The logical function F is given by the expression. Logic and true sets
Job source: Solution 2437. Unified State Exam 2017. Computer Science. V.R. Leschiner. 10 options.
Task 2. The logical function F is given by the expression . Determine which column of the truth table of the function F corresponds to each of the variables x, y, z.
In your answer, write the letters x, y, z in the order in which their corresponding columns appear (first - the letter corresponding to the 1st column, then - the letter corresponding to the 2nd column, then - the letter corresponding to the 3rd column) . Write the letters in the answer in a row; there is no need to put any separators between the letters.
Solution.
Let us rewrite the expression for F taking into account the priorities of the operations of negation, conjunction and disjunction:
.
Consider the 4th row of the table (1,1,0)=0. From this we can see that the third place must be either the variable y or the variable z, otherwise the second bracket will contain 1, which will lead to the value F=1. Now consider the 5th row of the table (0,0,1)=1. Since x must be in the first or second place, the first parenthesis will give 1 only when y is in the 3rd place. Considering that the second bracket is always equal to 0, then F=1 is obtained due to the 1 in the first bracket. Thus, we found that y is in 3rd place. Finally, consider the 7th row of the table (1,0,1)=0. Here y=1 and for F=0 it is necessary to have z=0 and x=1, therefore, x is in the 1st place, and z is in the second.
Analysis of task 2 of the Unified State Exam 2017 in computer science from the demo version project. This is a task of a basic level of difficulty. Approximate time to complete the task is 3 minutes.
Tested content elements: ability to construct truth tables and logical circuits. Content elements tested on the Unified State Exam: statements, logical operations, quantifiers, truth of statements.
Task 2:
Logic function F is given by the expression x /\¬ y /\ (¬ z \/ w).
The figure shows a fragment of the truth table of the function F containing All F true.
Determine which column of the function's truth table F each of the variables corresponds w, x, y, z.
Write the letters in your answer w, x, y, z in the order in which the corresponding columns appear (first - the letter corresponding to the first column; then - the letter corresponding to the second column, etc.) Write the letters in the answer in a row, there is no need to put any separators between the letters.
Example. If the function were given by the expression ¬ x \/ y, depending on two variables: x And y, and a fragment of its truth table was given, containing All sets of arguments for which the function F true.
Then the first column would correspond to the variable y, and the second column is a variable x. The answer should have written: yx.
Answer: ________
x /\¬ y /\ (¬ z \/ w)
A conjunction (logical multiplication) is true if and only if all statements are true. Therefore the variable X 1 .
Thus, the variable x corresponds to the column with variable 3.
Variable ¬y the column containing the value must match 0 .
A disjunction (logical addition) of two statements is true if and only if at least one statement is true.
Disjunction ¬z\/w in this line will be true only if z=0, w=1.
Thus, the variable ¬z corresponds to column with variable 1 (1 column), variable w corresponds to the column with variable 4 (column 4).
Logic function F is given by the expression x/\ ¬y/\ (¬z\/ w).
The figure shows a fragment of the truth table of the function F containing All sets of arguments for which the function F true.
Determine which column of the function's truth table F each of the variables corresponds w, x, y, z.
Write the letters in your answer w, x, y, z in the order they come
their corresponding columns (first – the letter corresponding to the first
column; then the letter corresponding to the second column, etc.) Letters
In your answer, write in a row, put no separators between letters.
no need.
Demo version of the Unified State Examination USE 2017 – task No. 2
Solution:
A conjunction (logical multiplication) is true if and only if all statements are true. Therefore the variable X 1 .
Variable ¬y must match the column in which all values are equal 0 .
A disjunction (logical addition) of two statements is true if and only if at least one statement is true.
Disjunction ¬z\/y z=0, w=1.
Thus, the variable ¬z w corresponds to the column with variable 4 (column 4).
Answer: zyxw
Demo version of the Unified State Examination USE 2016 – task No. 2
Logic function F is given by the expression (¬z)/\x \/ x/\y. Determine which column of the truth table of the function F corresponds to each of the variables x, y, z.
In your answer, write the letters x, y, z in the order in which their corresponding columns appear (first - the letter corresponding to the 1st column; then - the letter corresponding to the 2nd column; then - the letter corresponding to the 3rd column) . Write the letters in the answer in a row; there is no need to put any separators between the letters.
Example. Let an expression x → y be given, depending on two variables x and y, and a truth table:
Then the 1st column corresponds to the variable y, and the 2nd column
corresponds to the variable x. In the answer you need to write: yx.
Solution:
1. Let's write it down for this expression in simpler notation:
¬z*x + x*y = x*(¬z + y)
2. Conjunction (logical multiplication) is true if and only if all statements are true. Therefore, so that the function ( F) was equal to one ( 1 ), it is necessary that each factor be equal to one (1 ). Thus, when F=1, variable X must match the column in which all values are equal 1 .
3. Consider (¬z + y), at F=1 this expression is also equal to 1 (see point 2).
4. Disjunction (logical addition) of two statements is true if and only if at least one statement is true.
Disjunction ¬z\/y in this line will be true only if
- z = 0; y = 0 or y = 1;
- z = 1; y = 1
5. Thus, the variable ¬z corresponds to column with variable 1 (1 column), variable y
Answer: zyx
KIM Unified State Examination Unified State Exam 2016 (early period)– task No. 2
The logical function F is given by the expression
(x /\ y /\¬z) \/ (x /\ y /\ z) \/ (x /\¬y /\¬z).
The figure shows a fragment of the truth table of the function F, containing all sets of arguments for which the function F is true. Determine which column of the truth table of the function F corresponds to each of the variables x, y, z.
In your answer, write the letters x, y, z in the order in which their corresponding columns appear (first - the letter corresponding to the first column; then - the letter corresponding to the second column, etc.) Write the letters in the answer in a row, no separators There is no need to put it between letters.
R solution:
Let's write the given expression in simpler notation:
(x*y*¬z) + (x*y*z) + (x*¬y*¬z)=1
This expression is true when at least one of (x*y*¬z), (x*y*z), (x*¬y*¬z) equals 1. Conjunction (logical multiplication) is true if and only if when all statements are true.
At least one of these disjunctions x*y*¬z; x*y*z; x*¬y*¬z will be true only if x=1.
Thus, the variable X corresponds to the column with variable 2 (column 2).
Let y- variable 1, z- prem.3. Then, in the first case x*¬y*¬z will be true in the second case x*y*¬z, and in the third x*y*z.
Answer: yxz
The symbol F denotes one of the following logical expressions from three arguments: X, Y, Z. A fragment of the truth table of the expression F is given (see the table on the right). Which expression matches F?
| X | Y | Z | F |
| 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
1) X ∧ Y ∧ Z 2) ¬X ∨ Y ∨¬Z 3) X ∧ Y ∨ Z 4) X ∨ Y ∧ ¬Z
Solution:
1) X ∧ Y ∧ Z = 1.0.1 = 0 (does not match on 2nd line)
2) ¬X ∨ Y ∨¬Z = ¬0 ∨ 0 ∨ ¬0 = 1+0+1 = 1 (does not match on the 1st line)
3) X ∧ Y ∨ Z = 0.1+0 = 0 (does not match on the 3rd line)
4) X ∨ Y ∧ ¬Z (corresponds to F)
X ∨ Y ∧ ¬Z = 0 ∨ 0 ∧ ¬0 = 0+0.1 = 0
X ∨ Y ∧ ¬Z = 1 ∨ 0 ∧ ¬1 = 1+0.0 = 1
X ∨ Y ∧ ¬Z = 0 ∨ 1 ∧ ¬0 = 0+1.1 = 1
Answer: 4
Given a fragment of the truth table of the expression F. Which expression corresponds to F?
| A | B | C | F |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
1) (A → ¬B) ∨ C 2) (¬A ∨ B) ∧ C 3) (A ∧ B) → C 4) (A ∨ B) → C
Solution:
1) (A → ¬B) ∨ C = (1 → ¬0) ∨ 0 = (1 → 1) + 0 = 1 + 0 = 1 (does not match on 2nd line)
2) (¬A ∨ B) ∧ C = (¬1 ∨ 0) ∧ 1 = (0+0).1 = 0 (does not match on the 3rd line)
3) (A ∧ B) → C = (1 ∧ 0) → 0 = 0 → 0 = 1 (does not match on 2nd line)
4) (A ∨ B) → C (corresponds to F)
(A ∨ B) → C = (0 ∨ 1) → 1 = 1
(A ∨ B) → C = (1 ∨ 0) → 0 = 0
(A ∨ B) → C = (1 ∨ 0) → 1 = 1
Answer: 4
A logical expression is given that depends on 6 logical variables:
X1 ∨ ¬X2 ∨ X3 ∨ ¬X4 ∨ X5 ∨ X6
How many different sets of variable values are there for which the expression is true?
1) 1 2) 2 3) 63 4) 64
Solution:
False expression only in 1 case: X1=0, X2=1, X3=0, X4=1, X5=0, X6=0
X1 ∨ ¬X2 ∨ X3 ∨ ¬X4 ∨ X5 ∨ X6 = 0 ∨ ¬1 ∨ 0 ∨ ¬1 ∨ 0 ∨ 0 = 0
There are 2 6 =64 options in total, which means true
Answer: 63
A fragment of the truth table of the expression F is given.
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | F |
| 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 |
Which expression matches F?
1) x1 ∨ x2 ∨ ¬x3 ∨ x4 ∨ ¬x5 ∨ x6 ∨ ¬x7
2) x1 ∨ ¬x2 ∨ x3 ∨ ¬x4 ∨ ¬x5 ∨ x6 ∨ x7
3) x1 ∧ ¬x2 ∧ x3 ∧ ¬x4 ∧ x5 ∧ ¬x6 ∧ x7
4) x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7
Solution:
1) x1 ∨ x2 ∨ ¬x3 ∨ x4 ∨ ¬x5 ∨ x6 ∨ ¬x7 = 0 + 1 + … = 1 (does not match on the 1st line)
2) x1 ∨ ¬x2 ∨ x3 ∨ ¬x4 ∨ ¬x5 ∨ x6 ∨ x7 = 0 + 0 + 0 + 0 + 0 + 1 + 0 = 1 (does not match on the 1st line)
3) x1 ∧ ¬x2 ∧ x3 ∧ ¬x4 ∧ x5 ∧ ¬x6 ∧ x7 = 1.0. ...= 0 (does not match on 2nd line)
4) x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7 (corresponds to F)
x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7 = 1.1.1.1.1.1.1 = 1
x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7 = 0. … = 0
Answer: 4
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | x8 | F |
| 0 | 1 | 1 | ||||||
| 1 | 0 | 1 | 0 | |||||
| 1 | 0 | 1 |
What expression can F be?
1) x1 ∧ ¬x2 ∧ x3 ∧ ¬x4 ∧ x5 ∧ x6 ∧ ¬x7 ∧ ¬x8
2) ¬x1 ∨ x2 ∨ x3 ∨ ¬x4 ∨ ¬x5 ∨ ¬x6 ∨ ¬x7 ∨ x8
3) ¬x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ x5 ∧ ¬x6 ∧ ¬x7 ∧ ¬x8
4) ¬x1 ∨ ¬x2 ∨ ¬x3 ∨ ¬x4 ∨ ¬x5 ∨ ¬x6 ∨ ¬x7 ∨ ¬x8
Solution:
1) x1 ∧ ¬x2 ∧ x3 ∧ ¬x4 ∧ x5 ∧ x6 ∧ ¬x7 ∧ ¬x8 = x1 . ¬x2. 0 . ... = 0 (does not match on 1st line)
2) ¬x1 ∨ x2 ∨ x3 ∨ ¬x4 ∨ ¬x5 ∨ ¬x6 ∨ ¬x7 ∨ x8 (corresponds to F)
3) ¬x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ x5 ∧ ¬x6 ∧ ¬x7 ∧ ¬x8 = … ¬x7 ∧ ¬x8 = … ¬1 ∧ ¬x8 = … 0 ∧ ¬x8 = 0 (does not match on 1- th line)
4) ¬x1 ∨ ¬x2 ∨ ¬x3 ∨ ¬x4 ∨ ¬x5 ∨ ¬x6 ∨ ¬x7 ∨ ¬x8 = ¬x1 ∨ ¬x2 ∨ ¬x3 … = ¬1 ∨ ¬x2 ∨ ¬0 .. = 1 (not matches on the 2nd line)
Answer: 2
Given is a fragment of the truth table for the expression F:
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | F |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |
Find the minimum possible number of different rows in the complete truth table of this expression in which the value x5 matches F.
Solution:
Minimum possible number of distinct rows in which the value x5 matches F = 4
Answer: 4
Given is a fragment of the truth table for the expression F:
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | x8 | F |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
Find the maximum possible number of distinct rows in the complete truth table of this expression in which the value x6 does not coincide with F.
Solution:
Maximum possible number = 2 8 = 256
The maximum possible number of different rows in which the value x6 does not match F = 256 – 5 = 251
Answer: 251
Given is a fragment of the truth table for the expression F:
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | F |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |
Find the maximum possible number of different rows of the complete truth table of this expression in which the value ¬x5 ∨ x1 coincides with F.
Solution:
1+0=1 – does not match F
0+0=0 – does not match F
0+0=0 – does not match F
0+1=1 – coincides with F
1+0=1 – coincides with F
2 7 = 128 – 3 = 125
Answer: 125
Each Boolean expression A and B depends on the same set of 6 variables. In the truth tables, each of these expressions has exactly 4 units in the value column. What is the minimum possible number of ones in the value column of the truth table of the expression A ∨ B?
Solution:
Answer: 4
Each Boolean expression A and B depends on the same set of 7 variables. In the truth tables, each of these expressions has exactly 4 units in the value column. What is the maximum possible number of ones in the value column of the truth table of the expression A ∨ B?
Solution:
Answer: 8
Each Boolean expression A and B depends on the same set of 8 variables. In the truth tables, each of these expressions has exactly 5 units in the value column. What is the minimum possible number of zeros in the value column of the truth table of the expression A ∧ B?
Solution:
2 8 = 256 – 5 = 251
Answer: 251
Each Boolean expression A and B depends on the same set of 8 variables. In the truth tables, each of these expressions has exactly 6 units in the value column. What is the maximum possible number of zeros in the value column of the truth table of the expression A ∧ B?
Solution:
Answer: 256
The Boolean expressions A and B each depend on the same set of 5 variables. There are no matching rows in the truth tables of both expressions. How many ones will be contained in the value column of the truth table of the expression A ∧ B?
Solution:
There are no matching rows in the truth tables of both expressions.
Answer: 0
The Boolean expressions A and B each depend on the same set of 6 variables. There are no matching rows in the truth tables of both expressions. How many ones will be contained in the value column of the truth table of the expression A ∨ B?
Solution:
Answer: 64
Each of the Boolean expressions A and B depends on the same set of 7 variables. There are no matching rows in the truth tables of both expressions. What is the maximum possible number of zeros in the value column of the truth table of the expression ¬A ∨ B?
Solution:
A=1,B=0 => ¬0 ∨ 0 = 0 + 0 = 0
Answer: 128
Each of the Boolean expressions F and G contains 7 variables. There are exactly 8 identical rows in the truth tables of the expressions F and G, and exactly 5 of them have a 1 in the value column. How many rows of the truth table for the expression F ∨ G contain a 1 in the value column?
Solution:
There are exactly 8 identical rows, and exactly 5 of them have a 1 in the value column.
This means that exactly 3 of them have a 0 in the value column.
Answer: 125
The logical function F is given by the expression (a ∧ ¬c) ∨ (¬b ∧ ¬c). Determine which column of the truth table of the function F corresponds to each of the variables a, b, c.
| ? | ? | ? | F |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
In your answer, write the letters a, b, c in the order in which their corresponding columns appear.
Solution:
(a . ¬c) + (¬b . ¬c)
When c is 1, F is zero so the last column is c.
To determine the first and second columns, we can use the values from the 3rd row.
(a . 1) + (¬b . 1) = 0
Answer: ABC
The logical function F is given by the expression (a ∧ c)∨ (¬a ∧ (b ∨ ¬c)). Determine which column of the truth table of the function F corresponds to each of the variables a, b, c.
Based on the fact that when a=0 and c=0, then F=0, and the data from the second row, we can conclude that the third column contains b.
Answer: cab
The logical function F is given by x ∧ (¬y ∧ z ∧ ¬w ∨ y ∧ ¬z). The figure shows a fragment of the truth table of the function F, containing all sets of arguments for which the function F is true. Determine which column of the truth table of the function F corresponds to each of the variables x, y, z, w.
| ? | ? | ? | ? | F |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 |
In your answer, write the letters x, y, z, w in the order in which their corresponding columns appear.
Solution:
x ∧ (¬y ∧ z ∧ ¬w ∨ y ∧ ¬z)
x. (¬y . z . ¬w . y . ¬z)
Based on the fact that at x=0, then F=0, we can conclude that the second column contains x.
Answer: wxzy
Demonstration version of the Unified State Exam 2019 – task No. 2
Misha filled out the truth table of the function (¬x /\ ¬y) \/ (y≡z) \/ ¬w, but only managed to fill out a fragment of three different lines, without even indicating which column of the table corresponds to each of the variables w, x ,
y, z.
Determine which table column each variable w, x, y, z corresponds to.
In your answer, write the letters w, x, y, z in the order in which their corresponding columns appear (first the letter corresponding to the first column; then the letter corresponding to the second column, etc.). Letters
In your answer, write in a row; there is no need to put any separators between the letters.
Example. If the function were given by the expression ¬x \/ y, depending on two variables, and the table fragment would look like
then the first column would correspond to the variable y, and the second column would correspond to the variable x. The answer should have been written yx.
(¬x ¬y)+(y≡z)+¬w=0
w=1 w must be true; w - last
y and z must be different, so before the latter, it is x. the first two are y and z or z and y.
y and x cannot be false at the same time. The first is z.
Answer: zyxw
Demonstration version of the Unified State Exam 2018 – task No. 2
The logical function F is given by the expression ¬x \/ y \/ (¬z /\ w). The figure shows a fragment of the truth table of the function F, containing all sets of arguments for which the function F is false. Determine which column of the truth table of function F corresponds to each of the variables w, x, y, z
In your answer, write the letters w, x, y, z in the order in which their corresponding columns appear (first - the letter corresponding to the first column; then - the letter corresponding to the second column, etc.) Write the letters in the answer in a row, There is no need to put any separators between letters. Example. If the function were given by the expression ¬x\/y, depending on two variables: x and y, and a fragment of its truth table was given, containing all sets of arguments for which the function is true.
Then the first column would correspond to the variable y, and the second column would correspond to the variable x. The answer should have been written: yx.
Answer: xzwy
Logic function F is given by the expression x/\ ¬y/\ (¬z\/ w).
The figure shows a fragment of the truth table of the function F containing All sets of arguments for which the function F true.
Determine which column of the function's truth table F each of the variables corresponds w, x, y, z.
Write the letters in your answer w, x, y, z in the order they come
their corresponding columns (first – the letter corresponding to the first
column; then the letter corresponding to the second column, etc.) Letters
In your answer, write in a row, put no separators between letters.
no need.
Demonstration version of the Unified State Exam 2017 - task No. 2
Solution:
A conjunction (logical multiplication) is true if and only if all statements are true. Therefore the variable X 1 .
Variable ¬y must match the column in which all values are equal 0 .
A disjunction (logical addition) of two statements is true if and only if at least one statement is true.
Disjunction ¬z\/y z=0, w=1.
Thus, the variable ¬z w corresponds to the column with variable 4 (column 4).
Answer: zyxw
Demonstration version of the Unified State Exam 2016 - task No. 2
Logic function F is given by the expression (¬z)/\x \/ x/\y. Determine which column of the truth table of the function F corresponds to each of the variables x, y, z.
In your answer, write the letters x, y, z in the order in which their corresponding columns appear (first - the letter corresponding to the 1st column; then - the letter corresponding to the 2nd column; then - the letter corresponding to the 3rd column) . Write the letters in the answer in a row; there is no need to put any separators between the letters.
Example. Let an expression x → y be given, depending on two variables x and y, and a truth table:
Then the 1st column corresponds to the variable y, and the 2nd column
corresponds to the variable x. In the answer you need to write: yx.
Solution:
1. Let's write the given expression in simpler notation:
¬z*x + x*y = x*(¬z + y)
2. Conjunction (logical multiplication) is true if and only if all statements are true. Therefore, so that the function ( F) was equal to one ( 1 ), each factor must be equal to one ( 1 ). Thus, when F=1, variable X must match the column in which all values are equal 1 .
3. Consider (¬z + y), at F=1 this expression is also equal to 1 (see point 2).
4. Disjunction (logical addition) of two statements is true if and only if at least one statement is true.
Disjunction ¬z\/y in this line will be true only if
- z = 0; y = 0 or y = 1;
- z = 1; y = 1
5. Thus, the variable ¬z corresponds to column with variable 1 (1 column), variable y
Answer: zyx
KIM Unified State Exam 2016 (early period)– task No. 2
The logical function F is given by the expression
(x /\ y /\¬z) \/ (x /\ y /\ z) \/ (x /\¬y /\¬z).
The figure shows a fragment of the truth table of the function F, containing all sets of arguments for which the function F is true. Determine which column of the truth table of the function F corresponds to each of the variables x, y, z.
In your answer, write the letters x, y, z in the order in which their corresponding columns appear (first - the letter corresponding to the first column; then - the letter corresponding to the second column, etc.) Write the letters in the answer in a row, no separators There is no need to put it between letters.
R solution:
Let's write the given expression in simpler notation:
(x*y*¬z) + (x*y*z) + (x*¬y*¬z)=1
This expression is true when at least one of (x*y*¬z), (x*y*z), (x*¬y*¬z) equals 1. Conjunction (logical multiplication) is true if and only if when all statements are true.
At least one of these disjunctions x*y*¬z; x*y*z; x*¬y*¬z will be true only if x=1.
Thus, the variable X corresponds to the column with variable 2 (column 2).
Let y- variable 1, z- prem.3. Then, in the first case x*¬y*¬z will be true in the second case x*y*¬z, and in the third x*y*z.
Answer: yxz
The symbol F denotes one of the following logical expressions from three arguments: X, Y, Z. A fragment of the truth table of the expression F is given (see the table on the right). Which expression matches F?
| X | Y | Z | F |
| 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
1) X ∧ Y ∧ Z 2) ¬X ∨ Y ∨¬Z 3) X ∧ Y ∨ Z 4) X ∨ Y ∧ ¬Z
Solution:
1) X ∧ Y ∧ Z = 1.0.1 = 0 (does not match on 2nd line)
2) ¬X ∨ Y ∨¬Z = ¬0 ∨ 0 ∨ ¬0 = 1+0+1 = 1 (does not match on the 1st line)
3) X ∧ Y ∨ Z = 0.1+0 = 0 (does not match on the 3rd line)
4) X ∨ Y ∧ ¬Z (corresponds to F)
X ∨ Y ∧ ¬Z = 0 ∨ 0 ∧ ¬0 = 0+0.1 = 0
X ∨ Y ∧ ¬Z = 1 ∨ 0 ∧ ¬1 = 1+0.0 = 1
X ∨ Y ∧ ¬Z = 0 ∨ 1 ∧ ¬0 = 0+1.1 = 1
Answer: 4
Given a fragment of the truth table of the expression F. Which expression corresponds to F?
| A | B | C | F |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
1) (A → ¬B) ∨ C 2) (¬A ∨ B) ∧ C 3) (A ∧ B) → C 4) (A ∨ B) → C
Solution:
1) (A → ¬B) ∨ C = (1 → ¬0) ∨ 0 = (1 → 1) + 0 = 1 + 0 = 1 (does not match on 2nd line)
2) (¬A ∨ B) ∧ C = (¬1 ∨ 0) ∧ 1 = (0+0).1 = 0 (does not match on the 3rd line)
3) (A ∧ B) → C = (1 ∧ 0) → 0 = 0 → 0 = 1 (does not match on 2nd line)
4) (A ∨ B) → C (corresponds to F)
(A ∨ B) → C = (0 ∨ 1) → 1 = 1
(A ∨ B) → C = (1 ∨ 0) → 0 = 0
(A ∨ B) → C = (1 ∨ 0) → 1 = 1
Answer: 4
A logical expression is given that depends on 6 logical variables:
X1 ∨ ¬X2 ∨ X3 ∨ ¬X4 ∨ X5 ∨ X6
How many different sets of variable values are there for which the expression is true?
1) 1 2) 2 3) 63 4) 64
Solution:
False expression only in 1 case: X1=0, X2=1, X3=0, X4=1, X5=0, X6=0
X1 ∨ ¬X2 ∨ X3 ∨ ¬X4 ∨ X5 ∨ X6 = 0 ∨ ¬1 ∨ 0 ∨ ¬1 ∨ 0 ∨ 0 = 0
There are 2 6 =64 options in total, which means true
Answer: 63
A fragment of the truth table of the expression F is given.
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | F |
| 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 |
Which expression matches F?
1) x1 ∨ x2 ∨ ¬x3 ∨ x4 ∨ ¬x5 ∨ x6 ∨ ¬x7
2) x1 ∨ ¬x2 ∨ x3 ∨ ¬x4 ∨ ¬x5 ∨ x6 ∨ x7
3) x1 ∧ ¬x2 ∧ x3 ∧ ¬x4 ∧ x5 ∧ ¬x6 ∧ x7
4) x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7
Solution:
1) x1 ∨ x2 ∨ ¬x3 ∨ x4 ∨ ¬x5 ∨ x6 ∨ ¬x7 = 0 + 1 + … = 1 (does not match on the 1st line)
2) x1 ∨ ¬x2 ∨ x3 ∨ ¬x4 ∨ ¬x5 ∨ x6 ∨ x7 = 0 + 0 + 0 + 0 + 0 + 1 + 0 = 1 (does not match on the 1st line)
3) x1 ∧ ¬x2 ∧ x3 ∧ ¬x4 ∧ x5 ∧ ¬x6 ∧ x7 = 1.0. ...= 0 (does not match on 2nd line)
4) x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7 (corresponds to F)
x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7 = 1.1.1.1.1.1.1 = 1
x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7 = 0. … = 0
Answer: 4
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | x8 | F |
| 0 | 1 | 1 | ||||||
| 1 | 0 | 1 | 0 | |||||
| 1 | 0 | 1 |
What expression can F be?
1) x1 ∧ ¬x2 ∧ x3 ∧ ¬x4 ∧ x5 ∧ x6 ∧ ¬x7 ∧ ¬x8
2) ¬x1 ∨ x2 ∨ x3 ∨ ¬x4 ∨ ¬x5 ∨ ¬x6 ∨ ¬x7 ∨ x8
3) ¬x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ x5 ∧ ¬x6 ∧ ¬x7 ∧ ¬x8
4) ¬x1 ∨ ¬x2 ∨ ¬x3 ∨ ¬x4 ∨ ¬x5 ∨ ¬x6 ∨ ¬x7 ∨ ¬x8
Solution:
1) x1 ∧ ¬x2 ∧ x3 ∧ ¬x4 ∧ x5 ∧ x6 ∧ ¬x7 ∧ ¬x8 = x1 . ¬x2. 0 . ... = 0 (does not match on 1st line)
2) ¬x1 ∨ x2 ∨ x3 ∨ ¬x4 ∨ ¬x5 ∨ ¬x6 ∨ ¬x7 ∨ x8 (corresponds to F)
3) ¬x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ x5 ∧ ¬x6 ∧ ¬x7 ∧ ¬x8 = … ¬x7 ∧ ¬x8 = … ¬1 ∧ ¬x8 = … 0 ∧ ¬x8 = 0 (does not match on 1- th line)
4) ¬x1 ∨ ¬x2 ∨ ¬x3 ∨ ¬x4 ∨ ¬x5 ∨ ¬x6 ∨ ¬x7 ∨ ¬x8 = ¬x1 ∨ ¬x2 ∨ ¬x3 … = ¬1 ∨ ¬x2 ∨ ¬0 .. = 1 (not matches on the 2nd line)
Answer: 2
Given is a fragment of the truth table for the expression F:
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | F |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |
Find the minimum possible number of different rows in the complete truth table of this expression in which the value x5 matches F.
Solution:
Minimum possible number of distinct rows in which the value x5 matches F = 4
Answer: 4
Given is a fragment of the truth table for the expression F:
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | x8 | F |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
Find the maximum possible number of distinct rows in the complete truth table of this expression in which the value x6 does not coincide with F.
Solution:
Maximum possible number = 2 8 = 256
The maximum possible number of different rows in which the value x6 does not match F = 256 - 5 = 251
Answer: 251
Given is a fragment of the truth table for the expression F:
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | F |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |
Find the maximum possible number of different rows of the complete truth table of this expression in which the value ¬x5 ∨ x1 coincides with F.
Solution:
1+0=1 - does not match F
0+0=0 - does not match F
0+0=0 - does not match F
0+1=1 - same as F
1+0=1 - same as F
2 7 = 128 — 3 = 125
Answer: 125
Each Boolean expression A and B depends on the same set of 6 variables. In the truth tables, each of these expressions has exactly 4 units in the value column. What is the minimum possible number of ones in the value column of the truth table of the expression A ∨ B?
Solution:
Answer: 4
Each Boolean expression A and B depends on the same set of 7 variables. In the truth tables, each of these expressions has exactly 4 units in the value column. What is the maximum possible number of ones in the value column of the truth table of the expression A ∨ B?
Solution:
Answer: 8
Each Boolean expression A and B depends on the same set of 8 variables. In the truth tables, each of these expressions has exactly 5 units in the value column. What is the minimum possible number of zeros in the value column of the truth table of the expression A ∧ B?
Solution:
2 8 = 256 — 5 = 251
Answer: 251
Each Boolean expression A and B depends on the same set of 8 variables. In the truth tables, each of these expressions has exactly 6 units in the value column. What is the maximum possible number of zeros in the value column of the truth table of the expression A ∧ B?
Solution:
Answer: 256
The Boolean expressions A and B each depend on the same set of 5 variables. There are no matching rows in the truth tables of both expressions. How many ones will be contained in the value column of the truth table of the expression A ∧ B?
Solution:
There are no matching rows in the truth tables of both expressions.
Answer: 0
The Boolean expressions A and B each depend on the same set of 6 variables. There are no matching rows in the truth tables of both expressions. How many ones will be contained in the value column of the truth table of the expression A ∨ B?
The logical function F is given by the expression (a ∧ ¬c) ∨ (¬b ∧ ¬c). Determine which column of the truth table of the function F corresponds to each of the variables a, b, c.
| ? | ? | ? | F |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
In your answer, write the letters a, b, c in the order in which their corresponding columns appear.
Solution:
(a . ¬c) + (¬b . ¬c)
When c is 1, F is zero so the last column is c.
To determine the first and second columns, we can use the values from the 3rd row.
(a . 1) + (¬b . 1) = 0
Answer: ABC
The logical function F is given by the expression (a ∧ c)∨ (¬a ∧ (b ∨ ¬c)). Determine which column of the truth table of the function F corresponds to each of the variables a, b, c.
| ? | ? | ? | F | |
| 0 | 0 | 0 | 1 | |
| 0 | 0 | 1 | 1 | |
| 0 | 1 | 0 | 0 | |
| 0 | 1 | 1 | 0 | |
| 1 | 0 | 0 | 0 | |
| 1 | 0 | 1 | 1 | |
| 1 | 1 | 0 | 1 | |
| 0 | ||||
| 1 | 1 | 1 |
Based on the fact that when a=0 and c=0, then F=0, and the data from the second row, we can conclude that the third column contains b.
Answer: cab
The logical function F is given by x ∧ (¬y ∧ z ∧ ¬w ∨ y ∧ ¬z). The figure shows a fragment of the truth table of the function F, containing all sets of arguments for which the function F is true. Determine which column of the truth table of the function F corresponds to each of the variables x, y, z, w.
| ? | ? | ? | ? | F |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 |
In your answer, write the letters x, y, z, w in the order in which their corresponding columns appear.
Solution:
x ∧ (¬y ∧ z ∧ ¬w ∨ y ∧ ¬z)
x. (¬y . z . ¬w . y . ¬z)
Based on the fact that at x=0, then F=0, we can conclude that the second column contains x.
Answer: wxzy
