1. **Problem Statement:** We have a variant of the 0-1 knapsack problem where each item $a_i$ can be rejected, fully accepted, or half-accepted. The capacity is $S$, and $V(T,m)$ is the optimal profit for the first $m$ items with capacity $T$. 2. **Key Definitions:** - $v_i$: value of item $i$ - $s_i$: size of item $i$ - Half-accepting means taking half the size and half the value. 3. **Recurrence Relation:** The optimal profit $V(T,m)$ is the maximum of: - Not taking item $m$: $V(T,m-1)$ - Taking full item $m$ if it fits: $V(T - s_m, m-1) + v_m$ - Taking half item $m$ if half fits: $V(T - \frac{s_m}{2}, m-1) + \frac{v_m}{2}$ 4. **Check each option:** - (a) uses min and subtracts $v_m$ and $v_m/2$ from $T$, which is incorrect because capacity reduces by size, not value, and we maximize profit, not minimize. - (b) matches the correct recurrence: max, subtract sizes $s_m$ and $s_m/2$ from $T$, add values $v_m$ and $v_m/2$. - (c) is a simplified version ignoring half-acceptance, but correct for classical 0-1 knapsack. - (d) uses min instead of max, incorrect. - (e) uses max but subtracts values $v_m$ from $T$, incorrect. - (f) $V(T,1) = 0$ if $s_1 > T$ is correct: if the first item doesn't fit, profit is zero. - (g) subtracts values $v_m$ from $T$, incorrect. - (h) is false because some choices are correct. - (i) $V(0,m) = 0$ is correct: zero capacity means zero profit. 5. **Final correct statements:** (b), (c), (f), and (i). **Answer:** $$ \boxed{\text{Correct: } b, c, f, i} $$