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This is a place chosen to make available some ideas (on puzzles, patterns, etc)
that seemed unsuitable for posting in a Forum, but some people may find interesting.
For difficult puzzles, possible solutions will not be shown (only hints).
For other puzzles, the solution wil be hidden so that a serious attempt can be made
without direct access to a solution.
(hopefully this will make the puzzles more interesting for newcomers)
(Guidelines for using this site here)
(useful to look up for the meaning of some unfamiliar concept)
I propose this newly created Hidoku puzzle (for a manual solution !)
.---------------------------.
7 | 16 __ __ __ __ __ 49|
6 | __ 19 __ __ __ __ __|
5 | __ 27 __ __ __ __ __|
4 | __ 24 __ __ __ __ 10|
3 | 29 __ __ __ __ __ __|
2 | __ __ __ 34 39 __ 42|
1 | __ __ 35 __ __ __ 01|
'---------------------------'
a b c d e f g
Update 1 (Sept 01, 2026)
The positions 10 and 16 makes
possible to get 4 numbers
(marked with *).
.---------------------------.
7 | 16 *15 __ __ __ __ 49|
6 |*17 19 __ __ __ __ __|
5 |*18 27 __ __ __ __ __|
4 |*28 24 __ __ __ __ 10|
3 | 29 __ __ __ __ __ __|
2 | __ __ __ 34 39 __ 42|
1 | __ __ 35 __ __ __ 01|
'---------------------------'
a b c d e f g
How to advance from here ?
Update 2 (Sept 3, 2026)
After 4 numbers, a huge barrier seems
to exist for the next one! We have to
look for weaknesses and the natural
place for that is the lower right
corner. I have found a strong but
very complex move:
(40)f3 -> the path from 2 to 20 leaves no room to 33
/
(41)g3
|| \
|| (40)f2 -> path 2->10, locked numbers (06,33,38)c2.d2.e2 and path 35->39 are in conflict
||
|| (02)f2 -> (40)e3, (43)g3!, and path 2 -> 10 is tight -> no place for 38
|| /
(41)f3 (43)f2 -> (40)e3, (44)g3, (45)f4, (38)d3!, and path 2 -> 10 leaves no place for 33
|| \ /
|| (02)f1 (3)f2 -> (40)e3, and path 3->10 leaves no room for 38
|| \ /
|| (43)g3
|| \
|| (3)e1!!,(4)d1!,(5)c2, (44)f4!, now path 35 -> 39 leaves no room for 33
||
(41)f2->(02)f1, (03)e1, (04)d1, (05)c2, (36)b2!, (37)c3,(38)d3, (33)e3, and path 5->10 is impossible.
Since those three possible places for 41 each gives a contradiction (each as described), we must have
5.(41)f1!!! [three due to the complexity, but I would happier with a simpler move]
6.(02)f2 7.(40)e1 8.(38)d1 9.(37)c2
So, we get
.---------------------------.
7 | 16 *15 __ __ __ __ 49|
6 |*17 19 __ __ __ __ __|
5 |*18 27 __ __ __ __ __|
4 |*28 24 __ __ __ __ 10|
3 | 29 __ __ __ __ __ __|
2 | __ __ *37 34 39 *02 42|
1 | __ __ 35 *38 *40 *41 01|
'---------------------------'
a b c d e f g
The situation has improved a lot,
it seems. How to continue ?
Created: February 12, 2024
Contact: sudo.jco.br@gmail.com