Cheryl's birthday

Cheryl's birthday is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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

Cheryl gave them a list of 10 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 and Bernard separately the month and the day of the birthday respectively.

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 program in your language to deduce, by successive elimination, Cheryl's birthday.

See the Wikipedia article of the same name.

Related task

• Sum and Product Puzzle

Common Lisp

<lang lisp> (defparameter *possible-dates*

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

(defun count-items (list)

 "returns a list of (item how-many-of-it)"
 (mapcar #'(lambda (item) (list item (count item list :test #'equal)))

(remove-duplicates list :test #'equal)))

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

 (let ((unique-date-parts (remove-if-not #'(lambda (part) (= (cadr part) 1))

(count-items (mapcar alist-look-at possible-dates)))))

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

(defun person (person possible-dates)

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

(defun cheryls-birthday (possible-dates)

 (person 'albert			;And this is the final search Albert does in stage 3) and finds the birthday

(person 'bernard ;This is the search Bernard does in stage 2) (set-difference possible-dates (person 'bernard possible-dates) ;This is the search Albert does at first on behalf of Bernard in stage 1) :key #'cdr))))

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

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

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 hash

my %months = < January February March April May June July

     August September October November December>.kv.pairup;

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

Cheryl's birthday is July 16.