Last week I asked the following questions:

- Which suits have all 13 cards appearing at least once?
- Assuming you answered “more than zero”, can we actually remove a suit (regardless of identity of face-down cards)?

Bart Wright had some vague intuition that it might be possible. Judging from his writing, I think he would have some valuable management skills to contribute to any company who is interested in hiring. Unfortunately, he failed the “specific/measurable/achievable/relevant/time-bound” test. Schistocerca Americana has found a solution: Diamonds is the only suit with each card appearing at least once. His solution is as follows:

Deal – Do Nothing

1st Draw – ed

2nd Draw – hj, hf, dh, da, eb

3rd Draw – fb, hf, cg, eg, ig

4th Draw – ja, ji, ji, jd, fj, eg, fg, ei, gi, ga, ch

5th Draw – da, ad, ab, ib, ab, jb, af, aj, aj, ij, fj, hd, hj

The result is shown below. Using cut-n-paste in Excel proves this solution is indeed valid with no illegal moves, sloppy explanations or typos.

To be honest, I didn’t try to solve this problem myself since I am currently working on another fun project that is unrelated to Spider Solitaire. Well done to Bart Wright and Schistocerca Americana for their excellent contribution to this blog.

Of course, we are interested in removing eight suits instead of one. Clearly it makes sense to look for easy turnovers and empty columns at the beginning of the game. But the above was not an exercise in futility. At least we know that it’s possible to remove a suit just by sheer power of information (i.e. knowing the identity of unseen cards) even without an empty column. Besides an aspiring player must (i) learn to analyse long move-sequences involving a large number of face-up cards when playing without undo (ii) learn to play the cards well even when there is no empty column 😊

Let us first focus on exposing as many turnovers as possible without dealing any cards from the stock. Experimentation shows it is easy enough to turn over many cards in the tableau, including all cards in column 2:

Further experimentation with undo leads to the following cheat sheet:

It’s time for another fun question: **how many rows do we need to deal from the stock to be sure of procuring an empty column** (assuming the worst possible permutation of unseen cards)?

Note that NaN may be a valid answer if this turns out to be impossible even allowing for dealing all cards from the stock.