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Discrete Mathematics
Reporter: Anton Kuznietsov, Kharkiv Karazin National University, Ukraine
In Problems
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Discrete mathematics and programming
Ideas from combination theory and graph theory
Algorithmic programming
Some problems in discrete
mathematics
Math packages and programming
are applied in are applied in
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Scheme of the presentation
Problems 1. Knights and Liars 2. Competing people 3. Search for the
culprit 4. Queens 5. Knight’s move 6. Pavement
Conclusions
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1. Knights & Liars
Suppose, we are on a certain island and have talked with three inhabitants A, B and C.
Each of them is either a knight or a liar. Knights always say truth, liars always lie.
Two of them (A and B) came out with the following suggestions:
A: We all are liars.
B: Exactly one of us is a knight.
Question: Who of the inhabitants A, B and C is a knight, and who is a liar? Write down the inhabitants’ propositions, using formulas of proposition calculus.
a = true A – knight
A: B:cba )()()( cbacbacba
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1. Knights & Liars - solutiona = true A – knight
A: B:cba )()()( cbacbacba
acb
c at least 2 said truth,↯
↯a
bb
Answer: B is the only knight, A and C are liars.
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2. Competing people
Four boys – Alex, Bill, Charles and Daniel – had a running-competition.
Next day they were asked: “Who and what place has taken?” The boys answered so: Alex: I wasn’t the first and the last. Bill: I wasn’t the last. Charles: I was the first. Daniel: I was the last. It is known, than three of these answers are true and one is false. Question: Who has told a lie? Who is the champion?
0 1 1 0
1 1 1 0
1 0 0 0
0 0 0 1
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2. Competing people - solution 0 1 1 0
1 1 1 0
1 0 0 0
0 0 0 1
1 0 0 1
1 1 1 0
1 0 0 0
0 0 0 1
0 1 1 0
1 1 1 0
1 0 0 0
1 1 1 0
0 1 1 0
1 1 1 0
0 1 1 1
0 0 0 1
0 1 1 0
0 0 0 1
1 0 0 0
0 0 0 1
A - liar B - liar C - liar D - liar0 1 1 0
1 1 1 0
0 1 1 1
0 0 0 1
0 1 1 0
1 1 1 0
0 1 1 1
0 0 0 1
Answer: Charles is a liar, Bill is the champion.
0 1 1 0
0 0 0 1
1 0 0 0
0 0 0 1
0 1 1 0
1 1 1 0
1 0 0 0
1 1 1 0
1 0 0 1
1 1 1 0
1 0 0 0
0 0 0 1
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3. Search for the culprit
Four people (A, B, C, D) are under suspicion of committing a crime. The following is ascertained:
If A and B are guilty, then the suspected C is also guilty.
If A is guilty, then B or C is also guilty. If C is the culprit, then D is also guilty. If A is innocent, then D is the culprit. Question: Is D guilty?
A A is guiltyCBA CBA
DC DA
(1)
(2)
(3)
(4)
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3. Search for the culprit - solution
CBA CBA
DC DA
(1)
(2)
(3)
(4)
D)3(
A ACB )2(
D)3(C)1(
B C
D)4(
Answer: D is guilty.
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4. Queens Dispose eight queens
on the chess-board so, that the queens don't threaten each other.
Find all variants of such arrangement.
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4. Queens - solution
1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0
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5. Knight’s moves
There is a chess-board of size n x n (n <= 10).
A knight stands initially on the field with coordinates (x0, y0).
The knight has to visit every field of the chess-board exactly once.
Find the sequence of knight’s moves (if it exists).
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5. Knight’s moves - solution
1 10 5 16 25 4 17 2 11 6 9 20 13 24 15 18 3 22 7 12 21 8 19 14 23
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6. Pavement
Roadmen have pavement plates of size 1x1 and 1x2.
How many ways are there to pave the road of size 2xN (1<=N<=1000)?
The plates 1x2 are made on factory so, that they can be placed only with the wide side lengthwise the road.
2 x N
1, 4, 9, 25, 64, 169, 441, …
N = 1, 2, 3, …
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6. Pavement2 x N
– the number of ways to pave the road.
Nx;21 NNN xxx ;1,0 10 FF 1 NN Fx
;2,1 21 xx
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1, 4, 9, 25, 64, 169, 441, …
6. Pavement2 x N
21NF
N = 1, 2, 3, …
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6. Pavement
Roadmen have only plates of size 1x2.
The plates can be placed both lengthwise and crosswise the road.
How many ways are there in this case?
2 x N
;2,1 21 yy
.1 NN Fy
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6. Pavement3 x N
Roadmen have only plates of size 1x2.
The plates can be placed both lengthwise and crosswise the road.
How many ways are there in this case?
1 < N < 1000. N is even.
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6. Pavement3 x N
11 2 mmm BAA
211 mmm BAB
;2
nm
,
mA - the required quantity;
1mB- the number of ways to pave this road:
Am = 3, 11, 41, 153, 571, 2131, 7953, …
m = 1, 2, 3, …
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Conclusions
Combination theory, graph theory, pounding
theory, Fibonacci numbers, Catalan
numbers
Algorithmic programming
Problems of logic, combination theory, graph
theory
Programming
are applied in are applied in
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Thank you for your kind attention!
Reporter: Anton Kuznietsov, Kharkiv Karazin National University, Ukraine