SudokuSolves/OneFiveNineSquared

Archived puzzle : https://discord.com/channels/709370620642852885/727980232220606564/920521771923427328

PUZZLE = https://kitiara.has.coffee/159sqFP
SOLUTION = https://f-puzzles.com/?id=y4cnnzzo

DISCUSSION = https://discord.com/channels/709370620642852885/741008014349172737/885149815078531112

THREE TYPES:
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STEP 1: Colouring

For normal 159 puzzles, people often colour in c1, c5 and c9 to remind them that those columns are special.
Since this is 159 squared, and the rule applies horizontally too, let's do the same as well to r1, r5 and r9

https://f-puzzles.com/?id=yyvtt3hc

STEP 2: Basic indexing

https://f-puzzles.com/?id=y2bxycfh

STEP 3: Basic reverse indexing

r1c4-5 can't be "5" or "6" (it would put a 1 in box 5 which already has a 1)
r5-6c1 can't be "4" or "5" (it would put a 1 in box 5 which already has a 1)
r1 & r3 on c9 can't be "1" or "2" (it would put a 9 in box 1 which already has a 9)
r9c1-2 can't be "1" or "3" (it would put a 9 in box 1 which already has a 9)
r5c5-6 can't be "7" or "9" (it would put a 5 in box 8 which already has a 5)
r7 & r9 on c5 can't be "6" (it would put a 5 in box 8 which already has a 5)

https://f-puzzles.com/?id=yxorjwe6

STEP 4: Corners

r1c1 can be "1", but it can't be a "2" or a "3" because that would put two 1s in box 1.
r5c5 can be "5", but it can't be a "4" or a "6" because that would put two 5s in box 5.
r9c9 can be "9", but it can't be a "7" or an "8" because that would put two 9s in box 9.

https://f-puzzles.com/?id=y4jjpljf

STEP 5: Intersections

Let's conduct a thought experiment. Let's look at just the intersections of r1, r5 & r9 with c1, c5 & c9

If we restrict ourselves to filling the intersections with only the digits 1, 5 and 9, how many valid combinations are there?

159 195 591 519 915 951
591 951 915 191 159 519
915 519 159 915 591 195

It turns out there are only 6 combinations permitted by sudoku rules, all of the form:

ABC
BCA
CAB

And only 3 of those are valid under the 159 squared rules:

195 519 951
951 191 519
519 915 195

Which, if we match with our possible digits:

https://f-puzzles.com/?id=y6opkfhf

Would give us a single possibility:

A=1, B=9, C=5

195
951
519

https://f-puzzles.com/?id=y3b8w4r5

But is it possible to have other digits at those intersections?

STEP 6: Investigation

We know, because Sam C-L published one, that there are grids which comply with 159² constraints, which have some other digits:

https://f-puzzles.com/?id=y4vljkrx

But are any of those 48 compatible with what we have?

STEP 7: Consequences

https://f-puzzles.com/?id=y56slgyd

By similar reasoning to step 3:

In row 1 we can't have:
2 or 3 in c1-3
7 or 8 in c4-6
4 or 6 in c5-9

In col 1 we can't have:
2 or 3 in r1-3
7 or 8 in r4-6
4 or 6 in r5-9

In row 5 we can't have:
7 or 8 in c1-3
4 or 6 in c4-6
2 or 3 in c5-9

In col 5 we can't have:
7 or 8 in r1-3
4 or 6 in r4-6
2 or 3 in r5-9

In row 9 we can't have:
4 or 6 in c1-3
2 or 3 in c4-6
7 or 8 in c5-9

In col 9 we can't have:
4 or 6 in r1-3
2 or 3 in r4-6
7 or 8 in r5-9

https://f-puzzles.com/?id=y2baymzq

STEP 7: Placements

This gives us six hidden singles and, for most of them, we can place a 1,5 or 9, and then a third digit

row 1; R1C4 → 3
r3c4 = 1
r3c1 = 4

column 1; R6C1 → 2
r6c2 = 1
r1c2 = 6

row 5; R5C6 → 2
r2c6 = 5
r2c5 = 6

column 5; R7C5 → 8
r7c8 = 5
r5c8 = 7

row 9; R9C2 → 8
r8c2 = 9
r8c9 = 2

column 9; R8C9 → 2

tidying up any eliminations then gives us:

https://f-puzzles.com/?id=yyokbyje

STEP 8: Pairs

col 1 : 37 pair in r7-8 eliminates 7 from r2c1
that then eliminates 8 from r1c3

row 5 : 34 pair in c2-3 eliminates 4 from r5c7

row 9 : 67 pair in c4,6 eliminates 6 from c7-8
box 9 : 34 pair eliminates 4 from r7c9

both of those then eliminates the 6 from r6c9
making a 48 pair in col 9 which eliminates the 8 from r3c9
and then the 7 from r1c7

https://f-puzzles.com/?id=yxut56um

STEP 9: More placements

r1c3 = 7, r7c3 = 1, r7c1 = 3, eliminate 3 from r8c1
r2c1 = 8, r2c8 = 1, r1c8 = 2, eliminate 2 from r1c7
r5c7 = 6, r6x7 = 5, r6c5 = 7, eliminate 7 from r4c5
r3c5 = 2, r3c2 = 5, r5c2 = 3, eliminate 3 from r5c3
r3c9 = 7, r3c7 = 9, r9c7 = 3, eliminate 3 from r9c8
r7c9 = 6, r7c6 = 9, r9c6 = 7, eliminate 7 from r9c4

https://f-puzzles.com/?id=y2kkwhyu

STEP 10: Third time lucky - final placements

r1c7 = 8, r8c7 = 1, r8c1 = 7
r5c3 = 4, r4c3 = 5, r4c5 = 3
r9c4 = 6, r6c4 = 9, r6c9 = 4
r9c8 = 4, r4c8 = 9, r4c9 = 8

That completes r1, r5, r9, c1, c5 & c9

https://f-puzzles.com/?id=y2lhnkta

STEP 11: Naked all the way

The remaining naked singles can be placed via straight sudoku in under 30 seconds:

  Naked single;  R2C2 → 2
  Naked single;  R2C4 → 7
  Naked single;  R2C7 → 4
  Naked single;  R3C3 → 3
  Naked single;  R3C6 → 8
  Naked single;  R3C8 → 6
  Naked single;  R4C2 → 7
  Naked single;  R4C4 → 4
  Naked single;  R4C7 → 2
  Naked single;  R6C3 → 8
  Naked single;  R6C6 → 6
  Naked single;  R6C8 → 3
  Naked single;  R7C2 → 4
  Naked single;  R7C4 → 2
  Naked single;  R7C7 → 7
  Naked single;  R8C3 → 6
  Naked single;  R8C6 → 3
  Naked single;  R8C8 → 8

https://f-puzzles.com/?id=y4h28a4d

STEP 12: Easter Egg

Where Q={1,5,9}; rXcY? ⊃ Z; For ∀ X,Y:
rZcY? ⊃ X except where X∈Q and Y∈Q'
rXcZ? ⊃ Y except where Y∈Q and X∈Q'

To see what this means, let's do a bit of colouring.

I'm going to make the 4 'windows' green, and also the 9 intersections:

https://f-puzzles.com/?id=yx8wv3oo

the remaining red bits I'm going to turn blue if vertical and yellow if horizontal:

https://f-puzzles.com/?id=y4z7zwvc

And, just for effect, let's make the diagonal from r1c1 to r9c9 be a darker shade of green:

https://f-puzzles.com/?id=y5gwtc6m

Via normal 159 squared rules, we have a "1" in r1c1 telling us that to put a "1" in, err, r1c1

well, looking at the dark green cells, we can see a similar rule could be applied to each of them too, if this were a 123456789² puzzle. But surely such a thing is impossible?

Let's take a random light green cell from a window, that doesn't contain a 1, 5 or 9, such as r2c4 which contains a "7"

If this were a 123456789² puzzle, then row 7 would index the vertical position (2) and column 7 would index the horizontal position (4). And if we look at r7c4 and at r2c7 we see... a "2" and a "4". Hang on, that's can't be right. Must be a coincidence. Try again:

r8c6 is "3", so we'd need r3c6 to index its vertical position as 8, and r8c3 to index its horizontal position as "6". Checking... it's true!

In other words, horizontal 123456789 rules apply to blue squares
vertical 123456789 rules apply to yellow squares
And, since green = blue + yellow...
Both sets of constraints, full 123456789² rules, apply to every single green square.

It wasn't a result I anticipated when setting this puzzle. It surprised the heck out of me.
It wasn't planned or selected for - it is genuine emergent behaviour from the 159 rules.
I don't understand the mathematics behind it. Why did placing just 3 digits on a sudoku grid cause this?
But it's beautiful, isn't it?