समस्या
Multi-head attention lets a Transformer attend to different things at once. With input X (n tokens × d columns) and weight matrices of size d × d:
- Q = X × w_q, K = X × w_k, V = X × w_v
- Split the d columns into
num_headsheads of d_k = d / num_heads consecutive columns (head h uses columns h×d_k to (h+1)×d_k - 1). - For each head, run scaled dot-product attention on its columns: for token i, score_j = (Q_i · K_j) / √d_k over that head's columns, softmax over all n tokens j, and output = Σ weight_j × V_j (that head's columns).
- Put the heads' outputs back in their columns (concatenate) and multiply by w_o.
Implement multi_head_attention(x, w_q, w_k, w_v, w_o, num_heads).
Input
x: n × d input matrix.
w_q, w_k, w_v, w_o: d × d weight matrices.
num_heads: Number of heads (d is divisible by it).
Output
Return the n × d output matrix.
सीमाएँ
1 ≤ n ≤ 30, 1 ≤ d ≤ 32, 1 ≤ num_heads ≤ d and d is divisible by num_heads
-5 ≤ each value ≤ 5
उदाहरण
इनपुट:
x = [
[1.0, 0.0],
[0.0, 1.0]
]
w_k = [
[1.0, 0.0],
[0.0, 1.0]
]
w_o = [
[1.0, 0.0],
[0.0, 1.0]
]
w_q = [
[0.0, 0.0],
[0.0, 0.0]
]
w_v = [
[1.0, 0.0],
[0.0, 1.0]
]
num_heads = 1
अपेक्षित आउटपुट:
expected = [
[0.5, 0.5],
[0.5, 0.5]
]इनपुट:
std::vector<std::vector<double>> x = {
{1.0, 0.0},
{0.0, 1.0}
};
std::vector<std::vector<double>> w_k = {
{1.0, 0.0},
{0.0, 1.0}
};
std::vector<std::vector<double>> w_o = {
{1.0, 0.0},
{0.0, 1.0}
};
std::vector<std::vector<double>> w_q = {
{0.0, 0.0},
{0.0, 0.0}
};
std::vector<std::vector<double>> w_v = {
{1.0, 0.0},
{0.0, 1.0}
};
long long num_heads = 1;
अपेक्षित आउटपुट:
std::vector<std::vector<double>> expected = {
{0.5, 0.5},
{0.5, 0.5}
};इनपुट:
double[][] x = new double[][] {
new double[] {1.0, 0.0},
new double[] {0.0, 1.0}
};
double[][] w_k = new double[][] {
new double[] {1.0, 0.0},
new double[] {0.0, 1.0}
};
double[][] w_o = new double[][] {
new double[] {1.0, 0.0},
new double[] {0.0, 1.0}
};
double[][] w_q = new double[][] {
new double[] {0.0, 0.0},
new double[] {0.0, 0.0}
};
double[][] w_v = new double[][] {
new double[] {1.0, 0.0},
new double[] {0.0, 1.0}
};
long num_heads = 1;
अपेक्षित आउटपुट:
double[][] expected = new double[][] {
new double[] {0.5, 0.5},
new double[] {0.5, 0.5}
};व्याख्या
w_q is all zeros, so every score is 0 and each token averages the value rows: V = X, so both rows become [0.5, 0.5].
इनपुट:
x = [
[2.0, 4.0]
]
w_k = [
[1.0, 0.0],
[0.0, 1.0]
]
w_o = [
[0.0, 1.0],
[1.0, 0.0]
]
w_q = [
[1.0, 0.0],
[0.0, 1.0]
]
w_v = [
[1.0, 0.0],
[0.0, 1.0]
]
num_heads = 2
अपेक्षित आउटपुट:
expected = [
[4.0, 2.0]
]इनपुट:
std::vector<std::vector<double>> x = {
{2.0, 4.0}
};
std::vector<std::vector<double>> w_k = {
{1.0, 0.0},
{0.0, 1.0}
};
std::vector<std::vector<double>> w_o = {
{0.0, 1.0},
{1.0, 0.0}
};
std::vector<std::vector<double>> w_q = {
{1.0, 0.0},
{0.0, 1.0}
};
std::vector<std::vector<double>> w_v = {
{1.0, 0.0},
{0.0, 1.0}
};
long long num_heads = 2;
अपेक्षित आउटपुट:
std::vector<std::vector<double>> expected = {
{4.0, 2.0}
};इनपुट:
double[][] x = new double[][] {
new double[] {2.0, 4.0}
};
double[][] w_k = new double[][] {
new double[] {1.0, 0.0},
new double[] {0.0, 1.0}
};
double[][] w_o = new double[][] {
new double[] {0.0, 1.0},
new double[] {1.0, 0.0}
};
double[][] w_q = new double[][] {
new double[] {1.0, 0.0},
new double[] {0.0, 1.0}
};
double[][] w_v = new double[][] {
new double[] {1.0, 0.0},
new double[] {0.0, 1.0}
};
long num_heads = 2;
अपेक्षित आउटपुट:
double[][] expected = new double[][] {
new double[] {4.0, 2.0}
};व्याख्या
With a single token each head attends only to itself, so the concatenated output is V = [2, 4]. w_o swaps the two columns: [4, 2].
कैसे हल करें
पहले खुद कोशिश करें, फिर एक-एक करके संकेत खोलें।
संकेत 1
Project the input into queries, keys, and values first, then split the feature dimension into equal head-sized chunks.
संकेत 2
Run the same attention procedure inside each head independently, using the smaller head dimension for scaling.
संकेत 3
Concatenate head outputs in their original column order before the final output projection, and keep matrix dimensions aligned carefully.
अपेक्षित जटिलता
O(n^2 * d + n * d^2) time, O(n * d) extra space
संदर्भ
यह चुनौती CodeUtility के लिए अनुकूलित की गई है। ऊपर दिए गए स्टेटमेंट और सीमाओं का पालन करें।