How to implement the Softmax derivative independently from any loss function?

Question

For a neural networks library I implemented some activation functions and loss functions and their derivatives. They can be combined arbitrarily and the derivative at the output layers just becomes the product of the loss derivative and the activation derivative.

However, I failed to implement the derivative of the Softmax activation function independently from any loss function. Due to the normalization i.e. the denominator in the equation, changing a single input activation changes all output activations and not just one.

Here is my Softmax implementation where the derivative fails the gradient checking by about 1%. How can I implement the Softmax derivative so that it can be combined with any loss function?

import numpy as np


class Softmax:

    def compute(self, incoming):
        exps = np.exp(incoming)
        return exps / exps.sum()

    def delta(self, incoming, outgoing):
        exps = np.exp(incoming)
        others = exps.sum() - exps
        return 1 / (2 + exps / others + others / exps)


activation = Softmax()
cost = SquaredError()

outgoing = activation.compute(incoming)
delta_output_layer = activation.delta(incoming) * cost.delta(outgoing)

Answer 1

Mathematically, the derivative of Softmax σ(j) with respect to the logit Zi (for example, Wi*X) is

where the red delta is a Kronecker delta.

If you implement this iteratively in python:

def softmax_grad(s):
    # input s is softmax value of the original input x. Its shape is (1,n) 
    # i.e.  s = np.array([0.3,0.7]),  x = np.array([0,1])

    # make the matrix whose size is n^2.
    jacobian_m = np.diag(s)

    for i in range(len(jacobian_m)):
        for j in range(len(jacobian_m)):
            if i == j:
                jacobian_m[i][j] = s[i] * (1 - s[i])
            else: 
                jacobian_m[i][j] = -s[i] * s[j]
    return jacobian_m

Test:

In [95]: x
Out[95]: array([1, 2])

In [96]: softmax(x)
Out[96]: array([ 0.26894142,  0.73105858])

In [97]: softmax_grad(softmax(x))
Out[97]: 
array([[ 0.19661193, -0.19661193],
       [-0.19661193,  0.19661193]])

If you implement in a vectorized version:

soft_max = softmax(x)    

# reshape softmax to 2d so np.dot gives matrix multiplication

def softmax_grad(softmax):
    s = softmax.reshape(-1,1)
    return np.diagflat(s) - np.dot(s, s.T)

softmax_grad(soft_max)

#array([[ 0.19661193, -0.19661193],
#       [-0.19661193,  0.19661193]])

Source: https://medium.com/intuitionmath/how-to-implement-the-softmax-derivative-independently-from-any-loss-function-ae6d44363a9d

Answer 2

The other answers are great, here to share a simple implementation of forward/backward, regardless of loss functions.

In the image below, it is a brief derivation of the backward for softmax. The 2nd equation is loss function dependent, not part of our implementation.

backward verified by manual grad checking.

import numpy as np

class Softmax:
    def forward(self, x):
        mx = np.max(x, axis=1, keepdims=True)
        x = x - mx  # log-sum-exp trick
        e = np.exp(x)
        probs = e / np.sum(np.exp(x), axis=1, keepdims=True)
        return probs

    def backward(self, x, probs, bp_err):
        dim = x.shape[1]
        output = np.empty(x.shape)
        for j in range(dim):
            d_prob_over_xj = - (probs * probs[:,[j]])  # i.e. prob_k * prob_j, no matter k==j or not
            d_prob_over_xj[:,j] += probs[:,j]   # i.e. when k==j, +prob_j
            output[:,j] = np.sum(bp_err * d_prob_over_xj, axis=1)
        return output

def compute_manual_grads(x, pred_fn):
    eps = 1e-3
    batch_size, dim = x.shape

    grads = np.empty(x.shape)
    for i in range(batch_size):
        for j in range(dim):
            x[i,j] += eps
            y1 = pred_fn(x)

            x[i,j] -= 2*eps
            y2 = pred_fn(x)

            grads[i,j] = (y1 - y2) / (2*eps)
            x[i,j] += eps
    return grads

def loss_fn(probs, ys, loss_type):
    batch_size = probs.shape[0]
    # dummy mse
    if loss_type=="mse":
        loss = np.sum((np.take_along_axis(probs, ys.reshape(-1,1), axis=1) - 1)**2) / batch_size
        values = 2 * (np.take_along_axis(probs, ys.reshape(-1,1), axis=1) - 1) / batch_size

    # cross ent
    if loss_type=="xent":
        loss = - np.sum( np.take_along_axis(np.log(probs), ys.reshape(-1,1), axis=1) ) / batch_size
        values = -1 / np.take_along_axis(probs, ys.reshape(-1,1), axis=1) / batch_size

    err = np.zeros(probs.shape)
    np.put_along_axis(err, ys.reshape(-1,1), values, axis=1)
    return loss, err

if __name__ == "__main__":
    batch_size = 10
    dim = 5

    x = np.random.rand(batch_size, dim)
    ys = np.random.randint(0, dim, batch_size)
    for loss_type in ["mse", "xent"]:
        S = Softmax()
        probs = S.forward(x)
        loss, bp_err = loss_fn(probs, ys, loss_type)

        grads = S.backward(x, probs, bp_err)

        def pred_fn(x, ys):
            pred = S.forward(x)
            loss, err = loss_fn(pred, ys, loss_type)
            return loss

        manual_grads = compute_manual_grads(x, lambda x: pred_fn(x, ys))

        # compare both grads
        print(f"loss_type = {loss_type}, grad diff = {np.sum((grads - manual_grads)**2) / batch_size}")

Answer 3

Just in case you are processing in batches, here is an implementation in NumPy (tested vs TensorFlow). However, I will suggest avoiding the associated tensor operations, by mixing the jacobian with the cross-entropy, which leads to a very simple and efficient expression.

def softmax(z):
  exps = np.exp(z - np.max(z))
  return exps / np.sum(exps, axis=1, keepdims=True)

def softmax_jacob(s):
  return np.einsum('ij,jk->ijk', s, np.eye(s.shape[-1])) \
       - np.einsum('ij,ik->ijk', s, s)

def np_softmax_test(z):
  return softmax_jacob(softmax(z))

def tf_softmax_test(z):
  z = tf.constant(z, dtype=tf.float32)
  with tf.GradientTape() as g:
    g.watch(z)
    a = tf.nn.softmax(z) 
  jacob = g.batch_jacobian(a, z)
  return jacob.numpy()

z = np.random.randn(3, 5)
np.all(np.isclose(np_softmax_test(z), tf_softmax_test(z)))

Answer 4

Here is a c++ vectorized version, using intrinsics ( 22 times (!) faster than the non-SSE version):

// How many floats fit into __m256 "group".
// Used by vectors and matrices, to ensure their dimensions are appropriate for 
// intrinsics.
// Otherwise, consecutive rows of matrices will not be 16-byte aligned, and 
// operations on them will be incorrect.
#define F_MULTIPLE_OF_M256 8


//check to quickly see if your rows are divisible by m256.
//you can 'undefine' to save performance, after everything was verified to be correct.
#define ASSERT_THE_M256_MULTIPLES
#ifdef ASSERT_THE_M256_MULTIPLES
    #define assert_is_m256_multiple(x)  assert( (x%F_MULTIPLE_OF_M256) == 0)
#else
    #define assert_is_m256_multiple (q) 
#endif


// usually used at the end of our Reduce functions,
// where the final __m256 mSum needs to be collapsed into 1 scalar.
static inline float slow_hAdd_ps(__m256 x){
    const float *sumStart = reinterpret_cast<const float*>(&x);
    float sum = 0.0f;

    for(size_t i=0; i<F_MULTIPLE_OF_M256; ++i){
        sum += sumStart[i];
    }
    return sum;
}



f_vec SoftmaxGrad_fromResult(const float *softmaxResult,  size_t size,  
                             const float *gradFromAbove){//<--gradient vector, flowing into us from the above layer
assert_is_m256_multiple(size);
//allocate vector, where to store output:
f_vec grad_v(size, true);//true: skip filling with zeros, to save performance.

const __m256* end   = (const __m256*)(softmaxResult + size);


for(size_t i=0; i<size; ++i){// <--for every row
    //go through this i'th row:
    __m256 sum =  _mm256_set1_ps(0.0f);

    const __m256 neg_sft_i  =  _mm256_set1_ps( -softmaxResult[i] );
    const __m256 *s  =  (const __m256*)softmaxResult;
    const __m256 *gAbove  =   (__m256*)gradFromAbove;

    for (s;  s<end; ){
        __m256 mul =  _mm256_mul_ps(*s, neg_sft_i);  //  sftmaxResult_j  *  (-sftmaxResult_i)
        mul =  _mm256_mul_ps( mul, *gAbove );

        sum =  _mm256_add_ps( sum,  mul );//adding to the total sum of this row.
        ++s;
        ++gAbove;
    }
    grad_v[i]  =  slow_hAdd_ps( sum );//collapse the sum into 1 scalar (true sum of this row).
}//end for every row

//reset back to start and subtract a vector, to account for Kronecker delta:
__m256 *g =  (__m256*)grad_v._contents;
__m256 *s =  (__m256*)softmaxResult;
__m256 *gAbove =  (__m256*)gradFromAbove;

for(s; s<end; ){
    __m256 mul = _mm256_mul_ps(*s, *gAbove);
    *g = _mm256_add_ps( *g, mul );
    ++s; 
    ++g;
}

return grad_v;

}

---

If for some reason somebody wants a simple (non-SSE) version, here it is:

inline static void SoftmaxGrad_fromResult_nonSSE(const float* softmaxResult,  
                                                 const float *gradFromAbove,  //<--gradient vector, flowing into us from the above layer
                                                 float *gradOutput,  
                                                 size_t count ){
    // every pre-softmax element in a layer contributed to the softmax of every other element
    // (it went into the denominator). So gradient will be distributed from every post-softmax element to every pre-elem.
    for(size_t i=0; i<count; ++i){
        //go through this i'th row:
        float sum =  0.0f;

        const float neg_sft_i  =  -softmaxResult[i];

        for(size_t j=0; j<count; ++j){
            float mul =  gradFromAbove[j] * softmaxResult[j] * neg_sft_i;
            sum +=  mul;//adding to the total sum of this row.
        }
        //NOTICE: equals, overwriting any old values:
        gradOutput[i]  =  sum;
    }//end for every row

    for(size_t i=0; i<count; ++i){
        gradOutput[i] +=  softmaxResult[i] * gradFromAbove[i];
    }
}

Answer 5

It should be like this: (x is the input to the softmax layer and dy is the delta coming from the loss above it)

    dx = y * dy
    s = dx.sum(axis=dx.ndim - 1, keepdims=True)
    dx -= y * s

    return dx

But the way you compute the error should be:

    yact = activation.compute(x)
    ycost = cost.compute(yact)
    dsoftmax = activation.delta(x, cost.delta(yact, ycost, ytrue)) 

Explanation: Because the delta function is a part of the backpropagation algorithm, its responsibility is to multiply the vector dy (in my code, outgoing in your case) by the Jacobian of the compute(x) function evaluated at x. If you work out what does this Jacobian look like for softmax [1], and then multiply it from the left by a vector dy, after a bit of algebra you'll find out that you get something that corresponds to my Python code.

[1] https://stats.stackexchange.com/questions/79454/softmax-layer-in-a-neural-network