

end show

-- showList :: [a] -> String on showList(xs)

   "[" & intercalateS(", ", map(my show, xs)) & "]"

end showList

-- showTuple :: Tuple -> String on showTuple(tpl)

   set ca to current application
   script
       on |λ|(n)
           set v to (ca's NSDictionary's dictionaryWithDictionary:tpl)'s objectForKey:(n as string)
           if v ≠ missing value then
               unQuoted(show(item 1 of ((ca's NSArray's arrayWithObject:v) as list)))
           else
               missing value
           end if
       end |λ|
   end script
   "(" & intercalateS(", ", map(result, enumFromToInt(1, length of tpl))) & ")"

end showTuple

-- snd :: (a, b) -> b on snd(tpl)

   if class of tpl is record then
       |2| of tpl
   else
       item 2 of tpl
   end if

end snd

-- Enough for small scale sorts. -- Use instead sortOn :: Ord b => (a -> b) -> [a] -> [a] -- which is equivalent to the more flexible sortBy(comparing(f), xs) -- and uses a much faster ObjC NSArray sort method -- sortBy :: (a -> a -> Ordering) -> [a] -> [a] on sortBy(f, xs)

   if length of xs > 1 then
       set h to item 1 of xs
       set f to mReturn(f)
       script
           on |λ|(x)
               f's |λ|(x, h) ≤ 0
           end |λ|
       end script
       set lessMore to partition(result, rest of xs)
       sortBy(f, |1| of lessMore) & {h} & ¬
           sortBy(f, |2| of lessMore)
   else
       xs
   end if

end sortBy

-- splitOn :: String -> String -> [String] on splitOn(pat, src)

   set {dlm, my text item delimiters} to ¬
       {my text item delimiters, pat}
   set xs to text items of src
   set my text item delimiters to dlm
   return xs

end splitOn

-- swap :: (a, b) -> (b, a) on swap(ab)

   if class of ab is record then
       Tuple(|2| of ab, |1| of ab)
   else
       {item 2 of ab, item 1 of ab}
   end if

end swap

-- take :: Int -> [a] -> [a] -- take :: Int -> String -> String on take(n, xs)

   set c to class of xs
   if list is c then
       if 0 < n then
           items 1 thru min(n, length of xs) of xs
       else
           {}
       end if
   else if string is c then
       if 0 < n then
           text 1 thru min(n, length of xs) of xs
       else
           ""
       end if
   else if script is c then
       set ys to {}
       repeat with i from 1 to n
           set v to xs's |λ|()
           if missing value is v then
               return ys
           else
               set end of ys to v
           end if
       end repeat
       return ys
   else
       missing value
   end if

end take

-- toLower :: String -> String on toLower(str)

   set ca to current application
   ((ca's NSString's stringWithString:(str))'s ¬
       lowercaseStringWithLocale:(ca's NSLocale's currentLocale())) as text

end toLower

-- Tuple (,) :: a -> b -> (a, b) on Tuple(a, b)

   {type:"Tuple", |1|:a, |2|:b, length:2}

end Tuple

-- tupleFromList :: [a] -> (a, a ...) on tupleFromList(xs)

   set lng to length of xs
   if 1 < lng then
       if 2 < lng then
           set strSuffix to lng as string
       else
           set strSuffix to ""
       end if
       script kv
           on |λ|(a, x, i)
               insertMap(a, (i as string), x)
           end |λ|
       end script
       foldl(kv, {type:"Tuple" & strSuffix}, xs) & {length:lng}
   else
       missing value
   end if

end tupleFromList

-- unQuoted :: String -> String on unQuoted(s)

   script p
       on |λ|(x)
           --{34, 39} contains id of x
           34 = id of x
       end |λ|
   end script
   dropAround(p, s)

end unQuoted

-- unzip :: [(a,b)] -> ([a],[b]) on unzip(xys)

   set xs to {}
   set ys to {}
   repeat with xy in xys
       set end of xs to |1| of xy
       set end of ys to |2| of xy
   end repeat
   return Tuple(xs, ys)

end unzip

-- words :: String -> [String] on |words|(s)

   set ca to current application
   (((ca's NSString's stringWithString:(s))'s ¬
       componentsSeparatedByCharactersInSet:(ca's ¬
           NSCharacterSet's whitespaceAndNewlineCharacterSet()))'s ¬
       filteredArrayUsingPredicate:(ca's ¬
           NSPredicate's predicateWithFormat:"0 < length")) as list

end |words|</lang>

Output:
"[('july', '16')]"

AutoHotkey

<lang AutoHotkey>oDates:= {"May" : [ 15, 16, 19]

           ,"Jun" : [                17, 18]
           ,"Jul" : [14,         16]
           ,"Aug" : [14,     15,     17]}

filter1(oDates) filter2(oDates) filter3(oDates) MsgBox % result := checkAnswer(oDates) return

filter1(ByRef oDates){  ; remove months that has a unique day in it.

   for d, obj in MonthsOfDay(oDates)
       if (obj.count() = 1)
           for m, bool in obj
               oDates.Remove(m)

}

filter2(ByRef oDates){  ; remove non-unique days from remaining months.

   for d, obj in MonthsOfDay(oDates)
       if (obj.count() > 1)
           for m, bool in obj
               for i, day in oDates[m]
                   if (day=d)
                       oDates[m].Remove(i)

}

filter3(ByRef oDates){  ; remove months that has multiple days from remaining months.

   oRemove := []
   for m, obj in oDates
       if obj.count() > 1
           oRemove.Push(m)
   for i, m in oRemove
       oDates.Remove(m)

}

MonthsOfDay(oDates){  ; create a list of months per day.

   MonthsOfDay := []
   for m, obj in oDates
       for i, d in obj
           MonthsOfDay[d, m] := 1
   return MonthsOfDay

}

checkAnswer(oDates){  ; check unique answer if any.

   if oDates.count()>1
       return false
   for m, obj in oDates
       if obj.count() > 1
           return false
       else 
           return m " " obj.1

}</lang>

Output:
Jul 16

AWK

<lang AWK>

  1. syntax: GAWK -f CHERYLS_BIRTHDAY.AWK [-v debug={0|1}]
  2. sorting:
  3. PROCINFO["sorted_in"] is used by GAWK
  4. SORTTYPE is used by Thompson Automation's TAWK

BEGIN {

   debug += 0
   PROCINFO["sorted_in"] = "@ind_num_asc" ; SORTTYPE = 1
   n = split("05/15,05/16,05/19,06/17,06/18,07/14,07/16,08/14,08/15,08/17",arr,",")
   for (i=1; i<=n; i++) { # move dates to a more friendly structure
     mmdd_arr[arr[i]] = ""
   }
   print("Cheryl offers these ten MM/DD choices:")
   cb_show_dates()
   printf("Cheryl then tells Albert her birth 'month' and Bernard her birth 'day'.\n\n")
   print("1. Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.")
   cb_filter1()
   print("2. Bernard: At first I don't know when Cheryl's birthday is, but I know now.")
   cb_filter2()
   print("3. Albert: Then I also know when Cheryl's birthday is.")
   cb_filter3()
   exit(0)

} function cb_filter1( i,j) {

   print("deduction: the month cannot have a unique day, leaving:")
   cb_load_arrays(4)
   for (j in arr1) {
     if (arr1[j] == 1) {
       if (debug) { printf("unique day %s\n",j) }
       arr3[arr2[j]] = ""
     }
   }
   cb_remove_dates()

} function cb_filter2( i,j) {

   print("deduction: the day must be unique, leaving:")
   cb_load_arrays(4)
   for (j in arr1) {
     if (arr1[j] > 1) {
       if (debug) { printf("non-unique day %s\n",j) }
       arr3[j] = ""
     }
   }
   cb_remove_dates("...")

} function cb_filter3( i,j) {

   print("deduction: the month must be unique, leaving:")
   cb_load_arrays(1)
   for (j in arr1) {
     if (arr1[j] > 1) {
       if (debug) { printf("non-unique month %s\n",j) }
       arr3[j] = ""
     }
   }
   cb_remove_dates()

} function cb_load_arrays(col, i,key) {

   delete arr1
   delete arr2
   delete arr3
   for (i in mmdd_arr) {
     key = substr(i,col,2)
     arr1[key]++
     arr2[key] = substr(i,1,2)
   }

} function cb_remove_dates(pattern, i,j) {

   for (j in arr3) {
     for (i in mmdd_arr) {
       if (i ~ ("^" pattern j)) {
         if (debug) { printf("removing %s\n",i) }
         delete mmdd_arr[i]
       }
     }
   }
   cb_show_dates()

} function cb_show_dates( i) {

   if (debug) { printf("%d remaining\n",length(mmdd_arr)) }
   for (i in mmdd_arr) {
     printf("%s ",i)
   }
   printf("\n\n")

} </lang>

Output:
Cheryl offers these ten MM/DD choices:
05/15 05/16 05/19 06/17 06/18 07/14 07/16 08/14 08/15 08/17

Cheryl then tells Albert her birth 'month' and Bernard her birth 'day'.

1. Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.
deduction: the month cannot have a unique day, leaving:
07/14 07/16 08/14 08/15 08/17

2. Bernard: At first I don't know when Cheryl's birthday is, but I know now.
deduction: the day must be unique, leaving:
07/16 08/15 08/17

3. Albert: Then I also know when Cheryl's birthday is.
deduction: the month must be unique, leaving:
07/16

C

Translation of: C#

<lang c>#include <stdbool.h>

  1. include <stdio.h>

char *months[] = {

   "ERR", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"

};

struct Date {

   int month, day;
   bool active;

} dates[] = {

   {5,15,true}, {5,16,true}, {5,19,true},
   {6,17,true}, {6,18,true},
   {7,14,true}, {7,16,true},
   {8,14,true}, {8,15,true}, {8,17,true}

};

  1. define UPPER_BOUND (sizeof(dates) / sizeof(struct Date))

void printRemaining() {

   int i, c;
   for (i = 0, c = 0; i < UPPER_BOUND; i++) {
       if (dates[i].active) {
           c++;
       }
   }
   printf("%d remaining.\n", c);

}

void printAnswer() {

   int i;
   for (i = 0; i < UPPER_BOUND; i++) {
       if (dates[i].active) {
           printf("%s, %d\n", months[dates[i].month], dates[i].day);
       }
   }

}

void firstPass() {

   // the month cannot have a unique day
   int i, j, c;
   for (i = 0; i < UPPER_BOUND; i++) {
       c = 0;
       for (j = 0; j < UPPER_BOUND; j++) {
           if (dates[j].day == dates[i].day) {
               c++;
           }
       }
       if (c == 1) {
           for (j = 0; j < UPPER_BOUND; j++) {
               if (!dates[j].active) continue;
               if (dates[j].month == dates[i].month) {
                   dates[j].active = false;
               }
           }
       }
   }

}

void secondPass() {

   // the day must now be unique
   int i, j, c;
   for (i = 0; i < UPPER_BOUND; i++) {
       if (!dates[i].active) continue;
       c = 0;
       for (j = 0; j < UPPER_BOUND; j++) {
           if (!dates[j].active) continue;
           if (dates[j].day == dates[i].day) {
               c++;
           }
       }
       if (c > 1) {
           for (j = 0; j < UPPER_BOUND; j++) {
               if (!dates[j].active) continue;
               if (dates[j].day == dates[i].day) {
                   dates[j].active = false;
               }
           }
       }
   }

}

void thirdPass() {

   // the month must now be unique
   int i, j, c;
   for (i = 0; i < UPPER_BOUND; i++) {
       if (!dates[i].active) continue;
       c = 0;
       for (j = 0; j < UPPER_BOUND; j++) {
           if (!dates[j].active) continue;
           if (dates[j].month == dates[i].month) {
               c++;
           }
       }
       if (c > 1) {
           for (j = 0; j < UPPER_BOUND; j++) {
               if (!dates[j].active) continue;
               if (dates[j].month == dates[i].month) {
                   dates[j].active = false;
               }
           }
       }
   }

}

int main() {

   printRemaining();
   // the month cannot have a unique day
   firstPass();
   printRemaining();
   // the day must now be unique
   secondPass();
   printRemaining();
   // the month must now be unique
   thirdPass();
   printAnswer();
   return 0;

}</lang>

Output:
10 remaining.
5 remaining.
3 remaining.
Jul, 16

C#

<lang csharp>public static class CherylsBirthday {

   public static void Main() {
       var dates = new HashSet<(string month, int day)> {
           ("May", 15),
           ("May", 16),
           ("May", 19),
           ("June", 17),
           ("June", 18),
           ("July", 14),
           ("July", 16),
           ("August", 14),
           ("August", 15),
           ("August", 17)
       };
       Console.WriteLine(dates.Count + " remaining.");
       //The month cannot have a unique day.
       var monthsWithUniqueDays = dates.GroupBy(d => d.day).Where(g => g.Count() == 1).Select(g => g.First().month).ToHashSet();
       dates.RemoveWhere(d => monthsWithUniqueDays.Contains(d.month));
       Console.WriteLine(dates.Count + " remaining.");
       //The day must now be unique.
       dates.IntersectWith(dates.GroupBy(d => d.day).Where(g => g.Count() == 1).Select(g => g.First()));
       Console.WriteLine(dates.Count + " remaining.");
       //The month must now be unique.
       dates.IntersectWith(dates.GroupBy(d => d.month).Where(g => g.Count() == 1).Select(g => g.First()));
       Console.WriteLine(dates.Single());
   }
   

}</lang>

Output:
10 remaining.
5 remaining.
3 remaining.
(July, 16)

C++

Translation of: Go

<lang cpp>#include <algorithm>

  1. include <iostream>
  2. include <vector>

using namespace std;

const vector<string> MONTHS = {

   "Jan", "Feb", "Mar", "Apr", "May", "Jun",
   "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"

};

struct Birthday {

   int month, day;
   friend ostream &operator<<(ostream &, const Birthday &);

};

ostream &operator<<(ostream &out, const Birthday &birthday) {

   return out << MONTHS[birthday.month - 1] << ' ' << birthday.day;

}

template <typename C> bool monthUniqueIn(const Birthday &b, const C &container) {

   auto it = cbegin(container);
   auto end = cend(container);
   int count = 0;
   while (it != end) {
       if (it->month == b.month) {
           count++;
       }
       it = next(it);
   }
   return count == 1;

}

template <typename C> bool dayUniqueIn(const Birthday &b, const C &container) {

   auto it = cbegin(container);
   auto end = cend(container);
   int count = 0;
   while (it != end) {
       if (it->day == b.day) {
           count++;
       }
       it = next(it);
   }
   return count == 1;

}

template <typename C> bool monthWithUniqueDayIn(const Birthday &b, const C &container) {

   auto it = cbegin(container);
   auto end = cend(container);
   while (it != end) {
       if (it->month == b.month && dayUniqueIn(*it, container)) {
           return true;
       }
       it = next(it);
   }
   return false;

}

int main() {

   vector<Birthday> choices = {
       {5, 15}, {5, 16}, {5, 19}, {6, 17}, {6, 18},
       {7, 14}, {7, 16}, {8, 14}, {8, 15}, {8, 17},
   };
   // Albert knows the month but doesn't know the day.
   // So the month can't be unique within the choices.
   vector<Birthday> filtered;
   for (auto bd : choices) {
       if (!monthUniqueIn(bd, choices)) {
           filtered.push_back(bd);
       }
   }
   // Albert also knows that Bernard doesn't know the answer.
   // So the month can't have a unique day.
   vector<Birthday> filtered2;
   for (auto bd : filtered) {
       if (!monthWithUniqueDayIn(bd, filtered)) {
           filtered2.push_back(bd);
       }
   }
   // Bernard now knows the answer.
   // So the day must be unique within the remaining choices.
   vector<Birthday> filtered3;
   for (auto bd : filtered2) {
       if (dayUniqueIn(bd, filtered2)) {
           filtered3.push_back(bd);
       }
   }
   // Albert now knows the answer too.
   // So the month must be unique within the remaining choices.
   vector<Birthday> filtered4;
   for (auto bd : filtered3) {
       if (monthUniqueIn(bd, filtered3)) {
           filtered4.push_back(bd);
       }
   }
   if (filtered4.size() == 1) {
       cout << "Cheryl's birthday is " << filtered4[0] << '\n';
   } else {
       cout << "Something went wrong!\n";
   }
   return 0;

}</lang>

Output:
Cheryl's birthday is Jul 16

Common Lisp

<lang lisp>

Author
Amir Teymuri, Saturday 20.10.2018

(defparameter *possible-dates*

 '((15 . may) (16 . may) (19 . may)
   (17 . june) (18 . june)
   (14 . july) (16 . july)
   (14 . august) (15 . august) (17 . august)))

(defun unique-date-parts (possible-dates &key (alist-look-at #'car) (alist-r-assoc #'assoc))

 (let* ((date-parts (mapcar alist-look-at possible-dates))	 

(unique-date-parts (remove-if #'(lambda (part) (> (count part date-parts) 1)) date-parts)))

   (mapcar #'(lambda (part) (funcall alist-r-assoc part possible-dates))
   	    unique-date-parts)))

(defun person (person possible-dates)

 "Who's turn is it to think?"
 (case person
   ('albert (unique-date-parts possible-dates :alist-look-at #'cdr :alist-r-assoc #'rassoc))
   ('bernard (unique-date-parts possible-dates :alist-look-at #'car :alist-r-assoc #'assoc))))

(defun cheryls-birthday (possible-dates)

 (person 'albert			

(person 'bernard (set-difference possible-dates (person 'bernard possible-dates) :key #'cdr))))

(cheryls-birthday *possible-dates*) ;; => ((16 . JULY)) </lang>

D

<lang d>import std.algorithm.iteration : filter, joiner, map; import std.algorithm.searching : canFind; import std.algorithm.sorting : sort; import std.array : array; import std.datetime.date : Date, Month; import std.stdio : writeln;

void main() {

   auto choices = [
       // Month.jan
       Date(2019, Month.may, 15),
       Date(2019, Month.may, 16),
       Date(2019, Month.may, 19),  // unique day (1)
       Date(2019, Month.jun, 17),
       Date(2019, Month.jun, 18),  // unique day (1)
       Date(2019, Month.jul, 14),
       Date(2019, Month.jul, 16),  // final answer
       Date(2019, Month.aug, 14),
       Date(2019, Month.aug, 15),
       Date(2019, Month.aug, 17),
   ];
   // The month cannot have a unique day because Albert knows the month, and knows that Bernard does not know the answer
   auto uniqueMonths = choices.sort!"a.day < b.day".groupBy.filter!"a.array.length == 1".joiner.map!"a.month";
   // writeln(uniqueMonths.save);
   auto filter1 = choices.filter!(a => !canFind(uniqueMonths.save, a.month)).array;
   // Bernard now knows the answer, so the day must be unique within the remaining choices
   auto uniqueDays = filter1.sort!"a.day < b.day".groupBy.filter!"a.array.length == 1".joiner.map!"a.day";
   auto filter2 = filter1.filter!(a => canFind(uniqueDays.save, a.day)).array;
   // Albert knows the answer too, so the month must be unique within the remaining choices
   auto birthDay = filter2.sort!"a.month < b.month".groupBy.filter!"a.array.length == 1".joiner.front;
   // print the result
   writeln(birthDay.month, " ", birthDay.day);

}</lang>

Output:
jul 16

F#

<lang fsharp> //Find Cheryl's Birthday. Nigel Galloway: October 23rd., 2018 type Month = |May |June |July |August let fN n= n |> List.filter(fun (_,n)->(List.length n) < 2) |> List.unzip let dates = [(May,15);(May,16);(May,19);(June,17);(June,18);(July,14);(July,16);(August,14);(August,15);(August,17)] let _,n = dates |> List.groupBy snd |> fN let i = n |> List.concat |> List.map fst |> Set.ofList let _,g = dates |> List.filter(fun (n,_)->not (Set.contains n i)) |> List.groupBy snd |> fN let _,e = List.concat g |> List.groupBy fst |> fN printfn "%A" e </lang>

Output:
[[(July, 16)]]

Factor

<lang factor>USING: assocs calendar.english fry io kernel prettyprint sequences sets.extras ;

unique-by ( seq quot -- newseq )
   2dup map non-repeating '[ @ _ member? ] filter ; inline

ALIAS: day first ALIAS: month second

{

   { 15 5 } { 16 5 } { 19 5 } { 17 6 } { 18 6 }
   { 14 7 } { 16 7 } { 14 8 } { 15 8 } { 17 8 }

}

! the month cannot have a unique day dup [ day ] map non-repeating over extract-keys values '[ month _ member? ] reject

! of the remaining dates, day must be unique [ day ] unique-by

! of the remaining dates, month must be unique [ month ] unique-by

! print a date that looks like { { 16 7 } } first first2 month-name write bl .</lang>

Output:
July 16

Go

<lang go>package main

import (

   "fmt"
   "time"

)

type birthday struct{ month, day int }

func (b birthday) String() string {

   return fmt.Sprintf("%s %d", time.Month(b.month), b.day)

}

func (b birthday) monthUniqueIn(bds []birthday) bool {

   count := 0
   for _, bd := range bds {
       if bd.month == b.month {
           count++
       }
   }
   if count == 1 {
       return true
   }
   return false

}

func (b birthday) dayUniqueIn(bds []birthday) bool {

   count := 0
   for _, bd := range bds {
       if bd.day == b.day {
           count++
       }
   }
   if count == 1 {
       return true
   }
   return false

}

func (b birthday) monthWithUniqueDayIn(bds []birthday) bool {

   for _, bd := range bds {
       if bd.month == b.month && bd.dayUniqueIn(bds) {
           return true
       }
   }
   return false

}

func main() {

   choices := []birthday{
       {5, 15}, {5, 16}, {5, 19}, {6, 17}, {6, 18},
       {7, 14}, {7, 16}, {8, 14}, {8, 15}, {8, 17},
   }
   // Albert knows the month but doesn't know the day.
   // So the month can't be unique within the choices.
   var filtered []birthday
   for _, bd := range choices {
       if !bd.monthUniqueIn(choices) {
           filtered = append(filtered, bd)
       }
   }
   // Albert also knows that Bernard doesn't know the answer.
   // So the month can't have a unique day.
   var filtered2 []birthday
   for _, bd := range filtered {
       if !bd.monthWithUniqueDayIn(filtered) {
           filtered2 = append(filtered2, bd)
       }
   }
   // Bernard now knows the answer.
   // So the day must be unique within the remaining choices.
   var filtered3 []birthday
   for _, bd := range filtered2 {
       if bd.dayUniqueIn(filtered2) {
           filtered3 = append(filtered3, bd)
       }
   }
   // Albert now knows the answer too.
   // So the month must be unique within the remaining choices.
   var filtered4 []birthday
   for _, bd := range filtered3 {
       if bd.monthUniqueIn(filtered3) {
           filtered4 = append(filtered4, bd)
       }
   }
   if len(filtered4) == 1 {
       fmt.Println("Cheryl's birthday is", filtered4[0])
   } else {
       fmt.Println("Something went wrong!")
   }

}</lang>

Output:
Cheryl's birthday is July 16

Groovy

Translation of: Java

<lang groovy>import java.time.Month

class Main {

   private static class Birthday {
       private Month month
       private int day
       Birthday(Month month, int day) {
           this.month = month
           this.day = day
       }
       Month getMonth() {
           return month
       }
       int getDay() {
           return day
       }
       @Override
       String toString() {
           return month.toString() + " " + day
       }
   }
   static void main(String[] args) {
       List<Birthday> choices = [
               new Birthday(Month.MAY, 15),
               new Birthday(Month.MAY, 16),
               new Birthday(Month.MAY, 19),
               new Birthday(Month.JUNE, 17),
               new Birthday(Month.JUNE, 18),
               new Birthday(Month.JULY, 14),
               new Birthday(Month.JULY, 16),
               new Birthday(Month.AUGUST, 14),
               new Birthday(Month.AUGUST, 15),
               new Birthday(Month.AUGUST, 17)
       ]
       println("There are ${choices.size()} candidates remaining.")
       // The month cannot have a unique day because Albert knows the month, and knows that Bernard does not know the answer
       Set<Birthday> uniqueMonths = choices.groupBy { it.getDay() }
               .values()
               .findAll() { it.size() == 1 }
               .flatten()
               .collect { ((Birthday) it).getMonth() }
               .toSet() as Set<Birthday>
       def f1List = choices.findAll { !uniqueMonths.contains(it.getMonth()) }
       println("There are ${f1List.size()} candidates remaining.")
       // Bernard now knows the answer, so the day must be unique within the remaining choices
       List<Birthday> f2List = f1List.groupBy { it.getDay() }
               .values()
               .findAll { it.size() == 1 }
               .flatten()
               .toList() as List<Birthday>
       println("There are ${f2List.size()} candidates remaining.")
       // Albert knows the answer too, so the month must be unique within the remaining choices
       List<Birthday> f3List = f2List.groupBy { it.getMonth() }
               .values()
               .findAll { it.size() == 1 }
               .flatten()
               .toList() as List<Birthday>
       println("There are ${f3List.size()} candidates remaining.")
       if (f3List.size() == 1) {
           println("Cheryl's birthday is ${f3List.head()}")
       } else {
           System.out.println("No unique choice found")
       }
   }

}</lang>

Output:
There are 10 candidates remaining.
There are 5 candidates remaining.
There are 3 candidates remaining.
There are 1 candidates remaining.
Cheryl's birthday is JULY 16

Haskell

<lang haskell>{-# LANGUAGE OverloadedStrings #-}

import Data.List as L (filter, groupBy, head, length, sortOn) import Data.Map.Strict as M (Map, fromList, keys, lookup) import Data.Text as T (Text, splitOn, words) import Data.Maybe (fromJust) import Data.Ord (comparing) import Data.Function (on) import Data.Tuple (swap) import Data.Bool (bool)

data DatePart

 = Month
 | Day

type M = Text

type D = Text

main :: IO () main =

 print $
 -- The month with only one remaining day,
 --
 -- (A's month contains only one remaining day)
 -- (3 :: A "Then I also know")
 uniquePairing Month $
 -- among the days with unique months,
 --
 -- (B's day is paired with only one remaining month)
 -- (2 :: B "I know now")
 uniquePairing Day $
 -- excluding months with unique days,
 --
 -- (A's month is not among those with unique days)
 -- (1 :: A "I know that Bernard does not know")
 monthsWithUniqueDays False $
 -- from the given month-day pairs:
 --
 -- (0 :: Cheryl's list)
 (\(x:y:_) -> (x, y)) . T.words <$>
 splitOn
   ", "
   "May 15, May 16, May 19, June 17, June 18, \
             \July 14, July 16, Aug 14, Aug 15, Aug 17"

QUERY FUNCTIONS----------------------

monthsWithUniqueDays :: Bool -> [(M, D)] -> [(M, D)] monthsWithUniqueDays bln xs =

 let months = fst <$> uniquePairing Day xs
 in L.filter (bool not id bln . (`elem` months) . fst) xs

uniquePairing :: DatePart -> [(M, D)] -> [(M, D)] uniquePairing dp xs =

 let f =
       case dp of
         Month -> fst
         Day -> snd
 in (\md ->
        let dct = f md
            uniques =
              L.filter
                ((1 ==) . L.length . fromJust . flip M.lookup dct)
                (keys dct)
        in L.filter ((`elem` uniques) . f) xs)
      ((((,) . mapFromPairs) <*> mapFromPairs . fmap swap) xs)

mapFromPairs :: [(M, D)] -> Map Text [Text] mapFromPairs xs =

 M.fromList $
 ((,) . fst . L.head) <*> fmap snd <$>
 L.groupBy (on (==) fst) (L.sortOn fst xs)</lang>
Output:
[("July","16")]

J

Solution: <lang j>Dates=: <;._2 noun define 15 May 16 May 19 May 17 June 18 June 14 July 16 July 14 August 15 August 17 August )

getDayMonth=: |:@:(' '&splitstring&>) NB. retrieve lists of days and months from dates keep=: adverb def '] #~ u' NB. apply mask to filter dates

monthsWithUniqueDay=: {./. #~ (1=#)/. NB. list months that have a unique day isMonthWithoutUniqueDay=: (] -.@e. monthsWithUniqueDay)/@getDayMonth NB. mask of dates with a month that doesn't have a unique day

uniqueDayInMonth=: ~.@[ #~ (1=#)/. NB. list of days that are unique to 1 month isUniqueDayInMonth=: ([ e. uniqueDayInMonth)/@getDayMonth NB. mask of dates with a day that is unique to 1 month

uniqueMonth=: ~.@] #~ (1=#)/.~ NB. list of months with 1 unique day isUniqueMonth=: (] e. uniqueMonth)/@getDayMonth NB. mask of dates with a month that has 1 unique day</lang> Usage: <lang j> isUniqueMonth keep isUniqueDayInMonth keep isMonthWithoutUniqueDay keep Dates +-------+ |16 July| +-------+</lang>

Alternative Approach

The concepts here are the same, of course, it's just the presentation that's different.

<lang J>possible=: cut;._2 'May 15, May 16, May 19, June 17, June 18, July 14, July 16, August 14, August 15, August 17,'

Albert=: {."1 NB. Albert knows month Bernard=: {:"1 NB. Bernard knows day

NB. Bernard's understanding of Albert's first pass

 days=: {:"1 possible
 invaliddays=: (1=#/.~ days)#~.days
 months=: {."1 possible
 validmonths=: months -. (days e. invaliddays)#months
 possibleA=. (months e. validmonths)# possible

NB. Albert's understanding of Bernard's pass

 days=: {:"1 possibleA
 invaliddays=: (1<#/.~ days)#~.days
 possibleB=. (days e. days-.invaliddays)# possibleA

NB. our understanding of Albert's second pass

 months=: {."1 possibleB
 invalidmonths=: (1<#/.~months)#~.months
 echo ;:inv (months e. months -. invalidmonths)#possibleB</lang>

This gives us the July 16 result we were expecting

Java

Translation of: D

<lang java>import java.time.Month; import java.util.Collection; import java.util.List; import java.util.Set; import java.util.stream.Collectors;

public class Main {

   private static class Birthday {
       private Month month;
       private int day;
       public Birthday(Month month, int day) {
           this.month = month;
           this.day = day;
       }
       public Month getMonth() {
           return month;
       }
       public int getDay() {
           return day;
       }
       @Override
       public String toString() {
           return month + " " + day;
       }
   }
   public static void main(String[] args) {
       List<Birthday> choices = List.of(
           new Birthday(Month.MAY, 15),
           new Birthday(Month.MAY, 16),
           new Birthday(Month.MAY, 19),
           new Birthday(Month.JUNE, 17),
           new Birthday(Month.JUNE, 18),
           new Birthday(Month.JULY, 14),
           new Birthday(Month.JULY, 16),
           new Birthday(Month.AUGUST, 14),
           new Birthday(Month.AUGUST, 15),
           new Birthday(Month.AUGUST, 17)
       );
       System.out.printf("There are %d candidates remaining.\n", choices.size());
       // The month cannot have a unique day because Albert knows the month, and knows that Bernard does not know the answer
       Set<Month> uniqueMonths = choices.stream()
           .collect(Collectors.groupingBy(Birthday::getDay))
           .values()
           .stream()
           .filter(g -> g.size() == 1)
           .flatMap(Collection::stream)
           .map(Birthday::getMonth)
           .collect(Collectors.toSet());
       List<Birthday> f1List = choices.stream()
           .filter(birthday -> !uniqueMonths.contains(birthday.month))
           .collect(Collectors.toList());
       System.out.printf("There are %d candidates remaining.\n", f1List.size());
       // Bernard now knows the answer, so the day must be unique within the remaining choices
       List<Birthday> f2List = f1List.stream()
           .collect(Collectors.groupingBy(Birthday::getDay))
           .values()
           .stream()
           .filter(g -> g.size() == 1)
           .flatMap(Collection::stream)
           .collect(Collectors.toList());
       System.out.printf("There are %d candidates remaining.\n", f2List.size());
       // Albert knows the answer too, so the month must be unique within the remaining choices
       List<Birthday> f3List = f2List.stream()
           .collect(Collectors.groupingBy(Birthday::getMonth))
           .values()
           .stream()
           .filter(g -> g.size() == 1)
           .flatMap(Collection::stream)
           .collect(Collectors.toList());
       System.out.printf("There are %d candidates remaining.\n", f3List.size());
       if (f3List.size() == 1) {
           System.out.printf("Cheryl's birthday is %s\n", f3List.get(0));
       } else {
           System.out.println("No unique choice found");
       }
   }

}</lang>

Output:
There are 10 candidates remaining.
There are 5 candidates remaining.
There are 3 candidates remaining.
There are 1 candidates remaining.
Cheryl's birthday is JULY 16

JavaScript

<lang javascript>(() => {

   'use strict';
   // main :: IO ()
   const main = () => {
       const
           month = fst,
           day = snd;
       showLog(
           map(x => Array.from(x), (
               // The month with only one remaining day,
               // (A's month contains only one remaining day)
               // (3 :: A "Then I also know")
               uniquePairing(month)(
                   // among the days with unique months,
                   // (B's day is paired with only one remaining month)
                   // (2 :: B "I know now")
                   uniquePairing(day)(
                       // excluding months with unique days,
                       // (A's month is not among those with unique days)
                       // (1 :: A "I know that Bernard does not know")
                       monthsWithUniqueDays(false)(
                           // from the given month-day pairs:
                           // (0 :: Cheryl's list)
                           map(x => tupleFromList(words(strip(x))),
                               splitOn(/,\s+/,
                                   `May 15, May 16, May 19,
                                       June 17, June 18, July 14, July 16,
                                       Aug 14, Aug 15, Aug 17`
                               )
                           )
                       )
                   )
               )
           ))
       );
   };
   // monthsWithUniqueDays :: Bool -> [(Month, Day)] -> [(Month, Day)]
   const monthsWithUniqueDays = blnInclude => xs => {
       const months = map(fst, uniquePairing(snd)(xs));
       return filter(
           md => (blnInclude ? id : not)(
               elem(fst(md), months)
           ),
           xs
       );
   };
   // uniquePairing :: ((a, a) -> a) ->
   //      -> [(Month, Day)] -> [(Month, Day)]
   const uniquePairing = f => xs =>
       bindPairs(xs,
           md => {
               const
                   dct = f(md),
                   matches = filter(
                       k => 1 === length(dct[k]),
                       Object.keys(dct)
                   );
               return filter(tpl => elem(f(tpl), matches), xs);
           }
       );
   // bindPairs :: [(Month, Day)] -> (Dict, Dict) -> [(Month, Day)]
   const bindPairs = (xs, f) => f(
       Tuple(
           dictFromPairs(fst)(snd)(xs),
           dictFromPairs(snd)(fst)(xs)
       )
   );
   // dictFromPairs :: ((a, a) -> a) -> ((a, a) -> a) -> [(a, a)] -> Dict
   const dictFromPairs = f => g => xs =>
       foldl((a, tpl) => Object.assign(
           a, {
               [f(tpl)]: (a[f(tpl)] || []).concat(g(tpl).toString())
           }
       ), {}, xs);


   // GENERIC ABSTRACTIONS -------------------------------
   // Tuple (,) :: a -> b -> (a, b)
   const Tuple = (a, b) => ({
       type: 'Tuple',
       '0': a,
       '1': b,
       length: 2
   });
   // elem :: Eq a => a -> [a] -> Bool
   const elem = (x, xs) => xs.includes(x);
   // filter :: (a -> Bool) -> [a] -> [a]
   const filter = (f, xs) => xs.filter(f);
   // foldl :: (a -> b -> a) -> a -> [b] -> a
   const foldl = (f, a, xs) => xs.reduce(f, a);
   // fst :: (a, b) -> a
   const fst = tpl => tpl[0];
   // id :: a -> a
   const id = x => x;
   // intersect :: (Eq a) => [a] -> [a] -> [a]
   const intersect = (xs, ys) =>
       xs.filter(x => -1 !== ys.indexOf(x));
   // Returns Infinity over objects without finite length
   // this enables zip and zipWith to choose the shorter
   // argument when one is non-finite, like cycle, repeat etc
   // length :: [a] -> Int
   const length = xs =>
       (Array.isArray(xs) || 'string' === typeof xs) ? (
           xs.length
       ) : Infinity;
   // map :: (a -> b) -> [a] -> [b]
   const map = (f, xs) => xs.map(f);
   // not :: Bool -> Bool
   const not = b => !b;
   // showLog :: a -> IO ()
   const showLog = (...args) =>
       console.log(
           args
           .map(JSON.stringify)
           .join(' -> ')
       );
   // snd :: (a, b) -> b
   const snd = tpl => tpl[1];
   // splitOn :: String -> String -> [String]
   const splitOn = (pat, src) =>
       src.split(pat);
   // strip :: String -> String
   const strip = s => s.trim();
   // tupleFromList :: [a] -> (a, a ...)
   const tupleFromList = xs =>
       TupleN.apply(null, xs);
   // TupleN :: a -> b ...  -> (a, b ... )
   function TupleN() {
       const
           args = Array.from(arguments),
           lng = args.length;
       return lng > 1 ? Object.assign(
           args.reduce((a, x, i) => Object.assign(a, {
               [i]: x
           }), {
               type: 'Tuple' + (2 < lng ? lng.toString() : ),
               length: lng
           })
       ) : args[0];
   };
   // words :: String -> [String]
   const words = s => s.split(/\s+/);
   // MAIN ---
   return main();

})();</lang>

Output:
[["July","16"]]

Julia

<lang julia>const dates = [[15, "May"], [16, "May"], [19, "May"], [17, "June"], [18, "June"],

   [14, "July"], [16, "July"], [14, "August"], [15, "August"], [17, "August"]]

uniqueday(parr) = filter(x -> count(y -> y[1] == x[1], parr) == 1, parr)

  1. At the start, they come to know that they have no unique day of month to identify.

const f1 = filter(m -> !(m[2] in [d[2] for d in uniqueday(dates)]), dates)

  1. After cutting months with unique dates, get months remaining that now have a unique date.

const f2 = uniqueday(f1)

  1. filter for those of the finally remaining months that have only one date left.

const bday = filter(x -> count(m -> m[2] == x[2], f2) == 1, f2)[]

println("Cheryl's birthday is $(bday[2]) $(bday[1]).")

</lang>

Output:
Cheryl's birthday is July 16.

Kotlin

Translation of: Go

<lang scala>// Version 1.2.71

val months = listOf(

   "January", "February", "March", "April", "May", "June",
   "July", "August", "September", "October", "November", "December"

)

class Birthday(val month: Int, val day: Int) {

   public override fun toString() = "${months[month - 1]} $day"
   public fun monthUniqueIn(bds: List<Birthday>): Boolean {
       return bds.count { this.month == it.month } == 1 
   }
   public fun dayUniqueIn(bds: List<Birthday>): Boolean {
       return bds.count { this.day == it.day } == 1
   }
   public fun monthWithUniqueDayIn(bds: List<Birthday>): Boolean {
       return bds.any { (this.month == it.month) && it.dayUniqueIn(bds) }
   }

}

fun main(args: Array<String>) {

   val choices = listOf(
       Birthday(5, 15), Birthday(5, 16), Birthday(5, 19), Birthday(6, 17), 
       Birthday(6, 18), Birthday(7, 14), Birthday(7, 16), Birthday(8, 14),
       Birthday(8, 15), Birthday(8, 17)
   )
   // Albert knows the month but doesn't know the day.
   // So the month can't be unique within the choices.
   var filtered = choices.filterNot { it.monthUniqueIn(choices) }
   // Albert also knows that Bernard doesn't know the answer.
   // So the month can't have a unique day.
   filtered = filtered.filterNot { it.monthWithUniqueDayIn(filtered) }
   // Bernard now knows the answer.
   // So the day must be unique within the remaining choices.
   filtered = filtered.filter { it.dayUniqueIn(filtered) }
   // Albert now knows the answer too.
   // So the month must be unique within the remaining choices.
   filtered = filtered.filter { it.monthUniqueIn(filtered) }
   if (filtered.size == 1)
       println("Cheryl's birthday is ${filtered[0]}")
   else
       println("Something went wrong!")

}</lang>

Output:
Cheryl's birthday is July 16

Lua

<lang lua>-- Cheryl's Birthday in Lua 6/15/2020 db

local function Date(mon,day)

 return { mon=mon, day=day, valid=true }

end

local choices = {

 Date("May", 15), Date("May", 16), Date("May", 19),
 Date("June", 17), Date("June", 18),
 Date("July", 14), Date("July", 16),
 Date("August", 14), Date("August", 15), Date("August", 17)

}

local function apply(t, f)

 for k, v in ipairs(t) do
   f(k, v)
 end

end

local function filter(t, f)

 local result = {}
 for k, v in ipairs(t) do
   if f(k, v) then
     result[#result+1] = v
   end
 end
 return result

end

local function map(t, f)

 local result = {}
 for k, v in ipairs(t) do
   result[#result+1] = f(k, v)
 end
 return result

end

local function count(t) return #t end local function isvalid(k, v) return v.valid end local function invalidate(k, v) v.valid = false end local function remaining() return filter(choices, isvalid) end

local function listValidChoices()

 print("   " .. table.concat(map(remaining(), function(k, v) return v.mon .. " " .. v.day end), ", "))
 print()

end

print("Cheryl offers these ten choices:") listValidChoices()

print("1) Albert knows that Bernard also cannot yet know, so cannot be a month with a unique day, leaving:") apply(remaining(), function(k, v)

 if count(filter(choices, function(k2, v2) return v.day==v2.day end)) == 1 then
   apply(filter(remaining(), function(k2, v2) return v.mon==v2.mon end), invalidate)
 end

end) listValidChoices()

print("2) After Albert's revelation, Bernard now knows, so day must be unique, leaving:") apply(remaining(), function(k, v)

 local subset = filter(remaining(), function(k2, v2) return v.day==v2.day end)
 if count(subset) > 1 then apply(subset, invalidate) end

end) listValidChoices()

print("3) After Bernard's revelation, Albert now knows, so month must be unique, leaving only:") apply(remaining(), function(k, v)

 local subset = filter(remaining(), function(k2, v2) return v.mon==v2.mon end)
 if count(subset) > 1 then apply(subset, invalidate) end

end) listValidChoices()</lang>

Output:
Cheryl offers these ten choices:
   May 15, May 16, May 19, June 17, June 18, July 14, July 16, August 14, August 15, August 17

1) Albert knows that Bernard also cannot yet know, so cannot be a month with a unique day, leaving:
   July 14, July 16, August 14, August 15, August 17

2) After Albert's revelation, Bernard now knows, so day must be unique, leaving:
   July 16, August 15, August 17

3) After Bernard's revelation, Albert now knows, so month must be unique, leaving only:
   July 16

Mathematica / Wolfram Language

<lang Mathematica>opts = Tuples[{{"May"}, {15, 16, 19}}]~Join~Tuples[{{"June"}, {17, 18}}]~Join~Tuples[{{"July"}, {14, 16}}]~Join~Tuples[{{"August"}, {14, 15, 17}}]; monthsdelete = Select[GatherBy[opts, Last], Length /* EqualTo[1]]All, 1, 1; opts = DeleteCases[opts, {Alternatives @@ monthsdelete, _}] removedates = Catenate@Select[GatherBy[opts, Last], Length /* GreaterThan[1]]; opts = DeleteCases[opts, Alternatives @@ removedates] Select[GatherBy[opts, First], Length /* EqualTo[1]]</lang>

Output:
{{"July", 14}, {"July", 16}, {"August", 14}, {"August", 15}, {"August", 17}}
{{"July", 16}, {"August", 15}, {"August", 17}}
{{{"July", 16}}}

Nim

<lang Nim>import tables import sets import strformat

type Date = tuple[month: string, day: int]

const Dates = [Date ("May", 15), ("May", 16), ("May", 19), ("June", 17), ("June", 18),

              ("July", 14), ("July", 16), ("August", 14), ("August", 15), ("August", 17)]

const

 MonthTable: Table[int, HashSet[string]] =
               static:
                 var t: Table[int, HashSet[string]]
                 for date in Dates:
                   t.mgetOrPut(date.day, initHashSet[string]()).incl(date.month)
                 t
 DayTable: Table[string, HashSet[int]] =
             static:
               var t: Table[string, HashSet[int]]
               for date in Dates:
                 t.mgetOrPut(date.month, initHashSet[int]()).incl(date.day)
               t

var possibleMonths: HashSet[string] # Set of possible months. var possibleDays: HashSet[int] # Set of possible days.


  1. Albert: I don't know when Cheryl's birthday is, ...
  2. => eliminate months with a single possible day.

for month, days in DayTable.pairs:

 if days.len > 1:
   possibleMonths.incl(month)
  1. ... but I know that Bernard does not know too.
  2. => eliminate months with one day present only in this month.

for month, days in DayTable.pairs:

 for day in days:
   if MonthTable[day].len == 1:
     possibleMonths.excl(month)

echo fmt"After first Albert's sentence, possible months are {possibleMonths}."

  1. Bernard: At first I don't know when Cheryl's birthday is, ...
  2. => eliminate days with a single possible month.

for day, months in MonthTable.pairs:

 if months.len > 1:
   possibleDays.incl(day)
  1. ... but I know now.
  2. => eliminate days which are compatible with several months in "possibleMonths".

var impossibleDays: HashSet[int] # Days which are eliminated by this sentence. for day in possibleDays:

 if (MonthTable[day] * possibleMonths).len > 1:
   impossibleDays.incl(day)

possibleDays.excl(impossibleDays) echo fmt"After Bernard's sentence, possible days are {possibleDays}."

  1. Albert: Then I also know when Cheryl's birthday is.
  2. => eliminate months which are compatible with several days in "possibleDays".

var impossibleMonths: HashSet[string] # Months which are eliminated by this sentence. for month in possibleMonths:

 if (DayTable[month] * possibleDays).len > 1:
   impossibleMonths.incl(month)

possibleMonths.excl(impossibleMonths)

doAssert possibleMonths.len == 1 let month = possibleMonths.pop() echo fmt"After second Albert's sentence, remaining month is {month}..."

possibleDays = possibleDays * DayTable[month] doAssert possibleDays.len == 1 let day = possibleDays.pop() echo fmt"and thus remaining day is {day}."

echo "" echo fmt"So birthday date is {month} {day}."</lang>

Output:
After first Albert's sentence, possible months are {"August", "July"}.
After Bernard's sentence, possible days are {16, 17, 15}.
After second Albert's sentence, remaining month is July...
and thus remaining day is 16.

So birthday date is July 16

Perl

<lang perl>sub filter {

   my($test,@dates) = @_;
   my(%M,%D,@filtered);
   # analysis of potential birthdays, keyed by month and by day
   for my $date (@dates) {
       my($mon,$day) = split '-', $date;
       $M{$mon}{cnt}++;
       $D{$day}{cnt}++;
       push @{$M{$mon}{day}},  $day;
       push @{$D{$day}{mon}},  $mon;
       push @{$M{$mon}{bday}}, "$mon-$day";
       push @{$D{$day}{bday}}, "$mon-$day";
   }
   # eliminates May/Jun dates based on 18th and 19th being singletons
   if ($test eq 'singleton') {
       my %skip;
       for my $day (grep { $D{$_}{cnt} == 1 } keys %D) { $skip{ @{$D{$day}{mon}}[0] }++    }
       for my $mon (grep { ! $skip{$_}      } keys %M) { push @filtered, @{$M{$mon}{bday}} }
   # eliminates Jul/Aug 14th because day count > 1 across months
   } elsif ($test eq 'duplicate') {
       for my $day (grep { $D{$_}{cnt} == 1 } keys %D) { push @filtered, @{$D{$day}{bday}} }
   # eliminates Aug 15th/17th because day count > 1, within month
   } elsif ($test eq 'multiple') {
       for my $day (grep { $M{$_}{cnt} == 1 } keys %M) { push @filtered, @{$M{$day}{bday}} }
   }
   return @filtered;

}

  1. doesn't matter what order singleton/duplicate tests are run, but 'multiple' must be last;

my @dates = qw<5-15 5-16 5-19 6-17 6-18 7-14 7-16 8-14 8-15 8-17>; @dates = filter($_, @dates) for qw<singleton duplicate multiple>;

my @months = qw<_ January February March April May June July August September October November December>;

my ($m, $d) = split '-', $dates[0]; print "Cheryl's birthday is $months[$m] $d.\n";</lang>

Output:
Cheryl's birthday is July 16.

Phix

-- demo\rosetta\Cheryls_Birthday.exw
with javascript_semantics
sequence choices = {{5, 15}, {5, 16}, {5, 19}, {6, 17}, {6, 18},
                    {7, 14}, {7, 16}, {8, 14}, {8, 15}, {8, 17}}
 
sequence mwud = repeat(false,12)    -- months with unique days
 
for step=1 to 4 do
    sequence {months,days} = columnize(choices)
    bool impossible = false
    for i=length(choices) to 1 by -1 do
        integer {m,d} = choices[i]
        switch step do
            case 1:   mwud[m] += (sum(sq_eq(days,d))=1)
            case 2: impossible = mwud[m]
            case 3: impossible = (sum(sq_eq(days,d))!=1)
            case 4: impossible = (sum(sq_eq(months,m))!=1)
        end switch
        if impossible then
            choices[i..i] = {}
        end if
    end for
end for
?choices

Iterating backwards down the choices array simplifies element removal, or more accurately removes the need for "not increment i".
Step 1&2 is months with unique days, step 3 is days with unique months, step 4 is unique months.

Output:
{{7,16}}

functional/filter

(this can also be found in demo\rosetta\Cheryls_Birthday.exw)

with javascript_semantics
enum MONTH, DAY

function unique_month(sequence si, months)
    return sum(sq_eq(months,si[MONTH]))=1
end function
 
function unique_day(sequence si, days)
    return sum(sq_eq(days,si[DAY]))=1
end function

function month_without_unique_day(sequence si, months_with_unique_day)
    return not find(si[MONTH],months_with_unique_day)
end function

sequence choices = {{5, 15}, {5, 16}, {5, 19}, {6, 17}, {6, 18},
                    {7, 14}, {7, 16}, {8, 14}, {8, 15}, {8, 17}}
 
-- Albert knows the month but does not know the day.
-- So the month cannot be unique within the choices.
-- However this step would change nothing, hence omit it.
--  (obvs. non_unique_month() would be as above, but !=1)
--choices = filter(choices,non_unique_month,vslice(choices,MONTH))

-- Albert also knows that Bernard doesn't know the answer.
-- So the month cannot have a unique day.
sequence unique_days = filter(choices,unique_day,vslice(choices,DAY))
sequence months_with_unique_day = unique(vslice(unique_days,MONTH))

choices = filter(choices,month_without_unique_day,months_with_unique_day)

-- Bernard now knows the answer.
-- So the day must be unique within the remaining choices.
choices = filter(choices,unique_day,vslice(choices,DAY))

-- Albert now knows the answer too.
-- So the month must be unique within the remaining choices.
choices = filter(choices,unique_month,vslice(choices,MONTH))

if length(choices)!=1 then crash("Something went wrong!") end if

include builtins\timedate.e
timedate td = repeat(0,6)
{td[DT_MONTH],td[DT_DAY]} = choices[1]
printf(1,"Cheryl's birthday is %s\n", {format_timedate(td,"Mmmm ddth")})
Output:
Cheryl's birthday is July 16th

Python

Functional

Works with: Python version 3

<lang python>Cheryl's Birthday

from itertools import groupby from re import split


  1. main :: IO ()

def main():

   Derivation of the date.
   month, day = 0, 1
   print(
       # (3 :: A "Then I also know")
       # (A's month contains only one remaining day)
       uniquePairing(month)(
           # (2 :: B "I know now")
           # (B's day is paired with only one remaining month)
           uniquePairing(day)(
               # (1 :: A "I know that Bernard does not know")
               # (A's month is not among those with unique days)
               monthsWithUniqueDays(False)([
                   # 0 :: Cheryl's list:
                   tuple(x.split()) for x in
                   split(
                       ', ',
                       'May 15, May 16, May 19, ' +
                       'June 17, June 18, ' +
                       'July 14, July 16, ' +
                       'Aug 14, Aug 15, Aug 17'
                   )
               ])
           )
       )
   )


  1. ------------------- QUERY FUNCTIONS --------------------
  1. monthsWithUniqueDays :: Bool -> [(Month, Day)] -> [(Month, Day)]

def monthsWithUniqueDays(blnInclude):

   The subset of months with (or without) unique days.
   
   def go(xs):
       month, day = 0, 1
       months = [fst(x) for x in uniquePairing(day)(xs)]
       return [
           md for md in xs
           if blnInclude or not (md[month] in months)
       ]
   return go


  1. uniquePairing :: DatePart -> [(Month, Day)] -> [(Month, Day)]

def uniquePairing(i):

   Subset of months (or days) with a unique intersection.
   
   def go(xs):
       def inner(md):
           dct = md[i]
           uniques = [
               k for k in dct.keys()
               if 1 == len(dct[k])
           ]
           return [tpl for tpl in xs if tpl[i] in uniques]
       return inner
   return ap(bindPairs)(go)


  1. bindPairs :: [(Month, Day)] ->
  2. ((Dict String [String], Dict String [String])
  3. -> [(Month, Day)]) -> [(Month, Day)]

def bindPairs(xs):

   List monad injection operator for lists
      of (Month, Day) pairs.
   
   return lambda f: f(
       (
           dictFromPairs(xs),
           dictFromPairs(
               [(b, a) for (a, b) in xs]
           )
       )
   )


  1. dictFromPairs :: [(Month, Day)] -> Dict Text [Text]

def dictFromPairs(xs):

   A dictionary derived from a list of
      month day pairs.
   
   return {
       k: [snd(x) for x in m] for k, m in groupby(
           sorted(xs, key=fst), key=fst
       )
   }


  1. ----------------------- GENERIC ------------------------
  1. ap :: (a -> b -> c) -> (a -> b) -> a -> c

def ap(f):

   Applicative instance for functions.
   
   def go(g):
       def fxgx(x):
           return f(x)(
               g(x)
           )
       return fxgx
   return go


  1. fst :: (a, b) -> a

def fst(tpl):

   First component of a pair.
   
   return tpl[0]


  1. snd :: (a, b) -> b

def snd(tpl):

   Second component of a pair.
   
   return tpl[1]


if __name__ == '__main__':

   main()</lang>
Output:
[('July', '16')]

Racket

Translation of: Kotlin

<lang racket>#lang racket

(define ((is x #:key [key identity]) y) (equal? (key x) (key y)))

(define albert first) (define bernard second)

(define (((unique who) chs) date) (= 1 (count (is date #:key who) chs)))

(define (((unique-fix who-fix who) chs) date)

 (ormap (conjoin (is date #:key who-fix) ((unique who) chs)) chs))

(define-syntax-rule (solve <chs> [<act> <arg>] ...)

 (let* ([chs <chs>] [chs (<act> (<arg> chs) chs)] ...) chs))

(solve '((May 15) (May 16) (May 19) (June 17) (June 18)

        (July 14) (July 16) (August 14) (August 15) (August 17))
      
      ;; Albert knows the month but doesn't know the day.
      ;; So the month can't be unique within the choices.
      [filter-not (unique albert)]
      ;; Albert also knows that Bernard doesn't know the answer.
      ;; So the month can't have a unique day.
      [filter-not (unique-fix albert bernard)]
      ;; Bernard now knows the answer.
      ;; So the day must be unique within the remaining choices.
      [filter (unique bernard)]
      ;; Albert now knows the answer too.
      ;; So the month must be unique within the remaining choices
      [filter (unique albert)])</lang>
Output:
'((July 16))

Raku

(formerly Perl 6)

<lang perl6>my @dates =

   { :15day, :5month },
   { :16day, :5month },
   { :19day, :5month },
   { :17day, :6month },
   { :18day, :6month },
   { :14day, :7month },
   { :16day, :7month },
   { :14day, :8month },
   { :15day, :8month },
   { :17day, :8month }
  1. Month can't have a unique day

my @filtered = @dates.grep(*.<month> != one(@dates.grep(*.<day> == one(@dates».<day>))».<month>));

  1. Day must be unique and unambiguous in remaining months

my $birthday = @filtered.grep(*.<day> == one(@filtered».<day>)).classify({.<month>})\

   .first(*.value.elems == 1).value[0];
  1. convenience array

my @months = < January February March April May June July August September October November December>;

say "Cheryl's birthday is { @months[$birthday<month>] } {$birthday<day>}.";</lang>

Output:
Cheryl's birthday is July 16.

REXX

<lang rexx>/*REXX pgm finds Cheryl's birth date based on a person knowing the birth month, another */ /*──────────────────────── person knowing the birth day, given a list of possible dates.*/ $= 'May-15 May-16 May-19 June-17 June-18 July-14 July-16 August-14 August-15 August-17' call delDays unique('day') $= unique('day') $= unique('month') if words($)==1 then say "Cheryl's birthday is" translate($, , '-')

               else say "error in the program's logic."

exit 0 /*──────────────────────────────────────────────────────────────────────────────────────*/ unique: arg u 2, dups; #= words($); $$= $

           do    j=#  to 2  by -1
           if u=='D'     then parse value word($, j)  with  '-'   x
                         else parse value word($, j)  with   x   '-'
              do k=1  for j-1
              if u=='D'  then parse value word($, k)  with  '-'   y
                         else parse value word($, k)  with   y   '-'
              if x==y    then dups= dups k j
              end   /*k*/
           end      /*j*/
                                 do    d=#  for #  by -1
                                    do p=1  for words(dups)  until ?==d;  ?= word(dups,p)
                                    if ?==d  then  $$= delword($$, ?, 1)
                                    end   /*d*/
                                 end      /*d*/
        if words($$)==0  then return $
                         else return $$

/*──────────────────────────────────────────────────────────────────────────────────────*/ delDays: parse arg days; #= words(days)

          do    j=#   for #   by -1;  parse value word(days, j) with x '-';  ##= words($)
             do k=##  for ##  by -1;  parse value word($,    k) with y '-'
             if x\==y  then iterate;  $= delword($, k, 1)
             end   /*k*/
          end      /*j*/
        return $</lang>
output   when using the internal default input:
Cheryl's birthday is July 16

Ruby

Translation of: C#

<lang ruby>dates = [

   ["May", 15],
   ["May", 16],
   ["May", 19],
   ["June", 17],
   ["June", 18],
   ["July", 14],
   ["July", 16],
   ["August", 14],
   ["August", 15],
   ["August", 17],

]

print dates.length, " remaining\n"

  1. the month cannot have a unique day

uniqueMonths = dates.group_by { |m,d| d }

                   .select { |k,v| v.size == 1 }
                   .map { |k,v| v.flatten }
                   .map { |m,d| m }

dates.delete_if { |m,d| uniqueMonths.include? m } print dates.length, " remaining\n"

  1. the day must be unique

dates = dates .group_by { |m,d| d }

               .select { |k,v| v.size == 1 }
               .map { |k,v| v.flatten }

print dates.length, " remaining\n"

  1. the month must now be unique

dates = dates .group_by { |m,d| m }

               .select { |k,v| v.size == 1 }
               .map { |k,v| v }
               .flatten

print dates</lang>

Output:
10 remaining
5 remaining
3 remaining
["July", 16]

Rust

<lang rust> // This version is based on the Go version on Rosettacode

  1. [derive(PartialEq, Debug, Copy, Clone)]

enum Month {

   May,
   June,
   July,
   August,

}

  1. [derive(PartialEq, Debug, Copy, Clone)]

struct Birthday {

   month: Month,
   day: u8,

}

impl Birthday {

   fn month_unique_in(&self, birthdays: &[Birthday]) -> bool {
       birthdays
           .iter()
           .filter(|birthday| birthday.month == self.month)
           .count()
           == 1
   }
   fn day_unique_in(&self, birthdays: &[Birthday]) -> bool {
       birthdays
           .iter()
           .filter(|birthday| birthday.day == self.day)
           .count()
           == 1
   }
   fn month_with_unique_day_in(&self, birthdays: &[Birthday]) -> bool {
       birthdays
           .iter()
           .any(|birthday| self.month == birthday.month && birthday.day_unique_in(birthdays))
   }

}

fn solution() -> Option<Birthday> {

   let mut choices: Vec<Birthday> = vec![
       Birthday {
           month: Month::May,
           day: 15,
       },
       Birthday {
           month: Month::May,
           day: 16,
       },
       Birthday {
           month: Month::May,
           day: 19,
       },
       Birthday {
           month: Month::June,
           day: 17,
       },
       Birthday {
           month: Month::June,
           day: 18,
       },
       Birthday {
           month: Month::July,
           day: 14,
       },
       Birthday {
           month: Month::July,
           day: 16,
       },
       Birthday {
           month: Month::August,
           day: 14,
       },
       Birthday {
           month: Month::August,
           day: 15,
       },
       Birthday {
           month: Month::August,
           day: 17,
       },
   ];
   // Albert knows the month but doesn't know the day.
   // So the month can't be unique within the choices.
   let choices_copy = choices.clone();
   choices.retain(|birthday| !(&birthday.month_unique_in(&choices_copy)));
   // Albert also knows that Bernard doesn't know the answer.
   // So the month can't have a unique day.
   let choices_copy = choices.clone();
   choices.retain(|birthday| !(birthday.month_with_unique_day_in(&choices_copy)));
   // Bernard now knows the answer.
   // So the day must be unique within the remaining choices.
   let choices_copy = choices.clone();
   choices.retain(|birthday| birthday.day_unique_in(&choices_copy));
   // Albert now knows the answer too.
   // So the month must be unique within the remaining choices.
   let choices_copy = choices.clone();
   choices.retain(|birthday| birthday.month_unique_in(&choices_copy));
   if choices.len() == 1 {
       Some(choices[0])
   } else {
       None
   }

}

fn main() {

   match solution() {
       Some(solution) => println!("Cheryl's birthday is {:?}", solution),
       None => panic!("Didn't work!"),
   }

}

</lang>

Output:
Cheryl's birthday is Birthday { month: July, day: 16 }

Scala

Translation of: D

<lang scala>import java.time.format.DateTimeFormatter import java.time.{LocalDate, Month}

object Cheryl {

 def main(args: Array[String]): Unit = {
   val choices = List(
     LocalDate.of(2019, Month.MAY, 15),
     LocalDate.of(2019, Month.MAY, 16),
     LocalDate.of(2019, Month.MAY, 19),
     LocalDate.of(2019, Month.JUNE, 17),
     LocalDate.of(2019, Month.JUNE, 18),
     LocalDate.of(2019, Month.JULY, 14),
     LocalDate.of(2019, Month.JULY, 16),
     LocalDate.of(2019, Month.AUGUST, 14),
     LocalDate.of(2019, Month.AUGUST, 15),
     LocalDate.of(2019, Month.AUGUST, 17)
   )
   // The month cannot have a unique day because Albert knows the month, and knows that Bernard does not know the answer
   val uniqueMonths = choices.groupBy(_.getDayOfMonth)
     .filter(a => a._2.length == 1)
     .flatMap(a => a._2)
     .map(a => a.getMonth)
   val filter1 = choices.filterNot(a => uniqueMonths.exists(b => a.getMonth == b))
   // Bernard now knows the answer, so the day must be unique within the remaining choices
   val uniqueDays = filter1.groupBy(_.getDayOfMonth)
     .filter(a => a._2.length == 1)
     .flatMap(a => a._2)
     .map(a => a.getDayOfMonth)
   val filter2 = filter1.filter(a => uniqueDays.exists(b => a.getDayOfMonth == b))
   // Albert knows the answer too, so the month must be unique within the remaining choices
   val birthDay = filter2.groupBy(_.getMonth)
     .filter(a => a._2.length == 1)
     .flatMap(a => a._2)
     .head
   // print the result
   printf(birthDay.format(DateTimeFormatter.ofPattern("MMMM dd")))
 }

}</lang>

Output:
July 16

Scala-ish approach

Output:

See it yourself by running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).

Works with: Scala version 2.13

<lang Scala>object Cheryl_sBirthday extends App {

 private val possiblerDates = Set(
   Date("May", 15), Date("May", 16), Date("May", 19),
   Date("June", 17), Date("June", 18),
   Date("July", 14), Date("July", 16),
   Date("August", 14), Date("August", 15), Date("August", 17)
 )
 private def clou3: Date = {
   // Find the dates with ONE unique once and only occurrence of the day of the month.
   def onceDates[K](toBeExcluded: Set[Date], selector: Date => K): Seq[Date] =
     toBeExcluded.groupBy(selector).filter { case (_, multiSet) => multiSet.size == 1 }.values.flatten.toSeq
   // 1) Albert tells us that Bernard doesn't know the answer,
   // so we know the answer must be in months that does NOT have a same day of month.
   val uniqueMonths = onceDates(possiblerDates, (date: Date) => date.dayOfMonth).map(_.month)
   // Remove the dates with those months. The dates remain which has NOT those months.
   val clou1 = possiblerDates.filterNot(p => uniqueMonths.contains(p.month))
   // 2) Since Bernard now knows the answer, that tells us that the day MUST be unique among the remaining birthdays.
   val uniqueDays = onceDates(clou1, (date: Date) => date.dayOfMonth).map(_.dayOfMonth)
   // 3) Since Albert now knows the answer, that tells us the answer has to be unique by month.
   // First, as the first parameter, intersect clou1 (Albert) with uniqueDays (Bernard)
   onceDates(clou1.filter(date => uniqueDays.contains(date.dayOfMonth)), (date: Date) => date.month).head
 }
 case class Date(month: String, dayOfMonth: Int) {
   override def toString: String = s"${"🎂 " * 3}$dayOfMonth $month${" 🎂" * 3}"
 }
 println(clou3)

}</lang>

Sidef

Translation of: Raku

<lang ruby>func f(day, month) {

   Date.parse("#{day} #{month}", "%d %B")

}

var dates = [

   f(15, "May"),
   f(16, "May"),
   f(19, "May"),
   f(17, "June"),
   f(18, "June"),
   f(14, "July"),
   f(16, "July"),
   f(14, "August"),
   f(15, "August"),
   f(17, "August")

]   var filtered = dates.grep {

   dates.grep {
       dates.map{ .day }.count(.day) == 1
   }.map{ .month }.count(.month) != 1

}   var birthday = filtered.grep {

   filtered.map{ .day }.count(.day) == 1

}.group_by{ .month }.values.first_by { .len == 1 }[0]   say "Cheryl's birthday is #{birthday.fullmonth} #{birthday.day}."</lang>

Output:
Cheryl's birthday is July 16.

Swift

Translation of: Kotlin

<lang swift>struct MonthDay: CustomStringConvertible {

 static let months = [
   "January", "February", "March", "April", "May", "June",
   "July", "August", "September", "October", "November", "December"
 ]
 var month: Int
 var day: Int
 var description: String { "\(MonthDay.months[month - 1]) \(day)" }
 private func isUniqueIn(months: [MonthDay], by prop: KeyPath<MonthDay, Int>) -> Bool {
   return months.lazy.filter({ $0[keyPath: prop] == self[keyPath: prop] }).count == 1
 }
 func monthIsUniqueIn(months: [MonthDay]) -> Bool {
   return isUniqueIn(months: months, by: \.month)
 }
 func dayIsUniqueIn(months: [MonthDay]) -> Bool {
   return isUniqueIn(months: months, by: \.day)
 }
 func monthWithUniqueDayIn(months: [MonthDay]) -> Bool {
   return months.firstIndex(where: { $0.month == month && $0.dayIsUniqueIn(months: months) }) != nil
 }

}

let choices = [

 MonthDay(month: 5, day: 15),
 MonthDay(month: 5, day: 16),
 MonthDay(month: 5, day: 19),
 MonthDay(month: 6, day: 17),
 MonthDay(month: 6, day: 18),
 MonthDay(month: 7, day: 14),
 MonthDay(month: 7, day: 16),
 MonthDay(month: 8, day: 14),
 MonthDay(month: 8, day: 15),
 MonthDay(month: 8, day: 17)

]

// Albert knows the month, but not the day, so he doesn't have a gimmie month let albertKnows = choices.filter({ !$0.monthIsUniqueIn(months: choices) })

// Albert also knows that Bernard doesn't know, so it can't be a gimmie day let bernardKnows = albertKnows.filter({ !$0.monthWithUniqueDayIn(months: albertKnows) })

// Bernard now knows the birthday, so it must be a unique day within the remaining choices let bernardKnowsMore = bernardKnows.filter({ $0.dayIsUniqueIn(months: bernardKnows) })

// Albert knows the birthday now, so it must be a unique month within the remaining choices guard let birthday = bernardKnowsMore.filter({ $0.monthIsUniqueIn(months: bernardKnowsMore) }).first else {

 fatalError()

}

print("Cheryl's birthday is \(birthday)")</lang>

Output:
Cheryl's birthday is July 16

VBA

<lang vb>Private Sub exclude_unique_days(w As Collection)

   Dim number_of_dates(31) As Integer
   Dim months_to_exclude As New Collection
   For Each v In w
       number_of_dates(v(1)) = number_of_dates(v(1)) + 1
   Next v
   For i = w.Count To 1 Step -1
       If number_of_dates(w(i)(1)) = 1 Then
           months_to_exclude.Add w(i)(0)
           w.Remove i
       End If
   Next i
   For Each m In months_to_exclude
       exclude_month w, m
   Next m

End Sub Private Sub exclude_month(x As Collection, v As Variant)

   For i = x.Count To 1 Step -1
       If x(i)(0) = v Then x.Remove i
   Next i

End Sub Private Sub exclude_non_unique_days(w As Collection)

   Dim number_of_dates(31) As Integer
   For Each v In w
       number_of_dates(v(1)) = number_of_dates(v(1)) + 1
   Next v
   For i = w.Count To 1 Step -1
       If number_of_dates(w(i)(1)) > 1 Then
           w.Remove i
       End If
   Next i

End Sub Private Sub exclude_non_unique_months(w As Collection)

   Dim months As New Collection
   For Each v In w
       On Error GoTo 1
       months.Add v(0), v(0)
   Next v

1:

   For i = w.Count To 1 Step -1
       If w(i)(0) = v(0) Then
           w.Remove i
       End If
   Next i

End Sub Public Sub cherylsbirthday()

   Dim v As New Collection
   s = "May 15, May 16, May 19, June 17, June 18, July 14, July 16, August 14, August 15, August 17"
   t = Split(s, ",")
   For Each u In t
       v.Add Split(Trim(u), " ")
   Next u
   '1) Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.
   exclude_unique_days v
   '2) Bernard: At first I don't know when Cheryl's birthday is, but I know now.
   exclude_non_unique_days v
   '3) Albert: Then I also know when Cheryl's birthday is.
   exclude_non_unique_months v
   Debug.Print v(1)(0); " "; v(1)(1)

End Sub</lang>

Output:
July 16

Visual Basic .NET

Translation of: C#

<lang vbnet>Module Module1

   Structure MonDay
       Dim month As String
       Dim day As Integer
       Sub New(m As String, d As Integer)
           month = m
           day = d
       End Sub
       Public Overrides Function ToString() As String
           Return String.Format("({0}, {1})", month, day)
       End Function
   End Structure
   Sub Main()
       Dim dates = New HashSet(Of MonDay) From {
           New MonDay("May", 15),
           New MonDay("May", 16),
           New MonDay("May", 19),
           New MonDay("June", 17),
           New MonDay("June", 18),
           New MonDay("July", 14),
           New MonDay("July", 16),
           New MonDay("August", 14),
           New MonDay("August", 15),
           New MonDay("August", 17)
       }
       Console.WriteLine("{0} remaining.", dates.Count)
       ' The month cannot have a unique day.
       Dim monthsWithUniqueDays = dates.GroupBy(Function(d) d.day).Where(Function(g) g.Count() = 1).Select(Function(g) g.First().month).ToHashSet()
       dates.RemoveWhere(Function(d) monthsWithUniqueDays.Contains(d.month))
       Console.WriteLine("{0} remaining.", dates.Count)
       ' The day must now be unique.
       dates.IntersectWith(dates.GroupBy(Function(d) d.day).Where(Function(g) g.Count() = 1).Select(Function(g) g.First()))
       Console.WriteLine("{0} remaining.", dates.Count)
       ' The month must now be unique.
       dates.IntersectWith(dates.GroupBy(Function(d) d.month).Where(Function(g) g.Count() = 1).Select(Function(g) g.First()))
       Console.WriteLine(dates.Single())
   End Sub

End Module</lang>

Output:
10 remaining.
5 remaining.
3 remaining.
(July, 16)

Wren

Translation of: Kotlin

<lang ecmascript>var Months = [

   "January", "February", "March", "April", "May", "June",
   "July", "August", "September", "October", "November", "December"

]

class Birthday {

   construct new(month, day) {
       _month = month
       _day = day
   }
   month { _month }
   day   { _day }
   toString { "%(Months[_month-1]) %(day)" }
   monthUniqueIn(bds) { bds.count { |bd| _month == bd.month } == 1 }
   dayUniqueIn(bds)   { bds.count { |bd| _day == bd.day } == 1 }
   monthWithUniqueDayIn(bds) { bds.any { |bd| (_month == bd.month) && bd.dayUniqueIn(bds) } }

}

var choices = [

   Birthday.new(5, 15), Birthday.new(5, 16), Birthday.new(5, 19), Birthday.new(6, 17), 
   Birthday.new(6, 18), Birthday.new(7, 14), Birthday.new(7, 16), Birthday.new(8, 14),
   Birthday.new(8, 15), Birthday.new(8, 17)

]

// Albert knows the month but doesn't know the day. // So the month can't be unique within the choices. var filtered = choices.where { |bd| !bd.monthUniqueIn(choices) }.toList

// Albert also knows that Bernard doesn't know the answer. // So the month can't have a unique day. filtered = filtered.where { |bd| !bd.monthWithUniqueDayIn(filtered) }.toList

// Bernard now knows the answer. // So the day must be unique within the remaining choices. filtered = filtered.where { |bd| bd.dayUniqueIn(filtered) }.toList

// Albert now knows the answer too. // So the month must be unique within the remaining choices. filtered = filtered.where { |bd| bd.monthUniqueIn(filtered) }.toList

if (filtered.count == 1) {

   System.print("Cheryl's birthday is %(filtered[0])")

} else {

   System.print("Something went wrong!")

}</lang>

Output:
Cheryl's birthday is July 16

zkl

<lang zkl>dates:=T(T("May", 15), T("May", 16), T("May", 19),

        T("June",  17), T("June",  18), 

T("July", 14), T("July", 16), T("August",14), T("August",15), T("August",17) ); mDs:=dates.pump(Dictionary().appendKV); // "June":(15,16,19), ... dMs:=dates.pump(Dictionary().appendKV,"reverse"); // 15:"May", 16:"May", 19:"May", ...

// remove unique days (18,19) --> "July":(14,16),"August":(14,15,17) dMs.values.apply2('wrap(ms){ if(ms.len()==1) mDs.del(ms[0]) });

// find intersection of above days --> (14) fcn intersection(l1,l2){ l1.pump(List,l2.holds,'==(True),Void.Filter) } badDs:=mDs.values.reduce(intersection);

// --> July:(16),August:(15,17) --> ( ("July",(16)) ) theDay:=mDs.filter('wrap([(m,ds)]){ ds.removeEach(badDs).len()==1 });

// print birthday such that muliples are shown, if any println("Cheryl's birthday is ",theDay.flatten().flatten().concat(" "));</lang>

Output:
Cheryl's birthday is July 16