The Mathematics of Sudoku Tom Davis [email protected]http://www.geometer.org/mathcircles (Preliminary) October 12, 2008 1 Introduction Sudoku is a (sometimes addictive) puzzle presented on a square grid that is usually 9 × 9, but is sometimes 16 × 16 or other sizes. We will consider here only the 9 × 9 case, although most of what follows can be extended to larger puzzles. Sudoku puzzles can be found in many daily newspapers, and there are thousands of references to it on the internet. Most puzzles are ranked as to difficulty, but the rankings vary from puzzle designer to puzzle designer. Sudoku is an abbreviation for a Japanese phrase meaning ”the digits must remain single”, and it was in Japan that the puzzle first became popular. The puzzle is also known as ”Number Place”. Sudoku (although it was not originally called that) was apparently invented by Howard Garns in 1979. It was first published by Dell Magazines (which continues to do so), but now is available in hundreds of publications. i h g f e d c b a 1 2 3 4 5 6 7 8 9 4 8 9 4 6 7 5 6 1 4 2 1 6 5 5 8 7 9 4 1 7 8 6 9 3 4 5 9 6 3 7 2 4 1 Figure 1: An easy sudoku puzzle At the time of publication of this article, Sudoku is very popular, but it is of course difficult to predict whether it will remain so. It does have many features of puzzles that remain popular: puzzles are available of all degrees of difficulty, the rules are very simple, your ability to solve them improves with time, and it is the sort of puzzle where the person solving it makes con- tinuous progress toward a solution, as is the case with crossword puzzles. The original grid has some of the squares filled with the digits from 1 to 9 and the goal is to com- plete the grid so that every row, column and out- lined 3 × 3 sub-grid contains each of the digits exactly once. A valid puzzle admits exactly one solution. Figure 1 is a relatively easy sudoku puzzle. If you have never tried to solve one, attempt this one (using a pencil!) before you continue, and see what strategies you can find. It will probably take more time than you think, but you will get much better with practice. The solution appears in section 20. Sudoku is mathematically interesting in a variety of ways. Both simple and intricate logic can be 1
Sudoku is a (sometimes addictive) puzzle presented on a square grid that is usually9 × 9, but issometimes16× 16 or other sizes. We will consider here only the9× 9 case, although most of whatfollows can be extended to larger puzzles. Sudoku puzzles can be found in many daily newspapers,and there are thousands of references to it on the internet. Most puzzles are ranked as to difficulty,but the rankings vary from puzzle designer to puzzle designer.
Sudoku is an abbreviation for a Japanese phrase meaning ”thedigits must remain single”, and it wasin Japan that the puzzle first became popular. The puzzle is also known as ”Number Place”. Sudoku(although it was not originally called that) was apparentlyinvented by Howard Garns in 1979. Itwas first published by Dell Magazines (which continues to do so), but now is available in hundredsof publications.
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1 2 3 4 5 6 7 8 9
4 89 4 6 75 6 1 4
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6 3 7 24 1
Figure 1: An easy sudoku puzzle
At the time of publication of this article, Sudokuis very popular, but it is of course difficult topredict whether it will remain so. It does havemany features of puzzles that remain popular:puzzles are available of all degrees of difficulty,the rules are very simple, your ability to solvethem improves with time, and it is the sort ofpuzzle where the person solving it makes con-tinuous progress toward a solution, as is the casewith crossword puzzles.
The original grid has some of the squares filledwith the digits from1 to9 and the goal is to com-plete the grid so that every row, column and out-lined 3 × 3 sub-grid contains each of the digitsexactly once. A valid puzzle admits exactly onesolution.
Figure 1 is a relatively easy sudoku puzzle. Ifyou have never tried to solve one, attempt thisone (using a pencil!) before you continue, andsee what strategies you can find. It will probablytake more time than you think, but you will get much better with practice. The solution appears insection 20.
Sudoku is mathematically interesting in a variety of ways. Both simple and intricate logic can be
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applied to solve a puzzle, it can be viewed as a graph coloringproblem and it certainly has someinteresting combinatorial aspects.
We will begin by examining some logical and mathematical approaches to solving sudoku puzzlesbeginning with the most obvious and we will continue to more and more sophisticated techniques(see, for example, multi-coloring, described in section 9.2). Later in this article we will look at afew other mathematical aspects of sudoku.
A large literature on sudoku exists on the internet with a fairly standardized terminology, which wewill use here:
• A “square” refers to one of the81 boxes in the sudoku grid, each of which is to be filledeventually with a digit from1 to 9.
• A “block” refers to a3 × 3 sub-grid of the main puzzle in which all of the numbers mustappear exactly once in a solution. We will refer to a block by its columns and rows. Thusblockghi456 includes the squaresg4, g5, g6, h4, h5, h6, i4, i5 andi6.
• A “candidate” is a number that could possibly go into a squarein the grid. Each candidatethat we can eliminate from a square brings us closer to a solution.
• Many arguments apply equally well to a row, column or block, and to keep from having towrite “row, column or block” over and over, we may refer to it as a “virtual line”. A typicaluse of “virtual line” might be this: “If you know the values of8 of the9 squares in a virtualline, you can always deduce the value of the missing one.” In the9 × 9 sudoku puzzles thereare27 such virtual lines.
• Sometimes you would like to talk about all of the squares thatcannot contain the same numberas a given square since they share a row or column or block. These are sometimes called the“buddies” of that square. For example, you might say something like, “If two buddies of asquare have only the same two possible candidates, then you can eliminate those as candidatesfor the square.” Each square has 20 buddies.
2 Obvious Strategies
Strategies in this section are mathematically obvious, although searching for them in a puzzle maysometimes be difficult, simply because there are a lot of things to look for. Most puzzles rankedas “easy” and even some ranked “intermediate” can be completely solved using only techniquesdiscussed in this section. The methods are presented roughly in order of increasing difficulty for ahuman. For a computer, a completely different approach is often simpler.
2.1 Unique Missing Candidate
If eight of the nine elements in any virtual line (row, columnor block) are already determined, thefinal element has to be the one that is missing. When most of thesquares are already filled in thistechnique is heavily used. Similarly: If eight of the nine values are impossible in a given square,that square’s value must be the ninth.
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Figure 2: Candidate Elimination and Naked Singles
2.2 Naked Singles
For any given sudoku position, imagine listing all the possible candidates from1 to 9 in each unfilledsquare. Next, for every squareS whose value isv, erasev as a possible candidate in every squarethat is a buddy ofS. The remaining values in each square are candidates for thatsquare. Whenthis is done, if only a single candidatev remains in squareS, we can assign the valuev to S. Thissituation is referred to as a “naked single”.
In the example on the left in figure 2 the larger numbers in the squares represent determined values.All other squares contain a list of possible candidates, where the elimination in the previous para-graph has been performed. In this example, the puzzle contains three naked singles ate2 andh3(where a2 must be inserted), and ate8 (where a7 must be inserted).
Notice that once you have assigned these values to the three squares, other naked singles will appear.For example, as soon as the2 is inserted ath3, you can eliminate the2’s as candidates inh3’sbuddies, and when this is done,i3 will become a naked single that must be filled with8. The positionon the right side of figure 2 shows the same puzzle after the three squares have been assigned valuesand the obvious candidates have been eliminated from the buddies of those squares.
2.3 Hidden Singles
Sometimes there are cells whose values are easily assigned,but a simple elimination of candidatesas described in the last section does not make it obvious. If you reexamine the situation on the leftside of figure 2, there is a hidden single in squareg2 whose value must be5. Although at first glancethere are five possible candiates forg2 (1, 2, 5, 8 and9), if you look in column2 it is the uniquesquare that can contain a5. (The squareg2 is also a hidden single in the blockghi123.) Thus5 canbe placed in squareg2. The5 in squareg2 is “hidden” in the sense that without further examination,it appears that there are5 possible candidates for that square.
To find hidden singles look in every virtual line for a candidate that appears in only one of thesquares making up that virtual line. When that occurs, you’ve found a hidden single, and you can
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immediately assign that candidate to the square.
To check your understanding, make sure you see why there is another hidden single in squared9 infigure 2.
The techniques in this section immediately assign a value toa square. Most puzzles that are ranked“easy” and many that are ranked “intermediate” can be completely solved using only these methods.
The remainder of the methods that we will consider usually donot directly allow you to fill in asquare. Instead, they allow you to eliminate candidates from certain squares. When all but one ofthe candidates have been eliminated, the square’s value is determined.
3 Locked Candidates
Locked candidates are forced to be within a certain part of a row, column or block.
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Figure 3: Locked Candidates
Sometimes you can find a block where the onlypossible positions for a candidate are in one rowor column within that block. Since the blockmust contain the candidate, the candidate mustappear in that row or column within the block.This means that you can eliminate the candidateas a possibility in the intersection of that row orcolumn with other blocks.
A similar situation can occur when a numbermissing from a row or column can occur onlywithin one of the blocks that intersect that rowor column. Thus the candidate must lie on theintersection of the row/column and block andhence cannot be a candidate in any of the othersquares that make up the block.
Both of these situations are illustrated in fig-ure 3. The blockdef789 must contain a2, andthe only places this can occur are in squaresf7andf8: both in rowf . Therefore2 cannot be acandidate in any other squares in rowf , includ-
ing squaref5 (sof5 must contain a3). Similarly, the2 in block ghi456 must lie in column4 so2cannot be a candidate in any other squares of that column, includingd4.
Finally, the5 that must occur in column9 has to fall within the blockdef789 so 5 cannot be acandidate in any of the other squares in blockdef789, includingf7 andf8.
4 Naked and Hidden Pairs, Triplets, Quads, . . .
These are similar to naked singles, discussed in section 2.2, except that instead of having only onecandidate in a cell, you have the same two candidates in two cells (or, in the case of naked triplets,the same three candidates in three cells, et cetera).
A naked pair, triplet or quad must be in the same virtual line.A naked triplet’s three values must
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be the only values that occur in three squares (and similarly, a naked quad’s four values must be theonly ones occurring in four squares). When this occurs, those n squares must contain all and onlythosen values, wheren = 1, 2 or 3. Those values can be eliminated as candidates from any othersquare in that virtual line.
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Figure 4: A Naked Pair
Figure 4 shows how to use a naked pair. Insquaresa2 anda8 the only candidates that ap-pear are a2 and a7. That means that7 must bein one, and2 in the other. But then the2 and7cannot appear in any of the other squares in thatrow, so2 can be eliminated as a candidate ina3and both2 and7 can be eliminated as candidatesin a9.
For a naked pair, both squares must have exactly the same two candidates, but for naked triplets,quads, et cetera, the only requirement is that the three (or four) values be theonly values appearingin those squares in some virtual line. For example, if three entries in a row admit the following setsof candidates:{1, 3}, {3, 7} and{1, 7} then it is impossible for a1, 3 or 7 to appear in any othersquare of that row.
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Figure 5: A Naked Triple
Figure 5 contains a naked triple. In rowasquaresa2, a8 anda9 contain the naked tripleconsisting of the numbers1, 3 and7. Thus thosenumbers must appear in those squares in someorder. For that reason,1 and3 can be eliminatedas candidates from squaresa4 anda5.
Hidden pairs, triples and quads are related tonaked pairs, triples and quads in the same way that hidden singles are related to naked singles.In figure 6 consider rowi. The only squares in rowi in which the values1, 4 and8 appear are insquaresi1, i5 andi6. Therefore we can eliminate candidates2 and6 from squarei1 and candidate3 from i5.
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Figure 6: A Hidden Triple
Remember, of course, that although the threeexamples above illustrate the naked and hiddensets in a row, these sets can appear in any virtualline: a row, column, or block. There is also noreason that there could not be a naked or hiddenquintet, sextet, and so on, especially for versions
of sudoku on grids that are larger than9 × 9.
There is also the possibility of something called a remote naked pair, but we will discuss that later,in section 10.
5 Tuleja’s (or Mr. T’s) Theorem
A few months ago Alan Lipp communicated to me a very interesting observation made by his friend,Greg Tuleja, about Sudoku that can sometimes be very useful in solving a puzzle. Since many peoplehave trouble pronouncing “Tuleja”, Alan called it “Mr. T’s Theorem”. I’ll call it “Tuleja’s Theorem”here.
This is different from all the other strategies in this paperin that the theorem does not usually yield
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information directly that allows us to fill in a square or to eliminate a candidate, but rather provides anobservation that allows us to do so after a small (or sometimes large) number of logical conclusions.In this section we will illustrate and prove Tuleja’s Theorem, and will then go through a series ofexamples where deductions from the theorem allow us to advance our solution. It is hard to knowexactly where to place this section in the article, since some applications are trivial and some deep.
Consider the completed Sudoku puzzle illus-trated in Figure 7. The theorem applies to anyrow of blocks or any column of blocks and statesthat there are only two situations that can occur.
The first is illustrated in the row of blocksabc123, abc345 and abc789. In this case thenumbers1, 2 and3 occur in row1 of the firstblock, row 3 of the second, and row2 of thethird. If this situation occurs where the samethree numbers appear in different rows of allthree blocks, then all the other sets of three num-bers in rows will also repeat. So in this case,we’ve got 645, 456 and 546 as rows in thoseblocks, and also987, 798 and789. The orderof the numbers within the blocks is arbitrary. Inthe example, the same thing occurs in the rowof blocks labeled withg, h andi. If you reflectthis puzzle across the main diagonal, it is obvi-ous that the same thing can occur in columns.
If a set of three numbers is repeated in a row ofany two blocks it is clear that the same set must occur in the third block in the other row. Thenhowever an additional row is filled in, it forces those same numbers to appear as sets of three in rowsin the other blocks. In the example, suppose that the123 rows are filled in but everything else isblank. When we choose645 in theb123 row, it is clear that those numbers then must go in thea456and thus in thec789 rows. That leaves the numbers9, 8 and7 to fill the remaining rows.
That’s pretty simple, but Tuleja noticed that there isonly one other thing that can happen. Look atthe blocks in thedef row. Instead of all three numbers being repeated, there are three pairs of twonumbers that repeat:85, 41 and73 in this example. It will always happen that if the first situationdoes not occur, there will be three sets of two numbers that sit in the rows (or columns, obviously)as in the example.
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Figure 8: Proof of Tuleja’s Theorem
In the column of three blocks under123, thosepairs are69, 48 and35. As an exercise, makesure you can find the numbers that do this in thecolumns of blocks headed by456 and by789.
Since the exact numbers don’t matter, we’ll usenice sets like123 to make the proof easier to fol-low. We will also state the proof in terms of rowsof blocks, but the same reasoning can be used forcolumns of blocks.
Suppose123 appears in the top row of the firstblock. If the123 appears in another row of another block then all three sets ofnumbers repeat in
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all the blocks’ columns. If not, the1, 2 and3 must be split between two rows of the next block. Ifthe split is not three and zero, it has to be two and one, so one of the other rows in the second blockcontains two of the three numbers, say1 and2. Then the12 pair has to be in the remaining row ofthe third block. Figure 8 illustrates what might occur in this situation. In the figure, the numbers1,2 and3 are placed arbitrarily in the rows, but once the rows for1 and2 are determined, the rowswhere the3 must go are fixed as in the figure. There are no vertical dividerlines within the blocksto emphasize the fact that the horizonal order of the entriesmakes no difference in these arguments.
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Figure 9: Proof of Tuleja’s Theorem
Next, we observe that the number completing the12 set in the other rows cannot be3, and it cannotbe the same in both rows or it would force that same number in the top left row. Let us assume thatthe numbers filling the rows are4 and5 as in the left side of Figure 9. This, of course, forces rowsfor 4 and5 in the other blocks as illustrated in the example on the rightof the same figure.
From that, it is clear that whatever pair goes intob23 also has to go intoa89, and that, in turn, forcesthe numbers that go into the remaining slots. This completesthe proof of Tuleja’s Theorem.
Notice also that the other three numbers, the ones that are not in the repeated pairs, have to cyclethrough the remaining slots. In this article I will refer to those unmatched numbers as “floaters”. Inthe examples in Figure 9, the floaters are3, 4 and5.
5.1 Applications of Tuleja’s Theroem
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Figure 10: Example 1: Tuleja’s Theorem
In the examples that follow, in most cases, atleast, other standard techniques can be used tofill in numbers or eliminate candidates. The ex-amples simply illustrate how Tuleja’s theoremmight be used. Generally it is most useful whena fair number of squares have been filled sincesome data is required to figure out which num-bers form the pairs and which numbers are thefloaters.
We will begin with a simple example, illustratedin Figure 10.
Consider the second column of blocks. In thatcolumn, 73 is a pair and it is obvious thatwe have only pairs and not triples repeating.The number6 is obviously a floater (since it ismatched with73 in the center block), making41a pair. That means there must be a4 in eitherh5or i5, but it cannot go inh5, so a4 must go inthe circledi5. There is a faster way to get this,
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of course: the4 in column6 forces a4 in column1 of the central block, forcing a4 in h5 or i5.
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Figure 11: Example 2: Tuleja’s Theorem
Figure 11 illustrates another use of the theorem.
This time we consider the column of blocks un-der789. Since we have the numbers615 in col-umn8 of the top block and we have a75 in col-umn 7 of the middle block, we’ve got sets ofpairs and not triples in that column.
One of the pairs must be either61, 65 or 15 fromthe615 set, and those two numbers must go intothe squares marked with an “x”. Since there isalready a5 in the third block under column9,that eliminates65 and15, so the two “x” slotscontain a1 and a6. Thus a6 goes in the circledsquare and a1 in the other.
When you first try to apply Tuleja’s theorem,you will probably find that it is easy to makemistakes. A good way to practice is to do easySudoku puzzles on the internet or other placewhere you have easy access to the answers andas you work them, search for applications ofTuleja’s theorem before you try anything else.
When you think you’ve got one, check the answer to make sure your reasoning was correct. Whenyour answer is wrong, review your reasoning to learn where you made your mistake.
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Figure 12: Examples 3 and 4: Tuleja’s Theorem
The next two examples come from the same Sudoku puzzle. In thepuzzle on the left in Figure 12consider the lower row of blocks.92 is a pair, so4 and7 are floaters. In the middle lower block wehave563, so56, 53 or 63 must be a pair. But the4 in the lower left block tells us that6 is part ofthe pair, so it is either56 or 63. Since there is a5 in row i, the pair must be63, so a3 belongs in thecircled positioni1.
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After placing the3 in i1, we now have the configuration shown on the right in Figure 12.We aremissing8 and9 in the center block, so positionx must contain an8 or a 9. In the center row ofblocks,89 cannot be a pair since8 and9 go in two rows in the center block. In the center row of thecenter block, we can conclude that the pair is one of:38, 39, 48 or 49. It can’t be38 or 39 becauseof the right block in the central row of blocks. It also can’t be 49 since there is a9 in positionc2.Thus there is an8 in the circledd2 square.
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Figure 13: Example 5: Tuleja’s Theorem
The final example displayed in Figure 13. Con-sider the top row of blocks.
Since there is a2 in column9 there must be a2 inone of the squares marked with anx. Thereforeone of the squares markedy must also contain a2. Because we then have21 in two block rows,21 must be a pair, and7 a floater. (Clearly thereare only pairs and not triples in the row since wehave a721 and a73 in different block rows.)
Since7 is a floater and appears with a3 in therightmost block,3 is not a floater. The439 inthe leftmost block tells us that the pair is either34 or 39. Since3 appears with6 in the top rowof the middle block,6 is a floater, so85 is a pairand the only place the5 can go is in the circledsquare,12.
(Also note that since34 or 39 is a pair, and thereis a4 elsewhere in the middle block, thata5 mustbe9.)
6 X-Wings and Swordfish
An X-wing configuration occurs when the same candidate occurs exactly twice in two rows and inthe same columns of those two rows. (Or you can swap the words “rows” and “columns” in theprevious sentence.) In the configuration on the left in figure14 the candidate3 occurs exactly twicein rowsc andh and in those two rows, it appears in columns2 and7. It does not matter that thecandidate3 occurs in other places in the puzzle.
The squares where the X-wing candidate (3, in this case) can go form a rectangle, so a pair ofopposite corners of that rectangle must contain the candidate. In the example, this means that the3’s are either in bothc2 andh7 or they are in bothc7 andh2. Perhaps the fact that connecting thepossible pairs would form an ‘X’, like the X-wing fighters in the Star Wars movies gives this strategyits name.
In any case, since one pair of two corners must both contain the candidate, no other squares in thecolumns or rows that contain the corners of the rectangle cancontain that candidate. In the example,we can thus conclude that3 cannot be a candidate in squaresa7, f7 or i2.
A swordfish is like an X-wing except that there must be three rows with the three candidates appear-ing in only three columns. As was the case with naked and hidden triples, for a swordfish there is norequirement that the candidate be in all three positions. The candidate must occur three times, oncein each row, but since the occurrences in those rows are restricted to exactly three columns, all the
columns must be used as well. The reasoning is similar to thatused for the x-wing: once you find aswordfish configuration, the candidate cannot appear in any other squares of the three columns androws. Of course you can again swap the words “rows” and “columns” in the description above.
A sample swordfish configuration appears on the right in figure14. In this case, the candidate is7,and the columns that form the swordfish are2, 5 and8. In these columns the value7 appears onlyin rowsa, f andi. One7 must appear in each of these rows and each of the columns, so noothersquares in those rows and columns can contain a7. Thus the candidate7 can be eliminated fromsquaresa1, f1, f6, i6 andi9.
Of course there is nothing special about a swordfish configuration; “super-swordfish” with 4, 5, or6 candidates might be possible. They are rare but not particularly difficult to spot. The “super-swordfish” with4 rows and4 columns is sometimes called a “jellyfish”. If you are playingon astandard9 × 9 grid, the most complex situation you would need to look for would be a jellyfish,since if there were a5 × 5 super-swordfish, there would have to be in addition a4 × 4 or smallerswordfish in the remaining rows or columns. It’s too bad that there’s no real need for the5 × 5super-swordfish, since the usual name in the internet literature is so nice: it is called a “squirmbag”.
But let us see why this is so, for a particular situation. It will be clear how the argument can beextended to others.
Suppose that the candidate1 has been assigned to two squares. Then there are7 rows and columnsin which a1 hasnot been assigned. If we find a (4-row) jellyfish, we would like to show that theremust be a3-column (or simpler) swordfish. Assume that in each of rowsw, x, y andz the number1is a possible candidate only in columnsα, β, γ andδ. It may not be a candidate in all those columns,but in those four rows, it will never be outside those columns.
But that means that in theother three columns, the candidate1 will be missing from the rowsw, x,y andz, so it must appear only in the other three rows. That means there is at most a (3-column)swordfish.
Obviously there is nothing special about the7, 4 and3 in the argument above. If there aren availablerows and columns, and you find ak-row “swordfish”, there must be an(n−k)-column “swordfish”.
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7 The XY-Wing and XYZ-Wing
Sometimes a square has exactly two candidates and we are logically led to the same conclusion nomatter which of the two we assume to be the correct one. An “XY-wing” represents such a situation.This is a sort of “guess and check” strategy, but it only looksahead one step so it is easily done by ahuman.
f
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c
b
a
1 2 3 4 5 6
XY YZ
XZ *
Figure 15: XY-Wing
In the configuration in figure 15, suppose that there are two possible can-didates in squaresb2, b5 ande2. In the figure, the candidates are justcalledX , Y andZ. Consider the contents of squareb2. If X belongs inb2, then there must be aZ in e2 and thereforeZ cannot be a candidate ine5. But the other possibility is thatb2 contains aY . In this case,b5 mustbeZ and again,e5 cannot beZ. Thus no matter which of the two valuesgoes inb2, we can deduce thatZ is impossible in squaree5.
Now consider the configuration on the left in figure 16. If either X orY belongs in squarea2, Z cannot be a candidate in the three squaresindicated by asterisks. Similarly, in the configuration in the center in thesame figure,Z cannot be a candidate in two more squares indicated by asterisks.
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XY XZ
YZ * * *
+c
b
a
1 2 3 4 5 6
* XY * XZ
YZ
→c
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1 2 3 4 5 6
* XY * XZ
YZ * * *
Figure 16: XY-WingObviously, the two configurations on the left in figure 16 can be combined to make the configurationon the right in the same figure whereZ can be eliminated as a candidate in any of the squares markedwith an asterisk.
An example of an XY-wing in an actual puzzle appears in figure 18. Notice that in squaresd8 andf7 (both in the same block,def789) and in squared1 we have candidates{8, 9}, {3, 9} and{8, 3},respectively. No matter which of the two values appear ind8, a 3 must appear in eitherd1 or f7.Because of this, we can eliminate3 as a candidate from squaresd7, f1 andf2.
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b
a
1 2 3 4 5 6
YZ
XYZ * * XZ
Figure 17: XYZ-Wing
The XYZ-wing is a slight variation on the XY-wing. If you can find asquare that contains exactly the candidatesX , Y andZ and it has twobuddies, one of which has only candidatesX andZ while the other hasonly candidatesY andZ, then any square that is buddies withall three ofthose squares cannot admit the candidateZ. On the left side of figure 17this situation occurs, and candidateZ can be eliminated from squaresb2andb3.
8 XY-Chains
There is another way to look at the XY-wing. We can think of theexample in figure 15 as a sortof chain frome2 to b2 to b5. If the value ine2 is notZ, then it isX . Sinceb2 is a buddy ofe2this would forceb2 to beY , and sinceb5 is a buddy ofb2, thenb5 would be forced to beZ. Thereasoning can be reversed if we assumeb5 is notZ and we can conclude thatb2 must beZ. Thusexactly one ofb2 or b5 must beZ and the other is not. Any squares that are buddies of bothb2 andb5 (only e5 in the example in figure 15) cannot possibly beZ.
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Figure 18: XY-Wing Example (left), XY-Chain (right)
The interesting observation we can make from this is that there is no need for such a chain to be onlytwo steps long: it can be as long as we want, as long as the same candidate appears at both ends.When this occurs, we can eliminate that candidate from any ofthe squares that are buddies of bothsquares that are the endpoints of such a chain. We will call these XY-chains.
Consider the situation on the right in figure 18. Look at the following chain of squares linked inexactly the same way that the three squares in an XY-wing are linked: i8− e8− e2− e5− g5− h4.Each of the squares is a buddy of the next; each square contains only two possible candidates, andfinally, those two candidates match with one of the two candidates of the squares on either side ofit in the chain. Finally, the left-over candidate (1 in this example) is the same in squaresi8 andh4.By stepping through the chain we can conclude that ifi8 is not1 thenh4 is, and ifh4 is not1 theni4 is. Thus eitheri8 or h4 must be1, so squares that are buddies of bothi8 andh4 cannot be1 andwe can eliminate1 as a candidate from squaresh7, h8 andi4.
We can also note that naked pairs (discussed in section 4) area simpler version of this idea, butwhile an XY-wing is an XY-chain with three links, a naked pairis a chain with only two links. Butthe naked pair that contains a1 and a2 in each square is like two of these chains: one with1 at theendpoints and one with2 at the endpoints, so all1’s and2’s can be eliminated from squares that arebuddies of the two that make up the square. Similar reasoningcan be applied to see that a nakedtriple like 12, 23 and31 can be thought of as three different XY-chains where each pair is a differentset of endpoints.
9 Coloring and Multi-Coloring
Coloring and multi-coloring are techniques that eliminatecandidates based on logical chains ofdeduction. The coloring method, especially, is simple enough that it can be done by hand.
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9.1 Simple Coloring
Consider the example in figure 19 where we consider a few squares that admit the candidate1. Let’sassume for now that these are the only possible locations forthe candidate1 in the puzzle. Certainvirtual lines contain exactly two places where the candidate1 can go: rowsb andi, columns3 and6, and blockdef123. In each of these virtual lines, exactly one of the possible squares can contain a1 and once it is selected, the other cannot.
But this creates a sort of “chain” iff1 contains1, thene3 must not, and sincee3 must not,b3 must,sob6 must not,i6 must, andi9 must not. If, on the other hand,f1 does not contain a one, the sameseries of virtual line interactions will force an alternating set of conclusions and every square in thechain will be forced to have the opposite value.
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a
1 2 3 4 5 6 7 8 9
1+ 1–
1–1+
1+ 1–
Figure 19: Simple Coloring
In the figure we’ve marked the squares with+and− according to the assumption thatf1 doescontain a1, but of course it may be the case thatf1 does not contain1, and all the+ and− signswould be interchanged. Rather than using the“+” and “−” characters that could imply pres-ence or absence of a value it is better simply toimagine coloring each square in the chain blackor white, and either all the black squares have a1and all the white squares do not, or the opposite.
In most situations, not all of the squares in a puz-zle that admit a candidate can be colored: onlythose squares where the candidate appears ex-actly twice in some virtual line can be part of achain. If there are three candidates in a row, forexample, and one of them is colored, we cannotimmediately assign colors to the two others inthat row although we may be able to do so later,based on other links in the chain.
Suppose now that for some candidate you havediscovered such a chain1 and have colored it in this alternating manner.
It may be that there are additional squares where the candidate could possibly occur that do nothappen to lie in the colored chain. In figure 19, suppose square f1 is colored black and so squarei9 must be colored white. Consider the squaref9 that lies at the intersection off1’s row andi9’scolumn. Sincef1 andi9 have opposite colors, exactly one of themwill contain a1, and thereforeit is impossible for the squaref9 to contain a1, so1 can be eliminated as a possible candidate insquaref9.
This is probably easier to see with the concrete example displayed on the left in figure 20 where weconsider1 as a possible candidate. In rowd, d1 andd5 are the only occurrences of candidate1, sowe colord5 black andd1 white. Butd1 andf3 are the only possibilities for1 in block def123, sosinced1 is white,f3 is black. By similar reasoning, sincef3 is black,g3 andf8 are white. Sincef8is white,e7 is black, and sincee7 is black,c7 is white. That’s a pretty complicated chain, but here’swhat we’ve got: black:{d5, f3, e7} white: {d1, g3, f8, c7}. A grid that displays just the coloredsquares appears on the right in figure 20.
1For astute readers, it may not really be a chain, but it could be a tree, or even have loops, as long as the black/white
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BB WW
Figure 20: Simple Coloring
Squarec5 is at the intersection ofc7’s row andd5’s column, butc7 is white andd5 is black, so1cannot be a candidate in squarec5. Similarly, squareg5 is in the same row asg3 and same columnasd5 which are white and black, respectively, so1 also cannot be a candidate ing5.
Note that after making eliminations like this, it may be possible to extend the coloring to additionalsquares although that is not the case in figure 20.
There is nothing special about a row-column intersection. Any time two oppositely-colored squares“intersect” via virtual lines of any sort in another square,the candidate can be eliminated as a pos-sibility in that square. Sometimes a candidate can be assigned immediately to a square on the basisof coloring. Suppose that two squares of a chain are the same color, but lie in the same virtual line.If squares of that color contained the candidate, then two squares in the same virtual line wouldcontain it which is impossible, so the candidate can be immediately assigned to all squares of theother color. This situation is shown in figure 21. In that figure, the board on the left is colored forcandidate8 as shown on the right. Note that the black squares conflict in afew places: in column1, row c, and in blockabc123. This means that the candidate8 can be eliminated from every squarecolored black.
9.2 Multi-coloring
Sometimes a position can be colored for a particular candidate and multiple coloring chains exist,but none of them are usable to eliminate that candidate from other squares. If there are multiplechains, it is worth looking for a multi-coloring situation.
Consider the puzzle in figure 22. Assume that in the parts of the puzzle that are not shown there areno other places that the candidate1 can occur. When this diagram is colored, there are two coloringchains. Instead of using words like “black” and “white” we will used letters, likeA, B, C, a, b andc
where theA anda represent opposite colors, as do theB andb, C andc and so on. In figure 22 rowsa andc and in column3 there are only two possible locations for candidate1.
alternation is preserved.
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Figure 21: Simple Coloring: Chain Conflict
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1B
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1 1 1
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1 1
Figure 22: Multi-coloring
When this grid is colored, it will look something like this:squaresc1 and f3 have colorA and squarec3 has colora. Squarea2 has colorb and squarea5 has colorB. (Notethat the colors assigned are arbitrary. All that matters is thatsquaresc1 andf3 have the same color that is the opposite ofc3 and thata2 anda5 have opposite colors that are differentfrom the other assigned colors. Note that none of the othersquares with1 as a candidate can be colored, since all arein virtual lines with more than two squares that potentiallycould contain the candidate1.
If we consider the color “a” as standing for the sentence: “Ev-ery square containing the colora contains a1,” and so on,then we can write little logical expressions indicating therelationships among the various colorswhen they are interpreted as sentences. The obvious ones areof the form: “a = ¬ A” or “ A = ¬ a”(where the logical symbol “¬” means “not” and the symbol “=” means “is logically equivalent to”).In other words, ifa is true thenA is false, and vice-versa.
In this section, we will be performing what is known as boolean algebra2 on expressions involving“sentences” such asa, A, b, B and so on.
Although the values of non-opposite colors do not necessarily have anything to do with each other,in figure 22, the paira andb, for example, are linked, since they occur in the same block.If a is true,thenb cannot be, and vice-versa, but it may be true that botha andb are false. We will express thisrelationship as “a⊼b” and read it as “a excludesb”. Obviously, if a⊼b thenb⊼a3. Also, it is obviousin the configuration in figure 22 thatb⊼A.
Another way to think ofa⊼b is as “If a is true then so isB.” and at the same time, “Ifb is true thenso isA.” If a⊼b then at least one ofa or b must be false. Equivalently, ifa⊼b os true then at least
2See section 19 for a text on boolean algebra.3The truth table fora⊼b is equivalent to the “nand” (“A nand B” is the same as “not(A and B)”) logical operator that is
heavily used in computer hardware logic designs and is sometimes represented by the symbol “⊼”.
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one ofA or B must be true. This means that any square that is a buddy of two squares coloredAandB must not allow the candidate since one of the two squares coloredA or B must contain thecandidate. In figure 22, this means that1 cannot be a candidate in squaref5.
To condense all of the above into a single statement, we know that if a⊼b for some candidate thenany square that is buddies of bothA andB cannot contain that candidate.
Let us begin with a simple example of multicoloring displayed in figure 23. On the left is thecomplete situation, and on the right is a simplified version where only squares having the number6as a possible candidate are displayed.
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Figure 23: Simple Multi-coloring Example: Coloring on Right
In row b and column4 there are only two squares that admit candidate6, so we have colored allthose squares withC andc. In the same way, the two squares in column6 are colored withB andb,while A anda are used to color four squares that share, in pairs, rowg, column9 and blockghi789.
In this example, we note thata⊼b because instances of them lie in squaresf7 andf9 which arebuddies. Because of this, any square that is a simultaneous buddy of a square coloredA and of onecoloredB cannot allow6 as a candidate. In the figure, squarea1 is buddies of bothg1 anda7, so6cannot be a candidate in squarea1, so we can see in full puzzle on the left that3 can be assigned tosquarea1.
In the figure,B⊼c (since they lie in the blockabc789) andc⊼a (since they lie in columna) as well,but there are no squares that are simultaneous buddies of squares coloredb andC or of C andA sowe cannot use those exclusion relationships to help solve this puzzle.
Next, we will look at a multicoloring situation that is quitecomplex because much more can be donewith multicoloring. In complex situations, there may be many independent color chains with colorsA anda, B andb, C andc, and so on. When that occurs, we need to look for consequencesof thefollowing inference:
If a⊼b andB⊼c thena⊼c.
It’s not hard to see why: Ifa is true,b is not, soB is true, and the second exclusion implies thatc isnot. The reasoning is trivially reversed to show that ifc is true thena is not, so we obtaina⊼c.
Thus to do multi-coloring for a particular candidate, proceed as follows:
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• Construct all possible color chains for the diagram.
• Find all exclusionary relationships from pairs of colored squares that are buddies.
• Take the collection of relationships and complete it to its transitive closure using the idea thatif (a⊼b andB⊼c) thena⊼c.
• For every exclusionary pair in the transitive closure, find buddies of squares colored withcolors opposite to those in the pair, and eliminate the candidate as a possibilty from all ofthem.
Figure 24: Complex Multi-coloring Example: Coloring on Right
Let’s look at a very complex multi-coloring application. See figure 24 where only the presence ofsquares that admit the candidate9 are marked (all, of course, must admit other candidates). Ontheleft is the complete grid and on the right is a simplified version where only the squares admittingcandidate9 are shown, and all of the color chains are displayed. It is an excellent exercise to lookat the diagram on the right to make certain that you understand exactly how all the color chains areconstructed.
The next step in the application of multi-color is to find all the exclusionary pairs, and the initial listis displayed in table 1. Note that the “⊼” operation is commutative, so if you thinka⊼b should be inthe list and it is not, be sure to look forb⊼a as well.
Table 1: Direct Exclusions
A⊼E a⊼b D⊼e A⊼d A⊼C
A⊼c b⊼E A⊼D C⊼d
From these initial exclusions, a number of others can be deduced. For example, froma⊼b andA⊼d
we can conclude thatb⊼d. Note that to make this implication, we are implicitly usingthe fact thata⊼b andb⊼a are equivalent. Similarly, sinceA⊼E andD⊼e we can conlude thatA⊼D. See if youcan discover others before reading on.
In fact, if we make all such deductions, and then all deductions from those, and so on, there aretwelve additional exclusions that we find, and they are displayed in table 2
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Table 2: Derived Exclusions
b⊼d b⊼C b⊼c b⊼D C⊼e
A⊼A A⊼b b⊼e b⊼b A⊼e
E⊼b A⊼D
For most of them, we need to look for generalized intersections of the opposites of the exclusionaryvalues. For example, sinceA⊼e and there is ana in c1 and aE in a4, then9 cannot be a candidate insquaresc5 or c6. Also, since we’ve gotA⊼A andb⊼b we can conclude thata andB are true since ifit is impossible for a statement and itself to be true, the statement must be false.
10 Remote Naked Pairs
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a
1 2 3 4 5 61 2 1 2
1 2
*1 2
Figure 25: RemoteNaked Pairs
This technique is related both to naked pairs, simple coloring, and XY-chains.Sometimes there will be a series of squares with the same two candidates,and only those two candidates that form a chain in the same waythat weconsidered chains of single candidates in section 9.
Consider figure 25. In it, we have naked pairsa3 anda6, a6 andc5, andc5andd5, where each pair shares a row, column or block. By an alternatingcoloring argument, it should be clear thata3 andd5 are effectively a nakedpair: one of the two must contain the candidate1 and the other,2. Thus thesquare marked with an asterisk,d3 cannot contain either a1 or a2, and theycan be eliminated as possible candidates for squared3.
11 Unique Solution Constraints
If you know that the puzzle has a unique solution, which any reasonable puzzle should, sometimesthat information can eliminate some candidates. For example, let’s examine the example in figure 26.
In row c, columns4 and6, the only possible candidates are1 and2. But in rowg, columns4 and6,the candidates are1, 2 and8. We claim that8 must appear ing4 or g6. If it does not, then the fourcorners of the squarec4, c6, g4 andg6 will all have exactly the same two candidates,1 and2, so wecould assign the value1 to either pair of opposite corners, and both must yield validsolutions. Ifthere is a unique solution, this cannot occur, so one ofg4 or g6 must contain the value8. But if that’sthe case, squarei4 cannot be8, so the candidate8 can be eliminated from squarei4. In addition,since eitherg4 or g6 must be8, g8 cannot be8 since it is in the same row as the other two.
In the same figure, a similar situation appears in another place. See if you can find it. Hint: itcolumn-oriented instead of row-oriented.
Let’s go back and see exactly what is going on, and from that, we’ll be able to find a number oftechniques that are based on the same general idea. Figure 27shows a basic illegal block. Anythingat all can occur in the squares that are not circled, but note that an assignment of a2 or a7 to anyof the circled squares forces the values of the others in an alternating pattern. But any of the squarescan be assigned a2 or a7 and the resulting pattern will be legal, and this means thereare two validsolutions to the puzzle.
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Figure 26: Uniqueness Constraint
This means that if some assignment causes an il-legal block to be formed, that assignment is im-possible, and we can use that fact to eliminatecertain possibilities, as we did in the example infigure 26. Note that the four corners must notonly form a rectangle, but they must be arrangedso that two pairs of adjacent corners must liewithin the same blocks. If the four corners lie infour different blocks, then constraints from thosedifferent blocks can force the values one way orthe other.
Now let us examine some variations of thistheme. In the rest of the examples in this section,we’ll assume nothing about the empty squares.They may have values assigned to them or maybe empty. On the left in figure 28 we see some-thing that is almost the same as what we saw infigure 27 and the only thing that makes it legalis the presence of the possibility of a3 in squareb1. If it is not a3, then we would have the illegal
block, so there must be a3 in squareb1. Note that if, in the figure, squareb1 had contained the pos-sibilities 2, 3, 4, and7, at least the two possibilities2 and7 could still be eliminated as candidates,so only a3 or a4 could be entered in that square.
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Figure 27: An illegal block
The example in the middle of figure 28 is similar to the originalexample in this section except that the additional number occurs intwo different blocks instead of one. As before, at least one of thosesquares must contain the number (3 in this case), so the value3can be eliminated from any of the other squares in that row (rowb, in this case), but not in either of the blocks, since the one that isforced to be3 might be in the other block.
The example on the right in figure 28 illustrates another sortofdeduction that could be made. We know that at least one ofb1 andc1 must contain a number otherthan a2 or a7, but we don’t know which one. If we think of the combination ofthe two squares asa sort of unit, wedo know that this unit will contain either a3 or a4. This two-square unit, togetherwith squarea3 (which has3 and4 as its unique possibilities) means that no other square in the blockabc123 can contain a3 or a4. If the 34 square had been ina1 we could in addition eliminate3 and4 as candidates from any of the other squares in column1 outside the first block.
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Figure 28: Uniqueness ConsiderationsNote that we can have both a3 and a4 in either or both squaresb1 andc1 in this example on theright. As long as both occur, the argument holds. Also note that if the3’s and4’s appeared in rowband the entries in rowc were both27, and the34 square were in rowb we could eliminate any more3’s and4’s in that row.
This method is almost like guessing, but it is aform of guessing that is not too hard for a hu-man to do. There are various types of forcingchains, but the easiest to understand works onlywith cells that contain two candidates.
The idea is this: for each of the two-candidatecells, tentatively set the value of that cell to thefirst value and see if that forces any other two-candidate cells to take on a value. If so, findadditional two-candidate cells whose values areforced and so on until there are no more forcingmoves. Then repeat the same operation assum-ing that the original cell had the other value.
If, after making all possible forced moves withone assumption and with the other, there existsa cell that is forced to the same value, no matterwhat, then that must be the value for that cell.
As an example, consider the example in fig-ure 29, and let’s begin with cellb3 which cancontain either a1 or a3. If b3 = 1, theni3 = 3,soh2 = 9, soh4 = 1. On the other hand, ifb3 = 3 theni3 = 1 soi4 = 9 soh4 = 1. In other words,it doesn’t matter which value we assume thatb3 takes; either assumption leads to the conclusion thath4 = 1, so we can go ahead and assign1 to cellh4.
Note that XY-wings that we considered in section 7 are basically very short forcing chains.
13 Guessing
The methods above will solve almost every sudoku puzzle thatyou will find in newspapers, and infact, you will probably hardly ever need to use anything as complex as multi-coloring to solve suchpuzzles. But there do exist puzzles that do have a unique solution, but cannot be solved using all themethods above.
One method that will always work, although from time to time it needs to be applied recursively,is simply making a guess and examining the consequences of the guess. In a situation that seemsimpossible, choose a square that has more than one possible candidate, remember the situation,make a guess at the value for that square and solve the resulting puzzle. If you can solve it, great—you’re done. If that puzzle cannot be solved, then the guess you made must be incorrect, it can beeliminated as a candidate for that square, and you can returnto the saved puzzle and try to solve itwith one candidate eliminated.
Obviously, when you try to solve the puzzle after having madea guess, you may arrive at anothersituation where another guess is required, in which case a second level of guess must be made, andso on. But since the method always eventually eliminates candidates, you must arrive at the solution,if there is one. In computer science, this technique is knownas a recursive search. Figure 30 is anexample of such a puzzle that cannot be cracked with any of themethods discussed so far except for
20
guessing. The solution to this puzzle can be found in section20.
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1 2 3 4 5 6 7 8 91 2
5 61 2 1 2
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8 71 2 3
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34 5 6 2 8
1
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9 2 6 83
47
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1 347 9
Figure 30: A Very Hard Sudoku Puzzle
Guessing is a direct logical approach in thesense that we assume something is true, such as“squarei3 is a 3” and then we follow the con-sequences of that to see if it results in a contra-diction. A purely formal approach can also betaken. For every one of the81 squares, there is aset of18 statements about the square that can beeither true or false. These have the form “squarex has the valuei” or “squarex does not have thevaluei”, wherei is a number between1 and9.There are18 × 81 = 1458 of these assertions.
From the initial configuration, we know someare true and some are false, and it is possible tomake logical deductions from them. If we knowthat squarex is 4 then we know that ifi 6= 4 thestatements “x is i” are false, and the statements“x is noti” are all true except wheni = 4. Simi-larly, we can make logical conclusions about thebuddies ofx if we know that squarex has a cer-tain value. If all the buddies ofx cannot have avalue, thenx must have that value, and there are a few other similar rules of inference we can use toassign truth values to the1458 propositions. In principle, a person can search the list over and overand see if any of these rules of inference can be applied, and if so, apply them to assign yet moretruth values. Repeating this over and over will almost always succeed in solving the puzzle.
The method described above does not involve guessing, and works directly forward using onlylogical consequences, but it is not a reasonable way for a human to solve the puzzle. Computerprograms are great at this sort of analysis, but they may needto apply thousands of such inferencerules to take each step forward.
14 Equivalent Puzzles
There is no reason that the numbers1 through9 need to be used for a sudoku problem. We never doany arithmetic with them: they simply represent9 different symbols and solving the puzzle consistsof trying to place these symbols in a grid subject to various constraints.
In fact, the construction of a valid completed sudoku grid isequivalent to a graph-theoretic coloringproblem in the following sense. Imagine that every one of the81 squares is a vertex in a graph, andthere is an edge connecting every pair of vertices whose squares are buddies. Each vertex will beconnected to20 other vertices, so the sudoku graph will consist of81 · 20/2 = 810 edges. Findinga valid sudoku grid amounts to finding a way to color the vertices of the graph with nine differentcolors such that no two adjacent vertices share the same color.
Since the symbols do not matter, we could use the lettersA throughI or any other set of ninedistinct symbols to represent what is essentially the same sudoku puzzle. If we take a valid grid andexchange the numbers1 and2, this is also essentially the same puzzle. In fact, any permutation ofthe values1 through9 will also yield an equivalent puzzle, so there are9! = 362880 versions ofevery puzzle available simply by rearranging the digits.
21
If you were trying to calculate how many grids there are, a good approach would be to assume thatthe top row consists of the numbers1 through9 in order, to count the number of grids of that typethere are, and then to multiply that result by9! = 362880.
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5 4 32
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147 8
Figure 31: Essentially Equivalent Puzzles
In addition to simply rearranging the numbers, there are other things you could do to a puzzle thatwould effectively leave it the same. For example, you could exchange any two columns (or rows)of numbers, as long as the columns (or rows) pass through the same blocks. You can exchange anycolumn (or row) of blocks with another column (or row) of blocks. Finally, you can rotate the entriesin a grid by any number of quarter-turns, or you could mirror the grid across a diagonal.
Figure 31 shows some examples. If the puzzle on the left is theoriginal one, the one in the centershows what is obtained with a trivial rearrangement of the digits 1 through9 (the entries1, 2, . . . ,9 are replaced in the center version by4, 8, 1, 6, 5, 3, 7, 2 and9, respectively). The version on theright is also equivalent, but it is very difficult to see how itis related to the puzzle on the left.
One obvious mathematical question is then, how many equivalent puzzles are there of each sudokugrid in the sense above?
Another interesting mathematical question arises, and that is the following: given two puzzles thatare equivalent in the sense above, and given a sequence of steps toward the solution of one that areselected from among those explained in earlier chapters, will those same steps work to solve theother puzzle. In other words, if there is a swordfish positionin one, will we arrive at a differentswordfish in the other? The answer is yes, but how would you go about proving it?
Notice that the puzzle on the left (and in the center) in figure31 is symmetric in the sense that if youmark the squares where clues appear, they remain the same if the puzzle is rotated by 180 degreesabout the center. Other versions of symmetric puzzles couldbe obtained by mirroring the cluesquares horizontally or vertically. Most published puzzles have this form. This doesn’t necessarilymake them easier or harder, but it makes them look aesthetically better, in the same way that mostcrossword puzzles published in the United States are also symmetric.
Another interesting question is this: given a symmetric puzzle, how many equivalent versions of itare there?
22
15 Counting Sudoku Grids
A sudoku grid is a special case of a9 × 9 latin square4 with the additional constraint barring dupli-cates in the blocks. There are a lot of9 × 9 latin squares:
5524751496156892842531225600.
Bertram Felgenhauer (see the first article referenced in section 19) has counted the number of uniquesudoku grids using a computer, and his result has been verified by a number of other people, andthat number turns out to be much smaller, but also huge:
6670903752021072936960 = 220385171277042679711.
The number above includes all permutations of the numbers1 through9 in each valid grid, so if wedivide it by9! we obtain:
18383222420692992 = 21334277042679711,
which will be the number of inequivalent grids.
16 Magic Sudoku Grids
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1 2 3 4 5 6 7 8 9
4 7 29 8
1 66 5 2 7 8
8 97 8 2 3
8 12 9 4 68
Figure 32: Magic Sudoku PuzzleA latin square has all the digits in each row and column. A “magic square” is a latin square whereeach diagonal also contains all the digits. Is there such a thing as a magic sudoku grid? The answeris yes, and in fact there are a lot of them:4752, in fact, if we assume that the main diagonal containsthe digits in a fixed order. All4752 of the grids can be completed, and all of them in multiple ways.The puzzle presented in figure 32 is a standard sudoku puzzle,except that it is easier since it requiresthat each of the diagonals contains all the digits from1 to 9.
4A latin square is a grid where the only constraint is that there be no duplicate entries in any row or any column.
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17 Minimal Sudoku Puzzles
What is the minimal number of locations must be filled in an otherwise empty grid that will guaranteethat there is a unique solution? As of the time this paper was written, the answer to that question isstill unknown, but examples exist of puzzles that have only17 locations filled and do have a uniquesolution. Figure 33 shows such a puzzle on the left. Althoughthis puzzle contains the minimumamount of information in terms of initial clues, it is not, infact, a difficult puzzle. The puzzle to theright in the same figure contains18 clues, and is symmetric. This is the smallest known size for asymmetric sudoku puzzle.
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8 3 26 1
56 31 7
2 86
8 25 3 9
Figure 33: Minimal Puzzles
18 Constructing Puzzles and Measuring Their Difficulty
The difficulty of a sudoku puzzle has very little to do with thenumber of clues given initially.Usually, the difficulty ratings are given to indicate how hard it would be for a human to solve thepuzzle. A computer program to solve sudoku puzzles is almosttrivial to write: it merely needs tocheck if the current situation is solved, and if not, make a guess in one of the squares that is not yetfilled, remembering the situation before the guess. If that guess leads to a solution, great; otherwise,restore the grid to the state before the guess was made and make another guess.
The problem with the guessing scheme is that the stack of guesses may get to be twenty or thirtydeep and it is impossible for a human to keep track of this, buttrivial for a computer. A much moretypical method to evaluate the difficulty of a puzzle is relative to the sorts of solution techniques thatwere presented in the earlier sections of this article.
In this article, the techniques were introduced in an order that roughly corresponds to their difficultyfor a human. Any human can look at a row, column or block and seeif there is just one missingnumber and if so, fill it in, et cetera.
So to test the difficulty of a problem, a reasonable method might be this. Try, in order of increasingdifficulty, the various techniques presented in this article. As soon as one succeeds, make that move,and return to the beginning of the list of techniques. As the solution proceeds, keep track of thenumber of times each technique was used. At the end, you’ll have a list of counts, and the moretimes difficult techniques (like swordfish, coloring, or multi-coloring) were used, the more difficultthe puzzle was.
The rankings seen in newspapers generally require that the first couple of rankings (say beginningand intermediate) don’t use any technique other than those that yield a value to assign to a square oneach move. In other words, they require only obvious candidates, naked and hidden singles to solve.
24
Published puzzles almost never require guessing and backtracking, but the methods used to assign adegree of difficulty vary from puzzle-maker to puzzle-maker.
With a computer, it is easy to generate sudoku puzzles. Firstfind a valid solution, which can bedone easily by assigning a few random numbers to a grid and finding any solution. Next startremoving numbers (or pairs of numbers, if a symmetric puzzleis desired) and try to solve theresulting puzzle. If it has a unique solution, remove more numbers and continue. If not, replacethe previously-removed numbers and try again until a sufficient number of squares are empty. Thepuzzle’s difficulty can then be determined using the techniques described above. The entire processwill take only a fraction of a second, so one would not need to wait long to obtain a puzzle of anydesired degree of difficulty.
19 References
At the time of writing this article, the following are good resources for sudoku on the internet and inbooks:
• http://www.afjarvis.staff.shef.ac.uk/sudoku/: Felgenhauer’s paper that countspossible sudoku grids.
• http://www.geometer.org/puzzles: You can download the source code for the author’sprogram that solves sudoku puzzles and can generate the graphics used in this article.
• http://www.websudoku.com/: This page by Gideon Greenspan and Rachel Lee generatessudoku games of varying degrees of difficulty and allows you to solve the problem online.
• http://angusj.com/sudoku/: From this page you can download a program written byAngus Johnson that runs under Windows that will help you construct and solve sudoku prob-lems. In addition, the page points to a step-by-step guide for solving sudoku, similar to whatappears in this document.
• http://www.sadmansoftware.com/sudoku/index.html: This site by Simon Armstrongpoints to some nice descriptions of solution techniques, most of which are discussed in thisarticle.
• http://www.setbb.com/phpbb/index.php: This page is a forum for people who want tosolve and construct sudoku puzzles as well as for people who want to write computer programsto solve sudoku automatically.
• http://www.madoverlord.com/projects/sudoku.t: A downloadable program calledSudoku Susser by Robert Woodhead for the Mac, Windows and Linux that will solve almostany puzzle using logic alone. The distribution comes with great documentation as well, thatdescribes many of the techniques presented here and others besides.
• How to solve sudoku: A step-by-step guideby Robin Wilson, published by The InfiniteIdeas Company, Limited, Oxford, 2005.
• At the time of publication, there are literally hundreds of books filled with sudoku puzzles ofvarying degrees of difficulty available in any bookstore.
• Introduction to Boolean Algebrasby Philip Dwinger, published by Physica-Verlag, 1971.
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