Consider the syntax directed translation given by the following grammar and semantic rules. Here $N, I, F$ and $B$ are non-terminals. $N$ is the starting non-terminal, and $\#$, $\mathbf{0}$ and $\mathbf{1}$ are lexical tokens corresponding to input letters “#”, “0” and “1”, respectively. $X.val$ denotes the synthesized attribute (a numeric value) associated with a non-terminal $X$. $I_1$ and $F_1$ denote occurrences of $I$ and $F$ on the right hand side of a production, respectively. For the tokens $\mathbf{0}$ and $\mathbf{1}$, $\mathbf{0}.val = 0$ and $\mathbf{1}.val = 1$. $$\begin{array}{llll} N & \rightarrow & I \, \# \, F & N.val = I.val + F.val I & \rightarrow & I_1 \, B & I.val = (2 \, I_1.val) + B.val I & \rightarrow & B & I.val = B.val F & \rightarrow & B \, F_1 & F.val = \frac{1}{2}(B.val + F_1.val) F & \rightarrow & B & F.val = \frac{1}{2} B.val B & \rightarrow & \mathbf{0} & B.val = \mathbf{0}.val B & \rightarrow & \mathbf{1} & B.val = \mathbf{1}.val \end{array}$$ The value computed by the translation scheme for the input string $$10\#011$$ is ________. (Rounded off to three decimal places)

To find the computed value for the input string 10#011, we evaluate the synthesized attributes for the integer part ($I$) corresponding to 10 and the fractional part ($F$) corresponding to 011.

### 1. Integer Part: 10

• The leftmost bit 1 derives:

  $$B \rightarrow 1 \implies B.val = 1$$

  $$I \rightarrow B \implies I.val = B.val = 1$$

• The next bit 0 derives:

  $$B \rightarrow 0 \implies B.val = 0$$

  $$I \rightarrow I_1 \, B \implies I.val = (2 \times 1) + 0 = 2$$

Thus, $I.val = 2$.

---

### 2. Fractional Part: 011 Since $F \rightarrow B \, F_1$ is right-recursive, evaluation occurs bottom-up from right to left:

• For the rightmost bit 1 (3rd bit after #):

  $$B \rightarrow 1 \implies B.val = 1$$

  $$F \rightarrow B \implies F.val = \frac{1}{2} \times 1 = 0.5$$

• For the middle bit 1 (2nd bit after #):

  $$B \rightarrow 1 \implies B.val = 1$$

  $$F \rightarrow B \, F_1 \implies F.val = \frac{1}{2}(1 + 0.5) = \frac{1.5}{2} = 0.75$$

• For the leftmost bit 0 (1st bit after #):

  $$B \rightarrow 0 \implies B.val = 0$$

  $$F \rightarrow B \, F_1 \implies F.val = \frac{1}{2}(0 + 0.75) = \frac{0.75}{2} = 0.375$$

Thus, $F.val = 0.375$.

---

### 3. Final Value:

Rounded off to three decimal places, the value is 2.375 (official range: 2.374 to 2.376).