xinyin025/hello-algo
1
0
Fork 0
mirror of https://github.com/krahets/hello-algo.git synced 2026-08-20 23:41:00 +00:00
Code Issues Packages Projects Releases Wiki Activity

build

Browse Source
This commit is contained in:
krahets
2026-04-03 18:46:15 +08:00
parent 377736b1bd
commit 9d21ca86b0
352 changed files with 46563 additions and 11262 deletions
Download Patch File Download Diff File Expand all files Collapse all files
en/docs/chapter_greedy/fractional_knapsack_problem.md
+9 -9
View File
@@ -6,7 +6,7 @@ comments: true
!!! question
Given $n$ items, where the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with capacity $cap$. Each item can be selected only once, **but a portion of an item can be selected, with the value calculated based on the proportion of weight selected**, what is the maximum value of items in the knapsack under the limited capacity? An example is shown in Figure 15-3.
Given $n$ items, where the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with capacity $cap$. Each item can be selected only once, **but a fraction of an item may be selected, with its value proportional to the selected weight**. What is the maximum total value that can be placed in the knapsack under the capacity constraint? An example is shown in Figure 15-3.
![Example data for the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_example.png){ class="animation-figure" }
@@ -14,7 +14,7 @@ comments: true
The fractional knapsack problem is very similar overall to the 0-1 knapsack problem, with states including the current item $i$ and capacity $c$, and the goal being to maximize value under the limited knapsack capacity.
The difference is that this problem allows selecting only a portion of an item. As shown in Figure 15-4, **we can arbitrarily split items and calculate the corresponding value based on the weight proportion**.
The difference is that this problem allows selecting only a fraction of an item. As shown in Figure 15-4, **we can split an item arbitrarily and compute its value in proportion to the selected weight**.
1. For item $i$, its value per unit weight is $val[i-1] / wgt[i-1]$, referred to as unit value.
2. Suppose we put a portion of item $i$ with weight $w$ into the knapsack, then the value added to the knapsack is $w \times val[i-1] / wgt[i-1]$.
@@ -25,7 +25,7 @@ The difference is that this problem allows selecting only a portion of an item.
### 1.   Greedy Strategy Determination
Maximizing the total value of items in the knapsack **is essentially maximizing the value per unit weight of items**. From this, we can derive the greedy strategy shown in Figure 15-5.
Maximizing the total value in the knapsack **essentially means prioritizing items with higher value per unit weight**. From this observation, we can derive the greedy strategy shown in Figure 15-5.
1. Sort items by unit value from high to low.
2. Iterate through all items, **greedily selecting the item with the highest unit value in each round**.
@@ -37,7 +37,7 @@ Maximizing the total value of items in the knapsack **is essentially maximizing
### 2.   Code Implementation
We created an `Item` class to facilitate sorting items by unit value. We loop to make greedy selections, breaking when the knapsack is full and returning the solution:
We define an `Item` class so that items can be sorted by unit value. We then iterate through the sorted items greedily, stopping once the knapsack is full and returning the result:
=== "Python"
@@ -531,7 +531,7 @@ We created an `Item` class to facilitate sorting items by unit value. We loop to
end
```
The time complexity of built-in sorting algorithms is usually $O(\log n)$, and the space complexity is usually $O(\log n)$ or $O(n)$, depending on the specific implementation of the programming language.
Built-in sorting algorithms usually take $O(n \log n)$ time, and their space complexity is usually $O(\log n)$ or $O(n)$, depending on the specific implementation of the programming language.
Apart from sorting, in the worst case the entire item list needs to be traversed, **therefore the time complexity is $O(n)$**, where $n$ is the number of items.
@@ -539,13 +539,13 @@ 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 of `res`, but this solution does not include item $x$.
We use proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm produces an optimal value `res`, but the resulting 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 item $x$ has the highest unit value, 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 $x$**.
Now remove one unit of weight from any item in the knapsack and replace it with one unit of weight from item $x$. Since item $x$ has the highest unit value, the total value after the replacement must be greater than `res`. **This contradicts the assumption that `res` is optimal, proving that any optimal solution must include item $x$**.
For other items in this solution, we can also construct the above contradiction. In summary, **items with greater unit value are always better choices**, which proves that the greedy strategy is effective.
We can construct the same contradiction for the other items in the solution as well. In summary, **items with higher unit value are always the better choice**, which proves that the greedy strategy is effective.
As shown in Figure 15-6, if we view item weight and item unit value as the horizontal and vertical axes of a two-dimensional chart respectively, then the fractional knapsack problem can be transformed into "finding the maximum area enclosed within a limited horizontal axis range". This analogy can help us understand the effectiveness of the greedy strategy from a geometric perspective.
As shown in Figure 15-6, if we treat item weight and unit value as the horizontal and vertical axes of a two-dimensional chart, then the fractional knapsack problem can be viewed as "finding the maximum area enclosed within a bounded interval on the horizontal axis." This analogy helps explain the effectiveness of the greedy strategy from a geometric perspective.
![Geometric representation of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png){ class="animation-figure" }
en/docs/chapter_greedy/greedy_algorithm.md
+25 -25
View File
@@ -4,20 +4,20 @@ comments: true
# 15.1   Greedy Algorithm
<u>Greedy algorithm</u> is a common algorithm for solving optimization problems. Its basic idea is to make the seemingly best choice at each decision stage of the problem, that is, to greedily make locally optimal decisions in hopes of obtaining a globally optimal solution. Greedy algorithms are simple and efficient, and are widely applied in many practical problems.
<u>Greedy algorithm</u> is a common approach to solving optimization problems. Its basic idea is to choose the option that appears best at each decision stage, that is, to greedily make locally optimal decisions in the hope of obtaining a globally optimal solution. Greedy algorithms are simple and efficient, and are widely used in many practical problems.
Greedy algorithms and dynamic programming are both commonly used to solve optimization problems. They share some similarities, such as both relying on the optimal substructure property, but they work differently.
- Dynamic programming considers all previous decisions when making the current decision, and uses solutions to past subproblems to construct the solution to the current subproblem.
- Greedy algorithms do not consider past decisions, but instead make greedy choices moving forward, continually reducing the problem size until the problem is solved.
We will first understand how greedy algorithms work through the example problem "coin change". This problem has already been introduced in the "Complete Knapsack Problem" chapter, so I believe you are not unfamiliar with it.
We will first understand how greedy algorithms work through the example problem "coin change." This problem was already introduced in the "Complete Knapsack Problem" chapter, so it should already be familiar to you.
!!! question
Given $n$ types of coins, where the denomination of the $i$-th type of coin is $coins[i - 1]$, and the target amount is $amt$, with each type of coin available for repeated selection, what is the minimum number of coins needed to make up the target amount? If it is impossible to make up the target amount, return $-1$.
Given $n$ types of coins, where the denomination of the $i$-th type is $coins[i - 1]$, a target amount $amt$, and an unlimited number of coins of each type, what is the minimum number of coins needed to make up the target amount? If the target amount cannot be made up, return $-1$.
The greedy strategy adopted for this problem is shown in Figure 15-1. Given a target amount, **we greedily select the coin that is not greater than and closest to it**, and continuously repeat this step until the target amount is reached.
The greedy strategy for this problem is shown in Figure 15-1. Given a target amount, **we greedily choose the coin that does not exceed it and is closest to it**, repeating this step until the target amount is made up.
![Greedy strategy for coin change](greedy_algorithm.assets/coin_change_greedy_strategy.png){ class="animation-figure" }
@@ -330,28 +330,28 @@ The implementation code is as follows:
end
```
You might exclaim: So clean! The greedy algorithm solves the coin change problem in about ten lines of code.
You may find yourself exclaiming, "So clean!" The greedy algorithm solves the coin change problem in only about ten lines of code.
## 15.1.1 &nbsp; Advantages and Limitations of Greedy Algorithms
**Greedy algorithms are not only straightforward and simple to implement, but are also usually very efficient**. In the code above, if the smallest coin denomination is $\min(coins)$, the greedy choice loops at most $amt / \min(coins)$ times, giving a time complexity of $O(amt / \min(coins))$. This is an order of magnitude smaller than the time complexity of the dynamic programming solution $O(n \times amt)$.
**Greedy algorithms are not only straightforward to apply and easy to implement, but are also usually very efficient**. In the code above, if the smallest coin denomination is $\min(coins)$, the greedy selection loop runs at most $amt / \min(coins)$ times, giving a time complexity of $O(amt / \min(coins))$. This is an order of magnitude lower than the time complexity of the dynamic programming solution, $O(n \times amt)$.
However, **for certain coin denomination combinations, greedy algorithms cannot find the optimal solution**. Figure 15-2 provides two examples.
However, **for some coin denomination sets, greedy algorithms cannot find the optimal solution**. Figure 15-2 shows two examples.
- **Positive example $coins = [1, 5, 10, 20, 50, 100]$**: With this coin combination, given any $amt$, the greedy algorithm can find the optimal solution.
- **Negative example $coins = [1, 20, 50]$**: Suppose $amt = 60$, the greedy algorithm can only find the combination $50 + 1 \times 10$, totaling $11$ coins, but dynamic programming can find the optimal solution $20 + 20 + 20$, requiring only $3$ coins.
- **Negative example $coins = [1, 49, 50]$**: Suppose $amt = 98$, the greedy algorithm can only find the combination $50 + 1 \times 48$, totaling $49$ coins, but dynamic programming can find the optimal solution $49 + 49$, requiring only $2$ coins.
- **Positive example $coins = [1, 5, 10, 20, 50, 100]$**: With this coin set, the greedy algorithm can find the optimal solution for any $amt$.
- **Counterexample $coins = [1, 20, 50]$**: Suppose $amt = 60$. The greedy algorithm can only find the combination $50 + 1 \times 10$, using $11$ coins in total, whereas dynamic programming can find the optimal solution $20 + 20 + 20$ using only $3$ coins.
- **Counterexample $coins = [1, 49, 50]$**: Suppose $amt = 98$. The greedy algorithm can only find the combination $50 + 1 \times 48$, using $49$ coins in total, whereas dynamic programming can find the optimal solution $49 + 49$ using only $2$ coins.
![Examples where greedy algorithms cannot find the optimal solution](greedy_algorithm.assets/coin_change_greedy_vs_dp.png){ class="animation-figure" }
<p align="center"> Figure 15-2 &nbsp; Examples where greedy algorithms cannot find the optimal solution </p>
In other words, for the coin change problem, greedy algorithms cannot guarantee finding the global optimal solution, and may even find very poor solutions. It is better suited for solving with dynamic programming.
In other words, for the coin change problem, greedy algorithms cannot guarantee a globally optimal solution and may even produce very poor results. This problem is better solved with dynamic programming.
Generally, the applicability of greedy algorithms falls into the following two situations.
In general, greedy algorithms are applicable in the following two situations.
1. **Can guarantee finding the optimal solution**: In this situation, greedy algorithms are often the best choice, because they tend to be more efficient than backtracking and dynamic programming.
2. **Can find an approximate optimal solution**: Greedy algorithms are also applicable in this situation. For many complex problems, finding the global optimal solution is very difficult, and being able to find a suboptimal solution with high efficiency is also very good.
1. **The optimal solution can be guaranteed**: In this case, greedy algorithms are often the best choice because they tend to be more efficient than backtracking and dynamic programming.
2. **An approximately optimal solution can be found**: Greedy algorithms are also useful in this case. For many complex problems, finding the global optimal solution is very difficult, so efficiently finding a suboptimal solution is already a very good outcome.
## 15.1.2 &nbsp; Characteristics of Greedy Algorithms
@@ -366,30 +366,30 @@ Optimal substructure has already been introduced in the "Dynamic Programming" ch
We mainly explore methods for determining the greedy choice property. Although its description seems relatively simple, **in practice, for many problems, proving the greedy choice property is not easy**.
For example, in the coin change problem, although we can easily provide counterexamples to disprove the greedy choice property, proving it is quite difficult. If asked: **what conditions must a coin combination satisfy to be solvable using a greedy algorithm**? We often can only rely on intuition or examples to give an ambiguous answer, and find it difficult to provide a rigorous mathematical proof.
For example, in the coin change problem, although we can easily provide counterexamples to disprove the greedy choice property, proving that it holds is much harder. If asked, **under what conditions can a coin set be solved using a greedy algorithm**? We often can only rely on intuition or examples to give a vague answer, and it is difficult to provide a rigorous mathematical proof.
!!! quote
There is a paper that presents an algorithm with $O(n^3)$ time complexity for determining whether a coin combination can use a greedy algorithm to find the optimal solution for any amount.
There is a paper that presents an $O(n^3)$ algorithm for determining whether a coin set can be solved optimally by a greedy algorithm for any amount.
Pearson, D. A polynomial-time algorithm for the change-making problem[J]. Operations Research Letters, 2005, 33(3): 231-234.
## 15.1.3 &nbsp; Steps for Solving Problems with Greedy Algorithms
The problem-solving process for greedy problems can generally be divided into the following three steps.
The general process for solving greedy problems can be divided into the following three steps.
1. **Problem analysis**: Sort out and understand the problem characteristics, including state definition, optimization objectives, and constraints, etc. This step is also involved in backtracking and dynamic programming.
2. **Determine the greedy strategy**: Determine how to make greedy choices at each step. This strategy should be able to reduce the problem size at each step, ultimately solving the entire problem.
3. **Correctness proof**: It is usually necessary to prove that the problem has both greedy choice property and optimal substructure. This step may require mathematical proofs, such as mathematical induction or proof by contradiction.
1. **Problem analysis**: Sort out and understand the characteristics of the problem, including state definitions, optimization objectives, and constraints. This step also appears in backtracking and dynamic programming.
2. **Determine the greedy strategy**: Decide how to make a greedy choice at each step. This strategy should reduce the problem size step by step and ultimately solve the entire problem.
3. **Correctness proof**: It is usually necessary to prove that the problem has both greedy choice property and optimal substructure. This step may require mathematical tools such as induction or proof by contradiction.
Determining the greedy strategy is the core step in solving the problem, but it may not be easy to implement, mainly for the following reasons.
Determining the greedy strategy is the core step in solving such problems, but it may not be easy in practice, mainly for the following reasons.
- **Greedy strategies differ greatly between different problems**. For many problems, the greedy strategy is relatively straightforward, and we can derive it through some general thinking and attempts. However, for some complex problems, the greedy strategy may be very elusive, which really tests one's problem-solving experience and algorithmic ability.
- **Some greedy strategies are highly misleading**. When we confidently design a greedy strategy, write the solution code and submit it for testing, we may find that some test cases cannot pass. This is because the designed greedy strategy is only "partially correct", as exemplified by the coin change problem discussed above.
- **Greedy strategies vary greatly from problem to problem**. For many problems, the greedy strategy is fairly intuitive and can be derived through rough reasoning and experimentation. For some complex problems, however, the greedy strategy may be deeply hidden, which strongly tests one's problem-solving experience and algorithmic ability.
- **Some greedy strategies are highly deceptive**. We may confidently design a greedy strategy, write the solution code, and submit it, only to find that some test cases fail. This is because the designed greedy strategy is only "partially correct," as exemplified by the coin change problem discussed above.
To ensure correctness, we should rigorously mathematically prove the greedy strategy, **usually using proof by contradiction or mathematical induction**.
To ensure correctness, we should give a rigorous mathematical proof of the greedy strategy, **usually using proof by contradiction or mathematical induction**.
However, correctness proofs may also not be easy. If we have no clue, we usually choose to debug the code based on test cases, step by step modifying and verifying the greedy strategy.
However, correctness proofs can also be difficult. If we have no clear direction, we usually resort to debugging against test cases, revising and validating the greedy strategy step by step.
## 15.1.4 &nbsp; Typical Problems Solved by Greedy Algorithms
en/docs/chapter_greedy/index.md
+2 -2
View File
@@ -9,9 +9,9 @@ icon: material/head-heart-outline
!!! abstract
Sunflowers turn toward the sun, constantly pursuing the maximum potential for their own growth.
Sunflowers turn toward the sun, always seeking the fullest growth possible.
Through rounds of simple choices, greedy strategies gradually lead to the best answer.
Through successive simple choices, greedy strategies gradually lead to the optimal solution.
## Chapter contents
en/docs/chapter_greedy/max_capacity_problem.md
+18 -18
View File
@@ -6,56 +6,56 @@ comments: true
!!! question
Input an array $ht$, where each element represents the height of a vertical partition. Any two partitions in the array, along with the space between them, can form a container.
Given an array $ht$, where each element represents the height of a vertical partition. Any two partitions in the array, together with the space between them, can form a container.
The capacity of the container equals the product of height and width (area), where the height is determined by the shorter partition, and the width is the difference in array indices between the two partitions.
The capacity of the container equals the product of its height and width (that is, its area), where the height is determined by the shorter partition and the width is the difference between the array indices of the two partitions.
Please select two partitions in the array such that the capacity of the formed container is maximized, and return the maximum capacity. An example is shown in Figure 15-7.
Select two partitions in the array such that the capacity of the resulting container is maximized, and return that maximum capacity. An example is shown in Figure 15-7.
![Example data for the max capacity problem](max_capacity_problem.assets/max_capacity_example.png){ class="animation-figure" }
<p align="center"> Figure 15-7 &nbsp; Example data for the max capacity problem </p>
The container is formed by any two partitions, **therefore the state of this problem is the indices of two partitions, denoted as $[i, j]$**.
The container is formed by any two partitions, **so the state of this problem is the indices of the two partitions, denoted by $[i, j]$**.
According to the problem description, capacity equals height multiplied by width, where height is determined by the shorter partition, and width is the difference in array indices between the two partitions. Let the capacity be $cap[i, j]$, then the calculation formula is:
According to the problem statement, capacity equals height multiplied by width, where the height is determined by the shorter partition and the width is the difference between the array indices of the two partitions. Let the capacity be $cap[i, j]$; then we obtain the following formula:
$$
cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
$$
Let the array length be $n$, then the number of combinations of two partitions (total number of states) is $C_n^2 = \frac{n(n - 1)}{2}$. Most directly, **we can exhaustively enumerate all states** to find the maximum capacity, with time complexity $O(n^2)$.
Let the array length be $n$. Then the number of ways to choose two partitions (that is, the total number of states) is $C_n^2 = \frac{n(n - 1)}{2}$. The most straightforward approach is to **exhaustively enumerate all states** to find the maximum capacity, which has a time complexity of $O(n^2)$.
### 1. &nbsp; Greedy Strategy Determination
This problem has a more efficient solution. As shown in Figure 15-8, select a state $[i, j]$ where index $i < j$ and height $ht[i] < ht[j]$, meaning $i$ is the short partition and $j$ is the long partition.
This problem has a more efficient solution. As shown in Figure 15-8, consider a state $[i, j]$ where $i < j$ and $ht[i] < ht[j]$. In this case, $i$ is the shorter partition and $j$ is the taller partition.
![Initial state](max_capacity_problem.assets/max_capacity_initial_state.png){ class="animation-figure" }
<p align="center"> Figure 15-8 &nbsp; Initial state </p>
As shown in Figure 15-9, **if we now move the long partition $j$ closer to the short partition $i$, the capacity will definitely decrease**.
As shown in Figure 15-9, **if we now move the taller partition $j$ inward toward the shorter partition $i$, the capacity will definitely decrease**.
This is because after moving the long partition $j$, the width $j-i$ definitely decreases; and since height is determined by the short partition, the height can only remain unchanged ($i$ is still the short partition) or decrease (the moved $j$ becomes the short partition).
This is because after moving the taller partition $j$, the width $j-i$ definitely decreases. Since the height is determined by the shorter partition, the height can only stay the same ($i$ remains the shorter partition) or decrease ($j$ becomes the shorter partition after being moved).
![State after moving the long partition inward](max_capacity_problem.assets/max_capacity_moving_long_board.png){ class="animation-figure" }
<p align="center"> Figure 15-9 &nbsp; State after moving the long partition inward </p>
Conversely, **we can only possibly increase capacity by contracting the short partition $i$ inward**. Because although width will definitely decrease, **height may increase** (the moved short partition $i$ may become taller). For example, in Figure 15-10, the area increases after moving the short partition.
Conversely, **only by moving the shorter partition $i$ inward can the capacity possibly increase**. Although the width will definitely decrease, **the height may increase** (the moved partition at $i$ may be taller). For example, in Figure 15-10, the area increases after moving the shorter partition.
![State after moving the short partition inward](max_capacity_problem.assets/max_capacity_moving_short_board.png){ class="animation-figure" }
<p align="center"> Figure 15-10 &nbsp; State after moving the short partition inward </p>
From this we can derive the greedy strategy for this problem: initialize two pointers at both ends of the container, and in each round contract the pointer corresponding to the short partition inward, until the two pointers meet.
From this, we can derive the greedy strategy for this problem: initialize two pointers at the two ends, and in each round move the pointer corresponding to the shorter partition inward until the two pointers meet.
Figure 15-11 shows the execution process of the greedy strategy.
1. In the initial state, pointers $i$ and $j$ are at both ends of the array.
2. Calculate the capacity of the current state $cap[i, j]$, and update the maximum capacity.
3. Compare the heights of partition $i$ and partition $j$, and move the short partition inward by one position.
4. Loop through steps `2.` and `3.` until $i$ and $j$ meet.
3. Compare the heights of partitions $i$ and $j$, and move the pointer corresponding to the shorter partition inward by one position.
4. Repeat steps `2.` and `3.` until $i$ and $j$ meet.
=== "<1>"
![Greedy process for the max capacity problem](max_capacity_problem.assets/max_capacity_greedy_step1.png){ class="animation-figure" }
@@ -88,9 +88,9 @@ Figure 15-11 shows the execution process of the greedy strategy.
### 2. &nbsp; Code Implementation
The code loops at most $n$ rounds, **therefore the time complexity is $O(n)$**.
The code runs for at most $n$ rounds, **so the time complexity is $O(n)$**.
Variables $i$, $j$, and $res$ use a constant amount of extra space, **therefore the space complexity is $O(1)$**.
Variables $i$, $j$, and $res$ use only a constant amount of extra space, **so the space complexity is $O(1)$**.
=== "Python"
@@ -425,7 +425,7 @@ Variables $i$, $j$, and $res$ use a constant amount of extra space, **therefore
The reason greedy is faster than exhaustive enumeration is that each round of greedy selection "skips" some states.
For example, in state $cap[i, j]$ where $i$ is the short partition and $j$ is the long partition, if we greedily move the short partition $i$ inward by one position, the states shown in Figure 15-12 will be "skipped". **This means that the capacities of these states cannot be verified later**.
For example, in state $cap[i, j]$, suppose $i$ is the shorter partition and $j$ is the taller partition. If we greedily move the shorter partition $i$ inward by one position, the states shown in Figure 15-12 will be "skipped." **This means that their capacities can no longer be checked later**.
$$
cap[i, i+1], cap[i, i+2], \dots, cap[i, j-2], cap[i, j-1]
@@ -435,6 +435,6 @@ $$
<p align="center"> Figure 15-12 &nbsp; States skipped by moving the short partition </p>
Observing carefully, **these skipped states are actually all the states obtained by moving the long partition $j$ inward**. We have already proven that moving the long partition inward will definitely decrease capacity. That is, the skipped states cannot possibly be the optimal solution, **skipping them will not cause us to miss the optimal solution**.
A closer look shows that **these skipped states are exactly the states obtained by moving the taller partition $j$ inward**. We have already proven that moving the taller partition inward will definitely decrease the capacity. Therefore, none of the skipped states can be the optimal solution, **so skipping them does not cause us to miss the optimum**.
The above analysis shows that the operation of moving the short partition is "safe", and the greedy strategy is effective.
The above analysis shows that moving the shorter partition is a "safe" operation, and that the greedy strategy is effective.
en/docs/chapter_greedy/max_product_cutting_problem.md
+15 -15
View File
@@ -2,11 +2,11 @@
comments: true
---
# 15.4 &nbsp; Max Product Cutting Problem
# 15.4 &nbsp; Maximum Product Cutting Problem
!!! question
Given a positive integer $n$, split it into the sum of at least two positive integers, and find the maximum product of all integers after splitting, as shown in Figure 15-13.
Given a positive integer $n$, split it into the sum of at least two positive integers and find the maximum product of the resulting integers, as shown in Figure 15-13.
![Problem definition of max product cutting](max_product_cutting_problem.assets/max_product_cutting_definition.png){ class="animation-figure" }
@@ -24,11 +24,11 @@ $$
\max(\prod_{i=1}^{m}n_i)
$$
We need to think about: how large should the splitting count $m$ be, and what should each $n_i$ be?
We need to determine how many parts $m$ there should be and what each $n_i$ should be.
### 1. &nbsp; Greedy Strategy Determination
### 1. &nbsp; Determining the Greedy Strategy
Based on experience, the product of two integers is often greater than their sum. Suppose we split out a factor of $2$ from $n$, then their product is $2(n-2)$. We compare this product with $n$:
As a rule of thumb, the product of two integers is often greater than their sum. Suppose we split off a factor of $2$ from $n$; the resulting product is $2(n-2)$. We compare this product with $n$:
$$
\begin{aligned}
@@ -40,7 +40,7 @@ $$
As shown in Figure 15-14, when $n \geq 4$, splitting out a $2$ will increase the product, **which indicates that integers greater than or equal to $4$ should all be split**.
**Greedy strategy one**: If the splitting scheme includes factors $\geq 4$, then they should continue to be split. The final splitting scheme should only contain factors $1$, $2$, and $3$.
**Greedy strategy one**: If the splitting scheme contains a factor $\geq 4$, it should be split further. The final splitting scheme should contain only the factors $1$, $2$, and $3$.
![Splitting causes product to increase](max_product_cutting_problem.assets/max_product_cutting_greedy_infer1.png){ class="animation-figure" }
@@ -50,7 +50,7 @@ Next, consider which factor is optimal. Among the three factors $1$, $2$, and $3
As shown in Figure 15-15, when $n = 6$, we have $3 \times 3 > 2 \times 2 \times 2$. **This means that splitting out $3$ is better than splitting out $2$**.
**Greedy strategy two**: In the splitting scheme, there should be at most two $2$s. Because three $2$s can always be replaced by two $3$s to obtain a larger product.
**Greedy strategy two**: In the splitting scheme, there should be at most two $2$s, because three $2$s can always be replaced by two $3$s to obtain a larger product.
![Optimal splitting factor](max_product_cutting_problem.assets/max_product_cutting_greedy_infer2.png){ class="animation-figure" }
@@ -60,12 +60,12 @@ In summary, the following greedy strategies can be derived.
1. Input integer $n$, continuously split out factor $3$ until the remainder is $0$, $1$, or $2$.
2. When the remainder is $0$, it means $n$ is a multiple of $3$, so no further action is needed.
3. When the remainder is $2$, do not continue splitting, keep it.
4. When the remainder is $1$, since $2 \times 2 > 1 \times 3$, the last $3$ should be replaced with $2$.
3. When the remainder is $2$, do not split it further; keep it as is.
4. When the remainder is $1$, since $2 \times 2 > 1 \times 3$, replace the final $3$ and the remaining $1$ with two $2$s.
### 2. &nbsp; Code Implementation
As shown in Figure 15-16, we don't need to use loops to split the integer, but can use integer division to get the count of $3$s as $a$, and modulo operation to get the remainder as $b$, at which point we have:
As shown in Figure 15-16, we do not need loops to split the integer. Instead, we use integer division to obtain the number of $3$s, denoted by $a$, and the modulo operation to obtain the remainder $b$, giving:
$$
n = 3 a + b
@@ -390,7 +390,7 @@ Please note that for the edge case of $n \leq 3$, a $1$ must be split out, with
<p align="center"> Figure 15-16 &nbsp; Calculation method for max product cutting </p>
**The time complexity depends on the implementation of the exponentiation operation in the programming language**. Taking Python as an example, there are three commonly used power calculation functions.
**The time complexity depends on how exponentiation is implemented in the programming language**. Taking Python as an example, there are three commonly used ways to compute powers.
- Both the operator `**` and the function `pow()` have time complexity $O(\log⁡ a)$.
- The function `math.pow()` internally calls the C library's `pow()` function, which performs floating-point exponentiation, with time complexity $O(1)$.
@@ -399,8 +399,8 @@ Variables $a$ and $b$ use a constant amount of extra space, **therefore the spac
### 3. &nbsp; Correctness Proof
Using proof by contradiction, only analyzing the case where $n \geq 4$.
We use proof by contradiction and consider only the case where $n \geq 4$.
1. **All factors $\leq 3$**: Suppose the optimal splitting scheme includes a factor $x \geq 4$, then it can definitely continue to be split into $2(x-2)$ to obtain a larger (or equal) product. This contradicts the assumption.
2. **The splitting scheme does not contain $1$**: Suppose the optimal splitting scheme includes a factor of $1$, then it can definitely be merged into another factor to obtain a larger product. This contradicts the assumption.
3. **The splitting scheme contains at most two $2$s**: Suppose the optimal splitting scheme includes three $2$s, then they can definitely be replaced by two $3$s for a larger product. This contradicts the assumption.
1. **All factors $\leq 3$**: Suppose the optimal splitting scheme includes a factor $x \geq 4$. Then it can be further split into $2(x-2)$ to obtain a larger (or equal) product. This contradicts the assumption.
2. **The splitting scheme does not contain $1$**: Suppose the optimal splitting scheme includes a factor of $1$. Then it can be merged into another factor to obtain a larger product. This contradicts the assumption.
3. **The splitting scheme contains at most two $2$s**: Suppose the optimal splitting scheme includes three $2$s. Then they can be replaced by two $3$s, yielding a larger product. This contradicts the assumption.
en/docs/chapter_greedy/summary.md
+3 -3
View File
@@ -12,7 +12,7 @@ comments: true
- In the coin change problem, for certain coin combinations, greedy algorithms can guarantee finding the optimal solution; for other coin combinations, however, greedy algorithms may find very poor solutions.
- Problems suitable for solving with greedy algorithms have two major properties: greedy choice property and optimal substructure. The greedy choice property represents the effectiveness of the greedy strategy.
- For some complex problems, proving the greedy choice property is not simple. Relatively speaking, disproving it is easier, such as in the coin change problem.
- Solving greedy problems mainly consists of three steps: problem analysis, determining the greedy strategy, and correctness proof. Among these, determining the greedy strategy is the core step, and correctness proof is often the difficult point.
- The fractional knapsack problem, based on the 0-1 knapsack problem, allows selecting a portion of items, and therefore can be solved using greedy algorithms. The correctness of the greedy strategy can be proven using proof by contradiction.
- The max capacity problem can be solved using exhaustive enumeration with time complexity $O(n^2)$. By designing a greedy strategy to move the short partition inward in each round, the time complexity can be optimized to $O(n)$.
- Solving greedy problems mainly consists of three steps: problem analysis, determining the greedy strategy, and correctness proof. Among these, determining the greedy strategy is the core step, and correctness proof is often the main difficulty.
- The fractional knapsack problem, based on the 0-1 knapsack problem, allows selecting fractions of items, and therefore can be solved using greedy algorithms. The correctness of the greedy strategy can be proven using proof by contradiction.
- The max capacity problem can be solved using exhaustive enumeration with time complexity $O(n^2)$. By designing a greedy strategy to move the shorter side inward in each round, the time complexity can be optimized to $O(n)$.
- In the max product cutting problem, we successively derive two greedy strategies: integers $\geq 4$ should all continue to be split, and the optimal splitting factor is $3$. The code includes exponentiation operations, and the time complexity depends on the implementation method of exponentiation, typically being $O(1)$ or $O(\log n)$.