1) passes through two points on the plane, and only one straight line passes through.

Analysis: (for reference only): X and Y are two points on the plane,: Z is a straight line passing through X and Y, and: X and Y are the same.

2) Not all soldiers want to be generals, and soldiers who don't want to be generals are not necessarily good soldiers (a form, including full-name quantifiers and existential quantifiers).

Analysis: (for reference only): X is a soldier. X wants to be a general and X is a good soldier.

﹁

Second, fill in the blanks

1. Let a divide on A={ 1, 2, 3, 4, 5, 6, 7}, R={{ 1, 2}, {3, 4, 5}, {6, 7}}. Then the corresponding equivalence relation r). There are (128) symmetric and antisymmetric relations on set a. ..

Analysis: empty 1: the method of using cartesian product 2X2+3X3 +2X2 = 17; ;

Empty 2: Separable partial order relations are

Empty 3: Symmetric and antisymmetric, represented by a matrix, that is, the number on the diagonal consists of 0 and 1, and all other values are 0. So, there is = 128.

2. Two relations on the set A={a, B, C, d} are known = {

Analysis: the method of matrix is the simplest, and matrix multiplication can solve it. The answer is already in italics. Substitute the matrix according to the formula and multiply it by the matrix, as shown in the figure below.

3. A store provides three different pens, assuming that when customer Xiao Wang enters the store, there are at least five pens of each type. Then there are (2 1) ways for Xiao Wang to choose five pens.

Analysis: S is a multiple set with K class objects, and each element has infinite repetitions, so the number of R combinations of S is, so the answer to this question is = = 2 1.

4. Let Km, n be a complete bipartite graph, in which m and n vertices are divided into two parts, then the chromatic number of Km, n is (2).

Analysis: Theorem 1 A graph G is 2- colorable if and only if G is bipartite; Therefore, we know that the chromatic number of bipartite graphs is 2.

Theorem 2 Odd cycle and odd order wheel graphs are 3- chromatic graphs, and even order wheel graphs are 4- chromatic graphs.

5. Let the vertex set of tree T be V = {V 1, V2, ..., VN}, and the average degree of T be, please use D = (2/(2-d)) to represent the number of vertices of tree T..

Analysis: According to the average degree D, the total degree of the tree is nD, and the relationship between the number of vertices N and the number of edges K of the tree is k = n- 1, so there is an equation nD/2 = n- 1, which is simplified to n = 2/(2-D).

Third, the calculation problem

1. If the individual domain is {a, b, c}, write the following formula as a propositional logic formula.

Analysis: Single field {a, b, c} For logical propositional quantifiers, it is a conjunction operation of single field, while it is a disjunction operation of single field. Therefore,

2. Calculate the principal disjunctive normal form and principal conjunctive normal form of the following formula, and write out the solution steps. The results are expressed in concise form by minimax and minimax.

Analysis: There are two ways to solve this problem, one is to use truth table, and the other is formula conversion.

Method 1: first solve the problem with the truth table:

Then the principal disjunctive paradigm is

Lord conjunctive normal form is

Method 2: Formula derivation

The main disjunctive paradigm is =?

Therefore, Lord conjunctive normal form is

Fourth, answer questions.

1. Write a relation on set A, which is both an equivalence relation and a partial order relation, and briefly explain the characteristics of this relation.

Analysis: Let the set A={a, b, c}, and the equivalence relation satisfies the following conditions: reflexivity, symmetry, transitivity; The conditions that satisfy the partial order relation are reflexivity, antisymmetry and transitivity. In the condition, the relation R of A must satisfy the equivalence and partial order relation, that is, R must satisfy both symmetry and antisymmetry relation. Then r = {

2. Find the expression that satisfies the recursive relation, in which the initial conditions.

Analysis: This question examines the homogeneous recurrence relation of constant coefficients. The original formula in the problem is transformed into, so the characteristic equation of the formula is.

? . Find the characteristic root. If these three features do not have multiple roots, the general solution is:

What can I get by substituting three characteristic roots into the formula? . Substitution results in three equations.

Solving these three ternary linear equations: the solution of substitution?

3. Let the generating function of series be, series? The generating functions of are if, and, and.

Analysis: Do you have any questions to know? And, get =?

= ? Because,

Verb (abbreviation of verb) proves the problem

Prove the following identity:, which means to take I combinations of n elements.

Prove: to prove the original form

.

The equation holds, that is, the method of taking k+ 1 from n elements can be divided into two parts, one part contains 1 specific element A, and the other part does not contain A. ..

therefore

And so on:

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