Hard | Strategy |

N undercover agents have been found in don's lair. Less than half of them are terrorists and the rest are anti-terrorists. The nature of their job is so secret that there is no proof what so ever to testify who is who. Although each of them knows who was actual terrorist and who was anti because they worked in teams. A query consists of asking person i if person j is Anti. Anti will always speak truth but a terrorist may lie to confuse you. The goal is to find out one anti in fewest queries.

Hint

It can be done in less than N queries. End of a chain is testified by one anti somewhere in the middle.

Solution

Please note that this question cannot be solved with any algorithm if terrorists are more than or equal to anti, because terrorists can plan to always lie.

Solution by jadu, on CSE Blog, taking only N-1 queries

We will try to find a chain of persons (i1,i2,i3....im) such that each ij is queried about i(j+1) and answer is correct. Note that if such chain contains an anti then last person in the chain would be anti. So query person 1 about 2. If answer is yes query person 2 about 3. continue till answer is correct. suppose at some point person i when queried about person j says wrong, in that case remove person i and j from the chain, and continue query process by querying predecessor of i about successor of j. Here note that when we remove person i and j from the chain at-least one of them must be faulty. here for each person except the first one, we are querying once. Hence N-1 comparison in worst case. Think about the best case using this algorithm.

Solution by jadu, on CSE Blog, taking only N-1 queries

We will try to find a chain of persons (i1,i2,i3....im) such that each ij is queried about i(j+1) and answer is correct. Note that if such chain contains an anti then last person in the chain would be anti. So query person 1 about 2. If answer is yes query person 2 about 3. continue till answer is correct. suppose at some point person i when queried about person j says wrong, in that case remove person i and j from the chain, and continue query process by querying predecessor of i about successor of j. Here note that when we remove person i and j from the chain at-least one of them must be faulty. here for each person except the first one, we are querying once. Hence N-1 comparison in worst case. Think about the best case using this algorithm.

Source: Toad Puzzles

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