Discrete Mathematics Reporter: Anton Kuznietsov, Kharkiv Karazin National University, Ukraine In Problems
Dec 16, 2015
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