I made a little solver that seems to work (tested 2 different grids), but might not solve all grids. Here is the code:

hintstyle = Sequence[11, Red];
candstyle = Sequence[8, Gray];
fixedstyle = Sequence[22];
opchars = {Plus -> "+", Minus -> "-", Times -> "\[Times]", 
   Divide -> "\[Divide]"};

ClearAll[SquareSides, ClueText, DrawGrid]
SquareSides[n_, m_] := Module[{},
  {{{n, -m}, {n + 1, -m}},
   {{n, -m}, {n, -m - 1}},
   {{n + 1, -m}, {n + 1, -m - 1}},
   {{n, -m - 1}, {n + 1, -m - 1}}
   }
  ]
ClueText[cands_, {x_, y_}] := Module[{center, r = 0.2, cp},
  center = {x + 0.5, -y - 0.5};
  If[Length[cands] == 1,
   Text[Style[First[cands], fixedstyle], center]
   ,
   cp = CirclePoints[center, {r, \[Pi]/2}, n];
   cp = MapThread[
      Text[Style[#1, candstyle], #2] &, {Range[n], cp}][[cands]];
   cp
   ]
  ]
DrawGrid[cands_List, hints_List] := 
 Module[{hor, ver, all, rem, boxborders, topleft, hinttext, candpos, 
   candtext},
  hor = Flatten[Outer[List, Range[n], Range[n + 1]], 1];
  ver = Flatten[Outer[List, Range[n + 1], Range[n]], 1];
  all = Join[{#, # + {1, 0}} & /@ hor, {#, # + {0, 1}} & /@ ver];
  all[[All, All, 2]] *= -1;
  topleft = First[TakeSmallestBy[#, Total, 1]] & /@ hints[[All, 3]];
  topleft += 0.08;
  topleft[[All, 2]] *= -1;
  hinttext = ToString[#1] <> (#2 /. opchars) & @@@ hints;
  hinttext = 
   MapThread[
    Text[Style[#1, hintstyle], #2, {-1, 1}] &, {hinttext, topleft}];
  boxborders = Apply[SquareSides, hints[[All, 3]], {2}];
  boxborders = Flatten[#, 1] & /@ boxborders;
  boxborders = 
   Select[Tally[#], Last[#] == 1 &][[All, 1]] & /@ boxborders;
  boxborders = Flatten[boxborders, 1];
  rem = Complement[all, boxborders];
  candpos = Tuples[Range[n], 2];
  candtext = MapThread[ClueText, {Flatten[cands, 1], candpos}];
  Graphics[{{Gray, Thickness[0.005], Line /@ rem},
    hinttext,
    {Thickness[0.01], Line[boxborders]},
    candtext
    }, ImageSize -> 65 n]
  ]

ClearAll[ApplyHint, DivideFunc, MinusFunc]
DivideFunc[x_, y_] := Abs[Log[x/y]]
MinusFunc[x_, y_] := Abs[x - y]
ApplyHint[ans_Integer, op : Divide | Minus, cells_List] := 
 Module[{cns},
  If[Length[cells] == 2,
   Switch[op,
    Divide, ApplyHint[Log[ans], DivideFunc, cells],
    Minus, ApplyHint[ans, MinusFunc, cells]
    ]
   ,
   Print["Divide/Minus should have 2 cells!!"];
   Abort[];
   ]
  ]
ApplyHint[ans_, op : (Times | Plus | DivideFunc | MinusFunc), 
  cells_List] := Module[{cns},
  cns = Extract[cands, cells];
  cns = Select[Tuples[cns], op @@ # == ans &];
  (* TODO: not only should the 'ans' match, 
  but also we can not have the same values in the same row/column, 
  filter for that here\[Ellipsis] *)

  cns = DeleteDuplicates /@ (cns\[Transpose]);
  MapThread[(Part[cands, Sequence @@ #2] = #1) &, {cns, cells}]
  ]

ClearAll[DeleteCands]
DeleteCands[{n_, m_}, 
  del_] := (Part[cands, n, m] = Complement[Part[cands, n, m], del])

ClearAll[NakedSubset, SameValsCorrectnQ, NakedDelete]
SameValsCorrectnQ[cells_, n_Integer] := Module[{celldata},
  celldata = Extract[cands, cells];
  If[AllTrue[celldata, Length[#] == n &],
   Equal @@ (Sort /@ celldata)
   ,
   False
   ]
  ]
NakedSubset[m_Integer] := Module[{cellgroups},
  cellgroups = 
   Join[Outer[List, Range[n], Range[n]], 
    Outer[List, Range[n], Range[n]]\[Transpose]];
  Do[NakedSubset[m, cg], {cg, cellgroups}]
  ]
NakedSubset[m_Integer, cells_List] := Module[{subsets},
  subsets = Subsets[cells, {m}];
  subsets = Select[subsets, SameValsCorrectnQ[#, m] &];
  Do[NakedDelete[cells, ss], {ss, subsets}]
  ]
NakedDelete[cells_List, samesubs_List] := Module[{other, digits},
  other = Complement[cells, samesubs];
  digits = First[Extract[cands, samesubs[[{1}]]]];
  Do[DeleteCands[o, digits], {o, other}]
  ]

A puzzle has a size and some hints, and then we solve it:

n = 6;
hints = {
   {3, Plus, {{1, 1}}},
   {7, Plus, {{2, 1}, {1, 2}, {2, 2}}},
   {4, Minus, {{3, 1}, {4, 1}}},
   {16, Plus, {{5, 1}, {5, 2}, {6, 2}}},
   {2, Plus, {{6, 1}}},
   {60, Times, {{3, 2}, {4, 2}, {4, 3}}},
   {3, Minus, {{1, 3}, {1, 4}}},
   {108, Times, {{3, 3}, {2, 3}, {2, 4}}},
   {5, Plus, {{5, 3}}},
   {5, Plus, {{6, 3}, {6, 4}}},
   {11, Plus, {{3, 4}, {3, 5}, {4, 5}}},
   {2, Divide, {{4, 4}, {5, 4}}},
   {20, Times, {{1, 5}, {2, 5}, {2, 6}}},
   {9, Times, {{5, 5}, {5, 6}, {6, 5}}},
   {6, Plus, {{1, 6}}},
   {2, Divide, {{3, 6}, {4, 6}}},
   {5, Plus, {{6, 6}}}
   };

AbsoluteTiming[
 cands = ConstantArray[Range[n], {n, n}];
 old = False;
 While[old =!= cands,
  old = cands;
  AbsoluteTiming[ApplyHint @@@ hints;];
  NakedSubset /@ Range[Ceiling[n/2]];
  ];
 ]
DrawGrid[cands, hints]

Giving:

Or for the original problems:

n = 4;
hints = {
   {12, Times, {{1, 1}, {2, 1}, {2, 2}}},
   {2, Divide, {{3, 1}, {4, 1}}},
   {11, Plus, {{1, 2}, {1, 3}, {1, 4}, {2, 3}}},
   {2, Minus, {{3, 2}, {3, 3}}},
   {2, Plus, {{4, 2}}},
   {4, Plus, {{4, 3}, {4, 4}}},
   {3, Minus, {{2, 4}, {3, 4}}}
   };
AbsoluteTiming[
 cands = ConstantArray[Range[n], {n, n}];
 old = False;
 While[old =!= cands,
  old = cands;
  AbsoluteTiming[ApplyHint @@@ hints;];
  NakedSubset /@ Range[Ceiling[n/2]];
  ];
 ]
DrawGrid[cands, hints]

Giving:

Solves it within 10 milliseconds for me. It by no means solves all the KenKen puzzles, once should add the Hidden Subsets technique and eliminate possibilities that I commented in the code.

Cheers!

Very interesting, it's like sudoku, but a little bit more sophisticated. I added an extra step to your code which speeds it up 5X.

start=AbsoluteTime[];
Dimensions[p4=Permutations[Range[4]]]
Dimensions[a1=Flatten[Outer[List,p4,p4,p4,p4,1],3]]
Dimensions[a1=Select[a1,Total[#]=={10,10,10,10}&]]
Dimensions[a2=Select[a1,(And@@(And@@Unequal[Sequence@@#]&)/@Transpose[#])&]]
Dimensions[a3=Select[a2,#[[1,1]]*#[[1,2]]*#[[2,2]]==12&]]
Dimensions[a4=Select[a3,Abs[Log[#[[1,3]]/#[[1,4]]]]==Log[2]&]]
Dimensions[a5=Select[a4,#[[2,1]]+#[[3,1]]+#[[3,2]]+#[[4,1]]==11&]]
Dimensions[a6=Select[a5,Abs[#[[2,3]]-#[[3,3]]]==2&]]
Dimensions[a7=Select[a6,#[[2,4]]==2&]]
a7
AbsoluteTime[]-start

Note that I overwrite a1 on line 4. This substantially reduces the number of grids and the operation is very fast.

O btw, you can simply one of your filters:

Unequal[Sequence @@ #] 

can be

Unequal @@ #

Very nice idea though, but I don't think this will work for a 6*6 grid or even larger...

Can the puzzle itself be automatically and randomly generated ? Then a generated puzzle and its solution would make a complete application like an app at Demonstration Project.

Your method is an impressive implementation in Mathematica of how KenKen is done "by hand".

In[1]:= MemoryInUse[]

Out[1]= 28876504

In[2]:= Timing @ Dimensions[p4 = Permutations[Range[4]]]

Out[2]= {0., {24, 4}}

In[3]:= MemoryInUse[]

Out[3]= 38100432

In[4]:= Timing @ Dimensions[a1 = Tuples[p4, 4]] 

Out[4]= {0.015625, {331776, 4, 4}}

In[5]:= MemoryInUse[]

Out[5]= 80572336

In[7]:= Timing @ Dimensions[a2 = Intersection[a1, Transpose /@ a1]]

Out[7]= {0.078125, {576, 4, 4}}

In[8]:= MemoryInUse[]

Out[8]= 80657696

In[9]:= Clear[a1]

In[10]:= MemoryInUse[]

Out[10]= 41259120

Hidden Subsets can also be implemented by just looking for naked subsets larger than n/2. Change the line:

  NakedSubset /@ Range[Ceiling[n/2]];

to:

  NakedSubset /@ Range[n];

in order to 'implement' 'hidden subsets' (we are looking for large naked subsets, which is equivalent, might be slower though...).

This is indeed a lot faster; how about the memory consumption? The {331776, 4, 4} - array should take up some memory...

I've been able to speed up my code for doing the 4x4 KenKen by doing it row at a time

In[1]:= n = 4;

In[2]:= rn = Range[n];

In[3]:= colUnequal[l_List] :=  
 Select[l, And @@ (UnsameQ  @@@ Transpose[#]) &]

In[4]:= addRow[l_List, p_List] := 
 Join[#[[1]], {#[[2]]}] & /@ Tuples[{l, p}]

In[5]:= Timing @ Dimensions[p = Permutations[rn]]

Out[5]= {0., {24, 4}}

In[6]:= Timing @ 
 Dimensions[ps = Select[p, Abs[Log[#[[3]]/#[[4]]]] == Log[2] &]]

Out[6]= {0., {8, 4}}

In[7]:= Timing @ Dimensions[a2 = Tuples[{ps, p}] ]

Out[7]= {0., {192, 2, 4}}

In[8]:= Timing @ Dimensions[a2s = colUnequal[a2]]

Out[8]= {0., {72, 2, 4}}

In[9]:= Timing @ Dimensions[a2s1 = Select[a2s, #[[2, 4]] == 2 &]] 

Out[9]= {0., {12, 2, 4}}

In[10]:= Timing @ 
 Dimensions[a2s2 = Select[a2s1, #[[1, 1]]*#[[1, 2]]*#[[2, 2]] == 12 &]]

Out[10]= {0., {4, 2, 4}}

In[11]:= Timing @ Dimensions[a3 = addRow[a2s2, p]]

Out[11]= {0., {96, 3, 4}}

In[12]:= Timing @ Dimensions[a3s = colUnequal[a3]]

Out[12]= {0., {8, 3, 4}}

In[13]:= Timing @ 
 Dimensions[a3s1 = Select[a3s, Abs[#[[2, 3]] - #[[3, 3]]] == 2 &]]

Out[13]= {0., {3, 3, 4}}

In[14]:= Timing @ Dimensions[a4 = addRow[a3s, p]]

Out[14]= {0., {192, 4, 4}}

In[15]:= Timing @ Dimensions[a4s = colUnequal[a4]]

Out[15]= {0., {8, 4, 4}}

In[16]:= Timing @ 
 Dimensions[
  a4s1 = Select[
    a4s, #[[2, 1]] + #[[3, 1]] + #[[3, 2]] + #[[4, 1]] == 11 &]]

Out[16]= {0., {1, 4, 4}}

In[17]:= a4s1[[1]]

Out[17]= {{1, 3, 2, 4}, {3, 4, 1, 2}, {4, 2, 3, 1}, {2, 1, 4, 3}}