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

Corollary 35b is not valid (35b-> 17a)

And false any upstream pair as
#3r4c4 6r5c5 ->35b
#3r4c4 1r5c4 ->35b
#3r4c6 6r6c5 ->35b

=================== #24 r7c79
if(24r7c79)
Can’t be 24 in base must be ‘1’ + 2|4
triplet 357 r4c467 ->24c -> 36f same conflict

Here is the point where I mark a pause to optimize the next pair eliminations

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

status after post 1

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 still possible in base 12 14 17 24 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 #17a #35b -> 17a #3r4c4 6r5c5 ->35b #3r4c4 1r5c4 ->35b #3r4c6 6r6c5 ->35b #24 r7c79

new pair éliminations (and corollary invalid pairs)
mainly using the backbone a,b,c,d,e,f and the exocet

Code: Select all: interesting step #24c (17r4c8; 36r7c12) if (24c) must be 1r4c4 | 7 r4c6 if 1r7c4;7r4c8 -> must be 1r7c9 r7c7 is empty if 7r4c6,1r4c8 -> 4r7c6,2r7c4 again r7c7 empty

#24c
#56d (-> 24c)
#17e (-> 24c)
#2r4c1 6r5c3 ->24c
#4r4c2 6r6c3 ->24c

Code: Select all: ====#17r7c12 with 24r4c78 and 24r7c8 r4c7is empty

#2346r7c4679 #368r8c13r9c23

Code: Select all: ===== #14 in base 14r7c12 now 36r7c26 7r7c1 2r7c7 1r7c4 so 4r7c9;1r4c8 35r4c47,24r4c12 not valid

Now still possible in base 12 17 24

quick pairs éliminations

Code: Select all: ==== #2r4c1 3r4c4 ->25r4c12 37r4c57 39HPr5c1r6c2 4r5c3 6r6c3 6r6c6 conflict

Code: Select all: #1457r4c2678 ===== #4r4c2 3r4c6 ->25r4c12 31r4c57 6r5c3 6r5c5 conflict

this will be the end for thix post, next one will give a new set of pair eliminations now not too far from the first candidate clearing

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

this is my current best way to show it

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

if(3r4c6)
3r4c6 -> 6r5c5;5r4c12 -> 4r5c3 this is 2r4c1 5r4c2 4r4c7
4r4c7 -> 4r3c9 4r8cc8 89 r69c8
3r5c1 not valid (8r5c9r6c8)
8r8c1
no 3 column 1

if(5r4c6) slightly more complex
I show the PM after 5r4c6 and easy moves

Code: Select all: 2568 4568 7 |268 246 48 |1 3 9 2689 4689 24 |23689 2346 1 |7 26 5 269 1 3 |269 5 7 |8 246 24 --------------------------------------------------------- 23 34 8 |1 9 5 |24 7 6 569 7 456d |368 36 2 |3459 1 348 1 569 256d |4 7 38 |2359 289 238 --------------------------------------------------------- 7 36 9 |2 8 346h |34 5 1 38 2 1 |5 34 349 |6 489 7 4 3568 56e |7 1 369 |239 289 238

can be r4c12 23 or r4c12 34

if(23r4c12)

3r8c1
4r5c7->4r8c3
no solution for 34r8c5

if(34r4c12)

again the refreshed PM for an easier reading

Code: Select all: 2568 4568 7 |268 246 48 |1 3 9 2689 4689 24 |23689 2346 1 |7 26 5 269 1 3 |269 5 7 |8 246 24 --------------------------------------------------------- 3 4 8 |1 9 5 |2 7 6 569 7 56d |368 36 2 |3459 1 348 1 569 2 |4 7 38 |359 89 38 --------------------------------------------------------- 7 36 9 |2 8 346h |34 5 1 8 2 1 |5 34 349 |6 49 7 4 356 56e |7 1 369 |39 289 238

if(3r7c7)
9r9c7; 9r6c8; 5 r6c7 ; 6r7c2 r6c2 empty

if(4r7c7)

9r8c8;8r6c8
38r6c689 conflict

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 any eliminations.

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 14568 124567 |268 2467 4678 |1 3 9 26789 4689 2467 |23689 23467 1 |2 467 5 2679 1 3 |269 5 4679 |8 467 47 --------------------------------------------------------- 235c 345c 8 |135b 9 357b |3457 12 6 3569 7 456d |13568 136 2 |3459 1489 348 1 3569 256d |4 367 35678 |3579 2789 378 --------------------------------------------------------- 367f 36f 9 |1236 8 346h |347h 5 12 3578 2 157e |1359 134 3459 |6 4789 13478 4 3568 156e |7 1236 3569 |39 89 1238

can be 1|2 r4c8

if 1r4c8 2r7c9
1r4c8 -> 2r4c1 #4r7c2 4r5c3 7r4c6 4r7c7
then
1r7c4 4r7c6-> 6r5c5 5r6c3 17e not valid

if 2r4c8 1r7c9

The New Sudoku Players' Forum

Golden nugget updated path

Golden nugget updated path

Re: Golden nugget updated path

showing 17 in base not valid

show 17 in base not valid

Golden nugget updated path secon phase

#12 in base