Geodesic
Defragmentation of permutation tables with four columns
Diskretnaya Matematika, Tome 21 (2009) no. 4, pp. 95-104

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We consider a table with columns each of which contains each symbol of an alphabet $A$ precisely once; the remaining elements of a column are equal to zero. It is required to transform the table preserving the sets of elements in each row and each column to the form with consecutively placed elements of alphabet $A$ in each row. We give conditions for solvability of this problem. 
@article{DM_2009_21_4_a8,
     author = {A. M. Magomedov},
     title = {Defragmentation of permutation tables with four columns},
     journal = {Diskretnaya Matematika},
     pages = {95--104},
     publisher = {mathdoc},
     volume = {21},
     number = {4},
     year = {2009},
     language = {ru},
     url = {http://geodesic.mathdoc.fr/item/DM_2009_21_4_a8/}
}
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A. M. Magomedov. Defragmentation of permutation tables with four columns. Diskretnaya Matematika, Tome 21 (2009) no. 4, pp. 95-104. http://geodesic.mathdoc.fr/item/DM_2009_21_4_a8/