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Yudong Jin
84 lines
4.5 KiB
Markdown
84 lines
4.5 KiB
Markdown
# Exercises
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## Concept Review
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### Is Choosing the Largest Coin Always Best?
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The coin denominations are `[1, 7, 10]`, and the target amount is 14.
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<!-- numbered-subquestions -->
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1. Follow the rule "always choose the largest denomination that does not exceed the remaining amount," and write the coins selected.
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2. Is there a solution using fewer coins? If so, give one; otherwise, explain why not.
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3. Does this example show that the greedy strategy is correct for every set of coin denominations?
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??? success "Answer"
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1. The greedy strategy selects `10 + 1 + 1 + 1 + 1`, using 5 coins in total.
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2. There is a solution using fewer coins: `7 + 7`, which uses only 2 coins.
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3. No. This counterexample shows that, for arbitrary coin denominations, repeatedly choosing the largest currently available denomination does not necessarily minimize the number of coins.
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The largest immediate choice may prevent a better combination later.
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### Which Item Should Go into the Knapsack First?
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A knapsack with a capacity of 4 kilograms can hold the following items. A fraction of an item may be taken,
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and the value obtained is proportional to its weight:
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- Item A: weight 4 kilograms, value 20.
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- Item B: weight 3 kilograms, value 18.
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<!-- numbered-subquestions -->
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1. What is the value per kilogram of each item? Which item should be placed in the knapsack first?
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2. Fill the knapsack using the greedy strategy for the fractional knapsack problem. What is the final value?
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3. When items can be divided and the knapsack limits total weight, should items be compared by total value or by value per kilogram? Why?
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??? success "Answer"
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1. A has value `20 ÷ 4 = 5` per kilogram, while B has value `18 ÷ 3 = 6` per kilogram,
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so B, with the higher value per unit weight, should be placed in the knapsack first.
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2. First take all of B, using 3 kilograms of capacity and gaining a value of 18. With 1 kilogram of capacity remaining,
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take 1 kilogram of A, gaining a value of 5. The final value is `18 + 5 = 23`.
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3. The knapsack limits total weight, and items can be divided, so they should be compared by value per unit weight.
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Although A has a higher total value, its value per kilogram is lower than B's. Filling the knapsack with A first would yield only a value of 20.
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### Which Pointer Should Move Next?
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The partition heights are `[1, 8, 6, 2, 5]`. Use two pointers, one at each end, to find the maximum capacity.
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The capacity equals "the height of the shorter partition × the distance between the partitions' indices."
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<!-- numbered-subquestions -->
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1. Initially, the left pointer is at index 0 and the right pointer is at index 4. What is the current capacity? Which pointer should move next?
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2. After making the move chosen in Question 1, at which indices are the two pointers? What is the capacity now? Which pointer should move next?
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3. For the current pair of partitions, you could move either the pointer at the shorter partition or the pointer at the taller partition. Which move could still produce a greater capacity, and why?
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??? success "Answer"
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1. The current capacity is `min(1, 5) × (4 - 0) = 4`. The left partition is shorter, so move the left pointer.
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2. After the left pointer moves, the pointers are at indices 1 and 4. The current capacity is
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`min(8, 5) × (4 - 1) = 15`. The right partition is shorter, so move the right pointer next.
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3. Moving the pointer at the shorter partition is the only move that could still produce a greater capacity. If the taller partition's pointer moves, the distance certainly decreases while the height remains limited by the unmoved shorter partition,
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so the capacity can only stay the same or decrease. Only by moving the shorter partition can a taller partition possibly be found.
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## Programming Exercises
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### Fractional Knapsack
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You are given equal-length arrays `wgt` and `val`, where `wgt[i] > 0` and `val[i] >= 0`. The knapsack has capacity `cap >= 0`.
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There is only one of each item, but any fraction of an item may be placed in the knapsack.
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The value obtained is proportional to the fraction of the item's total weight that is included. Use a greedy algorithm
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and return the maximum total value the knapsack can hold as a real number.
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??? tip "Hints"
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1. First calculate each item's value per unit weight as val[i] / wgt[i], keeping the fractional part of the division
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2. Place items with higher value per unit weight into the knapsack first
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3. If the remaining capacity is less than the current item's weight, take exactly the fraction that fills the knapsack and stop
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