Fine-tuning CodeLlama-7B for Fortran Code Generation

Find Winner on a Tic Tac Toe Game

Tic-tac-toe is played by two players A and B on a 3 x 3 grid. The rules of Tic-Tac-Toe are:

Players take turns placing characters into empty squares ' '.
The first player A always places 'X' characters, while the second player B always places 'O' characters.
'X' and 'O' characters are always placed into empty squares, never on filled ones.
The game ends when there are three of the same (non-empty) character filling any row, column, or diagonal.
The game also ends if all squares are non-empty.
No more moves can be played if the game is over.

Given a 2D integer array moves where moves[i] = [rowi, coli] indicates that the ith move will be played on grid[rowi][coli]. return the winner of the game if it exists (A or B). In case the game ends in a draw return "Draw ". If there are still movements to play return "Pending ".

You can assume that moves is valid (i.e., it follows the rules of Tic-Tac-Toe), the grid is initially empty, and A will play first.

Example 1:

Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]] Output: "A " Explanation: A wins, they always play first.

Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]] Output: "B " Explanation: B wins.

Example 3:

Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]] Output: "Draw " Explanation: The game ends in a draw since there are no moves to make.

Constraints:

1 <= moves.length <= 9
moves[i].length == 2
0 <= rowi, coli <= 2
There are no repeated elements on moves.
moves follow the rules of tic tac toe.

Note: This problem is from LeetCode.

❌ Compiled

❌ Executed

❌ Correct

program tictactoe
    implicit none

    integer :: i, j, k, l, m, n
    integer :: moves(9, 2)
    character(len=1) :: grid(3, 3)
    character(len=1) :: winner

    ! Examples
    moves(:, :) = reshape([&
        & [0, 0], [2, 0], [1, 1], [2, 1], [2, 2],&
        & [0, 0], [1, 1], [0, 1], [0, 2], [1, 0], [2, 0]], shape(moves))

    do i = 1, size(moves, 1)
        call solve(moves(i, :), grid, winner)
        write(*, *) 'Move ', i, ':', moves(i, :)
        write(*, *) 'Grid:'
        do j = 1, size(grid, 1)
            write(*, '(3A)') ((grid(j, k), k = 1, size(grid, 2)), new_line(''))
        end do
        write(*, *) 'Winner: ', winner
    end do

contains

    subroutine solve(move, grid, winner)
        implicit none
        integer, intent(in) :: move(2)
        character(len=1), intent(inout) :: grid(3, 3)
        character(len=1), intent(out) :: winner

        integer :: row, col
        logical :: found

        row = move(1)
        col = move(2)

        if (grid(row, col) /= ' ') then
            stop 'Invalid move'
        end if

        grid(row, col) = 'X'

        found = .false.
        do i = 1, size(grid, 1)
            if (all(grid(i, :) == 'X')) then
                winner = 'A'
                found = .true.
                exit
            else if (all(grid(:, i) == 'X')) then
                winner = 'A'
                found = .true.
                exit
            end if
        end do

        if (.not. found) then
            do i = 1, size(grid, 1) - 2
                if (grid(i, i) == 'X' .and. grid(i + 1, i + 1) == 'X' .and. grid(i + 2, i + 2) == 'X') then
                    winner = 'A'
                    found = .true.
                    exit
                else if (grid(i, i) == 'X' .and. grid(i + 1, i + 1) == 'X' .and. grid(i + 2, i + 2) == 'X') then
                    winner = 'A'
                    found = .true.
                    exit
                end if
            end do
        end if

        if (.not. found) then
            do i = 1, size(grid, 1)
                if (all(grid(i, :) == 'O')) then
                    winner = 'B'
                    found = .true.
                    exit
                else if (all(grid(:, i) == 'O')) then
                    winner = 'B'
                    found = .true.
                    exit
                end if
            end do
        end if

        if (.not. found) then
            do i = 1, size(grid, 1) - 2
                if (grid(i, i) == 'O' .and. grid(i + 1, i + 1) == 'O' .and. grid(i + 2, i + 2) == 'O') then
                    winner = 'B'
                    found = .true.
                    exit
                else if (grid(i, i) == 'O' .and. grid(i + 1, i + 1) == 'O' .and. grid(i + 2, i + 2) == 'O') then
                    winner = 'B'
                    found = .true.
                    exit
                end if
            end do
        end if

        if (.not. found) then
            if (all

stderr

temp.f95:19:42:

   19 |             write(*, '(3A)') ((grid(j, k), k = 1, size(grid, 2)), new_line(''))
      |                                          1
Error: Expected a right parenthesis in expression at (1)
temp.f95:100:19:

  100 |             if (all
      |                   1
Error: Function ‘all’ requires an argument list at (1)
f951: Error: Unexpected end of file in ‘temp.f95’

❌ Compiled

❌ Executed

❌ Correct

PROGRAM TIC_TAC_TOE
  IMPLICIT NONE

  INTEGER, PARAMETER :: N = 3
  INTEGER, DIMENSION(N, N) :: grid
  INTEGER, DIMENSION(N, N) :: moves
  INTEGER :: i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
  INTEGER :: row, col
  INTEGER :: count
  INTEGER :: winner

  ! Initialize the grid
  grid = 0

  ! Read the moves
  DO i = 1, SIZE(moves, DIM=1)
    row = moves(i, 1)
    col = moves(i, 2)
    grid(row, col) = i
  END DO

  ! Check for a winner
  winner = 0
  DO i = 1, N
    ! Check rows
    count = 0
    DO j = 1, N
      IF (grid(i, j) /= 0) THEN
        count = count + 1
        IF (count == 3) THEN
          winner = grid(i, j)
          EXIT
        END IF
      END IF
    END DO
    IF (winner /= 0) EXIT

    ! Check columns
    count = 0
    DO j = 1, N
      IF (grid(j, i) /= 0) THEN
        count = count + 1
        IF (count == 3) THEN
          winner = grid(j, i)
          EXIT
        END IF
      END IF
    END DO
    IF (winner /= 0) EXIT

    ! Check diagonals
    count = 0
    DO j = 1, N
      IF (grid(j, j) /= 0) THEN
        count = count + 1
        IF (count == 3) THEN
          winner = grid(j, j)
          EXIT
        END IF
      END IF
    END DO
    IF (winner /= 0) EXIT

    count = 0
    DO j = 1, N
      IF (grid(j, N-j+1) /= 0) THEN
        count = count + 1
        IF (count == 3) THEN
          winner = grid(j, N-j+1)
          EXIT
        END IF
      END IF
    END DO
    IF (winner /= 0) EXIT
  END DO

  ! Check for a draw
  IF (winner == 0) THEN
    count = 0
    DO i = 1, N
      DO j = 1, N
        IF (grid(i, j) == 0) THEN
          count = count + 1
        END IF
      END DO
    END DO
    IF (count == 0) winner = -1
  END IF

  ! Print the winner
  IF (winner == 1) THEN
    PRINT *, "A "
  ELSE IF (winner == 2) THEN
    PRINT *, "B "
  ELSE IF (winner == -1) THEN
    PRINT *, "Draw "
  ELSE
    PRINT *, "Pending "
  END IF

END PROGRAM TIC_TAC_TOE

stderr

temp.f95:7:29:

    7 |   INTEGER :: i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
      |                             1
Error: Symbol ‘n’ at (1) already has basic type of INTEGER
temp.f95:16:6:

   16 |   DO i = 1, SIZE(moves, DIM=1)
      |      1
Error: Symbol ‘i’ at (1) has no IMPLICIT type
temp.f95:27:8:

   27 |     DO j = 1, N
      |        1
Error: Symbol ‘j’ at (1) has no IMPLICIT type

🌐 Data from online sources

def validateBinaryTreeNodes(n, leftChild, rightChild):
    parent = [-1] * n
    for i in range(n):
        if leftChild[i] != -1:
            if parent[leftChild[i]] != -1:
                return False
            parent[leftChild[i]] = i
        if rightChild[i] != -1:
            if parent[rightChild[i]] != -1:
                return False
            parent[rightChild[i]] = i

    root = -1
    for i in range(n):
        if parent[i] == -1:
            if root != -1:
                return False
            root = i

    return root != -1

Create an array parent to keep track of the parent of each node. Initialize each element to -1.
Iterate through each node i: a. Check the left child of i. If it's not -1, check if it already has a parent. If it does, return false (two parents for a node makes it an invalid tree). Set the parent of the left child to i. b. Do the same for the right child.
Iterate through the parent array to find a node without a parent (root). If a root is found and another node without a parent is found later, return false (more than one tree).
If a root is found, return true (a valid tree). Otherwise, return false (no tree found).

🌐 Data from online sources

bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
    vector<int> parent(n, -1);
    for (int i = 0; i < n; ++i) {
        if (leftChild[i] != -1) {
            if (parent[leftChild[i]] != -1) return false;
            parent[leftChild[i]] = i;
        }
        if (rightChild[i] != -1) {
            if (parent[rightChild[i]] != -1) return false;
            parent[rightChild[i]] = i;
        }
    }
    int root = -1;
    for (int i = 0; i < n; ++i) {
        if (parent[i] == -1) {
            if (root != -1) return false;
            root = i;
        }
    }
    return root != -1;
}

Create an array parent to keep track of the parent of each node. Initialize each element to -1.
Iterate through each node i: a. Check the left child of i. If it's not -1, check if it already has a parent. If it does, return false (two parents for a node makes it an invalid tree). Set the parent of the left child to i. b. Do the same for the right child.
Iterate through the parent array to find a node without a parent (root). If a root is found and another node without a parent is found later, return false (more than one tree).
If a root is found, return true (a valid tree). Otherwise, return false (no tree found).