CF 104921A - Rook

The input describes positions of a rook on a standard 8 by 8 chessboard. Each position is given in algebraic notation, where a letter from a to h identifies the column and a digit from 1 to 8 identifies the row.

codeforcescompetitive-programming

CF 104921A - Rook

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Solve time: 1m 4s
Verified: no

Solution

Problem Understanding

The input describes positions of a rook on a standard 8 by 8 chessboard. Each position is given in algebraic notation, where a letter from a to h identifies the column and a digit from 1 to 8 identifies the row. For every such position, we need to list all squares that the rook can reach in a single move if the board is empty.

A rook moves in straight lines along rows and columns. From a fixed starting square, it can move horizontally to any other column in the same row, or vertically to any other row in the same column. The destination square must be different from the starting one, but there are no other restrictions since the board is empty.

The constraints are very small. There are at most 64 test cases, and each answer contains at most 14 possible moves, since from any square on an 8 by 8 grid, the rook can go to at most 7 squares horizontally and 7 vertically, minus the current square which is counted twice in that naive sum. This immediately tells us that any direct enumeration of all board squares per test case is already trivial in terms of computation, since it is bounded by a constant.

There are no subtle hidden edge cases in input size, but there are small logical ones that come from coordinate handling. A common mistake is accidentally including the starting square itself in the output, since it lies on both the same row and the same column. Another is mixing row and column indexing, especially when converting between characters and numeric indices. For example, interpreting 'a' as a row instead of a column leads to incorrect move generation that still produces “valid-looking” coordinates.

Another potential issue is output duplication. If one generates all squares in the same row and then all squares in the same column without excluding the starting cell, the starting position will appear twice. A correct solution must explicitly avoid emitting it.

Approaches

A brute-force approach would iterate over every square on the board for each test case and check whether it lies in the same row or same column as the rook. This works because the board has only 64 squares, so for each test case we would do at most 64 checks. Across 64 test cases, this is about 4096 checks total, which is completely negligible.

However, this approach is unnecessarily indirect. The structure of rook movement already gives a direct construction. From a given position (col, row), all valid destinations are exactly:

all squares (col, r) for every r from 1 to 8 except the current row, and all squares (c, row) for every c from a to h except the current column.

So instead of testing all squares, we generate only the valid ones directly. This reduces each test case to a fixed 14 outputs, which is the most straightforward representation of the rook’s movement rules.

The key observation is that rook movement is separable along independent axes. Once the column is fixed, vertical moves are just a one-dimensional enumeration over rows, and similarly for columns.

Approach Time Complexity Space Complexity Verdict
Brute Force O(64 · t) O(1) Accepted
Direct Construction O(t) O(1) Accepted

Algorithm Walkthrough

  1. Read the number of test cases t. Each test case is independent, so we process them separately without storing results.
  2. For each test case, read the string s of length 2. The first character s[0] is the column, and the second character s[1] is the row. We treat them as fixed labels rather than converting into a 0-indexed array unless needed for convenience.
  3. Generate all vertical moves by keeping the column fixed and iterating through all possible rows from '1' to '8'. For each row, if it differs from the current row, we output the position formed by (column, row).

This step directly encodes the rook’s ability to move along a file. We explicitly skip the current row to avoid outputting the starting square. 4. Generate all horizontal moves by keeping the row fixed and iterating through all columns from 'a' to 'h'. For each column, if it differs from the current column, we output the position formed by (column, row).

This mirrors the same logic in the orthogonal direction. Again, we skip the starting column to avoid duplication. 5. Output order does not matter, so vertical and horizontal results can be printed in any sequence. This removes any need for sorting or structured ordering.

Why it works

The rook’s movement is completely characterized by equality in exactly one coordinate: either the row is fixed or the column is fixed. Every valid destination must satisfy exactly one of these constraints while differing in the other coordinate. By iterating over all possible values in both dimensions and excluding the starting square, we enumerate all and only valid moves. No other square satisfies the rook’s movement rule, so the construction is both complete and exact.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input().strip())
    for _ in range(t):
        s = input().strip()
        col, row = s[0], s[1]

        # vertical moves
        for r in "12345678":
            if r != row:
                sys.stdout.write(col + r + "\n")

        # horizontal moves
        for c in "abcdefgh":
            if c != col:
                sys.stdout.write(c + row + "\n")

if __name__ == "__main__":
    solve()

The solution directly decomposes rook movement into two independent loops. The first loop fixes the column and varies the row, which generates all vertical moves. The second loop fixes the row and varies the column, which generates all horizontal moves. The conditional checks ensure the starting square is excluded exactly once in each direction, preventing duplication.

Using sys.stdin.readline and sys.stdout.write avoids overhead from repeated print calls, though in this problem the performance difference is not critical due to the tiny input size. The representation remains purely character-based, avoiding any need for integer coordinate conversion.

Worked Examples

Example Trace 1

Input:

d5

We track generation of moves:

Step Column Row Generated
Start d 5 -
Vertical loop r=1 d 1 d1
Vertical loop r=5 d 5 skip
Vertical loop r=8 d 8 d8
Horizontal loop c=a a 5 a5
Horizontal loop c=d d 5 skip
Horizontal loop c=h h 5 h5

This confirms that the algorithm produces exactly all squares sharing row 5 or column d, excluding (d,5).

Example Trace 2

Input:

a1
Step Column Row Generated
Start a 1 -
Vertical loop r=1 a 1 skip
Vertical loop r=8 a 8 a8
Horizontal loop c=a a 1 skip
Horizontal loop c=h h 1 h1

This demonstrates correct handling of a corner square where only 14 moves still exist but many loops are skipped at boundaries.

The trace confirms that even when the rook is on an edge or corner, the logic remains consistent and no invalid coordinates are produced.

Complexity Analysis

Measure Complexity Explanation
Time O(t) Each test case generates at most 14 outputs using fixed-size loops over 8 rows and 8 columns
Space O(1) No auxiliary storage beyond constant variables

The solution runs comfortably within limits since even in the worst case of 64 test cases, the total number of printed lines is bounded by a small constant factor.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output

    import sys as _sys
    input = _sys.stdin.readline

    def solve():
        t = int(input().strip())
        for _ in range(t):
            s = input().strip()
            col, row = s[0], s[1]

            for r in "12345678":
                if r != row:
                    print(col + r)

            for c in "abcdefgh":
                if c != col:
                    print(c + row)

    solve()
    return output.getvalue().strip()

# provided sample-like case
assert run("1\nd5\n") == "\n".join([
"d1","d2","d3","d4","d6","d7","d8",
"a5","b5","c5","e5","f5","g5","h5"
]), "sample 1"

# corner position
assert run("1\na1\n") == "\n".join([
"a2","a3","a4","a5","a6","a7","a8",
"b1","c1","d1","e1","f1","g1","h1"
])

# center position
assert run("1\ne4\n") is not None

# repeated same position
assert run("3\nd5\nd5\nd5\n") is not None
Test input Expected output What it validates
a1 all moves along row 1 and column a corner handling and skipping self
d5 full 14-move set general correctness
repeated d5 same output repeated independence of test cases

Edge Cases

The main edge case is ensuring the starting square is excluded even though it lies on both a valid row and a valid column. For input d5, the vertical generation produces d5 when iterating row 5, and the horizontal generation also produces d5 when iterating column d. The implementation explicitly checks inequality in both loops, so both occurrences are skipped. The output therefore never includes the origin.

Another case is board boundaries, such as a1. In the vertical loop, only rows 2 to 8 are emitted since 1 is skipped. In the horizontal loop, only columns b to h are emitted. No out-of-bounds coordinates can occur because iteration is strictly over fixed valid character sets, so there is no arithmetic indexing that could drift outside the board.

A final subtle point is duplicate logic across directions. Since row and column loops are independent, they cannot interfere with each other, and there is no possibility of generating invalid hybrid coordinates like mismatched symbols.