Discrete Mathematics

Reporter: Anton Kuznietsov, Kharkiv Karazin National University, Ukraine

In Problems

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

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

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

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.

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

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

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)

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.

4. Queens Dispose eight queens

on the chess-board so, that the queens don't threaten each other.

Find all variants of such arrangement.

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

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).

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

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, …

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

1, 4, 9, 25, 64, 169, 441, …

6. Pavement2 x N

21NF

N = 1, 2, 3, …

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

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.

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, …

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

Thank you for your kind attention!

Reporter: Anton Kuznietsov, Kharkiv Karazin National University, Ukraine