As ever, even though it’s the very first in the series, Online Sudoku Workshop is brought to you by your friendly, neighborhood Crip Dyke.

This is a strange turn, I’m sure. Gender and sudoku? Well, yes. But not yet. Mostly I’m doing this because it’s fun to talk logic and sudoku and just how exciting it is to learn new things, but I will use this as a metaphor in discussing gender later. Come for the sudoku, stay for the sudoku. Come back again later for the gender. If you’re only in it for the gender metaphor, this will certainly be too long for you. In fact, it’s too long for a single post. I’ll start my puzzle-solving with some of the puzzle already done, and still I have to break this up into 3 posts. [Don’t worry, though. If the merchandizing revenues are good enough, I’ll go back and do the prequels so you can see how we arrived at the point we’ll begin our exciting puzzle-solving together.]

On the other hand, however, many people tend to be fascinated by Sudoku. PZ, I’m sure, has many regular readers far more skilled at the puzzles than I. But a great many people seem to run into a wall with sudoku, giving up on “logic” or thinking that sudoku experts are employing some tricky rules that aren’t available to the plebeian puzzle solvers. At its worst, those who are best at sudoku, and even average folk like me, can have our process separated out as some “magical thinking” that is, in fact, not logical at all. We’re going to demystify some of this in a manner not entirely dissimilar to the way the OGW series has tried to demystify certain things about gender, sex, and (to a much lesser extent) sexuality.

You won’t be dazzled by deep thinking here, but that’s part of the point: nearly anyone can do even the hardest puzzles. That most of us don’t isn’t an indication of a common inability to actually follow the logic when presented, or to perform it on one’s own once given an example or two. Like the type of formal logic practiced in university philosophy departments, these examples don’t even teach you any truly new trick. Each new logical “rule” is simply recognizing that one has performed steps 1-72 before, they come out the same way every time, so why not call steps 1-72 a proof, and then, whenever you find yourself needing to go in the same direction from a different starting point, write out step 72 as step 2, with a notation that you’re using the Crip Dyke’s Super-Genius, Completely Indispensable Rule For Sudoku Victory and skip steps 2-71. Anyone can do steps 2-71, but boy can it be tedious.

Imagine, if you will, reproving The Calculus every time instead of just subtracting one from the exponent and multiplying the variable by the original exponent. Yuck. Sure it can be done, and when you see it laid out all the steps do, in fact, make perfect sense. It doesn’t seem complicated. It follows logically. But starting out not having an idea of where you want to go, it’s hard to choose which logical step to take next or to see where these steps might go so that you feel as if you’re making progress. Thus relying on what has been done before (whether invented by you or read somewhere as a “tip” or “trick” or “Super-Genius, Completely Indispensable Rule”) rather than taking each tiny step one at a time, is what allows us to focus on the truly puzzling bits, the bits that are new in this particular group of digits and boxes. This allows the game to stay fresh and fun.

Unfortunately, skipping steps 2-71 makes it appear as if you really are adding rules or tricks on top of the basic game, rather than just using your experience and your memory of how the logic played out last time to move along a little faster. I won’t be skipping as many steps here so that people who aren’t familiar with the puzzles can see how experienced players manage to derive answers from what appears to be insufficient information (and thus appear to derive answers magically, or without logic and/or reason).

The rules of sudoku are easy, though I’ll provide them at the end of the post for those unfamiliar with the puzzles. Here, I just want to provide the barest-bones explanation:

Rule 1: each box may contain one and only one digit. Zero is not available, the only eligible digits are 1 through 9.

Rule 2: the game must finish with each row, square, and column containing one and only one copy of each digit 1 through 9 in its 9 boxes.

Rule 3: if any row already has a 1 (or a 4 or a 9) in it, you may not add another 1 (or 4 or 9) to any box in the same row, and likewise for any square or column.

Rule 4: if you can show based only on the first two rules that a particular digit is the only possible digit that can be written in without violating the first rule, then write it in.

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On to the puzzle

Let’s start a few steps in. We’ll be using images here: for those who have trouble displaying or reading images, they are unfortunately necessary here and there simply is no way to provide captions that contain all the relevant information. Any screen reader would be entirely insufficient: my apologies.

For those looking at the images, however, one can see two different colors used for the digits of the puzzle. The black/bold digits were provided as the puzzle’s initial conditions. The periwinkle digits that are less bold/intense are digits that I’ve filled in. If you want to try to follow along, you can go to websudoku.com and allow it to serve up a random puzzle. Right below the puzzle will be the difficulty level, the puzzle’s identifying number, and blue hyperlink text to “Select a puzzle”. Click that text, choose “hard” as your difficulty and enter in the puzzle number from the image: 3,524,140,531.

I had done a bit already when I decided to use this puzzle as my example.  The 8s in the top left and top right squares, respectively are easily derived by starting with the top right square.

 

Fair warning: the next bit is ultra-pedantic for those who have never encountered sudoku before. We’ll pick up speed very rapidly at the end of the pedantic section, but reading it can also give you a feel for the coordinate system before things move too fast, so you may not want to skip: your choice. 

At the beginning of the puzzle, it does not yet have an 8 in it. Look below it, and you see columns G and I each contain a black/bold 8 (one of the “given” digits with which you begin the puzzle), in squares middle-right and bottom-right respectively. So we can’t have any other 8s in columns G or I. The top right contains the portion of column G that includes boxes A:G -> C:G. Two of those begin the game empty (A:G and C:G), but neither can be filled with an 8 (and we wish to add an 8 to the top right square because of rule 2) because of the 8 in box D:G (the top left box of the middle-right square).

A:I -> C:I are the three boxes on the right of the top right square. None of them are filled by any givens at the start of the game, but we can fill none of these empty boxes with an 8 because they are all part of column I and column I already has an 8 in box G:I (the top right box of the bottom-right square). So no 8 in boxes A:I -> C:I per rule 3.

Column H, that runs through the middle of the top-right square has no 8 in it already. So we can (we must!) locate the top-right square’s 8 somewhere in its portion of column 8. Once there, of course, it will count as the 8 for both the square and the 8 of column H, and of course also for whatever row happens to contain it as well! The top box of column H (the top middle box of the top-right square) is already full. So no putting in 8 in box A:H per rule 1.

The middle box of the top right square (box B:H) is empty. We might be able to put an 8 there, but we’ve only satisfied the no duplicates condition (rule 3) for our square and column. Looking at the row (row B) we see an 8 in the top-middle square, in box B:E. Sadly, B:H is out.

However, since the top right square must contain an 8 in one of its boxes, since it cannot contain more than one digit in any box or more than one 8 in the square as a whole, and since there is only one box left in that square, the 8 for the top-right square must go in box C:H. Per rule 4, right it in.

End martinet-level pedantry.

Now we’re just on to the normal pedantry of trying to communicate logic with the english language.

The 8 in the top left square is easily located because we now have an 8 in rows B and C. While the top left square doesn’t have an 8, in boxes A:A and A:B it does have two given digits. So we’re left with A:C as the only possibility for our top-left square’s 8. All of the blue digits with which we start were derived using these simple rules. But let’s take a look at where our starting position quickly leads:

 

The middle-right box needed a 2 and a 5 to complete it. By looking at the 2 in box B:G, we eliminate any boxes in column G as possible homes for the 2 in the middle-right square. As there are only two boxes left unfilled in the square (not coincidentally since we have 2 digits left to place), we eliminate the empty box in row G (F:G) and put the 2 in F:I. Now F:G is the only empty box, we have only one digit left for the box, and we put in our 5.

Easy peasy, right? These are the baby steps.

 

The left-middle square has no 5, and has only three empty boxes in which to put a 5. The 5 we have in D:E eliminates 2 of the empty boxes, those in row D at the top of that square. You can see that I’ve highlighted the 5 after placing it in E:C.

So simple, right? These rules work really, really well! They call this puzzle hard? What for, this is all obvious! …except what do we do next?

 

Having played out most of the obvious moves, the next is less obvious. But still it is a mere application of the same rules we already have. The only changes in this new image are in the True Neutral middle square where we filled in 1, 6, & 8, starting with the highlighted 1.

In the image before this one, we knew that the bottom three boxes of that square (F:D -> F:F) cannot contain any of the digits from elsewhere in  row F. Ticking off digits, we easily found that only digits 1, 6, & 8 remain to fill those 3 boxes. We’re not surprised that 1, 6, & 8 appear nowhere else in the middle square, since they must appear in row F, don’t appear in the portions of row F in the left-middle and right-middle squares, and thus must appear in the middle square’s bottom row. If they appeared elsewhere as well, we’d have a problem.

We consult the columns. Maybe we’ll find a column from D to F with two of the three digits above or below the middle square, allowing us to place the third digit at the appropriate place in row F?

No such luck. In columns D to F, 1, 6, & 8 appear once each, in separate columns. Each digit has two possible locations in row F. Without pinning one down to eliminate one of the possibilities for one of the other two digits, we’re completely stuck.

So we move on to related work, hoping maybe if we can nail down a 1, 6, or 8 in either the top-middle square or the bottom middle square, we can finish off row F and get some sense of accomplishment. But there’s not enough information about any of these to pin any of them down to a specific box in the top-middle or bottom-middle squares.

One interesting thing that does jump out, however, is that in the top-middle square, we cannot pin the digit 1 down to a box, but we can pin it down to a column. B:A has a 1 in it, eliminating any boxes in row B from the top-middle square’s possible locations for a 1. This also eliminates column F in its entirety since column F’s only vacant box was in row B (box B:F). We already had noticed the 1 in column D (in the bottom-middle square, box G:D). We can’t narrow the location down further than to say that the 1 in the top middle square must be either in box A:E or in box C:E (B:E is full with one of our givens, and even if it wasn’t, row B is out because of the 1 in box B:A). But since A:E and C:E are both in column E, and we know that the top-middle square must have a 1 and so eventually we will fill in one of those boxes with a 1, we cannot put a 1 in F:E without creating an inevitable violation of rule 3 for column E.

The rules are exactly the same, but instead of waiting until we can pin the 1 down to either A:E or C:E, we simply acknowledge that we have enough information to disqualify F:E from containing a 1, even though we have not placed a 1 in any specific box in row F, in column E, or in the middle square. Knowing that we will place one in column E is enough.

Now of the three columns with which row F intersects in that middle square, we’ve eliminated two of them. Only one eligible column left, and so box F:F must contain the 1 for row F (and for the middle square, and for column F). I’ve highlighted that 1 in box F:F.

The 8 and 6 of row F are now quickly filled in, but we still haven’t filled in the 1 in column E that made fixing all three of these digits possible!

Though we’re not adding any rules, we now become acutely aware that rule 4 allows considerably more latitude than we thought: we managed to prove that a 1 must go in box F:F, using only rules 1-3, but with less information than we might have thought was required. We might have thought we had to pin a digit down to a specific box before we could use it to help prove the location of another digit, but we don’t.

The trick is to stop thinking of empty boxes as containing no information. They do contain information: spatial information. Each box has a row, column, and square to which it simultaneously belongs. These multiple, simultaneous memberships are facts about the boxes. Those facts, even if they don’t tell us everything we want to know, certainly tell us some things that can be useful in specific contexts. For instance, it is the common membership in the top-middle square that allow eliminating possible 1s from other boxes in the top-middle square to have any relevance to any other box in the top-middle square. Then it is the vertical relationship between box A:E and box C:E that gives us column information.

And that information is useful only because we can make use of other information: knowing what is NOT contained in a box is as valuable as knowing what is contained in the box. In fact, knowing a particular digit = YES for a particular box, is exactly the same as knowing a particular digit = NO for a particular box, eight times over. Rule out eight digits for a particular box, and you’re done whether or not you bother actually writing in the one digit the box may still legally contain.

Empty boxes are, in fact, full of information: there would be no way to solve anything but the easiest of sudoku otherwise. In fact, even the least practiced sudoku solvers ascribe information to empty boxes. Ruling out eight boxes for a particular digit forces the final box to contain that digit, and we only “know” that we have one box left if all those “empty” boxes contain a bit of information that says, “not that digit” along with the information about square, column, and row membership that tells us where and how far to run with that bit of information.

As a heuristic, we tend to believe if we haven’t made the right mark in the right box, we haven’t assigned any information to it. In fact, we retroactively tend to think of information pertaining to a 1 in column D as information about the 1 that was eventually placed in box F:F. “That’s the information that shows the 1 has to go in this box here,” rather than

That’s the information that allowed me to determine 2 more bits of information: no 1 in boxes A:D and B:D. Those two bits, combined with already known bits allowed me to eventually place a no 1 in box F:E.

Combined with no 1 in box F:D and, because they are full with other digits, a no 1 in boxes F:A -> F:C and F:G -> F:I, logic forced me to place a no 2, no 3, no …, no 9 in box F:F and a yes 1 in F:F, but all that other information still exists, is still attached to other boxes in other locations, and could still be used for other purposes. In fact, all of it was used for other purposes many times along the way.

Why don’t we remember that all the information exists separately of the 1 we finally placed in box F:F?

…because we are teleological creatures. We have a goal: we acquired the information we did for the purpose of putting a digit in a box. Once the goal is accomplished, the information loses its usefulness. Thus we only ever actually think about the information in association with our goals. If our goal is frustrated, humans tend to consider our information gathering (or other efforts towards that goal) to be wasted. But as so-called “basic research” or “fundamental research” repeatedly shows: we often don’t know what use information will be.

Pursuing the information without teleology, without purpose isn’t possible. We do the research because we’re curious or need a job or because even though we don’t know whether this will help us cure cancer or prove the existence of a god, we really, really hope it will. However, we can keep in mind our human failings and tendencies and try hard to hold on to what we learn along the way without disposing of information if it has already served, or if it can no longer serve, the purpose for which we originally sought it out.

Note that this isn’t hard. This isn’t a change of the rules. The three numbers on the right of row F ruled out 5, 9, & 2 for box F:F. The three numbers on the left of row F ruled out 3, 7 & 4. Box F:F had to have some digit, 1 to 9, and collectively the whole row F had to contain all those digits. The row already had 6, and perfectly according to the rules the final 3 boxes not only may contain, but must contain one each of 1, 6, & 8. Since neither F:D nor F:E could house the digit 1, F:F must contain that digit.

Nothing is hard about the logic or about the puzzle.

What is hard is setting aside our human tendency to place information in a particular teleological context. Not being able to localize the 1 of the top-middle square to a particular box, most of us will stop, frustrated. “Can’t figure out where the 1 is in the middle square until I know where the 1 goes in the top-middle square,” is a rule we impose that is not part of the puzzle/game. But self-imposed or not, the inability to think of a semi-localized 1 as useful in exactly the same ways as a fully localized 1 hampers our ability to reach our larger goals: in this case filling in all the boxes. In fact, the 1 that must be in A:E or  C:E (but who knows which?) is exactly as useful as a fully localized 1 for the purpose of ruling out box F:E, they just wouldn’t be useful at all for trying to make deductions about boxes A:G or C:G.

In this sense, the sudoku masters among us are operating with fewer rules, abandoning the de facto requirement so many of us impose on the data themselves, limiting their usefulness to the situations we decide are appropriate, not the situations in which logic dictates they impact an answer.

Ironically, even though we only wanted information about the 1 in the top-middle square in order to rule out boxes in row F, in judging the information only when it is successful in (fully) localizing a digit, we make it less useful for localizing digits when the digit the information successfully localizes isn’t the digit we expected to localize. It’s still progress towards our actual, ultimate goal, but we turn away from the information as unfruitful is it’s not fruitful in exactly the manner we expected.

Okay. That’s enough for now. I’ll move on to the rest of the puzzle in my next post.

 

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Terminology & Rules:

Each square has 9 boxes in a 3 by 3 grid. Each complete puzzle has 9 squares in a 3 by 3 grid. Thus each puzzle has 81 boxes in nine contiguous 3 by 3 grids (or one giant 9 by 9 grid). You can see this in the bold lines separating squares from each other below. But of course, as the squares are joined together, you get continuous rows and columns as well – 9 of each, with each row or column containing 9 Boxes. in a line, horizontal or vertical, respectively. Identifying these units – squares, rows, columns, and the boxes that make up each of the other 3 – is key because the challenge is to fill in a single digit in each empty box such that every square has one each of the digits 1 to 9 contained within. Likewise, every row must have one each of the digits 1 to 9; every column must have one each of the digits 1 to 9.

It is the comparison of which numbers are contained in a row to which are contained in a column to which are contained in a square that allows for correct deductions to be made from the information at hand. Follow four simple rules.

Rule 1: each box may contain one and only one digit. Zero is not available, the only eligible digits are 1 through 9.

Rule 2: the game must finish with each row, square, and column containing one and only one copy of each digit 1 through 9 in its 9 boxes.

Rule 3: if any row already has a 1 (or a 4 or a 9) in it, you may not add another 1 (or 4 or 9) to any box in the same row, and likewise for any square or column.

Rule 4: if you can show based only on the first three rules that a particular digit is the only possible digit that can be written in without violating the first rule, then write it in.

The 9 boxes across the top collectively make up row A. The 9 boxes whose tops each border one box of row A make up row B, and so on. The bottommost row is row I. The 9 boxes on the furthest left collectively make up column A. The 9 boxes who each share a left border with one box from column A collectively make up column B, and so on. The rightmost column is column I.

Specific boxes are named with the row first, then a colon, then the column. The very first box in the very topmost left (in this puzzle it has the digit 4 in black/bold) is box A:A. The first 3 boxes of rows A, with the first 3 of row B & first 3 of row C collectively make up one square which can be seen by the darker lines making up square-boundaries than make up box-boundaries that are internal to a square.

You can, of course, designate a square using coordinates like A:A-C + B:A-C + C:A-C, or abbreviated coordinates such as A:A to C:C. But I’ll routinely simply call that square “top left”. With top, middle, and bottom, and left, middle, and right, we get 9 combinations that easily identify specific squares. Middle-middle will just be called middle. I will attempt to refrain from referring to squares using Lawful Good, Chaotic Neutral, etc. If I fail, I of course blame the corrupting influence of my friends, retaining no responsibility myself.

 

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Where can I play sudoku?

The screen shots above were taken during my puzzle solving at websudoku.com. I have no idea if they’re the best, worst, or otherwise among sudoku sites, but they were tops in an english language search when I went to go ogle puzzle possibilities and I’ve never seen any reason to try anything different.

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To answer a question from Kaintuckee Bob, I have to add a couple more images. Bob’s question is in comment #13. I’ll keep these images left-justified as a visual signal that this isn’t part of the main post.

In the first image to follow, we have the puzzle much closer to the starting conditions. As always, the “givens” are in black/bold. My additions are in periwinkle. I’d say they were in blue, but it’s much more fun to say periwinkle and, TBH, it really does look periwinkle-colored to me.

 

Here we go:

The 8s were added first. Then on the far left side I saw the 1s in columns B and C (D:B and G:C respectively). Even if I hadn’t already added the 8 in box A:C, still the only place to put a 1 in the top-left square is box B:A. The 1s didn’t immediately lead anywhere else, so I grabbed some low-hanging 3s. That started with the middle-right square. The 3 given us in I:H in that bottom row way on the right knocks out any boxes as hosts for a 3 throughout column H. There are only three free boxes not in column H. Two of them, however, are in row F, and in box F:A we have another 3 in our givens. If we can’t have a 3 in row F or in column H, there is only 1 free box in the middle-right square for our 3: E:G.

With help from that 3, we turn to the top-right square. Here there’s a 3 in box A:B that knocks out any boxes in row A. In the top-middle square there is only one free box for a new digit in row C. But the 3 in box H:E in the bottom-middle square knocks out all of column E. With the top row and middle column knocked out as possible homes for 3s in the top-middle square, the only 2 remaining free squares are both in row B. Thus from our given there can be no 3 in the top-right square’s row A boxes and from the semi-localized 3 in the top-middle box we can rule out the top-right square’s row B boxes. Any 3 in the top-right square must be in one of the three boxes of that square that are part of row C. C:H is ruled out by the 3 below in column H at box I:H. C:G is ruled out by the 3 we placed in column G at E:G.  Therefore, only C:I is available for the 3.

The 3 in the bottom-left square (G:C) is utterly easy to get from the givens and not relevant now.

The next part is a bit tricky, so with what I wanted to talk about in mind, I actually deleted some of this before I went back and did it again later in the manner I showed you above. Unfortunately, I didn’t apparently save redacted screen shots, so you’ll have to bear with me learning a different way to find certain information that we were able to figure out under other conditions later.

The very bottom right box of the bottom-right square (and thus the bottom-right of the whole puzzle is I:I. We start with a 1 given us there. Knock out column I from the middle-right and top right squares as possible homes to any 1s for those squares.

Now look at the first 1 we placed, that didn’t immediately lead anywhere. However, it knocks out row B’s free boxes as potential homes to any 1s for the top-right square. With the top-right square’s row B and column I knocked out, there are only two possible homes remaining for a 1. Fortunately for us, they are both in column G. Leaving the top-right square, we move to the middle right. Consider column G knocked out because we know the top-right square’s 1 must be found in that column (even if we don’t know the box). Column I is knocked out by I:I. Row D is knocked out by the 1 given in D:B. The only remaining possible home for a 1 is E:H, so I filled it in.

 

So what next? The 7s in E:E and D:I. They tell us that the 7 in the left-middle square is not in row D or E. There’s only one free space, fill in a 7 in F:B. But how do I get from there to placing the highlighted 7, key to our next moves, in box C:C?

Easy. Complementary semi-localized digits.

WTF you ask?

Yeah. It’s like this: with the 1, 2, 3, 4 & 8 all filled in for the top-left square, we find that square needing these 4 digits: 5, 6, 7 & 9.

Looking nearby for those digits, we find a 6 and a 9 in the top-middle box, both in row C. Since row A is full up (as far as the top-left square is concerned), these two digits must both go in row B. Since there are only two boxes available in row B in that square, even though we can’t tell which box will ultimately contain the 6 and which the 9, with two boxes and two digits, we know that neither box B:B nor box B:C is available for any numbers other than 6 or 9. Now we have two boxes left in the top-left square, both in row C, box C:B and box C:C. The 7 we filled in a moment ago in the square below (box F:B) rules out any other 7s in column B, and thus rules out any 7 in C:B. The only place left for a 7 to go is C:C. Of course, that also means, even though it is not yet shown, that the only place to put a 5 in the top-left square is C:B. With 2.5.7.9. .6. .8.3 as the current status of our row C, we’re well positioned to move on.

Okay, the 5 is in box D:E is highlighted, but first note the 4 we’ve added in the same row. It’s in box D:H. How did we get that? There were only three free spaces in the middle-right square. Two of them were in the same row, row F. We had a 4 given to us in the middle-left square, box F:C. We knock out row F and the only place left for a 4 in that square is D:H. Write it in.

Then the beauty. Again we have complementary semi-localized digits: in the bottom two free spaces of the middle-right square (boxes F:G and F:I) we must place 2 & 5. If 2 ends up in F:G, 5 will end up in F:I and vice versa. But no matter what, we can certainly say that no 5s can exist in the middle square in row F. There’s a 5 at the bottom of column D and at the top of column F. The middle square can only have a 5 in column E, but E:E is taken by one of our givens and F:E is ruled out by the 5 we just semi-localized to row F in the middle-right square. D:E is the only place left to put the 5. I did so and highlighted it.

We always pay attention to somethings more than others. In this case, I actually  missed at first pass that there’s a 2 in column G in box B:G. I could have localized the 2 to box F:I (by ruling out F:G by virtue of B:G), thus forcing the 5 to the specific box F:G. Then the middle-right square would be completely full and we could see the 5 we’re using to localize the middle-square’s 5. I didn’t notice it right away, but that obviously wasn’t important. I had enough info for my purposes, I’d pick up the info about the 2 in B:G and make use of it later.

Kaintuckee Bob had already noticed that:

If you remove from the middle box the 5, 9, and 3 there are at least two places within that box where each number can go.
5 can only go in column E, but could go in row D or F.
9 can only go in row D, but could go in column E or F.
3 can only go in row D, but could go in column D or F.
As soon as you place the ‘5’, the rest follows naturally enough (leaves only one place for 9, which leaves only one place for 3), but it is non-obvious to me how you determined from the initial position that the 5 went in row D as opposed to row F.

Now that we’ve used the invisible 5 located on the bottom of the middle-right square to rule out row F for any 5s, Kaintuckee Bob’s logic takes over and fills in the rest of the answer to his question flawlessly.