How to eliminate/nominate candidates without guessing?
For the 3s it's quite easy because of the odd numbers but how to get further?!
For thoose who don't know the game, here's a short introduction:
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1$\begingroup$ The first column must include the 2, but after that it seems difficult. $\endgroup$– Jaap ScherphuisOct 9, 2018 at 5:56
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1$\begingroup$ This previous Rullo question includes a description of the rules. $\endgroup$– Jaap ScherphuisOct 9, 2018 at 7:48
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2$\begingroup$ what's the '20' on the lower left/right edges? $\endgroup$– JMPOct 9, 2018 at 11:35
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1$\begingroup$ At this point I would probably make a guess, and if that turned out wrong, backtrack. A good option to guess is to include the $4$ in the first column, third row from the bottom (sum $23$). The reason is that this immediately determines that column and row, leading to many forced moves, quickly leading either to a solution or a contradiction. $\endgroup$– Jaap ScherphuisOct 9, 2018 at 12:55
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1$\begingroup$ @Dewey823 4+3+4+2+3+2+2 =20. The "target" value for this row is 14. When you press the square with the 14 the tooltip appears which tells you the actual sum of all active candidates. If you press another square the tooltip appears for the other row... $\endgroup$– MatOct 10, 2018 at 7:08
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3 Answers
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Logic grid puzzles are generally supposed to have a unique solution. This Rullo puzzle has several solutions (unless there is some rule we are missing), so I would consider it an invalid puzzle. Here are four such solutions.
The top two solutions are what I got when I chose to include the 4 in the first column, 3rd row. This eventually (after I think one or two more guesses) led to a situation with only four undetermined cells arranged in a rectangle all with a 2. One diagonal pair had to be included, the other excluded, but both were possible.
The other two solutions are what I got when I chose to include the 4 in the first column, 6th row. This led to about 6 undetermined cells where there were two ways to choose from.
There may be even more solutions.
answered Oct 11, 2018 at 5:49
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Partial Answer
Added in green, in column one you need 6 more to make 16 and there is only one two
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$\begingroup$ I haven't had much time to look at it, but that 2 means both 4s in that row can't be on (but one can) $\endgroup$– Dewey823Oct 10, 2018 at 18:26
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I believe this is the solution:
Green means keep, red means don't
After getting to where you were I kind of made educated guesses based on areas that had rows and columns that needed to get rid of the same values.
