A river crossing problem!
7 families with a mother, a father and a daughter each, come to a river they need to cross. There is only one boat for two persons. Only fathers know how to sail the boat. A daughter can be alone only with one of her parents. How can these 7 families cross the river under these conditions? How many times will the river be crossed, less or more than 30?Parent tip
Explore different combinations and have a good time drawing, sketching, writing the solution. You can even make a paper models for each member of the story and solve the problem through play.
Let us denote the members of each of the family using letters F, M and D for father, mother and daughter respectively and number from 1 to 7 to indicate the family. So, M4 will be the mother from 4th family. Since fathers are only allowed to sail the boat, and daughters should be with one of their parents, here is one possible way for crossing the river:
At this moment all mothers with their daughters are on the other side of the river, and one possible way for fathers to cross the river is the following:
This is also an optimal plan for crossing the river under given conditions. The river is crossed 39 times.
This problem was inspired by Alexander Shapovalov's river crossing problems.
- F1 and M2 cross, M2 stays to the other side, F1 returns,
- F2 and D2 cross, D2 stays to the other side, F2 returns,
- F2 and M3 cross, M3 stays to the other side, F2 returns,
- F3 and D3 cross, D3 stays to the other side, F3 returns,
- F3 and M4 cross, M4 stays to the other side, F3 returns,
- F4 and D4 cross, D4 stays to the other side, F4 returns,
- F4 and M5 cross, M5 stays to the other side, F4 returns,
- F5 and D5 cross, D5 stays to the other side, F5 returns,
- F5 and M6 cross, M6 stays to the other side, F5 returns,
- F6 and D6 cross, D6 stays to the other side, F6 returns,
- F6 and M7 cross, M7 stays to the other side, F6 returns,
- F7 and D7 cross, D7 stays to the other side, F7 returns,
- F7 and M1 cross, M1 stays to the other side, F7 returns,
- F1 and D1 cross, D1 stays to the other side, F1 returns.
At this moment all mothers with their daughters are on the other side of the river, and one possible way for fathers to cross the river is the following:
- F1 and F2 cross, F2 stays to the other side, F1 returns,
- F1 and F3 cross, F3 stays to the other side, F1 returns,
- F1 and F4 cross, F4 stays to the other side, F1 returns,
- F1 and F5 cross, F5 stays to the other side, F1 returns,
- F1 and F6 cross, F6 stays to the other side, F1 returns,
- F1 and F7 cross, F1 and F7 stay to the other side.
This is also an optimal plan for crossing the river under given conditions. The river is crossed 39 times.
This problem was inspired by Alexander Shapovalov's river crossing problems.
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