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Problematics | Crossing the river
Chart out the travel route of three gangsters, first by boat with their sons, then alone on the road
A very old puzzle has been passed down from generation to generation in India, and many of us may have heard it first-hand rather than read it in a book. It involves a boatman who has to ferry across a tiger, a goat and a bundle of leaves, but his boat can carry only one of these at a time, and he can never leave the goat unattended with the tiger, and the leaves unattended with the goat. If you know the puzzle, you will also know how the boatman manages. Anyway, here is a variation.
#Puzzle 196.1
Three gangsters accompanied by their sons, all six of them expert rowers but none swimmers, want to cross a river by boat. The boat can accommodate only two of them at a time. Although the men are collaborators in many crimes, they are also capable of kidnapping one another’s sons. As such, none of the gangsters would want to leave his son with one or both of the other men. There is no fear in leaving two of the boys together alone, or if both their fathers are present. On the other hand, it is unsafe to leave one father with two boys, or two fathers with three boys, because the absent father’s son might be kidnapped. In short, if a boy is with one or more men, he is safe only if one of those men is his father. There can be times when a solitary man or a solitary boy crosses from either bank to the other.
Can you chart out the entire travel schedule? A number of trips will be required both ways, so it will help if you number them.
#Puzzle 196.2
When they are not with their sons, the three gangsters can be mutually cooperative. After committing a crime together, they need to get back to headquarters which is 52 km away. There is only one vehicle, a scooter stolen by Mr X, which travels at a fixed 20 km per hour. Mr Y’s walking speed is 5 km/h, and Mr Z’s is 4 km/hr. They work out a deal: Mr X will carry one of his cronies in the rear seat while the other walks. After travelling some distance, Mr X will drop his passenger who will walk the rest of the way. Mr X himself will return towards his other colleague, pick him up, and go back towards headquarters. They time it in such a way that all three of them reach their destination at the same time.
According to Henry Dudeney, my source for the original version of this puzzle, no one stops until the journey is complete, and no time is lost when dropping or picking up a passenger.
How do they manage it?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 195.1
Hi Kabir,
If all chips have their light faces up, the sum of the numbers shown = 0 + 1 + 2 + 3 + 4 + 5 = 15. Every chip with bold face up adds 5 to the sum, that is, if there are n chips with bold face up, the sum is 15 + 5n.
In step 1, the spectator flips 3 chips and then in step 2, the chips that were initially bold face up are flipped. The combined effect of these two steps is as follows.
Chips initially light face up and not flipped in step 1 remain light face up.
Chips initially light face up and flipped in step 1 become bold face up.
Chips initially bold face up and not flipped in step 1 become light face up.
Chips initially bold face up and flipped in step 1 become bold face up again.
Therefore, after step 2, three chips (those chosen to be flipped by the spectator in step 1) are bold face up and the other three are light face up. The sum of the numbers showing at this point is 15 + 5x3 = 30.
In step 3, three chips are flipped and covered while the other three remain uncovered. Suppose B is the number of bold face chips among those that were not flipped and are visible. It implies that 3 – B boldface chips and B lightface chips were flipped in step 3. Flipping a lightface chip adds 5 to the sum and flipping a boldface chip reduces the sum by 5. Therefore, the sum of the numbers on the six chips (three of them visible and three covered) should now be 30 + (5B) – 5(3 – B) = 15 + 10B. If S is the sum of the numbers on the chips visible, then the sum of the numbers on the covered chips is 15 + 10B – S, which can be easily calculated knowing B and S.
— Professor Anshul Kumar, New Delhi
Apart from Professor Anshul Kumar, only Vinod Mahajan has observed that, after the second step, the only three coins showing boldface up will be the same three coins flipped by the spectator in the first step. This is indirectly significant to the solution; the direct significance comes from the fact that three chips are boldface up and three are lightface up after two steps. After the third step, the covered and uncovered lots will have an equal number of chips that are boldface up (and therefore an equal number that are lightface up). Dr Sunita Gupta and Yadvendra Somra have shown this and derived the equation from there. A few other solutions have been a little longwinded, and a few others a little sketchy.
#Puzzle 195.2
Dr Sunita Gupta has sent a list of over 50 words containing the letters ABCDEF, and it is not possible to list them all. After pooling all the words sent by various readers, the commonest ones include BOLDFACE (very relevant to #195.1), BACKFIRED, BIFURCATED, FABRICATED, BACKFIELD and FORBIDDANCE.
Solved both puzzles: Professor Anshul Kumar (Delhi), Vinod Mahajan (Delhi), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Sabornee Jana (Mumbai), Shri Ram Aggarwal (Delhi), Ajay Ashok (Delhi), Shishir Gupta (Indore)
Solved Puzzle #195.2: Kanwarjit Singh (Chief Commissioner of Income-tax, retired), YK Munjal (Delhi)
