solve the puzzle how many liars?

Question

Each boy in a group of $20$ boys either always tells thuth or always tells a lie. These boys are sitting around a table. Each boy says that his neighbours are liars. Prove that at least $7$ out of $20$ must be truth tellers.

score 5 · Answer 1 · answered Mar 10 '14 at 08:25

5

If there are $\le 6$ truth tellers, then there exist $3$ liars sitting together by pigeon-hole principle.

Now the one in the middle of these three is telling the truth when he/she says "my neighbours are liars", a contradiction!

answered Mar 10 '14 at 08:25

Fei Gao

406

Would you mind explaining exactly what are the pigeons and what are the pigeonholes? – Marc van Leeuwen Mar 10 '14 at 08:26
@MarcvanLeeuwen Indeed $LLTLLTLLTLLTLLTLLTLL$ ... – Mark Bennet Mar 10 '14 at 08:31
@MarcvanLeeuwen pigeon: liars, which is 14; hole: sits between two consecutive truth tellers, which is 6; So there exist two consecutive truth tellers which have at least $\lceil 14/6 \rceil=3$ liars sitting between them. – Fei Gao Mar 10 '14 at 08:44
@FeiGao: Thanks, I understand now. The pigeonholes are the gaps between two truth tellers, of which there are at most $6$ (since the table is round) – Marc van Leeuwen Mar 12 '14 at 14:45
it is possible that k in proof given by me at last is equal to 7. Placing 7 truthtellers around the table, there would be 7 gaps between them. We put one liar in one gap and two liars in each other gaps. – Murtuza Vadharia Mar 13 '14 at 09:41

score 3 · Accepted Answer · edited Mar 14 '14 at 08:30

A liar tells that both his neighbours are liars. Therefore at least 1 neighbour of each liar is truthful. Therefore no three successive boys can be liars.

Each truth teller's both neighbour are liars, so no two successive boys are truthful.

Suppose there are $k$ truth tellers. Around the table, between pairs of successive truth tellers there are $k$ gaps. In each gap, there must be at least one liar (otherwise, the successive boys would be truthful) and at most two liars (otherwise, there would be $3$ successive liars). Therefore the number of liars is at least $k$ and at most $2k$. So total number of boys is at least $k + k =2k$ and at most $k+2k=3k$.

If $k\leq 6$, then $3k \leq 18$ but we have $20$ boys. Therefore $k$ is greater than or equal to $7$. Hence proved.

Marc van Leeuwen · Answer 3 · 2014-03-12T14:43:26.197

2

No three consecutive boys can be liars, or else the middle one would not say both his neighbours are liars. So every boy either tells the truth or has a neighbour that tells the truth (since the table is round, everybody has two neighbours). If there were at most $6$ truth tellers, that would account at most for those $6$, plus their $12$ neighbours; their might be overlap, but in any case this cannot cover all $20$ boys, contradiction

edited Mar 12 '14 at 14:43

answered Mar 10 '14 at 08:24

Marc van Leeuwen

115,048

1

its not a contradiction – Murtuza Vadharia Mar 12 '14 at 14:21
@MurtuzaVadharia: If there are at most $6$ truth tellers, at most $18$ people are either a truth teller or next to a truth teller. This contradicts the conclusion above there every one of the $20$ boys tells the truth or has a neighbour that tells the truth. – Marc van Leeuwen Mar 12 '14 at 14:43
1

it is possible that k in proof given by me at last is equal to 7. Placing 7 truthtellers around the table, there would be 7 gaps between them. We put one liar in one gap and two liars in each other gaps. – Murtuza Vadharia Mar 13 '14 at 09:43

solve the puzzle how many liars?

3 Answers3

Linked