Recall we asked the following questions:
- Suppose you were legally required to turn over all
threefour possible cards simultaneously. What would be your play? - Suppose after making your first move you were able to call the next card (both rank and suit). What would be your top three choices?
To answer the first question: the obvious moves are to build in-suit with the Q-J-T of Hearts and then put that on one of the Kings, followed by an off-suit J-T. The problem is that if we turn over a new Queen then we lose a turnover since our off-suit J-T cannot play onto the Q.
At this stage, it might occur to you that one can play the Q of Hearts onto a King, then add the J of SPADES and an off-suit Ten. Finally we can build in-suit with the J-T of Hearts. The difference is we only build in-suit once, but we get an extra turnover if any Queen appears. It’s not at all clear whether sacrificing a turnover like this is worthwhile, but the main point I wish to make is such a possibility exists – and you will find many more examples in your journey to Spider Solitaire mastery 😊
As for the second question, Bart correctly identified that Queen is the only rank that gains two turnovers. The best suit is clearly Spades, ensuring both turnovers are in-suit. Bart then wants the Queen of Hearts, but I prefer the Queen of Clubs since that means one of our extra turn-overs are in-suit. Bart then has Queen of Clubs as third choice, which I would agree with given he has already committed to the wrong Queen for 2nd choice 😊
Bart chose the QH because Hearts is closest to completion and exposing more Hearts will further that goal, even if some cards are duplicated (since we might need the “other” queen later). At this early stage, I don’t like committing to the “suit nearest to completion”. There are plenty more cards to come, and we might find e.g. the 9-8-7-6-5-4 of Diamonds turns up and suddenly we wish we focussed on Diamonds from the beginning. Of course if we do manage to land the Bart’s Quickie cheevo I am happy to be proven wrong.
In other words, let’s just focus on maximum turn-overs, in-suit builds and empty columns and once we have more cards in play it will become obvious which suit is closest to completion.
Well that’s enough pontification on the opening position for now. Let’s make some moves.
(eg = KH, ce = AC, ie = KD, ae=7D, fa=7C)
Well that was disappointing. We drew two Kings and an Ace, and only obtained one turnover more than our guaranteed minimum. Looks like Trevor’s Quickie isn’t happening any time soon. ☹ Note that I eschewed the Q-J of Hearts in-suit build for reasons described earlier.
So here we are, like it or not. At least the Kind of Diamonds landed on another King. How would you continue here? (HINT: consider all reasonable options before committing to a line of play).
Edifying Thoughts of a Spider Solitaire Addict 
As for the last move, leaving the jack-ten atomic isn’t something that would have occurred to me, but I see its advantages and actually feel pretty clear it is best. I learned a little bit, thanks!
As for “ask the genie”, we differed on whether queen of hearts or clubs was the better second choice, I certainly see the advantage of the in-suit king/queen club combo, and I guess on balance I think you’re right (how did I miss that?) That king-queen isn’t going anywhere, but it is the start of a club suit.
You say that suit-bunching isn’t a good idea. I agree that it is rarely important and only applies as a tie-breaker when there is hardly any other reason to choose. It’s a tiny, tiny factor. In this case you found a more important one. I certainly agree it would be folly to keep anything like “we’re now working hearts” in mind.
To the current situation, obviously turnovers are important. So is working towards a space, so we look hard at that column 6 with only 3 cards under it. Turns out these two goals are in harmony. We can uncover a card in either column 3 or 6, but only one of them because we have only one 2 and those columns both use it up with their ace move. Column 6 is better both because it is closer to getting an empty column, but much more because exposing that 7 is vital for the second turnover in column 2. The next question is whether to move that 2 of clubs first. Yes because it leaves column 3 atomic. But do we want to move the 3 first? Yes, to leave column 7 atomic.
Do we want to move the 4 first? Column 5 is nowhere near being atomic. It would expose a 10, but we’ll have another 10 exposed once we’re done with column 2, and the cost is using up a receiving 5 we might want later. Not moving the 4 of diamonds lets us retain in reserve the 5 of diamonds onto 6 of diamonds if it turns out to be in our interest (do we need to put something on a 6 or a king?).
We can’t go yet one step farther and move the 5 of diamonds first and build all the others on top of the 6 because if there’s anything out-of-suit on it, then we can no longer clear column b. So, we’re going to leave the 4 of diamonds where it is and let column 5 become even more of a junk pile.
(1) ge, 3 of hearts to 4 of diamonds
(2) ce, 2 of clubs to 3 of hearts,
(3) fe, ace of spades to 2 of hearts
(4) fi, 7 of clubs to 8 of clubs, uncovering a card.
(5) bi, 6 of diamonds to 7 of clubs
(6) ba, ten of clubs to jack of diamonds, uncovering another card.
So that’s as far as I got in the time I might allot in a real game. Now to think even harder… I can’t think of anything else. Columns 4, 5, 9 and 10 are all locked up with kings and not going anywhere. 1 and 8 are fairly active but we can’t help them out this turn. We can avoid making 8 worse… 7 of clubs onto 8 of clubs is better than 8 of spades, both because it’s in suit and it junks up column 8 less. Columns 2, 3, 6 and 7 are the active columns here, and I think we’ve done as well with them as we could. We leave 2 atomic and uncover cards in the other two.
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