# LSTM candidate and output gate

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/formulas/42

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/formulas/42/

The candidate is proposed new cell content, so it may be negative. The output gate is a positive sigmoid multiplier that controls how much of the activated cell is exposed. Candidate content and gate values therefore have different roles.



$$
\widetilde C_t=\tanh(W_C[h_{t-1},x_t]+b_C)
$$





$$
o_t=\sigma(W_o[h_{t-1},x_t]+b_o)
$$



## LSTM numerical when gates must first be calculated

**Question.** At an LSTM time step, xₜ = 1, hₜ₋₁ = 0 and Cₜ₋₁ = 0.2. The affine preactivations of the forget, input and output gates are all 0. The candidate preactivation is 0.5. Calculate the activated gates, candidate cell content, new cell state and new hidden state.

### Given

x = 1; old hidden state = 0; old cell = 0.2.

Forget/input/output preactivations are all 0.

Candidate preactivation = 0.5.

### To find

Activate all gates and candidate, then find C and h.

### Formula — which one and why

**Activate raw gate scores first.** A preactivation is an unprocessed score. Convert forget/input/output scores with sigmoid and candidate scores with tanh before using the cell-update equations.

z\_f,z\_i,z\_o = gate preactivations; z\_c = candidate preactivation. An already activated gate must not be passed through sigmoid again.



$$
f=\sigma(z_f),\quad i=\sigma(z_i),\quad o=\sigma(z_o)
$$





$$
\widetilde C=\tanh(z_c)
$$



**Yaad rakhne ka cue:** Raw score pehle activate; activated gate ko dobara sigmoid nahi.

**LSTM cell and hidden state.** Multiply old memory by the forget gate; multiply candidate content by the input gate; add those two contributions. Apply tanh to the new cell and multiply by the output gate.

f,i,o = activated forget/input/output gates; C\_old = previous cell; C\_tilde = candidate content; C = new cell; h = visible hidden state. ⊙ means elementwise multiplication.



$$
C=f\odot C_{\mathrm{old}}+i\odot\widetilde C
$$





$$
h=o\odot\tanh C
$$



**Yaad rakhne ka cue:** Purana rakho + naya lo; phir tanh aur output gate.

### Solution — step by step

### Distinguish preactivations from gate outputs

The affine transformations have already been evaluated in this question. Their outputs are preactivations, so activation functions are still required. Sigmoid is used for f, i and o; tanh is used for the candidate.



$$
\sigma(a)=\frac1{1+e^{-a}}
$$





$$
C_t=f_tC_{t-1}+i_t\widetilde C_t,\qquad h_t=o_t\tanh(C_t)
$$



![A gate preactivation of zero becomes a sigmoid gate output of 0.5. Use that activated gate value in the cell update.](https://deep-learning-study-room.vercel.app/assets/figures/v_lstm_gates_first.svg)

Diagram: A gate preactivation of zero becomes a sigmoid gate output of 0.5. Use that activated gate value in the cell update.. Labels: When the question supplies gate raw scores | When the question supplies gate raw scores | A gate raw score is not yet a gate value. | Gate scores | f, i, o scores | = 0 | Apply sigmoid | Each gate | becomes 0.5 | Candidate | score | 0.5 → tanh → | 0.462117 | Then update | Compute cell, | then hidden | output | Sigmoid(0) = 0.5. Do not replace a zero raw score with a closed gate.

A gate preactivation of zero becomes a sigmoid gate output of 0.5. Use that activated gate value in the cell update. Tap the diagram to enlarge.

**Step 1: activate the three gates.**



$$
f_t=i_t=o_t=\sigma(0)=\frac1{1+e^0}=0.5
$$



**Step 2: activate the candidate preactivation.**



$$
\widetilde C_t=\tanh(0.5)\approx0.462117157
$$



**Step 3: calculate the old and new cell contributions separately.**



$$
f_tC_{t-1}=0.5(0.2)=0.1
$$





$$
i_t\widetilde C_t=0.5\tanh(0.5)\approx0.231058579
$$



**Step 4: calculate the updated cell.**



$$
C_t=0.1+0.5\tanh(0.5)\approx0.331058579
$$



**Step 5: calculate the hidden output from the updated cell.**



$$
h_t=0.5\tanh(C_t)\approx0.159735818
$$



**Final answer:** fₜ = iₜ = oₜ = 0.5; C̃ₜ ≈ 0.462117157; Cₜ ≈ 0.331058579; hₜ ≈ 0.159735818. Retain unrounded calculator values between steps. A gate preactivation of 0 is different from a supplied gate output of 0: the former becomes 0.5 under sigmoid, while the latter would suppress its contribution completely.

### LSTM gates from raw score 0

**Question. An LSTM has forget, input and output gate raw scores all equal to 0, candidate raw score 0, and old cell 1. Calculate all gates, candidate, new cell and hidden state.**

### Given

Raw scores z\_f = z\_i = z\_o = z\_candidate = 0.

Old cell C\_old = 1; gate activation sigmoid; candidate activation tanh.

### To find

f,i,o,candidate,C\_new,h\_new.

### Formula — which one and why

Raw gate scores require sigmoid first. The candidate uses tanh; do not apply sigmoid to it.

f retains old memory, i admits candidate, o controls exposed hidden output; C is cell memory.



$$
f=i=o=\sigma(z)
$$





$$
\widetilde C=\tanh(z)
$$





$$
C_{\mathrm{new}}=fC_{\mathrm{old}}+i\widetilde C
$$





$$
h_{\mathrm{new}}=o\tanh(C_{\mathrm{new}})
$$



**Yaad rakhne ka cue:** Gates sigmoid; candidate tanh; cell mein add; hidden pe output gate.

### Solution — step by step

**Step 1: calculate the shared gate value.**



$$
\sigma(0)=\frac1{1+e^{-0}}\approx0.500000000
$$



**Step 2: calculate the candidate.**



$$
\widetilde C=\tanh(0)\approx0.000000000
$$



**Step 3: retain the old-cell part.**



$$
fC_{\mathrm{old}}=(0.500000000)(1)\approx0.500000000
$$



**Step 4: admit the new content.**



$$
i\widetilde C=(0.500000000)(0.000000000)\approx0.000000000
$$



**Step 5: add the memory contributions.**



$$
C_{\mathrm{new}}=0.500000000+(0.000000000)\approx0.500000000
$$



**Step 6: apply tanh to the cell.**



$$
\tanh(C_{\mathrm{new}})\approx0.462117157
$$



**Step 7: apply the output gate.**



$$
h_{\mathrm{new}}=(0.500000000)(0.462117157)\approx0.231058579
$$



**Final answer:** f=i=o≈0.500000000; candidate≈0.000000000; cell≈0.500000000; hidden≈0.231058579.

### LSTM gates from raw score 1

**Question. An LSTM has forget, input and output gate raw scores all equal to 1, candidate raw score 1, and old cell -0.5. Calculate all gates, candidate, new cell and hidden state.**

### Given

Raw scores z\_f = z\_i = z\_o = z\_candidate = 1.

Old cell C\_old = -0.5; gate activation sigmoid; candidate activation tanh.

### To find

f,i,o,candidate,C\_new,h\_new.

### Formula — which one and why

Raw gate scores require sigmoid first. The candidate uses tanh; do not apply sigmoid to it.

f retains old memory, i admits candidate, o controls exposed hidden output; C is cell memory.



$$
f=i=o=\sigma(z)
$$





$$
\widetilde C=\tanh(z)
$$





$$
C_{\mathrm{new}}=fC_{\mathrm{old}}+i\widetilde C
$$





$$
h_{\mathrm{new}}=o\tanh(C_{\mathrm{new}})
$$



**Yaad rakhne ka cue:** Gates sigmoid; candidate tanh; cell mein add; hidden pe output gate.

### Solution — step by step

**Step 1: calculate the shared gate value.**



$$
\sigma(1)=\frac1{1+e^{-1}}\approx0.731058579
$$



**Step 2: calculate the candidate.**



$$
\widetilde C=\tanh(1)\approx0.761594156
$$



**Step 3: retain the old-cell part.**



$$
fC_{\mathrm{old}}=(0.731058579)(-0.5)\approx-0.365529289
$$



**Step 4: admit the new content.**



$$
i\widetilde C=(0.731058579)(0.761594156)\approx0.556769941
$$



**Step 5: add the memory contributions.**



$$
C_{\mathrm{new}}=-0.365529289+(0.556769941)\approx0.191240652
$$



**Step 6: apply tanh to the cell.**



$$
\tanh(C_{\mathrm{new}})\approx0.188942847
$$



**Step 7: apply the output gate.**



$$
h_{\mathrm{new}}=(0.731058579)(0.188942847)\approx0.138128289
$$



**Final answer:** f=i=o≈0.731058579; candidate≈0.761594156; cell≈0.191240652; hidden≈0.138128289.

## Drill 43 Calculate gates before the cell

## Question

**Question.** Gate preactivations zf = zi = zo = 0, candidate preactivation 0.5, Cprevious = 0.2. Find gates, C, h.

## Answer — step by step

### Given

Raw gate scores: zf = zi = zo = 0.

Raw candidate score = 0.5; old cell C = 0.2.

### To find

Activate gates/candidate, then calculate the new cell and hidden state.

### Formula — which one and why

**Activate raw gate scores first.** A preactivation is an unprocessed score. Convert forget/input/output scores with sigmoid and candidate scores with tanh before using the cell-update equations.

z\_f,z\_i,z\_o = gate preactivations; z\_c = candidate preactivation. An already activated gate must not be passed through sigmoid again.



$$
f=\sigma(z_f),\quad i=\sigma(z_i),\quad o=\sigma(z_o)
$$





$$
\widetilde C=\tanh(z_c)
$$



**Yaad rakhne ka cue:** Raw score pehle activate; activated gate ko dobara sigmoid nahi.

**LSTM cell and hidden state.** Multiply old memory by the forget gate; multiply candidate content by the input gate; add those two contributions. Apply tanh to the new cell and multiply by the output gate.

f,i,o = activated forget/input/output gates; C\_old = previous cell; C\_tilde = candidate content; C = new cell; h = visible hidden state. ⊙ means elementwise multiplication.



$$
C=f\odot C_{\mathrm{old}}+i\odot\widetilde C
$$





$$
h=o\odot\tanh C
$$



**Yaad rakhne ka cue:** Purana rakho + naya lo; phir tanh aur output gate.

### Solution — step by step

### Meaning and formulas

The question supplies raw gate scores, called preactivations. Convert them to gates using sigmoid first. Convert the candidate's score using tanh.



$$
f=\sigma(z_f),\quad i=\sigma(z_i),\quad o=\sigma(z_o),\quad \widetilde C=\tanh(z_C)
$$





$$
\sigma(z)=\frac1{1+e^{-z}},\quad C=fC_{\mathrm{previous}}+i\widetilde C,\quad h=o\tanh(C)
$$



### Step 1: turn each zero score into a gate



$$
\sigma(0)=\frac1{1+e^0}=\frac12=0.5
$$



Thus f = i = o = 0.5. A zero score is not a zero gate.

### Step 2: calculate candidate content



$$
\widetilde C=\tanh(0.5)=0.462117157
$$



### Step 3: combine retained memory and new content



$$
fC_{\mathrm{previous}}=0.5\times0.2=0.1
$$





$$
i\widetilde C=0.5\times0.462117157=0.231058579
$$





$$
C=0.1+0.231058579=0.331058579
$$



### Step 4: compute the visible hidden state



$$
\tanh(C)=\tanh(0.331058579)\approx0.319471636
$$





$$
h=0.5\times0.319471636=0.159735818
$$



**Final answer:** gates 0.5 each; candidate ≈ 0.462117157; C ≈ 0.331058579; h ≈ 0.159735818.

## Drill 50 Three step LSTM retention and replacement

## Question

**Question.** C₀ = 0, output gate 0.7 at all steps. Supplied(f, i, candidate) are(1,.5,.8), (.9,0,−.2), (.2,.5,−.6). Find C and h at each step.

## Answer — step by step

### Given

Initial C₀ = 0; output gate o = 0.7 at all three steps.

Supplied (f,i,candidate): (1,0.5,0.8), (0.9,0,−0.2), (0.2,0.5,−0.6).

### To find

Calculate C₁,h₁,C₂,h₂,C₃,h₃ in order.

### Formula — which one and why

**LSTM cell and hidden state.** Multiply old memory by the forget gate; multiply candidate content by the input gate; add those two contributions. Apply tanh to the new cell and multiply by the output gate.

f,i,o = activated forget/input/output gates; C\_old = previous cell; C\_tilde = candidate content; C = new cell; h = visible hidden state. ⊙ means elementwise multiplication.



$$
C=f\odot C_{\mathrm{old}}+i\odot\widetilde C
$$





$$
h=o\odot\tanh C
$$



**Yaad rakhne ka cue:** Purana rakho + naya lo; phir tanh aur output gate.

### Solution — step by step

### Meaning and formulas

At each time step use the previous **cell state C**, not the previous visible output h, in the memory equation. The output gate is 0.7 at every step.



$$
C_t=f_tC_{t-1}+i_t\widetilde C_t,\qquad h_t=o_t\tanh(C_t)
$$



**Step 1: Time 1: admit new content.**



$$
f_1C_0=1(0)=0,\qquad i_1\widetilde C_1=0.5(0.8)=0.4
$$





$$
C_1=0+0.4=0.4
$$





$$
h_1=0.7\tanh(0.4)=0.7(0.379948962)=0.265964274
$$



**Step 2: Time 2: retain memory and admit no new content.**



$$
f_2C_1=0.9(0.4)=0.36,\qquad i_2\widetilde C_2=0(-0.2)=0
$$





$$
C_2=0.36+0=0.36
$$





$$
h_2=0.7\tanh(0.36)=0.7(0.345214034)=0.241649824
$$



**Step 3: Time 3: retain a small part and add negative content.**



$$
f_3C_2=0.2(0.36)=0.072,\qquad i_3\widetilde C_3=0.5(-0.6)=-0.3
$$





$$
C_3=0.072-0.3=-0.228
$$





$$
h_3=0.7\tanh(-0.228)=0.7(-0.224129674)=-0.156890772
$$



**Final cell states:** 0.4, 0.36, −0.228. **Final hidden states:** approximately 0.265964274, 0.241649824, −0.156890772. The supplied gate value 1 is an idealized limit, used exactly as given.