Say we have an $N\times M$ binary nonogram, where the $i^{th}$ row has to sum to $n_i$, and the $i^{th}$ column has to sum to $m_i,$ so that we have constraints $(n_1,\dots,n_N)$ for rows and constraints $(m_1,\dots,m_M)$ for columns. Say we have a solution to this nonogram, $U$.
Is there some $N\times M$ nonogram solution $V$ with rows that sum to: $(n_{\max},\dots,n_{\max})$ and columns that sum to $(m_{\max},\dots,m_{\max})$, with $n_{\max} = \max (n_1,\dots,n_N),\ m_{\max}= \max(m_1,\dots,m_M)$ where erasing cells of $V$ results in $U$?
I strongly suspect the answer is trivially 'yes.' I feel like we can take $U$ and somehow algorithmically fill it in in some special way to get $V$, but am not sure how to do so.
No, because $Nm_{max}$ may not be equal to $Mn_{max}$.
For example consider a 2 by 2 grid with a black top half.