MathandSudoku

ExploringSudokuboardsthroughgraphtheory,grouptheory,andcombinatorics

KyleOddson

UnderthedirectionofDr.JohnCaughman

Math401PortlandStateUniversity

Winter2016

AbstractEncodingSudokupuzzlesaspartiallycoloredgraphs,westateandproveAkman’stheorem[1]regardingtheassociatedpartialchromaticpolynomial[2];wecountthe4x4 sudoku boards, in total and fundamentally distinct; we count the diagonallydistinct4x4sudokuboards;andweclassifyandenumeratethedifferentstructuretypesof4x4boards.IntroductionSudoku is a logic-based puzzle game relating to Latin squares [3]. In the mostcommon size, 9x9, each row, column, and marked 3x3 block must contain thenumbers 1 through 9. A sudoku board can be formed for any n∈ ℕ, with theresultingboardhavingnnxnblocksandtotalsize𝑛!x 𝑛!.Inthisinvestigation,wewillusen=2forourboards,for4x4sudoku.Weselectedthissize,asopposedtothestandard9x9,foreaseofcalculation:thereexistsomewhereontheorder10!"9x9boards[4].Aswillbeshown,thereexistagreatdealfewer4x4boards.I.SudokuPuzzlesandBoardsasGraphs,andPartialChromaticPolynomialsConsidering a sudoku board as a mathematical object, it is useful to encode acompletedboardasagraph:eachvertexcorrespondstoacellintheboard,andtwodistinctverticesareadjacentiffthetwocellssharearow,column,ornxnblock.

Letasudokuboardbeacompletedsudokupuzzle,sothateachofthe𝑛!cellscontainsadigit.Letasudokupuzzlebeapartiallycompletedsudokuboard:thatis,atmost𝑛! − 1cellscontaindigits.Anycellcontainingadigitcanthusbeencodedasa colored vertex. If a board is properly solved—i.e., the rules of sudoku arerespectedandeverycellcontainsadigit—thenthegraphhasapropercoloring.Anypuzzle, then,corresponds toapartialcoloringofat leastoneboard.Awell-formedpuzzle corresponds to exactly one board: a well-formed puzzle has a uniquesolution. This is not the case for every puzzle; indeed, a puzzle that is not well-formedmaynothaveasolution,andwillnotcorrespondtoanyboard.

Note:individualcellsshallbereferredtobytheir(𝑟𝑜𝑤, 𝑐𝑜𝑙𝑢𝑚𝑛)coordinates.Anequivalenceclassofa cell shall compriseall cells currentlymarkedwith,or toinclude,thesameentryasthecell.Take,forexample,thepuzzle:

2 4

3 2

2 3 4 1

4 1 2 3

Figure1:A4x4sudokupuzzle.

Asapartiallycoloredgraph,wehaveequivalenceclasses:"1" = 1, 1 = 1, 1 , 3, 4 ;"2" = 1, 2 = 1, 2 , 2, 4 , 3, 1 , 4, 3 ;"3" = 2, 1 = 2, 1 , 3, 2 , 4, 4 ;"4" = 1, 4 = { 1, 4 , 3, 3 , 4, 1 }.Any board correspond to this puzzle will be a completion of this puzzle. Anyproperly colored board corresponding to the coloring of this puzzle will beconsistentwiththepartialcoloringofthepuzzle.

According to the standard rules of sudoku, only𝑛!colors, or the numbers{1,… ,𝑛} may be used to complete the coloring/board. However, the onlyrequirementforapropercoloringtobeconsistentwiththepartialcoloringisthatthe already-colored vertices retain their colors (this also implies that the existingequivalenceclassesand independentsetsretaintheircurrentmembers).Thentheminimum number of colors required for a proper, consistent, completion of anypartial coloring is equal to the number of colors present in the partial coloring,which is the same as the number of distinct digits appearing in the puzzle. Themaximumnumberofcolorswhichmayappearinaproper,consistent,completionofa partial coloring is equal to thenumber of blank cells plus thenumber of colorsappearinginthepuzzle:tobeconsistent,apropercompletionmustnotchangeanyof the colors used, and so will require at least that many colors. Going beyondstandardsudokuplay,eachblankcellmayreceiveanunuseddistinctdigit.Thenthegreatestnumberof colors foranyproper, consistent completion is the sumof thenumberofdistinctcluesandthenumberofemptycells.

Anytwoverticeswiththesamecolorshareanequivalenceclass;furthertheyare in the same independent set. Note that there is a one-to-one correspondencebetween independent sets and color classes. Any un-colored vertex (empty cell)maybeplacedintoanexistingindependentset,providedthatthecoloringremainsproper (the rules of sudoku are respected), or it may be placed into a newindependent set. Call a proper completion of the partial coloring generic if it ismerely a partitioning of the vertices into independent sets: that is, no colorsassignedtotheemptycellswhiletheyareplacedintocolorclasses.

Using the technique of deletion-contraction, the chromatic polynomial of agraph can be found:𝜒(𝐺, 𝑘), where𝜒(𝐺, 𝑘)equals the number of proper vertexcoloringsofthegraphGusingatmostkcolors.Thistechniquecanalsobeusedonapartiallycoloredgraph,togeneratethepartialchromaticpolynomial.IfweobservethesamepuzzleasFig.1,weseethatcell(1, 3)mustavoidtwocolors,thoseofcolorclasses“2”and“4”.Morespecifically, thecell(1, 3)cannotshareacolorclasswithequivalenceclasses 1, 2 & [ 1, 4 ].

2 4

3 2

2 3 4 1

4 1 2 3

Weshallencodetheemptycellsasagraph(withtheverticeslabeledwiththeirgridcoordinates):

Weshallnowencodeitasapartiallycoloredgraph,replacingthevertexlabelswiththecolorclasseseachvertexmustavoidforapropercoloring:

When applying deletion-contraction, any vertex formed by contracting an edgesharestheadjacencies,andthusthecoloringrestrictions,ofthepreviouslydistinctvertices.Thentherestrictedcolorclassesofacontracted-edgevertexcorrespondtothe union of the restrictions of the distinct vertices. We now apply deletion-contraction to our sample partial coloring, using the recursion formula for thechromaticpolynomial,𝜒 𝐺, 𝑘 = 𝜒 𝐺 − 𝑒, 𝑘 − 𝜒(𝐺 ∙ 𝑒, 𝑘).Note:weshorten𝜒 𝐺, 𝑘 to𝜒 𝐺 .

= –

= – – +

= –

– +

– + +For

,weseethatvertexCmustavoidfourcolors;thenforak-coloring, it can receive any of𝑘 − 4colors. Vertex B

mustavoidcolors“2”and“4”,aswellaswhichevercolorisplacedonvertexC;thenBcanreceiveanyof𝑘 − 3colors.VertexAmustavoidcolors“2”,“3”,and“4”,andsocan receive any of𝑘 − 3colors. Then the term for this portion of the chromaticpolynomial is 𝑘 − 4 𝑘 − 3 !. Applying this reasoning to every term,we find thatchromaticpolynomialwithrestrictionsofourpuzzleis:𝜒 𝐺, 𝑘 = 𝑘 − 3 ! − 𝑘 − 3 ! 𝑘 − 4 − 𝑘 − 4 𝑘 − 3 ! + 𝑘 − 4 !

− 𝑘 − 4 𝑘 − 3 ! + 𝑘 − 4 𝑘 − 3 + 𝑘 − 4 !𝜒 𝐺, 𝑘 = 𝑘! − 15𝑘! + 87𝑘! − 230𝑘 + 233Inorderforanycoloringtobeconsistentwiththepuzzle,kmustbeatleastaslargeasthenumberofdistinctcolorsalreadyused:here,𝑘 ≥ 4.

Wesee that𝜒 4 = 1; indeed, thispuzzlehasonepossiblecompletion in linewithstandardsudokurules:

→

Wealsoseethat𝜒 5 = 8.Thisindicatesthatif5colors,orthedigits {1,… , 5}weretobeallowed,thenthereare8possibleconsistentboards:5 2 3 4 5 2 1 4 5 2 3 4

3 4 1 2 3 4 5 2 3 4 5 2

2 3 4 1 2 3 4 1 2 3 4 1

4 1 2 3 4 1 2 3 4 1 2 3

1 2 3 4 1 2 5 4 1 2 3 4

3 5 1 2 3 5 1 2 3 4 5 2

2 3 4 1 2 3 4 1 2 3 4 1

4 1 2 3 4 1 2 3 4 1 2 3

1 2 5 4 1 2 3 4

3 4 1 2 3 4 1 2

2 3 4 1 2 3 4 1

4 1 2 3 4 1 2 3

2 4

3 2

2 3 4 1

4 1 2 3

1 2 3 4

3 4 1 2

2 3 4 1

4 1 2 3

Notethataproperk-coloringisapropercoloringusingatmostkcolors:allowing5colorsdoesnotpreventa4-coloring.Then𝜒(5),thenumberofcompletionsusingatmost 5 digits, is equal to𝜒(4), the number of completions using atmost 4 digits(here, there is one), plus the number of completions using exactly 5 digits (here,thereareseven).

Letnnowbethenumberofverticesinthegraphencodingofthesudokugrid(for9x9,𝑛 = 81; for4x4,𝑛 = 16); lettbethenumberverticesalreadycolored(orthenumberofcluesappearinginthepuzzle);let𝜆!bethenumberofdistinctcolorsin the partial coloring (or the number of distinct digits among the clues). In ourexample, ,𝑛 = 16; 𝑡 = 12; 𝜆! = 4.WenowstateAkman’s theorem[1]: “LetG beagraphwithn vertices, andC beapartialpropercoloringoftverticesofGusingexactly𝜆!colors.Define𝑝!,!(𝜆)tobethenumberofwaysCcanbecompletedtoaproper𝜆-coloringofG.Thenfor𝜆 ≥ 𝜆!,theexpression𝑝!,!(𝜆)isamonicpolynomialofdegree𝑛 − 𝑡.”Proof:Followingourearlierdiscussion,thepartialpropercoloring,C,oftvertices,isthepuzzlewithtclues,𝜆!ofwhicharedistinctdigits.For𝜆 = 𝜆!,nonewcolorsmaybeadded (nonewdigitsmaybeused).That is, thereshall remainexactly𝜆!colorclassesinthecompletedboard;sincethesecolorsarealreadyassigned,thereisonewaycolorassignment. If theboardhas𝜆 = 𝜆! + 1independentsets, thensincethepuzzlealreadyhas𝜆!independentsets,allcellsbelongingtoanyexistingcolorclassmustretain thealready-assignedcolor;anycellsbelongingto thenewcolorclasswillreceivethe𝜆 − 𝜆! = 1newcolor.

Aspreviouslydiscussed,themaximumnumberofcolorclassesappearinginany board is the sum of the number of already appearing color classes and thenumberofemptycells.Thenthetotalnumberofcolorclasses/independentsets ishere𝜆! ≤ 𝜆 ≤ 𝑛 − 𝑡 + 𝜆!. If we let𝑟 = 𝜆 − 𝜆!, the number of new color classes,then0 ≤ 𝑟 ≤ 𝑛 − 𝑡.

We apply induction to𝜆 > 𝜆! : for each additional new color class, allpreviously assigned colors must be avoided, and we find a falling factorial ofpossiblecolorchoicesforeachoftheadditionalrcolorclasses,upto𝑟 = 𝑛 − 𝑡.Since𝑟 = 0is included (and applies to nonew colors), then there are 𝜆 − 𝜆! (𝜆 − 𝜆! −1) ∙∙∙ (𝜆 − 𝜆! − 𝑟 + 1)ways of applying colors to the color classes for the boardpartitionwhereineachblankcellisinitsowncolorclass.

Then we have determined the number of ways of assigning colors to the𝜆! + 𝑟color classes of a given partition of the board. Call the number of ways of

2 4

3 2

2 3 4 1

4 1 2 3

partitioning the𝑛 − 𝑡empty cells into exactly r color classes𝑚!(𝐺,𝐶). For eachdistinct partitioning of the puzzle and empty cells into color classes, there are𝜆 − 𝜆! (𝜆 − 𝜆! − 1) ∙∙∙ (𝜆 − 𝜆! − 𝑟 + 1)waysofassigningcolors.Thenwhere𝜒(𝑘),the chromatic polynomial evaluated at k, indicates the number of complete,consistent coloringsof a𝜆!-partial-coloringusingatmostk colors,𝜒(𝑘)equals thesum of the number of completions using exactly𝜆!colors, exactly𝜆! + 1colors,…,exactly k colors. Then 𝜒 𝑘 = 𝑚! 𝐺,𝐶 𝜆 − 𝜆! 𝜆 − 𝜆! − 1 ∙∙∙ 𝜆 − 𝜆! − 𝑟 +!!!

!!!1 = 𝑝!,!(𝜆). Each term is a polynomial of degree r; the last term is of degree𝑟 = 𝑛 − 𝑡 , and correspoinds to the complete colorings using exactly𝑛 − 𝑡 newcolors, or partitioning the empty cells into exactly𝑛 − 𝑡new color classes. Sincethereareexactly𝑛 − 𝑡emptycells/un-coloredvertices,thenthereisonlyonesuchpartition.Andsinceeachtermofdegreer, for0 ≤ 𝑟 ≤ 𝑛 − 𝑡, thesumoftheterms,𝑝!,! ,ismonicofdegree𝑛 − 𝑡,andcorrespondsdirectlytothechromaticpolynomialofthesudokugraphwithinitialcoloringrestraintsontheemptycells.◊II.Enumerationof4x4SudokuBoardsA.TotalEnumeration:Claim:∃288total4x4sudokuboards.Proof:Pickanarbitraryfilling-inoftheupperleft2x2block:Turningtowardthelowerrightblock,werecognizethat“d”mustappearinoneofthe4spots{(3,3),(3,4),(4,3),(4,4)}.Weconditionfourcases:

I 1 2 3 4

II 1 2 3 4

1 a b

1 a b

2 c d

2 c d

3 d

3 d

4

4

1 2 3 4

1 a b

2 c d

3

4

III 1 2 3 4

IV 1 2 3 4

1 a b

1 a b

2 c d

2 c d

3

3

4 d

4 d

Foreachcase,wesubscriptthepossibleentriesforallremainingemptycells:

I 1 2 3 4

II 1 2 3 4

1 a bc d

1 a bd c

2 c dab ab

2 c dab ab

3b ac

dac

3b ac ac

d

4d ac ab abc

4d ac abc ab

III 1 2 3 4

IV 1 2 3 4

1 a bc d

1 a bd c

2 c dab ab

2 c dab ab

3d ac ab abc

3d ac abc ab

4b ac

dac

4b ac ac

d

Wenotethatforeachcase,thereisasinglecellwiththreepossibleentries:{I:(4,4),II:(4,3),III:(3,4),IV:(3,3)}.Anycellwithasingleoptionisfilled,andweconditionsubcases:

Ia 1 2 3 4

Ib 1 2 3 4

Ic 1 2 3 4

1 a b c d

1 a b c d

1 a b c d

2 c dab ab

2 c dab ab

2 c dab ab

3 bac

dac

3 bac

dac

3 bac

dac

4 dac ab

a

4 dac ab

b

4 dac ab

c

IIa 1 2 3 4

IIb 1 2 3 4

IIc 1 2 3 4

1 a b d c

1 a b d c

1 a b d c

2 c dab ab

2 c dab ab

2 c dab ab

3 bac ac

d

3 bac ac

d

3 bac ac

d

4 dac

aab

4 dac

bab

4 dac

cab

IIIa 1 2 3 4

IIIb 1 2 3 4

IIIc 1 2 3 4

1 a b c d

1 a b c d

1 a b c d

2 c dab ab

2 c dab ab

2 c dab ab

3 dac ab

a

3 dac ab

b

3 dac ab

c

4 bac

dac

4 bac

dac

4 bac

dac

IVa 1 2 3 4

IVb 1 2 3 4

IVc 1 2 3 4

1 a b d c

1 a b d c

1 a b d c

2 c dab ab

2 c dab ab

2 c dab ab

3 dac

aab

3 dac

bab

3 dac

cab

4 bac ac

d

4 bac ac

d

4 bac d

ac

Recognizing that these cases limit the options for the remaining cells, we fill outaccordingtotherules:Ia 1 2 3 4

Ib 1 2 3 4

Ic 1 2 3 4

1 a b c d

1 a b c d

1 a b c d

2 c d a b

2 c d b a

2 c d a b

3 b a d c

3 b a d c

3 b c d a

4 d c b a

4 d c a b

4 d a b c

IIa 1 2 3 4

IIb 1 2 3 4

IIc 1 2 3 4

1 a b d c

1 a b d c

1 a b d c

2 c d b a

2 c d a b

2 c d b a

3 b a c d

3 b a c d

3 b c a d

4 d c a b

4 d c b a

4 d a c b

IIIa 1 2 3 4

IIIb 1 2 3 4

IIIc 1 2 3 4

1 a b c d

1 a b c d

1 a b c d

2 c d a b

2 c d b a

2 c d a b

3 d c b a

3 d c a b

3 d a b c

4 b a d c

4 b a d c

4 b c d a

IVa 1 2 3 4

IVb 1 2 3 4

IVc 1 2 3 4

1 a b d c

1 a b d c

1 a b d c

2 c d b a

2 c d a b

2 c d b a

3 d c a b

3 d c b a

3 d a c b

4 b a c d

4 b a c d

4 b c a d

Weseethateachofthe4casesresultsin3singularoutcomes,basedonanarbitraryorderingoftheupperleftblock.As(a,b,c,d)canrepresentanypermutationofthenumbers1 through4, thereare4! = 24possibleorderingsof theupper leftblock,eachwith12distinctoutcomes. ∴ ∃ 4! ∙ 12 = 24 ∙ 12 = 288completed4x4sudokuboards.◊

Itisinterestingtonotethatofthe576order4Latinsquares[3],exactlyhalfofthemare4x4sudokuboards.B.DistinctEnumeration:Claim:∃2fundamentallydifferent4x4sudokuboards.Proof:LetSbethesetofall2884x4sudokuboards.Any𝑥,𝑦 ∈ 𝑆willbeconsideredequivalent if anygroupactionorcombinationofgroupactionson𝑥yield(s)𝑦. Let𝐺 = < 𝐵,𝐶, 𝑟!, 𝑟!, 𝑐!, 𝑐!,𝐷! > ∪ 𝑆!,thegroupgeneratedbythelistedactionsandthegroupofpermutationson{1,… , 4},thelistedactionsdefinedthusly:–Bplacesthefirsttworows(inorder)inthepositionofthelasttworows,andthelasttworows(inorder)inthepositionofthefirsttworows;–Cplaces the first twocolumns(inorder) in thepositionof the last twocolumns,andthelasttwocolumns(inorder)inthepositionofthefirsttwocolumns;–𝑟!swapstherowpositionsofrows1and2;–𝑟!swapstherowpositionsofrows3and4;–𝑐!swapsthecolumnpositionsofrows1and2;–𝑐!swapsthecolumnpositionsofcolumns3and4;–𝐷!indicatesthesymmetriesofthesquare:{𝐼,𝑅,𝑅!,𝑅!,𝐹,𝐹𝑅,𝐹𝑅!,𝐹𝑅!}Itcanbeobservedthat,givenalegalsudokuboard,thelegalityispreservedbyeachoftheseactions:forthesymmetriesofthesquare,allrows,columns,and2x2blocksarepreserved,whilecolumnsmaybecomerowsandviceversa,andtheorderofallrows and/or columnsmay be reversed; for the first six actions, any 2x2 block ispreserved, and if the permutation of a row or column is altered, so too are thepermutations of all intersecting rows/columns. Further, given that functioncompositionisassociative,anycombinationoftheseactionsalsopreserveslegality. Weconsideranytwosudokuboardsequivalentiffanycompositionoftheseactions and/or𝑆!transforms one to the other. Of the 288 total boards, we foundthat a boardwith any given upper left blockwas permutable to 24 others. Thenthereareatmost288÷24=12non-equivalentboards.Weexaminethe12subcasesfound,andfindthatsomeareindeedtransformationsofothers:

=IIa=IIIa

=IVa

=IIb

Ia 1 2 3 4

1 2 3 4

1 a b c d

1 a b d c

2 c d a b

𝑐!→ 2 c d b a

3 b a d c

3 b a c d

4 d c b a

4 d c a b

Ia 1 2 3 4

1 2 3 4

1 a b c d

1 a b c d

2 c d a b 𝑟!→

2 c d a b

3 b a d c

3 d c b a

4 d c b a

4 b a d c

IIIa 1 2 3 4

1 2 3 4

1 a b c d

1 a b d c

2 c d a b 𝑐!→

2 c d b a

3 d c b a

3 d c a b

4 b a d c

4 b a c d

Ib 1 2 3 4

1 2 3 4

1 a b c d

1 a b d c

2 c d b a 𝑐!→

2 c d a b

3 b a d c

3 b a c d

4 d c a b

4 d c b a

=IIIb

=IVb

=IIc=IIIc

Ib 1 2 3 4

1 2 3 4

1 a b c d

1 a b c d

2 c d b a 𝑟!→

2 c d b a

3 b a d c

3 d c a b

4 d c a b

4 b a d c

IIIb 1 2 3 4

1 2 3 4

1 a b c d

1 a b d c

2 c d b a 𝑐!→

2 c d a b

3 d c a b

3 d c b a

4 b a d c

4 b a c d

Ic 1 2 3 4

1 2 3 4

1 a b c d

1 a b d c

2 c d a b 𝑐!→

2 c d b a

3 b c d a

3 b c a d

4 d a b c

4 d a c b

Ic 1 2 3 4

1 2 3 4

1 a b c d

1 a b c d

2 c d a b 𝑟!→

2 c d a b

3 b c d a

3 d a b c

4 d a b c

4 b c d a

=IVc

Wehavethusfoundthatthereareatmostthreeequivalenceclassesofboards.Byrepresentative,theyare:

A 1 2 3 4

B 1 2 3 4

C 1 2 3 4

1 a b c d

1 a b c d

1 a b c d

2 c d a b

2 c d b a

2 c d a b

3 b a d c

3 b a d c

3 b c d a

4 d c b a

4 d c a b

4 d a b c

We note that our transformations are actions of a group, with the operation offunction composition; then combinations of transformations are themselves validtransformations:

B 1 2 3 4

1 2 3 4

1 2 3 4

1 a b c d

1 a c b d 1 a b c d

2 c d b a 𝐹𝑅!→

2 b d a c 𝑏𝑐 →

2 c d a b

3 b a d c

3 c b d a 3 b c d a

4 d c a b

4 d a c b 4 d a b c

=C

IIIc 1 2 3 4

1 2 3 4

1 a b c d

1 a b d c

2 c d a b 𝑐!→

2 c d b a

3 d a b c

3 d a c b

4 b c d a

4 b c a d

Thenweseethatthereareatmosttwoequivalenceclassesof4x4boards:

A 1 2 3 4

B 1 2 3 4

1 a b c d

1 a b c d

2 c d a b

2 c d b a

3 b a d c

3 b a d c

4 d c b a

4 d c a b

NoticethattheonlydifferencesbetweenboardsAandBarethecellpairsin

row2,columns3&4,androw4,columns3&4.InordertotransformfromAtoB(and since G is a group, if such a transformation exists then an inversetransformation exists from B to A), we need to make the following cell changes:2, 3 ↔ 2, 4 ; 4, 3 ↔ (4, 4) . If this is a valid transformation, then it is acombination of the generating actions; further, if it is a valid action, then whenapplied to a legal board, legality is preservedunder the action.However, observethat this action, performed on a legal board, renders the board illegal (observecolumns3and4):

1 2 3 4

1 2 3 4

1 a c b d

1 a c b d

2 b d a c →2 b d c a

3 c b d a

3 c b d a

4 d a c b

4 d a b c

Thenthisactionisnotalegaltransformation,andthusnotacombinationofanyofthegenerationactions,andthusnotatransformationbelongingtoourgroup.Thentheredonotexistanytransformations 𝐴 ↔ [𝐵].∴ ∃ 2fundamentallydifferent4x4sudokuboards.◊

Wealsoseethat 𝐴 = 24 ∙ 4 = 96, and 𝐵 = 24 ∙ 8 = 192.

III.Diagonal4x4SudokuBoardsA:DistinctEnumeration:We turnourattention todiagonallydistinct sudokuboards: thosewith theaddedconstraintthateachofthemaindiagonalsmustalsocontainthenumbers{1,2,3,4}.(SeeAkman [1] for a proof of existence of diagonally distinct boards for any sizesudoku.)Claim:Thereisonlyonedistinctdiagonal4x4sudokuboard.Proof:Giventhattherearetwodistinctboardsoverall,thentheremustbeatmost2distinctdiagonalboards.WeseethatboardIIIa,of[A],satisfies:

IIIa 1 2 3 4

1 a b c d

2 c d a b

3 d c b a

4 b a d c

However,[B]doesnotreadilypresentsuchaboard.Observethatdiagonality

is a structural feature, and does not change under permutation: For any diagonalwherenotallof{a,b,c,d}arepresent,nopermutation{a,b,c,d}→ 𝑆!canaddanyof themissing element(s) (aspermutations are one-to-one). Equivalence class [B]wasfoundbycollapsing8subcases,alltransformabletoeachother.The8subcaseseachrepresented24boards,allachievablebypermutation.Thensincenoneofthe8subcasesare themselvesnotdiagonal, thennoneof thepermutations representedbythe8subcasesarethemselvesdiagonal.Thereare288boardstotal;96arein[A],leaving192in[B].Asobserved,thereare8boardsstructurallynon-diagonal,eachrepresenting 24 non-diagonal permutations. Then [B] contains8 ∙ 24 = 192non-diagonal boards⇒[B] contains no diagonal boards. Then since a single diagonalboardhasbeenfoundin[A],allotherdiagonalboardsmustbereachablebysomecombinationoftransformationsandpermutations.∴ ∃ 1fundamentallydistinct4x4sudokuboard.◊

B.TotalEnumeration:Claim:∃ 48totaldiagonal4x4sudokuboards.Proof:ObservethatboardsIIaandIIIa,ofequivalenceclass[A],arebothdiagonal,whereas boards Ia and IVa are not. Each of IIa and IIIa represent 24 boards bypermutation;thenthereareatleast24 ∙ 2 = 48diagonalboards.

In order to prove that there are atmost 48 diagonal boards, we turn ourattention to the different structures of the 4x4 boards, and which group actionspreserveorbreakthosestructures.C.Structuresof4x4BoardStructuresandGroupActions:Itisobviousthatallof𝑆!aswellastheactionsof𝐷!preservediagonalityofaboard.We focus on{𝐵,𝐶, 𝑟!, 𝑟!, 𝑐!, 𝑐!}. Observe that underR, a 90º clockwise rotation, allrows become columns and all columns become rows. Then under R,𝐵 ≡ 𝐶, 𝑟! ≡𝑐!, 𝑟! ≡ 𝑐!.Thenwemayfocusonlyon{𝐵,𝐶, 𝑟!, 𝑟!}.ConsiderthediagonalboardIIIa:

IIIa 1 2 3 4

1 a b c d

2 c d a b

3 d c b a

4 b a d c

Under𝑟!,thisistransformedthusly:

IIIa

a b c d

c d a b

c d a b 𝑟!→

a b c d

d c b a

d c b a

b a d c

b a d c

Clearly,diagonalityisnotpreserved.Andunder𝑟!:

IIIa

a b c d

a b c d

c d a b 𝑟!→

c d a b

d c b a

b a d c

b a d c

d c b a

Again,diagonalityisnotpreserved.However,diagonality ispreservedunderB (andthusunderC):Forthetwo

cells (4, 1), (3, 2), in diagonal board these are two of the four distinct digits. Theothertwoofthefourmustbepresent,insomeorder,incells(1,4)and(2,3).Thesameistrueoftheothertwoblocks, fortheupperleft- lowerrightdiagonal.Thenthedigits inthefourcells(3,1),(4,2),(1,3),(2,4),mustalsobethefourdistinctdigits.UndertheactionB,thecells(3,1)and(4,2)becomethetoptwodigitsoftheupper-leftlowerrightdiagonal,whilecells(1,3)and(2,4)becomethelowertwodigits of the diagonal. Similarly for the upper right- lower left diagonal. Then Bpreservesdiagonality;byRor𝑅!,sotoodoesC.

Considerthetwonon-diagonalrepresentativesofequivalenceclass[A]:

Ia

IVa

a b c d

a b d c

c d a b

c d b a

b a d c

d c a b

d c b a

b a c d

Justas{𝐷!, 𝑆!,𝐵,𝐶}preservediagonality,thesameargumentholdsfortheseactionspreserving non-diagonality. Indeed, the actions{𝑟!, 𝑟!, 𝑐!, 𝑐!}are the only actionswhichwilltransformdiagonalboardstonon-diagonalboardsandviceversa.

Consider the diagonal entries of the two boards above. Board Ia has asentriesonitsmaindiagonals{a,d};boardIvahasentries{ad},{bc}.Wewillclassifyboardstructuresbasedontheirdiagonalentries.BoardIaandits24permutationshave a total of 2 distinct digits along the main diagonals. Board IVa and its 24permutationshave2distinctdigitsalongeachofitsmaindiagonals.BoardsIIaandIIIa,beingthediagonalboards,ofcoursehaveallofthedistinctdigitsalongeachoftheir main diagonals. We see, then that equivalence class [A] can be furtherstructuredintothreeequivalencesubclasses:[Ia],whichcontains24boards,eachofwhichhaveatotalof2diagonaldigits;[IIa]=[IIIa],whichcontains48boards,eachof which are diagonal; and [IVa], which contains 24 boards, each of which has 2distinctdigitsalongeachofitsdiagonals.

Theseaccountforall96oftheboardsof[A];astherearenodiagonalboardsin [B], then there are at most 48 diagonal boards; given the lower bound foundearlier,thenthereareexactly48diagonal4x4boards.◊Wehaveseenboardstructures,allofequivalenceclass[A],whichcontaineither2or4distinctdigitsalongthediagonals.Whatoftheboardsof[B]?Indeed,alloftheboardsof[B]eachcontain3distinctdigitsalongeachdiagonal. Itcanbeseen(attheendofsectionII.A)thateachofthe8representativesofequivalenceclass[B]=[C]havethreedistinctdigitsalongeachmaindiagonal.Itisclear that no permutation of𝑆!will change the number of distinct digits along adiagonal.Here,weshowthatthese8boards(andtheirpermutations)areallofthepossible4x4sudokuboardswith3distinctdigitsacrossthediagonals. Weconditionboardsbasedon theupper left- lowerrightdiagonal.Givenasequence _ _ | _ _,we see that there are four structural possibilities: the first cellmatchesthethird,thefirstcellmatchesthefourth,thesecondcellmatchesthethird,the second cellmatches the fourth;with each of the two remaining cells distinct.Withoutlossofgenerality,wenotethepossibilities:

1) a 2) a

b b

a c

c a

3) a 4) a

b b

b c

c b

Recognizing that these cases limit the options for the remaining cells, we fill outaccordingtotherules:

1) acd bcd bd

2) acd bd bcd

cd

bcd

a cd

b acd

bcd

cd abd

bd

a cbd

bd

abd

c bcd cd bd

a

3) acd cd

b 4) acd

bcd

cd

bacd ad

cd

bad acd

acd acd

bad

bad

cad

bad ad

c cd acd ad

b

Wepickanarbitrarysinglecellwithtwooptions,condition,andfillout:

1a) a c b d 1b) a d c b

d b c a c b d a

c d a b b c a d

b a d c d a b c

2a) a c d b 2b) a d b c

d b a c c b a d

b a c d d a c b

c d b a b c d a

3a) a c d b 3b) a d c b

d b c a c b a d

c a b d d c b a

b d a c b a d c

4a) a c b d 4b) a d b c

d b a c c b d a

b d c a b a c d

c a d b d c a b

Thoughtheseboardsdonotinitiallyseemfamiliar,weseethatunderthefollowingpermutations, these boards are, in fact, the exact representatives of equivalenceclass[B]:1a): 𝑏𝑑𝑐 →IIc; 1b): 𝑏𝑑 →IIIb; 2a): 𝑏𝑑𝑐 →IIIC; 2b): 𝑏𝑑 →IIB;3a) 𝑏𝑑𝑐 →IB; 3b): 𝑏𝑑 →IC; 4a): 𝑏𝑑𝑐 →IVB; 4b): 𝑏𝑑 →IVC.

Thenwehave8boardstructures,eachrepresenting24permutations; thenweseethatthereareexactly192possibleboardswithexactlythreedistinctdigitsalongeachmaindiagonal.Asthisistheexactsizeof[B],werecognizethestructureinherenttotheequivalenceclass.

Wenowseethereasonforthetwoequivalenceclasses:nopermutation∈ 𝑆!can change the number of digits along a diagonal, and any of the group actiontransformations must change an even number of digits when counting along thediagonals.Further,wenowseeadistinctivewaytoclassifyboards:thatofcountingthedistinctdigitsalongeachmaindiagonal.◊

ConclusionAvariationonLatinsquares,Sudokuasapuzzlewasinventedinthelate1970s[5].Aswithmanyplayfuldiversions,thoughitsoriginswerenotmathematical,sudokuboards as mathematical objects hold great opportunities for study. Indeed, it issurprisinghowmuchcanbefoundinasudokuboard,andthemanywaysitcanbestudied. Certainly the puzzle maker(s) who created “Number Place” had nointention that aboardbe encodedas a graph, or that chromaticpolynomialsmayindicatethenumberofvalidsolutionsforagivenpuzzle. Regardingthechromaticpolynomialasrelatestosudokugraphs,thepartialchromaticpolynomialisanobjectdeservingfurtherstudy.Whileithasbeenshownthatthispolynomialismonicofknowndegree,ourinvestigationhasnotshedlightonitsfurtherproperties,thosesharedwithordifferentfromtheclassicchromaticpolynomial.For instance, iseverypowerof𝜆present(uptothedegree)?Mustthesignsofthetermsalternate?Whatisthemeaning,ifany,oftheconstantterm?Whydoes this polynomial have a constant term at all? These questions, beyond ourscope,piquethecuriosity. We have used 4x4 boards as our object of study; in the world of sudoku,thesearesmallboards,andinthemathematicalstudyofsudoku,4x4isthetrivialcase.Givenanopportunity, this author intends to study sudoku further, including9x9boards;inthemeantime,weshallnotforgetthatthisfascinatingmathematicalobjectisalsoanenjoyablepuzzle.

Sources[1]Akman,Fusun."PartialChromaticPolynomialsandDiagonallyDistinctSudokuSquares."[0804.0284]PartialChromaticPolynomialsandDiagonallyDistinctSudokuSquares.ArXiv.org,2July2008.Web.26Feb.2016.[2] West, Douglas Brent. Introduction to Graph Theory. Upper Saddle River, NJ:PrenticeHall,2001.Print.[3]Weisstein,EricW."LatinSquare."FromMathWorld--AWolframWebResource.http://mathworld.wolfram.com/LatinSquare.html[4]Felgenhauer,Bertram,andFrazerJarvis."MathematicsofSudokuI-Universityof Sheffield." Mathematics of Sudoku I. University of Sheffield, UK, 25 Jan. 2006.Web.26Feb.2016.[5]Smith,David."SoYouThoughtSudokuCamefromtheLandoftheRisingSun..."TheGuardian.GuardianNewsandMedia,14May2005.Web.26Feb.2016.FurtherReadingDavis, Tom. "The Mathematics of Sudoku." The Mathematics of Sudoku. 13 Sept.2012.Web.26Feb.2016.<http://www.geometer.org/mathcircles/sudoku.pdf>.Herzberg, Agnes M., and H. Ram Murty. "Sudoku Squares and ChromaticPolynomials."NoticesoftheAMS54.6(2007):708-17.Web.26Feb.2016.Rosenhouse, Jason, and Laura Taalman.TakingSudokuSeriously:TheMathbehindtheWorld'sMostPopularPencilPuzzle.Oxford:OxfordUP,2011.Print."Sudoku Variations (puzzles)." Web log post. Grandmaster Puzzles. Ed. ThomasSnyder.Web.26Feb.2016.<http://www.gmpuzzles.com/blog/category/sudoku/>.