Tower of Hanoi – Inverted Solution:
If the tower consisted of only 1 disk, the puzzle could be completed in only 1 move. If the tower consisted of 2 disks, the puzzle could be completed in 3 moves. if the tower consisted of 3 disks, the puzzle could be completed in 7 moves. For each of these cases, the number of moves is . Thus for the 5 disk model, the puzzle could be completed in moves.
The Travelling Salesman Solution:
The distances between the various cities, as measured in centimeters on the model, are shown on the table below.
|Olympia||Salem||Boise||Helena||Sacramento||Carson City||Salt Lake City||Cheyenne||Denver||Santa Fe||Phoenix|
|Salt Lake City||60.2||56.8||30.5||39.6||46.9||37.7||35.3||36.2||44.7||46.2|
Selecting Salt Lake City as the starting city, a Mathematica computer program ran for 2 days to arrive at the path that generated the shortest distance of 276.7 cm. That path is
Salt Lake City – Carson City – Sacramento – Salem – Olympia – Boise – Helena – Cheyenne – Denver – Santa Fe – Phoenix
The path from Salt Lake City with the second shortest length of 281.2 cm is
Salt Lake City – Cheyenne – Denver – Santa Fe – Phoenix – Carson City – Sacramento – Salem – Olympia – Boise – Helena
If the starting city is Boise, the shortest path of 276.6 cm is
Boise – Helena – Olympia – Salem – Sacramento – Carson City – Salt Lake City – Cheyenne – Denver – Santa Fe – Phoenix
The path from Boise with the second shortest length of 284.6 cm is
Boise – Helena – Olympia – Salem – Carson City – Sacramento – Salt Lake City – Cheyenne – Denver – Santa Fe – Phoenix
The Which is Fastest Solution:
The How Far Out Solution:
Start by placing the plates directly on top of each other with one edge of the plates in line with the edge of the platform. Slide the top plate as far as possible. Next slide the second plate as far as possible. Do the same for the third, fourth and fifth plates.
Let’s look at this numerically knowing that the plates are 24 cm square. Thus, we can slide the top plate 12 cm (1/2 of its length). The second plate can be slid 6 cm (1/4 of its length). The top plate now has 18 cm beyond the edge of the platform or 3/4 of its length. We can slide the third plate 4 cm (1/6 of its length) and the fourth plate 3 cm (1/8 of its length). The top plate has now been slid 12 + 6 + 4 + 3 = 25 cm and is 1 cm beyond the platform.
By moving these plates as described above we have moved the top plate 1/2 + 1/4 + 1/6 + 1/8 of its length. You may notice that each of the fractions has a common factor of 1/2. If we factor out the 1/2 we are left with = 1/2(1 + 1/2 + 1/3 + 1/4). If this pattern is continued with 8 plates then the maximum distance that can be achieved is 1/2(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8) of its 24 cm length which is about 32.6 cm or 8.6 cm beyond the edge of the platform.
If this pattern is extended, you may notice that we are working with the harmonic series (1 + 1/2 + 1/3 + 1/4 + 1/5 + … + 1/n) which does not converge. Even if we multiply this sum by 1/2 as we have done above, it will still not converge. This means that with the right number of plates, we could extend the top plate to be any distance beyond the end of the platform.