The five members of the Family and their five dogs (each family member owned one
ID: 3122136 • Letter: T
Question
The five members of the Family and their five dogs (each family member owned one of the dogs) were hiking when they encountered a river to cross. They rented a boat that could hold three living things: people or dogs. Unfortunately, the dogs were temperamental. Each was comfortable only with its owner and could not be near another person, not even momentarily, unless its owner was present. The dogs could be with other dogs, however, without their owners. The crossing would have been impossible except that Lisa’s dog had attended a first-rate obedience school and knew how to operate the boat. No other dogs were that well educated.
Please use WORK BACKWARDS method to solve this problem and do not copy paste from any web site such as https://sg.answers.yahoo.com/question/index?qid=20090909060712AAAaMgR
https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.495636.html
https://answers.yahoo.com/question/index?qid=20070411160559AAPQ7z2
Thanks.
Explanation / Answer
We must split the dogs from their owner:
Dog 1 can drive the boat:
Dog 1 can take dogs 2 & 3 across, and dog 1 can come back.
We can either take person 2 and 3 or 2 dogs.
case1:
Person 2 and 3 cross.Person 2 and dog 2 must come back.
Case2:
Dog 1 takes dog4 across and comes back. Now we have splits with 3 dogs (2,3,4) across. Person 2,3 ,4can cross making all dog happy. One pair must come back.
It beats case1 and me follow case2.
AFTER:
Persons/dogs 3 and4 can across.
case1: Bring dog 1 back.
Person 1-5 can across and dog 4 across. Dog 1,2,3 cross. Dog 1 returns.
Person 1-5 are across. Dogs2-4 are across.last 2 dogs cross.
case 2: Bring dog 1 and 4 back.
Persons 1-5 are across. Dogs 1-2-3 cross. Dog 1 returns.
Persons 1-5 , dogs 2,3 are cross. Last three dogs are cross.
Can follow both cases, as both cases yield same number of trips