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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
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解释的不错
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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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关键要素. u- A1 F9 K* s7 L5 B. T
1. **基线条件(Base Case)**0 T% s& y! S( s  r2 v+ y# I
   - 递归终止的条件,防止无限循环
7 W- e+ z: X5 G; t! J6 q( |" ]   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1. e6 q, Z6 O. z  r+ A5 `

& e7 W! m: b! e7 Z2. **递归条件(Recursive Case)**
+ o. a$ l( B' [9 f  v   - 将原问题分解为更小的子问题5 ~; _- \# n2 s$ N( e
   - 例如:n! = n × (n-1)!* b; H( G2 V1 O; k4 K

; x0 A5 J! X1 L 经典示例:计算阶乘. M( u3 \6 E  L0 G; Z
python! w7 @( D$ @& v4 k
def factorial(n):
; }+ Y: u2 {5 n! K    if n == 0:        # 基线条件
" _7 u% t+ N! ?; y5 O7 c        return 12 {! u5 C: F. H4 S) |# X
    else:             # 递归条件/ j4 Q% p! n. A  v6 I- v5 R) O5 \
        return n * factorial(n-1)7 q. f) d! x3 I4 b/ R, d
执行过程(以计算 3! 为例):9 \6 z8 t* _1 G7 s  @& o& O2 i: A  L# B
factorial(3)- v* B  \$ t9 V. n- k
3 * factorial(2)6 `3 ^! S3 R) a# s( \
3 * (2 * factorial(1))
( C4 I$ V  g7 G3 * (2 * (1 * factorial(0)))" W: J& a& w% N
3 * (2 * (1 * 1)) = 63 @) c/ I: c) Z5 l) |; X
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递归思维要点
4 g6 A/ z) T$ `% c9 W1. **信任递归**:假设子问题已经解决,专注当前层逻辑
# j  `6 H2 R( F! `2 |2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
$ T1 K- c) a' o3. **递推过程**:不断向下分解问题(递), e6 l6 V% C$ M/ u8 |( W' Z' z
4. **回溯过程**:组合子问题结果返回(归)! U' R' E- x* ~2 X/ v2 d
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注意事项) F+ @) v9 `: e
必须要有终止条件
. \1 C, o5 t7 X; M- A递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
1 k( w' y( I7 e2 D8 O' Z% N  h某些问题用递归更直观(如树遍历),但效率可能不如迭代3 u+ q* W* p# F; p( _7 Y0 F
尾递归优化可以提升效率(但Python不支持)3 q9 c9 u0 s! e0 w  \0 k" d

' {3 D2 f+ l4 r  A8 ^. c7 A 递归 vs 迭代
- e( l! K0 y( x' u* m% T|          | 递归                          | 迭代               |
+ w, p# U8 e8 M  Y|----------|-----------------------------|------------------|
0 k) X2 K4 G8 Z8 E8 T$ L. T4 Q| 实现方式    | 函数自调用                        | 循环结构            |
" e* c* r# r( p6 }" f" s9 ]| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |6 v) k. B2 J: }/ ~& o
| 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         |9 f9 i3 R. [  m7 u. W
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
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1 b8 {1 {0 v7 @( h2 K  ?! e3 t7 P 经典递归应用场景/ G% P8 a$ h+ h8 C6 B3 i% w
1. 文件系统遍历(目录树结构)' P$ ]3 }+ L& u3 ~) P3 C) C0 c
2. 快速排序/归并排序算法$ k9 p" H" P4 w
3. 汉诺塔问题( ~! }  h; N; O4 ~' _( F
4. 二叉树遍历(前序/中序/后序)1 c/ o, B/ i( Q. d: `  u+ W) o2 d
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,
) U  ]& \9 W$ D* R1 J我推理机的核心算法应该是二叉树遍历的变种。( o5 C1 U7 x: h' z
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:
* V8 k: b4 G/ h$ R& A4 N7 W0 YKey Idea of Recursion
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A recursive function solves a problem by:
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    Breaking the problem into smaller instances of the same problem.: t7 f6 j- {+ G" T

7 X& I- c5 M* R  Y$ W% M2 W    Solving the smallest instance directly (base case).* k$ T; n0 b4 }4 b- \
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    Combining the results of smaller instances to solve the larger problem.: H& y! w( r+ i3 ~+ c

' Y, K. j$ Y, U- Y2 O- R1 uComponents of a Recursive Function
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6 V4 U6 T4 M7 Z, r6 W2 Z    Base Case:
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        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.
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$ t6 s7 d' E: V$ X        It acts as the stopping condition to prevent infinite recursion.
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7 `0 o% C( @4 ?0 I0 s. P. u8 p        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.9 \- C# e' r$ o; D# b8 j6 J( ^

( K  g( w* J4 S; |9 I8 V    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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Example: Factorial Calculation. S( O+ l5 R0 x8 W

* G. v0 |: z1 l! w5 \2 fThe 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:
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    Base case: 0! = 1
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% h* Y! v" j* b8 u/ r    Recursive case: n! = n * (n-1)!5 t, d+ s8 N# p  M/ Q5 ~% X% L

4 Z# H! w  I. M$ V; n3 E% {Here’s how it looks in code (Python):
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  R; z% [" d' S7 Udef factorial(n):
" Q& p. O. f: H# p) Y; Z    # Base case
/ u! M: D" D# w  [' n& M    if n == 0:  O' c, m9 ]& r# D
        return 1% ?; P2 d" d* O" M8 J
    # Recursive case+ n* }. U; N. _4 X
    else:- E/ s& t; t6 I, v& Y0 x
        return n * factorial(n - 1)
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# Example usage/ A  [8 }* [7 c6 w+ D
print(factorial(5))  # Output: 120
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How Recursion Works
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. Y- U0 e9 s& p: ?    The function keeps calling itself with smaller inputs until it reaches the base case.
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7 O, J1 }7 y# j% r: F- c8 \' R    Once the base case is reached, the function starts returning values back up the call stack.
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    These returned values are combined to produce the final result.
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For factorial(5):
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/ z0 s5 }5 K  ifactorial(5) = 5 * factorial(4)
5 b+ x, h3 U. D. G4 e6 F8 }factorial(4) = 4 * factorial(3)
& U7 s9 q% ~) b" q/ d7 R2 \0 y8 Xfactorial(3) = 3 * factorial(2)
) ]; @% Y. ^$ }7 ofactorial(2) = 2 * factorial(1)
, O3 G/ j0 Q$ o% `) v. P' T+ Xfactorial(1) = 1 * factorial(0)+ ^( R% _+ q" f: G$ F8 w; N5 i
factorial(0) = 1  # Base case5 \. E1 n) M: ~7 a

" D% a. B' {' ^  O7 jThen, the results are combined:
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6 l. S; H* \; P4 E# {, r+ k7 Ufactorial(1) = 1 * 1 = 1
9 L: k/ T9 U! f* `6 rfactorial(2) = 2 * 1 = 2
& W% m; J% R0 t% z3 p7 c2 [& gfactorial(3) = 3 * 2 = 6
0 v* T& ~9 R, L8 Y+ Sfactorial(4) = 4 * 6 = 24
0 ?: H! B' _3 v  I  ?! W, w6 Gfactorial(5) = 5 * 24 = 120) N! V' S8 C3 J6 F0 N
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Advantages of Recursion
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/ X, J$ ^0 F$ `1 D. I1 o    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).1 ^* X8 @* S- C1 S( W

: E' v8 G+ s2 Z5 k    Readability: Recursive code can be more readable and concise compared to iterative solutions.
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1 B; s/ Y3 A9 E+ o* [; C' DDisadvantages of Recursion
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0 c6 T$ W% U9 I2 i& x2 j" P    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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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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( R% d; j7 ?. W( {6 W, s& @4 e: GWhen to Use Recursion0 O2 D  |- \3 @' @; Q

0 h: s! H* A' ^3 U    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
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) k; @, ?+ Y( d2 W: t+ y% @    Problems with a clear base case and recursive case." X5 n$ o0 ]. p" [  w! p, e+ h4 X  L  h

. ?6 e  }8 |" L3 [) G: rExample: Fibonacci Sequence' F$ B9 o/ f/ X# J4 Q4 v
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The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
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    Base case: fib(0) = 0, fib(1) = 1
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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python" w4 Y6 a, {2 w* E- H
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5 ]$ h+ e0 z! ^def fibonacci(n):
' Z1 R0 R: e" A; i/ U    # Base cases  f) m. K- V% z) X* F( j
    if n == 0:
4 g& E' o5 |0 q) w( G, d        return 05 U- y0 d# s5 w' S
    elif n == 1:; f& d; C  o: ]
        return 1+ o9 g: H3 F+ c8 W. K5 H
    # Recursive case
* m6 j4 w. v1 R* y9 y1 f    else:
: \( B* ]+ q3 E$ _6 U        return fibonacci(n - 1) + fibonacci(n - 2)  L8 y" h3 [  {/ e1 Y5 M
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# Example usage- D  i; P2 x6 ~4 K  Y
print(fibonacci(6))  # Output: 8: U. O3 P) M% I* d( }2 s
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Tail Recursion! I+ d0 t+ R8 o2 P6 I% |% k% n

0 Z* |: A! v1 R: jTail 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).
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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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