FastPrepHandwritten Softmax
Problem · Array

Handwritten Softmax

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Problem statement

Given a nonempty double array logits, return its softmax probability array in the same order.

Let m be the maximum logit. For every index i, compute:

probability[i] = exp(logits[i] - m) / sum(exp(logits[j] - m))

Subtracting m is required for numerical stability and does not change the mathematical softmax. Implement the loops directly without a machine-learning or numerical-array library.

Function

softmax(logits: double[]) → double[]

Examples

Example 1

logits = [1.0,2.0,3.0]return = [0.09003057317038046,0.24472847105479764,0.6652409557748218]

After subtracting the maximum 3, normalize [exp(-2), exp(-1), 1] by their sum.

Example 2

logits = [1000.0,1000.0]return = [0.5,0.5]

Equal logits have equal exponentials after maximum subtraction, so they split the probability mass evenly.

Example 3

logits = [5.0]return = [1.0]

A one-element array has all of the probability mass at its only index.

Constraints

  • 1 <= logits.length <= 200
  • -1000 <= logits[i] <= 1000
  • Every logit is finite: no value is NaN or infinity.
  • The result is compared element by element with absolute tolerance 1e-12.

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public double[] softmax(double[] logits) {
    // write your code here
}
logits[1.0,2.0,3.0]
expected[0.09003057317038046,0.24472847105479764,0.6652409557748218]
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