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comments: true
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# 15.2 Fractional knapsack problem
!!! question
Given $n$ items, the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with a capacity of $cap$. Each item can be chosen only once, **but a part of the item can be selected, with its value calculated based on the proportion of the weight chosen**, what is the maximum value of the items in the knapsack under the limited capacity? An example is shown in Figure 15-3.
{ class="animation-figure" }
Figure 15-3 Example data of the fractional knapsack problem
The fractional knapsack problem is very similar overall to the 0-1 knapsack problem, involving the current item $i$ and capacity $c$, aiming to maximize the value within the limited capacity of the knapsack.
The difference is that, in this problem, only a part of an item can be chosen. As shown in Figure 15-4, **we can arbitrarily split the items and calculate the corresponding value based on the weight proportion**.
1. For item $i$, its value per unit weight is $val[i-1] / wgt[i-1]$, referred to as the unit value.
2. Suppose we put a part of item $i$ with weight $w$ into the knapsack, then the value added to the knapsack is $w \times val[i-1] / wgt[i-1]$.
{ class="animation-figure" }
Figure 15-4 Value per unit weight of the item
### 1. Greedy strategy determination
Maximizing the total value of the items in the knapsack essentially means maximizing the value per unit weight. From this, the greedy strategy shown in Figure 15-5 can be deduced.
1. Sort the items by their unit value from high to low.
2. Iterate over all items, **greedily choosing the item with the highest unit value in each round**.
3. If the remaining capacity of the knapsack is insufficient, use part of the current item to fill the knapsack.
{ class="animation-figure" }
Figure 15-5 Greedy strategy of the fractional knapsack problem
### 2. Code implementation
We have created an `Item` class in order to sort the items by their unit value. We loop and make greedy choices until the knapsack is full, then exit and return the solution:
=== "Python"
```python title="fractional_knapsack.py"
class Item:
"""物品"""
def __init__(self, w: int, v: int):
self.w = w # 物品重量
self.v = v # 物品价值
def fractional_knapsack(wgt: list[int], val: list[int], cap: int) -> int:
"""分数背包：贪心"""
# 创建物品列表，包含两个属性：重量、价值
items = [Item(w, v) for w, v in zip(wgt, val)]
# 按照单位价值 item.v / item.w 从高到低进行排序
items.sort(key=lambda item: item.v / item.w, reverse=True)
# 循环贪心选择
res = 0
for item in items:
if item.w <= cap:
# 若剩余容量充足，则将当前物品整个装进背包
res += item.v
cap -= item.w
else:
# 若剩余容量不足，则将当前物品的一部分装进背包
res += (item.v / item.w) * cap
# 已无剩余容量，因此跳出循环
break
return res
```
=== "C++"
```cpp title="fractional_knapsack.cpp"
/* 物品 */
class Item {
public:
int w; // 物品重量
int v; // 物品价值
Item(int w, int v) : w(w), v(v) {
}
};
/* 分数背包：贪心 */
double fractionalKnapsack(vector &wgt, vector &val, int cap) {
// 创建物品列表，包含两个属性：重量、价值
vector- items;
for (int i = 0; i < wgt.size(); i++) {
items.push_back(Item(wgt[i], val[i]));
}
// 按照单位价值 item.v / item.w 从高到低进行排序
sort(items.begin(), items.end(), [](Item &a, Item &b) { return (double)a.v / a.w > (double)b.v / b.w; });
// 循环贪心选择
double res = 0;
for (auto &item : items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (double)item.v / item.w * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "Java"
```java title="fractional_knapsack.java"
/* 物品 */
class Item {
int w; // 物品重量
int v; // 物品价值
public Item(int w, int v) {
this.w = w;
this.v = v;
}
}
/* 分数背包：贪心 */
double fractionalKnapsack(int[] wgt, int[] val, int cap) {
// 创建物品列表，包含两个属性：重量、价值
Item[] items = new Item[wgt.length];
for (int i = 0; i < wgt.length; i++) {
items[i] = new Item(wgt[i], val[i]);
}
// 按照单位价值 item.v / item.w 从高到低进行排序
Arrays.sort(items, Comparator.comparingDouble(item -> -((double) item.v / item.w)));
// 循环贪心选择
double res = 0;
for (Item item : items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (double) item.v / item.w * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "C#"
```csharp title="fractional_knapsack.cs"
/* 物品 */
class Item(int w, int v) {
public int w = w; // 物品重量
public int v = v; // 物品价值
}
/* 分数背包：贪心 */
double FractionalKnapsack(int[] wgt, int[] val, int cap) {
// 创建物品列表，包含两个属性：重量、价值
Item[] items = new Item[wgt.Length];
for (int i = 0; i < wgt.Length; i++) {
items[i] = new Item(wgt[i], val[i]);
}
// 按照单位价值 item.v / item.w 从高到低进行排序
Array.Sort(items, (x, y) => (y.v / y.w).CompareTo(x.v / x.w));
// 循环贪心选择
double res = 0;
foreach (Item item in items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (double)item.v / item.w * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "Go"
```go title="fractional_knapsack.go"
/* 物品 */
type Item struct {
w int // 物品重量
v int // 物品价值
}
/* 分数背包：贪心 */
func fractionalKnapsack(wgt []int, val []int, cap int) float64 {
// 创建物品列表，包含两个属性：重量、价值
items := make([]Item, len(wgt))
for i := 0; i < len(wgt); i++ {
items[i] = Item{wgt[i], val[i]}
}
// 按照单位价值 item.v / item.w 从高到低进行排序
sort.Slice(items, func(i, j int) bool {
return float64(items[i].v)/float64(items[i].w) > float64(items[j].v)/float64(items[j].w)
})
// 循环贪心选择
res := 0.0
for _, item := range items {
if item.w <= cap {
// 若剩余容量充足，则将当前物品整个装进背包
res += float64(item.v)
cap -= item.w
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += float64(item.v) / float64(item.w) * float64(cap)
// 已无剩余容量，因此跳出循环
break
}
}
return res
}
```
=== "Swift"
```swift title="fractional_knapsack.swift"
/* 物品 */
class Item {
var w: Int // 物品重量
var v: Int // 物品价值
init(w: Int, v: Int) {
self.w = w
self.v = v
}
}
/* 分数背包：贪心 */
func fractionalKnapsack(wgt: [Int], val: [Int], cap: Int) -> Double {
// 创建物品列表，包含两个属性：重量、价值
var items = zip(wgt, val).map { Item(w: $0, v: $1) }
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort { -(Double($0.v) / Double($0.w)) < -(Double($1.v) / Double($1.w)) }
// 循环贪心选择
var res = 0.0
var cap = cap
for item in items {
if item.w <= cap {
// 若剩余容量充足，则将当前物品整个装进背包
res += Double(item.v)
cap -= item.w
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += Double(item.v) / Double(item.w) * Double(cap)
// 已无剩余容量，因此跳出循环
break
}
}
return res
}
```
=== "JS"
```javascript title="fractional_knapsack.js"
/* 物品 */
class Item {
constructor(w, v) {
this.w = w; // 物品重量
this.v = v; // 物品价值
}
}
/* 分数背包：贪心 */
function fractionalKnapsack(wgt, val, cap) {
// 创建物品列表，包含两个属性：重量、价值
const items = wgt.map((w, i) => new Item(w, val[i]));
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort((a, b) => b.v / b.w - a.v / a.w);
// 循环贪心选择
let res = 0;
for (const item of items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (item.v / item.w) * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "TS"
```typescript title="fractional_knapsack.ts"
/* 物品 */
class Item {
w: number; // 物品重量
v: number; // 物品价值
constructor(w: number, v: number) {
this.w = w;
this.v = v;
}
}
/* 分数背包：贪心 */
function fractionalKnapsack(wgt: number[], val: number[], cap: number): number {
// 创建物品列表，包含两个属性：重量、价值
const items: Item[] = wgt.map((w, i) => new Item(w, val[i]));
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort((a, b) => b.v / b.w - a.v / a.w);
// 循环贪心选择
let res = 0;
for (const item of items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (item.v / item.w) * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "Dart"
```dart title="fractional_knapsack.dart"
/* 物品 */
class Item {
int w; // 物品重量
int v; // 物品价值
Item(this.w, this.v);
}
/* 分数背包：贪心 */
double fractionalKnapsack(List wgt, List val, int cap) {
// 创建物品列表，包含两个属性：重量、价值
List
- items = List.generate(wgt.length, (i) => Item(wgt[i], val[i]));
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort((a, b) => (b.v / b.w).compareTo(a.v / a.w));
// 循环贪心选择
double res = 0;
for (Item item in items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += item.v / item.w * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "Rust"
```rust title="fractional_knapsack.rs"
/* 物品 */
struct Item {
w: i32, // 物品重量
v: i32, // 物品价值
}
impl Item {
fn new(w: i32, v: i32) -> Self {
Self { w, v }
}
}
/* 分数背包：贪心 */
fn fractional_knapsack(wgt: &[i32], val: &[i32], mut cap: i32) -> f64 {
// 创建物品列表，包含两个属性：重量、价值
let mut items = wgt
.iter()
.zip(val.iter())
.map(|(&w, &v)| Item::new(w, v))
.collect::>();
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort_by(|a, b| {
(b.v as f64 / b.w as f64)
.partial_cmp(&(a.v as f64 / a.w as f64))
.unwrap()
});
// 循环贪心选择
let mut res = 0.0;
for item in &items {
if item.w <= cap {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v as f64;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += item.v as f64 / item.w as f64 * cap as f64;
// 已无剩余容量，因此跳出循环
break;
}
}
res
}
```
=== "C"
```c title="fractional_knapsack.c"
/* 物品 */
typedef struct {
int w; // 物品重量
int v; // 物品价值
} Item;
/* 分数背包：贪心 */
float fractionalKnapsack(int wgt[], int val[], int itemCount, int cap) {
// 创建物品列表，包含两个属性：重量、价值
Item *items = malloc(sizeof(Item) * itemCount);
for (int i = 0; i < itemCount; i++) {
items[i] = (Item){.w = wgt[i], .v = val[i]};
}
// 按照单位价值 item.v / item.w 从高到低进行排序
qsort(items, (size_t)itemCount, sizeof(Item), sortByValueDensity);
// 循环贪心选择
float res = 0.0;
for (int i = 0; i < itemCount; i++) {
if (items[i].w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += items[i].v;
cap -= items[i].w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (float)cap / items[i].w * items[i].v;
cap = 0;
break;
}
}
free(items);
return res;
}
```
=== "Kotlin"
```kotlin title="fractional_knapsack.kt"
/* 物品 */
class Item(
val w: Int, // 物品
val v: Int // 物品价值
)
/* 分数背包：贪心 */
fun fractionalKnapsack(wgt: IntArray, _val: IntArray, c: Int): Double {
// 创建物品列表，包含两个属性：重量、价值
var cap = c
val items = arrayOfNulls
- (wgt.size)
for (i in wgt.indices) {
items[i] = Item(wgt[i], _val[i])
}
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sortBy { item: Item? -> -(item!!.v.toDouble() / item.w) }
// 循环贪心选择
var res = 0.0
for (item in items) {
if (item!!.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v
cap -= item.w
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += item.v.toDouble() / item.w * cap
// 已无剩余容量，因此跳出循环
break
}
}
return res
}
```
=== "Ruby"
```ruby title="fractional_knapsack.rb"
[class]{Item}-[func]{}
[class]{}-[func]{fractional_knapsack}
```
=== "Zig"
```zig title="fractional_knapsack.zig"
[class]{Item}-[func]{}
[class]{}-[func]{fractionalKnapsack}
```
??? pythontutor "Code Visualization"
Apart from sorting, in the worst case, the entire list of items needs to be traversed, **hence the time complexity is $O(n)$**, where $n$ is the number of items.
Since an `Item` object list is initialized, **the space complexity is $O(n)$**.
### 3. Correctness proof
Using proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm yields a maximum value `res`, but the solution does not include item $x$.
Now remove a unit weight of any item from the knapsack and replace it with a unit weight of item $x$. Since the unit value of item $x$ is the highest, the total value after replacement will definitely be greater than `res`. **This contradicts the assumption that `res` is the optimal solution, proving that the optimal solution must include item $x$**.
For other items in this solution, we can also construct the above contradiction. Overall, **items with greater unit value are always better choices**, proving that the greedy strategy is effective.
As shown in Figure 15-6, if the item weight and unit value are viewed as the horizontal and vertical axes of a two-dimensional chart respectively, the fractional knapsack problem can be transformed into "seeking the largest area enclosed within a limited horizontal axis range". This analogy can help us understand the effectiveness of the greedy strategy from a geometric perspective.
{ class="animation-figure" }
Figure 15-6 Geometric representation of the fractional knapsack problem