There are $ n \ge 3 $ puddings in a room. If a pudding $ A $ hates a pudding $ B $, then $ B $ hates $ A $ as well. Suppose the following two conditions holds: 1. Given any four puddings, there are two puddings who like each other. 2. For any positive integer $ m $, if there are $ m $ puddings who like each other, then there exists $ 3 $ puddings (from the other $ n-m $ puddings) that hate each other. Find the smallest possible value of $ n $.
Problem
Source: 2019 Taiwan TST Round 2
Tags: combinatorics
01.04.2020 11:09
Is it $n=5$? I might've misunderstood the question but, $n=5$ should satisfy. Consider this, A1 , A2, A3, A4, A5 be the 5 puddings. A1 likes A2, A5 A3,A4,A5 hate each other. Won't this satisfy?
01.04.2020 11:56
$n=5$ doesn't work. @above for your graph condition 1 isn't satisfied for A2,A3,A4,A5 We have $n \leq 13$. Here is an example for $13$: Let $a,b,c,d,e,f,g,h,i,j,k,l,m$ be the vertices. Then the graph $G$ with those vertices and the following set of edges: $ab,ac,ad,ae,ah,ak,be,bf,bg,ch,ci,cj,dk,dl,dm,hk,hl,hm$ $bh,bi,bj,bk,bl,bm,ce,cf,cg,ck,cl,cm,de,df,dg,dh,di,dj,ik,il,im$ $eh,ei,ej,ek,el,em,fh,fi,fj,fk,fl,fm,gh,gi,gj,gk,gl,gm,jk,jl,jm$ satisfies both the conditions. (Here, an edge represents liking)
09.06.2020 21:36
So, is there any $n <13$?
01.08.2020 00:16
@above heptagon works n=6 bad
17.10.2020 23:55
For $n=7$ denote the puddings by $A_1;...;A_7$ where an edge represent "like" Mod $7$ join $A_i$ with $A_{i+3}$ and $A_{i+4}$
24.08.2022 12:41
The answer is 7. To prove for n<7 is that we use every degree less than 4