Question 7
When we analyze the worst time complexity of an algorithm, it is worth observing how the computation time increases as , the size of input data, increases. The Landau’s notation, which is often used for representing an asymptotic upper bound of ignoring the constant factor, is defined as the following. For a positive function , if there exists a positive constant , and
holds, then
is defined to hold.
-
and are defined to be equal if and are equivalent for any . For each of (A) to (C), find all formulae in the box below that are equal to it. If there is no such formula, just write “none”.
(A) (B) (C)
- For each proposition below, determine with proof whether it holds or not. Note that and are positive functions.
(A) and are equal.
(B) If holds, then .
当我们分析一个算法的最坏时间复杂度时,观察计算时间 随输入数据大小 的增加而增加是很有价值的。Landau 的 表示法,通常用于表示忽略常数因子的 的渐近上界,定义如下。对于一个正函数 ,如果存在一个正常数 ,且
成立,则定义
成立。
-
如果 和 对任何 都等价,则定义 和 是相等的。对于 (A) 到 (C) 的每一个,找到下面框中等于它的所有公式。如果没有这样的公式,只写“none”。
(A) (B) (C)
- 对于下面的每个命题,确定是否成立,并提供证明。注意 和 是正函数。
-
(A) 和 是相等的。
-
(B) 如果 成立,则 。
Question 8
Let be the set of non-zero complex two-dimensional vectors. Let be a 2 by 2 real symmetric matrix, and be the unit matrix.
-
Find all the eigenvalues of .
-
Under the assumption of , answer i) and ii).
i) Let be the matrix whose first and second columns consist of the eigenvectors and for the eigenvalues and , respectively. Show that is invertible and satisfies .
ii) Prove that the set and are equal.
-
For each of the statements A), B), and C), answer the conditions on matrix elements for the statement to hold.
A) Every can be expressed as with some .
B) No can be expressed as with some .
C) At least one can be expressed as with some .
设 为非零复二维向量的集合。设 为一个 2×2 的实对称矩阵, 为单位矩阵。
-
找出 的所有特征值 。
-
在假设 的条件下,回答 i) 和 ii)。
i) 设 为一个矩阵,其第一列和第二列分别由特征值 和 的特征向量 和 组成。证明 是可逆的,并且满足 。
ii) 证明集合 和 是相等的。
-
对于每个陈述 A), B), 和 C),回答矩阵元素 的条件使该陈述成立。
A) 每个 都可以表示为 ,其中 。
B) 没有 可以表示为 ,其中 。
C) 至少有一个 可以表示为 ,其中 。
Question 9
An array of real numbers, , is called a heap if for , where is the quotient of integer divided by 2 (e.g., ). A heap can be treated as a binary tree such that is the root node and is the parent node of . Answer the following questions.
- Explain whether array is a heap or not.
- Show that if is a heap, is also a heap for any .
- Suppose that is a heap but is not. Describe a procedure that exchanges elements in to make it a heap in worst-case time.
- Suppose that is a heap. Prove that is maximum among all elements in . Suppose that after replacing with , is not a heap. Describe a procedure that exchanges elements in to make it a heap in worst-case time.
- Describe an algorithm that exchanges elements in to make it a heap in worst-case time.
- Using the procedures defined above, describe an algorithm that sorts array in worst-case time.
一个包含 个实数的数组 ,如果满足 对于 ,其中 是整数 除以 2 的商(例如,),则称其为堆。堆可以被看作是一个二叉树,其中 是根节点, 是 的父节点。回答以下问题。
- 解释数组 是否是一个堆。
- 证明如果 是一个堆,那么 也是一个堆,对于任何 。
- 假设 是一个堆,但 不是。描述一种在 最坏情况下,将 中的元素交换使其成为堆的方法。
- 假设 是一个堆。证明 是 中最大的元素。假设在用 替换 后, 不是一个堆。描述一种在 最坏情况下,将 中的元素交换使其成为堆的方法。
- 描述一种在 最坏情况下,将 中的元素交换使其成为堆的算法。
- 使用上述定义的方法,描述一种在 最坏情况下,对数组 进行排序的算法。
Question 10
A bipartite graph is an undirected graph whose nodes can be divided into two groups such that any two nodes in a group are not connected by an edge. Answer the following questions about bipartite graphs.
- Write the incidence matrix of the following graph .
graph TB
subgraph BLACK_NODES
v1
v2
v3
v4
end
subgraph WHITE_NODES
v5
v6
v7
v8
end
v1 --- v6
v1 --- v8
v2 --- v5
v2 --- v6
v3 --- v7
v3 --- v8
v4 --- v5
v4 --- v8
-
A matching in a graph is a set of edges without common nodes. A perfect matching is a matching that covers all nodes of the graph. List all perfect matchings of .
-
Prove that a tree is always a bipartite graph.
-
Prove that a bipartite graph does not contain a circuit with an odd number of edges.
二分图是一个无向图,其节点可以分为两组,使得组内的任意两个节点不通过边连接。回答以下关于二分图的问题。
- 写出以下图 的关联矩阵。
graph TB
subgraph BLACK_NODES
v1
v2
v3
v4
end
subgraph WHITE_NODES
v5
v6
v7
v8
end
v1 --- v6
v1 --- v8
v2 --- v5
v2 --- v6
v3 --- v7
v3 --- v8
v4 --- v5
v4 --- v8
-
图中的一个匹配是一组没有公共节点的边。完美匹配是覆盖图中所有节点的匹配。列出图 的所有完美匹配。
-
证明树总是一个二分图。
-
证明二分图不包含奇数条边的回路。
Question 11
For an arbitrary random variable that takes values in non-negative integers, we define the probability generating function of as . Here, denotes the expected value of , denotes the probability that assumes value , and represents a real number.
- Show the following equalities (a), (b). (a) (b)
In the following, and are mutually independent, identically distributed random variables that take values in non-negative integers.
-
For a positive integer , we define a random variable . Show .
-
We define a random variable . Here,
Show .
(Hint: for a positive integer )
- For , , calculate and .
对于一个取非负整数值的任意随机变量 ,我们定义 的概率生成函数 为 。其中, 表示 的期望值, 表示 取值为 的概率, 表示一个实数。
- 证明以下等式 (a), (b)。 (a) (b)
在下列情形中, 和 是相互独立的同分布随机变量,取非负整数值。
-
对于一个正整数 ,我们定义一个随机变量 。证明 。
-
我们定义一个随机变量 。 其中,
证明 。
(提示: 对于一个正整数 )
- 对于 , ,计算 和 。
Question 12
Suppose that the maximum score of the global alignment of the two sequences, and , is calculated by a dynamic programming using the following iterations (A).
Here we assume that , and that is the score of alignment of and .
We also assume that the same score of is given to any gap of length .
Answer the following problems.
-
Show the general form of .
-
In order to get the maximum score of the alignment as after iterations using (A) for and , answer what initial values should be assigned to the following variables.
- for
- for
You may suppose that
\begin{aligned}
M(i, j) &= \max[M(i-1, j-1), X_1(i-1, j-1), Y_1(i-1, j-1),X_2(i-1, j-1), Y_2(i-1, j-1)] + s(x_i, y_j), \
X_1(i, j) &= \max[M(i-1, j) - d_1, X_1(i-1, j) - e_1], \
Y_1(i, j) &= \max[M(i, j-1) - d_1, Y_1(i, j-1) - e_1], \
X_2(i, j) &= \max[M(i-1, j) - d_2, X_2(i-1, j) - e_2], \
Y_2(i, j) &= \max[M(i, j-1) - d_2, Y_2(i, j-1) - e_2].
\end{aligned}
--- 假设通过使用以下迭代 (A) 计算两个序列 $x = x_1, \ldots, x_m$ 和 $y = y_1, \ldots, y_n$ 的全局比对的最大得分。\begin{aligned}
M(i, j) &= \max[M(i-1, j-1), X(i-1, j-1), Y(i-1, j-1)] + s(x_i, y_j), \
X(i, j) &= \max[M(i-1, j) - d, X(i-1, j) - e], \
Y(i, j) &= \max[M(i, j-1) - d, Y(i, j-1) - e].
\end{aligned} \tag{A}
这里我们假设 $d > 0$,$e > 0$ 并且 $s(x_i, y_j)$ 是 $x_i$ 和 $y_j$ 的比对得分。 我们还假设对于任意长度为 $k$ 的间隙给出了相同的得分 $g(k)$。 回答以下问题。 1. 展示 $g(k)$ 的一般形式。 2. 为了在使用 (A) 迭代后获得最大比对得分 $\max[M(m, n), X(m, n), Y(m, n)]$,回答应为以下变量分配什么初始值。 - $X(i, 0)$ 对于 $i = 1, \ldots, m$ - $Y(0, j)$ 对于 $j = 1, \ldots, n$ 你可以假设\begin{aligned}
M(0, 0) &= 0, \
X(0, 0) &= Y(0, 0) = -\infty, \
M(i, 0) &= Y(i, 0) = -\infty \quad (i = 1, \ldots, m), \
M(0, j) &= X(0, j) = -\infty \quad (j = 1, \ldots, n).
\end{aligned}
3. 对于生物序列的比对,通常我们设置 $d > e$。简要描述原因。 4. 为了更准确地处理长间隙的得分,让我们考虑使用以下迭代进行比对。展示当 $d_2 > d_1$,$e_2 > e_1$ 时长度为 $k$ 的间隙得分的一般形式。注意对 $k$ 值的分类是必要的。\begin{aligned}
M(i, j) &= \max[M(i-1, j-1), X_1(i-1, j-1), Y_1(i-1, j-1), X_2(i-1, j-1), Y_2(i-1, j-1)] + s(x_i, y_j), \
X_1(i, j) &= \max[M(i-1, j) - d_1, X_1(i-1, j) - e_1], \
Y_1(i, j) &= \max[M(i, j-1) - d_1, Y_1(i, j-1) - e_1], \
X_2(i, j) &= \max[M(i-1, j) - d_2, X_2(i-1, j) - e_2], \
Y_2(i, j) &= \max[M(i, j-1) - d_2, Y_2(i, j-1) - e_2].
\end{aligned}