Problème
The softmax function turns a list of scores (logits) into probabilities. Language models use it to choose the next word:
softmax(x)_i = e^(x_i) / (e^(x_1) + e^(x_2) + … + e^(x_n))
Large scores make e^x overflow, so first subtract the largest score from every score. The result is mathematically the same.
Implement softmax(logits).
Input
logits: A list of scores.
Output
Return the list of probabilities, in the same order. They are all between 0 and 1 and add up to 1.
Contraintes
1 ≤ length ≤ 1,000
-1,000 ≤ each score ≤ 1,000
Exemples
Entrée :
logits = [1.0, 2.0, 3.0]
Sortie attendue :
expected = [0.090031, 0.244728, 0.665241]Entrée :
std::vector<double> logits = {1.0, 2.0, 3.0};
Sortie attendue :
std::vector<double> expected = {0.090031, 0.244728, 0.665241};Entrée :
double[] logits = new double[] {1.0, 2.0, 3.0};
Sortie attendue :
double[] expected = new double[] {0.090031, 0.244728, 0.665241};Explication
e¹ ≈ 2.718, e² ≈ 7.389 and e³ ≈ 20.086 add up to 30.193, so the probabilities are about [0.090031, 0.244728, 0.665241].
Entrée :
logits = [1000.0, 1000.0]
Sortie attendue :
expected = [0.5, 0.5]Entrée :
std::vector<double> logits = {1000.0, 1000.0};
Sortie attendue :
std::vector<double> expected = {0.5, 0.5};Entrée :
double[] logits = new double[] {1000.0, 1000.0};
Sortie attendue :
double[] expected = new double[] {0.5, 0.5};Explication
Computing e^1000 directly overflows. Subtracting the maximum gives [0, 0], and e⁰ / (e⁰ + e⁰) = 0.5 for both.
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.