CF 102697158 - Dot Product

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Solve time: 58s
Verified: yes

Solution

Problem Understanding

The task is ordinary matrix multiplication. The input describes two rectangular matrices, m1 and m2. The first four integers give the width and height of the two matrices, in that order. The rows of m1 are provided first, followed immediately by all rows of m2.

Suppose m1 has height H1 and width W1, while m2 has height H2 and width W2. For matrix multiplication to be defined, the number of columns of m1 must equal the number of rows of m2, so W1 = H2. The result has H1 rows and W2 columns. Its entry at row i and column j is obtained by taking row i of m1, column j of m2, multiplying corresponding values, and adding those products.

For example, with

m1 = [1 2]
     [3 4]

m2 = [5 6]
     [7 8]

the top-left result is 1 * 5 + 2 * 7 = 19, while the bottom-right result is 3 * 6 + 4 * 8 = 50. The complete result is consequently

19 22
43 50

The published statement does not give explicit numerical upper bounds for the matrix dimensions or element values. What it does specify is a one-second time limit and 256 MB of memory. Since the direct matrix multiplication algorithm performs one multiplication and one addition for every triple consisting of a result row, a shared dimension position, and a result column, its running time is O(H1 * W1 * W2). With no stronger dimension constraints available in the statement, there is no justified subquadratic or optimized matrix multiplication requirement to infer. The intended solution is the standard triple-loop multiplication.

There are several input situations that can expose mistakes in a careless implementation. A matrix with only one row or one column is a simple boundary case. For example,

1 1 1 1
7
6

produces

because there is exactly one pair of values to multiply. An implementation that assumes at least two rows or columns can fail here.

A rectangular matrix is another common source of indexing errors. For example,

produces

22 28

because the single row [1, 2, 3] is multiplied against each column of the second matrix. Code that accidentally treats the first dimension as the number of rows rather than the width will read the matrices incorrectly.

Zero values also matter because the result can contain zero even when the matrices themselves are non-empty. For example,

produces

0 0

A careless implementation that initializes each result cell from the first product rather than from zero can get such cases wrong.

Negative values are another useful boundary check. For example,

produces

-5 -7

since the two entries are 1 * 3 + (-2) * 4 = -5 and 1 * 4 + (-2) * 5 only if the second matrix has a second row, which this particular input does not. The valid rectangular interpretation is determined by the dimensions, so dimensions and row lengths must always be kept consistent when constructing tests.

Approaches

The direct approach follows the mathematical definition of matrix multiplication exactly. For every result cell (i, j), iterate through the shared dimension k and add m1[i][k] * m2[k][j] to the answer. This is correct because the definition of matrix multiplication is precisely that dot product between the corresponding row and column.

If m1 has dimensions H1 × W1 and m2 has dimensions W1 × W2, there are H1 * W2 result cells. Computing one cell examines all W1 shared positions, so the total number of multiplication operations is exactly H1 * W1 * W2. The same order of additions is performed. For square matrices of size n, this becomes n^3 operations.

The brute-force description and the optimal solution are effectively the same here. There is no additional structure in the matrices that the problem asks us to exploit, and the statement provides no special restrictions such as sparsity, binary values, or repeated queries. The key observation is simply that each result entry is independent once the two input matrices have been read. We can compute every entry directly from its corresponding row and column.

An implementation can either explicitly construct the transpose of m2 and take row-row dot products, or access m2[k][j] directly inside the innermost loop. The latter avoids storing another matrix and is already straightforward enough for the dimensions expected by this basic problem.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	`O(H1 * W1 * W2)`	`O(H1 * W1 + W1 * W2 + H1 * W2)`	Accepted
Optimal	`O(H1 * W1 * W2)`	`O(H1 * W1 + W1 * W2 + H1 * W2)`	Accepted

Here "brute force" and "optimal" have the same asymptotic complexity because the problem directly asks for the complete matrix product. The useful optimization is implementation clarity and avoiding unnecessary work, not a different asymptotic algorithm.

Algorithm Walkthrough

Read W1, H1, W2, and H2. The width is the number of values in each row, while the height is the number of rows, so m1 has H1 rows and m2 has H2 rows. Matrix multiplication requires W1 = H2.
Read H1 rows, each containing W1 values, and store them as m1. Keeping the rows intact makes accessing m1[i][k] direct.
Read H2 rows, each containing W2 values, and store them as m2. We need direct access to m2[k][j] while computing result cell (i, j).
Create a result matrix with H1 rows and W2 columns, initially filled with zero. Every result entry is a sum of products, so zero is the correct starting value even when some matrix values are negative.
For every result row i, iterate over every result column j. These two indices completely identify one cell of the output matrix.
For the chosen (i, j), iterate k from 0 through W1 - 1 and add m1[i][k] * m2[k][j] to the result. The index k selects matching components from row i of the first matrix and column j of the second matrix.
After all k values have been processed, the result cell contains exactly the dot product of the selected row and column. Repeat for every (i, j) and print the completed matrix.

Why it works

The invariant for the innermost loop is that after processing positions 0 through k - 1, the current result cell contains

m1[i][0] * m2[0][j] + ... + m1[i][k-1] * m2[k-1][j].

Initially no positions have been processed, so the sum is zero. Each iteration adds the next required product, preserving the invariant. After all W1 positions have been processed, the cell contains the complete dot product of row i of m1 and column j of m2. Since the algorithm performs this independently for every output position, every cell of the produced matrix has its mathematically correct value.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    w1, h1, w2, h2 = map(int, input().split())

    m1 = [list(map(int, input().split())) for _ in range(h1)]
    m2 = [list(map(int, input().split())) for _ in range(h2)]

    result = [[0] * w2 for _ in range(h1)]

    for i in range(h1):
        for j in range(w2):
            total = 0
            for k in range(w1):
                total += m1[i][k] * m2[k][j]
            result[i][j] = total

    sys.stdout.write(
        "\n".join(" ".join(map(str, row)) for row in result)
    )

if __name__ == "__main__":
    solve()

The first line is read as width, height for each matrix. This order follows the actual input format, which gives width before height. Converting those values into w1, h1, w2, and h2 prevents the common mistake of reversing rows and columns.

The first list comprehension reads exactly h1 rows for m1, and each row is expected to contain w1 values. The second does the same for m2. Because the second matrix has h2 = w1 rows, m2[k] is valid for every k used by the multiplication loop.

The result matrix has h1 rows and w2 columns because matrix multiplication preserves the row count of the first matrix and the column count of the second. Every cell starts at zero before its dot product is accumulated.

The innermost expression uses m1[i][k] * m2[k][j]. The first index pair walks across a row of m1, while the second accesses a column of m2. Swapping either of these indices is a classic matrix multiplication bug.

Python integers have arbitrary precision, so there is no integer overflow issue from the multiplication or accumulation.

The final join produces exactly one output row per result row, with spaces between values. It also avoids an unnecessary trailing space at the end of each line.

Worked Examples

Sample 1

The input contains two 2 × 2 matrices.

For each result position, the algorithm accumulates the products from the shared dimension.

`i`	`j`	`k`	`m1[i][k]`	`m2[k][j]`	Accumulated value
0	0	0	1	5	5
0	0	1	2	7	19
0	1	0	1	6	6
0	1	1	2	8	22
1	0	0	3	5	15
1	0	1	4	7	43
1	1	0	3	6	18
1	1	1	4	8	50

The final matrix is

19 22
43 50

The trace demonstrates the core invariant: every result cell starts at zero and receives exactly one product for every position in the shared dimension.

Example 2

Consider a rectangular multiplication:

Here m1 is 2 × 3 and m2 is 3 × 2, so the result is 2 × 2.

`i`	`j`	`k`	Product	Accumulated value
0	0	0	`1 * 7 = 7`	7
0	0	1	`2 * 9 = 18`	25
0	0	2	`3 * 11 = 33`	58
0	1	0	`1 * 8 = 8`	8
0	1	1	`2 * 10 = 20`	28
0	1	2	`3 * 12 = 36`	64
1	0	0	`4 * 7 = 28`	28
1	0	1	`5 * 9 = 45`	73
1	0	2	`6 * 11 = 66`	139
1	1	0	`4 * 8 = 32`	32
1	1	1	`5 * 10 = 50`	82
1	1	2	`6 * 12 = 72`	154

The output is

58 64
139 154

This example exercises the distinction between width and height. The first matrix has three values per row but only two rows, while the second has two values per row and three rows. The loop bounds follow those dimensions rather than assuming that the matrices are square.

Complexity Analysis

Measure	Complexity	Explanation
Time	`O(H1 * W1 * W2)`	There are `H1 * W2` result cells and each requires `W1` multiply-add operations.
Space	`O(H1 * W1 + W1 * W2 + H1 * W2)`	The two input matrices and the result matrix are stored.

The published problem page gives a one-second time limit and 256 MB memory limit, but does not expose explicit numerical dimension bounds. The implementation uses the standard cubic-style matrix multiplication for the given rectangular dimensions, with no unnecessary auxiliary matrix.

Test Cases

The statement provides one sample. Since the public statement does not publish numerical maximum dimensions, a genuinely maximum-size test cannot be specified from the available constraints. The stress case below instead uses a substantially larger rectangular multiplication while preserving the required dimensions.

import sys
import io

def solve():
    input = sys.stdin.readline

    w1, h1, w2, h2 = map(int, input().split())

    m1 = [list(map(int, input().split())) for _ in range(h1)]
    m2 = [list(map(int, input().split())) for _ in range(h2)]

    result = [[0] * w2 for _ in range(h1)]

    for i in range(h1):
        for j in range(w2):
            total = 0
            for k in range(w1):
                total += m1[i][k] * m2[k][j]
            result[i][j] = total

    return "\n".join(" ".join(map(str, row)) for row in result)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(inp)
    try:
        return solve()
    finally:
        sys.stdin = old_stdin

# Provided sample
assert run("""2 2 2 2
1 2
3 4
5 6
7 8
""") == """19 22
43 50""", "sample 1"

# Minimum-size matrices
assert run("""1 1 1 1
7
6
""") == """42""", "minimum-size case"

# All-zero values
assert run("""2 2 2 2
0 0
0 0
5 6
7 8
""") == """0 0
0 0""", "zero matrix"

# All-equal values
assert run("""3 2 2 3
2 2 2
2 2 2
3 3
3 3
3 3
""") == """18 18
18 18""", "all-equal values"

# Rectangular dimensions and negative values
assert run("""3 1 2 3
1 -2 3
4 5
6 -7
8 9
""") == """16 46""", "rectangular and negative values"

# Stress case: 30 x 40 multiplied by 40 x 20.
# The expected result is computed independently from the simple pattern.
h1, w1, h2, w2 = 30, 40, 40, 20
m1 = [[1] * w1 for _ in range(h1)]
m2 = [[2] * w2 for _ in range(h2)]

stress_lines = [f"{w1} {h1} {w2} {h2}"]
stress_lines += [" ".join(map(str, row)) for row in m1]
stress_lines += [" ".join(map(str, row)) for row in m2]

expected_row = " ".join(["80"] * w2)
expected = "\n".join([expected_row] * h1)

assert run("\n".join(stress_lines) + "\n") == expected, "rectangular stress case"

print("All tests passed.")

Test input	Expected output	What it validates
`1 1 1 1 / 7 / 6`	`42`	Smallest possible matrix product and boundary indices
`2 2 2 2` with zero first matrix	`0 0 / 0 0`	Correct zero initialization
`3 2 2 3` with all values equal	`18 18 / 18 18`	Full traversal of the shared dimension
`3 1 2 3` with mixed signs	`16 46`	Rectangular matrices and negative values
`30 40 20 40` with constant values	Every entry `80`	Larger rectangular multiplication and indexing

Edge Cases

A 1 × 1 multiplication reduces the entire algorithm to a single iteration. For

1 1 1 1
7
6

the loops select i = 0, j = 0, and k = 0. The accumulator becomes 7 * 6 = 42, so the output is

There are no special cases in the implementation because the ordinary loop boundaries already handle the smallest dimensions.

A rectangular matrix catches code that confuses width and height. Consider

For result cell (0, 0), the algorithm accesses (1, 7), (2, 9), and (3, 11) across the shared dimension, producing 58. The result dimensions are 2 × 2, not 3 × 3 or 2 × 3. The output is

58 64
139 154

The loop structure derives these dimensions directly from h1 and w2, so no square-matrix assumption is made.

A zero-valued matrix checks the initialization of the accumulator. With

every product is zero. Each accumulator remains zero through the complete inner loop, producing

0 0
0 0

If an implementation initialized an accumulator from an input product rather than from zero, this case could expose the error.

Finally, the dimensions must be interpreted consistently. For a 3 × 1 matrix multiplied by a 3 × 2 matrix, the input must be written as

The first matrix has one row containing three values, and the second has three rows containing two values each. The first output entry is 1 * 4 + 2 * 6 + 3 * 8 = 40, while the second is 1 * 5 + 2 * 7 + 3 * 9 = 46. The output is

40 46

The implementation gets the boundary exactly right because k ranges over range(w1), while the second matrix is indexed as m2[k][j]. This is the central indexing relationship in matrix multiplication.