Freecell: how many can you move?

zudensternen

Here's a little exercise I've been wanting to do for a while. Say you have a column of consecutive cards you want to move to another (non-empty) column. How many you can move in one click depends on how many open free cells and/or empty columns there are. The attached shows what those numbers are.

CattM

I'm not trying to compete just for me..I can't see well so I'm slow...how do you get back on?/

cellborn

You can work out a formula... its probably already published out there somewhere

The max will be less if you want to move to an empty column

Also max could be more if you can build on some of the other non-empty columns as temporary storage but that's just by luck

the 3rd example, 1 free and 1 empty, is the simplest example of what i call a split move. You move 2, then 2, then 2. that's 6 single card moves to accomplish moving the sequence of 4

jim

The max will be less if you want to move to an empty column

Yep!

Also max could be more if you can build on some of the other non-empty columns as temporary storage but that's just by luck

Yes, but we don't support doing this automatically. It didn't seem worth it, and we try to stick to only doing obvious moves automatically.

the 3rd example, 1 free and 1 empty, is the simplest example of what i call a split move. You move 2, then 2, then 2. that's 6 single card moves to accomplish moving the sequence of 4

Yep and we charge you for all 6 moves from a single drag. We should really keep track of "actions" (you did 1 action here) and "moves" (which resulted in 6 moves) separately, but alas, not yet.

@zudensternen, I think you might reconsider your 2-empty 0-free case. We say you can move 4, not 3. We agree with your n-free 1-empty results.

You can work out a formula

Yep and @zudensternen should be able to see it for at least the n-free 1-empty cases. The 0-empty n-free cases are slightly harder. Then you can combine them!

its probably already published out there somewhere

For sure. In particular the n-empty 0-free case is equivalent to The Tower of Hanoi. Do try solving it yourself before looking!