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No-op puzzle https://www.calcudoku.org/forum/viewtopic.php?f=16&t=126 |
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Author: | sneaklyfox [ Sat Dec 10, 2011 11:32 pm ] |
Post subject: | Re: No-op puzzle |
clm wrote: But in a "3" three-cell cage (L-shaped in a 6x6 puzzle) [1,1,3] could be another solution with the ":" operator, we go to a different situation, the cage "3" itself would admit several combinations, but the ambiguity we are talking about refers to the situation with the same operands, like with the [1,2,6] in "3", when the "-" and the ":" can be applied indifferently, as jomapil mentions and as you mention with the pair [1,2] in a "2" cage, otherwise, sneaklyfox, we could create additional confusion in this discussion. In the "3" three-cell cage (L-shaped in 6x6), [1,1,3] would not be a possibility if we want to retain operator unicity because it could be done using the : operator or x operator. I'm just arguing for not having operator unicity as I think it limits possibilities. |
Author: | clm [ Sun Dec 11, 2011 1:45 am ] |
Post subject: | Re: No-op puzzle |
sneaklyfox wrote: clm wrote: But in a "3" three-cell cage (L-shaped in a 6x6 puzzle) [1,1,3] could be another solution with the ":" operator, we go to a different situation, the cage "3" itself would admit several combinations, but the ambiguity we are talking about refers to the situation with the same operands, like with the [1,2,6] in "3", when the "-" and the ":" can be applied indifferently, as jomapil mentions and as you mention with the pair [1,2] in a "2" cage, otherwise, sneaklyfox, we could create additional confusion in this discussion. In the "3" three-cell cage (L-shaped in 6x6), [1,1,3] would not be a possibility if we want to retain operator unicity because it could be done using the : operator or x operator. I'm just arguing for not having operator unicity as I think it limits possibilities. Yes, you are right, for the unicity of the operator the only possibility is, in your example, "3" = [1, 1, 5] I had probably the mind in other place, all this interesting discussion has been useful in some way clarifying that it is better for this type of puzzle not forcing the unicity of the operator so that in a case like that, for a "3" cage (3-cell L-shape), we may consider "3" = [1, 1, 3], [1, 1, 5], [1, 2, 6] and forget the operator so arranging the numbers according to the rest of the puzzle. In the normal puzzles we imagine which numbers produce the result shown in the cage with that defined operator, in the no-op puzzle we must think in the possible numbers together with any possible operator that may produce that "blind" result. I am thinking too that increasing the number of the operators, using not only the arithmetic but the full set we have till the moment (that is, including the bitwise OR, the mod, the exponentiation, ... and even starting the numbers in 0) we may obtain very complex, diverse and difficult puzzles. |
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