Unique Rectangle Solution for February 25, 2007

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 Sudoku Techniques.

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


PUzzle start

A few Unique Possibilities are available here. They are included in the next puzzle picture.

Easy cell solutions


Getting to UP 29

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 ...


Hidden pairs 67, Naked pair 67, Locked 4s

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
      • =>h23≠4
    • Naked pair 67 at i17
      • =>i2≠6 & i3≠7

Triple 135


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:

  • c1 = 1% box & row

Almost Unique Rectangle


AUR 29

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


Hidden Pair 29

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


Sideways skyscraper 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


A solution

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

  1. Start at 23 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
  2. Locked 8's at e789 forbids e2346=8 UP 27
  3. Locked 4's at e78 forbids e1236=4 UP 29
    1. Hidden pair 67 at ei1 forbids i1=1
    2. Hidden pair 67 at g8i7 forbids g8=14, i7=138
    3. Pair 67 at i17 forbids i3=7, i2=6
    4. Locked 4's at h78 forbids h23=4
    5. Triple 135 at h2,i23 forbids c2,h1,h2=1 and forbids g2,gh3=3 and forbids h3=5 UP 30
    1. AUR 29 at b13, h13 forbids b3=29
    2. Hidden pair 29 at b1a3 forbids a3=45 UP 34
  6. 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:

  1. Start at 23 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
  2. Locked 8's at e789 forbids e2346=8 UP 27
  3. Locked 4's at e78 forbids e1236=4 UP 29
    1. Hidden pair 67 at ei1 forbids i1=1
    2. Hidden pair 67 at g8i7 forbids g8=14, i7=138
    3. Pair 67 at i17 forbids i3=7, i2=6
    4. Locked 4's at h78 forbids h23=4
    5. Triple 135 at h2,i23 forbids c2,h1,h2=1 and forbids g2,gh3=3 and forbids h3=5 UP 30
  5. Differences start below
    1. Locked 9's at c456 forbids a45,b6=9
    2. fc on 5's: c456 == c2 -- h2 == h4 forbids a4=5
    3. triple 128 at a478 forbids a3=2, a59=8, a9=1
    4. Locked 2's at b13 forbids b68=2
    5. Y wing style: b2=4 == g2=4 -- g2=6 == g8=6 -- b8=6 == b8=8 forbids b2=8 UP 31
  6. 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
    1. +4+2+2 = 20 total native strong sets for AUR solution
    2. +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
    1. +.15+.06 = .40 for AUR solution
    2. +.03+.03+.14 = .39 for Proof
  • Maximum depth
    1. 4 at step 5.1 for AUR solution
    2. 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.

4 Comments
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Steve  From Ohio    Supporting Member
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Last weekend, my two youngest sons barely failed to qualify for the state championship Junior High wrestling tournament. One placed fifth at the district tournament, the other placed sixth. Placers one through four automatically qualified for the state tournament.
This weekend, I am taking them both to a 'second chance' tournament that can also qualify them for the State tournament. Hopefully, they both will do well.
I leave shortly for about a 3 hour drive with a van load full of seventh and eighth grade wrestlers. My sanity is a likely casualty.

to them all!
Dave  From Minnesota
Similar path, but much less efficient:
1) Start at 23. UP 26
2) naked triple 148 at e789, forbids e1=4, e23=48, e6=148. UP 29.
3a) hidden pair 67 at ei1, forbids i1=1.
3b) hidden pair 59 at ab9, forbids a9=18, b9=38
3c) locked 3's at bc7, forbids hi7=3
3d) naked triple at aeh7, forbids b7=8, ci7=18.
3e) pair 67 at i17, forbids i2=6, i3=7
3f) hidden pair 67 at g8i7, forbids g8=14.
3g) locked 4's at h78, forbids h23=4.
3h) naked triple 135 at h2i23, forbids g2=13, g3=3, h1=1, h3=35. UP 30.
4a) locked 8's at h789, forbid h4=8
4b) locked 8's at ac4, forbid a5, b6, c56=8.
4c) c2=5 == c2=8 -- b2=8 == b8=8 -- b8=6 == g8=6 -- g8=7 == g3=7 -- e3=7 == e3=3 -- i3=3 == i3=5, forbids ab3,hi2=5. UP 81.
ap  From india
hello Guru..pranam!(greetings)well thank you for descending to my level and make the proffs of late more comprehensive..well i solved todays puzzle as per your instructions..do i understand that,irrespective of the fourth corners status,i.e it may have more than one possibility,if the other three corners are sure possibility of a particular candidate,it can be identified as a unique rectangle??or in other words to observe a unique rectangle we must have minimum three corners with same possibility??
wonder if i have conveyed myself clear..
Steve  From Ohio    Supporting Member
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Hi AP!
There are many ways to configure using an almost Unique Rectangle to elimimate potential candidates.

In almost all of them, if one considers the forbidden state:
Pair xy in columns (*,**), rows (#,##) and boxes (?,??) and sets that state to false, then any solving activity that causes that forbidden state to occur is forbidden, as the puzzle then would have multiple solutions.

Common errors:
1) It is essential that the four corners are contained in exactly two distinct boxes, rows and columns.(boxes being the items often overlooked)
2)It is essential that the two candidates, xy, are possible candidates in all four corners of the proposed forbidden rectangle.

In the case of this puzzle, where 3 corners are already limited to the 2 candidates each - one can safely eliminate both of those candidates from the remaining corner - but only because both of those candidates were still possible there!

In other unique rectangle configurations, other possible eliminations can occur.

Uniqueness considerations are a bit different from standard eliminations, as they require that the forbidden state be a possible state.

Hopefully that makes some sense.
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