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Puzzling

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    $\begingroup$ Is the solution unique? It looks there are two solutions where '6' and '16' could be swapped to get from one to the other. $\endgroup$
    – hexomino
    Commented Mar 5 at 17:44
  • 1
    $\begingroup$ 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? $\endgroup$
    – CrSb0001
    Commented Mar 5 at 17:48
  • 2
    $\begingroup$ For $3\times3$, the minimum is $2$ (not $3$): \begin{matrix} . & . & . \\ . & . & 9 \\ 7 & . &. \\ \end{matrix} $\endgroup$
    – RobPratt
    Commented Mar 6 at 17:41
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    $\begingroup$ For $3\times4$, the minimum is $2$ (not $4$): \begin{matrix}.&.&.&.\\.&.&.&.\\.&.&11&6\\ \end{matrix} $\endgroup$
    – RobPratt
    Commented Mar 7 at 0:01
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    $\begingroup$ I'm pretty sure this works for 3 x 7: \begin{array} {c} . & . & 16 & . & 5 & . & . \\ . & . & . & . & . & . & . \\ . & 18 & . & . & . & . & 3 \\ \end{array} $\endgroup$
    – Nitrodon
    Commented Mar 7 at 15:05