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Discrete Mathematics Reporter: Anton Kuznietsov, Kharkiv Karazin National University, Ukraine In Problems.

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Presentation on theme: "Discrete Mathematics Reporter: Anton Kuznietsov, Kharkiv Karazin National University, Ukraine In Problems."— Presentation transcript:

1 Discrete Mathematics Reporter: Anton Kuznietsov, Kharkiv Karazin National University, Ukraine In Problems

2 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

3 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

4 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:

5 1. Knights & Liars - solution a = true  A – knight A:B: at least 2 said truth, ↯ ↯ b Answer: B is the only knight, A and C are liars.

6 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

7 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 - liarB - liarC - liarD - liar 0 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

8 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 guilty (1) (2) (3) (4)

9 3. Search for the culprit - solution (1) (2) (3) (4) A B C Answer: D is guilty.

10 4. Queens Dispose eight queens on the chess-board so, that the queens don't threaten each other. Find all variants of such arrangement.

11 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

12 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).

13 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

14 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, …

15 6. Pavement2 x N – the number of ways to pave the road.

16 1, 4, 9, 25, 64, 169, 441, … 6. Pavement2 x N N = 1, 2, 3, …

17 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

18 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.

19 6. Pavement3 x N, - the required quantity; - the number of ways to pave this road: A m = 3, 11, 41, 153, 571, 2131, 7953, … m = 1, 2, 3, …

20 Conclusions Combination theory, graph theory, pounding theory, Fibonacci numbers, Catalan numbers Algorithmic programming Problems of logic, combination theory, graph theory Programming are applied in

21 Thank you for your kind attention! Reporter: Anton Kuznietsov, Kharkiv Karazin National University, Ukraine


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