Challenge 1: A 5-Queens Solution, Verified by Hand — Possible Solution
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queens_demo.pl:

:- use_module(library(clpfd)).

queens(N, Queens) :-
    length(Queens, N),
    Queens ins 1..N,
    safe_queens(Queens),
    label(Queens).

safe_queens([]).
safe_queens([Q|Qs]) :-
    safe_queens(Qs, Q, 1),
    safe_queens(Qs).

safe_queens([], _, _).
safe_queens([Q|Qs], Q0, D0) :-
    Q0 #\= Q,
    abs(Q0 - Q) #\= D0,
    D1 #= D0 + 1,
    safe_queens(Qs, Q0, D1).

Query:

    ?- queens(5, Qs), !.
    Qs = [1, 3, 5, 2, 4].   % one valid solution; exact result depends on the solver's search order

Verifying [1, 3, 5, 2, 4] by hand (column, row) pairs: (1,1) (2,3) (3,5) (4,2) (5,4)

Rows distinct: 1, 3, 5, 2, 4 -- all five different. OK.

Diagonal check (|row difference| must not equal |column difference| for every pair):
  (1,1)-(2,3): |1-3|=2, cols differ by 1 -- 2 =/= 1, OK.
  (1,1)-(3,5): |1-5|=4, cols differ by 2 -- 4 =/= 2, OK.
  (1,1)-(4,2): |1-2|=1, cols differ by 3 -- 1 =/= 3, OK.
  (1,1)-(5,4): |1-4|=3, cols differ by 4 -- 3 =/= 4, OK.
  (2,3)-(3,5): |3-5|=2, cols differ by 1 -- 2 =/= 1, OK.
  (2,3)-(4,2): |3-2|=1, cols differ by 2 -- 1 =/= 2, OK.
  (2,3)-(5,4): |3-4|=1, cols differ by 3 -- 1 =/= 3, OK.
  (3,5)-(4,2): |5-2|=3, cols differ by 1 -- 3 =/= 1, OK.
  (3,5)-(5,4): |5-4|=1, cols differ by 2 -- 1 =/= 2, OK.
  (4,2)-(5,4): |2-4|=2, cols differ by 1 -- 2 =/= 1, OK.

Every pair passes both checks -- no shared row, no shared diagonal, and no shared column
by construction (one row value per column index). [1, 3, 5, 2, 4] is a genuinely valid
5-queens placement.

WHY THIS WORKS AS AN ANSWER
------------------------------
This runs the chapter's own queens/2 predicate unmodified for N=5, then
manually re-verifies every one of the 10 row/diagonal pairs by hand rather
than just trusting the solver's output, directly confirming the CLP(FD)
constraints actually encode the real N-Queens rules correctly.