Tag: undo

Playing with the “Undo” Awesome Superpower.

In this hand I wanna set the task of winning a game with undo. Normally I would view undoing moves as a cardinal sin – equivalent to Mark Goodliffe’s infamous bifurcation strategy when live-solving Sudoku. but I will allow myself this luxury for an important reason: I needed undo to get my paper published when proving that a certain Spider Solitaire was biased (or at least there was good reason to believe so). Therefore, the U-bomb will not be considered a rude four-letter word and there will be no attempt to encrypt it with a rot-13 cypher.

Our goal is to win the following deal with the luxury of undo. I will not attempt to optimise my score. Also, there will be no cheevo considerations. Note that Microsoft Windows does not offer the player of explicitly restarting a hand: the best we can do is repeatedly press undo until we reach the start (Some folk have complained about this, but I have seen much worse bugs from other servers. Hence, I will avoid the Microsoft-bashing bandwagon for now). At least Microsoft allows undo of every move, including removing a suit or dealing a new row. Other programs may be less luxurious in that regard.

You may have recognised this deal from my previous blog posts. I deliberately did this since a random deal should be easily won with the undo superpower – but since I lost rather badly without undo I would expect this particular deal would not be a walkover.

When playing with undo I assume we have the luxury of card-tracking (this is equivalent to tile-tracking for serious Scrabble players). A card-tracking sheet will indicate the identity of known cards in the starting position. This would look something like the following:

I will use four different colours green/blue/red/black for C/D/H/S respectively. This colour scheme is often used in poker.

SANITY CHECK: the cards in the first four columns are all different suits. If this colour scheme is inconvenient (e.g. for people with red-green colour blindness) please let me know in the comments!

We will start with a warm-up question: what is the minimum number of face-up cards we are guaranteed if undo is allowed and we don’t care about losing 1 point for every move or undo?

NOTE: For purposes of this exercise, we will pretend we have conveniently forgotten about my previous blog posts. This means e.g. the answer is not X, where X is the number of face-up cards when I conceded the game in my previous post.

Game on (19 Feb)

Okay, now we can be justified in complaining about our bad 76,85,67,75. We got 4 kings in round 3 and round 4 yields only one turn-over in column 3 or column 8 (exercise for the reader!). At this stage, the game is almost certainly lost, and although it is possible to search for the best chance (no matter how slight), I would rather discuss the possibility of winning if 85,78,68,79 was allowed – but only because

  • somebody commented on my earlier post, asking if 85,78,68,79 was cheating
  • She is the only person to comment on any of my posts, if we ignore the Evil Villain who writes in Russian and is obviously trying to entice me into watching 80,79,82,78,79,71,82,65,80,72,73,67 videos (I’ve had plenty of likes for my silly stories however).

I prefer to play without 85,78,68,79 because the game can almost always be won (just like Freecell). However, I will allow exceptions if the player is very very smart at math and wants to write a paper on Spider. For instance, if we believe the game is rigged then we need to determine the identity of all face-down cards so we can test Random Move algorithms on a particular hand. At the time of writing, Ninja Monkey can only play well at the one-suit level but this time he has learnt a trick or two at the highest difficulty level.

So assuming this game is lost without 85,78,68,79, our new question is: how easy is it to establish the identity of all face down cards with 85,78,68,79?

If you have any experience and use 85,78,68,79 a lot, you would know the power of empty columns. For instance in the start position if you had a “free cell” you are guaranteed 10 turn-overs with 85,78,68,79 even if the starting hand was five Kings and five Aces! That’s a lot better than 1 turn-over without 85,78,68,79.

As a simple exercise for the reader, go back to (i) an earlier post when I had an empty column (ii) my original start position. How many guaranteed turnovers do I get if 85,78,68,79 is allowed?