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Scaled Dot-Product Attention

Implement the attention operation at the heart of Transformers and large language models.

O que você vai aprender

  • Score queries against keys
  • Apply softmax to get attention weights
  • Mix values with those weights

Problema

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Attention lets each position in a sequence look at every other position and mix their information. It is the core of the Transformer architecture used by large language models:

Attention(Q, K, V) = softmax(Q × Kᵀ / √d) × V

For each query row q:

  1. Score every key row k: score = (q · k) / √d, where d is the number of columns of K.
  2. Turn the scores into weights with softmax (subtract the largest score before taking e^x).
  3. The output row is the weighted sum of the value rows: Σ weight_j × values[j].

Implement attention(queries, keys, values).

Input

queries: n rows, each with d numbers.

keys: m rows, each with d numbers.

values: m rows (one per key), each with v numbers.

Output

Return the n × v output matrix.

Restrições

1 ≤ n, m ≤ 50, 1 ≤ d, v ≤ 32

-10 ≤ each value ≤ 10

Exemplos

Exemplo 1
Entrada:
keys = [
    [1.0, 0.0],
    [1.0, 0.0]
]
values = [
    [2.0, 4.0],
    [6.0, 8.0]
]
queries = [
    [1.0, 0.0]
]

Saída esperada:
expected = [
    [4.0, 6.0]
]
Entrada:
std::vector<std::vector<double>> keys = {
    {1.0, 0.0},
    {1.0, 0.0}
};
std::vector<std::vector<double>> values = {
    {2.0, 4.0},
    {6.0, 8.0}
};
std::vector<std::vector<double>> queries = {
    {1.0, 0.0}
};

Saída esperada:
std::vector<std::vector<double>> expected = {
    {4.0, 6.0}
};
Entrada:
double[][] keys = new double[][] {
    new double[] {1.0, 0.0},
    new double[] {1.0, 0.0}
};
double[][] values = new double[][] {
    new double[] {2.0, 4.0},
    new double[] {6.0, 8.0}
};
double[][] queries = new double[][] {
    new double[] {1.0, 0.0}
};

Saída esperada:
double[][] expected = new double[][] {
    new double[] {4.0, 6.0}
};

Explicação

Both keys give the same score, so the softmax weights are 0.5 and 0.5. The output is the average of the value rows: [(2 + 6) / 2, (4 + 8) / 2] = [4, 6].

Exemplo 2
Entrada:
keys = [
    [3.0, 1.0],
    [-2.0, 5.0],
    [7.0, 7.0]
]
values = [
    [1.0],
    [2.0],
    [6.0]
]
queries = [
    [0.0, 0.0]
]

Saída esperada:
expected = [
    [3.0]
]
Entrada:
std::vector<std::vector<double>> keys = {
    {3.0, 1.0},
    {-2.0, 5.0},
    {7.0, 7.0}
};
std::vector<std::vector<double>> values = {
    {1.0},
    {2.0},
    {6.0}
};
std::vector<std::vector<double>> queries = {
    {0.0, 0.0}
};

Saída esperada:
std::vector<std::vector<double>> expected = {
    {3.0}
};
Entrada:
double[][] keys = new double[][] {
    new double[] {3.0, 1.0},
    new double[] {-2.0, 5.0},
    new double[] {7.0, 7.0}
};
double[][] values = new double[][] {
    new double[] {1.0},
    new double[] {2.0},
    new double[] {6.0}
};
double[][] queries = new double[][] {
    new double[] {0.0, 0.0}
};

Saída esperada:
double[][] expected = new double[][] {
    new double[] {3.0}
};

Explicação

A zero query scores 0 against every key, so attention is spread evenly: (1 + 2 + 6) / 3 = 3.

Referências

Este desafio foi adaptado para o CodeUtility. Siga o enunciado e as restrições acima.

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