/* 

  Borrowed books in Picat.

  From Adrian Groza "Modelling Puzzles in First Order Logic"
  See https://www.researchgate.net/publication/374588335_Measuring_reasoning_capabilities_of_ChatGPT
  """
  Puzzle 53. Borrowed books

  Four children, borrowed a specific book from the library. Each book has a unique sub-
  ject and number of pages. Find out in which month each child got the book and their
  number of pages. You know that:

  1. If Janet got a History book, then the Fantasy book has 350 pages.
  2. The Thriller is 50 pages larger than the book borrowed in May, but 50 pages shorter
     than Anthony's choice.
  3. A girl borrowed a book in May, the other girl, Meredith, got a History book.
  4. Janet didn't get the shortest book of 250 pages.
  5. Exactly one of the customers shares the initial in the name and in the month he/she
     got a book.
  6. The fantasy is 50 pages larger than Mark's book.
  7. In June a Fantasy book wasn't borrowed, nor a Romance one.
  8. The largest book and the Romance one have been borrowed in consecutive months.
  9. If Mark borrowed a book in the first month (i.e. April), then the Fantasy book is the
     largest one with 400 pages long.

  (taken from Brainzilla-www.brainzilla.com)
  """

  There are two possible solutions

  Meredith History  June  250
  Janet    Romance  May   300
  Mark     Thriller July  350
  Anthony  Fantasy  April 400
  LargestBook = 400

  Meredith History  July  250
  Janet    Romance  May   300
  Mark     Thriller June  350
  Anthony  Fantasy  April 400
  LargestBook = 400


  The difference are the months for Meredith and Mark (June or July).


  This program was created by Hakan Kjellerstrand, hakank@gmail.com
  See also my Picat page: http://www.hakank.org/picat/

*/

import cp.

main => go.

go =>
  N = 4,
  
  Names = new_list(N),
  Names :: 1..N,
  Names = [Anthony,Janet,Meredith,Mark],
  NamesS = ["Anthony","Janet","Meredith","Mark"],

  Subject = new_list(N),
  Subject :: 1..N,
  Subject = [History,Fantasy,Thriller,Romance],
  SubjectS = ["History","Fantasy","Thriller","Romance"],  

  % Keep pages constant for symmetry breaking
  Pages = new_list(N),
  Pages = 1..N, 
  Pages = [P250,_P300,P350,P400],
  PagesS = [250,300,350,400],

  Month = new_list(N),
  Month :: 1..N,
  Month = [April,May,June,July],
  MonthS = ["April","May","June","July"],

  all_different(Names),
  all_different(Subject),
  all_different(Pages),
  all_different(Month),

  LargestBook #= max(Pages),

  % 1. If Janet got a History book, then the Fantasy book has 350 pages.
  (Janet #= History) #=> Fantasy #= P350,
  
  % 2. The Thriller is 50 pages larger than the book borrowed in May,
  %    but 50 pages shorter than Anthony's choice.
  May + 1 #= Thriller,
  Thriller + 1 #= Anthony,

  % 3. A girl borrowed a book in May, the other girl, Meredith, got
  %    a History book.
  May #= Janet,
  Meredith #= History,
  
  % 4. Janet didn't get the shortest book of 250 pages.
  Janet #!= P250,
  
  % 5. Exactly one of the customers shares the initial in the name and
  %    in the month he/she got a book.
  sum([April #= Anthony,May #= Meredith, May #= Mark, June #= Janet, July #= Janet]) #= 1,
  
  % 6. The fantasy is 50 pages larger than Mark's book.
  Fantasy #!= Mark,
  Fantasy #= Mark + 1,
    
  % 7. In June a Fantasy book wasn't borrowed, nor a Romance one.
  June #!= Fantasy,
  June #!= Romance,
  
  % 8. The largest book and the Romance one have been borrowed
  %    in consecutive months.
  Romance #!= LargestBook,

  (LargestBook #= April #/\ Romance #= May)
  #\/ (LargestBook #= May   #/\ Romance #= April) 
  
  % #\/ (LargestBook #= May   #/\ Romance #= June)  % Not possible, see clue 7
  #\/ (LargestBook #= June  #/\ Romance #= May) 
  
  #\/ (LargestBook #= June  #/\ Romance #= July),
  % #\/ (LargestBook #= July  #/\ Romance #= June), % Not possible, see clue 7


  % 9. If Mark borrowed a book in the first month (i.e. April), then the
  %    Fantasy book is the largest one with 400 pages long.
  (Mark #= April) #=> (Fantasy #= LargestBook #/\ LargestBook #= P400),

  Vars = Names ++ Subject ++ Pages ++ Month ++ [LargestBook],
  solve(Vars),

  foreach(I in 1..N),
    element(Nn,Names,I),  
    element(S,Subject,I),
    element(M,Month,I),
    element(P,Pages,I),
    printf("%-8s %-8s %-5s %d\n",NamesS[Nn],SubjectS[S],MonthS[M],PagesS[P])
  end,
  println('LargestBook'=PagesS[LargestBook]),
  nl,
  fail,

  nl.