An Introduction to Tabled Logic Programming with Picat

Picat is a new logic-based programming language. In many ways, Picat is similar to Prolog, especially B-Prolog, but it has functions in addition to predicates, pattern-matching instead of unification in predicate heads, list comprehensions and optional destructive assignment. Knowing some Prolog helps when learning Picat but is by no means required.

According to the authors of the language, Picat is an acronym for:

Picat has a lot of interesting features, such as constraint logic programming support and interfaces to various solvers. In this article, I focus on one aspect of Picat: tabling and a tabling-based planner module.

First, let's install and learn some basics of Picat. Installing Picat is easy; you can download precompiled binaries for 32- and 64-bit Linux systems, as well as binaries for other platforms. If you want to compile it yourself, C source code is available under the Mozilla Public License. The examples here use Picat version 1.2, but newer or slightly older versions also should work.

After the installation, you can run picat from a command line and see Picat's prompt:

Picat 1.2, (C) picat-lang.org, 2013-2015.
Picat>

You can run commands (queries) interactively with this interface.

Let's start with the mandatory “Hello, World”:

Picat> println("Hello, World!").
Hello, World!
yes

No real surprises here. The yes at the end means that Picat successfully executed the query.

For the next example, let's assign 2 to a variable:

Picat> X = 2.
X = 2
yes

Note the uppercase letter for the variable name; all variable names must start with a capital letter or an underscore (the same as in Prolog).

Next, assign symbols to the X variable (symbols are enclosed in single quotes; for many symbols, quotes are optional, and double-quoted strings, like the “Hello, World!” above, are lists of symbols):

Picat> X = a.
X = a
yes
Picat> X = 'a'.
X = a
yes

For capital-letter symbols, single quotes are mandatory (otherwise it will be treated as a variable name):

Picat> X = A.
A = X
yes
Picat> X = 'A'.
X = 'A'
yes

Note that the variable X in different queries (separated by a full stop) are completely independent different variables.

Next, let's work with lists:

Picat> X = [1, 2, 3, a].
X = [1,2,3,a]
yes

Lists are heterogeneous in Picat, so you can have numbers as the first three list elements and a symbol as the last element.

You can calculate the results of arithmetic expressions like this:

Picat> X = 2 + 2.
X = 4
yes
Picat> X = [1, a, 2 + 2].
X = [1,a,4]
yes

Picat> X = 2, X = X + 1.
no

This probably is pretty surprising for you if your background is in mainstream imperative languages. But from the logic point of view, it makes prefect sense: X can't be equal to X + 1.

Using := instead of = produces a more expected answer:

Picat> X = 2, X := X + 1.
X = 3
yes

The destructive assignment operator := allows you to override Picat's usual single-assignment “write once” policy for variables. It works in a way you'd expect from an imperative language.

You can use the [|] notation to get a “head” (the first element) and a “tail” (the rest of the elements) of a list:

Picat> X = [1, 2, 3], [Tail | Head] = X.
X = [1,2,3]
Tail = 1
Head = [2,3]
yes

You can use the same notation to add an element to the beginning of a list:

Picat> X = [1, 2, 3], Y = [0 | X].
X = [1,2,3]
Y = [0,1,2,3]
yes
Picat> X = [1, 2, 3], X := [0 | X].
X = [0,1,2,3]
yes

The first example creates a new variable Y, and the second example reuses X with the assignment operator.

Although it's handy to be able to run small queries interactively to try different things, for larger programs, you probably will want to write the code to a file and run it as a script.

To learn some of Picat's syntactical features, let's create a program (script) for a TPK algorithm. TPK is an algorithm proposed by D. Knuth and L. Pardo to show the basic syntax of a programming language beyond the “Hello, World!” program. The algorithm asks a user to enter 11 real numbers (a0...a10). After that, for i = 10...0 (in that order), the algorithm computes the value of an arithmetic function y = f(ai), and outputs a pair (i, y), if y <= 400 or (i, TOO LARGE) otherwise.

f(T) = sqrt(abs(T)) + 5 * T**3.
main =>
    N = 11,
    As = to_array([read_real() : I in 1..N]),
    foreach (I in N..-1..1)
        Y = f(As[I]),
        if Y > 400 then
            printf("%w TOO LARGE\n", I)
        else
            printf("%w %w\n", I, Y)
        end
    end.

First, the code defines a function to calculate the value of f (a function in Picat is a special kind of a predicate that always succeeds with a return value). The main predicate follows (main is a default name for the predicate that will be run during script execution). The code uses list comprehension (similar to what you have in Python, for example) to read the 11 space-separated real numbers into an array As. The foreach loop iterates over the numbers in the array; I goes from 11 to 1 with the step -1 (in Picat, array indices are 1-based). The loop body calculates the value of y for every iteration and prints the result using an “if-then-else” construct. printf is similar to the corresponding C language function; %w can be seen as a “wild card” control sequence to output values of different types.

You can save this program to a file with the .pi extension (let's call it tpk.pi), and then run it using the command picat tpk.pi. Input 11 space-separated numbers and press Enter.

Now that you have some familiarity with the Picat syntax and how to run the scripts, let's proceed directly to tabling. Tabling is a form of automatic caching or memoization—results of previous computations can be stored to avoid unnecessary recomputation.

You can see the benefits of tabling by comparing two versions of a program that calculates Fibonacci numbers with and without tabling.

Listing 2 shows a naive recursive Fibonacci implementation in Picat.

fib(0) = 1.
fib(1) = 1.
fib(N) = F =>
  N > 1,
  F = fib(N - 1) + fib(N - 2).

main =>
  N = read_int(),
  println(fib(N)).

This naive implementation works, but it has an exponential running time. Computing the 37th Fibonacci number takes more than two seconds on my machine:

$ time echo 37 | picat fib_naive.pi
39088169
real	0m2.604s

Computing the 100th Fibonacci number would take this program forever!

But, you can add just one line (table) at the beginning of the program to see a dramatic improvement in running time.

Now you can get not only 37th Fibonacci number instantly, but even the 1,337th (and the answer will not suffer from overflow, because Picat supports arbitrary-length integers).

Effectively, with tabling, you can change the asymptotic running time from exponential to linear.

An even more useful feature is “mode-directed” tabling. Using it you can instruct Picat to store the minimal or the maximal of all possible answers for a non-deterministic goal. This feature is very handy when implementing dynamic programming algorithms. However, that topic is beyond the scope of this article; see Picat's official documentation to learn more about mode-directed tabling.

Picat also has a tabling-based planner module, which can be used to solve artificial intelligence planning problems. This module provides a higher level of abstraction and declarativity.

To use the module, an application programmer has to specify action and final predicates.

The final predicate, in its simplest form, has only one parameter—the current state—and succeeds if the state is final.

The action predicate usually has several clauses—one for each possible action or group of related actions. This predicate has four parameters: current state, new state, action name and action cost.

Let's build a maze-solver using the planner module. The maze-solving program will read a maze map from the standard input and output the best sequence of steps to get to the exit. Here is an example map:

5 5
@.#..
=.#..
.##..
.#X..
.|...

The first line contains the dimensions of the maze: the number of rows R and columns C.

Next, R lines describe the rows of the maze. Here is the description of the map symbols:

The hero can move up, down, left and right (no diagonals) to any open cell (a cell without a wall or a closed door). The hero can pick up keys and open doors. Opening a door and moving into a newly open cell is considered one action. To open a door, the hero must have a key.

Because this is a magic maze, the key disappears after it opens a door. All keys are identical, so opening a door basically just decreases the number of keys the hero has by one.

The goal is to reach the exit using the minimum amount of energy. Moving to an open cell costs one energy unit, picking up a key costs one energy unit, and opening a door and moving to the cell previously occupied by that door costs two energy units.

Let's represent a state for this problem as a tuple (R, C, (ExitI, ExitJ), Walls, Doors, Keys, K, (HeroI, HeroJ)):

Let's first do some boring work of translating a textual representation of a maze to an initial state in the format defined before.

The main predicate is an imperative procedure in constructing an initial state from a textual representation of a maze: you read the input line by line, symbol by symbol, and then construct the lists of walls, doors and keys, as well as record the coordinates of the hero and the exit.

main =>
  R = read_int(), C = read_int(),
  Walls = [], Doors = [], Keys = [],
  (ExitI, ExitJ) = (_, _),
  (HeroI, HeroJ) = (_, _),
  foreach (I in 1..R)
    Line = read_line(),
    foreach (J in 1..C)
      Char = Line[J],
      if Char == '@' then
        HeroI := I, HeroJ := J
      end,
      if Char == 'X' then
        ExitI := I, ExitJ := J
      end,
      if Char == '#' then
        Walls := [(I, J) | Walls]
      end,
      if Char == '|' then
        Doors := [(I, J) | Doors]
      end,
      if Char == '=' then
        Keys := [(I, J) | Keys]
      end
    end
  end,
  InitState = (R, C, (ExitI, ExitJ),
               Walls, Doors, Keys,
               0, (HeroI, HeroJ)),
  println(InitState).

Let's save this program to maze_read.pi, the maze description from above to maze.txt, and run the program (the output is split into several lines for clarity):

$ picat maze_read.pi < maze.txt
5,5,
(4,3),
[(4,2),(3,3),(3,2),(2,3),(1,3)],
[(5,2)],
[(2,1)],
0,1,1

So, you have the dimensions of the maze (5 by 5), the coordinates of the exit (4, 3), the list of the coordinates of all five walls, a one-element list of the closed doors and a one-element list of the keys available for picking up. The hero has 0 keys and starts in cell (1, 1).

Now that you have your state, you can define some predicates to solve the problem. First, the final predicate for the planner module:

final((_, _, (I, J), _, _, _, _, (I, J))) =>
  true.

The state is final when the hero is in the cell with the same coordinates as the exit cell. Variables with name _ are throw-away, “don't care” variables that are not required to have any specific value (Picat invents a different name for each _ behind the scenes, so they don't have to be equal either).

Next, describe the action to take a key if the hero is in a cell with one:

action(State, NewState, Action, Cost) ?=>
  (R, C, (ExitI, ExitJ), Walls, Doors, Keys,
    K, (HeroI, HeroJ)) = State,
    select((HeroI, HeroJ), Keys, NewKeys),
    Action = $take_key(HeroI, HeroJ),
    Cost = 1,
    NewState = (R, C, (ExitI, ExitJ),
                Walls, Doors, NewKeys,
                K + 1, (HeroI, HeroJ)).

First you decompose the state into components, and then you try to select a key with the current coordinates of the hero from the Keys list. If there is such a key, this will succeed, and the rest of the keys will be assigned to “NewKeys”; otherwise, select fails, and Picat will break the execution of this action clause.

The name of the action is take_key, with the coordinates of the event in the parentheses (the $ instructs Picat to treat it literally, like a string, and not to try to execute as a function), and the cost is one energy unit.

The new state is almost the same as the old state, except that the number of keys the hero has is increased by one, and the current key no longer is available to pick up.

Besides picking up keys, there are two more possible actions: moving to an empty cell and moving to a cell with a door after opening it. It's a good idea to combine both these actions into one clause, because they share a lot of code used to select a new hero position and check whether it's within the maze boundary:

action(State, NewState, Action, Cost) =>
  (R, C, (ExitI, ExitJ), Walls, Doors, Keys,
    K, (HeroI, HeroJ)) = State,
  (
    Di = 0, Dj = 1
  ;
    Di = 0, Dj = -1
  ;
    Di = 1, Dj = 0
  ;
    Di = -1, Dj = 0
  ),
  NewHeroI = HeroI + Di,
  NewHeroJ = HeroJ + Dj,
  NewHeroI >= 1, NewHeroI <= R,
  NewHeroJ >= 1, NewHeroJ <= C,
  (
    % move to open cell
    not membchk((NewHeroI, NewHeroJ), Walls),
    not membchk((NewHeroI, NewHeroJ), Doors),
    Action = $move(NewHeroI, NewHeroJ),
    Cost = 1,
    NewState = (R, C, (ExitI, ExitJ),
                Walls, Doors, Keys,
                K, (NewHeroI, NewHeroJ))
  ;
    % open a door and move to that cell
    K > 0,
    select((NewHeroI, NewHeroJ), Doors, NewDoors),
    Action = $open(NewHeroI, NewHeroJ),
    Cost = 2,
    NewState = (R, C, (ExitI, ExitJ),
                Walls, NewDoors, Keys,
                K - 1, (NewHeroI, NewHeroJ))
  ).

Again, first you decompose the state into the components. Next, you try all possible new positions for the hero with non-deterministic disjunction: ;.

A position must be within the maze boundaries: I must be from 1 to R, and J must be from 1 to C. After that, there are two possibilities: move to an open cell, or open a door and move to that cell.

Moving to an open cell is possible only if there isn't a wall or a closed door at the desired position. Two not membchk lines verify this condition. If the condition is met, the action name is move, and the cost is one energy unit. The only change in the state is the hero's position.

Opening an door is possible if there is a door at the position and the hero has at least one key. The select line here is similar to the line for the take action, but now you select a door instead of a key. If the conditions are met, the action name is open, and the cost is two energy units. The new state is almost the same as the old state, but the door is removed from the list of doors, the number of keys the hero has is decreased by one, and the hero has moved to a new position.

To use the defined final and action predicates and find the plan, you need to change println(InitState) to best_plan_unbounded(InitState, Plan), println(Plan) in the main from the maze_read.pi program. (Note: best_plan_unbounded is one of the predicates of the planner module for finding best plans. This particular version uses memory to avoid re-exploring states, converting tree search in the space of all possible plans to graph search.)

Listing 4 shows the complete maze program.

import planner.

action(State, NewState, Action, Cost) ?=>
  (R, C, (ExitI, ExitJ), Walls, Doors, Keys,
    K, (HeroI, HeroJ)) = State,
    select((HeroI, HeroJ), Keys, NewKeys),
    Action = $take_key(HeroI, HeroJ),
    Cost = 1,
    NewState = (R, C, (ExitI, ExitJ),
                Walls, Doors, NewKeys,
                K + 1, (HeroI, HeroJ)).

action(State, NewState, Action, Cost) =>
  (R, C, (ExitI, ExitJ), Walls, Doors, Keys,
    K, (HeroI, HeroJ)) = State,
  (
    Di = 0, Dj = 1
  ;
    Di = 0, Dj = -1
  ;
    Di = 1, Dj = 0
  ;
    Di = -1, Dj = 0
  ),

  NewHeroI = HeroI + Di,
  NewHeroJ = HeroJ + Dj,
  NewHeroI >= 1, NewHeroI <= R,
  NewHeroJ >= 1, NewHeroJ <= C,

  (
    % move to open cell
    not membchk((NewHeroI, NewHeroJ), Walls),
    not membchk((NewHeroI, NewHeroJ), Doors),
    Action = $move(NewHeroI, NewHeroJ),
    Cost = 1,
    NewState = (R, C, (ExitI, ExitJ),
                Walls, Doors, Keys,
                K, (NewHeroI, NewHeroJ))
  ;
    % open a door and move to that cell
    K > 0,
    select((NewHeroI, NewHeroJ), Doors, NewDoors),
    Action = $open(NewHeroI, NewHeroJ),
    Cost = 2,
    NewState = (R, C, (ExitI, ExitJ),
                Walls, NewDoors, Keys,
                K - 1, (NewHeroI, NewHeroJ))
  ).

final((_, _, (I, J), _, _, _, _, (I, J))) =>
  true.

main =>
  R = read_int(), C = read_int(),
  Walls = [], Doors = [], Keys = [],
  (ExitI, ExitJ) = (_, _),
  (HeroI, HeroJ) = (_, _),
  foreach (I in 1..R)
    Line = read_line(),
    foreach (J in 1..C)
      Char = Line[J],
      if Char == '@' then
        HeroI := I, HeroJ := J
      end,
      if Char == 'X' then
        ExitI := I, ExitJ := J
      end,
      if Char == '#' then
        Walls := [(I, J) | Walls]
      end,
      if Char == '|' then
        Doors := [(I, J) | Doors]
      end,
      if Char == '=' then
        Keys := [(I, J) | Keys]
      end
    end
  end,
  InitState = (R, C, (ExitI, ExitJ),
               Walls, Doors, Keys,
               0, (HeroI, HeroJ)),
  best_plan_unbounded(InitState, Plan),
  println(Plan).

After running it for the maze used above, you get an optimal plan (list of actions) to solve the maze (the output is split into several lines for clarity):

$ picat maze.pi < maze.txt
[
 move(2,1),
 take_key(2,1),
 move(3,1),
 move(4,1),
 move(5,1),
 open(5,2),
 move(5,3),move(4,3)
]

You can try to run this program with inputs of various sizes and with different features. For example, this input requires the hero to take a key to the right, then go left to get more keys, and then go right again to the exit:

1 10 ==|=|@=||X

Of course, you can improve the maze program in many different ways:

Overall, Picat looks like a really good starting point for a journey into the realm of logic-based programming languages. It provides many of the goodies Prolog has, such as non-determinism and built-in depths-first search with backtracking, and at the same time, Picat allows you to fall back to convenient imperative concepts like destructive assignments and loops. Higher-level features, like tabling and the planner module, provide ways to write concise, declarative and efficient programs.