Systems of logical equations. Logics. Logic functions. Solving Equations
Truth table - a table containing all possible combinations of input variables and their corresponding output values.
The truth table contains 2n rows, where n is the number of input variables, and n+m are columns, where m are the output variables.
Instruction. When entering from the keyboard, use the following conventions:
For example, the logical expression abc+ab~c+a~bc must be entered like this: a*b*c+a*b=c+a=b*cTo enter data in the form of a logical diagram, use this service.
Logic function input rules
- Use the + sign instead of v (disjunction, OR).
- Before the logical function, you do not need to specify the function designation. For example, instead of F(x,y)=(x|y)=(x^y) you would simply type (x|y)=(x^y) .
- Maximum amount variables is 10 .
The design and analysis of computer logic circuits is carried out with the help of a special section of mathematics - the algebra of logic. In the algebra of logic, three main logical functions can be distinguished: "NOT" (negation), "AND" (conjunction), "OR" (disjunction).
To create any logical device, it is necessary to determine the dependence of each of the output variables on the current input variables, such a dependence is called a switching function or a function of the logic algebra.
A logic algebra function is called fully defined if all 2 n of its values are given, where n is the number of output variables.
If not all values are defined, the function is called partially defined.
A device is called logical if its state is described using a function of the algebra of logic.
The following methods are used to represent the logic algebra function:
In algebraic form, it is possible to construct a diagram of a logical device using logical elements. 
Figure 1 - Diagram of a logical device
All operations of the algebra of logic are defined truth tables values. The truth table determines the result of performing an operation for all possible x logical values of the original statements. The number of options that reflect the result of applying the operations will depend on the number of statements in the logical expression. If the number of statements in the logical expression is N, then the truth table will contain 2 N rows, since there are 2 N different combinations of possible argument values.
Operation NOT - logical negation (inversion)
The logical operation is NOT applied to a single argument, which can be either a simple or a complex logical expression. The result of the operation is NOT the following:- if the original expression is true, then the result of its negation will be false;
- if the original expression is false, then the result of its negation will be true.
not A, Ā, not A, ¬A, !A
The result of the negation operation is NOT determined by the following truth table:
| A | not A |
| 0 | 1 |
| 1 | 0 |
The result of the negation operation is true when the original statement is false, and vice versa.
Operation OR - logical addition (disjunction, union)
The logical OR operation performs the function of combining two statements, which can be either a simple or a complex logical expression. Statements that are initial for a logical operation are called arguments. The result of the OR operation is an expression that will be true if and only if at least one of the original expressions is true.Designations used: A or B, A V B, A or B, A||B.
The result of the OR operation is determined by the following truth table:
The result of the OR operation is true when A is true, or B is true, or both A and B are true, and false when both A and B are false.
Operation AND - logical multiplication (conjunction)
The logical operation AND performs the function of the intersection of two statements (arguments), which can be either a simple or a complex logical expression. The result of the AND operation is an expression that is true if and only if both of the original expressions are true.Symbols used: A and B, A Λ B, A & B, A and B.
The result of the AND operation is determined by the following truth table:
| A | B | A and B |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
The result of operation AND is true if and only if statements A and B are both true, and false in all other cases.
Operation "IF-THEN" - logical consequence (implication)
This operation connects two simple logical expressions, of which the first is a condition, and the second is a consequence of this condition.Applied designations:
if A, then B; A attracts B; if A then B; A → B.
Truth table:
| A | B | A→B |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
The result of the operation of consequence (implication) is false only when premise A is true and conclusion B (consequence) is false.
Operation "A if and only if B" (equivalence, equivalence)
Applicable designation: A ↔ B, A ~ B.Truth table:
| A | B | A↔B |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
Modulo 2 addition operation (XOR, exclusive or, strict disjunction)
Notation used: A XOR B, A ⊕ B.Truth table:
| A | B | A⊕B |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
The result of an equivalence operation is true only if both A and B are both true or both false.
Precedence of logical operations
- Actions in brackets
- Inversion
- Conjunction (&)
- Disjunction (V), Exclusive OR (XOR), modulo 2 sum
- Implication (→)
- Equivalence (↔)
Perfect disjunctive normal form
Perfect disjunctive normal form of a formula(SDNF) is a formula equivalent to it, which is a disjunction of elementary conjunctions, which has the following properties:- Each logical term of the formula contains all the variables included in the function F(x 1 ,x 2 ,...x n).
- All logical terms of the formula are different.
- No logical term contains a variable and its negation.
- No logical term in a formula contains the same variable twice.
For each function, SDNF and SKNF are uniquely defined up to a permutation.
Perfect conjunctive normal form
Perfect conjunctive normal form of a formula (SKNF) is a formula equivalent to it, which is a conjunction of elementary disjunctions that satisfies the following properties:- All elementary disjunctions contain all variables included in the function F(x 1 ,x 2 ,...x n).
- All elementary disjunctions are distinct.
- Each elementary disjunction contains a variable once.
- No elementary disjunction contains a variable and its negation.
How to Solve Some Problems in Sections A and B of the Computer Science Exam
Lesson number 3. Logics. Logic functions. Solving Equations
A large number of USE tasks are devoted to the logic of propositions. To solve most of them, it is enough to know the basic laws of propositional logic, knowledge of the truth tables of logical functions of one and two variables. I will give the basic laws of propositional logic.
- Commutativity of disjunction and conjunction:
a ˅ b ≡ b ˅ a
a^b ≡ b^a - The distributive law regarding disjunction and conjunction:
a ˅ (b^c) ≡ (a ˅ b) ^(a ˅ c)
a ^ (b ˅ c) ≡ (a ^ b) ˅ (a ^ c) - Negative negation:
¬(¬a) ≡ a - Consistency:
a ^ ¬a ≡ false - Exclusive third:
a ˅ ¬a ≡ true - De Morgan's laws:
¬(a ˅ b) ≡ ¬a ˄ ¬b
¬(a ˄ b) ≡ ¬a ˅ ¬b - Simplification:
a ˄ a ≡ a
a ˅ a ≡ a
a ˄ true ≡ a
a ˄ false ≡ false - Absorption:
a ˄ (a ˅ b) ≡ a
a ˅ (a ˄ b) ≡ a - Replacing the implication
a → b ≡ ¬a ˅ b - Change of identity
a ≡ b ≡(a ˄ b) ˅ (¬a ˄ ¬b)
Representation of logical functions
Any logical function of n variables - F(x 1 , x 2 , ... x n) can be defined by a truth table. Such a table contains 2 n sets of variables, for each of which the value of the function on this set is specified. This method is good when the number of variables is relatively small. Even for n > 5, the representation becomes poorly visible.
Another way is to define the function by some formula, using well-known fairly simple functions. The system of functions (f 1 , f 2 , … f k ) is called complete if any logical function can be expressed by a formula containing only functions f i .
The system of functions (¬, ˄, ˅) is complete. Laws 9 and 10 are examples of how implication and identity are expressed through negation, conjunction, and disjunction.
In fact, the system of two functions is also complete - negation and conjunction or negation and disjunction. Representations follow from De Morgan's laws that allow expressing a conjunction through negation and disjunction and, accordingly, expressing a disjunction through negation and conjunction:
(a ˅ b) ≡ ¬(¬a ˄ ¬b)
(a ˄ b) ≡ ¬(¬a ˅ ¬b)
Paradoxically, a system consisting of only one function is complete. There are two binary functions - anticonjunction and antidisjunction, called Pierce's arrow and Schaeffer's stroke, representing a hollow system.
The basic functions of programming languages usually include identity, negation, conjunction and disjunction. In the tasks of the exam, along with these functions, there is often an implication.
Let's look at a few simple tasks related to logical functions.
Task 15:
A fragment of the truth table is given. Which of the three given functions corresponds to this fragment?
| x1 | x2 | x3 | x4 | F |
| 1 | 1 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 |
- (X 1 → X 2) ˄ ¬ X 3 ˅ X 4
- (¬X 1 ˄ X 2) ˅ (¬X 3 ˄ X 4)
- ¬ X 1 ˅ X 2 ˅ (X 3 ˄ X 4)
Feature number 3.
To solve the problem, you need to know the truth tables of basic functions and keep in mind the priorities of operations. Let me remind you that conjunction (logical multiplication) has a higher priority and is performed before disjunction (logical addition). When calculating, it is easy to see that the functions with numbers 1 and 2 on the third set have the value 1 and for this reason do not correspond to the fragment.
Task 16:
Which of the following numbers satisfies the condition:
(digits, starting with the most significant digit, go in descending order) → (number - even) ˄ (lowest digit - even) ˄ (highest digit - odd)
If there are several such numbers, indicate the largest.
- 13579
- 97531
- 24678
- 15386
The condition is satisfied by the number 4.
The first two numbers do not satisfy the condition for the reason that the lowest digit is odd. A conjunction of conditions is false if one of the terms of the conjunction is false. For the third number, the condition for the highest digit is not met. For the fourth number, the conditions imposed on the minor and major digits of the number are met. The first term of the conjunction is also true, since an implication is true if its premise is false, which is the case here.
Problem 17: Two witnesses testified as follows:
First Witness: If A is guilty, then B is certainly guilty, and C is innocent.
Second witness: Two are guilty. And one of the remaining ones is definitely guilty and guilty, but I can’t say exactly who.
What conclusions about the guilt of A, B, and C can be drawn from the evidence?
Answer: It follows from the testimony that A and B are guilty, but C is innocent.
Solution: Of course, the answer can be given based on common sense. But let's look at how this can be done strictly and formally.
The first thing to do is to formalize the statements. Let's introduce three Boolean variables, A, B, and C, each of which is true (1) if the corresponding suspect is guilty. Then the testimony of the first witness is given by the formula:
A → (B ˄ ¬C)
The testimony of the second witness is given by the formula:
A ˄ ((B ˄ ¬C) ˅ (¬B ˄ C))
The testimonies of both witnesses are assumed to be true and represent the conjunction of the corresponding formulas.
Let's build a truth table for these readings:
| A | B | C | F1 | F2 | F 1 F 2 |
| 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 |
The summary evidence is true in only one case, leading to a clear answer - A and B are guilty, and C is innocent.
It also follows from the analysis of this table that the testimony of the second witness is more informative. Only two things follow from the truth of his testimony. possible options A and B are guilty and C is innocent, or A and C are guilty and B is innocent. The testimony of the first witness is less informative - there are 5 different options that correspond to his testimony. Together, the testimonies of both witnesses give an unequivocal answer about the guilt of the suspects.
Logic equations and systems of equations
Let F(x 1 , x 2 , …x n) be a logical function of n variables. The logical equation is:
F(x 1, x 2, ... x n) \u003d C,
The constant C has the value 1 or 0.
A logical equation can have from 0 to 2n different solutions. If C is equal to 1, then the solutions are all those sets of variables from the truth table on which the function F takes the value true (1). The remaining sets are solutions of the equation for C equal to zero. We can always consider only equations of the form:
F(x 1 , x 2 , …x n) = 1
Indeed, let the equation be given:
F(x 1 , x 2 , …x n) = 0
In this case, you can go to the equivalent equation:
¬F(x 1 , x 2 , …x n) = 1
Consider a system of k logical equations:
F 1 (x 1, x 2, ... x n) \u003d 1
F 2 (x 1, x 2, ... x n) \u003d 1
F k (x 1 , x 2 , …x n) = 1
The solution of the system is a set of variables on which all equations of the system are satisfied. In terms of logical functions, to obtain a solution to the system of logical equations, one should find a set on which the logical function Ф is true, representing the conjunction of the original functions F:
Ф = F 1 ˄ F 2 ˄ … F k
If the number of variables is small, for example, less than 5, then it is not difficult to build a truth table for the function Ф, which allows you to say how many solutions the system has and what are the sets that give solutions.
In some tasks of the Unified State Examination on finding solutions to a system of logical equations, the number of variables reaches the value of 10. Then building a truth table becomes an almost unsolvable task. Solving the problem requires a different approach. For an arbitrary system of equations, there is no general way, other than enumeration, that allows solving such problems.
In the problems proposed in the exam, the solution is usually based on taking into account the specifics of the system of equations. I repeat, apart from enumeration of all variants of a set of variables, there is no general way to solve the problem. The solution must be built based on the specifics of the system. It is often useful to carry out a preliminary simplification of a system of equations using known laws of logic. Another useful technique for solving this problem is as follows. We are not interested in all sets, but only those on which the function Ф has the value 1. Instead of constructing a complete truth table, we will build its analogue - a binary decision tree. Each branch of this tree corresponds to one solution and specifies a set on which the function Ф has the value 1. The number of branches in the decision tree coincides with the number of solutions to the system of equations.
What is a binary decision tree and how it is built, I will explain with examples of several tasks.
Problem 18
How many different sets of values of boolean variables x1, x2, x3, x4, x5, y1, y2, y3, y4, y5 are there that satisfy a system of two equations?
Answer: The system has 36 different solutions.
Solution: The system of equations includes two equations. Let's find the number of solutions for the first equation, depending on 5 variables – x 1 , x 2 , …x 5 . The first equation can in turn be considered as a system of 5 equations. As has been shown, the system of equations actually represents a conjunction of logical functions. The converse statement is also true - the conjunction of conditions can be considered as a system of equations.
Let's build a decision tree for the implication (x1→ x2), the first term of the conjunction, which can be considered as the first equation. Here is what the graphic representation of this tree looks like:
The tree has two levels equation variables. The first level describes the first variable X 1 . Two branches of this level reflect the possible values of this variable - 1 and 0. At the second level, the branches of the tree reflect only those possible values of the variable X 2 for which the equation takes the value true. Since the equation defines an implication, the branch on which X 1 has the value 1 requires that X 2 has the value 1 on that branch. The branch on which X 1 has the value 0 generates two branches with X 2 values equal to 0 and 1 The constructed tree specifies three solutions, on which the implication X 1 → X 2 takes the value 1. On each branch, the corresponding set of values of the variables is written out, which gives the solution to the equation.
These sets are: ((1, 1), (0, 1), (0, 0))
Let's continue building the decision tree by adding the following equation, the following implication X 2 → X 3 . The specificity of our system of equations is that each new equation of the system uses one variable from the previous equation, adding one new variable. Since the variable X 2 already has values in the tree, then on all branches where the variable X 2 has the value 1, the variable X 3 will also have the value 1. For such branches, the construction of the tree continues to the next level, but no new branches appear. The only branch where the variable X 2 has the value 0 will give a branch into two branches, where the variable X 3 will get the values 0 and 1. Thus, each addition of a new equation, given its specificity, adds one solution. Original first equation:
(x1→x2) /\ (x2→x3) /\ (x3→x4) /\ (x4→x5) = 1
has 6 solutions. Here is what the complete decision tree for this equation looks like:
The second equation of our system is similar to the first:
(y1→y2) /\ (y2→y3) /\ (y3→y4) /\ (y4→y5) = 1
The only difference is that the equation uses Y variables. This equation also has 6 solutions. Since every variable solution X i can be combined with every variable solution Y j , then total number solutions is 36.
Note that the constructed decision tree gives not only the number of solutions (according to the number of branches), but also the solutions themselves, written out on each branch of the tree.
Problem 19
How many different sets of values of boolean variables x1, x2, x3, x4, x5, y1, y2, y3, y4, y5 are there that satisfy all of the following conditions?
(x1→x2) /\ (x2→x3) /\ (x3→x4) /\ (x4→x5) = 1
(y1→y2) /\ (y2→y3) /\ (y3→y4) /\ (y4→y5) = 1
(x1→ y1) = 1
This task is a modification of the previous task. The difference is that another equation is added that relates the X and Y variables.
From the equation X 1 → Y 1 it follows that when X 1 has the value 1 (one such solution exists), then Y 1 has the value 1. Thus, there is one set on which X 1 and Y 1 have the values 1. When X 1 equal to 0, Y 1 can have any value, both 0 and 1. Therefore, each set with X 1 equal to 0, and there are 5 such sets, corresponds to all 6 sets with Y variables. Therefore, the total number of solutions is 31 .
Problem 20
(¬X 1 ˅ X 2) ˄ (¬X 2 ˅ X 3) ˄ (¬X 3 ˅ X 4) ˄ (¬X 4 ˅ X 5) ˄ (¬X 5 ˅ X 1) = 1
Solution: Remembering the basic equivalence, we write our equation as:
(X 1 → X 2) ˄ (X 2 → X 3) ˄ (X 3 → X 4) ˄ (X 4 → X 5) ˄ (X 5 → X 1) = 1
The cyclic chain of implications means that the variables are identical, so our equation is equivalent to:
X 1 ≡ X 2 ≡ X 3 ≡ X 4 ≡ X 5 = 1
This equation has two solutions when all X i are either 1 or 0.
Problem 21
(X 1 → X 2) ˄ (X 2 → X 3) ˄ (X 3 → X 4) ˄ (X 4 → X 2) ˄ (X 4 → X 5) = 1
Solution: Just as in Problem 20, we pass from cyclic implications to identities by rewriting the equation in the form:
(X 1 → X 2) ˄ (X 2 ≡ X 3 ≡ X 4) ˄ (X 4 → X 5) = 1
Let's build a decision tree for this equation: 
Problem 22
How many solutions does the following system of equations have?
((X 1 ≡X 2) ˄ (X 3 ≡X 4)) ˅(¬(X 1 ≡X 2) ˄ ¬(X 3 ≡X4)) = 0
((X 3 ≡X 4) ˄ (X5 ≡X 6)) ˅(¬(X 3 ≡X 4) ˄ ¬(X5 ≡X 6)) = 0
((X5 ≡X 6) ˄ (X 7 ≡X 8)) ˅(¬(X5 ≡X 6) ˄ ¬(X 7 ≡X8)) = 0
((X 7 ≡X 8) ˄ (X9 ≡X 10)) ˅(¬(X 7 ≡X 8) ˄ ¬(X9 ≡X10)) = 0
Answer: 64
Solution: Let's go from 10 variables to 5 variables by introducing the following change of variables:
Y 1 = (X 1 ≡ X 2); Y 2 \u003d (X 3 ≡ X 4); Y 3 = (X 5 ≡ X 6); Y 4 \u003d (X 7 ≡ X 8); Y 5 \u003d (X 9 ≡ X 10);
Then the first equation will take the form:
(Y 1 ˄ Y 2) ˅ (¬Y 1 ˄ ¬Y 2) = 0
The equation can be simplified by writing it as:
(Y 1 ≡ Y 2) = 0
Passing to the traditional form, we write the system after simplifications in the form:
¬(Y 1 ≡ Y 2) = 1
¬(Y 2 ≡ Y 3) = 1
¬(Y 3 ≡ Y 4) = 1
¬(Y 4 ≡ Y 5) = 1
The decision tree for this system is simple and consists of two branches with alternating variable values:
Returning to the original X variables, note that each value of the Y variable corresponds to 2 values of the X variables, so each solution in the Y variables generates 2 5 solutions in the X variables. Two branches generate 2 * 2 5 solutions, so the total number of solutions is 64.
As you can see, each task for solving a system of equations requires its own approach. A common trick is to perform equivalent transformations to simplify the equations. A common technique is the construction of decision trees. The applied approach partially resembles the construction of a truth table with the peculiarity that not all sets of possible values of variables are constructed, but only those on which the function takes the value 1 (true). Often in the proposed problems there is no need to build a complete decision tree, since already at the initial stage it is possible to establish the regularity of the appearance of new branches at each next level, as was done, for example, in problem 18.
In general, problems for finding solutions to a system of logical equations are good mathematical exercises.
If the problem is difficult to solve manually, then you can entrust the solution of the problem to the computer by writing an appropriate program for solving equations and systems of equations.
It is easy to write such a program. Such a program will easily cope with all the tasks offered in the exam.
Oddly enough, but the task of finding solutions to systems of logical equations is also difficult for a computer, it turns out that a computer has its limits. The computer can easily cope with tasks where the number of variables is 20-30, but it will start to think for a long time on larger tasks. The point is that the function 2 n that specifies the number of sets is an exponent that grows rapidly with n. So fast that a normal personal computer can't handle a task with 40 variables in a day.
C# program for solving logical equations
It is useful to write a program for solving logical equations for many reasons, if only because it can be used to check the correctness of your own solution to the USE test problems. Another reason is that such a program is an excellent example of a programming problem that meets the requirements for category C problems in the USE.
The idea of constructing a program is simple - it is based on a complete enumeration of all possible sets of variable values. Since the number of variables n is known for a given logical equation or system of equations, the number of sets is also known - 2 n , which need to be sorted out. Using the basic functions of the C# language - negation, disjunction, conjunction and identity, it is easy to write a program that, for a given set of variables, calculates the value of a logical function corresponding to a logical equation or system of equations.
In such a program, you need to build a cycle by the number of sets, in the body of the cycle, by the set number, form the set itself, calculate the value of the function on this set, and if this value is equal to 1, then the set gives a solution to the equation.
The only difficulty that arises in the implementation of the program is related to the task of forming the set of variable values itself by the set number. The beauty of this task is that this seemingly difficult task, in fact, comes down to a simple task that has already arisen repeatedly. Indeed, it is enough to understand that the set of values of variables corresponding to the number i, consisting of zeros and ones, represents the binary representation of the number i. So that difficult task obtaining a set of values of variables by the number of the set is reduced to the well-known problem of converting a number into a binary system.
This is how the C# function that solves our problem looks like:
///
/// program for counting the number of solutions
/// logical equation (system of equations)
///
///
/// logical function - method,
/// whose signature is set by the DF delegate
///
/// number of variables
///
static int SolveEquations(DF fun, int n)
bool set = new bool[n];
int m = (int)Math.Pow(2, n); //number of sets
int p = 0, q = 0, k = 0;
//Full enumeration by the number of sets
for (int i = 0; i< m; i++)
//Formation of the next set — set,
//given by the binary representation of the number i
for (int j = 0; j< n; j++)
k = (int)Math.Pow(2, j);
//Calculate function value on set
To understand the program, I hope that the explanations of the idea of the program and the comments in its text will suffice. I will dwell only on the explanation of the heading of the above function. The SolveEquations function has two input parameters. The fun parameter specifies a logical function corresponding to the equation or system of equations being solved. The n parameter specifies the number of variables in the fun function. As a result, the SolveEquations function returns the number of solutions of the logical function, that is, the number of sets on which the function evaluates to true.
For schoolchildren, it is customary when for some function F(x) the input parameter x is a variable of arithmetic, string or boolean type. In our case, a more powerful design is used. The SolveEquations function refers to higher-order functions - functions of type F(f), whose parameters can be not only simple variables, but also functions.
The class of functions that can be passed as a parameter to the SolveEquations function is defined as follows:
delegate bool DF(bool vars);
This class includes all functions that are passed as a parameter a set of values of boolean variables specified by the vars array. The result is a Boolean value representing the value of the function on this set.
In conclusion, I will give a program in which the SolveEquations function is used to solve several systems of logical equations. The SolveEquations function is part of the following ProgramCommon class:
class ProgramCommon
delegate bool DF(bool vars);
static void Main(string args)
Console.WriteLine("Function And Solutions - " +
SolveEquations(FunAnd, 2));
Console.WriteLine("Function has 51 solutions - " +
SolveEquations(Fun51, 5));
Console.WriteLine("Function has 53 solutions - " +
SolveEquations(Fun53, 10));
static bool FunAnd(bool vars)
return vars && vars;
static bool Fun51(bool vars)
f = f && (!vars || vars);
f = f && (!vars || vars);
f = f && (!vars || vars);
f = f && (!vars || vars);
f = f && (!vars || vars);
static bool Fun53(bool vars)
f = f && ((vars == vars) || (vars == vars));
f = f && ((vars == vars) || (vars == vars));
f = f && ((vars == vars) || (vars == vars));
f = f && ((vars == vars) || (vars == vars));
f = f && ((vars == vars) || (vars == vars));
f = f && ((vars == vars) || (vars == vars));
f = f && (!((vars == vars) || (vars == vars)));
Here is what the results of the solution for this program look like:
10 tasks for independent work
- Which of the three functions are equivalent:
- (X → Y) ˅ ¬Y
- ¬(X ˅ ¬Y) ˄ (X → ¬Y)
- ¬X ˄ Y
- A fragment of the truth table is given:
| x1 | x2 | x3 | x4 | F |
| 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
Which of the three functions corresponds to this fragment:
- (X 1 ˅ ¬X 2) ˄ (X 3 → X 4)
- (X 1 → X 3) ˄ X 2 ˅ X 4
- X 1 ˄ X 2 ˅ (X 3 → (X 1 ˅ X 4))
- The jury consists of three people. The decision is made if the chairman of the jury votes for it, supported by at least one of the jury members. Otherwise, no decision is made. Build a logical function that formalizes the decision making process.
- X wins over Y if four coin tosses come up heads three times. Define a boolean function describing payoff X.
- Words in a sentence are numbered starting from one. A sentence is considered well-formed if the following rules are met:
- If an even numbered word ends in a vowel, then the next word, if it exists, must begin with a vowel.
- If an odd numbered word ends in a consonant, then the next word, if it exists, must start with a consonant and end with a vowel.
Which of the following sentences are correct: - Mom washed Masha with soap.
- The leader is always a model.
- The truth is good, but happiness is better.
- How many solutions does the equation have:
(a ˄ ¬ b) ˅ (¬a ˄ b) → (c ˄ d) = 1 - List all solutions of the equation:
(a → b) → c = 0 - How many solutions does the following system of equations have:
X 0 → X 1 ˄ X 1 → X 2 = 1
X 2 → X 3 ˄ X 3 → X 4 = 1
X 5 → X 6 ˄ X 6 → X 7 = 1
X 7 → X 8 ˄ X 8 → X 9 = 1
X 0 → X 5 = 1 - How many solutions does the equation have:
((((X 0 → X 1) → X 2) → X 3) → X 4) → X 5 = 1
Answers to tasks:
- The functions b and c are equivalent.
- The fragment corresponds to function b.
- Let the boolean variable P take the value 1 when the chairman of the jury votes "for" the decision. Variables M 1 and M 2 represent the opinion of the jury members. The logical function that specifies the adoption of a positive decision can be written as follows:
P ˄ (M 1 ˅ M 2) - Let the boolean variable P i take on the value 1 when i-th throw the coin comes up heads. The logical function that defines the payoff X can be written as follows:
¬((¬P 1 ˄ (¬P 2 ˅ ¬P 3 ˅ ¬P 4)) ˅
(¬P 2 ˄ (¬P 3 ˅ ¬P 4)) ˅
(¬P 3 ˄ ¬P 4)) - Offer b.
- The equation has 3 solutions: (a = 1; b = 1; c = 0); (a = 0; b = 0; c = 0); (a=0; b=1; c=0)
This material contains a presentation that presents methods for solving logical equations and systems of logical equations in task B15 (No. 23, 2015) of the Unified State Examination in Informatics. It is known that this task is one of the most difficult among the tasks of the exam. The presentation can be useful when conducting lessons on the topic "Logic" in specialized classes, as well as in preparation for passing the exam.
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Solution of task B15 (system of logical equations) Vishnevskaya M.P., MAOU "Gymnasium No. 3" November 18, 2013, Saratov
Task B15 is one of the most difficult in the exam in computer science !!! Skills are checked: to transform expressions containing logical variables; describe in natural language the set of values of logical variables for which a given set of logical variables is true; count the number of binary sets that satisfy the given conditions. The most difficult, because there are no formal rules on how to do this, guesswork is required.
What not to do without!
What not to do without!
Conventions conjunction: A /\ B , A B , AB , А &B, A and B disjunction: A \ / B , A + B , A | B , A or B negation: A , A, not A equivalent: A B, A B, A B XOR: A B , A xor B
Variable substitution method How many different sets of values of boolean variables x1, x2, ..., x9, x10 exist that satisfy all of the following conditions: ((x1 ≡ x2) \/ (x3 ≡ x4)) /\ (¬(x1 ≡ x2) \/ ¬(x3 ≡ x4)) = 1 ((x3 ≡ x4) \/ (x5 ≡ x6)) /\ (¬(x3 ≡ x4) \/ ¬(x5 ≡ x6)) = 1 ((x5 ≡ x6 ) \/ (x7 ≡ x8)) /\ (¬(x5 ≡ x7) \/ ¬(x7 ≡ x8)) = 1 ((x7 ≡ x8) \/ (x9 ≡ x10)) /\ (¬(x7 ≡ x8) \/ ¬(x9 ≡ x10)) = 1 The answer does not need to list all the different sets x1, x2, …, x9, x10 under which the given system of equalities is satisfied. As an answer, you must indicate the number of such sets (demo version 2012)
Solution Step 1. Simplify by changing variables t1 = x1 x2 t2 = x3 x4 t3 = x5 x6 t4 = x7 x8 t5 = x9 x10 After simplification: (t1 \/ t2) /\ (¬t1 \/ ¬ t2) =1 (t2 \/ t3) /\ (¬t2 \/ ¬ t3) =1 (t3 \/ t4) /\ (¬t3 \/ ¬ t4) =1 (t4 \/ t5) /\ ( ¬t4 \/ ¬ t5) =1 Consider one of the equations: (t1 \/ t2) /\ (¬t1 \/ ¬ t2) =1 Obviously, it =1 only if one of the variables is 0 and the other is 1. We use the formula to express the XOR operation in terms of conjunction and disjunction: (t1 \/ t2) /\ (¬t1 \/ ¬ t2) = t1 t2 = ¬(t1 ≡ t2) =1 ¬(t1 ≡ t2) =1 ¬( t2 ≡ t3) =1 ¬(t3 ≡ t4) =1 ¬(t4 ≡ t5) =1
Step2. Analysis of the system .to. tk = x2k-1 ≡ x2k (t1 = x1 x2 ,….), then each value tk corresponds to two pairs of values x2k-1 and x2k , for example: tk =0 corresponds to two pairs - (0,1) and (1, 0) , and tk =1 are the pairs (0,0) and (1,1).
Step3. Counting the number of solutions. Each t has 2 solutions, the number of t is 5. Thus for variables t there are 2 5 = 32 solutions. But each t corresponds to a pair of solutions x, i.e. the original system has 2*32 = 64 solutions. Answer: 64
Partial solution elimination method How many different sets of values of logical variables x1, x2, …, x5, y1,y2,…, y5 exist that satisfy all the following conditions: (x1→ x2)∧(x2→ x3)∧(x3→ x4 )∧(x4→ x5) =1; (y1→ y2)∧(y2→ y3)∧(y3→ y4) ∧(y4→ y5) =1; y5→ x5 =1. The answer does not need to list all the different sets x1, x2, ..., x5, y 1, y2, ..., y5, under which this system of equalities is satisfied. As an answer, you must indicate the number of such sets.
Solution. Step 1. Consistent solution equations x1 1 0 x2 1 0 1 x3 1 0 1 1 x4 1 0 1 1 1 x5 1 0 1 1 1 1 each of the implications is true. The implication is false only in one case, when 1 0, in all other cases (0 0, 0 1, 1 1) the operation returns 1. Let's write this in the form of a table:
Step 1. Sequential solution of equations Т.о. 6 sets of solutions for х1,х2,х3,х4,х5 are received: (00000), (00001), (00011), (00111), (01111), (11111). Arguing similarly, we conclude that for y1, y2, y3, y4, y5 there is the same set of solutions. Because these equations are independent, i.e. there are no common variables in them, then the solution to this system of equations (without taking into account the third equation) will be 6 * 6 = 36 pairs of “xes” and “yes”. Consider the third equation: y5→ x5 =1 Pairs are the solution: 0 0 0 1 1 1 Pair is not a solution: 1 0
Let's compare the obtained solutions. Where y5 =1, x5=0 do not fit. there are 5 such pairs. The number of solutions of the system: 36-5= 31. Answer: 31 It took combinatorics!!!
Dynamic programming method How many different solutions does the logical equation x 1 → x 2 → x 3 → x 4 → x 5 → x 6 = 1 have, where x 1, x 2, ..., x 6 are logical variables? The answer does not need to list all the different sets of variable values for which this equality holds. As an answer, you need to indicate the number of such sets.
Solution Step1. Analysis of the condition On the left side of the equation, implication operations are sequentially written, the priority is the same. Rewrite: ((((X 1 → X 2) → X 3) → X 4) → X 5) → X 6 = 1 NB! Each next variable does not depend on the previous one, but on the result of the previous implication!
Step2. Revealing the pattern Consider the first implication, X 1 → X 2. Truth table: X 1 X 2 X 1 → X 2 0 0 1 0 1 1 1 0 0 1 1 1 From one 0 we got 2 ones, and from 1 we got one 0 and one 1. Only one 0 and three 1, this is the result of the first operation.
Step2. Revealing a pattern Connecting x 3 to the result of the first operation, we get: F(x 1 ,x 2) x 3 F(x 1 ,x 2) x 3 0 0 1 0 1 1 1 0 0 1 1 1 1 0 0 1 1 1 1 0 0 1 1 1 Out of two 0s - two 1s, out of each 1 (there are 3) one each 0 and 1 (3 + 3)
Step 3. Derivation of the formula you can make formulas for calculating the number of zeros N i and the number of ones E i for an equation with i variables: ,
Step 4. Filling in the table Let's fill in the table for i = 6 from left to right, calculating the number of zeros and ones using the above formulas; the table shows how the next column is built according to the previous one: : number of variables 1 2 3 4 5 6 Number of zeros N i 1 1 3 5 11 21 Number of ones E i 1 2*1+1= 3 2*1+3= 5 11 21 43 Answer: 43
Method using simplifications of logical expressions How many different solutions does the equation have ((J → K) → (M N L)) ((M N L) → (¬ J K)) (M → J) = 1 where J , K, L, M, N are logical variables? The answer does not need to list all the different sets of values J , K, L, M and N for which this equality holds. As an answer, you need to indicate the number of such sets.
Solution Note that J → K = ¬ J K We introduce a change of variables: J → K=A, M N L =B We rewrite the equation taking into account the change: (A → B) (B → A) (M → J)=1 4. (A B) (M → J)= 1 5. Obviously, A B for the same values of A and B 6. Consider the last implication M → J =1 This is possible if: M= J=0 M=0, J=1 M=J=1
Solution A B , then With M=J=0 we get 1 + K=0. There are no solutions. With M=0, J=1 we get 0 + K=0, K=0, and N and L - any, 4 solutions: ¬ J K = M N L K N L 0 0 0 0 0 1 0 1 0 0 1 one
Solution 10. With M=J=1 we get 0+K=1 *N * L , or K=N*L, 4 solutions: 11. Total has 4+4=8 solutions Answer: 8 K N L 0 0 0 0 0 1 0 1 0 1 1 1
Sources of information: O.B. Bogomolova, D.Yu. Usenkov. B15: new tasks and new solution // Informatics, No. 6, 2012, p. 35 – 39. K.Yu. Polyakov. Logical Equations // Informatics, No. 14, 2011, p. 30-35. http://ege-go.ru/zadania/grb/b15/, [Electronic resource]. http://kpolyakov.narod.ru/school/ege.htm, [Electronic resource].
Let be a logical function of n variables. The logical equation is:
The constant C has the value 1 or 0.
A logical equation can have from 0 to various solutions. If C is equal to 1, then the solutions are all those sets of variables from the truth table on which the function F takes the value true (1). The remaining sets are solutions of the equation for C equal to zero. We can always consider only equations of the form:
Indeed, let the equation be given:
In this case, you can go to the equivalent equation:
Consider a system of k logical equations:
The solution of the system is a set of variables on which all equations of the system are satisfied. In terms of logical functions, to obtain a solution to the system of logical equations, one should find a set on which the logical function Ф is true, representing the conjunction of the original functions:
If the number of variables is small, for example, less than 5, then it is not difficult to construct a truth table for the function , which allows you to say how many solutions the system has and what are the sets that give solutions.
In some tasks of the Unified State Examination on finding solutions to a system of logical equations, the number of variables reaches the value of 10. Then building a truth table becomes an almost unsolvable task. Solving the problem requires a different approach. For an arbitrary system of equations, there is no general way, other than enumeration, that allows solving such problems.
In the problems proposed in the exam, the solution is usually based on taking into account the specifics of the system of equations. I repeat, apart from enumeration of all variants of a set of variables, there is no general way to solve the problem. The solution must be built based on the specifics of the system. It is often useful to carry out a preliminary simplification of a system of equations using known laws of logic. Another useful technique for solving this problem is as follows. We are not interested in all sets, but only those on which the function has the value 1. Instead of building a complete truth table, we will build its analogue - a binary decision tree. Each branch of this tree corresponds to one solution and specifies the set on which the function has the value 1. The number of branches in the decision tree coincides with the number of solutions to the system of equations.
What is a binary decision tree and how it is built, I will explain with examples of several tasks.
Problem 18
How many different sets of values of boolean variables x1, x2, x3, x4, x5, y1, y2, y3, y4, y5 are there that satisfy a system of two equations?
Answer: The system has 36 different solutions.
Solution: The system of equations includes two equations. Let's find the number of solutions for the first equation depending on 5 variables - . The first equation can in turn be considered as a system of 5 equations. As has been shown, the system of equations actually represents a conjunction of logical functions. The converse statement is also true - the conjunction of conditions can be considered as a system of equations.
Let's build a decision tree for the implication () - the first term of the conjunction, which can be considered as the first equation. Here is what the graphic image of this tree looks like
The tree consists of two levels according to the number of variables in the equation. The first level describes the first variable . Two branches of this level reflect the possible values of this variable - 1 and 0. At the second level, the branches of the tree reflect only those possible values of the variable for which the equation takes the value true. Since the equation defines an implication, the branch on which it has a value of 1 requires that it has a value of 1 on that branch. The branch on which it has a value of 0 generates two branches with values equal to 0 and 1. The constructed tree defines three solutions, on where the implication takes the value 1. On each branch, the corresponding set of values of the variables is written out, which gives a solution to the equation.
These sets are: ((1, 1), (0, 1), (0, 0))
Let's continue building the decision tree by adding the following equation, the following implication. The specificity of our system of equations is that each new equation of the system uses one variable from the previous equation, adding one new variable. Since the variable already has values in the tree, then on all branches where the variable has a value of 1, the variable will also have a value of 1. For such branches, the construction of the tree continues to the next level, but no new branches appear. The only branch where the variable has the value 0 will give a branch into two branches, where the variable will get the values 0 and 1. Thus, each addition of a new equation, given its specificity, adds one solution. Original first equation:
has 6 solutions. Here is what the complete decision tree for this equation looks like:
The second equation of our system is similar to the first:
The only difference is that the equation uses Y variables. This equation also has 6 solutions. Since each variable solution can be combined with each variable solution, the total number of solutions is 36.
Note that the constructed decision tree gives not only the number of solutions (according to the number of branches), but also the solutions themselves, written out on each branch of the tree.
Problem 19
How many different sets of values of boolean variables x1, x2, x3, x4, x5, y1, y2, y3, y4, y5 are there that satisfy all of the following conditions?
This task is a modification of the previous task. The difference is that another equation is added that relates the X and Y variables.
It follows from the equation that when it has the value 1 (one such solution exists), then it has the value 1. Thus, there is one set on which and have the values 1. When equal to 0, it can have any value, both 0 and and 1. Therefore, each set with equal to 0, and there are 5 such sets, corresponds to all 6 sets with variables Y. Therefore, the total number of solutions is 31.
Problem 20
Solution: Remembering the basic equivalence, we write our equation as:
The cyclic chain of implications means that the variables are identical, so our equation is equivalent to:
This equation has two solutions when all are either 1 or 0.
Problem 21
How many solutions does the equation have:
Solution: Just as in Problem 20, we pass from cyclic implications to identities by rewriting the equation in the form:
Let's build a decision tree for this equation:
Problem 22
How many solutions does the following system of equations have?
Lesson topic: Solving logical equations
Educational - learning how to solve logical equations, the formation of skills and abilities for solving logical equations and building a logical expression according to the truth table;Educational - create conditions for the development of cognitive interest of students, promote the development of memory, attention, logical thinking;
Educational : contribute to the education of the ability to listen to the opinions of others, education of will and perseverance to achieve the final results.
Lesson type: combined lesson
Equipment: computer, multimedia projector, presentation 6.
During the classes
Repetition and updating of basic knowledge. Examination homework(10 minutes)
In the previous lessons, we got acquainted with the basic laws of the algebra of logic, learned how to use these laws to simplify logical expressions.
Let's check the homework on simplifying logical expressions:
1. Which of the following words satisfies the logical condition:
(first consonant → second consonant)٨ (last letter vowel → penultimate letter vowel)? If there are several such words, indicate the smallest of them.
1) ANNA 2) MARIA 3) OLEG 4) STEPAN
Let us introduce the notation:
A is the first letter of a consonant
B is the second letter of a consonant
S is the last vowel
D - penultimate vowel
Let's make an expression:
Let's make a table:
2. Indicate which logical expression is equivalent to the expression
Let's simplify the writing of the original expression and the proposed options:
3. A fragment of the truth table of the expression F is given:
What expression corresponds to F?
Let's determine the values of these expressions for the specified values of the arguments:
Familiarization with the topic of the lesson, presentation of new material (30 minutes)
We continue to study the basics of logic and the topic of our today's lesson "Solving logical equations." After studying this topic, you will learn the basic ways to solve logical equations, get the skills to solve these equations by using the language of logic algebra and the ability to compose a logical expression on the truth table.
1. Solve the logical equation
(¬K M) → (¬L M N)=0
Write your answer as a string of four characters: the values of the variables K, L, M, and N (in that order). So, for example, line 1101 corresponds to K=1, L=1, M=0, N=1.
Solution:
Let's transform the expression(¬K M) → (¬L M N)
The expression is false when both terms are false. The second term is equal to 0 if M=0, N=0, L=1. In the first term, K = 0, since M = 0, and 
.
Answer: 0100
2. How many solutions does the equation have (indicate only the number in your answer)?
Solution: transform the expression
(A+B)*(C+D)=1
A+B=1 and C+D=1
Method 2: compiling a truth table
3 way: construction of SDNF - a perfect disjunctive normal form for a function - a disjunction of complete regular elementary conjunctions.Let's transform the original expression, open the brackets in order to get the disjunction of the conjunctions:
(A+B)*(C+D)=A*C+B*C+A*D+B*D=
Let's supplement the conjunctions to complete conjunctions (the product of all arguments), open the brackets:
Consider the same conjunctions:
As a result, we obtain an SDNF containing 9 conjunctions. Therefore, the truth table for this function has a value of 1 on 9 rows out of 2 4 =16 sets of variable values.
3. How many solutions does the equation have (indicate only the number in your answer)?
Let's simplify the expression:
,
3 way: construction of SDNF
Consider the same conjunctions:
As a result, we get an SDNF containing 5 conjunctions. Therefore, the truth table for this function has a value of 1 on 5 rows of 2 4 =16 sets of variable values.
Building a logical expression according to the truth table:
for each row of the truth table containing 1, we compose the product of the arguments, and the variables equal to 0 are included in the product with negation, and the variables equal to 1 - without negation. The desired expression F will be composed of the sum of the products obtained. Then, if possible, this expression should be simplified.
Example: the truth table of an expression is given. Build a logical expression.
Solution:3. Homework (5 minutes)
Solve the equation:
How many solutions does the equation have (answer only the number)?
According to the given truth table, make a logical expression and
simplify it.
