The New Sudoku Players' Forum

by **champagne** » Thu Jan 01, 2026 8:30 am

Golden Nugget

Code: Select all: .......39.....1..5..3.5.8....8.9...6.7...2...1..4.......9.8..5..2....6..4..7.....

11,9 11,9 11,3 tax Golden-Nugget

Long ago (2008) I published on my website, now dead, a full solution for this very hard puzzle.
Recently, I was curious to see how “yzfwsf” ’s program would solve it. It is still unsolved, so I wanted to update my old solution and post it here.
I start with the first steps, needed later. As far as I could see, I have a lot of work to optimize the next steps, they will come in due time, likely in several more posts

The start PM for the puzzle

Code: Select all: 25678 14568 124567 |268 2467 4678 |1247 3 9 26789 4689 2467 |23689 23467 1 |247 2467 5 2679 1469 3 |269 5 4679 |8 12467 1247 --------------------------------------------------------- 235 345 8 |135 9 357 |123457 1247 6 3569 7 456 |13568 136 2 |13459 1489 1348 1 3569 256 |4 367 35678 |23579 2789 2378 --------------------------------------------------------- 367 136 9 |1236 8 346 |12347 5 12347 3578 2 157 |1359 134 3459 |6 14789 13478 4 13568 156 |7 1236 3569 |1239 1289 1238

Junior Exocet:Base Cells-r1c7,r2c7;Target Cells-r4c8,r4c9,r7c8,r7c9;Cross Cells-r347c123456
Target Cells Check: r7c9<>3

abi loop #27r12c7;#47r12c7 => #7r1c7
still possible in base 12 14 17 24

The PM after the JE

Code: Select all: 25678 14568 124567 |268 2467 4678 |124 3 9 26789 4689 2467 |23689 23467 1 |247 2467 5 2679 1469 3 |269 5 4679 |8 12467 1247 --------------------------------------------------------- 235c 345c 8 |135b 9 357b |123457a 1247a 6 3569 7 456d |13568 136 2 |13459 1489 1348 1 3569 256d |4 367 35678 |23579 2789 2378 --------------------------------------------------------- 367f 136f 9 |1236 8 346h |12347h 5 1247h 3578 2 157e |1359 134 3459 |6 14789 13478 4 13568 156e |7 1236 3569 |1239 1289 1238

The pairs of cells marked a,b,c,d,e,f play a key role.
17a =>35b =>24c => 56d => 17e => 36f
17f =>56e =>24d => 35c => 17b => 24a

In my path, the first part is to established false pairs needed later.

=======================> the main one is #17a
If 17a, (=>36f)
can’t be 17 in base so must be ‘1’+ 2|4 (+7r4c7)
‘h’ is now 24, not valid

[color=#0000BF][b]
Corollary 35b is not valid (35b-> 17a)

EDIT updated cleaned unused pairs not valid

by **champagne** » Fri Jan 02, 2026 5:11 am

[url][/url]after post 1, still possible in base 12 14 17 24
===========================================
interesting step #24c

if((24c), after the key clearing

Code: Select all: 25678 14568 124567 |268 2467 4678 |12 3 9 26789 4689 2467 |23689 23467 1 |27 2467 5 2679 1469 3 |269 5 4679 |8 12467 1247 --------------------------------------------------------- 2 4 8 |135b 9 357b |35 17 6 3569 7 56 |13568 136 2 |4 189 138 1 3569 56 |4 367 35678 |23579 2789 2378 -------------------------------------------------------- 36 36 9 |12 8 4 |127 5 127 3578 2 7 |1359 13 3459 |6 14789 13478 4 13568 1 |7 1236 3569 |39 289 238

if 1r7c4;7r4c8 -> must be 1r7c9
r7c7 is empty

if 7r4c6,1r4c8
->27r7c79;1r7c4;3r8c5;3r2c4;5r4c4;3r4c7;8r5c9
now 6r5c4;6r6c5 conflict

#24c
#56d (-> 24c)
#17e (-> 24c)

====================================================

#17f (#17r7c12)
with 24r4c78 and 24r7c8 r4c7is empty

=====================================================

#14 in base 14r7c12

Code: Select all: ===== now 36r7c26 7r7c1 2r7c7 1r7c4 so 4r7c9;1r4c8 35r4c47,24r4c12 not valid

========================================================
Now still possible in base 12 17 24

EDIT revised the clearing of 24c cleaned unused

by **champagne** » Fri Jan 02, 2026 8:30 am

At this point, showing 17 in base not valid seems a good idea but...
17 in base -> 17r3c12 or 17r3c26

17r3c12 -> 17e not valid, so must be 17r3c26
as #35r4c46, this bring us immediately to

Code: Select all: 25678 14568 124567 |268 2467 4678 |1 3 9 26789 4689 2467 |23689 23467 1 |7 2467 5 2679 1 3 |269 5 7 |8 12467 1247 --------------------------------------------------------- 235c 345c 8 |1 9 35 |123457a 7 6 3569 7 456d |13568 136 2 |13459 1489 1348 1 3569 256d |4 367 35678 |23579 2789 2378 --------------------------------------------------------- 367f 136f 9 |1236 8 346h |12347h 5 1 3578 2 157e |1359 134 3459 |6 14789 13478 4 13568 156e |7 1236 3569 |1239 1289 1238

I am looking for a relatively quick conflict to see this step as the good one here

EDIT
if 17 in base is not valid, last valid pairs in base are 12 and 24
So
7 is not in base
2 is in base

This is a huge step in the solution path, so IMO, this is worth a relatively hard step
Next post will show my best current elimination, I'll try to improve it later

by **champagne** » Sat Jan 03, 2026 8:17 am

can not be 17r3c12 -> 17e not valid

1r3c2;7r3c6 after basics the PM looks like

Code: Select all: 2568 4568 7 |268 246 468 |1 3 9 2689 4689 246 |23689 2346 1 |7 26 5 269 1 3 |269 5 7 |8 246 24 --------------------------------------------------------- 235c 345c 8 |1 9 35 |24 7 6 3569 7 456d |3568 36 2 |3459 1 348 1 3569 256d |4 7 3568 |2359 289 238 --------------------------------------------------------- 7 36 9 |236 8 346h |234 5 1 358 2 15 |35 134 3459 |6 489 7 4 3568 156e |7 1236 3569 |239 289 238

can not be 2r4c1
2r4c1->4r4c7;4c8c8;8c8c1;569r123c13r5c1;9r6c2
;4r3c9;26r23c8
now 8r5c9 8r6c8 conflict

refreshed PM without 2r4c1

Code: Select all: |2568 568 7 |68 246 468 |1 3 9 | |2689 689 4 |3689 236 1 |7 26 5 | |269 1 3 |69 5 7 |8 246 24 | ++++++++++++++++++++++++++++++++++++++++++++++ |35 4 8 |1 9 35 |2 7 6 | |3569 7 56 |3568 36 2 |3459 1 348 | |1 3569 2 |4 7 3568 |359 89 38 | ++++++++++++++++++++++++++++++++++++++++++++++ |7 36 9 |2 8 346 |34 5 1 | |358 2 15 |35 134 3459 |6 489 7 | |4 3568 156 |7 136 3569 |39 289 238 |

if 3 r4c6 -> 5r4c1;6r5c5-> r5c3 empty

without 3r4c6

Code: Select all: |2568 568 7 |68 246 468 |1 3 9 | |2689 689 4 |3689 236 1 |7 26 5 | |269 1 3 |69 5 7 |8 246 24 | ++++++++++++++++++++++++++++++++++++++++++++++ |3 4 8 |1 9 5 |2 7 6 | |3569 7 56 |3568 36 2 |3459 1 348 | |1 3569 2 |4 7 3568 |359 89 38 | l ++++++++++++++++++++++++++++++++++++++++++++++ |7 36 9 |2 8 346 |4 5 1 | |58 2 1 |5 134 3459 |6 489 7 | |4 358 156 |7 1 9 |3 289 238 | 5r4c6 5r5c4 1r8c3 1r9c5 3r4c1 #3r8c1

xw 6r56c25 + 3r8c46 ->9r9c6 "6r9c3;3r9c7;4r8c7
now 6r9c3;6r6c2;5r5c3
9r5c7
r5c1 is empty

we have covered all the tree, can not be 17 in base
now only possible in base 12 and 24
no 7 in base, 2 in base, we restart with a new pm after many eliminations.

EDIT changed the path for #17 in base

by **champagne** » Sun Jan 04, 2026 8:41 am

we are now with this reduced PM

Code: Select all: 25678 14568 124567 |268 2467 4678 |124 3 9 26789 4689 2467 |23689 23467 1 |24 467 5 2679 1469 3 |269 5 4679 |8 1467 147 --------------------------------------------------------- 235c 345c 8 |135b 9 357b |13457 124 6 3569 7 456d |13568 136 2 |13459 1489 1348 1 3569 256d |4 367 35678 |3579 2789 378 --------------------------------------------------------- 367f 136f 9 |1236 8 346h |1347h 5 124 3578 2 157e |1359 134 3459 |6 14789 13478 4 13568 156e |7 1236 3569 |139 189 1238

are still valid in base 12 24

These impossible pairs seem the most usefull for the next steps

#35b
#24c
#56d
#17e
#17f

In the next step, we show that "1" can not be in the base

by **champagne** » Mon Jan 05, 2026 8:00 am

with 12 in base after easy moves, we are here

Code: Select all: |25678 4568 24567 |268 2467 4678 |1 3 9 | |6789 4689 467 |3689 3467 1 |2 467 5 | |2679 1 3 |269 5 4679 |8 467 47 | ++++++++++++++++++++++++++++++++++++++++++++++++++++++ |235 345 8 |135 9 357 |3457 12 6 | |3569 7 456 |13568 136 2 |3459 1489 348 | |1 3569 256 |4 367 35678 |3579 2789 378 | ++++++++++++++++++++++++++++++++++++++++++++++++++++++ |367 36 9 |12 8 346 |347 5 12 | |3578 2 157 |1359 134 3459 |6 4789 13478 | |4 3568 156 |7 1236 3569 |39 289 1238 |

can be 1|2 r4c8

if(1r4c8)
1r4c8 -> 2r4c1 #4r4c2 4r5c3 7r4c6 4r7c6
then
1r7c4 4r7c6 6r9c56 3r8c4-> 6r6c5 5r6c3 17e not valid

if 2r4c8 1r7c9 next post
EDIT changed the conflict for 1r4c8

by **champagne** » Tue Jan 06, 2026 11:09 am

If 2r4c8 1r7c9
this is longer to get the conflict, so I keep it in a separate post
first the PM after basics

Code: Select all: |5678 4568 4567 |68 2 4678 |1 3 9 | |6789 4689 467 |368 3467 1 |2 467 5 | |2 1 3 |69 5 4679 |8 467 47 | +++++++++++++++++++++++++++++++++++++++++++++++++ |35 345 8 |1 9 357 |3457 2 6 | |3569 7 456 |3568 36 2 |3459 1 348 | |1 3569 2 |4 367 35678 |3579 789 378 | +++++++++++++++++++++++++++++++++++++++++++++++++ |367 36 9 |2 8 346 |347 5 1 | |3578 2 157 |359 134 3459 |6 4789 3478 | |4 3568 156 |7 136 3569 |39 89 2 |

if 4r5c3 7r4c6 #7r5c5 4r4c8 4r7c6 4r2c5
no 7 in column 5

if 4r4c2

#17e -> 47 r12c3 -> triplet 568 r1c124 -> #8r1c6;8r6c6 #8r5c4
356r5c345 -> 9r5c1;4r5c7 4r7c6 #4r8c5
136r589c5->7r6c5 #7r4c6
this pushes us here

Code: Select all: |568 568 47 |68 2 47 |1 3 9 | |689 689 47 |368 3467 1 |2 467 5 | |2 1 3 |69 5 4679 |8 467 47 | +++++++++++++++++++++++++++++++++++++++++++++++++ |35 4 8 |1 9 35a |357 2 6 | |9 7 56a |356a 36a 2 |4 1 348 | |1 3569 2 |4 7 8 |3579 789 378 | +++++++++++++++++++++++++++++++++++++++++++++++++ |367 36 9 |2 8 4 |347 5 1 | |3578 2 15 |359 13 3459 |6 4789 3478 | |4 3568 156 |7 136 3569 |39 89 2 |

6r9c56-> #6r9c3
in rcells marked 'a' must be r5c6=6 #6r5c3
conflict 15r589c3

now we have 24 in base and a new PM
Next step will be to clear the wrong case in targets 2r4c8 4r7c9 with then 2 key cells assigned

by **champagne** » Sat Jan 10, 2026 10:24 am

After no 1 in base, we have 24 in base
To reach an assignment, we must clear the wrong case 2r4c8 4r7c9.
Then, the puzzle is still rated skfr 8.4. This explains in some ways that an easier path by passing this step is very hard to build.

Here is my PM at this point

Code: Select all: 25678 14568 124567 |268 2467 4678 |24 3 9 26789 4689 2467 |23689 23467 1 |24 67 5 2679 469 3 |269 5 4679 |8 167 17 --------------------------------------------------------- 235c 345c 8 |135b 9 357b |1357 24 6 3569 7 456d |13568 136 2 |1359 1489 138 1 3569 256d |4 367 35678 |3579 2789 378 --------------------------------------------------------- 367f 136f 9 |1236 8 346h |137h 5 24 3578 2 157e |1359 134 3459 |6 1789 13478 4 13568 156e |7 1236 3569 |139 189 1238

Trying directly to show 2r4c8 4r7c9 seems difficult.
I'll show first
#2r4c8 4r7c9 3r4c4
#2r4c8 4r7c9 5r4c4

The PM in the exocet area partially solved using basic for 2r4c8 4r7c9

Code: Select all: 5678 1568 14567 |68 267 678 |24 3 9 789 89 47 |389 237 1 |24 6 5 2 69 3 |69 5 4 |8 17 17 --------------------------------------------------------- 35c 4 8 |135b 9 357b |1357 2 6 3569 7 56d |13568 136 2 |1359 4 138 1 3569 2 |4 367 35678 |3579 789 378 --------------------------------------------------------- 367f 136f 9 |2 8 36h |137h 5 4 3578 2 157e |1359 4 359 |6 1789 1378 4 13568 156e |7 136 3569 |139 189 2

with #17r7c12 -> 3|6 r7c12 -> #3r7c7
with #17r89c3 -> 5|6 r89c3 -> 147 r12c3
and the PM to show not valid

Code: Select all: 5678 1568 147 |68 267 678 |24 3 9 789 89 47 |389 237 1 |24 6 5 2 69 3 |69 5 4 |8 17 17 --------------------------------------------------------- 35c 4 8 |135b 9 357b |1357 2 6 3569 7 56d |13568 136 2 |1359 4 138 1 3569 2 |4 367 35678 |3579 789 378 --------------------------------------------------------- 367f 136f 9 |2 8 36h |17h 5 4 3578 2 157e |1359 4 359 |6 1789 1378 4 13568 156e |7 136 3569 |139 189 2

===============if 3r4c4 3r2c5
#35r4c46->7r4c6
27 locked in r1c5 in column 5

=============== if 5r4c4
my best contradiction
row 4 is 348597126
7r7c7 1r7c2 -> box 7

Code: Select all: 619 827 436

then 9r9c7 5r8c6 no room for "9" in column 6

now in the PM to prove false we have r7c4=1 -> 1r9c5
can't be 7r8c1, one of 17 r7c12 must be there

Code: Select all: 5678 1568 147 |68 267 78 |24 3 9 789 89 47 |389 237 1 |24 6 5 2 69 3 |69 5 4 |8 17 17 --------------------------------------------------------- 35c 4 8 |1 9 357b |357 2 6 3569 7 56d |3568 36 2 |1359 4 138 1 3569 2 |4 367 3578 |3579 789 378 --------------------------------------------------------- 367f 136f 9 |2 8 36h |17h 5 4 358 2 17e |359 4 359 |6 1789 1378 4 3568 56e |7 1 3569 |39 89 2

by **champagne** » Mon Jan 12, 2026 9:24 am

This step is short but gives the 2 key assigned cells
(can't be 7r8c1, one of 17 r7c12 must be there)

Code: Select all: 5678 1568 147 |68 267 678 |24 3 9 789 89 47 |389 237 1 |24 6 5 2 69 3 |69 5 4 |8 17 17 --------------------------------------------------------- 35 4 8 |1 9 357 |357 2 6 3569 7 56 |3568 36 2 |1359 4 138 1 3569 2 |4 367 35678 |3579 789 378 --------------------------------------------------------- 367 136 9 |2 8 36h |17 5 4 358 2 17 |359 4 359 |6 1789 1378 4 3568 56 |7 1 3569 |39 89 2

if 1r8c3
1r1c2;5r1c1;7r7c1;3r4c1;8r8c1;9r2c1;6r5c1;
5r5c3;9r6c2

8r8c1->8r9c8
7r7c1->7r46c7 789r6c8 is empty

if 7r8c3
7r7c7;7r4c6;1r5c7; 4r2c3;1r1c3

68r1c46->9r3c4
36r56c5->3r2c4;
->5r8c4; 8r5c4
->6r3c2;6r5c13;3r5c5
cell r5c9 empty

the main job is over
can't be 2r4c8 4r7c9
last is 4r4c8 2r7c9 assigned

the puzzle is now skfr 8.4, but we have already solved the exocet pattern.
I have to clean unused steps in the first posts.
next and last post will be the path derived from "yzfwsf" proposal for the new puzzle

.......39.....1..5..3.5.8....8.9..46.7...2...1..4.......9.8..52.2....6..4..7.....

All this has been done without the help of a computer, I have no tool so far to work on invalid PMs. I am sure that much shorter contradictions can be found.
This path fits with ageneral strategy suggested years ago for puzzles having the exocet pattern.
It is much better than what I posted in 2008

by **champagne** » Tue Jan 13, 2026 9:18 am

I intended first to use skfr of "yzfwsf" solvers to set up the last steps.
In both cases, we stille ahve plenty of smaller steps.

I'll finish using a one step smally harder step having applied the kite 7r4c67 7r36c9 -> #7r3c9

The PM is then

Code: Select all: 5678 1568 127 |68 467 678 |24 3 9 6789 689 27 |3689 3467 1 |24 67 5 679 4 3 |2 5 69 |8 167 17 --------------------------------------------------------- 2 35c 8 |15b 9 357b |1357 4 6 3569 7 4 |13568 136 2 |1359 189 138 1 3569 56d |4 367 35678 |3579 2 378 --------------------------------------------------------- 367f 136f 9 |136 8 4 |137h 5 2 358 2 157e |1359 13 359 |6 1789 4 4 13568 156e |7 2 3569 |139 189 138

if 27r12c3
1r1c2;5r1c1
can not be 17e 7r7c1;7r8c8;8r8c1
threat of invalid pattern 247r12c357 -> 3|6r12c5 -> triplet 136 in c5 7r6c5;7r1c6
then the pattern is solved and to clarify this a PM after key points

Code: Select all: 5 1 2 |68 6 7 |4 3 9 689 689 7 |3689 4 1 |24 67 5 69 4 3 |2 5 69 |8 167 17 --------------------------------------------------------- 2 35c 8 |15b 9 357b |1357 4 6 3569 7 4 |13568 136 2 |1359 189 138 1 3569 56d |4 7 35678 |3579 2 378 --------------------------------------------------------- 7 136f 9 |136 8 4 |13 5 2 8 2 15 |1359 13 359 |6 7 4 4 13568 156e |7 2 3569 |139 189 138

now r5c1=3;r4c2=5 r4c4=1 r5c5 is empty

r1c3=1 BTTE

The puzzle is solved, with several unused steps and the door open for better conflicts.
I'll revised the first posts to clean the path

by **champagne** » Fri Jan 16, 2026 9:28 am

Have started the cleaning of the path in the first posts.
One error in #24c seen; another conflict shown now

next will be the revision( if possible) of #17 in base

by **champagne** » Fri Jan 23, 2026 9:16 am

I checked the path and made several adjustments, typo or errors or better conflicts.
The edited path seems correct.

This is the end of my work on this very very hard sudoku
I tried to see what "yzfwsf" ' code could do with "platinum blonde", another very old puzzle.
I have seen interesting points, so, i'll post it in a separate thread in the next weeks.

The New Sudoku Players' Forum

Golden nugget updated path

Golden nugget updated path

showing #24c #17e #14 in base

showing 17 in base not valid

show 17 in base not valid

Golden nugget updated path secon phase

#12 in base

2r4c8 1r7c9 not valid

Re: Golden nugget updated path

Re: Golden nugget updated path

Re: Golden nugget updated path

Re: Golden nugget updated path

Re: Golden nugget updated path