Timeline for What is the minimum amount of numbers needed to create a 4x4 Hidato with a unique solution?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10 at 15:40 | comment | added | RobPratt | For $3\times5$, the minimum is $3$ (not $6$): \begin{matrix}.&.&2&9&.\\.&.&11&.&.\\.&.&.&.&.\\\end{matrix} | |
| Mar 7 at 15:08 | comment | added | CrSb0001 | @Nitrodon Oh my, I did not think that that would be possible! But at least it is :) | |
| Mar 7 at 15:05 | comment | added | Nitrodon | I'm pretty sure this works for 3 x 7: \begin{array} {c} . & . & 16 & . & 5 & . & . \\ . & . & . & . & . & . & . \\ . & 18 & . & . & . & . & 3 \\ \end{array} | |
| Mar 7 at 0:01 | comment | added | RobPratt | For $3\times4$, the minimum is $2$ (not $4$): \begin{matrix}.&.&.&.\\.&.&.&.\\.&.&11&6\\ \end{matrix} | |
| Mar 6 at 17:45 | comment | added | CrSb0001 | @RobPratt Oh my, that's actually impressive! Good job on finding that! | |
| Mar 6 at 17:41 | comment | added | RobPratt | For $3\times3$, the minimum is $2$ (not $3$): \begin{matrix} . & . & . \\ . & . & 9 \\ 7 & . &. \\ \end{matrix} | |
| Mar 6 at 16:32 | vote | accept | CrSb0001 | ||
| Mar 6 at 1:40 | history | became hot network question | |||
| Mar 5 at 18:20 | answer | added | Dr Xorile | timeline score: 23 | |
| Mar 5 at 18:19 | answer | added | tehtmi | timeline score: 5 | |
| Mar 5 at 18:01 | comment | added | hexomino | I think your comment is correct, you can alter the diagram in this way to get a unique solution from 7 clues. | |
| Mar 5 at 17:51 | answer | added | hexomino | timeline score: 7 | |
| Mar 5 at 17:48 | comment | added | CrSb0001 | Wait a sec, does that mean I have accidentally answered my own question? because then we could remove the 5 to get a 4x4 Hidato puzzle with a unique solution with only 7 clues? | |
| Mar 5 at 17:46 | comment | added | CrSb0001 | @hexomino ah right my bad. If it were to be unique with 8 numbers, I would actually have to remove the 7 and then put in a 6. My bad. | |
| Mar 5 at 17:44 | comment | added | hexomino | Is the solution unique? It looks there are two solutions where '6' and '16' could be swapped to get from one to the other. | |
| Mar 5 at 17:39 | history | asked | CrSb0001 | CC BY-SA 4.0 |