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MAT270 discrete mathematics课程简介
Discrete Mathematics is designed to meet the needs not only of students majoring in computer science but of a wider audience, especially students in mathematics and science. The course provides tools for formal reasoning. Topics include counting rules, propositional and first-order logic, set theory, functions (with an emphasis on recursive functions), partial order and equivalence relations, Boolean algebra, and switching circuits. Graphs and trees are also introduced. With an emphasis on communication skills, students are required to interpret, describe, discuss, and justify conclusions based on logical reasoning. While the particular focus of the course is on reasoning related to computer programs, no knowledge of programming is required.
Advisory: It is advisable to have knowledge in a course equivalent to MAT-121: College Algebra with a grade of C or better to succeed in this course. Students are responsible for making sure that they have the necessary knowledge.
PREREQUISITES
Discrete Mathematics is a fundamental course that provides a solid foundation for computer science, mathematics, and science students. The course covers a wide range of topics, including counting rules, logic, set theory, functions, relations, Boolean algebra, and graph theory. The emphasis on formal reasoning and communication skills makes this course valuable to all students.
To succeed in this course, it is advisable to have a good understanding of college algebra, as it forms the basis of many concepts covered in the course. Students should ensure that they have the necessary prerequisite knowledge before enrolling in the course.
The course provides tools for formal reasoning, which are essential for students pursuing computer science or related fields. These tools enable students to analyze and design algorithms, write correct programs, and prove the correctness of programs.
Overall, discrete mathematics is a challenging but rewarding course that prepares students for further study in computer science, mathematics, and science. The skills and knowledge gained in this course are valuable not only in academia but also in industry and other fields.
MAT270 discrete mathematics HELP（EXAM HELP， ONLINE TUTOR）
- A stack of cans (see Figure 1) refers to an arrangement of $n$ (equal-sized) cans in rows such that (i) the cans in each row form a single continuous block, and (ii) any can, except the ones in the bottom row, touches exactly two cans from the row beneath it. If the bottom row contains $k$ cans, this is a $(n, k)$ stack of cans. Figure 1 shows a $(28,12)$ stack of cans.
Your article should give a formula (as a function of $k$ ) for the number $s_k$ of stack of cans with exactly $k$ cans in the bottom row (and an arbitrary total number of cans). Your article should at least contain:
- A generating function for $\left(s_k\right)_{k \geq 0}$,
- A closed-form formula for $s_k$ for any $k \geq 0$,
- A relation between $s_k$ and another well-known sequence of numbers.
If you can find a bijective proof for $s_k$, you could also provide it. This is however not compulsory and should not replace any of the items above.
Let $S(x)$ be the generating function for the sequence ${s_k}{k\geq 0}$, i.e., $S(x)=\sum{k\geq 0} s_kx^k$. We can derive a recursive formula for $S(x)$ as follows:
Consider a $(n,k+1)$ stack of cans. If we remove the bottom row of $k+1$ cans, we are left with a $(n-k-1,k)$ stack of cans. Conversely, given any $(n-k-1,k)$ stack of cans, we can add a row of $k+1$ cans to the bottom to obtain a $(n,k+1)$ stack of cans. Therefore, we have the recurrence relation:
s_{k+1}=\sum_{n=k+1}^{\infty}s_{k}(n-k-1),sk+1=n=k+1∑∞sk(n−k−1),
where $s_k(n-k-1)$ represents the number of $(n,k+1)$ stacks of cans obtained by adding a row of $k+1$ cans to the top of a $(n-k-1,k)$ stack of cans.
Multiplying both sides of the recurrence relation by $x^{k+1}$ and summing over $k\geq 0$, we get:
\begin{aligned} \sum_{k\geq 0}s_{k+1}x^{k+1} &= \sum_{k\geq 0}\sum_{n=k+1}^{\infty}s_k(n-k-1)x^{k+1}\\ &= \sum_{n\geq 1}\sum_{k=0}^{n-1}s_k(n-k-1)x^{k+1}\\ &= \sum_{n\geq 1}\sum_{j=1}^{n-1}s_{n-j-1}jx^{n}\\ &= x\sum_{n\geq 1}\sum_{j=0}^{n-1}s_{n-j-1}jx^{n-1}\\ &= x\sum_{n\geq 0}\sum_{j=0}^{n}s_{n-j}jx^{n}\\ &= x\left(\sum_{n\geq 0}s_nx^n\right)\left(\sum_{j\geq 0}jx^j\right)\\ &= x\left(\sum_{n\geq 0}s_nx^n\right)\left(\frac{x}{(1-x)^2}\right). \end{aligned}k≥0∑sk+1xk+1=k≥0∑n=k+1∑∞sk(n−k−1)xk+1=n≥1∑k=0∑n−1sk(n−k−1)xk+1=n≥1∑j=1∑n−1sn−j−1jxn=xn≥1∑j=0∑n−1sn−j−1jxn−1=xn≥0∑j=0∑nsn−jjxn=x(n≥0∑snxn)(j≥0∑jxj)=x(n≥0∑snxn)((1−x)2x).
Therefore,
S(x)=\frac{x}{(1-x)^2}\sum_{k\geq 0}s_kx^k.S(x)=(1−x)2xk≥0∑skxk.
Solving for $\sum_{k\geq 0}s_kx^k$ in terms of $S(x)$, we obtain:
\sum_{k\geq 0}s_kx^k = x(1-x)S(x).k≥0∑skxk=x(1−x)S(x).
Since $s_0=1$, we have $\sum_{k\geq 0}s_k=1$, which implies that $S(1)=1$. Therefore, we have:
\sum_{k\geq 0}s_kx^k = \frac{x(1-x)}{(1-x)^2-x}=\frac{x}{1-2x+x^2}.k≥0∑skxk=(1−x)2−xx(1−x)=1−2x+x2x.
Using partial fraction decomposition, we can write.
- Recall the inclusion / exclusion formula, let $A_1, A_2, \ldots, A_n$ be events, and
$$
S_k:=\sum_{1 \leq i_1<i_2<\cdots<i_k \leq n} \mathbb{P}\left(A_{i_1} \bigwedge A_{i_2} \bigwedge \ldots \bigwedge A_{i_k}\right) .
$$
The inclusion/exclusion formula gives
$$
\mathbb{P}\left(\bigvee_{i=1}^n A_i\right)=\sum_{j=1}^n(-1)^{j-1} S_j
$$
In this problem we will develop and use an extension of it.
- Show that for all odd $k$ in ${1, \ldots, n}$,
$$
\mathbb{P}\left(\bigvee_{i=1}^n A_i\right) \leq \sum_{j=1}^k(-1)^{j-1} S_j,
$$
and describe how the union bound is a special case of this. - Show that for all even $k$ in ${1, \ldots, n}$,
$$
\mathbb{P}\left(\bigvee_{i=1}^n A_i\right) \geq \sum_{j=1}^k(-1)^{j-1} S_j
$$
The inclusion-exclusion principle is a powerful tool in probability theory, used to calculate the probability of the union of events. The standard form of the principle provides an exact formula for the probability of the union of $n$ events $A_1, A_2, \ldots, A_n$:
\mathbb{P}\left(\bigvee_{i=1}^n A_i\right)=\sum_{k=1}^n(-1)^{k-1} \sum_{1 \leq i_1<i_2<\cdots<i_k \leq n} \mathbb{P}\left(A_{i_1} \bigwedge A_{i_2} \bigwedge \ldots \bigwedge A_{i_k}\right) ,P(i=1⋁nAi)=k=1∑n(−1)k−11≤i1<i2<⋯<ik≤n∑P(Ai1⋀Ai2⋀…⋀Aik),
where $S_k$ is the sum over all $k$-fold intersections of events $A_i$. In this article, we will extend the inclusion-exclusion principle to provide tighter bounds on the probability of the union of events, based on the parity of the size of the intersection.
Let $A_1, A_2, \ldots, A_n$ be events, and let $S_k$ be as defined above. We will prove two separate inequalities based on the parity of $k$.
For odd $k$, we have:
\begin{align*} \mathbb{P}\left(\bigvee_{i=1}^n A_i\right) &= \sum_{k=1}^n (-1)^{k-1} S_k + (-1)^n \mathbb{P}\left(\bigwedge_{i=1}^n A_i\right) \ &\leq \sum_{k=1}^n (-1)^{k-1} S_k, \end{align*}
where the inequality follows from the fact that $\mathbb{P}\left(\bigwedge_{i=1}^n A_i\right) \geq 0$.
To see why this inequality holds, note that for odd $k$, the sum over all $k$-fold intersections in the inclusion-exclusion formula alternates in sign, starting with a positive sign. Therefore, the first $k$ terms in the sum are positive, and the remaining terms are negative. In particular, the sum up to $k$ is greater than or equal to the sum up to $k+1$, which implies the inequality.
The union bound is a special case of this inequality, where $k=1$. In this case, we have:
\mathbb{P}\left(\bigvee_{i=1}^n A_i\right) \leq \mathbb{P}\left(\bigvee_{i=1}^n A_i\right) + \mathbb{P}\left(\bigwedge_{i=1}^n \overline{A_i}\right) = 1,P(i=1⋁nAi)≤P(i=1⋁nAi)+P(i=1⋀nAi)=1,
where $\overline{A_i}$ denotes the complement of $A_i$. Therefore, the probability of the union of events is always less than or equal to 1.
Textbooks
• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.
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