Continuing the game from last week, we have reached this position when even my dad knows how to remove the Diamond suit on the next move.
How would you continue here?
First of all, I present Grasshopper’s solution. This deserves to be seen in full, and also illustrates how one can manipulate cards in the endgame without the help of an empty column – a fundamental skill for playing well at Spider Solitaire.
Easy stuff first, do the Dia run
Do a Club run in Col 1
Shift Heart J to Col 3
.ic (Is that a super move? Do I need to notate as such?)
Swap out Q’s Col 9&10
.ia, ji. aj
Fill void with Club K Q and cover with J from Col 2
Shift Club 8 in Col 9 to Col 7, move Club 10, 9-5 and 4-2 to Col 1
.ig, ia, ca, ga
Dig out the Club A and do Club run
.ib, ci, ca
If the above is accurate we have done a Club run and have maintained our void in Col 1. If not, you can stop reading now.
Surely we have to do a Spade run next, the location of the last Dia 4 is unknown, the second Heart run has needed turnovers strewn across 5 columns and if we solve this thing my guess is hearts clear last, and lastly the Club 4 is trapped by the Dia K.
So going forward from the point of removing the Club run I should think we should set things up so that, ideally, after the last draw we can make the Spade run and have two voids.
I intentionally did not join the Dia K&Q in Col 9&10 with the idea that we needed a “Clean Queen” in Col 10 to accept the upcoming Club J from Col 1. If we are going for a Spade run next and if it is possible, we will gain a queen in Col 4, and a Club queen at that! So lets start by swapping out the queens in Col 9&10 and keeping one hand on the ZKey in case we fail in the spade run.
.ia, ji, aj
Fill the void in Col 1 with the Spade 8 from Col 3 and distribute the J 10
.ca, ci, cd
Take the last draw.
Join the Club J 10 9 then do the spade run
.da, ha, cd, hd, gd
Create some voids
.ad, ae, fb, fi, cb Three Voids
This turns out to be completely correct. But before going on, a few general considerations are in order:
It’s hard to visualise moves even with Excel Cheatsheet, but I believe it is of immense benefit to the student to practice visualising long sequences of moves. If you’re reading this blog and don’t have the luxury of having the same position in your Spider Solitaire program, one suggestion is to write down a sequence of moves, then check your answer with Excel Cheatsheet 😊
Removing complete suits is often undesirable because they “can be used as lubricant” for other cards, as eloquently expressed by Bart. For instance, going back to the initial position (score = 70) one possibility is dg,hd,ha,fh,hf,hj,gj,gh to clean up column 8 to some extent. However, the tesuji of not removing a suit is much rarer when playing without undo.
Going back to the game:
It is trivial to determine the identity of the last two unknown cards. Of course, we remember to shift the Happy Stars of Bethlehem down a notch to indicate all cards have been dealt.
With four suits removed and three empty columns we are almost surely headed for victory. Of course, there is one particular Spider Solitaire server that taught me never to take anything for granted – but here we have no reason to suspect foul play. Let us finish off with a fun question:
- What is the minimum number of moves required to remove a fifth suit – which could be Clubs Hearts Spades or Diamonds?
To make it slightly more challenging, try to visualise your moves without the help of Excel Cheatsheet 😊
4 thoughts on “Game on (25 July 2021)”
This problem is simple enough I think I can see a solution.
The suit we want has to be clubs. We already have Q-9 and 8-5, and no other suit has anywhere near that in terms of in-suit sequences pre-built.
We’re going to dig column i out completely, that’s the basic approach.
ha,ih,jh builds A-3 clubs, A2 diamonds is in space a
ic,ac — uses space but gets one back so now we have 2 spaces, c has 3spades/A-2 diamonds
ia — JQK diamonds to space a, uncovers 2 of hearts, only one space left.
if fills last space with 2 of hearts, uncovers 4 clubs.
ig,ib gives us a space back as we clear column i.
ji exposes king clubs (using last space), now just put pieces together
dj,gj,hj and the clubs are done.
That’s 13 in all.
Can we do better? 6 moves are for putting club pieces together, so that can’t be shortened in any way. 3 moves are to dig to expose the 4 of clubs. One move to expose king of clubs (moving the ace of clubs was already counted as putting club pieces together). One to expose 3 of clubs. That’s 11. That leaves A2 diamonds onto a 3 to retrieve a space, 4 diamonds onto 5 of hearts to retrieve another space. We need both of those spaces. That’s 13.
I didn’t use the Cheatsheet, but you may note I did write after each step what was in the spaces. I suppose that’s cheating in a way. I mean, you could get rid of your cheat sheet and simply write down in text where things are after every move. But I figure I will get the “most improved” award since lately I haven’t had any solution at all.
Again I am honored to be showcased. But remember, I have a great teacher and a classmate willing and eager to share his knowledge.
Solution first, digressions to follow.
Assignment: From position 342, “What is the minimum number of moves required to remove a fifth suit – which could be Clubs Hearts Spades or Diamonds?”
Further parameter: Use only the two visuals given by Master Chi-Yuen. We retire the AncientExcel CheeterSheeter ………. for now.
Just looking at the location of the unturned spades and hearts and diamonds I will rule them out by “feel” … just 2 much diggn’ 2 do.
So my answer is a club run in 12 moves.
Uncover the Club K in Col 10 and cover with the Q-9 and 8-5
.ji, ja, dj, gj
This exposes a 4 in Col 4. Cover that 4 with the Spade 3 from Col 9 and Dia 2-1 from Col 8 while building he Club 3-A in Col 8
.if, id, hd, fh,
Do the club run
.ic, if, ij, hj
So a club run in 12 moves, if I am correct.
I probably would not proceed as such. I would think about guarding my voids better. Moving the Heart 4 from Col 4 to Col 2 (after moving the Club Q from Col 4) would be an automatic. But if we did arrive in this position, we could gain a void in Col 10
.je, jc, jf, jg
And often a single void in the endgame is all we need.
So here I am, “seeing” 16 moves into the mist to do a run and gain a void. Wow. Who am I and what have I done with HopHop??
Master Chi-Yuen, when you did a left hand turn and led us down this lane my reaction was, “But this is not where I want to go! Study 4SSS Sans ZKey want I (Yodaspeak).” But with you casting us into increasingly complicated positions along with the insights I have gained from Esteemed Scholar Bart I look at what I am capable of doing now as far as seeing card manipulation goes vs. my vision-‘bilities three months ago and I feel that I have made some progress.
I also need to thank the Happy Stars of Bethlehem for keeping me on the proper start square.
I agree with SA’s shorter solution. My thinking was somehow frozen out of the idea of moving the queen of clubs early to expose a very useful card underneath. But perhaps, inspired by his innovation, I can redeem myself by doing him one better.
My different idea is to put the 2 of diamonds into a space early, allowing the “ih” move to be simultaneously an uncovering move and a club-joining move. So there’s no need to pull 2A of clubs back out of a space.
First 4 moves are as SA did.
.ji, ja (Q of hearts into first space)
.hc (2 of diamonds into second space)
.ih, id, if (K of diamonds into last space)
.id, ij, hj
Trying once again (sigh) to prove it is the shortest, as before 6 moves are required to join the club pieces together. Overall, 7 moves are required to uncover the needed cards (1 in column 8, 4 in column 9, and 2 in column 10). However, we can combine joining with uncovering moves with ji and ih. Those are the only times the cards that we move to uncover something are also part of our club suit (a requirement for this kind of dual-purpose move). So 7+6-2 = 11.
In my first answer I had also allowed two moves to “recover spaces” but with more efficiency we don’t need any such moves.
But of course, nicely done. I officially change my answer to mirror that of Esteemed Scholar Bart, 11.