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C00010 00004 T H E S P I D E R P R O G R A M
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S P I D E R
SPIDER is a particularly challenging double-deck solitaire. Unlike most
solitaires, it provides extraordinary opportunites for the skillful player
to overcome bad luck in the deal by means of careful analysis and complex
manipulations. The SPIDER program does not actually play the game for you,
but rather takes the place of the cards (by displaying the tableau on the
DataDisc screen) and keeping a record of the game so that you can follow
out long lines of play without losing track of where you started from.
(Besides, it's a neat display hack.)
Page 3 of this documentation file describes the rules of the game; page 4
describes the use of the program; page 5 gives some "Spider problems" to
provide a feel for the complexity that can arise (the problems are based on
positions that arose in actual play).
(C) Copyright 1979 Donald R. Woods
R U L E S F O R S P I D E R
THE INITIAL TABLEAU. Shuffle together two decks of cards (104 cards in
all) and deal ten cards face down in a row. Deal three more rows face down
on the first. Next deal one card face down on each of four piles
(traditionally the leftmost four, but it doesn't matter; for aesthetic
reasons some people prefer the first, fourth, seventh, and tenth piles),
for a total of 44 face-down cards. Finally, deal one card face up on each
pile. These 54 cards comprise the initial tableau. NOTE: In the
description below, the card at the "bottom" of a pile is taken to be the
last one dealt or played onto that pile, as opposed to the card that is
underneath all the others in the pile. Thus the "bottom" card is the one
displayed bottommost on the screen by the Spider program.
BUILDING. All building is done in the tableau, there being no separate
foundations. On the bottom card of a pile may be placed any card of
next-lower rank, regardless of suit. (Cards rank King (highest), Queen,
Jack, 10, 9, . . . , 3, 2, Ace.) The bottom card of a pile is always
available to be moved, as is any sequence of cards at the bottom of a pile
that are consecutive and ascending in rank and of the same suit.
For example, suppose the 6, 5, and 4 of hearts are together at the bottom
of a pile, with the 4 bottommost. They may be moved as a unit, or the 4
can be moved by itself, or the 5 and 4 can be moved without moving the 6.
If the 6, 5, and 4 were moved onto a 7 of hearts at the bottom of some
other pile, the four cards could then be moved as a unit onto any 8; if the
7 were not a heart, however, then once the hearts were placed upon it it
would not be available to be moved until the hearts had been moved from it
(to another 7 or into a space; see below).
When all face-up cards have been removed from a pile, the bottom face-down
card is turned up and becomes available for play.
SPACES. When all cards have been moved away from some pile, the resulting
empty pile is called a "space" or "hole". Any card or sequence of cards
available for moving may be moved into a space. A King, or sequence headed
by a King, can be moved ONLY into a space, and once moved there can never
be moved out (except into another space, which doesn't accomplish anything)
unless it is being removed entirely as described below.
THE HAND. The 50 cards not dealt initially form the "hand". Whenever you
wish (typically, whenever you get stuck), you may deal a new row of ten
cards from the hand face-up upon the piles. NOTE: You are not allowed to
do this if you have any spaces. You must first fill them in. Notice that
these additional deals tend to introduce discontinuities in the piles; that
is, you can get cards covering others that are not next-higher in rank.
If you get stuck after having dealt the last of the five additional deals,
you have lost.
OBJECT OF PLAY. When you have assembled a complete suit of thirteen cards,
in sequence from King down to Ace, at the bottom of a pile, you may remove
the thirteen cards from the tableau entirely. Cards so removed are never
brought back into play; thus it is not always desirable to remove a suit
when you have the opportunity (though it usually is), since it may pay to
keep it around to aid in manipulating the other cards of that suit (recall
that there are 26 cards in each suit). The game is won if you manage to
remove all eight suits.
T H E S P I D E R P R O G R A M
The SPIDER program requires a DataDisc. When started up (via "R SPIDER"),
it will ask for a file name. If you have saved a game in progress (or have
copied one of the tableaux from the following page into a file) you may
specify the file name and pick up in the middle of the game; otherwise type
just a carriage return and the program will "shuffle" and deal a new game.
After the initial file-name question, the program enters the main command
loop, wherein it prompts for input with a "*". Erroneous input, including
attempts at making illegal moves, will result in a breedle. All commands
are terminated with a carriage return or, if you prefer, altmode. The
commands currently available are:
<n> <m> Move as many cards as possible from the bottom of column
<n> to the bottom of column <m>. (The columns are numbered
from 1 to 10 and are labelled on the screen for reference.)
In giving this command, you can separate the two numbers
with space, comma, or any other convenient non-numeric
character. Note that, unless column <m> is a space, there
is at most one choice for the number of cards to move. If
column <m> IS a space and you wish to move fewer than the
maximum, use the next command.
<n> <m> <k> Move <k> cards from the bottom of column <n> to column <m>.
<n> Remove a completed suit from the bottom of column <n>.
-<k> Back up over <k> moves. If you attempt to back up over
turning up a card or dealing a new round (i.e., over a move
that gained you information), you will be asked to confirm
doing so. Confirmation brands you as a cheater!
(null) Refresh the display.
A Automatically save the game (as if an F command had been
given) in file SPIDER.TMP after every <k>th subsequent move.
Disabled by <k> being non-positive (e.g., if <k> is null).
D Deal a new round. (All spaces must be filled.)
F Ask for a file name and save the game. All positions from
the initial deal through to the current one are included in
the file. The file format is not particularly bright, and
the file is thus rather large for well-progressed games;
the files are not recommended for permanent storage. The
line editor is loaded with a default file name SPIDER.<k>,
where <k> is the number of moves made so far.
L Look for a particular card in the tableau. Follow the L
with a space and then the rank and suit of a card (in that
order). The rank is either a number from 2 to 10 or one of
the letters A, J, Q, or K; the suit is one of C, D, H, S.
(Case of letters is, as in all commands, immaterial.) The
program will let you know where, if anywhere, the specified
card occurs among the face-up cards in the tableau.
P <n> Print the cards in column <n> in text form (in case the
display is too scrunched to read, which happens when about
30 or so cards get stuffed onto one pile). In the text
form, sequences of consecutive cards in a single suit are
clearly identified.
Q Quit (you'll be asked to confirm).
S Show status. The display will be replaced with a status
summary indicating how many deals are left, which completed
suits have been removed, how many face-down cards remain,
how many face-up cards there are and how they are
distributed among the suits, and which suits have enough
cards face-up to form a complete sequence of thirteen.
Typing any character restores the tableau display.
T Ask for a file name and write the current tableau in text
form (essentially the same as in the examples on the next
page, but without special identification of spaces). The
line editor is loaded with a default file name TBLEAU.<k>,
where <k> is the number of moves made so far.
X Ask for a file name and write the current tableau as a POX
source. You can POX the file directly or, if you know a
little about POX, you can merge the file with other POX
stuff. The line editor is loaded with a default file name
SP<k>.POX, where <k> is the number of moves made so far.
You'll also be asked what size font to use: M (medium) is
CRD165 and anything else gets you CRD219.
? or H Display a summary of this list of commands.
If you win, the program will exit automatically. If you lose, you'll have
to use the Q command to quit.
E X A M P L E S A N D C O N U N D R U M S
In all the examples etc. below, the tableau will be given here in the file
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, the
remaining pages of this file contain tableaux in Spider-program file
format. Thus, for instance, to look at the position given in the first
example, you would say:
.COPY FOO←SPIDER.DON[UP,DOC](6)
.R SPIDER
(it asks for file name)
FOO
The page number to copy will be given before each example.
===========================================================================
(Page 6)
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 --
6d 2c 4s Jh 7c
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.
===========================================================================
(Page 7)
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 (space) -- 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, 4-3d onto the 6s in
column 3, 2c out of the space and back onto the 4-3d, and finally the 7-5d
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 page 7 and see for yourself.)
===========================================================================
(Page 8)
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):
-- -- (space) -- 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:
-- -- (space) -- 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.)
===========================================================================
(Page 9)
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 -- (space) --
-- -- 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 4, 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.)
===========================================================================
(Page 10)
In the tableau below, one set of diamonds has been removed, and 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 to 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. 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 (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, 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 (space) 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 (space) 9d Kc Qs Ac (space) 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, bbut you might as well
maximise your chances anyway! (One way to add spice to an obviously-won
game is to attempt, after the last deal (or even the next-to-last), to play
only "natural" moves, i.e., moving cards onto other cards of the same suit.
If you're REALLY ahead of the game, try not using any spaces!)
===========================================================================
(Page 11)
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? (Page 12 has
the Spider file 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 (space) 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 plce 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:
(space) 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 (space) -- 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, 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:
(space) (space) 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:
(space) 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:
(space) (space) 9h 9c Kd 9d As (space) -- 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 (space) 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.
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