目前，我已经实施了以下受NIPS 2016 发布的 N-Pair Loss启发的解决方案：

import torch
from torch import nn
from matplotlib import pyplot as plt
import seaborn as sn


class NPairsLoss(nn.Module):
    """
    The N-Pairs Loss.
    It measures the loss given predicted tensors x1, x2 both with shape [batch_size, hidden_size],
    and target tensor y which is the identity matrix with shape  [batch_size, batch_size].
    """

    def __init__(self):
        super(NPairsLoss, self).__init__()
        self.ce = nn.CrossEntropyLoss()

    def show(self, similarity_scores):
        sn.heatmap(similarity_scores.detach().numpy(), annot=True, annot_kws={'size': 7}, vmin=-1.0, vmax=1.0)
        plt.show()

    def similarities(self, x1, x2):
        """
        Calculates the cosine similarity matrix for every pair (i, j),
        where i is an embedding from x1 and j is another embedding from x2.

        :param x1: a tensors with shape [batch_size, hidden_size].
        :param x2: a tensors with shape [batch_size, hidden_size].
        :return: the cosine similarity matrix with shape [batch_size, batch_size].
        """
        x1 = x1 / torch.norm(x1, dim=1, keepdim=True)
        x2 = x2 / torch.norm(x2, p=2, dim=1, keepdim=True)
        return torch.matmul(x1, x2.t())

    def forward(self, predict, target):
        """
        Computes the N-Pairs Loss between the target and predictions.
        :param predict: the prediction of the model,
        Contains the batches x1 (image embeddings) and x2 (description embeddings).
        :param target: the identity matrix with shape  [batch_size, batch_size].
        :return: N-Pairs Loss value.
        """
        x1, x2 = predict
        predict = self.similarities(x1, x2)
        self.show(predict)
        # by construction the probability distribution must be concentrated on the diagonal of the similarities matrix.
        # so, Cross Entropy can be used to measure the loss.
        return self.ce(predict, target)

但是，由于这种损失，模型最终会收敛到所有密集向量彼此相等的情况。通过执行以下代码片段可以看到：

batch_size=7
hidden_size=768
def m_model(scenario=0):
    if scenario == 0: # all equal all
        p1 = torch.ones((batch_size, hidden_size)) 
        p2 = p1
    elif scenario == 1: # all different all
        p1 = torch.ones((batch_size, hidden_size))
        p2 = -1*p1
    else: # desired case
        p1 = torch.rand((batch_size, hidden_size))
        p2=p1

    return p1, p2

predict = m_model(scenario=0)
target = torch.arange(batch_size)
loss = NPairsLoss(1)

print("Loss:", loss(predict, target))
# Loss: tensor(1.9459), using scenario=0
# Loss: tensor(1.9459), using scenario=1
# Loss: tensor(1.7364), using scenario=2

关于如何惩罚相似度矩阵具有所有相同值的这些场景的任何建议？