The New Sudoku Players' Forum

[Ruud]

Tarek,

the quad is formed by the 8 non-overlapping cells of the Windoku corners region and the Girandola region.

These cells are r5c1289 and r1289c5. Because we need 2 sets of the same 4 digits, r5c1289 must contain 4 different digits in these cells and r1289c5 must also contain these 4 digits. Furthermore, these digits are paired, so r5c19 = r28c5 and r5c28 = r19c5.

The other quads are in r5c3467 and r3467c5. Both contain the same 4 digits. Because r5c46 can see r46c5, these are also paired, so r5c46 = r37c5 and r5c37 = r46c5.

Here is the solution for your G1 with the first quads in green and the others in blue.

971|423|685
835|976|241
246|851|379
312|748|596
769|315|428
584|692|137
197|534|862
623|189|754
458|267|913

In SudoCue, I use the following names for the Windoku regions:
- LT Pane, RT Pane, LB Pane, RB Pane
- Left Stiles (r159c234)
- Right Stiles (r159c678)
- Top Stiles (r234c159)
- Bottom Stiles (r678c159)
- Corners (r159c159)

Finally, unless I made a serious mistake, your last puzzle has 5 solutions.

Ruud

[tarek]

Thanx Ruud

So we are talking about 4 sectors each having a 4cell constraint
theseare diveded into 2 groups with identical sectors

The subdicision to 2*2 is due to the fact that share a row & a sector


I checked the My last puzzle: the result came back as 1 solution Minimal puzzle.... your Last puzzle comes back as 1 solution Not minimal (r8c3 & r9c1 being superflous)

tarek

[Edit: A bug in program produced the above results]

[Ruud]

These are the 5 solutions reported by that stubborn program of mine:

Code: Select all

001000000093000040000000082000000000000000000000000000710000000030000510000000200

521864397893217645674593182386172954247935861159486723712658439438729516965341278
421678359893215647657493182368147925175329468249856731712534896936782514584961273
421678359893215647657493182368147925175329468249856731716532894932784516584961273
421678359893215647657493182368147925175329468942856731716532894239784516584961273
421678953893215647657439182968147325175923468342856791716592834239784516584361279

Where did I mess up?

Ruud

[tarek]

There was also this remote possibility that the problem from the start was at my end .

It was at my end ... . I mislabelled one of the cells in one of the sectors giving my self an extra constraint.

It's good that you discovered this .... This could have easily kept going on for ages.

Problem fixed now. back to business ..... will post a a valid puzzle when I run my program following some few tests.

Thanx,

tarek

[tarek]

This is a replacement for my last multi solution puzzle:

This should be minimal, human solvable ? time will tell

Code: Select all

 . . . | . . 8 | . . . 
 . 7 . | . . . | . 4 . 
 . . . | . . . | . . . 
-------+-------+------
 1 . . | . 4 . | . . . 
 . . . | 7 . 9 | . . . 
 . . . | . 1 . | . . 3 
-------+-------+------
 . . . | . . . | . . . 
 . 5 . | . . . | . 2 . 
 . . . | 1 . . | . . .

I am optimizing the solver so that I would post puzzles that I would provide a walkthrough solution if needed. sometime next week hopefully.

tarek

[HATMAN]

JSudoku does it with just 7 fishes - so almost certainly human solvable. I've taken the day off and am on my first bottle of wine (after many guinesses and a large G&T) - so there is little likelyhood of me doing it.

[tarek]

this should be human solvable, hopefully it could provide a challenge

Code: Select all

tarek-girandola3
.............4.......8.5.....7...8...1.....6...9...2.....3.6.......9.............
 . . . | . . . | . . . 
 . . . | . 4 . | . . . 
 . . . | 8 . 5 | . . . 
-------+-------+------
 . . 7 | . . . | 8 . . 
 . 1 . | . . . | . 6 . 
 . . 9 | . . . | 2 . . 
-------+-------+------
 . . . | 3 . 6 | . . . 
 . . . | . 9 . | . . . 
 . . . | . . . | . . . 

tarek

[HATMAN]

G3 is solvable without any big vanilla however careful use of the Law of Leftovers is needed.

If you leave the 4 out it is still unique but not human-solvable and the T&E is quite long.




7 8 3 2 6 1 4 9 5
6 5 2 9 4 3 1 7 8
9 4 1 8 7 5 6 2 3
2 3 7 6 5 9 8 4 1
4 1 5 7 8 2 3 6 9
8 6 9 1 3 4 2 5 7
5 7 8 3 2 6 9 1 4
1 2 4 5 9 8 7 3 6
3 9 6 4 1 7 5 8 2

[Ruud]

In G3, the outer quads allow immediate placement of r1c5, r5c1, c5c9 and r9c5.
After that, only pairs, triples and locked candidates are needed to solve it.

Taking out the 4 leaves a decision whether 2 or 4 belongs to the outer quads. Heavy T&E is the result...

Ruud

[tarek]

yes,

As you already pointed out, the 5 intersecting cells between the girandola & corner groups leve 8 non intersecting cells 4 of them are in row 5 the remaining 4 are in column 5.

This forces an obvious LOL between r5c28 & r19c5, similarly between r5c19 & r28c5.

this allows the easy placement of r19c5 & r5c19. It is a windoku from there.

the next puzzle will give the girandola group a more active role

tarek

[HATMAN]

Girandola 4: Double Windows AK with Flowers

There is a Girandola in here but I've used it with Double Windows which does not give the conflict with the WinDoku Corners group. (Note the eight windows give ten more hidden groups - 4 windows and 6 stiles)

I have added old lace and two other symmetrical groups (r5c5 is in all of them) which looks a bit like flower petals to me: so I calling the structure flowers.

In addition it is Anti-King (Touchless).

I have made the clues as minimally lexographic as I can. If non-consecutive or other sequencing is allowed you can do better.

This formulation is also X – if you can prove this you can use it (I could not see how to prove it).

If you use all the groups carefully you can avoid the fishes!




Edited: thanks Squirrel

[Squirrel]

I assume that r2c5 should be blue?

And would the proof be that the 4 colour regions are contained entirely within the two diagonals and r5 and c5? I'm not sure if that's enough to say for sure that the two diagonals are also regions.

[HATMAN]

Squirell

You are correct about the colour - I'll correct it tomorow.

The proof appears more complicated - the flowers + AK will allow a repeat at r1c1 and r8c8 adding a few windows stops this.

Thanks

Maurice

[tarek]

Nice variant HATMAN .... too many constraints (lovely centre cell)

here is the a proposed solution to G3

G3 Solution

Non-single moves:

.............4.......8.5.....7...8...1.....6...9...2.....3.6.......9.............

b7/c1 Line-Box
r12346c1<>3

b7/w3 Box-Window
r68c4<>4

257 is a naked triple in c4
r2c4<>2 r4c4<>2<>5 r6c4<>5<>7 r9c4<>2<>5<>7

r1/b3 Line-Box
r3c78<>4

149 is a hidden triple in r1
r1c6<>2<>3<>7 r1c7<>3<>5<>7 r1c8<>2<>3<>5<>7<>8

b3/c9 Box-Line
r5679c9<>5

c8/b9 Line-Box
r79c9<>8

58 is a hidden double in c9
r1c9<>2<>3<>7 r2c9<>1<>2<>3<>7

r1/b1 Line-Box
r23c23<>3

r9/b9 Line-box
r7c8 r8c78<>5

783261495652943178941875623237659841415782369869134257578326914124598736396417582

The variant with double windows definitely suits the girandola group better than the windoku

This poses a naming issue:

Vanilla + Girandola = Girandola
windoku + Girandola = ??? (Gwindoku )
double windows = ??? (windoku2)
double windows +Girandola = ??? (Gwindoku2)

any suggestions ??!!!

HATMAN,

I also looked at the double windows hidden constraints: I only spotted 2 boxes & 3 stiles.

tarek

[HATMAN]

Tarek

I'd go with the Gwindoku2

For the wide windows r234c123 etc. there are two hidden windows (r234c456 r678c456) and 3 stiles (r159c123 r159c456 r159c789) the same is true for the high windows r123c234 etc. (r456c234 r456c678 r123c159 r456c159 r789c159)

Maurice