The following illustrated solution for the
Tough Sudoku of May 25, 2009 employs
at least some steps of some interest, perhaps, to some. Although this sudoku can be solved without using the
uniqueness deduction illustrated below, in my opinion the uniqueness deduction simplifies the solution.
This Sudoku probably solves with fewer steps. The steps chosen below are primarily for the purpose of teaching
and illustration. I realize it is a bit grandiose of me to assume that I might do either well.
If this is your first visit to this blog, welcome! Unfortunately, if you are a first time visitor, this page may seem
like it is written in a different language. Well, it is. Previous blog pages may be helpful. Links to these pages are
found to the right, under Sudoku Techniques. The earliest posts are at the bottom, and if you have never perused
the intricacies of our special coded language here, you may wish to start close to beginning. Then again, you may just decide to
jump in and have fun decoding! The collection of sudoku strategies and sudoku techniques used in this puzzle are
previously described. If I were not so lazy, I could provide specific links.......
Many steps not illustrated are possible.
For those whom do not wish to struggle through the illustrations, I have written the steps
taken in this solution at the bottom of the page in sudoku.com.au notation.
Five Unique Possibilities (UPs) are available here.
- (4)c7 % row
(6)f6 % box
(4)f4 % box
(4)e2 % box
(7)e3 % box
- translation: Hidden single (4) at c7, as it is the only (4) possible in the row
The Puzzle at UP 28 Unique Possibilities to 28 given or solved cells
Two easy steps are available here:
- (np35)d46 =>
- np is shorthand for Naked Pair
- (nt189)g3h13 =>
- nt is shorthand for Naked Triple
Step 2: A short AIC (Alternating Inference Chain) using candidates 5 & 6
Above, I have illustrated a short AIC using candidates 5 & 6. The only difficult portion of this chain
is using the grouping of candidate 5 at ac5. The type of puzzle mark-up that I prefer highlights this type of
grouping, as I often find it useful. To highlight this group, I underline the 5 at g5, as it is the other end of
the almost bilocal group. The chain could be written as:
- (5)ac5 = (5-6)g5 = (6)i5 - (6)i1 = (6-5)a1 = (5)c3 => c46≠5
One may note that many of the steps shown here could have been done in a different order, such as the next step.
Step 3: Another short AIC
This step could be seen easily, perhaps for some more easily, using mostly naked sets. Above,
I have illustrated the following chain using hidden sets:
- (3)f2= (3-5)e1 = (5-6)a1 = (6-2)i1 = (2)d1 => f2≠2
Using mostly naked sets, one could find:
- (2)d1 = (nt139)bdh1 - (3)e1 = (3)f2 => f2≠2
Step 4: A continuous loop, or wrap-around chain, using candidates 2 & 5.
Short wrap-around chains such as the one illustrated below are my absolute favorite find while solving
a sudoku. They illustrate the power of AIC's best, and avoid having to consider the same information multiple
times to achieve multiple eliminations. But, enough with the inane preaching!
The first two steps have set up this step. The 2's at d78 above are eliminated because of the locked candidate
(2) at f89, and normally would not be considered to be a result of such a chain. Please note that it matters not
where one would start the following chain. Each potential starting point is completely arbitrary! I have been
told many times by my wife that I am also completely arbitrary, but I think her meaning is slightly different.
- (5)c3 = (5-2)c5 = (2)c8 - (2)f8 = (2-5)f9 = (5)f3 loop =>
- (5=2)c5 => c5≠13
- (2)c8 = (2)f8 => gi8≠2
- (2=5)f9 => f9≠13
- (5)f3 = (5)c3 => nothing, as this is already a native condition of the current puzzle
Step 5: Another short chain that finally yields an elimination which allows some cells to be solved.
Usually, if one wishes to minimize the number of steps, steps that result in solving some cells are the quickest
path. Oftentimes, however, the path has less to do with solving those cells, and more to do what is
forbidden by the new cell solutions. This is precisely the case with the following step, which is powerful because
it eventually eliminates (3) from g7. Here is the step:
- (2=5)f9 - (5=1)f3 - (1)g3 = (1)h13 - (1)h9 = (1)i9 =>
- (2)g7 % box
- (8)d7 % row
- UP 30
My step 8 is available now. However, first I will dispense of a couple of other candidates to get more bang for the buck.
Again, this is another one of those arbitrary decisions.
Step 6: A Ywing Style step using only hidden strengths
A typical short hidden Y Wing Style is illustrated to the left:
- (1)c4 = (1-2)b5 = (2-5)c5 = (5)c3 =>
Step 7: Another Ywing Style, this time using a mixture of hidden and naked strengths.
- (5)e1 = (5-1)f3 = (1)gh3 - (1=9)h1 =>
One could now use the locked 9's at d12. Doing so, or not doing so, will not matter after the next step.
Step 8: Uses a Almost Unique Rectangle Strong Inference Set in a chain.
The chain below is the longest chain of this solution. It has been available since 3 was removed from g7.
Oftentimes, the use of hidden strengths in conjunction with a Uniqueness situation can be powerful, as illustrated below.
- The (AUR35) at dg46 =>
- At least one of g46 is not (35) =>
- At least one of these is true: [(3)g8, (35)g5] - note the grouping of (35)g5
- (3)g8 = (35-6)g5 = (6)g2 - (6)c2 = (6)c8 - (6)e8 = (np39)e78 =>
- f8 ≠3 =>
- A cascade of naked and hidden singles to UP 50
The grouping (35-6)g5 merely means that (3)g5 and/or (5)g5 are weakly linked, or forbid, or prevent, (6)g5.
Step 9: A naked pair
A lowly naked pair suffices to knock out the rest of this puzzle.
- Start 23, UP 28
- (5)ac5=(5-6)g5=i5-i1=(6-5)a1=(5)c3 => c46≠5
- (3)f2=(3-5)e1=(5-6)a1=(6-2)i1=(2)d1 =>f2≠2
- (2)f9=f8-c8=(2-5)c5=c3-f3=(5)f9 loop => d78.gi8≠2, c5.f9≠13
- (2=5)f9-(5=1)f3-gh3=h1-h9=(1)i9 => ghi8≠1, i9≠2, UP 30
- (1)c4=(1-2)b5=(2-5)c5=(5)c3 => c3≠1
- (9)h1=h3-(9=5)c3-f3=(5)e1 => e1≠9
- (AUR35)dg46 => sis[(3)g8, (35)g5]
- (3)g8=(35-6)g5=g2-c2=c8-(6)e8=(np39)e78 => f8≠3, UP 50
- (np35)dg6 => ac6≠35, UP 81