IS CS-2019S2-06

题目来源：Problem 6 日期：2024-08-03 题目主题：CS-算法-决策问题

解题思路

这道题目涉及三个决策问题 A、B 和 C，分别关于数组中的两数之和、三数之和为零、以及平面上三点共线的判定。我们需要设计算法、分析复杂度，并且证明问题之间的关系。解题过程中会用到双指针、哈希表、代数技巧等方法。

Solution

1. Algorithm for Problem A

To solve Problem A in $O(n)$ time, we can use the two-pointer technique, taking advantage of the fact that the array is sorted.

def solve_problem_A(S, c):
    left, right = 0, len(S) - 1
    while left < right:
        current_sum = S[left] + S[right]
        if current_sum == c:
            return True
        elif current_sum < c:
            left += 1
        else:
            right -= 1
    return False

This algorithm works as follows:

1. Initialize two pointers, left at the beginning and right at the end of the array.
2. While left is less than right: a. Calculate the sum of S[left] and S[right]. b. If the sum equals c, we’ve found a solution, return True. c. If the sum is less than c, increment left to increase the sum. d. If the sum is greater than c, decrement right to decrease the sum.
3. If we exit the loop without finding a solution, return False.

Time complexity: $O(n)$, as we traverse the array at most once. Space complexity: $O(1)$, as we only use two pointers.

Proof of Correctness

We will prove this by contradiction. Suppose the greedy algorithm fails to find a solution $(s_i,s_j)$ such that $s_i+s_j=c$, even though such a solution exists.

Let $(s_p,s_q)$ be a pair that sums to $c$, where $p<q$. Consider the moment in our algorithm when the left pointer is at some position $l\le p$ and the right pointer is at some position $r\ge q$.

1. If $l=p$ and $r=q$, the algorithm would have found the solution, contradicting our assumption.

2. If $l<p$ or $r>q$, we have three cases: a) If $S[l]+S[r]=c$, the algorithm would have found a solution, contradicting our assumption. b) If $S[l]+S[r]<c$, the algorithm moves the left pointer. This is the correct move because:

   • We know $S[p]+S[q]=c$
   • And $S[l]+S[r]<c$
   • Since $S$ is sorted, $S[l]\le S[p]$ and $S[r]\ge S[q]$
   • Therefore, $S[l]+S[q]\le S[p]+S[q]=c$
   • This means no element to the left of or at $l$ can sum with $S[q]$ to $c$
   • So, we must move the left pointer to find a potential solution.

3. If $S[l]+S[r]>c$, the algorithm moves the right pointer. This is the correct move because:

   • We know $S[p]+S[q]=c$
   • And $S[l]+S[r]>c$
   • Since $S$ is sorted, $S[l]\le S[p]$ and $S[r]\ge S[q]$
   • Therefore, $S[p]+S[r]\ge S[p]+S[q]=c$
   • This means no element to the right of or at $r$ can sum with $S[p]$ to $c$
   • So, we must move the right pointer to find a potential solution.

In all cases, the algorithm makes the correct decision to move towards the solution $(s_p,s_q)$. This contradicts our initial assumption that the algorithm fails to find a solution when one exists.

Therefore, if a solution exists, the greedy algorithm will always find it. If the algorithm doesn’t find a solution, we can conclude that no solution exists.

2. Algorithm for Problem B

To solve Problem B in $O(n^2)$ time, we can use a hash table to store the negation of the sum of two elements, then check if the third element exists in the hash table.

def solve_problem_B(T):
    T = sorted(T)  # Sort the array first
    for i in range(len(T) - 2):
        if i > 0 and T[i] == T[i-1]:
            continue  # Skip duplicates
        left, right = i + 1, len(T) - 1
        while left < right:
            total = T[i] + T[left] + T[right]
            if total == 0:
                return True
            elif total < 0:
                left += 1
            else:
                right -= 1
    return False

This algorithm works as follows:

1. Sort the array T (this takes $O(n\log n)$ time, which is dominated by $O(n^2)$).
2. Iterate through the array with index i from $0$ to $n-3$.
3. For each i, use two pointers left and right to find if there exist j and k such that $T[i]+T[j]+T[k]=0$.
4. If found, return True. If not found after checking all possibilities, return False.

Time complexity: $O(n^2)$, as we have a nested loop structure. Space complexity: $O(1)$ if we don’t count the space for sorting, $O(n)$ if we do.

3. Expressing c in terms of a and b

To express $c$ using $a$ and $b$ for the points $(a,a^3)$, $(b,b^3)$, and $(c,c^3)$ being collinear:

The points are collinear if the slope between any two pairs of points is the same. The slope between $(a,a^3)$ and $(c,c^3)$ is:

$$\frac{c^3-a^3}{c-a}=c^2+ca+a^2$$

The slope between $(b,b^3)$ and $(c,c^3)$ is:

$$\frac{c^3-b^3}{c-b}=c^2+cb+b^2$$

Setting the slopes equal gives:

$$c^2+ca+a^2=c^2+cb+b^2$$

Simplifying, we get:

$$a^2+ac=b^2+bc$$

Solving for $c$:

$$c=\frac{a^2-b^2}{b-a}=-a-b$$

4. Proving no $O(n)$ algorithm for Problem C

We will prove this by contradiction. Assume there exists an $O(n)$ algorithm for Problem C. We’ll show that this would imply an $O(n)$ algorithm for Problem B, contradicting the given assumption.

Proof:

1. Suppose we have an $O(n)$ algorithm $A_C$ for Problem C.

2. Given an instance of Problem B with set $T=\{t_1,t_2,\dots ,t_n\}$, we can transform it into an instance of Problem C as follows: For each $t_i\in T$, create a point $(t_i,t_i^3)$ in the 2D plane. Let this set of points be $P=\{(t_1,t_1^3),(t_2,t_2^3),\dots ,(t_n,t_n^3)\}$.

3. Now, we claim that there exist $i,j,k$ such that $t_i+t_j+t_k=0$ if and only if the corresponding points $(t_i,t_i^3),(t_j,t_j^3),(t_k,t_k^3)$ are collinear.

   To prove this claim:

   • If $t_i+t_j+t_k=0$, then $t_k=-(t_i+t_j)$.
   • The points $(t_i,t_i^3),(t_j,t_j^3),(-(t_i+t_j),-(t_i+t_j{)}^3)$ are collinear because they satisfy the equation we derived in Question 3.
   • Conversely, if three points $(a,a^3),(b,b^3),(c,c^3)$ are collinear, then from our result in Question 3, we know that one of them must be the negative sum of the other two.

4. Therefore, we can solve Problem B by:

   • Transforming the input of B to an input of C (takes $O(n)$ time)
   • Running algorithm $A_C$ on this input (takes $O(n)$ time by assumption)
   • Interpreting the result of $A_C$ as the result for B (takes $O(1)$ time)

5. This would give us an $O(n)$ algorithm for Problem B, which contradicts our assumption that no such algorithm exists.

Therefore, our initial assumption must be false, and there cannot exist an $O(n)$ algorithm for Problem C under the given conditions.

知识点

两数之和双指针哈希表复杂度分析反证法

难点思路

问题 3 和问题 4 可能是较难的部分。问题 3 需要仔细的代数推导，而问题 4 需要构造一个巧妙的转化来建立问题 B 和问题 C 之间的联系。

解题技巧和信息

1. 对于排序数组的两数之和问题，双指针方法通常是最优解。
2. 对于三数之和问题，可以固定一个数，然后在剩余数中寻找两数之和，这样可以将复杂度从 O(n^3) 降低到 O(n^2)。
3. 在处理几何问题时，利用点斜式方程或者向量的方法通常很有帮助。
4. 在证明算法复杂度下界时，归约（reduction）是一个强大的工具。如果我们可以在线性时间内将一个已知的 ” 困难 ” 问题转化为目标问题，那么目标问题就不可能有比这个 ” 困难 ” 问题更优的解法。

重点词汇

two-pointer technique 双指针技巧

hash table 哈希表

collinearity 共线性

quadratic equation 二次方程

proof by contradiction 反证法

reduction 归约

参考资料

1. Introduction to Algorithms (CLRS), Chapter 2 (Getting Started) and Chapter 33 (Computational Geometry)
2. Algorithm Design Manual by Steven Skiena, Chapter 2 (Algorithm Analysis) and Chapter 17 (Computational Geometry)
3. Competitive Programmer’s Handbook by Antti Laaksonen, Chapter 5 (Complete Search)