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zh-hant/docs/chapter_computational_complexity/exercises.md
+678 -32
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@@ -11,33 +11,287 @@ comments: true
下面兩段程式碼都計算 $1 + 2 + \dots + n$（設 $n \ge 1$）。請把 `n` 設為 4，
按照程式實際執行的順序回答問題，然後比較兩種寫法的效率。
```python
def sum_iter(n):
s = 0
for i in range(1, n + 1):
s += i
return s
=== "Python"
def sum_recur(n):
if n == 1:
return 1
return n + sum_recur(n - 1)
```
```python title="complexity_exercises.py"
def sum_iter(n: int) -> int:
"""迭代求和"""
res = 0
for i in range(1, n + 1):
res += i
return res
def sum_recur(n: int) -> int:
"""遞迴求和"""
if n == 1:
return 1
return n + sum_recur(n - 1)
```
=== "C++"
```cpp title="complexity_exercises.cpp"
/* 迭代求和 */
int sumIter(int n) {
int res = 0;
for (int i = 1; i <= n; ++i) {
res += i;
}
return res;
}
/* 遞迴求和 */
int sumRecur(int n) {
if (n == 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "Java"
```java title="complexity_exercises.java"
/* 迭代求和 */
int sumIter(int n) {
int res = 0;
for (int i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 遞迴求和 */
int sumRecur(int n) {
if (n == 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "C#"
```csharp title="complexity_exercises.cs"
/* 迭代求和 */
int SumIter(int n) {
int res = 0;
for (int i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 遞迴求和 */
int SumRecur(int n) {
if (n == 1) {
return 1;
}
return n + SumRecur(n - 1);
}
```
=== "Go"
```go title="complexity_exercises.go"
/* 迭代求和 */
func sumIter(n int) int {
res := 0
for i := 1; i <= n; i++ {
res += i
}
return res
}
/* 遞迴求和 */
func sumRecur(n int) int {
if n == 1 {
return 1
}
return n + sumRecur(n-1)
}
```
=== "Swift"
```swift title="complexity_exercises.swift"
/* 迭代求和 */
func sumIter(n: Int) -> Int {
var res = 0
for i in 1 ... n {
res += i
}
return res
}
/* 遞迴求和 */
func sumRecur(n: Int) -> Int {
if n == 1 {
return 1
}
return n + sumRecur(n: n - 1)
}
```
=== "JS"
```javascript title="complexity_exercises.js"
/* 迭代求和 */
function sumIter(n) {
let res = 0;
for (let i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 遞迴求和 */
function sumRecur(n) {
if (n === 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "TS"
```typescript title="complexity_exercises.ts"
/* 迭代求和 */
function sumIter(n: number): number {
let res = 0;
for (let i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 遞迴求和 */
function sumRecur(n: number): number {
if (n === 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "Dart"
```dart title="complexity_exercises.dart"
/* 迭代求和 */
int sumIter(int n) {
int res = 0;
for (int i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 遞迴求和 */
int sumRecur(int n) {
if (n == 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "Rust"
```rust title="complexity_exercises.rs"
/* 迭代求和 */
fn sum_iter(n: i32) -> i32 {
let mut res = 0;
for i in 1..=n {
res += i;
}
res
}
/* 遞迴求和 */
fn sum_recur(n: i32) -> i32 {
if n == 1 {
return 1;
}
n + sum_recur(n - 1)
}
```
=== "C"
```c title="complexity_exercises.c"
/* 迭代求和 */
int sumIter(int n) {
int res = 0;
for (int i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* 遞迴求和 */
int sumRecur(int n) {
if (n == 1) {
return 1;
}
return n + sumRecur(n - 1);
}
```
=== "Kotlin"
```kotlin title="complexity_exercises.kt"
/* 迭代求和 */
fun sumIter(n: Int): Int {
var res = 0
for (i in 1..n) {
res += i
}
return res
}
/* 遞迴求和 */
fun sumRecur(n: Int): Int {
if (n == 1) {
return 1
}
return n + sumRecur(n - 1)
}
```
=== "Ruby"
```ruby title="complexity_exercises.rb"
### 迭代求和 ###
def sum_iter(n)
res = 0
for i in 1..n
res += i
end
res
end
### 遞迴求和 ###
def sum_recur(n)
return 1 if n == 1
n + sum_recur(n - 1)
end
```
<!-- numbered-subquestions -->
1. 執行 `sum_iter(4)` 時，每輪迴圈結束後，變數 `s` 的值分別是多少？
2. 執行 `sum_recur(4)` 時，會依次呼叫哪些函式？從最深的一層開始返回時，結果怎樣得到？
1. 輸入 `n = 4` 執行迭代函式時，每輪迴圈結束後，累加變數 `res` 的值分別是多少？
2. 輸入 `n = 4` 執行遞迴函式時，參數 `n` 會依次取哪些值？從最深的一層開始返回時，結果怎樣得到？
3. 兩種寫法的時間複雜度和空間複雜度分別是多少？結合第 1、2 問的執行過程說明理由。
??? success "參考答案"
1. 迴圈變數 `i` 依次為 `1、2、3、4`，每輪結束後，`s` 依次變為
`1、3、6、10`，所以 `sum_iter(4)` 返回 10。
1. 迴圈變數 `i` 依次為 `1、2、3、4`，每輪結束後，`res` 依次變為
`1、3、6、10`，所以迭代函式返回 10。
2. 函式依次呼叫
`sum_recur(4) → sum_recur(3) → sum_recur(2) → sum_recur(1)`。
`sum_recur(1)` 返回 1，隨後各層依次得到 `2 + 1 = 3`、`3 + 3 = 6`、`4 + 6 = 10`。
2. 參數 `n` 依次為 `4 → 3 → 2 → 1`。
最深一層返回 1，隨後各層依次得到 `2 + 1 = 3`、`3 + 3 = 6`、`4 + 6 = 10`。
在最深處，4 次函式呼叫都尚未結束。
3. 兩段程式碼都進行與 $n$ 成正比的迴圈或呼叫，因此時間複雜度均為 $O(n)$ 。
@@ -51,22 +305,414 @@ def sum_recur(n):
以下三個程式碼片段的輸入均為正整數 $n$ 。請按時間複雜度從低到高排序，並寫出各自的複雜度。
```python
# 片段一
s = 0
for i in range(n):
s += i
=== "Python"
# 片段二
s = 0
for i in range(n):
for j in range(i, n):
s += j
```python title="complexity_exercises.py"
def linear_loop(n: int) -> int:
"""線性階迴圈"""
res = 0
for i in range(n):
res += i
return res
# 片段三
while n > 1:
n = n // 2
```
def quadratic_loop(n: int) -> int:
"""平方階迴圈"""
res = 0
for i in range(n):
for j in range(i, n):
res += j
return res
def logarithmic_loop(n: int) -> int:
"""對數階迴圈"""
while n > 1:
n //= 2
return n
```
=== "C++"
```cpp title="complexity_exercises.cpp"
/* 線性階迴圈 */
int linearLoop(int n) {
int res = 0;
for (int i = 0; i < n; ++i) {
res += i;
}
return res;
}
/* 平方階迴圈 */
int quadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; ++i) {
for (int j = i; j < n; ++j) {
res += j;
}
}
return res;
}
/* 對數階迴圈 */
int logarithmicLoop(int n) {
while (n > 1) {
n /= 2;
}
return n;
}
```
=== "Java"
```java title="complexity_exercises.java"
/* 線性階迴圈 */
int linearLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方階迴圈 */
int quadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 對數階迴圈 */
int logarithmicLoop(int n) {
while (n > 1) {
n /= 2;
}
return n;
}
```
=== "C#"
```csharp title="complexity_exercises.cs"
/* 線性階迴圈 */
int LinearLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方階迴圈 */
int QuadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 對數階迴圈 */
int LogarithmicLoop(int n) {
while (n > 1) {
n /= 2;
}
return n;
}
```
=== "Go"
```go title="complexity_exercises.go"
/* 線性階迴圈 */
func linearLoop(n int) int {
res := 0
for i := 0; i < n; i++ {
res += i
}
return res
}
/* 平方階迴圈 */
func quadraticLoop(n int) int {
res := 0
for i := 0; i < n; i++ {
for j := i; j < n; j++ {
res += j
}
}
return res
}
/* 對數階迴圈 */
func logarithmicLoop(n int) int {
for n > 1 {
n /= 2
}
return n
}
```
=== "Swift"
```swift title="complexity_exercises.swift"
/* 線性階迴圈 */
func linearLoop(n: Int) -> Int {
var res = 0
for i in 0 ..< n {
res += i
}
return res
}
/* 平方階迴圈 */
func quadraticLoop(n: Int) -> Int {
var res = 0
for i in 0 ..< n {
for j in i ..< n {
res += j
}
}
return res
}
/* 對數階迴圈 */
func logarithmicLoop(n: Int) -> Int {
var n = n
while n > 1 {
n /= 2
}
return n
}
```
=== "JS"
```javascript title="complexity_exercises.js"
/* 線性階迴圈 */
function linearLoop(n) {
let res = 0;
for (let i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方階迴圈 */
function quadraticLoop(n) {
let res = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 對數階迴圈 */
function logarithmicLoop(n) {
while (n > 1) {
n = Math.floor(n / 2);
}
return n;
}
```
=== "TS"
```typescript title="complexity_exercises.ts"
/* 線性階迴圈 */
function linearLoop(n: number): number {
let res = 0;
for (let i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方階迴圈 */
function quadraticLoop(n: number): number {
let res = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 對數階迴圈 */
function logarithmicLoop(n: number): number {
while (n > 1) {
n = Math.floor(n / 2);
}
return n;
}
```
=== "Dart"
```dart title="complexity_exercises.dart"
/* 線性階迴圈 */
int linearLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方階迴圈 */
int quadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 對數階迴圈 */
int logarithmicLoop(int n) {
while (n > 1) {
n ~/= 2;
}
return n;
}
```
=== "Rust"
```rust title="complexity_exercises.rs"
/* 線性階迴圈 */
fn linear_loop(n: i32) -> i32 {
let mut res = 0;
for i in 0..n {
res += i;
}
res
}
/* 平方階迴圈 */
fn quadratic_loop(n: i32) -> i32 {
let mut res = 0;
for i in 0..n {
for j in i..n {
res += j;
}
}
res
}
/* 對數階迴圈 */
fn logarithmic_loop(mut n: i32) -> i32 {
while n > 1 {
n /= 2;
}
n
}
```
=== "C"
```c title="complexity_exercises.c"
/* 線性階迴圈 */
int linearLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
res += i;
}
return res;
}
/* 平方階迴圈 */
int quadraticLoop(int n) {
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
res += j;
}
}
return res;
}
/* 對數階迴圈 */
int logarithmicLoop(int n) {
while (n > 1) {
n /= 2;
}
return n;
}
```
=== "Kotlin"
```kotlin title="complexity_exercises.kt"
/* 線性階迴圈 */
fun linearLoop(n: Int): Int {
var res = 0
for (i in 0 until n) {
res += i
}
return res
}
/* 平方階迴圈 */
fun quadraticLoop(n: Int): Int {
var res = 0
for (i in 0 until n) {
for (j in i until n) {
res += j
}
}
return res
}
/* 對數階迴圈 */
fun logarithmicLoop(n: Int): Int {
var value = n
while (value > 1) {
value /= 2
}
return value
}
```
=== "Ruby"
```ruby title="complexity_exercises.rb"
### 線性階迴圈 ###
def linear_loop(n)
res = 0
for i in 0...n
res += i
end
res
end
### 平方階迴圈 ###
def quadratic_loop(n)
res = 0
for i in 0...n
for j in i...n
res += j
end
end
res
end
### 對數階迴圈 ###
def logarithmic_loop(n)
n /= 2 while n > 1
n
end
```
??? success "參考答案"
zh-hant/docs/chapter_divide_and_conquer/exercises.md
+205 -12
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@@ -29,23 +29,216 @@ comments: true
下面的遞迴函式用分治計算 $x^n$：
```python
def fast_pow(x, n):
if n == 0:
return 1
half = fast_pow(x, n // 2)
if n % 2 == 0:
return half * half
return half * half * x
```
=== "Python"
用它計算 `fast_pow(3, 5)`：
```python title="fast_power.py"
def fast_pow(x: int, n: int) -> int:
"""快速冪"""
if n == 0:
return 1
half = fast_pow(x, n // 2)
if n % 2 == 0:
return half * half
return half * half * x
```
=== "C++"
```cpp title="fast_power.cpp"
/* 快速冪 */
int fastPow(int x, int n) {
if (n == 0) {
return 1;
}
int half = fastPow(x, n / 2);
if (n % 2 == 0) {
return half * half;
}
return half * half * x;
}
```
=== "Java"
```java title="fast_power.java"
/* 快速冪 */
int fastPow(int x, int n) {
if (n == 0) {
return 1;
}
int half = fastPow(x, n / 2);
if (n % 2 == 0) {
return half * half;
}
return half * half * x;
}
```
=== "C#"
```csharp title="fast_power.cs"
/* 快速冪 */
int FastPow(int x, int n) {
if (n == 0) {
return 1;
}
int half = FastPow(x, n / 2);
if (n % 2 == 0) {
return half * half;
}
return half * half * x;
}
```
=== "Go"
```go title="fast_power.go"
/* 快速冪 */
func fastPow(x, n int) int {
if n == 0 {
return 1
}
half := fastPow(x, n/2)
if n%2 == 0 {
return half * half
}
return half * half * x
}
```
=== "Swift"
```swift title="fast_power.swift"
/* 快速冪 */
func fastPow(x: Int, n: Int) -> Int {
if n == 0 {
return 1
}
let half = fastPow(x: x, n: n / 2)
if n % 2 == 0 {
return half * half
}
return half * half * x
}
```
=== "JS"
```javascript title="fast_power.js"
/* 快速冪 */
function fastPow(x, n) {
if (n === 0) {
return 1;
}
const half = fastPow(x, Math.floor(n / 2));
if (n % 2 === 0) {
return half * half;
}
return half * half * x;
}
```
=== "TS"
```typescript title="fast_power.ts"
/* 快速冪 */
function fastPow(x: number, n: number): number {
if (n === 0) {
return 1;
}
const half = fastPow(x, Math.floor(n / 2));
if (n % 2 === 0) {
return half * half;
}
return half * half * x;
}
```
=== "Dart"
```dart title="fast_power.dart"
/* 快速冪 */
int fastPow(int x, int n) {
if (n == 0) {
return 1;
}
int half = fastPow(x, n ~/ 2);
if (n % 2 == 0) {
return half * half;
}
return half * half * x;
}
```
=== "Rust"
```rust title="fast_power.rs"
/* 快速冪 */
fn fast_pow(x: i32, n: i32) -> i32 {
if n == 0 {
return 1;
}
let half = fast_pow(x, n / 2);
if n % 2 == 0 {
return half * half;
}
half * half * x
}
```
=== "C"
```c title="fast_power.c"
/* 快速冪 */
int fastPow(int x, int n) {
if (n == 0) {
return 1;
}
int half = fastPow(x, n / 2);
if (n % 2 == 0) {
return half * half;
}
return half * half * x;
}
```
=== "Kotlin"
```kotlin title="fast_power.kt"
/* 快速冪 */
fun fastPow(x: Int, n: Int): Int {
if (n == 0) {
return 1
}
val half = fastPow(x, n / 2)
if (n % 2 == 0) {
return half * half
}
return half * half * x
}
```
=== "Ruby"
```ruby title="fast_power.rb"
### 快速冪 ###
def fast_pow(x, n)
return 1 if n == 0
half = fast_pow(x, n / 2)
return half * half if n.even?
half * half * x
end
```
令 `x = 3`、`n = 5`，用這個函式計算：
<!-- numbered-subquestions -->
1. 遞迴呼叫時，參數 `n` 依次變成哪些值？
2. 從最深層開始返回時，各層依次返回什麼值？
3. 為什麼要先儲存 `half`，而不是把 `fast_pow(x, n // 2)` 寫兩遍？
3. 為什麼要先把遞迴結果儲存為 `half`，而不是在乘法兩邊各呼叫一次相同的子問題？
??? success "參考答案"
@@ -54,7 +247,7 @@ def fast_pow(x, n):
2. `n = 0` 時返回 1；`n = 1` 時返回 $1×1×3=3$；
`n = 2` 時返回 $3×3=9$；`n = 5` 時返回 $9×9×3=243$。
3. 如果把 `fast_pow(x, n // 2)` 在乘法兩邊各寫一次，兩次遞迴會計算完全相同的子問題。
3. 如果在乘法兩邊各呼叫一次相同的子問題，兩次遞迴會進行完全相同的計算。
先把結果儲存為 `half`，每層就只遞迴一次，遞迴深度約為 $\log n$；
呼叫兩次會造成大量重複計算。