高级算法 (Fall 2016)/''Lovász'' Local Lemma and 组合数学 (Fall 2017)/Problem Set 1: Difference between pages

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(Created page with "=Random Search for <math>k</math>-SAT= We start by giving the definition of <math>k</math>-CNF and <math>k</math>-SAT. {{Theorem|Definition (exact-<math>k</math>-CNF)| :A logi...")
 
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=Random Search for <math>k</math>-SAT=
*每道题目的解答都要有<font color="red" size=4>完整的解题过程</font>。中英文不限。
We start by giving the definition of <math>k</math>-CNF and <math>k</math>-SAT.
{{Theorem|Definition (exact-<math>k</math>-CNF)|
:A logic expression <math>\phi</math> defined on <math>n</math> Boolean variables <math>x_1,x_2,\ldots,x_n\in\{\mathrm{true},\mathrm{false}\}</math> is said to be a '''conjunctive normal form (CNF)''' if <math>\phi</math> can be written as a conjunction(AND) of '''clauses''' as <math>\phi=C_1\wedge C_2\wedge\cdots\wedge C_m</math>, where each clause <math>C_i=\ell_{i_1}\vee \ell_{i_2}\vee\cdots\vee\ell_{i_k}</math> is a disjunction(OR) of '''literals''', where every literal <math>\ell_j</math> is either a variable <math>x_i</math> or the negation <math>\neg x_i</math> of a variable.
:*We call a CNF formula a '''exact-<math>k</math>-CNF''' if every clause consists of ''exact'' <math>k</math> ''distinct'' literals.
}}
For example:
:<math>
(x_1\vee \neg x_2 \vee \neg x_3)\wedge (\neg x_1\vee \neg x_3\vee x_4)\wedge (x_1\vee x_2\vee x_4)\wedge (x_2\vee x_3\vee \neg x_4)
</math>
is an exact-<math>3</math>-CNF formula.


;Remark
== Problem 1 ==
:The notion of exact-<math>k</math>-CNF is slightly more restrictive than the <math>k</math>-CNF, where each clause consists of ''at most'' <math>k</math> variables. The discussion of the subtle differences between these two definitions can be found [https://en.wikipedia.org/wiki/Boolean_satisfiability_problem#3-satisfiability here].
#有<math>k</math>种不同的明信片，每种明信片有无限多张，寄给<math>n</math>个人，每人一张，有多少种方法？
#有<math>k</math>种不同的明信片，每种明信片有无限多张，寄给<math>n</math>个人，每人一张，每个人必须收到不同种类的明信片，有多少种方法？
#有<math>k</math>种不同的明信片，每种明信片有无限多张，寄给<math>n</math>个人，每人收到<math>r</math>张不同的明信片（但不同的人可以收到相同的明信片），有多少种方法？
#只有一种明信片，共有<math>m</math>张，寄给<math>n</math>个人，全部寄完，每个人可以收多张明信片或者不收明信片，有多少种方法？
#有<math>k</math>种不同的明信片，其中第<math>i</math>种明信片有<math>m_i</math>张，寄给<math>n</math>个人，全部寄完，每个人可以收多张明信片或者不收明信片，有多少种方法？


A logic expression <math>\phi</math> is said to be '''satisfiable''' if there is an assignment of values of true or false to the variables <math>\boldsymbol{x}=(x_1,x_2,\ldots,x_n)</math> so that <math>\phi(\boldsymbol{x})</math> is true. For a CNF <math>\phi</math>, this mean that there is a truth assignment that satisfies all clauses in <math>\phi</math> simultaneously.
== Problem 2 ==
Find the number of ways to select <math>2n</math> balls from <math>n</math> identical blue balls, <math>n</math> identical red balls and <math>n</math> identical green balls.
* Give a combinatorial proof for the problem.
* Give an algebraic proof for the problem.


The '''exact-<math>k</math>-satisfiability (exact-<math>k</math>-SAT)''' problem is that given as input an exact-<math>k</math>-CNF formula <math>\phi</math> decide whether <math>\phi</math> is satisfiable.
== Problem 3 ==
{{Theorem|exact-<math>k</math>-SAT|
*一个长度为<math>n</math>的“山峦”是如下由<math>n</math>个"/"和<math>n</math>个"\"组成的，从坐标<math>(0,0)</math>到<math>(0,2n)</math>的折线，但任何时候都不允许低于<math>x</math>轴。例如下图：
:'''Input:''' an exact-<math>k</math>-CNF formula <math>\phi</math>.
:'''Output:''' whether <math>\phi</math> is satisfiable.
}}
It is well known that <math>k</math>-SAT is '''NP-complete''' for any <math>k\ge 3</math>. The same also holds for the exact-<math>k</math>-SAT.


== Satisfiability of exact-<math>k</math>-CNF==
    /\
Inspired by the Lovasz local lemma, we now consider the dependencies between clauses in a CNF formula.
  /  \/\/\    /\/\
  /        \/\/    \/\/\
  ----------------------
:长度为<math>n</math>的“山峦”有多少？


Given a CNF formula <math>\phi</math> defined over Boolean variables <math>\mathcal{X}=\{x_1,x_2,\ldots,x_n\}</math> and a clause <math>C</math> in <math>\phi</math>, we use <math>\mathsf{vbl}(C)\subseteq\mathcal{X}</math> to denote the set of variables that appear in <math>C</math>.
*一个长度为<math>n</math>的“地貌”是由<math>n</math>个"/"和<math>n</math>个"\"组成的，从坐标<math>(0,0)</math>到<math>(0,2n)</math>的折线，允许低于<math>x</math>轴。长度为<math>n</math>的“地貌”有多少？
For a clause <math>C</math> in a CNF formula <math>\phi</math>, its '''degree''' <math>d(C)=|\{D\neq C\mid \mathsf{D}\cap\mathsf{C}\neq\emptyset\}|</math> is the number of other clauses in <math>\phi</math> that share variables with <math>C</math>. The '''maximum degree''' <math>d</math> of a CNF formula <math>\phi</math> is <math>d=\max_{C\text{ in }\phi}d(C)</math>.


By the Lovasz local lemma, we almost immediately have the following theorem for the satisfiability of exact-<math>k</math>-CNF with bounded degree.
== Problem 4==
{{Theorem|Theorem|
李雷和韩梅梅竞选学生会主席，韩梅梅获得选票 <math>p</math> 张，李雷获得选票 <math>q</math> 张，<math>p>q</math>。我们将总共的 <math>p+q</math> 张选票一张一张的点数，有多少种选票的排序方式使得在整个点票过程中，韩梅梅的票数一直高于李雷的票数？等价地，假设选票均匀分布的随机排列，以多大概率在整个点票过程中，韩梅梅的票数一直高于李雷的票数。
:Let <math>\phi</math> be an exact-<math>k</math>-CNF formula with maximum degree <math>d</math>. If <math>d+1\le 2^{k}/\mathrm{e}</math> then <math>\phi</math> is always satisfiable.
}}
{{Proof|
Let <math>X_1,X_2,\ldots,X_n</math> be Boolean random variables sampled uniformly and independently from <math>\{\text{true},\text{false}\}</math>. We are going to show that <math>\phi</math> is satisfied by this random assignment with positive probability. Due to the probabilistic method, this will prove the existence of a satisfying assignment for <math>\phi</math>.


Suppose there are <math>m</math> clauses <math>C_1,C_2,\ldots,C_m</math> in <math>\phi</math>. Let <math>A_i</math> denote the bad event that <math>C_i</math> is not satisfied by the random assignment <math>X_1,X_2,\ldots,X_n</math>. Clearly, each <math>A_i</math> is dependent with at most <math>d</math> other <math>A_j</math>'s. And our goal is to show that
==Problem 5==
:<math>\Pr\left[\bigwedge_{i=1}^m\overline{A_i}\right]>0</math>.
A <math>2\times n</math> rectangle is to be paved with <math>1\times 2</math> identical blocks and <math>2\times 2</math> identical blocks. Let <math>f(n)</math> denote the number of ways that can be done. Find a recurrence relation for <math>f(n)</math>, solve the recurrence relation.


Recall that in an exact-<math>k</math>-CNF, every clause <math>C_i</math> consists of exact <math>k</math> variable, and is violated by precisely one local assignment among all <math>2^k</math> possibilities. Thus the probability that <math>C_i</math> is not satisfied is <math>\Pr[A_i]=2^{-k}</math>.
== Problem 6 ==
* 令<math>s_n</math>表示长度为<math>n</math>，没有2个连续的1的二进制串的数量，即
*:<math>s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}|</math>。
:求 <math>s_n</math>。


Assuming that <math>d+1\le 2^{k}/\mathrm{e}</math>, i.e. <math>\mathrm{e}(d+1)2^{-k}\le 1</math>, by the Lovasz local lemma (symmetric case), we have
*令<math>t_n</math>表示长度为<math>n</math>，没有3个连续的1的二进制串的数量，即
:<math>\Pr\left[\bigwedge_{i=1}^m\overline{A_i}\right]>0</math>.
*:<math>t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}|</math>。
}}
*#给出计算<math>t_n</math>的递归式，并给出足够的初始值。
*#计算<math>t_n</math>的生成函数<math>T(x)=\sum_{n\ge 0}t_n x^n</math>，给出生成函数<math>T(x)</math>的闭合形式。


== The random search algorithm ==
注意：只需解生成函数的闭合形式，无需展开。
The above theorem basically says that for a CNF if every individual clause is easy to satisfy and is dependent with few other clauses then the CNF should be always satisfiable. However, the theorem only states the existence of a satisfying solution, but does not specify a way to find this solution. Next we give a simple randomized algorithm and prove it can find the satisfying solution efficiently under a slightly stronger assumption than the Lovasz local lemma.


Given as input a CNF formula <math>\phi</math> defined on Boolean variables <math>\mathcal{X}=\{x_1,x_2,\ldots,x_n\}</math>, recall that for a clause <math>C</math> in a CNF <math>\phi</math>, we use <math>\mathsf{vbl}(C)\subseteq\mathcal{X}</math> to denote the set of variables on which <math>C</math> is defined.
== Problem 7 ==
 
Let <math>a_n</math> be a sequence of numbers satisfying the recurrence relation:
The following algorithm is due to Moser in 2009. The algorithm consists of two components: the main function ''Solve''() and a sub-routine ''Fix''().
:<math>p a_n+q a_{n-1}+r a_{n-2}=0</math>
 
with initial condition <math>a_0=s</math> and <math>a_1=t</math>, where <math>p,q,r,s,t</math> are constants such that <math>{p}+q+r=0</math>, <math>p\neq 0</math> and <math>s\neq t</math>. Solve the recurrence relation.
{{Theorem
|Solve(CNF <math>\phi</math>)|
:Pick values of <math>x_1,x_2\ldots,x_n</math> uniformly and independently at random;
:While there is an unsatisfied clause <math>C</math> in <math>\phi</math>
:: '''Fix'''(<math>C</math>);
}}
 
The sub-routine '''Fix()''' is a recursive procedure:
{{Theorem
|Fix(Clause <math>C</math>)|
:Replace the values of variables in <math>\mathsf{vbl}(C)</math> with new uniform and independent random values;
:While there is ''unsatisfied'' clause <math>D</math> (including <math>C</math> itself) that <math>\mathsf{vbl}(C)\cap \mathsf{vbl}(D)\neq \emptyset</math>
:: '''Fix'''(<math>D</math>);
}}
 
It is an amazing discovery that this simple algorithm works well as long as the condition of Lovasz local lemma is satisfied. Here we prove a slightly weaker statement for the convenience of analysis.
 
{{Theorem|Theorem|
:Let <math>\phi</math> be an exact-<math>k</math>-CNF formula with maximum degree <math>d</math>.
:There is a <math>d_0=\Theta(2^{k})</math> such that if <math>d+1< d_0</math> then the algorithm ''Solve''(<math>\phi</math>) finds a satisfying assignment for <math>\phi</math> in time <math>O(n+km\log m)</math> with high probability.
}}
Note that in the Lovasz local lemma, the above <math>d_0</math> is <math>d_0=2^{k}/\mathrm{e}</math>. So this theorem archives asymptotically the same bound as the Lovasz local lemma.
 
The analysis is based on a technique called '''''entropy compression'''''. This is a very clever idea and may be very different from what you might have seen so far about algorithm analysis.
We first give a high-level description of this idea:
* We use <math>\mathsf{Alg}(r, \phi)</math> to abstractly denote an algorithm <math>\mathsf{Alg}</math> running on an input <math>\phi</math> with random bits <math>r\in\{0,1\}^*</math>. For an algorithm with no access to the random bits <math>r</math>, once the input <math>\phi</math> is fixed, the behavior of the algorithm as well as its output is ''deterministic''. But for ''randomized'' algorithms, the behavior of <math>\mathsf{Alg}(r, \phi)</math> is a random variable even when the input <math>\phi</math> is fixed.
* Fix an arbitrary (worst-case) input <math>\phi</math>. We try to construct a ''succinct representation'' <math>c</math> of the behavior of <math>\mathsf{Alg}(r, \phi)</math> in such a manner that the random bits <math>r</math> can always be fully recovered from this succinct representation <math>c</math>. In other words, <math>\mathsf{Alg}(r, \phi)</math> gives an encoding (a 1-1 mapping) of the random bits <math>r</math> to a succinct representation <math>c</math>.
* It is a fundamental law that random bits cannot be compressed significantly by any encoding. Therefore if a longer running time of <math>\mathsf{Alg}(r, \phi)</math> would imply that the random bits <math>r</math> can be encoded to a succinct representation <math>c</math> which is much shorter than <math>r</math>, then we prove the running time of the algorithm <math>\mathsf{Alg}(r, \phi)</math> cannot be too long.
:* A natural way to reach this last contradiction is to have the following situation: As the running time of <math>\mathsf{Alg}(r, \phi)</math> grows, naturally both lengths of random bits <math>r</math> and the succinct representation <math>c</math> of the behavior of <math>\mathsf{Alg}(r, \phi)</math> grow. So if the former grows much faster than the latter as the running time grows, then a large running time may cause the length of <math>r</math> significantly greater than the length of <math>c</math>.
 
---------
We now proceed to the analysis of <math>\text{Solve}(\phi)</math> and prove the above theorem.
 
From now on we assume that the input instance <math>\phi</math> is an arbitrarily fixed exact-<math>k</math>-CNF formula with maximum degree <math>d</math>:
*Let <math>T</math> denote the total number of time the function <math>\text{Fix}()</math> is being called (including both the calls in <math>\text{Solve}(\phi)</math> and the recursive calls).
*Let <math>t=\min\{T,2^m\}</math>.
Note that <math>T</math> and <math>t</math> are random variables which depend solely on the random bits used by the algorithm (after the input <math>\phi</math> being fixed).
We are going to show that <math>t=O(m\log m)</math> with high probability, which means <math>T=O(m\log m)</math> with high probability.
 
The reason we consider <math>t=\min\{T,2^m\}</math> is that we are not sure whether <math>T</math> is finite or infinite, so we "truncate" <math>T</math> if it becomes too large. This truncation threshold is quite arbitrary as long as it is significantly  greater than <math>m\log m</math>.
 
We start by making the following simple observations:
*A clause <math>C</math> in <math>\phi</math> will be satisfied after  '''Fix'''(<math>C</math>) returned and will remain as being satisfied afterwards.
*At the moment a '''Fix'''(<math>C</math>) being called, the clause  <math>C</math> must be unsatisfied.
 
{{Theorem|Proposition|
# There are at most <math>m</math> top-level callings to  '''Fix'''(), where <math>m</math> is the total number of clauses in <math>\phi</math>.
# If a '''Fix'''(<math>C</math>) is being called, the values of variables in <math>C</math> are uniquely determined.
}}
 
 
----------
We consider a restrictive case.
 
Let <math>X_1,X_2,\ldots,X_m\in\{\mathrm{true},\mathrm{false}\}</math> be a set of ''mutually independent'' random variables which assume boolean values. Each event <math>A_i</math> is an AND of at most <math>k</math> literals (<math>X_i</math> or <math>\neg X_i</math>). Let <math>v(A_i)</math> be the set of the <math>k</math> variables that <math>A_i</math> depends on. The probability that none of the bad events occurs is
:<math>
\Pr\left[\bigwedge_{i=1}^n \overline{A_i}\right].
</math>
In this particular model, the dependency graph <math>D=(V,E)</math> is defined as that <math>(i,j)\in E</math> iff <math>v(A_i)\cap v(A_j)\neq \emptyset</math>.
 
Observe that <math>\overline{A_i}</math> is a clause (OR of literals). Thus, <math>\bigwedge_{i=1}^n \overline{A_i}</math> is a '''<math>k</math>-CNF''', the CNF that each clause depends on <math>k</math> variables.
The probability
:<math>
\bigwedge_{i=1}^n \overline{A_i}>0
</math>
means that the the <math>k</math>-CNF <math>\bigwedge_{i=1}^n \overline{A_i}</math> is satisfiable.
 
The satisfiability of <math>k</math>-CNF is a hard problem. In particular, 3SAT (the satisfiability of 3-CNF) is the first '''NP-complete''' problem (the Cook-Levin theorem). Given the current suspect on '''NP''' vs '''P''', we do not expect to solve this problem generally.
 
However, the condition of the Lovasz local lemma has an extra assumption on the degree of dependency graph. In our model, this means that each clause shares variables with at most <math>d</math> other clauses. We call a <math>k</math>-CNF with this property a <math>k</math>-CNF with bounded degree <math>d</math>.
 
Therefore, proving the Lovasz local lemma on the restricted forms of events as described above, can be reduced to the following problem:
;Problem
:Find a condition on <math>k</math> and <math>d</math>, such that any <math>k</math>-CNF with bounded degree <math>d</math> is satisfiable.
 
In 2009, Moser comes up with the following procedure solving the problem. He later generalizes the procedure to general forms of events. This not only gives a beautiful constructive proof to the Lovasz local lemma, but also provides an efficient randomized algorithm for finding a satisfiable assignment for a number of events with bounded dependencies.
 
Let <math>\phi</math> be a <math>k</math>-CNF of <math>n</math> clauses with bounded degree <math>d</math>,  defined on variables <math>X_1,\ldots,X_m</math>. The following procedure find a satisfiable assignment for <math>\phi</math>.
 
{{Theorem
|Solve(<math>\phi</math>)|
:Pick a random assignment of <math>X_1,\ldots,X_m</math>.
:While there is an unsatisfied clause <math>C</math> in <math>\phi</math>
:: '''Fix'''(<math>C</math>).
}}
 
The sub-routine '''Fix''' is defined as follows:
{{Theorem
|Fix(<math>C</math>)|
:Replace the variables in <math>v(C)</math> with new random values.
:While there is unsatisfied clause <math>D</math> that <math>v(C)\cap v(D)\neq \emptyset</math>
:: '''Fix'''(<math>D</math>).
}}
 
The procedure looks very simple. It just recursively fixes the unsatisfied clauses by randomly replacing the assignment to the variables.
 
We then prove it works.
 
===Number of top-level callings of Fix ===
In '''Solve'''(<math>\phi</math>), the subroutine '''Fix'''(<math>C</math>) is called. We now upper bound the number of times it is called (not including the recursive calls).
 
Assume '''Fix'''(<math>C</math>) always terminates.
:;Observation
::Every clause that was satisfied before '''Fix'''(<math>C</math>) was called will still remain satisfied and <math>C</math> will also be satisfied after '''Fix'''(<math>C</math>) returns.
 
The observation can be proved by induction on the structure of recursion.  Since there are <math>n</math> clauses, '''Solve'''(<math>\phi</math>) makes at most <math>n</math> calls to '''Fix'''.
 
We then prove that '''Fix'''(<math>C</math>) terminates.
 
=== Termination of Fix ===
The idea of the proof is to '''reconstruct''' a random string.
 
Suppose that during the running of '''Solve'''(<math>\phi</math>), the '''Fix''' subroutine is called for <math>t</math> times (including all the recursive calls).
 
Let <math>s</math> be the sequence of the random bits used by '''Solve'''(<math>\phi</math>). It is easy to see that the length of <math>s</math> is <math>|s|=m+tk</math>, because the initial random assignment of <math>m</math> variables takes <math>m</math> bits, and each time of calling '''Fix''' takes <math>k</math> bits.
 
We then reconstruct <math>s</math> in an alternative way.
 
Recall that '''Solve'''(<math>\phi</math>) calls '''Fix'''(<math>C</math>) at top-level for at most <math>n</math> times. Each calling of '''Fix'''(<math>C</math>) defines a recursion tree, rooted at clause <math>C</math>, and each node corresponds to a clause (not necessarily distinct, since a clause might be fixed for several times). Therefore, the entire running history of '''Solve'''(<math>\phi</math>) can be described by at most <math>n</math> recursion trees.
 
:;Observation 1
::Fix a <math>\phi</math>. The <math>n</math> recursion trees which capture the total running history of '''Solve'''(<math>\phi</math>) can be encoded in <math>n\log n+t(\log d+O(1))</math> bits.
Each root node corresponds to a clause. There are <math>n</math> clauses in <math>\phi</math>. The <math>n</math> root nodes can be represented in <math>n\log n</math> bits.
 
The smart part is how to encode the branches of the tree. Note that '''Fix'''(<math>C</math>) will call '''Fix'''(<math>D</math>) only for the <math>D</math> that shares variables with <math>C</math>. For a k-CNF with bounded degree <math>d</math>, each clause <math>C</math> can share variables with at most <math>d</math> many other clauses. Thus, each branch in the recursion tree can be represented  in <math>\log d</math> bits. There are extra <math>O(1)</math> bits needed to denote whether the recursion ends. So totally  <math>n\log n+t(\log d+O(1))</math> bits are sufficient to encode all <math>n</math> recursion trees.
 
:;Observation 2
::The random sequence <math>s</math> can be encoded in <math>m+n\log n+t(\log d+O(1))</math> bits.
 
With <math>n\log n+t(\log d+O(1))</math> bits, the structure of all the recursion trees can be encoded. With extra <math>m</math> bits, the final assignment of the <math>m</math>
variables is stored.
 
We then observe that with these information, the sequence of the random bits <math>s</math> can be reconstructed from backwards from the final assignment.
 
The key step is that a clause <math>C</math> is only fixed when it is unsatisfied (obvious), and an unsatisfied clause <math>C</math> must have exact one assignment (a clause is OR of literals, thus has exact one unsatisfied assignment). Thus, each node in the recursion tree tells the <math>k</math> random bits in the random sequence <math>s</math> used in the call of the Fix corresponding to the node. Therefore, <math>s</math> can be reconstructed from the final assignment plus at most <math>n</math> recursion trees, which can be encoded in at most <math>m+n\log n+t(\log d+O(1))</math> bits.
 
The following theorem lies in the heart of the '''Kolmogorov complexity'''. The theorem states that random sequence is '''incompressible'''.
{{Theorem
|Theorem (Kolmogorov)|
:For any encoding scheme , with high probability, a random sequence <math>s</math> is encoded in at least <math>|s|</math> bits.
}}
 
Applying the theorem, we have that with high probability,
:<math>m+n\log n+t(\log d+O(1))\ge |s|=m+tk</math>.
Therefore,
:<math>
t(k-O(1)-\log d)\le n\log n.
</math>
In order to bound <math>t</math>, we need
:<math>k-O(1)-\log d>0</math>,
which hold for <math>d< 2^{k-\alpha}</math> for some constant <math>\alpha>0</math>. In fact, in this case, <math>t=O(n\log n)</math>, the running time of the procedure is bounded by a polynomial!
 
=== Back to the local lemma ===
We showed that for <math>d<2^{k-O(1)}</math>, any <math>k</math>-CNF with bounded degree <math>d</math> is satisfiable, and the satisfied assignment can be found within polynomial time with high probability. Now we interprete this in a language of the local lemma.
 
Recall that the symmetric version of the local lemma:
{{Theorem
|Theorem (The local lemma: symmetric case)|
:Let <math>A_1,A_2,\ldots,A_n</math> be a set of events, and assume that the following hold:
:#for all <math>1\le i\le n</math>, <math>\Pr[A_i]\le p</math>;
:#the maximum degree of the dependency graph for the events <math>A_1,A_2,\ldots,A_n</math> is <math>d</math>, and
:::<math>ep(d+1)\le 1</math>.
:Then
::<math>\Pr\left[\bigwedge_{i=1}^n\overline{A_i}\right]>0</math>.
}}
Suppose the underlying probability space is a number of mutually independent uniform random boolean variables, and the evens <math>\overline{A_i}</math> are clauses defined on <math>k</math> variables. Then,
:<math>
p=2^{-k}
</math>
thus, the condition <math>ep(d+1)\le 1</math> means that
:<math>
d<2^{k}/e
</math>
which means the Moser's procedure is asymptotically optimal on the degree of dependency.

Revision as of 13:14, 17 September 2017

  • 每道题目的解答都要有完整的解题过程。中英文不限。

Problem 1

  1. 有[math]\displaystyle{ k }[/math]种不同的明信片，每种明信片有无限多张，寄给[math]\displaystyle{ n }[/math]个人，每人一张，有多少种方法？
  2. 有[math]\displaystyle{ k }[/math]种不同的明信片，每种明信片有无限多张，寄给[math]\displaystyle{ n }[/math]个人，每人一张，每个人必须收到不同种类的明信片，有多少种方法？
  3. 有[math]\displaystyle{ k }[/math]种不同的明信片，每种明信片有无限多张，寄给[math]\displaystyle{ n }[/math]个人，每人收到[math]\displaystyle{ r }[/math]张不同的明信片（但不同的人可以收到相同的明信片），有多少种方法？
  4. 只有一种明信片，共有[math]\displaystyle{ m }[/math]张，寄给[math]\displaystyle{ n }[/math]个人，全部寄完，每个人可以收多张明信片或者不收明信片，有多少种方法？
  5. 有[math]\displaystyle{ k }[/math]种不同的明信片，其中第[math]\displaystyle{ i }[/math]种明信片有[math]\displaystyle{ m_i }[/math]张，寄给[math]\displaystyle{ n }[/math]个人，全部寄完，每个人可以收多张明信片或者不收明信片，有多少种方法？

Problem 2

Find the number of ways to select [math]\displaystyle{ 2n }[/math] balls from [math]\displaystyle{ n }[/math] identical blue balls, [math]\displaystyle{ n }[/math] identical red balls and [math]\displaystyle{ n }[/math] identical green balls.

  • Give a combinatorial proof for the problem.
  • Give an algebraic proof for the problem.

Problem 3

  • 一个长度为[math]\displaystyle{ n }[/math]的“山峦”是如下由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的，从坐标[math]\displaystyle{ (0,0) }[/math]到[math]\displaystyle{ (0,2n) }[/math]的折线，但任何时候都不允许低于[math]\displaystyle{ x }[/math]轴。例如下图：
   /\
  /  \/\/\    /\/\
 /        \/\/    \/\/\
 ----------------------
长度为[math]\displaystyle{ n }[/math]的“山峦”有多少？
  • 一个长度为[math]\displaystyle{ n }[/math]的“地貌”是由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的，从坐标[math]\displaystyle{ (0,0) }[/math]到[math]\displaystyle{ (0,2n) }[/math]的折线，允许低于[math]\displaystyle{ x }[/math]轴。长度为[math]\displaystyle{ n }[/math]的“地貌”有多少？

Problem 4

李雷和韩梅梅竞选学生会主席，韩梅梅获得选票 [math]\displaystyle{ p }[/math] 张，李雷获得选票 [math]\displaystyle{ q }[/math] 张，[math]\displaystyle{ p\gt q }[/math]。我们将总共的 [math]\displaystyle{ p+q }[/math] 张选票一张一张的点数，有多少种选票的排序方式使得在整个点票过程中，韩梅梅的票数一直高于李雷的票数？等价地，假设选票均匀分布的随机排列，以多大概率在整个点票过程中，韩梅梅的票数一直高于李雷的票数。

Problem 5

A [math]\displaystyle{ 2\times n }[/math] rectangle is to be paved with [math]\displaystyle{ 1\times 2 }[/math] identical blocks and [math]\displaystyle{ 2\times 2 }[/math] identical blocks. Let [math]\displaystyle{ f(n) }[/math] denote the number of ways that can be done. Find a recurrence relation for [math]\displaystyle{ f(n) }[/math], solve the recurrence relation.

Problem 6

  • 令[math]\displaystyle{ s_n }[/math]表示长度为[math]\displaystyle{ n }[/math]，没有2个连续的1的二进制串的数量，即
    [math]\displaystyle{ s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}| }[/math]。
求 [math]\displaystyle{ s_n }[/math]。
  • 令[math]\displaystyle{ t_n }[/math]表示长度为[math]\displaystyle{ n }[/math]，没有3个连续的1的二进制串的数量，即
    [math]\displaystyle{ t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}| }[/math]。
    1. 给出计算[math]\displaystyle{ t_n }[/math]的递归式，并给出足够的初始值。
    2. 计算[math]\displaystyle{ t_n }[/math]的生成函数[math]\displaystyle{ T(x)=\sum_{n\ge 0}t_n x^n }[/math]，给出生成函数[math]\displaystyle{ T(x) }[/math]的闭合形式。

注意：只需解生成函数的闭合形式，无需展开。

Problem 7

Let [math]\displaystyle{ a_n }[/math] be a sequence of numbers satisfying the recurrence relation:

[math]\displaystyle{ p a_n+q a_{n-1}+r a_{n-2}=0 }[/math]

with initial condition [math]\displaystyle{ a_0=s }[/math] and [math]\displaystyle{ a_1=t }[/math], where [math]\displaystyle{ p,q,r,s,t }[/math] are constants such that [math]\displaystyle{ {p}+q+r=0 }[/math], [math]\displaystyle{ p\neq 0 }[/math] and [math]\displaystyle{ s\neq t }[/math]. Solve the recurrence relation.

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