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Sunday 18-Oct-2020

... by: Marek

I've had a look at recent WUs and found many occurrences of such a pattern.
WUs #417, #408, #407 and #405 even have its extensions.
I've posted a description of each of them under those puzzles.

Sunday 18-Oct-2020

... by: Frans Goosens

With trial and error

Combination B4=137 and F9=345

************************************************************
B4=1 F9=3 ( A1=4 ) Solved,


#422--------------------------Solution
000 000 201-------------485 679 231
090 050 860-------------293 154 867
600 020 090-------------617 328 495

000 003 002-------------569 783 142
000 900 000-------------328 941 576
700 060 980-------------741 265 983

006 090 750-------------136 892 754
850 400 000-------------852 437 619
970 000 000-------------974 516 328

Total solving time is : 13 sec.
Number of logical steps is : 185

Saturday 17-Oct-2020

... by: numpl_npm

6A4 9A6 9D3 9H9

JE2 [1234]HJ3 B1 F2 (i.e. AB1=HJ3 A1=F2)

5A1 8E3.

| a _ 5 | 6 _ 9 | 2 _ 1 | A
| b 9 _ | _ 5 _ | 8 6 _ | B
| 6 _ _ | _ 2 _ | _ 9 _ | C

| _ _ 9 | _ _ 3 | _ _ 2 | D
| _ _ 8 | 9 _ _ | _ _ _ | E
| 7 a _ | _ 6 _ | 9 8 _ | F

| _ _ 6 | _ 9 _ | 7 5 _ | G
| 8 5 _ | 4 _ _ | _ _ 9 | H
| 9 7 _ | _ _ _ | _ _ _ | J

(ab in HJ3)

(a)G9 (a)C7
| a=3 -> (a)BC4 (a)G9 (a)C7
| a=4 -> (a)G9 (a)C7

Then solved with basics only.

Saturday 17-Oct-2020

... by: numpl_npm

Sorry. The previous post is incorrect.

Saturday 17-Oct-2020

... by: numpl_npm

6A4 9A6 9D3 9H9

JE2 [1234]HJ3 B1 F2 (i.e. AB1=HJ3 A1=F2)

5A1 8E3.

| a _ 5 | 6 _ 9 | 2 _ 1 | A
| b 9 _ | _ 5 _ | 8 6 _ | B
| 6 _ _ | _ 2 _ | _ 9 _ | C

| _ _ 9 | _ _ 3 | _ _ 2 | D
| _ _ 8 | 9 _ _ | _ _ _ | E
| 7 a _ | _ 6 _ | 9 8 _ | F

| _ _ 6 | _ 9 _ | 7 5 _ | G
| 8 5 _ | 4 _ _ | _ _ 9 | H
| 9 7 _ | _ _ _ | _ _ _ | J

(ab in HJ3)

By Marek's nice idea,

(a)G9 (a)C7
| a=3 -> (a)BC4 (a)G9 (a)C7
| a=4 -> (a)G9 (a)C7

-(b)C469 (b)A8
| b=1 -> -(b)C9 and (b)F46 (b)G46 -(b)C46
| b=2 -> -(b)C469
| b=3 -> -(b)C6 and (b)F9 (b)G4 -(b)C49
| b=4 -> -(b)C4 and (b)G9 (b)F6 -(b)C69

(ab)A8.C7 57BC9

Then solved with basics.

Saturday 17-Oct-2020

... by: solver_delta

485 679 231
293 154 867
617 328 495
569 783 142
328 941 576
741 265 983
136 892 754
852 437 619
974 516 328

Saturday 17-Oct-2020

... by: Marek

JE (1234) HJ3 F2 B1:

Direct eliminations:
-12F2 (false in the mirror node)
-5A1, -8DE2 (non-base digits in mirror nodes, 6 is already locked in both)

Looking at the puzzle, we are close to having an additional cross line (rC).
C18 non-base givens
C2 in sight of both targets
C3 in sight of the base
C469 cannot contain a true base digit, since they lie in the cover houses for each base candidate included in them.
Therefore C57 contain the base digits.
+2HJ3 B1 E2
-1HJ3
-5C7 +5C9 => Naked Pair 34F29: -34F346

The digit (3 or 4) that is true in the base, is true in F2 (therefore false in F9) and true in C7.
Remote Pair 34C7F9: -34B9DE7


stte

Saturday 17-Oct-2020

... by: numpl_npm

Many similar puzzles inspire me how to solve those puzzles.

Saturday 17-Oct-2020

... by: numpl_npm

6A4 9A6 9D3 9H9

JE2 [1234]HJ3 B1 F2 (i.e. AB1=HJ3 A1=F2)

5A1 8E3

| _ _ 5 | 6 _ 9 | 2 _ 1 | A
| _ 9 _ | _ 5 _ | 8 6 _ | B
| 6 _ _ | _ 2 _ | _ 9 _ | C

| _ _ 9 | _ _ 3 | _ _ 2 | D
| _ _ 8 | 9 _ _ | _ _ _ | E
| 7 _ _ | _ 6 _ | 9 8 _ | F

| _ _ 6 | _ 9 _ | 7 5 _ | G
| 8 5 _ | 4 _ _ | _ _ 9 | H
| 9 7 _ | _ _ _ | _ _ _ | J

4A1
| 8J5 -> 8G9 then 3AJ5=[3A5 -3A1|3J5 -3HJ3 -3A1]=-3A1 4A1
| 7H5 -> 7A8 7E9 68GJ9 then 3AJ5=[3A5 -3A1|3J5 -3HJ3 -3A1]=-3A1 4A1
| 13HJ5 -> 7H6 6H7 6E9 7DE8 7A5 7E8 then 3AHJ8=[3A8 -3A1|3HJ8 -3HJ3 -3A1]=-3A1 4A1

Saturday 17-Oct-2020

... by: Cartman1337

Your solved solved this one, so shouldn't be posted as "unsolvable".

Additionally, I'd like to see some variation in the puzzles. I guess you have a long list of puzzles, and just post them in sorted order (rather than a preferred random order, which would give some variation from week to week). As it is, there's always a corner box with missing candidates 1/2/3/4, and at least one Exocet elimination on the same candidates. Most times there will always be an 8 in the top row of the conjoining box to avoid a naked pair unique rectangle problem on two of the aforementioned candidates, i.e. G4 became 8 here, and I could tell straight away that J9 would be 8 because of the other candidates in H/J3 and G4/6, as so many times before in these puzzles. As such, I'd love to see the puzzles chosen in a random order rather than an ordered selection, to provide some other variations on the candidates from week to week, and not always have to be about 1, 2, 3, 4 and 8...

Saturday 17-Oct-2020

... by: numpl_npm

6A4 9A6 9D3 9H9

JE2 [1234]HJ3 B1 F2 (i.e. AB1=HJ3 A1=F2)

With basics only,
 3A1 & 3F2 -> contra.
 4A1 & 4F2 -> solved.

Saturday 17-Oct-2020

... by: James Havard

Top box row 49 subs and 64 seconds and bottom box row 37 subs and 47 seconds. Picking one of the subs from the top box row and one from the bottom box row to get A1=4 with G2=3 to get a singles solution.

Saturday 17-Oct-2020

... by: James Havard

263 subs and 272 seconds for E1=3 with D7=1 for a basics solution.

Saturday 17-Oct-2020

... by: BobW

JE2: (1234)HJ3, B1 F2 => -12F2 -2EF3 -5A1 -8DE2
Then many chains to solution.

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