Consider the following grammar along with translation rules. $$\begin{array}{ll} S \rightarrow S_1 \text{ \# } T & \{S_{\bullet val} = S_{1\bullet val} * T_{\bullet val}\} S \rightarrow T & \{S_{\bullet val} = T_{\bullet val}\} T \rightarrow T_1 \% R & \{T_{\bullet val} = T_{1\bullet val} \div R_{\bullet val}\} T \rightarrow R & \{T_{\bullet val} = R_{\bullet val}\} R \rightarrow id & \{R_{\bullet val} = id_{\bullet val}\} \end{array}$$ Here # and % are operators and $id$ is a token that represents an integer and $id_{\bullet val}$ represents the corresponding integer value. The set of non-terminals is $\{S, T, R, P\}$ and a subscripted non-terminal indicates an instance of the non-terminal. Using this translation scheme, the computed value of $S_{\bullet val}$ for root of the parse tree for the expression $20\#10\%5\#8\%2\%2$ is __________.

From the grammar and translation rules, we can determine the operator precedence and associativity:

1. Precedence:

   • Non-terminal $T$ is derived from $S$, which means the operator % (evaluated as division $\div$) has higher precedence than # (evaluated as multiplication $*$).

2. Associativity:

   • Both productions $S \rightarrow S_1 \text{ \# } T$ and $T \rightarrow T_1 \% R$ are left-recursive, meaning both operators # and % are left-associative.

Given the expression:

Evaluating by syntax-directed translation:

• Subtree for the first operand of #:

  $$T_1 = 20$$

  $$S_1 = T_1 = 20$$

• Subtree for the second operand of # ($10 \% 5$):

  $$T_2 = 10 \div 5 = 2$$

• Evaluate the first # (left-associative):

  $$S_2 = S_1 * T_2 = 20 * 2 = 40$$

• Subtree for the third operand ($8 \% 2 \% 2$, left-associative division):

  $$T_3 = (8 \div 2) \div 2 = 4 \div 2 = 2$$

• Evaluate the second # at the root:

  $$S_{\bullet val} = S_2 * T_3 = 40 * 2 = 80$$

Thus, the computed value at the root is 80.