The following is an illustrated solution for the
Tough Sudoku of February 25, 2007.
There are many ways to tackle this one. The primary focus of this solution is to illustrate
using an Almost Unique Rectangle.
You may need to refer to previous blog pages to understand
this solution. Links to these pages are found to the right, under Previous Entries.
At many times during this illustration, there are other steps available. It is not the goal
of this page to show every possible step, but rather to illustrate steps that, taken together,
unlock this puzzle
The information on the following blog pages is required to understand this solution:
The illustrations of forbidding chains (also called Alternating Inference Chains or AIC)
used on this blog page will share the same key:
- black lines = strong links
- red lines = weak links
- candidates crossed out in red = candidates proven false
Many steps that are possible will not be shown to keep the page as short as possible.
However, every step that is shown can be justified by considering only the previous illustrated
steps.
Puzzle at start
A few Unique Possibilities are available here. They are included in the next puzzle picture.
Easy cell solutions
After
- d7 = 2% box
- d4 = 7% column & box
- d9 = 6% column
One has:
- Locked 4's at e78
- Locked 8's at e789
Thus sans Possibility Matrix:
- f3 = 8% box
- f1 = 4% box
- f4 = 6% column
Many Hidden Pairs plus ...
There are many hidden pairs availabe here, still without using the possibility matrix.
Illustrated to the left are:
- Hidden pair 67 at ei1
- Hidden pair 67 at g8, i7
- thus
- Locked 4's at h78
- Naked pair 67 at i17
Triple 135
Using the previous information, one can now find the triple 135 at h2, i23 by entering
in the possibilities. This justifies eliminating all the indicated candidates.
One can solve one more cell:
Almost Unique Rectangle
The highlit 29's form three corners of a possible Unique Rectangle. In order to use this
technique, it is absolutely required that
- One acknowledges that it is assumed that the puzzle has no more than one solution
- That both 29 are possible candidates at the fourth corner: b3
The logic for this step can be viewed many ways. Here is one way to look at it:
- Suppose that b3≠45
- => b3 = 29
- => pair 29 in the following houses
- boxes b2 & h2
- columns b & h
- rows 1 & 3
- => nothing can happen on the puzzle grid to prefer 2 over 9 at any of the
four cells: bh13
- => the puzzle either has no solutions, or it has at least two solutions
- Since it is assumed that the puzzle has exactly one solution, b3=45
Please note that I have been careful not to call this blog page a proof. Since this step
assumes uniqueness of solution, this is a solution, not a proof. From this point forward, all
the steps taken are tainted by this assumption. After finishing the puzzle, I will not have
proven that the puzzle has a unique solution.
Hidden Pair 29 or Naked triple 458
Highlit on the left are the remaining possible locations for 29 in box b2. Thus, 45 cannot
exist at a3. One can also use the naked triple at b23, c2 to make the same eliminations.
Some more Unique possibilities are now available:
- a5 = 4% column
- d5 = 8% cell
- d6 = 4% cell
- i6 = 8% column
Coloring on 5s
There are several possible steps available considering only the 5s. Illustrated to the left
is the following fc on 5s:
- b3 == i3 -- i5 == c5 => b6,c2≠5
This starts a cascade of Unique Possibities leading to a solved puzzle, beginning with:
- c2 = 8% cell
- a4 = 8% box and row
- % cell to the end (naked singles to the end)
Solved Puzzle
This is one possible solution to this puzzle. It is not certain at this point that it is
the only solution. Nevertheless, using the AUR step certainly made finding this solution easy.
Solution as shown
- Start at 23 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
- Locked 8's at e789 forbids e2346=8 UP 27
- Locked 4's at e78 forbids e1236=4 UP 29
- Hidden pair 67 at ei1 forbids i1=1
- Hidden pair 67 at g8i7 forbids g8=14, i7=138
- Pair 67 at i17 forbids i3=7, i2=6
- Locked 4's at h78 forbids h23=4
- Triple 135 at h2,i23 forbids c2,h1,h2=1 and forbids g2,gh3=3 and forbids h3=5 UP 30
- AUR 29 at b13, h13 forbids b3=29
- Hidden pair 29 at b1a3 forbids a3=45 UP 34
- fc on 5's: b3 == i3 -- i5 == c5 forbids c2,b6=5 UP 81
Proof
The following is a proof of unique solution for this puzzle:
- Start at 23 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
- Locked 8's at e789 forbids e2346=8 UP 27
- Locked 4's at e78 forbids e1236=4 UP 29
- Hidden pair 67 at ei1 forbids i1=1
- Hidden pair 67 at g8i7 forbids g8=14, i7=138
- Pair 67 at i17 forbids i3=7, i2=6
- Locked 4's at h78 forbids h23=4
- Triple 135 at h2,i23 forbids c2,h1,h2=1 and forbids g2,gh3=3 and forbids h3=5 UP 30
- Differences start below
- Locked 9's at c456 forbids a45,b6=9
- fc on 5's: c456 == c2 -- h2 == h4 forbids a4=5
- triple 128 at a478 forbids a3=2, a59=8, a9=1
- Locked 2's at b13 forbids b68=2
- Y wing style: b2=4 == g2=4 -- g2=6 == g8=6 -- b8=6 == b8=8 forbids b2=8 UP 31
- Locked 5's at c456 forbids a5b6=5 UP 81
If you have been following this blog, there is nothing very difficult about the
proof.
Thus, one does not
need to use the Almost Unique Rectangle. It is really up to each
individual whether to allow using Uniqueness of Solution in their bag of tricks to solve sudoku
puzzles.
Comparison of the two listed paths
- Sets: 1+1+2+2+2+1+3=12 for both paths steps 1 through 4
- +4+2+2 = 20 total native strong sets for AUR solution
- +1+2+3+1+3+1 = 23 total native strong sets for Proof
- Ratings: .03+.09+.07 = .19 for both paths steps 1 through 4
- +.15+.06 = .40 for AUR solution
- +.03+.03+.14 = .39 for Proof
- Maximum depth
- 4 at step 5.1 for AUR solution
- 3 at multiple steps for Proof
Thus, about the same difficulty for each path. For many, the AUR soltuion is probably the
easier,
softer, way. However, the latter Proof does not use a possibly
controversial Sudoku rule.
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