Happy New Game – Round 5

22 thoughts on “Happy New Game – Round 5”

1. Score 481.

   Before looking at the monkey’s move and GM’s hint, the best I had was getting a single space and using it up to get a turnover in column c. It was a rather short plan.

   Then I looked at GM’s comment. So that’s a video of you? Handsome fellow! So I managed to stretch my YouTube skills to see you’ve got other videos, at least 3 of which pertain to Spider. So when I have a little time, I should look at them. If not having done so makes me an ill-prepared student, chalk it up to ignorance about how to check out someone in the modern online world.

   Inspired by GM’s hint to take our time, I looked back a little harder, and found a way to get 2 spaces:

   db,fd,fe,df,eg,cg,ec.

   Two spaces gives us lots of plays. Ultimately I can’t think of anything worthwhile to be gained by digging deep into one of the other columns (f,g,h or i), i.e. using the spaces to absorb rank breaks. The best I see are 2 guaranteed turnovers from column c. But we can clean up first, since we have 2 spaces but won’t for long.

   hc,hd,hj,ae,de,c(1)-h(1),eh,cf(superx2), g(2)-h(2), f(3)-h(5), f(2)-g(2), gi.

   There’s so much chance of getting a single step wrong above that I’ll give what’s at the ends of the relevant columns when I’m done.

   a — unchanged
   b — TH, 9S
   c — K clubs
   d — space
   e — space
   f — misc A-J “garbage”
   g — K hearts
   h — still has the “7 gap”, but below is simply 3-6 of spades.
   i — Complete club suit, except QS instead of QC.
   j — 2A of spades.

   Notable here is that column h is well-positioned to gobble up a 7. It’s also in a good position to support taking a heart suit off soon — all the lower cards are in the first few columns. Anyway, we could proceed here to uncover with cd, ca.

   But I think I see something better, which I’ll do instead. The first several moves are to create an atomic 9T of diamonds in the space instead of 9S/TH. Move garbage A-8 in f over to column “a” — there’s enough to do this, borrowing from spades or clubs as needed. Then: fe (2 cards), bf. 9T in column e is atomic.

   Now have column h gobble up the 7 from column b: hd, bh, dh

   Now we put it all together: cd, bc, cf (uncover). So now we have a card uncovered in c plus column b remaining as a space.

   With one space, and borrowing from hearts, spades and clubs, I can move the A-7 of hearts from column f over into column A and the assorted junk to column f instead. And why not? It would simplify what might come next.

   Now we have a turnover. What might happen next? To get anal, going through the remaining ranks of that turnover:

   A — bad
   2 — column h absorbs
   3 — bad for now
   4 (none)
   5 — maybe ab, ac, ae — organizing column a and retaining space, might be able to remove heart suit
   6 — regain extra space from column j, lots of options
   7 — bad
   8 (none)
   9 (none)
   10 — ac, ab, ae
   J — ec
   Q — all are good, Q clubs very good, can remove suit and gain back 2nd space
   K – bad

   5 cards are bad (A, 3, 7, K)
   3 cards are “temporizing” (2, J)
   5 temporize plus potential (5, T, other Qs)
   4 very good (6, Q clubs)

   But to just be non-anal, just wait until the card turns over and see what happens next.

   Chances are excellent that GM sees something even better, but I guess he won’t tell us yet. But obviously handling this position right could have a huge effect on whether we win or not.

   I’ll confess I worked this out with real cards on a real card table — the equivalent of what IM Bug has called his “cheater sheeter”. No way I could keep all those moves in my head. I’ll call it the “senior citizen exemption.” Call it cheating if you will.

   It certainly looked grim when I first saw the new deal, but now I think we have a fighting chance.

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   1. Hi Bart,

      Well done on finding the 2 turnovers. Somehow Monkey failed to find even one turnover.
      When I ran the Monkey algorithm several more times it did manage to find the 2nd turnover. I agree
      we have a fighting chance – one dream card could put us in very good shape.

      I found one typo in your move sequence. Your move “ae” should instead read “he”. Otherwise
      it’s mathematically impossible for column a to be unchanged, as you claimed. 🙂

      I assume your full move sequence is given below (stopping after the first turnover).
      Please let me know if this is incorrect:

      db,fd,fe,df,eg,cg,ec (get 2 spaces)
      hc,hd,hj,he,de,c1=h1,eh,cf,g2=h2,f3=h5,f2=g2,gi (tidy up suits)
      fa,fe(2 cards), bg,hd,bh,dh (insert 7h into column 8)
      cd,bc,cf (turn over column 3)

      As for the senior citizen exemption, I prefer to use Microsoft Paint to copy the game state and
      move cards around. Doesn’t look great but it gets the job done.

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      1. You corrected a typo of mine, but I believe you introduced one of your own. After “(2 cards)” it should read bf instead of bg, which is impossible. When I was a programmer, the compiler never accepted “you know what I mean” but I think we humans can do that. Interesting that you felt comfortable replacing “Move garbage A-8 in f over to column “a” — there’s enough to do this, borrowing from spades or clubs as needed” with a simple “fa”. That’s one heck of a supermove. Not that my explanation is one iota more precise, but I spent longer waving my hands in the air.

         You should make a video of using Microsoft Paint to move cards around. I can’t quite picture it. “Paint” was one very early program in the Windows system. My recollection is it allowed you to do lots of stuff that looked cool but was basically useless — for what I needed or could figure out at the time.

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      2. Blockchain has been updated. Unfortunately I can’t add a video on Microsoft Paint unless I update my WordPress account (which I am not yet willing to do – if I get more readership then I might reconsider).

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2. I missed the two void capability. I should not have missed it.

   Additionally I have no chance of following the beautiful sequences put forth by Esteemed Scholar Bart. I await the new configuration and then maybe I can see how “there” got “here”.

   As far as Esteemed Scholar Bart using his manual CheeterSheeter we are here to have fun and learn. Methinks it is just fine to do what needs to be done. Our original premise here was and is to get better at 4SSS SansZKey, but if we stray and state we stray there should not be a problem. I hope that makes sense, I don’t trust my communication abilities much these days.

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   1. Blockchain has been updated. Unfortunately I can’t add a video on Microsoft Paint (as requested by Bart) unless I update my WordPress account (which I am not yet willing to do – if I get more readership then I might reconsider).

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3. Score 406

   The obvious “baseline” move is to put the ten of hearts in the space. We get one turnover and hope for the best. If it’s not good, game over. I see a way to also remove hearts, giving a second turnover. It also uses up our last space, like the baseline plan, but we get 2 chances to get a good card, not just one. We could remove the hearts with the king in column g instead of the one in h, but that 4 under the g king is useless at present. We’re better off taking our chances on a random card.

   We need the 10-9 of clubs on the jack of hearts in column h to pull this off at all — the 8 of clubs has to land on something in-suit, allowing us to manipulate those resulting 8-10 units with a single space. But once we’ve done that we can do further clean-up. By careful pre-positioning, we can reassemble the clubs J-A so that if one of the queens of clubs comes up we can remove clubs even if we have no spaces. Who would have thought that it was important beforehand to swap the 8 of spades out of column e and replace it with the 8 of clubs in column i?

   Implementing the initial “a7=f7” felt a bit like the Tower of Hanoi. The way I would do it without looking for optimizations would take dozens of moves. It’s a good time to be in a scoring system that doesn’t count number of moves. Intermediate steps would be getting A-5 of hearts in column a, then doing more to get the 3-6 of spades there, then more to get the a-7 of clubs there, and then swapping that out with the A-7 of hearts in f (which has been broken apart and put back together a few times along the way). Having full sequences of low hearts, spades, and clubs outside of the column we want to move (and no diamonds in the crucial area) is vital for moving such a miscellaneous column as “a” with only a single space.

   In all this shuffling, that A hearts/2 diamonds duo is kind of like a housefly. If you want to get at something under it, it will just get out of the way by flying to the other available 3 without being asked.

   So, the prep work: a7=f7, ie, h8=i2, e7=i5,

   The heart of the plan: ac, ab, ah

   A small interruption from our sponsors (separate small tweaks): e5=f7, e5=f3. This removes two suit breaks, and while it leaves column e far from atomic, it has just 3 pieces (ozonic?) for when/if it’s time to reclaim it as a space.

   Continuing: ih, h10=i3, ia(3 cards), hi, ce(!), ch (turns c).
   Unless something dramatic happens, we continue with eh to remove the hearts.

   The reason for “ce” is that in all the programs I’ve seen, a suit vanishes as soon as it’s complete, even if you don’t want it to. “ce” keeps the hearts from vanishing when we move the 10 in c onto the J in h. Maybe there’s nothing that could happen that would want us to still have hearts on the board instead of removing them? I’m not clever enough to see any such guarantee.

   Now, looking below “Decision Time”. The monkey’s plan isn’t as good. You then said, “The real question is what happens after [ac,ab,ae]”, but I bet you were subtly misdirecting us, intentionally or not, to steer us away from vital prep work that has to happen BEFORE that.

   There’s an old joke about an incident between a mathematician and an engineer.
   E: You’re so impractical I bet you don’t know how to fry eggs.
   M: Of course I do. Take the pan from the shelf and put it on the burner, put in the eggs, and cook.
   E: But what if the pan is already on the burner?
   M (thinks briefly): Put the pan back onto the shelf and you’ve reduced it to the original problem!

   If you were carrying out these steps with a real program, you’d probably find large inefficiencies in executing those group instructions like “e7=i5”, moving things back and forth unnecessarily. But I figure that a simpler exposition beats an efficient implementation.

   Final answer:

   a7=f7, ie, h8=i2, e7=i5, ac, ab, ah, e5=f7, e5=f3, ih, h10=i3, ia(3 cards), hi, ce, ch (turns c).

   More likely than not there’s some small error or two in there, and I authorize GM to “do what I meant” rather than what I said.

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   1. This seems to make sense, but ia(3 cards) should be ia(10 cards)

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      1. You’re right I made a mistake, but it isn’t corrected the way you assumed. This stuff is hard for me to keep straight, and in this case “do what I mean” wasn’t enough…

         You’re right that “ia(3 cards)” is wrong. But instead of your proposal of “ia(10 cards), the actual intended answer is “ha(3 cards)” AND omit the following “hi”.

         Revised final sequence of moves:

         a7=f7, ie, h8=i2, e7=i5, ac, ab, ah, e5=f7, e5=f3, ih, h10=i3, ha(3 cards), ce, ch (turns c).

         To try to find another way of expressing it that should cross-check, this is where things stand in key columns when I’m done:

         Column 1: 89T of spades
         Column 2: 4 diamonds
         Column 3: we await the card that is turned over
         Column 4: K clubs (unchanged)
         Column 5: A2 hearts, 3456 spades, 78 clubs, 9T diamonds
         Column 6: AH, 2D, 3C, 4C, 5H, 6S, 7H, 8H, 9S, TH, JC, KS, 6D
         Column 7: Unchanged (KH, 4H, etc.)
         Column 8: 3456789TJQK hearts
         Column 9: Unchanged after a lot of intermediate activity: A23456789TJ clubs, QH, KC, etc.
         Column 10: Unchanged

         At least that’s what I think until you find the next mistake…

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      2. Blockchain is updated, as per Bug’s request

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4. Don’t wait on me, I am in a fog and incapable of these lengthy visions.

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   1. Blockchain is updated

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5. Score 360

   I’m happy with your “trivial” call on the last move.

   We keep getting cards that are kind of middling. They don’t put us out of our misery, and don’t open up new vistas of progress. By my count we now have 3 bad turnovers, 5 “temporizing” ones, and 6 very good ones.

   Trying to dig in column f or g give us a net loss of spaces with no net benefit at this point. We might soon want to put the king in a space to access column i, but we’d rather do that when we can get back a space from the one we use than just a turnover, as it would be now.

   Spades are difficult to remove, but that’s not necessarily such a bad thing. With the two kings and the only showing J buried in 3 different columns, and no 7s at all, a mile is as good as a miss. “Aha! Now we get to remove the spades!” is going to come late enough (if it does) that we will have won the game already.

   We have two spaces available to us, and we’re basically still in the position of putting things spaces one at a time and hoping for good turnovers. I’d rather put the 6 in the space first, since we already have a 5 in a space (column j). It results in more in-suit sequences too. And it opens up the moving the i king to a space without net loss of a space.

   We don’t have to do “id” right away, but if we keep limping along we’ll want it done before we fill the last space. Unless both of the other queens come up before either of the club queens do, we won’t be blocked from removing a club suit from d and be forced to do it in “i” instead, but we might want to put the king from i in a space if our last 6 comes up, or maybe the king from g if the last 3 comes up, so best to have atomic kings in those columns.

   Final answer: bc, id, eb, he

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   1. bc, id, eb, he works for me

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      1. Blockchain updated. Next card could see a loss, a flurry of trivial decisions leading to a win, or somewhere in between.

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6. Score 354

   tl;dr — skip to “final answer” below.

   K was the very worst of the 3 bad cards. It’s a lot harder to win now, even if the next card or two is ideal.

   ce is appealing. It keeps more “potential energy” in the form of rank breaks we could use to our advantage.

   However, another slant on this is that things are likely to be tight, and getting better organized may be harder later even if we turn over good cards, so we close off some possibilities now as the price of getting organized. Especially after continuations, I think this is not true.

   jb, (removes 2 as a “stay alive” card)
   de, (borrow some club lubricant)
   jd
   ij (lose space to king — still a bad thing without good payback)
   ia, ba (reclaim space)
   e8=i1,
   e3=i10, (these 2 notable club/diamond “combing” moves)
   d5=e5, return club lubricant
   ig, (now line up some kings and queens
   di,
   ag (make a atomic again)
   e5=b
   ce (uncover)

   Removing spades is now not so far-fetched now, if the other J and one of the 7s comes up. Still, on balance, I favor the simple move as it keeps more possibilities alive:

   initial question answer: ce

   It looks like monkey went for a “lite” version of the clean-up.

   Now GM suggests continuations.

   (((((((((((((( First, after my complex possibility:

   AD = loss
   2H = loss <—- worse
   3D = ce, uncover c
   5C = loss
   6H = bc, hb, uncover h
   7S = ca, uncover c <—- worse
   JS = ac, i11-a, ci, uncover c
   JH – ac, ha, uncover h
   QC — bc, ic, get SPACE and uncover c
   other Q — put on K, uncover C

   Compared to what's below, the complex case is worse in 3 cases, and slightly better in 1. ))))))))))))))))))

   Second, after simple initial move:

   Final answer: "ce" now

   Continuation chart:

   AD = loss
   2H = be, ce, uncover c
   3D = ce, uncover c
   5C = loss
   6H = jb, jc, hj, uncover h.
   7S = ca, bc, uncover c plus SPACE
   JS = ac, ca, uncover c
   JH – ac, ca, uncover h
   QC — dc, dh, get SPACE and uncover c
   other Q — put on K, uncover c

   How to continue after that gets complicated, though if we first got a jack, then get a queen, then we can move the jack onto the queen, and uncover but also get a SPACE from column a and a SPACE where the queen was…

   Chances seem pretty bad. I think we would need to get 2 spaces pretty soon.

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   1. Two very good turnovers in column 3 makes a lot of difference! It’s looking like game over … in the positive sense of the word 🙂

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7. Score 451

   Those certainly were two good cards! I’m surprised that just two cards could put us at (or very close to) a guaranteed win.

   I’ved noted over the years that there are a couple approaches to games. If you’re in it purely for the skill of it, then you quit as soon as you know you’ve won. If you’ve got a more child-like playful approach, you keep moving the cards around until you win. Spider is very satisfying that way — even if you weren’t keeping a won/loss record I would be inclined to do it. So here GM nudges us to declare we’ve won without watching the cards bounce or fly or whatever they do in this version. I suppose making all those happy in-suit sequences culminating in all 8 suits gone is a hard thing to convey in our medium anyway… Just noting it. I’ve noted it in myself in old computer games like Civilization 2 and Master of Orion 1. I know I’m winning but I want to keep conquering everyone…

   Losing in fact would be stunningly unlikely, so I’m happy to see what the monkey does.

   But here GM poses what is to me an intricate challenge.

   I’ve tried going through the possibilities, and it looks to me like there’s always a devious run of uncovers to thwart us. I’m probably wrong, but I’ll go through them.

   First, to claim our 3 spaces, reversibly, pile up the 10-3 of spades in column a. We do in fact have the cards for removing the spade suit, but as I had predicted earlier I think the cost is still too high.

   Which column to dig in? 7 looks good at first, but I think it is a dead end, as we lose spaces to the king and 4. Then the jack can go on the queen if we could get there, but we have only one space. That 8 of spades goes in the space, what next? It seems like there has to be a run of pure hearts to replace it with, and all we have is the 87, and no place to stash the smaller cards, even if they were all spades. I’m not sure we’re out of the woods even if we can do that and move the A-3 to uncover, but I’ll leave it there unless a workaround is found.

   8 is fine as far as it goes but we have “holes without cards” problems. We can easily find ourselves with the king in one space, uncovering the ace. That leaves us just two others. That’s pretty similar to the other columns, where an ace taking up one of our spaces is part of getting thwarted.

   9 leaves us with a single space after we put both kings in spaces. I think if we uncover the ace and then 5 it sinks us.

   Column 6 seems most promising to me. The Jack goes on the queen, then two spaces are occupied with king and 6. Tha 6 was interesting because it turns both 5 and 7 into good cards. However, suppose we get A, which we put in the space, and then J of hearts. We redeem our space in column 10 but use up the receiving 6. Put the jack of hearts in the space, uncovering a 5, and then what? We can recover a space by putting the A-10 of spades on the jack, but what to do next with that one space? We can arrange for the king of diamonds in 8 to have nothing on it, so putting it in the space reveals… the 6 of hearts. Putting a 5 on a 6 we open another space. We can put either of the kings from columns 7 or 9 in, allowing us to get one step further — but it still looks to me like it’s not quite enough.

   It is I would guess more likely than not that I missed something, but I’m hoping GM can spot it and tell me what it is.

   I look forward to the post mortem too.

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   1. I too would attack Col 6

      Do you think this game is mathematically won with best play regardless of the permutation of unseen cards?

      Not Sure. Seems like I have lost a fair bunch of games when the two or three bad cards out of ten were first in line, and also I have won a fair bunch of games due to more than two or three really good draws at the end.

      If you think this game has no further interest, are you happy to claim a moral victory (having come this far!) and let Ninja Monkey play out the remaining moves?

      Master Chi-Yuen, when you are in a position of leadership, lead.

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      1. Monkey has managed to achieve the victory – but his endgame technique might need a bit of work. In any case, well done Team!

         Also congrats to both Bug and Bart for their efforts for trying to prove the game was mathematically won before the monkey took over. The point of the exercise is not to get an exact answer, but rather to give the reader a feel for when a game is getting close to mathematically won.

         Summary and post-mortem coming soon to a place near u

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8. I realized that the dead end I thought I had diagnosed in column 7 yesterday is not actually a problem. I asked what happened after we put the 8 of spades in the space. What can happen instead of finding a heart suit right away is that a sequence of spades up to the 7 can get put on the 8, the 8 of hearts under it onto the 9 of hearts in column 7, that spade sequence up to the 8 onto the 9 of spades we exposed, then with heart/spade swaps we can do what’s needed.

   So we’re on to how to get out of the woods from there. I find it darn complicated. The first card will be the ace, to take up our last legitimate space… But if the next two cards were jacks, we can have pre-arranged to have pure sets of hearts and of spades leading up to the 10 that we can just put on them and then put them back in the space. For 6 of hearts we can break up column 10 to find a place for that 5 to go, for 7 of spades we have an 8 to put it on… what if the 7 came up first? Could that then keep us from handling a heart jack? No, because we can make a space from the spades onto the heart jack… My best prep features 4 of diamonds, 3 of spades, and 2A of diamonds in the space before we take off the heart 4 (just to give an example of the precision of the prep).

   I certainly wouldn’t stake my life on all the secondary combinations working out, but it looks to me like they all work!

   It is a proof on the theory of of “twisty passages all different”, and I’d be delighted to see a more elegant summary.

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   1. Monkey has managed to achieve the victory – but his endgame technique might need a bit of work. In any case, well done Team!

      Also congrats to both Bug and Bart for their efforts for trying to prove the game was mathematically won before the monkey took over. The point of the exercise is not to get an exact answer, but rather to give the reader a feel for when a game is getting close to mathematically won.

      Summary and post-mortem coming soon to a place near u

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