FreeCell long runs: how each empty column doubles your move

You have a run of six cards in FreeCell and the right card to put it on, yet the game will not move it. One move later, after a single card has gone somewhere else, the whole run moves at once. Nothing about the run changed; the space around it did.

The rule behind it

Officially, FreeCell moves one card at a time. Card Lounge moves a whole run in one action whenever there would be enough free cells and empty columns to move it card by card. The limit is:

capacity = (empty free cells + 1) × 2^(empty columns)

An empty column that the run is moving into does not count towards that power.

Two things follow:

• Free cells add. Each empty free cell raises the first factor by one.
• Empty columns multiply. Each empty column, other than the destination, doubles the whole result.

A run moved this way counts as a single move in your statistics, however many cards it contains.

The ladder of capacities

Here is the limit for a run moving onto a card in another column. Each line is a number of empty free cells, and the four figures are for 0, 1, 2 and 3 empty columns.

• 0 free cells: 1, 2, 4, 8
• 1 free cell: 2, 4, 8, 16
• 2 free cells: 3, 6, 12, 24
• 3 free cells: 4, 8, 16, 32
• 4 free cells: 5, 10, 20, 40

Read along a line and each step doubles. Read down and each step adds the same amount: one card with no empty column, two with one, four with two. A run built down in alternating colours can never be longer than 13 cards, from King to Ace, so figures above 13 are room you will never need.

When the destination is an empty column

A run moving into an empty column uses that column as its target, so the column cannot also serve as storage. The figure drops one step to the left on the ladder:

• 2 empty free cells and no other empty column: (2 + 1) × 1 = 3 cards.
• 2 empty free cells and one other empty column: (2 + 1) × 2 = 6 cards.
• 4 empty free cells and no other empty column: (4 + 1) × 1 = 5 cards.

So with two free cells and one empty column you can move six cards onto a card but only three into the empty column itself. If the run you want there is too long, move just the top part: pick up the card where the part should start, and it travels with everything above it. With a limit of 3, the run 10♦ 9♣ 8♥ 7♠ 6♥ 5♣ can go into an empty column only as 7♠ 6♥ 5♣.

Worked example: empty a column first

Two of your four free cells are taken, by the K♣ and the 3♠. The columns that matter, top card last:

Col 1:  ... 8♠ 4♦ 10♦ 9♣ 8♥ 7♠ 6♥ 5♣
Col 2:  ... 3♥ J♠
Col 3:  9♥
Col 4:  ... J♦ 10♠

You want the six-card run 10♦ 9♣ 8♥ 7♠ 6♥ 5♣ on the J♠ in column 2. The colours fit, but with two empty free cells and no empty column the limit is (2 + 1) × 1 = 3, so the game will not move six.

1. Move the 9♥ from column 3 onto the 10♠ in column 4, a red 9 on a black 10. Column 3 is now empty.
2. The limit is now (2 + 1) × 2 = 6. Move the whole run onto the J♠ in one action.

Column 1 now ends with the 4♦, and column 3 is still empty, because the game only borrows the space for the move. Had you wanted the six cards in column 3 itself, the limit there would have been 3.

Spending a free cell to empty a column

If the last card in a column has no tableau home, a free cell can still pay. With three empty free cells and no empty column the limit is 4. Park the single card of a one-card column in a free cell and you have two empty free cells and one empty column: (2 + 1) × 2 = 6.

The same trade with all four free cells empty raises the limit from 5 to (3 + 1) × 2 = 8. That trade pays, as long as the parked card has a way out later. For the wider cost of parking cards, see FreeCell cell economy.

Try it

Work out each limit.

1. One empty free cell and two empty columns, moving onto a card.
2. The same, but moving into one of the two empty columns.
3. Three empty free cells and no empty column; then you empty a one-card column by moving its card onto another column. What is the limit now, onto a card?
4. All four free cells full and three empty columns, moving onto a card.

Answers: (1 + 1) × 4 = 8. Then (1 + 1) × 2 = 4, because the destination does not count. Then (3 + 1) × 2 = 8, double the 4 you had before. Finally (0 + 1) × 8 = 8: even with every free cell full, three empty columns move eight cards.

Try the numbers on a real deal in FreeCell, and look up terms such as free cell and run in the glossary.