How to use previous output and hidden states from LSTM for the attention mechanism? - tensorflow

Question

I am currently trying to code the attention mechanism from this paper: "Effective Approaches to Attention-based Neural Machine Translation", Luong, Pham, Manning (2015). (I use global attention with the dot score).
However, I am unsure on how to input the hidden and output states from the lstm decode. The issue is that the input of the lstm decoder at time t depends on quantities that I need to compute using the output and hidden states from t-1.
Here is the relevant part of the code:
with tf.variable_scope('data'):
prob = tf.placeholder_with_default(1.0, shape=())
X_or = tf.placeholder(shape = [batch_size, timesteps_1, num_input], dtype = tf.float32, name = "input")
X = tf.unstack(X_or, timesteps_1, 1)
y = tf.placeholder(shape = [window_size,1], dtype = tf.float32, name = "label_annotation")
logits = tf.zeros((1,1), tf.float32)
with tf.variable_scope('lstm_cell_encoder'):
rnn_layers = [tf.nn.rnn_cell.LSTMCell(size) for size in [hidden_size, hidden_size]]
multi_rnn_cell = tf.nn.rnn_cell.MultiRNNCell(rnn_layers)
lstm_outputs, lstm_state = tf.contrib.rnn.static_rnn(cell=multi_rnn_cell,inputs=X,dtype=tf.float32)
concat_lstm_outputs = tf.stack(tf.squeeze(lstm_outputs))
last_encoder_state = lstm_state[-1]
with tf.variable_scope('lstm_cell_decoder'):
initial_input = tf.unstack(tf.zeros(shape=(1,1,hidden_size2)))
rnn_decoder_cell = tf.nn.rnn_cell.LSTMCell(hidden_size, state_is_tuple = True)
# Compute the hidden and output of h_1
for index in range(window_size):
output_decoder, state_decoder = tf.nn.static_rnn(rnn_decoder_cell, initial_input, initial_state=last_encoder_state, dtype=tf.float32)
# Compute the score for source output vector
scores = tf.matmul(concat_lstm_outputs, tf.reshape(output_decoder[-1],(hidden_size,1)))
attention_coef = tf.nn.softmax(scores)
context_vector = tf.reduce_sum(tf.multiply(concat_lstm_outputs, tf.reshape(attention_coef, (window_size, 1))),0)
context_vector = tf.reshape(context_vector, (1,hidden_size))
# compute the tilda hidden state \tilde{h}_t=tanh(W[c_t, h_t]+b_t)
concat_context = tf.concat([context_vector, output_decoder[-1]], axis = 1)
W_tilde = tf.Variable(tf.random_normal(shape = [hidden_size*2, hidden_size2], stddev = 0.1), name = "weights_tilde", trainable = True)
b_tilde = tf.Variable(tf.zeros([1, hidden_size2]), name="bias_tilde", trainable = True)
hidden_tilde = tf.nn.tanh(tf.matmul(concat_context, W_tilde)+b_tilde) # hidden_tilde is [1*64]
# update for next time step
initial_input = tf.unstack(tf.reshape(hidden_tilde, (1,1,hidden_size2)))
last_encoder_state = state_decoder
# predict the target
W_target = tf.Variable(tf.random_normal(shape = [hidden_size2, 1], stddev = 0.1), name = "weights_target", trainable = True)
logit = tf.matmul(hidden_tilde, W_target)
logits = tf.concat([logits, logit], axis = 0)
logits = logits[1:]
The part inside the loop is what I am unsure of. Does tensorflow remember the computational graph when I overwrite the variable "initial_input" and "last_encoder_state"?

Answer 1

I think your model will be much simplified if you use tf.contrib.seq2seq.AttentionWrapper with one of implementations: BahdanauAttention or LuongAttention.
This way it'll be possible to wire the attention vector on a cell level, so that cell output is already after attention applied. Example from the seq2seq tutorial:
cell = LSTMCell(512)
attention_mechanism = tf.contrib.seq2seq.LuongAttention(512, encoder_outputs)
attn_cell = tf.contrib.seq2seq.AttentionWrapper(cell, attention_mechanism, attention_size=256)
Note that this way you won't need a loop of window_size, because tf.nn.static_rnn or tf.nn.dynamic_rnn will instantiate the cells wrapped with attention.
Regarding your question: you should distinguish python variables and tensorflow graph nodes: you can assign last_encoder_state to a different tensor, the original graph node won't change because of this. This is flexible, but can be also misleading in the result network - you might think that you connect an LSTM to one tensor, but it's actually the other. In general, you shouldn't do that.