Question 10
Captain Crook and his crew divide 10 gold bricks. Each visible chest contains at most one brick, and chains connecting the chests form a tree. Captain and crew alternately take a chest, Captain first. Whenever possible, the chosen chest must be adjacent to a previously taken chest; otherwise any chest may be taken. Captain may therefore start anywhere. The source diagrams illustrate this rule.
Can the crew arrange the tree so that Captain obtains at most one brick? If not, how can they minimize his possible gain? Explain a complete strategy.
In the first illustrated chain, three gold chests alternate with two empty ones; starting at the left gold chest gives Captain all three bricks. In the second illustrated tree, Captain can guarantee two bricks by starting at the leftmost or middle gold chest. In the third illustrated tree, Captain can get only one brick. The original illustrations below show these three arrangements.
Can the crew arrange the tree so that Captain obtains at most one brick? If not, how can they minimize his possible gain? Explain a complete strategy.
In the first illustrated chain, three gold chests alternate with two empty ones; starting at the left gold chest gives Captain all three bricks. In the second illustrated tree, Captain can guarantee two bricks by starting at the leftmost or middle gold chest. In the third illustrated tree, Captain can get only one brick. The original illustrations below show these three arrangements.