So - there can only be a quicker method if there's a quicker way to move a tower of two discs than the one we described above. Method is only quicker if there is a quicker way to move n-2 discs, and then n-3 and so on, all the way down to 2. So our new method can only be quicker if there is a quicker way to move the top n-1 discs. However this quicker method works, there's one move that you always have to make: you have to move the last disc at the bottom of peg 1 to another peg. So m n =īut is there any other way moving an n-tower one that requires less moves? Let's suppose that there is. How does m n depend on m n-1? Well, moving the n-tower consisted of moving the n-1-tower twice plus one extra move. Write m n for the number of moves we need to do this. The proof just described gives a recipe for moving a tower of height n. This chain goes on forever and so any tower, no matter how tall, can be moved without breaking the So, since we know how to move the tower of height 2 - that was easy - we now have a way of moving a tower of height 3, then of height 4, and then height 5, etc, etc. What this argument tells us is that once we know how to move a tower of height n-1, we can move a tower of height n. Once the top n-1 discs are sitting on peg 2, we move the largest disc to peg 3 and then transfer the n-1-tower from peg 2 to peg 3 in the same way as before - we've moved the whole tower of height n! The fact that the bottom disc keeps sitting on peg 1 all the way through makes no difference, as it is bigger than all the others anyway, so we can place any disc we want on top of it. Now we can move the top n-1 discs to peg 2 without ever placing a larger one onto a smaller one: we simply use the same method as we would if there were only n-1 discs in total. On peg 1, so that pegs 2 and 3 are empty. Now imagine that we have a tower of height n and that we know how to solve the game for a tower of height n-1. We've already seen that it's possible to move a tower of height one, two or three. Our first task is to prove that it is possible to move a tower of any height (that is one made up of any number of discs) from one peg to another without breaking the rules. With n discs, in other words that there is no quicker way? The solution Can you express m n in terms of m n-1? Can you show that m n is in fact the minimum number of moves needed to solve the puzzle Write m n for the number of moves you need to solve the puzzle with n discs according to this recipe. Your proof now gives you a recipe for solving the puzzle for any number of discs. Now assume that you can move a tower of n-1 discs: does this help you to move a tower of n discs? If yes, then why does this prove that the puzzle can be solved for any number of discs? (A proof that uses this way of thinking is an example of the principle of We've already seen that the puzzle can be solved for n equal to 2 and 3. Hint: Suppose that you have n discs in total. But can you show that it is possible to move a tower consisting of any number of A little thought will show you that moving a tower of three discs is possible also. If you've only got two discs in total, then the puzzle is easy: move the top disc from peg 1 to peg 2, then move the bottom disc from peg 1 to peg 3, and finally move the smaller disc on peg 2 onto the larger disc on peg 3 - done.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |