Skip to content
GitLab
Projects
Groups
Snippets
Help
Loading...
Help
Help
Support
Community forum
Keyboard shortcuts
?
Submit feedback
Contribute to GitLab
Sign in
Toggle navigation
why3
Project overview
Project overview
Details
Activity
Releases
Repository
Repository
Files
Commits
Branches
Tags
Contributors
Graph
Compare
Issues
119
Issues
119
List
Boards
Labels
Service Desk
Milestones
Merge Requests
16
Merge Requests
16
Operations
Operations
Incidents
Packages & Registries
Packages & Registries
Container Registry
Analytics
Analytics
Repository
Value Stream
Wiki
Wiki
Snippets
Snippets
Members
Members
Collapse sidebar
Close sidebar
Activity
Graph
Create a new issue
Commits
Issue Boards
Open sidebar
Why3
why3
Commits
a32e6807
Commit
a32e6807
authored
Jun 22, 2011
by
JeanChristophe Filliâtre
Browse files
Options
Browse Files
Download
Email Patches
Plain Diff
documentation: solution for VSTTE'10 problem 4
parent
8ba811e5
Changes
3
Hide whitespace changes
Inline
Sidebyside
Showing
3 changed files
with
255 additions
and
12 deletions
+255
12
doc/whyml.tex
doc/whyml.tex
+246
2
examples/programs/vstte10_queens.mlw
examples/programs/vstte10_queens.mlw
+5
6
examples/programs/vstte10_queens/why3session.xml
examples/programs/vstte10_queens/why3session.xml
+4
4
No files found.
doc/whyml.tex
View file @
a32e6807
...
...
@@ 466,15 +466,259 @@ the loop invariant.
\subsection
{
Problem 4: NQueens
}
The fourth problem is probably the most challenging one.
We have to verify the implementation of a program which solves the
$
N
$
queens puzzle: place
$
N
$
queens on an
$
N
\times
N
$
chess board so that no queen can capture another one with a
legal move.
The program should return a placement if there is a solution and
indicates that there is no solution otherwise. A placement is a
$
N
$
element array which assigns the queen on row
$
i
$
to its column.
Thus we start our module by importing arithmetic and arrays:
\begin{verbatim}
module NQueens
use import int.Int
use import module array.Array
\end{verbatim}
The code is a simple backtracking algorithm, which tries to put a queen
on each row of the chess board, one by one (there is basically no
better way to solve the
$
N
$
queens puzzle).
A building block is a function which checks whether the queen on a
given row may attack another queen on a previous row. To verify this
function, we first define a more elementary predicate, which expresses
that queens on row
\texttt
{
pos
}
and
\texttt
{
q
}
do no attack each other:
\begin{verbatim}
logic consistent
_
row (board: array int) (pos: int) (q: int) =
board[q] <> board[pos] and
board[q]  board[pos] <> pos  q and
board[pos]  board[q] <> pos  q
\end{verbatim}
Then it is possible to define the consistency of row
\texttt
{
pos
}
with respect to all previous rows:
\begin{verbatim}
logic is
_
consistent (board: array int) (pos: int) =
forall q:int. 0 <= q < pos > consistent
_
row board pos q
\end{verbatim}
Implementing a function which decides this predicate is another
matter. In order for it to be efficient, we want to return
\texttt
{
False
}
as soon as a queen attacks the queen on row
\texttt
{
pos
}
. We use an exception for this purpose and it carries the
row of the attacking queen:
\begin{verbatim}
exception Inconsistent int
\end{verbatim}
The check is implemented by a function
\verb

check_is_consistent

,
which takes the board and the row
\texttt
{
pos
}
as arguments, and scans
rows from 0 to
\texttt
{
pos1
}
looking for an attacking queen. As soon
as one is found, the exception is raised. It is caught immediately
outside the loop and
\texttt
{
False
}
is returned. Whenever the end of
the loop is reached,
\texttt
{
True
}
is returned.
\begin{verbatim}
let check
_
is
_
consistent (board: array int) (pos: int) =
{
0 <= pos < length board
}
try
for q = 0 to pos  1 do
invariant
{
forall j:int. 0 <= j < q > consistent
_
row board pos j
}
let bq = board[q] in
let bpos = board[pos] in
if bq = bpos then raise (Inconsistent q);
if bq  bpos = pos  q then raise (Inconsistent q);
if bpos  bq = pos  q then raise (Inconsistent q)
done;
True
with Inconsistent q >
assert
{
not (consistent
_
row board pos q)
}
;
False
end
{
result=True <> is
_
consistent board pos
}
\end{verbatim}
The assertion in the exception handler is a cut for SMT solvers.
This first part of the solution is given Figure~
\ref
{
fig:NQueens1
}
.
\begin{figure}
\centering
\begin{verbatim}
module NQueens
use import int.Int
use import module array.Array
logic consistent
_
row (board: array int) (pos: int) (q: int) =
board[q] <> board[pos] and
board[q]  board[pos] <> pos  q and
board[pos]  board[q] <> pos  q
logic is
_
consistent (board: array int) (pos: int) =
forall q:int. 0 <= q < pos > consistent
_
row board pos q
exception Inconsistent int
let check
_
is
_
consistent (board: array int) (pos: int) =
{
0 <= pos < length board
}
try
for q = 0 to pos  1 do
invariant
{
forall j:int. 0 <= j < q > consistent
_
row board pos j
}
let bq = board[q] in
let bpos = board[pos] in
if bq = bpos then raise (Inconsistent q);
if bq  bpos = pos  q then raise (Inconsistent q);
if bpos  bq = pos  q then raise (Inconsistent q)
done;
True
with Inconsistent q >
assert
{
not (consistent
_
row board pos q)
}
;
False
end
{
result=True <> is
_
consistent board pos
}
\end{verbatim}
\vspace*
{
2em
}
\hrulefill
\caption
{
Solution for VSTTE'10 competition problem 4 (1/2).
}
\label
{
fig:NQueens1
}
\end{figure}
We now proceed with the verification of the backtracking algorithm.
The specification requires us to define the notion of solution, which
is straightforward using the predicate
\verb

is_consistent

above.
However, since the algorithm will try to complete a given partial
solution, it is more convenient to define the notion of partial
solution, up to a given row. It is even more convenient to split it in
two predicates, one related to legal column values and another to
consistency of rows:
\begin{verbatim}
logic is
_
board (board: array int) (pos: int) =
forall q:int. 0 <= q < pos > 0 <= board[q] < length board
logic solution (board: array int) (pos: int) =
is
_
board board pos and
forall q:int. 0 <= q < pos > is
_
consistent board q
\end{verbatim}
The algorithm will not mutate the partial solution it is given and,
in case of a search failure, will claim that there is no solution
extending this prefix. For this reason, we introduce a predicate
comparing two chess boards for equality up to a given row:
\begin{verbatim}
logic eq
_
board (b1 b2: array int) (pos: int) =
forall q:int. 0 <= q < pos > b1[q] = b2[q]
\end{verbatim}
The search itself makes use of an exception to signal a successful search:
\begin{verbatim}
exception Solution
\end{verbatim}
The backtracking code is a recursive function
\verb

bt_queens

which
takes the chess board, its size, and the starting row for the search.
The termination is ensured by the obvious variant
\texttt
{
npos
}
.
\begin{verbatim}
let rec bt
_
queens (board: array int) (n: int) (pos: int) variant
{
npos
}
=
\end{verbatim}
The precondition relates
\texttt
{
board
}
,
\texttt
{
pos
}
, and
\texttt
{
n
}
and requires
\texttt
{
board
}
to be a solution up to
\texttt
{
pos
}
:
\begin{verbatim}
{
length board = n and 0 <= pos <= n and solution board pos
}
label Init:
\end{verbatim}
We place a label
\texttt
{
Init
}
immediately after the precondition to
be able to refer to the value of
\texttt
{
board
}
in the prestate.
Whenever we reach the end of the chess board, we have found a solution
and we signal it using exception
\texttt
{
Solution
}
:
\begin{verbatim}
if pos = n then raise Solution;
\end{verbatim}
Otherwise we scan all possible positions for the queen on row
\texttt
{
pos
}
with a
\texttt
{
for
}
loop:
\begin{verbatim}
for i = 0 to n  1 do
\end{verbatim}
The loop invariant states that we have not modified the solution
prefix so far, and that we have not found any solution that would
extend this prefix with a queen on row
\texttt
{
pos
}
at a column below
\texttt
{
i
}
:
\begin{verbatim}
invariant
{
eq
_
board board (at board Init) pos and
forall b:array int. length b = n > is
_
board b n >
eq
_
board board b pos > 0 <= b[pos] < i > not (solution b n)
}
\end{verbatim}
Then we assign column
\texttt
{
i
}
to the queen on row
\texttt
{
pos
}
and
we check for a possible attack with
\verb

check_is_consistent

. If
not, we call
\verb

bt_queens

recursively on the next row.
\begin{verbatim}
board[pos] < i;
if check
_
is
_
consistent board pos then bt
_
queens board n (pos + 1)
done
\end{verbatim}
This completes the loop and function
\verb

bt_queens

as well.
The postcondition is twofold: either the function exits normally and
then there is no solution extending the prefix in
\texttt
{
board
}
,
which has not been modified;
or the function raises
\texttt
{
Solution
}
and we have a solution in
\texttt
{
board
}
.
\begin{verbatim}
{
eq
_
board board (old board) pos and
forall b:array int. length b = n > is
_
board b n >
eq
_
board board b pos > not (solution b n)
}
 Solution >
{
solution board n
}
\end{verbatim}
Solving the puzzle is a simple call to
\verb

bt_queens

, starting the
search on row 0. The postcondition is also twofold, as for
\verb

bt_queens

, yet slightly simpler.
\begin{verbatim}
let queens (board: array int) (n: int) =
{
0 <= length board = n
}
bt
_
queens board n 0
{
forall b:array int. length b = n > is
_
board b n > not (solution b n)
}
 Solution >
{
solution board n
}
\end{verbatim}
This second part of the solution is given Figure~
\ref
{
fig:NQueens2
}
.
With the help of a few auxiliary lemmas  not given here but available
from
\why
's sources  the verification conditions are all discharged
automatically, including the verification of the lemmas themselves.
\begin{figure}
\centering
\begin{verbatim}
logic is
_
board (board: array int) (pos: int) =
forall q:int. 0 <= q < pos > 0 <= board[q] < length board
logic solution (board: array int) (pos: int) =
is
_
board board pos and
forall q:int. 0 <= q < pos > is
_
consistent board q
logic eq
_
board (b1 b2: array int) (pos: int) =
forall q:int. 0 <= q < pos > b1[q] = b2[q]
exception Solution
let rec bt
_
queens (board: array int) (n: int) (pos: int) variant
{
n  pos
}
=
{
length board = n and 0 <= pos <= n and solution board pos
}
label Init:
if pos = n then raise Solution;
for i = 0 to n  1 do
invariant
{
eq
_
board board (at board Init) pos and
forall b:array int. length b = n > is
_
board b n >
eq
_
board board b pos > 0 <= b[pos] < i > not (solution b n)
}
board[pos] < i;
if check
_
is
_
consistent board pos then bt
_
queens board n (pos + 1)
done
{
(* no solution *)
eq
_
board board (old board) pos and
forall b:array int. length b = n > is
_
board b n >
eq
_
board board b pos > not (solution b n)
}
 Solution >
{
(* a solution *)
solution board n
}
let queens (board: array int) (n: int) =
{
0 <= length board = n
}
bt
_
queens board n 0
{
forall b:array int. length b = n > is
_
board b n > not (solution b n)
}
 Solution >
{
solution board n
}
end
\end{verbatim}
\vspace*
{
2em
}
\hrulefill
\caption
{
Solution for VSTTE'10 competition problem 4.
}
\label
{
fig:NQueens
}
\caption
{
Solution for VSTTE'10 competition problem 4
(2/2)
.
}
\label
{
fig:NQueens
2
}
\end{figure}
\subsection
{
Problem 5: Amortized Queue
}
The last problem consists in verifying the implementation of a
...
...
examples/programs/vstte10_queens.mlw
View file @
a32e6807
...
...
@@ 49,7 +49,7 @@ module NQueens
exception Inconsistent int
let check_is_consistent (board: array int)
pos
=
let check_is_consistent (board: array int)
(pos: int)
=
{ 0 <= pos < length board }
try
for q = 0 to pos  1 do
...
...
@@ 80,28 +80,27 @@ module NQueens
exception Solution
let rec bt_queens (board: array int)
n pos
variant { n  pos } =
let rec bt_queens (board: array int)
(n: int) (pos: int)
variant { n  pos } =
{ length board = n and 0 <= pos <= n and solution board pos }
label Init:
if pos = n then raise Solution;
for i = 0 to n  1 do
invariant {
length board = n and
eq_board board (at board Init) pos and
eq_board board (at board Init) pos and
forall b:array int. length b = n > is_board b n >
eq_board board b pos > 0 <= b[pos] < i > not (solution b n) }
board[pos] < i;
assert { eq_board board (at board Init) pos };
if check_is_consistent board pos then bt_queens board n (pos+1)
done
{ (* no solution *)
length board = n and
eq_board board (old board) pos and
eq_board board (old board) pos and
forall b:array int. length b = n > is_board b n >
eq_board board b pos > not (solution b n) }
 Solution >
{ (* a solution *)
solution board n }
let queens (board: array int)
n
=
let queens (board: array int)
(n: int)
=
{ 0 <= length board = n }
bt_queens board n 0
{ forall b:array int. length b = n > is_board b n > not (solution b n) }
...
...
examples/programs/vstte10_queens/why3session.xml
View file @
a32e6807
...
...
@@ 43,14 +43,14 @@
<result
status=
"valid"
time=
"0.60"
/>
</proof>
</goal>
<goal
name=
"WP_parameter bt_queens"
expl=
"correctness of parameter bt_queens"
sum=
"
b59f9ab1ac42b3755b8e06f7dc04fc2a
"
proved=
"true"
expanded=
"true"
>
<goal
name=
"WP_parameter bt_queens"
expl=
"correctness of parameter bt_queens"
sum=
"
90c2b563e0abbc3b4a70fec2410b7f1b
"
proved=
"true"
expanded=
"true"
>
<proof
prover=
"cvc3"
timelimit=
"20"
edited=
""
obsolete=
"false"
>
<result
status=
"valid"
time=
"0.
9
6"
/>
<result
status=
"valid"
time=
"0.
3
6"
/>
</proof>
</goal>
<goal
name=
"WP_parameter queens"
expl=
"correctness of parameter queens"
sum=
"
fa29a0090876c2c2eae4d8423017ff3c
"
proved=
"true"
expanded=
"true"
>
<goal
name=
"WP_parameter queens"
expl=
"correctness of parameter queens"
sum=
"
67746ab73fbf4bab85d11e7011676a38
"
proved=
"true"
expanded=
"true"
>
<proof
prover=
"altergo"
timelimit=
"20"
edited=
""
obsolete=
"false"
>
<result
status=
"valid"
time=
"0.0
3
"
/>
<result
status=
"valid"
time=
"0.0
5
"
/>
</proof>
</goal>
</theory>
...
...
Write
Preview
Markdown
is supported
0%
Try again
or
attach a new file
.
Attach a file
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Cancel
Please
register
or
sign in
to comment