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|Transforming the regular puzzles in no-op puzzles
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|Author:||clm [ Sun Aug 28, 2016 11:24 pm ]|
|Post subject:||Transforming the regular puzzles in no-op puzzles|
Transforming regular puzzles in no-op puzzles to increase the level of difficulty.
Suppose that we could have a new feature in the page permitting to individually remove the operators in any selected puzzle. Since the structure is correct and the solution is unique it is obvious that, after removing the operators, we could always find at least that original solution, that is, a solution exists, with some particularities that I will comment soon.
First it must be said that doing this increases the level of difficulty. I have proved this feature with the regular 4x4’s, 5x5,’s, 6x6’s and 7x7’s, as well as with selected puzzles from the books, and I believe that the difficulty increases “exponentially” with the size, provided that we assume only algebraic operators (a totally different thing is if we include the mod, parity, zeros, exponentiation, etc., as a possibility, let’s think for instance in a 8x8, this is simply terrible). As an example you can try today’s Aug 28, 2016, 5x5 difficult, and you will confirm that after a “hard” but beautiful analysis you will find the same, and only the same, configuration of operators initially given (except, as we well know, of course, the cage “2” in e1-e2, since being the operands 1 and 2 it can be “2x” besides the given “2:”).
This new feature, consider it as a suggestion, would probably make much more interesting many small puzzles to the advanced players (perhaps a little tired of the “too easy” 4x4’s, 5x5’ or 6x6’s, which solution looks very “straight forward” most days).
Inversely, an important question arises: Given a particular distribution of cages and “results” in a no-op puzzle obtained in this way ¿How many possible solutions exist? ¿It is there a unique solution? ¿Are there more?. I am not referring to the operators themselves (we know that the operators are not unique) but to the structure of the numbers, the possible permutations between them.
I am afraid this is a very difficult question to be answered at this moment and it would require a lot of reasoning, analysis and computer testing.
Note: After removing the operators, thus producing a no-op puzzle, from any given regular and proper puzzle (with a unique solution), we can find one or more cages that could admit different operators, this is due to the non-unicity of the operators. For instance, “+” or ”x”, in a 3-cell in-line cage “6”, with the operands [1,2,3]; “x” or “:” in cages “n” of the type [1,n]; “-” or “:” like in a 3-cell cage “3” with operands [1,2,6] or in a 2-cell cage “4” with operands [2,4].
But, generally speaking, “-” and “x” or “+” and “:” are not compatible.
The exception to the above is the puzzle “from 0” that has special circumstances since here we can find up to three different operators for the same cage, for instance, a cage “0”, 3-cell L-shape, with operands [0,4,4] where the cage could be “0-“, “0x” or “0:”. Even, cages “n” (2-cell) of the type [0,n] admit “+” and “-".
In a 2-cell cage “1” in a no-op puzzle we always place a “-”, but if we had a “no-op mod puzzle”, a cage “1” could be “1mod”, so in this case we could find cages where a “-” or a “mod” could be set.
Finally, when dealing with a puzzle with negative numbers, symmetric or non-symmetric, which contains zeroes, we can find “0+”, “0-”, “0x” and “0:” (all four algebraic operators) for the same cage like, for instance, in a 3-cell L-shape cage with operands [-2,0,2] with the 0 in the angle.
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