IV. Examples and Conundrums

In all the examples below, the tableau will be given here in the text as best as can be done within the limits of the character set. For those who want to see the positions more graphically via the program, each example also starts with a line of apparent garbage that, if selected, can be read using the [File: Resume from Selection] command to display the position. (The line is quite long and may wrap around onto more than one line when you display it; you have to select it all to restore the position.)

Some of these examples are extremely complex. Novice players may wish to step through just the first example, to get an idea of how to play, and save the other examples for later. The final example is a deck that is especially easy to win with (unless you're trying to win with all eight suits still in the tableau), so you might try that one to boost your confidence if you're having a lot of trouble getting anywhere.


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Here, to start you off, is an example of the beginning of a game. We'll step through it and look at the rationale behind the recommended moves. Here's the initial tableau:

 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     --     --     --     --     --     --
 --    10d     As     --     3h     9s     --     Jh     Qh     --
 6d                   4s                   3c                   7c

The two primary rules of thumb to bear in mind throughout the game, and particularly at the start, are (1) try to get a space, and (2) keep your options open. The first rule should be fairly clear; the second leads to a few common strategic decisions. First, given the choice, make a "natural" move instead of an "unnatural" one, where a natural move is one that brings together two cards of the same suit. This keeps our options open by allowing us to move the newly combined cards as a unit should we turn up an appropriate card. Second, given the choice, move a card (or pile) that has more than one place it can go. This keeps our options open by allowing us to move it to the other place if for some reason we want to dig into the pile sitting in the first location. Third, work from the top down. Thus we move a 9 onto a 10 before moving an 8 onto the 9 (unless the latter move is natural while the former is not), since once we move an unnatural 8 onto the 9 we won't be able to move the 9. Now, with these ideas in mind, let's look at the play of the above tableau.

Our highest-ranking move is Jack onto Queen, and it's also our only natural move, so it wins for sure. We move the Jh from column 8 to column 9, and in this particular game we chance to turn up a 6s in column 8. Now we have no natural moves. We could try for the space by moving the 6s to column 10, but that move isn't going to go away, so instead we go from the top down by moving the 10d from 2 to 9. This time we turn up a 4c. No hesitation about this one! We move the 3c from 7 to 2. (Note that we still have the 4s onto which we can, eventually, move the 3h, so we're not giving up our option of digging into pile 5. But even if we didn't have the other 4, making the natural move would be the better play.) In column 7 the card turned up is a 2c, which we promptly move to column 2, turning up a 10h. The tableau is now:

 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     --     --     --     --     --     --
 --     4c     --     --     --     --    10h     6s     --     --
 --     3c     As     --     3h     9s            Qh     --     7c
 6d     2c            4s                                 Jh    10d
                                                               

Having once again run out of natural moves, we revert to working from the top down, and move the 9s from 6 to 9. This also follows the rule of moving a pile that has more than one place to go; if we find ourselves interested in digging through column 9 we can move the 9s to column 7 instead. But for now, since column 7 looks like a more likely place to dig, we'll bury column 9 a bit more. In column 6 we turn up a Kc. Since we have no place to move the 10d from column 9, we are unable to get pile 9 moved onto the newly revealed King. Them's the breaks.

Continuing from the top down, we decide it's time to move a 6 onto the 7c. Which 6 should we move? Neither is natural, but the one in column 8 looks like a better one to move since we're only 3 cards away from getting a space in that column. So we move the 6s from 8 to 10 and turn up a 6c. We're getting low on things to do now; we can move the 3h or the As. Going by the top-down rule, we move the 3h from 5 to 4, turning up a 2h, which we move onto the 3h (now in column 4). This time we turn up a 9d:

 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     9d     --     --     6c     --     --
 --     4c     --     --             Kc    10h           --     --
 --     3c     As     --                          Qh      --
 6d     2c            4s                     Jh     7c
              3h                    10d     6s
              2h                     9s

We could now move the 9d from 5 to 7, but instead we choose to move the As from column 3, since there are two places to put it. Column 4 is already unnatural, so we'll move it there. The card turned up is the other As. We could move this Ace onto the other deuce, but this would lose us our option of moving the first Ace there should we want to dig into column 4, so we'll let the top-down rule take precedence and move the 9d. But let's not be hasty! Instead of moving the 9d from 5 to 7, we'll move the 9s from 9 to 7 and then move the 9d from 5 to 9; this puts the 9d with a 10d, which it can't hurt to do. This time we turn up a Qh. Since we're so close to a space now, we keep going by moving the Qh from 5 to 6, turning up a 10d:

 --     --     --     --    10d     --     --     --     --     --
 --     --     --     --         --     --     --     --     --
 --     --     --     --         --     --     6c     --     --
 --     4c     As     --         Kc    10h         --     --
 --     3c         --         Qh     9s         Qh     --
 6d     2c         4s                     Jh     7c
             3h                    10d     6s
             2h                     9d
             As

Only one move left to try: we move the As from 3 to 2, turning up a 7h. Once again, we shuffle things around a bit so keep as many piles natural as possible; we move the 6s from 10 to 3 and the 6c from 8 to 10, turning up a 5d. We move the 5d from 8 to 1 (natural) and turn up a 3s:

 --     --     --     --    10d     --     --     3s     --     --
 --     --     --     --         --     --         --     --
 --     --     7h     --         --     --         --     --
 --     4c     6s     --         Kc    10h         --     --
 --     3c         --         Qh     9s         Qh     --
 6d     2c         4s                     Jh     7c
 5d     As         3h                    10d     6c
             2h                     9d
             As

We have no more moves (aside from useless maneuvers such as moving the 9d from 9 to 5), so it's now time to deal a new round. We never did get a space, but we got two piles down to a single card each, so we are quite likely to get a space soon after the new deal. This game is going somewhat better than average and will very likely be won with proper play. If you actually do get a space in the first round, you're doing particularly well.


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Now, for your first "Spider problem", here is a relatively simple position. In the tableau shown below, what should you do? First off, what are your options? On what should you base your choice? (After the tableau is the "solution", so don't read further until you're ready!)

10h (sp) -- Ad -- Qc -- 3s Qh -- -- 7d -- Jh -- 2s -- -- 6d Kh 10d -- -- 8s 5d Qc 6c -- 7d 7s Qs Jc 4c Qd 6d 6s Js 10c 3d Jd 5d 10d 9c 2d 10h 4d 8c Js 3d 7c 10s 2c 6c 9s As 5c 8h 4c 7h 3c 6h 2c 5h Ac 4s Qd 3h 10c 10s 9c 8c 8s 7h 7s 6h 6s 5h 5s 4h 4s 3h 3s 2h 2s Ah

Solution:

First, the options. There's no way to get through column 5 or 7 to turn up a new card. (This should be pretty obvious; we'll save detailed analyses of this sort of thing for cases where it's not as clear.) Nor does it do us any good to dig into column 4 or 6. We don't have any complete suits showing, so there's no way we can try to put one together. That leaves three fairly simple options: (1) we could move the 8-6s from column 3 into the space, turning up a new card, (2) we could dig through column 10 (moving the Ace onto a deuce, the 2c into the space, 5-3d onto the 6s in column 3, 2c out of the space and back onto the 5-3d, and finally the 7-6d into the space) and turn up a new card there, or (3) we could fill in the space and deal a new round.

It's usually a good idea to turn up more cards when possible rather than bury everything under a new deal, so we'll discount the third option. That leaves us with the choice of which column to dig through, 3 or 10. The two are equally close to becoming new spaces (three face-down cards each), so that's not a consideration here. Let's consider what the face-down card might be that will be revealed. If it's a Jack, 4, or King, we can get back the space (which we'll have lost in the process of getting to the new card). If it's a 9 or 8, we MIGHT get the space back right away; it depends on whether we moved the 8 (from column 3) or the 7 (from column 10) into the space. Looking at the tableau, we see there are five 8's visible, but only three 9's. Thus it's more likely we'll turn up a 9, so we should go for column 3. (Sorry for all this gory detail, but this is after all intended as an introductory example.)

So it looks like the best thing to do is move the 8-6s from column 3 into the space. But wait! Suppose the card turned up isn't a Jack, 4, King, or 9, and furthermore isn't an Ace or 5 (which we would be able to move elsewhere immediately)? Is there anything we can do ahead of time to hedge our bets? Yes! We can move the spade Ace from column 10 to column 5, then use the space to swap the deuces in columns 6 and 10 (move one deuce into the space, move the other deuce to the other column, and move the first deuce out of the space). Now column 10 contains just the 7 through deuce of diamonds, and if we chance to turn up an 8 in column 3 we can move the 7-2d onto it. Note that we have to do this BEFORE we move the 8-6s into the space, since we need the space to swap the deuces. In fact, in the game where this particular tableau arose, the card turned up in column 3 was the diamond 8. The preparations made in column 10 eventually produced not one but TWO spaces! (Play it out using the program and see for yourself.)


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Turn up another face-down card WITHOUT dealing more cards or "using up" the space. (You may, of course, use the space, so long as you are sure you can get it back no matter what the card turned up turns out to be.) Note that there are enough clubs and hearts showing to form complete sets of those suits. Here's the tableau (again, the solution follows the tableau):

 --     --    (sp)     --     Qh     --     --     --     --     --
 --     --         Ks     Jh     --     Jc     --     --     --
 --     --         2s     7h     --     8h     --     --     Kh
 --     --         As         As     7h     Kc     --     Qc
 Jc     Kd         9h         9h         Qc     Kc     Jh
10h     9d         8d         8s         Jd     7s     Js
 9c     8d                 7c         5s     6s     3d
 8s     7d                 6c         4h     Qh     2c
 7d     3h                 5c         3c     Js     Ad
 6h     2h                 4h         2c     8c     8h
 5d     Ah                 3s         Ac
 3d     2d                 2s        10s
 2d                     5c         9s
 Ah                     6d         8c
 9s                     5h         7s
10c                             6s
                             5s
                             4c

Solution:

First, we ascertain that we can't get a second space. The only place where we might be able to do so is column 5, and to move the Q-Jh we need to find a King that doesn't already have a Queen on it. (We'll call this a "free King", for short.) There are three free Kings, but the one in column 9 is useless since we need another free King to get to it, and those in columns 2 and 4 are inaccessible since there are no free 3's. Hence, whatever we do, we have to do it using only the one space.

Next, can we remove a complete set of clubs or hearts? Well, hearts are out, because the only Kh showing is in column 10, and the only 10h is in column 1, and getting to each of them requires that we move a 3 onto a free 4. Since there's only one free 4 (in column 8), we lose. How about clubs? They don't work out, either, but the proof is trickier. The only 9c is in column 1 and getting to it will require our sole free 4. Thus we can't use the Qc in column 10, and must instead use the Qc from column 8. To reach it we need a free 6; we have exactly one free 6, namely in column 9. We CAN get to this 6, without losing the space, by a fairly convoluted sequence of moves. You may want to figure out how it can be done before reading on. . . . Ready? Okay, proceed as follows: 7h from 5 to 10, 10c from 1 to 5, 8c from 9 to 1, Js from 9 to 3 (into the space), 10c from 5 to 3, Jh from 5 to 9, 10c from 3 to 9, Js from 3 to 5, 10c from 9 to 5, Q-Jh from 9 to 3, 7-6s from 9 to 4, and finally Q-Jh from 3 to 9, getting the space back.

Having determined that we can, if desired, obtain a free 6, let's get back to the question of the clubs. The only 7c is in column 6, and getting to it requires a free 6. But we need the free 6 to get to the Qc as well. So we again lose. We are thus reduced to uncovering a card without removing any suits and without getting any more spaces. Which column is it to be? It obviously can't be a column containing a King, since (given that we can't remove any completed suits) the only place a King can go is into the space. And it can't be column 1 or 7, since that would require a free Queen, and there isn't any. So it must be column 6. We can get through that column by first digging through to the free 6 as described earlier, and then playing: 5h from 6 to 4, 6d from 6 to 10, 5c from 6 to 10, 3-2s from 6 to 3, 4h from 6 to 4, 3-2s from 3 to 4, 7-5c from 6 to 1. The tableau now looks like this:

 --     --    (sp)     --     Qh     --     --     --     --     --
 --     --         Ks     Js     --     Jc     --     --     --
 --     --         2s    10c     --     8h     --     --     Kh
 --     --         As         As     7h     Kc     --     Qc
 Jc     Kd         9h         9h         Qc     Kc     Jh
10h     9d         8d         8s         Jd     Qh     Js
 9c     8d         7s                 5s     Jh     3d
 8s     7d         6s                 4h         2c
 7d     3h         5h                 3c         Ad
 6h     2h         4h                 2c         8h
 5d     Ah         3s                 Ac         7h
 3d     2d         2s                10s         6d
 2d                             9s         5c
 Ah                             8c
 9s                             7s
 8c                             6s
 7c                             5s
 6c                             4c
 5c

Once again, it's time to make contingency plans. If we just move the 9h-8s onto the 10c and the As onto the 2s, we could be in trouble if we turn up a King. The lone space won't be sufficient for us to be able to move the stuff out of column 5 onto the King. So we undo some of what we did in the course of getting the free 6: Jh from 9 to 3, 10c from 5 to 3, Js from 5 to 9, 10c from 3 to 9, Jh from 3 to 5. While we're at it, it can't hurt to move the 4c from 8 to 1, and in a moment we'll match the 8s with a 9s, too. We now proceed: 8s from 6 to 3, 9h from 6 to 9, 8-5c from 1 to 9, 8s from 3 to 1, and finally As from 6 to 4. (Once again, preparation pays off; in the game where this took place, the card turned up was indeed a King.)


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Again, complete sets of clubs and hearts are available. Without dealing any more cards or turning up any face-down cards, remove a set of clubs AND a set of hearts (not necessarily in that order). Can you remove them in the other order?

 --     --     8c     Ks     Kh     5c     Kc     --    (sp)     --
 --     --         Qh     Qh             --         --
 --     --         Jc     Jh             --         Kh
 --     --        10c    10c             Kc         Qc
 Jc     Kd         9c     9h             Qc         Jh
10h     9d         8c     8d             Jd         Js
 9s     8d         7c     7d            10h         3d
 8s     7d         6c     6d             9h         2c
 7s     3h         5c     5d             8h         Ad
 6s     2h         4c     4h             7h         8h
 5s     Ah         3c     3d             6h         7h
 4s     2d         2c     2d             5h
 3c     Ah         Ac                 4h
 2s                             3h
 As

Solution:

The clubs look like the better bet, since the Jack through Ace are already assembled and there's a King-Queen in column 8. Let's see what can be done. Since there are no free 9's or 6's, we have to remove the first completed suit without the benefit of any additional spaces. Since we are also short on free 4's, this means we can't use the Qc in column 10. That seems okay; the one in column 8 looks easier to get to anyhow. All we have to do is move the Jd somewhere (along with the 10-3h). There are no free Queens, so the Jack will have to move into the space (or some other Jack must move into the space to free up a Queen). But we can't move the Jd anywhere while the hearts are there, and the only free Jack is in column 10 where we can't get at it. We could move the 10-3h into the space, but then what do we do with the Jack? Looks like the clubs aren't going to work after all.

Let's try the hearts. It looks like we'll have the same problem, since we have to move the 10c from column 5 somewhere else to clear off the K-Jh. The only place we can move the 10c is the space, and to do that we have to do something about the 9h attached to the 10c. Since we don't have any free 10's, what can we do? The idea is to use the space to swap things around such that the sequences of a single suit are where we need them most. We do it as follows: First we get the 4h out of the way by moving 3-2d from 5 to 9, 4h from 5 to 6, and 3-2d from 9 to 6. Then we move 8-5d from 5 to 9, 8-3h from 8 to 5, 8-5d from 9 to 8, 9-3h from 5 to 9, 9-Ac from 4 to 5, 9-3h from 9 to 4, 8-5d from 8 to 9, 8-3h from 4 to 8, 8-5d from 9 to 4 (we certainly have made a mess of all those nice clubs in column 4, haven't we?), 10-Ac from 5 to 9, 10-3h from 8 to 5, 10-Ac from 9 to 8. Now we can move the Ah from 2 to 6, 2d from 2 to 9, and 2-Ah from 2 to 5 to complete the hearts. The 2d comes out of the space and back to column 2, and removing the hearts gives us a second space. With two spaces we have no trouble straightening the clubs back out and completing a set.

Note that, rather than removing the completed set of clubs from column 8, we should pile a Q-Ac into column 7 and remove the suit from there. We can always move the Kc from column 8 into the newly created space in column 7 if we wish, but by getting the space we keep our options open. Note also that, had there been a 10d around, we might have been able to pull the same trick with the Jd in column 8 as we did with the 10c in column 5; since there wasn't, though, we had to go after the hearts first.


C576XV@Ra`MgY>9XFSnF/^j?Pe:O`5J_0EZo@Uj?Pe:O`5J_0EZo@Uj?Pe:O`5J_0EZo@Uj?Pe:O`5J_0EZo@Uj?Pe:O`5J_0EZo@Uj?P 3>0f9NC7I^ H=Zn HUUgKl1YC4I^SDYR PFeM2 B1D[ C:^>LPV;] MAK\1F[lAVk In the tableau below, there are two deals (20 cards) remaining. What do you do?

 Kh     Ks     Kh     Qh     Jc     Qd     Js     9s    10s     6d
 Qh    10h     Qs     Ac    10c     Jh     Kc     9c     Kd     Qs
 Jd     9d     8s     Ac     9h    10h     Qc     Kc     Qd     Js
10s     8d     7h     4s     8h     6s     Jc     9h     Jd
 9s         6h     3h         5s    10c     8h    10d
 8s         5h             4h     9c     7h     9d
 7s         4h             3h     8c     6h     8c
 6c         8d             2h     7c     5d     7s
 5c         7d             As     6c     4s     6s
 4c         6d                 5c         5s
         5d                 4c
         4d                 3c
         3d                 2c
         2d                 3s
         Ad
         7d

Solution:

If you grabbed at the opportunity to remove a set of clubs from column 7, shame on you! Once you do that, you're stuck, and except for a little bit of "naturalising" (swapping cards so as to create longer runs in a single suit) you can do nothing but deal out another round. In fact, it is possible to remove TWO suits, get two spaces, and straighten out almost all the suit changes before running out of steam. But let's start by looking at the various options available to see what our reasoning should be.

Ignoring for the moment whether we can remove any suits, can we get a space? Obviously, we can't get a space in any column containing a King, which leaves columns 4, 5, 6, and 10. Column 5 is hopeless because the only free Queens (columns 3 and 4) cannot be reached without either two free 9's or two free deuces. We have one free deuce (column 7) and both free 9's are trapped under a King. Getting through any of the other columns requires a free King. The King in column 8 cannot be reached without the temporary use of a free 6, and the only such 6 is under a Queen and thus requires another free King to reach. On the other hand, we CAN reach the King in column 2. But in order to do so we need temporary use of a free 10, which means we'll have had to move the 6s from column 6 onto our only free 7. Thus, though we might be able to move the Qs out of column 10, we wouldn't have any place to move the 6d. And, since we lack a free Queen on which to park the Jh from column 6, the free King won't be enough to get through that column. Column 4 we've already identified as being hopeless. Thus we conclude that we can't get a space without removing any suits.

Now what? If we remove the clubs from column 7, all we get is a free Jack, which we have no use for, and which we can't move out of the way. We can't remove a set of spades since we have no 2s showing, and likewise there's no Ah visible. We can't remove diamonds (at least, not first) because we can't get to the Kd without a free 9 on which to park the 8c, and both free 9's are trapped under a Kc. So it looks like the first order of business is to remove a set of clubs from column 8, if possible.

That's a big "if"; it looks rather hopeless. Digging through to the Kc requires temporary use of a free 5 and 6, and permanent use of a free 10. Well, a moment ago we said we might be able to reach the 6d in column 10. Can we in fact do so? Yes, if we're VERY careful! Once we uncover the Ks in column 2, we want to be able to move the Q-Js from column 10, so we can't afford to put the 10h from column 2 on top of them. That means we have to find a 10s to move to column 10, thereby freeing up a different Jack. Such a 10 is in column 1, and we'd better move it BEFORE using up the free 7! Without further ado, here we go: 6-4c from 1 to 3, 10-7s from 1 to 10, 3s from 7 to 8, As from 6 to 7, 4-2h from 6 to 9, 6-5s from 6 to 10 (conveniently the correct suit), 9-8d from 2 to 6, 10h from 2 to 1, Q-5s from 10 to 2, 4-3s from 8 to 2, 5d from 8 to 10, 9-6h from 8 to 1, As from 7 to 9, Q-2c from 7 to 8, 3h from 4 to 3, 4s from 4 to 10, Ac from 4 to 8, and voila! we remove a set of clubs from column 8. Here's the new tableau:

 Kh     Ks     Kh     Qh     Jc     Qd     Js     9s    10s     6d
 Qh     Qs     Qs     Ac    10c     Jh     Kc     9c     Kd     5d
 Jd     Js     8s         9h    10h             Qd     4s
10h    10s     7h         8h     9d             Jd
 9h     9s     6h             8d            10d
 8h     8s     5h                         9d
 7h     7s     4h                         8c
 6h     6s     8d                         7s
     5s     7d                         6s
     4s     6d                         5s
     3s     5d                         4h
         4d                         3h
         3d                         2h
         2d                         As
         Ad
         7d
         6c
         5c
         4c
         3h

Unfortunately, though we've uncovered a free 9, there's no longer anything we can do with it, because we've added some crud to column 9, not to mention column 3. Furthermore, if we're going to form a set of diamonds, we'd best not lose track of the stuff in column 3. So let's go back to the very beginning and try again, this time keeping things a bit more available: 7d from 3 to 5, 6-4c from 1 to 5, 10-7s from 1 to 10, 3s from 7 to 8, As from 6 to 7, 4-2h from 6 to 9, 6-5s from 6 to 10, 9-8d from 2 to 6, 10h from 2 to 1, Q-5s from 10 to 2, 4-3s from 8 to 2, 5d from 8 to 10, 9-6h from 8 to 1, 4-2h from 9 to 10, As from 7 to 10, Q-2c from 7 to 8, 3h from 4 to 5, 4s from 4 to 9, Ac from 4 to 8, and again we are able to remove a set of clubs from column 8. Now, however, the tableau looks like this:

 Kh     Ks     Kh     Qh     Jc     Qd     Js     9s    10s     6d
 Qh     Qs     Qs     Ac    10c     Jh     Kc     9c     Kd     5d
 Jd     Js     8s         9h    10h             Qd     4h
10h    10s     7h         8h     9d             Jd     3h
 9h     9s     6h         7d     8d            10d     2h
 8h     8s     5h         6c                 9d     As
 7h     7s     4h         5c                 8c
 6h     6s     8d         4c                 7s
     5s     7d         3h                 6s
     4s     6d                         5s
     3s     5d                         4s
         4d
         3d
         2d
         Ad

Now we can get out a set of diamonds as follows: 7-4s from 9 to 6, 8c from 9 to 8, 8-Ad from 3 to 9, and remove the diamonds:

 Kh     Ks     Kh     Qh     Jc     Qd     Js     9s    10s     6d
 Qh     Qs     Qs     Ac    10c     Jh     Kc     9c         5d
 Jd     Js     8s         9h    10h         8c         4h
10h    10s     7h         8h     9d                 3h
 9h     9s     6h         7d     8d                 2h
 8h     8s     5h         6c     7s                 As
 7h     7s     4h         5c     6s
 6h     6s             4c     5s
     5s             3h     4s
     4s
     3s

Well, looking better, but it's still not obvious where we can find a space. Based on the "free" cards, there are only two possibilities: either we use the 10s in column 9 to uncover the 9s in column 8 and thus the Qs in column 3 and thus get a space in column 5, or we dig through column 6. Column 5 doesn't work because we need temporary use of a free 7. It may be that, by judicious advance planning, we could have arranged to have a clearer path through this column (e.g., by building column 1's hearts in column 5), but we needn't go back for a third try because we can get through column 6 anyway. We'll need temporary use of an 8, 10, and Queen, and getting the Queen will require permanent use of the 10, so we have to do things in the right order again. Moreover, we have to be careful not to build anything new in column 6 that will be hard to move out later. So we do it this way: 7-4s from 6 to 8, 9-8d from 6 to 9, 7-4s from 8 to 9, 9-6h from 1 to 6, 9-8c from 8 to 1, 7-4h from 3 to 1, 8s from 3 to 8, J-6h from 6 to 3, Q-3s from 2 to 7 (to prepare for later), and Qd from 6 to 2. And here we are:

 Kh     Ks     Kh     Qh     Jc    (sp)     Js     9s    10s     6d
 Qh     Qd     Qs     Ac    10c         Kc     8s     9d     5d
 Jd         Jh         9h         Qs         8d     4h
10h        10h         8h         Js         7s     3h
 9c         9h         7d        10s         6s     2h
 8c         8h         6c         9s         5s     As
 7h         7h         5c         8s         4s
 6h         6h         4c         7s
 5h                 3h         6s
 4h                         5s
                         4s
                         3s

Without detailing the exact moves from here on, the next steps should probably be something like this: Jc-3h from column 5 onto Qd in column 2, getting second space. Swap 9-8d in column 9 with 9-8s in column 8 to form 10-4s in one chunk. Move Kc-3s from column 7 into a space and get the space back by moving 10-4s onto Js. Move Ac from column 4 into a space and get the space back by moving J-6h from column 3 onto Qh and J-4s onto Qs. After a bit more "naturalising", you should have a tableau something like:

 Kh     Ks     Kh    (sp)     9d     Kc     Qs     Ac    (sp)     6c
 Qh     Qs     Qh         8d     Qd     Js             5c
 Jc     Js     Jh         7d     Jd    10s             4c
10c    10s    10h         6d    10h     9s
 9c     9s     9h         5d     9h     8s
 8c     8s     8h         4h     8h     7s
     7s     7h         3h     7h     6s
     6s     6h             6h     5s
     5s     5h                 4s
     4s     4h
     3s     3h
         2h
         As

This is as much straightening out as you can accomplish with the cards available, so it's finally time to deal another round. But first, you have to fill in the spaces (them's the rules!). So how should you fill them in? It's largely a matter of personal preference, but one likely possibility is to move the Q-Jd into one space and the 10-6h into the other. The reason for this is that there's already a Queen in a space, so by creating a free King you have an extra chance at a space early in the next round. Granted that you are almost certain to win at this point, but you might as well maximise your chances anyway!


3^SD>\QBCTSZ`23d9N@7I^SGYN 1Ti>Sd9Mc4IZ NM ZnhZM =1f9NFCfhOcV;\] UH8.

3^SD>\QBCTSZ`23d9N@7I^SGYN 1Ti>Sd9Mc4IZ NM ZnhZM =1f9NFCfhOcV;\] UH8.

Can a set of spades be removed WITHOUT first getting a space or dealing any more cards? If so, how? If not, prove it! Would it make any difference if the Js in column 7 were swapped with the Jd in column 9? (The second "garbage" line above is for this modified tableau.) Finally, given that it can be done, remove a set of hearts (as usual, without dealing any more cards). What is the minimum number of other suits that must be removed in order to do so? Does the order of the two face-down cards matter?

 Kd     Jc     9h     9c     Qh     9d     Qc     Ks     --     Qs
 Qd    10c    10s     8c     Jc     Qh     Kc     Js     --     Kc
 Jd     9c         5s    10h     Jh     Qs     2d     Jd     3c
10d     8c         7d     9h     Ad     Js     Ac     7d
 9d     7c         Kd     8h        10d     7h     6h
 8d     6c             6s         9s         5c
 7h     5c             5s         8s         4h
 6d     4s             4s         7s         3h
 5d     3s             3s         6s         2d
 4d     2s             2c         Qd         Ah
 3d     As             6h                 4c
 2h     As             5d                 3c
 Ah                 4d                 2c
 8h                 9s                 Ac
                                 7c

Solution:

This is a complicated one, so take a deep breath! (If you didn't find it complicated, then perhaps you weren't thorough in your analysis. Unless you (a) decided the spades could not be removed without getting a space, (b) realised that swapping the Jacks affects this, and (c) considered the 9d in column 1 at some point in your proof, your analysis is incomplete.)

First let's consider the problem of putting together a set of spades. We begin by finding all the pieces. The only Ks is in column 8; the only Qs we can possibly get to without a space is in column 7. (Actually, we shouldn't be too hasty; if we could remove a set of clubs without getting a space, we could reach the Qs in column 10. But in moving the Qs we'd create a space, whether we needed it or not; and besides, the only Qc is in column 7 with the other Kc in the way.) In digging to the Ks and Qs we'll reach both Jacks, so they shouldn't be a problem. The 10s is in column 3, and the remaining spades are at various depths in columns 2, 5, and 7. Can we pull all these cards together?

To get to the Ks we need a free 8, a free 3, and a free Queen (even though we may end up using the Js from column 8, we need some place to put it in order to get to the King). The 8 in column 4 is inaccessible unless we can remove a set of diamonds, which in turn is impossible without a space since the 7d in column 4 is inaccessible and likewise for the 7d in column 9 due to the absence of free 5's. But we have a free 8 in column 1 and another in column 5 (if we can reach it), so there's no problem with that. We also have exactly one free 3, and one free Queen. So far so good. Can we reach the Qs in column 7? That requires a free King, which is no problem. It also requires someplace to move the 9-6s and the 10d. This should pose no problem either. Note that, though we need a free 10 and a free Jack for this, we don't "use up" those free cards by moving the 9-6s and 10d, since we uncover another 10 and Jack to become new free cards. However, notice that we're eventually going to have to reach the 5s in column 5, and this will use up the free 10. So we have to dig through column 7 before that. In fact, we have to move the 10d out of column 7 before moving the 9s out of column 5, because once we move the latter we'll have 9's on all the 10's, and the 10d won't be movable. Or will it? If we could put a 9d onto the 10d (freeing up a different 10), we could move the 9s onto the newly freed 10 and still be able to move the 10d. Let's assume for the moment that this is impossible (we'll prove it later, but don't want to digress too far here). To repeat, then, we need to move the 10d out of column 7 before moving the 9s out of column 5. Where does the 10d go? The free Jack in column 9 is inaccessible without a free 5, and the other free Jacks (in columns 6 and 8) each require a free deuce (even though the Jack in column 8 doesn't require us to use up the deuce permanently). The only free deuce is in column 5, and we can't get to it without moving the 9s. So we're stuck!

Now let's follow out that digression and make sure we can't get a 9d onto the 10d. We certainly can't use the 9d in column 6, since that would create a space, which is verboten. In order to reach the 9d in column 1, we'd have to move the 8h. If we put it onto the 9s in column 5, we would then be unable to move that 9s later on (we have only one free 9 available; as we'll see later, we can't get to the one in column 3 without moving the 9s from column 5). If we moved the 8-6s from column 7 onto the 9s in column 5, and then moved the 8h onto the newly freed 9s, we wouldn't be able to move THAT 9s later, so we either wouldn't be able to reach the Qs (if we had left the 9s in column 7 when we put the 8h on it) or else we'd be unable to reach the 10s (if we had moved the 9s there first). So, although we might be able to get the 9d from column 1 onto the 10d in column 7, by the time we did so we'd have made a hopeless mess out of the spades. The conclusion from all this is that it's impossible to remove a set of spades without first getting a space.

Now, what if the Jacks were swapped as described? In that case, we wouldn't need a free Jack on which to park the 10d; we could move the J-10d as a unit. So the plan is to move the Qd out of column 7, followed by the 9-6s and J-10d. Then we can use up the free 10 by moving the 9s out of column 5 and finish bringing together the spades. The complete sequence is: Qd from 7 to 4, 9-6s from 7 to 3, J-10d from 7 to 4, 9s from 5 to 4, 5-4d from 5 to 3, 7h from 8 to 1, 6h from 5 to 1, 5-4d from 3 to 1, 2c from 5 to 10, 5-3s from 5 to 3, Ac from 8 to 10, and now we have to be careful not to move the 2d from column 8 onto the spades in column 3, so instead we move 3s from 3 to 1, 2d from 8 to 1, Js from 8 to 7, Q-Js from 7 to 8, 10-4s from 3 to 8, As from 2 to 1, and 3-As from 2 to 8. Voila!

That was for warm-up; what about removing the set of hearts? The first step is easy: we look around to see where all the hearts are and find that the King and 5 are missing. Hence these must be the two face-down cards. It remains to be seen whether their order is significant.

In the course of discussing the spades, we observed that we cannot remove a set of diamonds or clubs without first getting a space, and we also proved the same thing for the spades. Since we can't get past the 4c in column 9 without a space, it looks like our first order of business is getting one. Columns 1, 4, 7, 8, 9, and 10 are out, for obvious reasons. Column 3 looks like the best bet, but in order to move the 10s we need a free Jack, and that in turn requires a free deuce, and THAT requires that we move the 9s from column 5 onto the 10s. Thus, by the time we manage to move the 10s, we'll no longer have a free 10 on which to put the 9h to get the space. Column 6 is similarly hopeless; in order to move the Ad we need to use up the free 10. Column 2 is out of the question since there's no place to put the 4-As. That leaves column 5.

To get through column 5 we need to use up a 10, two 7's, and a King, and we also need temporary use of a 6, 3, Jack, and Queen. Getting the Jack will be no trouble once we've gotten to the 2c, and getting the 6 just needs another free King, which we can get from either column 8 or column 10. Let's use the one in column 10; the only thing we have to watch out for is that if we wait too long to uncover that King (in particular, if we wait until we need it to put the Qh on to clear the space), we may find the 3c is immovable due to our having moved stuff onto it in the meanwhile. So we have to move the 3c onto the 4d at some early opportunity. Here we go: 9s from 5 to 3, Qd from 7 to 4, 5-4d from 5 to 7, 6h from 5 to 8, 2c from 5 to 10, 6-3s from 5 to 9, 3-2c from 10 to 7, Ad from 6 to 7, 10-8h from 5 to 6, Jc from 5 to 4, and finally Qh from 5 to 10. The tableau now looks like this:

 Kd     Jc     9h     9c    (sp)     9d     Qc     Ks     --     Qs
 Qd    10c    10s     8c         Qh     Kc     Js     --     Kc
 Jd     9c     9s     5s         Jh     Qs     2d     Jd     Qh
10d     8c         7d        10h     Js     Ac     7d
 9d     7c         Kd         9h    10d     7h     6h
 8d     6c         Qd         8h     9s     6h     5c
 7h     5c         Jc             8s         4h
 6d     4s                     7s         3h
 5d     3s                     6s         2d
 4d     2s                     5d         Ah
 3d     As                     4d         4c
 2h     As                     3c         3c
 Ah                         2c         2c
 8h                         Ad         Ac
                                 7c
                                 6s
                                 5s
                                 4s
                                 3s

Where do we go from here? Well, we're trying to minimise the number of suits (other than hearts) removed, so let's see if we can get the hearts out right away. We would need to dig through column 9; to do that we would have to move the 4-Ac into the space (or onto a free 5; we'll come back to this), after which we would have no place to move the 4-3h. If we could get a free 5 without using up the space, we might fare better, but the only free 5 is in column 4, and to get to it we must put the Kd into the space (remember we're assuming we're not going to remove any other suits) and we have no free 10 with which to restore the space via column 4. Nor can we get any more spaces; all columns contain Kings or 9's or Aces, and there are no free 10's or deuces, so digging through any pile would cost us the space, and would get us at most one space in return. Thus we conclude that we must remove another suit before the hearts. Which suit is it to be?

It can't be clubs. To reach the 10c (in column 2) we must move the first As into the space, since there are no free deuces anywhere. Having done so, we have no place to move the 4-As. (We have already noted that getting to the free 5 costs us the space.) On the other hand, we CAN remove either diamonds or spades. (If you thought you HAD to remove the diamonds, you might want to take a moment to study the above tableau and figure out how to remove the spades instead.) Let's look at the diamonds first. Most of them are already in column 1; all we need to dredge up are the 7, 2, and Ace. We'll ignore the diamonds in column 9 (we know we can't reach the 7d there, and the 2d is less accessible than that in column 8), and proceed thusly: 7-6h from 8 to 6, Ac from 8 to 5, Ad from 7 to 8, Ac from 5 to 7, 8h from 1 to 3, 2-Ah from 1 to 9, 2-Ad from 8 to 1, Jc from 4 to 10, K-Qd from 4 to 5, 6-Ad from 1 to 4, 7h from 1 to 3, and 7-Ad from 4 to 1. Removing the diamonds from column 1 would give us this position:

(sp)     Jc     9h     9c     Kd     9d     Qc     Ks     --     Qs
    10c    10s     8c     Qd     Qh     Kc     Js     --     Kc
     9c     9s     5s         Jh     Qs         Jd     Qh
     8c     8h            10h     Js         7d     Jc
     7c     7h             9h    10d         6h
     6c                 8h     9s         5c
     5c                 7h     8s         4h
     4s                 6h     7s         3h
     3s                     6s         2d
     2s                     5d         Ah
     As                     4d         4c
     As                     3c         3c
                         2c         2c
                         Ac         Ac
                                 7c
                                 6s
                                 5s
                                 4s
                                 3s
                                 2h
                                 Ah

Now, before we pursue this any further, let's go back and see how we can remove the spades instead. If we try to do so in the straightforward manner, we run into trouble. Presumably we would uncover the Ks in column 8 by moving the 7-6h onto an 8 and the 2d-Ac onto a 3 (probably swapping the Ac/Ad as we did in the previous paragraph). We would then move the Js out of column 8 and bring in a pile of spades from columns 7 (Q-J, 8-6), 3 (10-9), and 9 (5-3), piling them all onto the King. But then we'd be unable to get to the 2s in column 2. (Once we moved the first As into the space, we'd be unable to swap the 2-As with the 2-Ad (or whatever) blocking off the 3s in column 8.) The way out of this bind is to wait until the last minute to move anything onto the 3s, such that when we do it's the 2-As, and thus we won't need the space afterward. Here's how we can do it: 7-6h from 8 to 6, Ac from 8 to 5, Ad from 7 to 8, Ac from 5 to 7, 3-Ac from 7 to 5, 5-4d from 7 to 6, 3-Ac from 5 to 6, 8-6s from 7 to 3, 9s from 7 to 5, 10d from 7 to 4, 9s from 5 to 4, 10-6s from 3 to 7, 5-3s from 9 to 7. Now we're ready to go: As from 2 to 5, 2-As from 2 to 7, 2-Ad from 8 to 2, Js from 8 to 10, and Q-As from 7 to 8. Removing the suit gives this tableau:

 Kd     Jc     9h     9c     As     9d     Qc    (sp)     --     Qs
 Qd    10c         8c         Qh     Kc         --     Kc
 Jd     9c         5s         Jh             Jd     Qh
10d     8c         7d        10h             7d     Js
 9d     7c         Kd         9h             6h
 8d     6c         Qd         8h             5c
 7h     5c         Jc         7h             4h
 6d     4s        10d         6h             3h
 5d     3s         9s         5d             2d
 4d     2d                 4d             Ah
 3d     Ad                 3c             4c
 2h                     2c             3c
 Ah                     Ac             2c
 8h                                 Ac
                                 7c
                                 6s

Now, which of these two positions (resulting from removing either diamonds or spades) is better with regard to our ultimate goal -- the hearts? Well, in the tableau immediately above (with the spades removed), we still can't get through column 9 (same reasoning as before), nor can we get any more spaces (column 2 is the only chance, but we can't get through it). And since we can't get through column 2, we can't remove a set of clubs yet, so all we can do is remove a set of diamonds. If that's the case, we might as well have removed the diamonds first and then seen whether we could do without removing the spades! So we'll use the earlier tableau and proceed from there.

Now we can dig through column 9 and turn up a new card, but we'll lose the space in the process, because we've got only one free 8 left. Furthermore, to get to that free 8 we must use up our only free 6, so no matter which heart gets turned up we won't be able to move it, nor can it possibly get us the space back. Furthermore, we still can't get any additional spaces (short of removing more suits) due to the lack of free 10's and deuces. Thus we can't get out a set of hearts yet, but we're getting closer!

What next? We can now remove either spades or clubs. Either way we end up getting a new space. Removing the clubs has the advantage that it digs all the way to the 4-Ac in column 9, so let's try that approach. We'll start by dredging out the Qc: 3-Ac from 7 to 1, 5-4d from 7 to 6, 3-Ac from 1 to 6, 9-6s from 7 to 1, 10d from 7 to 10, 9-6s from 1 to 10, Js from 8 to 5, Q-Js from 7 to 8, Kc from 7 to 1, Qc from 7 to 1. Now we finish the job: As from 2 to 7, 4-As from 2 to 4, J-5c from 2 to 1, 2-Ah from 9 to 2, 6-3s from 9 to 3, 2-Ah from 2 to 3, 5-As from 4 to 10, 7c from 9 to 4, 4-Ac from 9 to 1. Removing the clubs from column 1 yields:

(sp)    (sp)     9h     9c     Kd     9d     As     Ks     --     Qs
        10s     8c     Qd     Qh         Qs     --     Kc
         9s     7c     Js     Jh         Js     Jd     Qh
         8h            10h             7d     Jc
         7h             9h             6h    10d
         6s             8h             5c     9s
         5s             7h             4h     8s
         4s             6h             3h     7s
         3s             5d             2d     6s
         2h             4d             Ah     5s
         Ah             3c                 4s
                     2c                 3s
                     Ac                 2s
                                     As

Surely two spaces will suffice! Except that now we've used up the last of the free 8's, so both the 7d and the Jd will cost us spaces (we can move the Jd onto the Qs in column 10, but that too costs us a space). If the 5h turns up, we'll be stuck, but what if we get the Kh? Then, with a bit of judicious planning, we can move the Qs out of column 10 onto the Kh. (The planning involves putting a Js on the Qs so the Jd can go elsewhere.) But the lone space won't be enough to get the Kh off of column 9, once the Q-Js are placed with it. So we must plan even further and leave a Q-Jh to be picked up by the Kh. This is our only hope of getting the hearts out (without removing the spades), so let's see how it works out: 9-As from 10 to 1, 10d from 10 to 5, 9-As from 1 to 5, 3-Ac from 6 to 1, 5-4d from 6 to 2, 10-6h from 6 to 8, 3-Ac from 1 to 2, Jc from 10 to 1, Jh from 6 to 10, Jc from 1 to 6. Now we've got the Jh with the Qh that we can move. (We can't move the Qh in column 6 since that would cost us a space.) Continuing: 5-As from 5 to 8, 3-Ac from 2 to 1, 5-4d from 2 to 5, 3-Ac from 1 to 5, Ah from 9 to 1, 2d from 9 to 2, 2-Ah from 3 to 9, 2d from 2 to 3, Ah from 1 to 3, 4-Ah from 9 to 1, 5c from 9 to 2, 6h from 9 to 4, 5c from 2 to 4, 4-Ah from 1 to 4, Q-Jh from 10 to 1, Kc from 10 to 2, Qs from 10 to 2, 7d from 9 to 10, Jd from 9 to 2, and we assume the Kh is turned up. We move Q-Jh from 1 to 9 and reach the following position:

(sp)     Kc     9h     9c     Kd     9d     As     Ks     --     7d
     Qs    10s     8c     Qd     Qh         Qs     Kh
     Jd     9s     7c     Js     Jc         Js     Qh
         8h     6h    10d            10h     Jh
         7h     5c     9s             9h
         6s     4h     8s             8h
         5s     3h     7s             7h
         4s     2h     6s             6h
         3s     Ah     5d             5s
         2d         4d             4s
         Ah         3c             3s
                 2c             2s
                 Ac             As

Unfortunately, despite our best preparations, we will be unable to combine the hearts once we move the K-Jh into the space and turn up the 5h. We could go back and try removing the spades instead of the clubs earlier, but it wouldn't help. We must remove both the spades AND the clubs (and the diamonds) before removing the hearts. We can't get the spades together starting with the above tableau -- we can't get through column 3 with only one space. So we'll back up to the previous tableau and proceed thusly: 2-Ah from 3 to 1, 6-3s from 3 to 4, 8-7h from 3 to 2, 10-9s from 3 to 8, 8-7h from 2 to 3, Ah from 9 to 2, 2d from 9 to 4, Ah from 2 to 4, 2-Ah from 1 to 9, 8-As from 10 to 8, and remove the spades. We now have this:

(sp)    (sp)     9h     9c     Kd     9d     As    (sp)     --     Qs
         8h     8c     Qd     Qh             --     Kc
         7h     7c     Js     Jh             Jd     Qh
             6s        10h             7d     Jc
             5s         9h             6h    10d
             4s         8h             5c     9s
             3s         7h             4h
             2d         6h             3h
             Ah         5d             2h
                     4d             Ah
                     3c
                     2c
                     Ac

With THREE spaces we should have no trouble! Then again, considering how careful we had to be to even come close using two spaces, perhaps we should be cautious! If we just start dumping things into spaces we may find we don't have enough spaces to move things around once we know what we want moved. So we'll start by gathering what hearts we have: 3-Ac from 6 to 1, 5-4d from 6 to 2, 6h from 6 to 3, 5-4d from 2 to 3, 3-Ac from 1 to 3, 4-Ah from 9 to 1, 5c from 9 to 2, 6h from 9 to 6, 5c from 2 to 6, 4-Ah from 1 to 6. Now, if we stuff the 7d and Jd from column 9 into a pair of spaces, and the 5h turns up, we can move 4-Ah from 6 to 9, 5c from 6 to 8, 5-Ah from 9 to 6, and Q-Ah from 6 onto the newly revealed Kh. But if the Kh is the first card turned up, we'll be in rough shape. So let's prepare for that contingency just as we did in our earlier attempt. We move 9s from 10 to 1, 10d from 10 to 5, 9s from 1 to 5, 4-Ah from 6 to 1, 5c from 6 to 2, J-6h from 6 to 8, Jc from 10 to 6, J-6h from 8 to 10, 10-6h from 10 to 6, 5c from 2 to 6, 4-Ah from 1 to 6, Q-Jh from 10 to 8, Kc from 10 to 1, Qs from 10 to 1, 7d from 9 to 10, and here we are:

 Kc    (sp)     9h     9c     Kd     9d     As     Qh     --     7d
 Qs         8h     8c     Qd     Qh         Jh     --
         7h     7c     Js     Jc             Jd
         6h     6s    10d    10h
         5d     5s     9s     9h
         4d     4s         8h
         3c     3s         7h
         2c     2d         6h
         Ac     Ah         5c
                     4h
                     3h
                     2h
                     Ah

No matter which heart is revealed when we move the Jd from 9 to 1, we will be able to finish combining the hearts.


Ae[bFBe`0XjJA71dJjS2;b0Q9[YT01:>4HQoE;]7OIIXJIB432/DZcKfD>>CSO_H>Zgi[ed:5NPgamO?=mKT\gBR^c=Gd@O^@TM>]QdcD :gKni/^ f15K/8 LRb;/e 8f=HX/D UPH2/> J?B>/H Pa?n0/o ggKE/R oQfL/j ?`9Al/^.

In case you're wondering what it takes to finish a game with all eight completed suits still sitting in the tableau, here's a deck (again, encountered in actual play) that makes it possible. The "solution" is left as an exercise.

This deck can also be useful as a confidence boost for novices who are having trouble winning at all, since it is relatively easy to win from this position if you are willing to remove suits as you complete them.

 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     --     --     --     --     --     --
 --     --     --     --     --     --     --     --     --     --
 --     9s     Kd     --     4s     8h     --     Qd     7s     --
10d            Jd            Qh            9d


[Copyright (c) 1989, Donald R. Woods and Sun Microsystems, Inc.]


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