This article explains backup scenarios when using network storage. We review the so-called “3-2-1 strategy” and summarize the methods you can use to send NAS data to the cloud

2 Comments

Anyway, I'm coming to 127. I figure it as follows:

For a 1 disk problem: 1 disk move.

For a 2 disk problem:

Move the smallest disk to an empty

Move the bottom to another empty

Move the smallest onto the other

3 moves

For a 3 disk problem:

Move the top two to an empty:3 moves

Move the bottom to an empty

Move the two onto the bottom:3 moves

7 moves

So the number of moves for n disks looks like the number of moves for a stack one less times 2, plus one:

moves for stack of n=2*(moves for stack of n-1)+1

So we have:

1 - 1

2 - 3

3 - 7

4 - 15

5 - 31

6 - 63

7 - 127

That's almost the same you have. Are we defining the problem the same? Are we counting exactly the same thing? What do you think?

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Title | # Comments | Views | Activity |
---|---|---|---|

Picking random number | 8 | 178 | |

C++ check and remove last word from a string | 5 | 212 | |

Raspberry Pi 3 to send text message | 9 | 147 | |

NEED HELP WITH VISUAL STUDIO 2017 (beginner) | 6 | 65 |

Join the community of 500,000 technology professionals and ask your questions.