# Drill 44 GRU reset then mix

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/read/17/tee-c17-s44

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/section/tee-c17-s44/

## Question

**Question.** hprevious = 0.4, r = 0.5, z = 0.25, x = 1, W = 0.2, U = 0.5, b = 0. Use candidate = tanh(Wx+U(r × hprevious)) and h = (1−z) hprevious+z candidate.

## Answer — step by step

### Given

Old hidden state = 0.4; reset r = 0.5; update z = 0.25.

x = 1; W = 0.2; U = 0.5; b = 0.

Use h = (1−z)h\_old + z×candidate.

### To find

Find reset contribution, candidate and new hidden state.

### Formula — which one and why

**GRU reset and state mixture.** The reset gate controls old-state information entering the candidate. The stated update convention then mixes the original old state and the candidate.

r = reset gate; z = update gate; h\_old = previous state; h\_tilde = candidate. Here z weights the candidate; some questions instead define z to weight the old state.



$$
\widetilde h=\tanh(Wx+U(r\odot h_{\mathrm{old}})+b)
$$





$$
h=(1-z)\odot h_{\mathrm{old}}+z\odot\widetilde h
$$



**Yaad rakhne ka cue:** Reset candidate ko control; update gate old-new mixture ko control.

### Solution — step by step

### Meaning and given convention

The reset gate r scales the old state while making the candidate. The update gate z controls the amount of candidate used in the final mixture **under this question's convention**. Some books reverse the roles of z and 1 − z, so copy the supplied formula.



$$
\widetilde h=\tanh\left(Wx+U(rh_{\mathrm{previous}})+b\right)
$$





$$
h=(1-z)h_{\mathrm{previous}}+z\widetilde h
$$



### Step 1: reset the old state for candidate construction



$$
rh_{\mathrm{previous}}=0.5\times0.4=0.2
$$



### Step 2: form the candidate's raw score



$$
Wx=0.2\times1=0.2
$$





$$
U(rh_{\mathrm{previous}})=0.5\times0.2=0.1
$$





$$
a=0.2+0.1+0=0.3
$$



### Step 3: apply tanh



$$
\widetilde h=\tanh(0.3)=0.291312612
$$



### Step 4: mix the original old state with the candidate



$$
1-z=1-0.25=0.75
$$





$$
h=0.75(0.4)+0.25(0.291312612)
$$





$$
h=0.3+0.072828153=0.372828153
$$



**Final answer: h ≈ 0.372828153.** The final old-state term uses 0.4, not the reset value 0.2.