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标题: 突然想到让deepseek来解释一下递归 [打印本页]

作者: 密银    时间: 2025-1-29 14:16
标题: 突然想到让deepseek来解释一下递归
本帖最后由 密银 于 2025-1-29 14:19 编辑 : B# n7 N  K6 \, f( C1 V
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解释的不错
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- c$ ^  ]5 a& J递归是一种通过将问题分解为更小的同类子问题来解决问题的方法。它的核心思想是:**函数直接或间接地调用自身**,直到满足终止条件。
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3 r! {6 H- e! G3 U6 e4 N( _  D$ ~/ a; P 关键要素
0 }6 P# `' u& z! ~; d* p1. **基线条件(Base Case)**# L3 k+ a/ q9 _3 A$ c% v/ ~, S
   - 递归终止的条件,防止无限循环
1 L" Y% q2 a/ q9 O% K0 l) J   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1* B. O; w5 l( a" `7 T( t' T
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2. **递归条件(Recursive Case)**+ m6 L  o% Z4 O: h. |+ g3 w( v/ v
   - 将原问题分解为更小的子问题2 S, L9 |0 g' c, t
   - 例如:n! = n × (n-1)!
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经典示例:计算阶乘
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def factorial(n):
. o" P& F, @7 l    if n == 0:        # 基线条件
# f0 ^* f# @5 F3 ^        return 1% `" Z. r; K( C' ?) p
    else:             # 递归条件
% t* y( b+ U& U( X        return n * factorial(n-1): c+ a& W. s# T
执行过程(以计算 3! 为例):
! r/ ?' f( W9 {) K! ]) @* m3 x3 Z- vfactorial(3)
* ]- W; v2 K9 E3 |; }0 X3 * factorial(2)
" x* c" k* a# J& C3 * (2 * factorial(1)); Z5 D( E' H. P+ u
3 * (2 * (1 * factorial(0)))- }- ~9 T# j. O, h) y
3 * (2 * (1 * 1)) = 6
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  v6 _5 S8 j- x4 ?: A' q: q 递归思维要点) s) C  m  f2 G/ P" c* W* \; {
1. **信任递归**:假设子问题已经解决,专注当前层逻辑, k$ y: Y6 `' h  U
2. **栈结构**:每次调用都会创建新的栈帧(内存空间)) i9 J# {- W7 W0 a( u; s( \* D* {
3. **递推过程**:不断向下分解问题(递)& W0 [. K/ J  @
4. **回溯过程**:组合子问题结果返回(归), C$ k5 w1 `+ l: G5 L( O- B

% z7 C0 ^; _' O) ^* A% B# K 注意事项0 c$ N* m! f' j/ c. `8 }. f
必须要有终止条件% B. Y7 f  }! e$ g5 F3 K
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)5 D& y- }# y; T: g8 S
某些问题用递归更直观(如树遍历),但效率可能不如迭代
5 c2 \$ I8 |6 E2 T8 l/ a2 B尾递归优化可以提升效率(但Python不支持)
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2 P& a- P! v0 o 递归 vs 迭代
4 V) ]" S' ]: Z0 j5 x2 S4 V|          | 递归                          | 迭代               |
/ q# o5 F: V- |9 ^|----------|-----------------------------|------------------|. B; H1 a, v# R0 o+ I) ]7 s9 n
| 实现方式    | 函数自调用                        | 循环结构            |  H$ n9 S0 ~- [
| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |2 w% y8 G& X: w6 S  C9 x) _% p* [" p
| 代码可读性  | 对符合递归思维的问题更直观                | 线性流程更直接         |( [- c$ z8 E) \: [3 ?
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |8 g7 l( @/ Q8 F1 \9 p0 j: v3 r& z
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经典递归应用场景; R2 R/ n2 J: w
1. 文件系统遍历(目录树结构)8 x& I$ x$ F5 t
2. 快速排序/归并排序算法
. Q# c1 ~- `0 x3. 汉诺塔问题. e' k7 R/ i' E+ u) V$ M7 H( x% h
4. 二叉树遍历(前序/中序/后序)
0 X- O$ u  e$ P: w6 _5. 生成所有可能的组合(回溯算法)
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试着用递归思维想象:你站在一面镜子前,镜子里有无数个逐渐变小的你在照镜子,这就是递归的直观体现。但记住每个"分身"最终都要有结束的时刻,这就是基线条件的重要性。
作者: testjhy    时间: 2025-1-30 00:07
挺好,递归思维要点与我能够回忆起来我当时写递归程序的思路很一致,,或者被它唤醒,9 n9 N0 _, X1 {4 Y' E7 ?3 |
我推理机的核心算法应该是二叉树遍历的变种。' ]* z& Y% q2 s4 l, t0 i5 T
另外知识系统的推理机搜索深度(递归深度)并不长,没有超过10层的,如果输入变量多的话,搜索宽度很大,但对那时的286-386DOS系统,计算压力也不算大。
作者: nanimarcus    时间: 2025-2-2 00:45
Recursion in programming is a technique where a function calls itself in order to solve a problem. It is a powerful concept that allows you to break down complex problems into smaller, more manageable subproblems. Here's a detailed explanation:3 A7 G* |8 ?! J# ]* r
Key Idea of Recursion6 \* P. `# r* F$ X% s8 G

( a" z1 u1 g7 a, h# N% LA recursive function solves a problem by:. u/ i" P& H9 |! O& D& P  r* @

  @6 x8 D2 E4 s9 I2 {3 k( ^    Breaking the problem into smaller instances of the same problem.
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$ L" L; n' w! A+ L) u* \8 p0 {    Solving the smallest instance directly (base case).
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    Combining the results of smaller instances to solve the larger problem.3 ?! {& n. D# f

! ]# J7 L; c5 j# K2 _3 q8 wComponents of a Recursive Function4 r1 m8 d3 P: Y. t- p$ S7 P$ T1 C

) L0 b3 e: N( {    Base Case:: g1 X" k: G3 G  n! T
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        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.% @1 M3 q- }, I: _0 F

! b' z8 g% o8 A0 C1 D2 }* v% m* K        It acts as the stopping condition to prevent infinite recursion.
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        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.
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    Recursive Case:
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        This is where the function calls itself with a smaller or simpler version of the problem.
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        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
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9 J% e; k; j  z6 \% f. c7 o9 \3 ^* QExample: Factorial Calculation
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& m; c/ Q8 G: E5 [" S  R# Y9 ~0 ?The factorial of a number n (denoted as n!) is the product of all positive integers less than or equal to n. It can be defined recursively as:# B+ }* z/ Q, x1 U7 }, O% p' l

; M7 i. a. f* K, g    Base case: 0! = 1
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  @2 Q9 t% W% a    Recursive case: n! = n * (n-1)!
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Here’s how it looks in code (Python):
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def factorial(n):
3 n8 N  H9 d# P* @' C3 U+ {    # Base case3 ^1 C# A, t* n7 L+ i2 ~+ f
    if n == 0:
2 f% P/ S0 t/ c5 }) W        return 1
" S+ s6 S9 e. n6 `) s    # Recursive case
% l. H: T3 V; f' r    else:5 B+ w$ ?) n3 i6 v, h
        return n * factorial(n - 1)
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" F  `; x& |: ~0 F& m+ k3 y# Example usage( o6 }* ~; V! x0 g
print(factorial(5))  # Output: 120
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How Recursion Works
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7 `  u0 @) Y, q: `7 W* G    The function keeps calling itself with smaller inputs until it reaches the base case.
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% [2 ^, a" m& \* e5 {. i( Y; E    Once the base case is reached, the function starts returning values back up the call stack.$ O$ ]+ f  z3 ]! z8 X) U

1 N& `; e  r! O2 H  |1 V    These returned values are combined to produce the final result.
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! X9 q6 v4 }. \4 L; u8 w) \! ?3 kFor factorial(5):
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9 \4 k4 N' \9 D5 u* t3 [factorial(5) = 5 * factorial(4)* |. P+ y& K. {# S. w3 e
factorial(4) = 4 * factorial(3)
. N" [- i* a3 dfactorial(3) = 3 * factorial(2), T8 t( l* [# m9 |
factorial(2) = 2 * factorial(1)) I* \9 T4 `# \" F
factorial(1) = 1 * factorial(0), B1 _' v( |% l' E* M3 ]
factorial(0) = 1  # Base case
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Then, the results are combined:
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1 b/ ~. z) O+ E4 H+ u; x$ hfactorial(1) = 1 * 1 = 1
+ E+ E0 P( m, k& Z1 Gfactorial(2) = 2 * 1 = 2
3 K4 L6 m# H% K: Y& Ufactorial(3) = 3 * 2 = 6$ d4 N) X  Z: k6 [! ?" p; G
factorial(4) = 4 * 6 = 24
7 m$ v8 p7 w+ D6 l/ ?8 N" F8 pfactorial(5) = 5 * 24 = 120
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  K- }  a; Y, iAdvantages of Recursion0 D8 L" n' j7 ~5 Y5 v& q
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    Simplicity: Recursive solutions are often more intuitive and easier to write for problems that have a natural recursive structure (e.g., tree traversals, divide-and-conquer algorithms).* L# l5 K# Q* x' l

- p7 C  i6 I) s$ Z4 e    Readability: Recursive code can be more readable and concise compared to iterative solutions., `% @9 h7 Z4 _# v: s
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Disadvantages of Recursion
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6 B5 v0 B$ w3 j% a! Q5 V% i    Performance Overhead: Each recursive call adds a new layer to the call stack, which can lead to high memory usage and potential stack overflow for deep recursion.
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0 v0 f% `' Q! b8 G    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).3 x/ H/ X/ O9 x( M3 ?1 Y6 g
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When to Use Recursion
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    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).# v4 A3 j3 d4 L; s' D7 a

$ n- [- N) \6 H3 L9 I/ t5 E    Problems with a clear base case and recursive case.1 w  R# p1 ~# Y" h- M- i- t$ A

) I2 v9 j/ K2 ~3 i% MExample: Fibonacci Sequence
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6 A* x! G3 m6 I7 f" ?6 L- MThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:8 g7 S) _, P0 @% L( o

2 }2 Q: G( v: ^5 f+ g3 @    Base case: fib(0) = 0, fib(1) = 1' j/ s5 u2 `1 @0 Q7 ?  g' N
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)' l% h. k! Z4 s& R. ~

0 U" E% u. A/ i% C' ppython
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% ~4 G; A1 j& |' S; Sdef fibonacci(n):4 M! }- r: k5 O; U
    # Base cases
& q0 d5 |# Q+ s! E- ?    if n == 0:' I  Y# N3 l& H
        return 0
0 g: q0 H0 K/ [4 B( ?' I4 K, O  ?" E8 g3 C    elif n == 1:
6 ?2 S4 O. h9 B3 B0 `: n, h2 Y        return 1
  T/ Q& \! J* x! @0 c8 B    # Recursive case
; H8 u8 x. F# M+ t0 h. I    else:
% F3 ]# t4 p7 z        return fibonacci(n - 1) + fibonacci(n - 2)
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# Example usage6 |  |9 S1 G. {; f* H9 ^
print(fibonacci(6))  # Output: 8% I! j$ I1 M# Q. ~# e
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Tail Recursion3 \% ~2 u1 P3 {3 n. X
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Tail recursion is a special case of recursion where the recursive call is the last operation in the function. Some programming languages optimize tail-recursive functions to avoid stack overflow, but not all languages (e.g., Python does not optimize tail recursion)." {! @% ^5 m' f
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In summary, recursion is a fundamental concept in programming that allows you to solve problems by breaking them into smaller, self-similar subproblems. It’s important to define a base case to avoid infinite recursion and to understand the trade-offs between recursion and iteration.
作者: nanimarcus    时间: 2025-2-2 00:47
我还让Deepseek 给我讲讲Linux Kernel Driver 现在的开发流程,让一个老同志复习复习,快忘光了。




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