Päevaülesanne

Juhuslik

Harjutuskomplekt

Balti Tee 1993 · Ülesanne 14

Kombinatoorika

A square is divided into 16 equal squares, obtaining the set of 25 different vertices. What is the least number of vertices one must remove from this set, so that no 4 points of the remaining set are the vertices of any square with sides parallel to the sides of the initial square?

Muuda valikut

Kui oled valmis

Ülevaatematerjal muutub kättesaadavaks järgmise päevaülesannete komplektiga.

Ülevaade

Teemad

Värvimised ja konfiguratsioonid · Dirichlet’ printsiip ja ekstremaalargumendid

Lahendused

Lahendus

Solution:

The example in Figure 3a demonstrates that it suffices to remove 8 vertices to "destroy" all squares. Assume now that we have managed to do that by removing only 6 vertices. Denote the horizontal and vertical lines by A,B,…,EA, B, \ldots, E and 1,2,…,51,2, \ldots, 5 respectively. Obviously, one of the removed vertices must be a vertex of the big square - let this be vertex A1A 1. Then, in order to "destroy" all the squares shown in Figure 3 b−e3 \mathrm{~b}-\mathrm{e} we have to remove vertices B2,C3,D4,D2B 2, C 3, D 4, D 2 and B4B 4. Thus we have removed 6 vertices without having any choice but a square shown in Figure 3f3 \mathrm{f} is still left intact.

Diagram for the mathnet 00xp 1 of bw-1993-14. a\mathrm{a} Diagram for the mathnet 00xp 1 of bw-1993-14. b\mathrm{b} Diagram for the mathnet 00xp 1 of bw-1993-14. c Diagram for the mathnet 00xp 1 of bw-1993-14. d Diagram for the mathnet 00xp 1 of bw-1993-14. e Diagram for the mathnet 00xp 1 of bw-1993-14. f Figure 3

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