Problema
A loss function turns a model's mistakes into one number to minimize. Implement compute_loss(kind, predictions, targets) for three common losses, where n is the number of predictions:
"mse"(mean squared error): average of (p - t)²"mae"(mean absolute error): average of |p - t|"binary_cross_entropy": -average of [t × ln(p) + (1 - t) × ln(1 - p)]. First clip each prediction p to the range [0.0000001, 0.9999999] so the logarithm is never ln(0).
Input
kind: "mse", "mae" or "binary_cross_entropy".
predictions: The model's predictions.
targets: The correct values (same length). For cross-entropy they are 0 or 1 and predictions are between 0 and 1.
Output
Return the loss as a decimal number.
Restricciones
1 ≤ length ≤ 1,000
-1,000 ≤ each value ≤ 1,000 (0 ≤ p ≤ 1 and t ∈ {0, 1} for cross-entropy)
Ejemplos
Entrada:
kind = "mse"
targets = [3.0, -0.5, 2.0]
predictions = [2.5, 0.0, 2.0]
Salida esperada:
expected = 0.166667Entrada:
std::string kind = "mse";
std::vector<double> targets = {3.0, -0.5, 2.0};
std::vector<double> predictions = {2.5, 0.0, 2.0};
Salida esperada:
double expected = 0.166667;Entrada:
string kind = "mse";
double[] targets = new double[] {3.0, -0.5, 2.0};
double[] predictions = new double[] {2.5, 0.0, 2.0};
Salida esperada:
double expected = 0.166667;Explicación
Squared errors: 0.5² = 0.25, 0.5² = 0.25 and 0² = 0. Their average is 0.5 / 3 ≈ 0.166667.
Entrada:
kind = "binary_cross_entropy"
targets = [1.0, 0.0]
predictions = [0.9, 0.2]
Salida esperada:
expected = 0.164252Entrada:
std::string kind = "binary_cross_entropy";
std::vector<double> targets = {1.0, 0.0};
std::vector<double> predictions = {0.9, 0.2};
Salida esperada:
double expected = 0.164252;Entrada:
string kind = "binary_cross_entropy";
double[] targets = new double[] {1.0, 0.0};
double[] predictions = new double[] {0.9, 0.2};
Salida esperada:
double expected = 0.164252;Explicación
-(ln(0.9) + ln(1 - 0.2)) / 2 = -(-0.105361 - 0.223144) / 2 ≈ 0.164252. Confident correct predictions give a small loss.
Cómo abordarlo
Inténtalo primero por tu cuenta y luego abre las pistas una a una.
Pista 1
Each loss compares predictions to targets element by element and then averages the per-example errors.
Pista 2
For cross-entropy, clamp predictions away from 0 and 1 before taking logarithms.
Pista 3
Remember that absolute value and squaring are different, and cross-entropy uses the target as a weight on each log term.
Complejidad esperada
O(n) time, O(1) extra space
Referencias
Este reto está adaptado para CodeUtility. Sigue el enunciado y las restricciones indicadas arriba.