Continuing from last week:
This is our cheat sheet so far:
One possible game state that can be reached from the start is the following:
Thanks to Bart Wright and S.A. for their input this week. Unfortunately, I’ve been busy with other stuff. To be more specific, I have discovered a truly remarkable idea for solving the problem of cheating in online chess tournaments. Unfortunately, this blog is not large enough to contain it. Besides, the tagline of the whole blog is “Spider Solitaire, the whole Spider Solitaire and nothing but the Spider Solitaire”, so I would also need to find some obscure connection to justify including this idea. On top of that, some of the comments failed the TL;DR test but the main gist seems to be “we can turn over new cards in Columns 1 and 6”.
S.A. gets bonus points for mentioning that we soon get to find out if Microsoft Solitaire allows scores to go into negative numbers, or whether the score remains at zero for the remainder of the game (and undo becomes free). I guess there is also the third option that the game terminates after 500 moves and we never get to find out if this game is winnable. Let’s hope it doesn’t come to that. But at least we have a “cliffhanger” which means Bart, S.A. and Sebastian are morally obliged to read the next few posts. Win-win for everybody 😊
Since we know the next deal is J6254A388K, it is relatively straightforward to determine the next unknown card in columns 1 and 6 starting from the above game state (NB: the next unknown card is determined by the cheat sheet, not the screenshot of the current game state).
Our luck is in: it is relatively straightforward to obtain a void in either Column 1 or Column 6 after dealing a row of ten cards. And yes, we have our first Nine 😊
It is not hard to see we have three guaranteed empty columns – so we should be able to determine the identity of almost every face down card. If you have played many games with undo, you are probably familiar with this pleasant situation.
QUESTION: What is the minimum guaranteed turnovers, starting from the above game state (score=109), playing with undo but not dealing any more rows from the stock?
BTW, in other news, Big Shiny Red Question Mark has been fully reinstated with all rights and privileges therein (thanks to Schistocerca Americana for the heads-up). But I’ve been told by a reliable source that BSRQM is still recovering from the mental scars of being Rick-Rolled. I guess there are things that 100,000,000,000 rupees can’t buy.
Edifying Thoughts of a Spider Solitaire Addict 
The problem seems too big and complicated for me to get a handle on. The last time I said that and read SA’s solution, it got me oriented enough that I could come up with something. Maybe he’ll have such a catalytic effect again.
Also I’d like to verify one point about what the puzzle is. When you speak of “guaranteed turnovers”, that does not mean identifying cards in order to put them in a spreadsheet, right? What you mean is cards turned over in the actual game state, which will of course include many whose identities we already know because of Undo. And many such cards we turned over in past forays might not be turned over now “for real” if they are under (say) a pair of kings.
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Guaranteed turnovers does indeed mean identifying cards in order to put them in a spreadsheet
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Esteemed Scholar Bart I totally agree given the two tools Master Chi-Yuen (formerly known as Master T) it becomes a mountain of an assignment. Again I used my AncientExcel CheeterSheeter to make a hybrid of the two at position 109. It shown the cards as shown at 109and also the known cards above ……………. ……………… ……………….
WHOA, HOLD THE BUS………… I just went back to my AE CS and I think I have totally blown the assignment
I need some more time, I’ll be back tomorrow or Wed.
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Master T I shall attempt to promptly present this weeks assignment so those active participants (and also valued unseen readers) whom have a full life can quickly get the facts, “Just the facts, Ma’am” (Shout out to Sergeant Joe Friday). But it is hard to think linearly when you have been Rick-Rolled by a trusted instructor. And furthermore the problem that presents itself to Rick-Rollers is that they sometimes give away unknowns. Such as we now know our Master Teacher’s full name is Trevor Chi-Yuen Tao, International Chess Master, Mathematical Genius, Classical Music Composer, Research Scientist, y mucho mas. I should think that we are all in agreement that Trevor Tao is a really cool name and I quickly glommed onto Master T as a proper name for this endeavor. But if I may, Trevor Chi-Yuen Tao is like in a whole ‘nother league. The handle “Master T”, while respectful, suddenly seems so Men in Black-ish. So if permitted I would change my reference from Master T to Master Chi-Yuen which just drips of Shaolin-ness and total respect.
But I digress.
We have six columns with face down cards, we can peek-a-boo with five of them:
Col 3
ab, ca, cj, aj, ca, ce
Col 4
NoCanDo The 5, 10, and Q can be placed on existing piles. We need voids for the K and two 4’s, so three blanks, all at the same time. Cols 1 and 7 are easy to empty, Col 6 goes fine until we reach the 5 when things come to a screeching stop seeing as the sole six available is reserved for the 5 in Col 4.
Col 5
dh, ed, ab, ea, eb
Col 8
ab, ha, he, hb, hd
Col 9
ie, ab, gb, gd, ge, ic, ia, id
Col 10
ab, gb, gd, gi, ja, cg, ca, ga, cb, jg, jc
As always I hope I done good and if jot my plan is to claim typos. In order to speed the reeders on their way I am typing really really fast and this inevitaably leads to missteaks.
My final though is that when we realize that this hand is either a W or L we should indeed drive the score to zero and beyond just for the sake of knowledge.
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Well done on your digression. You have indeed correctly identified my real name.
I consider myself a “known person” in a number of communities and I am not surprised someone was able to put the pieces together and work out who I am. For the same reason, I didn’t bother putting any effort to hide my real identity.
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It is an honor to be in your presence, Sir.
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tl;dr: 13
OK, now looking at the problem the way it’s intended, this feels more manageable. As GM noted, we can make three spaces. We use up our 9, our 8, and both 4s to do that, and we might as well. There’s nothing subtle going on — a space is money in the bank, fungible with regard to any other possibilities in this setup. Sometimes by not getting a space you can get more receiving cards, but not here. That is assuming we start with the board (109) as given without any initial undos. When I am done making the spaces, new cards on top in non-spaces are 2, 3, and 7. They could be wherever it is convenient.
Column 3. cj cd ca cb, cg. Got remaining 3 cards.
Column 4. dh, dj to clean stuff off the king, then da, da, db, dg, got one card, but still one left. No magic way to find homes for 4s, the two 5s are alread occupied.
Column 5. ea supermove, eb, eg — got remaining 2 cards
Column 8. Only two turnovers. Getting the 3 spaces used up our 9 and both 4s. so ha,hb (1 turnover),hg (2nd turnover). 2 of the 4 flipped.
Column 9. ha,hb exposes king. hg exposes 2. he (“e” or wherever that 3 went) exposes one card, but one left.
Column 10. ja,jb,jg, exposes one card under the 7, but 2 are left. However, if we first do a ca supermove after putting the first king in space “a”, and put that exposed ace on our 2 (wherever it went), that exposes an extra 8. So we have a home for the 7 and can get an extra one there.
For cols:3,4,5,8,9,10
Gotten: 3+1+2+2+1+2 = 11
Missed: 0+1+0+2+1+1 = 5
Next question… Should we be using undo before we start here? It is possible that in making one of the spaces we used up 2 receiving cards, for instance.
Dead end: Column 4. Could we have 3 spaces and still have a receiving 5? Or 2 spaces and two receiving 5s? Our spaces as we start in columns 1, 2, and 7. We have a choice as to where one of our spaces is: it could be column 2 or 6. Both require using up our 9 to hold the 8s. But what if we make spaces in 2 and 7 only? Almost works, but to expose that 5 in column 6 we need to use up the 9, which prevents column 2…
Column 8. This time it does help to clear column 6 instead of 2. Because when we’re done we have an extra receiving 4. That gives us a place for the 3, so we can put one more card in a space, and get another turnover. So 8 gets 3 turnovers. PLUS 1.
Column 9. Can’t do any magic to find a home for the 8, since the precious 9 was used to make one of the spaces. Can we somehow have an extra queen for the jack? A perfectly good queen got plastered by a king in the deal, so if we have a jack already on that queen before we deal, we can preserve the queen it came from. Looking back to the 141 board, it looks like we could move the ten/jack from column 3 to column 10. We have plenty of 10s so we won’t miss that one when it gets covered by a king. That gives us a home for the jack, so we can uncover one more card there in column 9. PLUS 1.
Column 10. No hope for the kings, and we already found a place to put the 7. Can’t see how to improve it.
So, final answer: “13”.
The only cards we can’t turn are the bottom ones in columns 4, 8, and 10.
As always, an error or oversight is more likely than not.
I looked at SA’s answer, and we do seem to be on different pages somehow. I assumed as a starting point that we had made the 3 spaces, and he does not. That’s the first immediate difference.
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Esteemed Scholar Bart I offer you a smile and a bow to show my appreciation of your excellent work this week. Nicely done, Sir. I learn from your organized mind.
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