The previously posted message canbe slightly complicated namely: find a chess knight route starting from a fixed chess cell, ending in another one (also fixed) thus, the knight passes by every cell once and only once.
The solution also follows from cyclical route previously described. Indeed, if, say, the knight goes e5-d3 instead of e5-d7 and travels thereafter 64-63-62-...-48 (count down) the last point will be d7 in this case. It means, the last route cell is changeble. Repeating the procedure an many times as it is needed we may replace the end cell in any position of the chess board.
The proving is as follows: let us assume that there is a chess board cell where we cannot replace the knight route end by the procedure described above. Assume also that this cell has a number N in current route configuration. It means that from the cell number N-1 we are not able to go directly on 64. Hence, cell 64 cannot be in two knight steps from the cell where the route end is impossible. In other words, if the knight route cannot end in one cell it cannot end in all cells of the same color.
The solution also follows from cyclical route previously described. Indeed, if, say, the knight goes e5-d3 instead of e5-d7 and travels thereafter 64-63-62-...-48 (count down) the last point will be d7 in this case. It means, the last route cell is changeble. Repeating the procedure an many times as it is needed we may replace the end cell in any position of the chess board.
The proving is as follows: let us assume that there is a chess board cell where we cannot replace the knight route end by the procedure described above. Assume also that this cell has a number N in current route configuration. It means that from the cell number N-1 we are not able to go directly on 64. Hence, cell 64 cannot be in two knight steps from the cell where the route end is impossible. In other words, if the knight route cannot end in one cell it cannot end in all cells of the same color.
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