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Defragmentation of permutation tables with four columns
Published/Copyright:
January 11, 2010
Abstract
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.
Received: 2008-11-25
Published Online: 2010-01-11
Published in Print: 2009-December
© de Gruyter 2009
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- The key space of the McEliece–Sidelnikov cryptosystem
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- Finding and estimating the number of repetition-free Boolean functions over the elementary basis in the form of a convergent series
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Articles in the same Issue
- The key space of the McEliece–Sidelnikov cryptosystem
- On the game-theoretical approach to the analysis of authentication codes
- Finding and estimating the number of repetition-free Boolean functions over the elementary basis in the form of a convergent series
- Problems on independence systems solvable by the greedy algorithm
- Defragmentation of permutation tables with four columns
- Nondegenerate colourings in the Brooks theorem
