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 Very Difficult Logic Problems Dotted Dwarfs Goto page Previous  1, 2
 Fri Feb 19, 2016 8:53 pm  by ArmyMP84 I agree with elcanrebat's solution above. Each dwarf counts the number of visible red dots, adds one, then assembles that many days later. All reds will assemble on the correct day, and all blues will assemble the day after the correct day (and after the event is complete).
 Wed Jul 13, 2016 7:06 am  by Dzallen Interestingly enough, if all the dwarfs are completely rational, there is no need for them to make any strategy whatsoever! Consider if there was only 1 red dot. The Dwarf with the red dot would realize immediately by seeing all blue dots, and hence appear on day 1. If there were 2 red dots, the two respective dwarfs would each only see 1 other red dot. However, when they realize that no-one appeared on the first day, they would deduce that there cannot be only 1 dot, i.e. they must have a red dot themselves. Hence they would appear together on the second day. If there were 3 red dots, the respective dwarfs see only 2 other red dots. But when no-one appears on the second day, they would deduce that there cannot be only 2 dots, i.e. they have a red dot themselves. Hence the three would appear on the third day. In general, if a dwarf realizes that no-one appeared on the day of the number of red dots he sees, he would deduce that he must have a red dot, and appear on the following day. It seems that some have ignored the fact that 'any other entry upon the meeting point of any dwarf results in failure'. Other solutions are of course viable as well, but this can be done without any planning or communication at all!Last edited by Dzallen on Mon Jul 25, 2016 2:22 am; edited 1 time in total
 Fri Jul 15, 2016 9:32 am  by bathfilms Simplest answer: Every dwarf must think to kill any dwarf with a blue dot! However, Dzallen's (Wed Jul 13, 2016 7:06 am) answer is ideal.
 Tue Oct 04, 2016 9:50 pm  by sjsimmons I agree with the solution that's been offered--everyone adds 1 to the number of red dots they see and then all the red dot people will come on the right day. It seems to me there's another solution, since the problem only asks that the number of dwarves with a red dot return on the correct day. They can all line up in a straight line. If the dwarf in front of you has a red dot move left. If the dwarf in front of you has a blue dot, move right. The dwarf in front then goes to the back of the line, looks at the dot in front of him, and makes the appropriate move. The group on the left would meet again in the number of days represented by the number of people in the group on the left. If only the red-dot dwarves are to return, then you can learn the color of your own dot by how the person behind you moved. Then those dwarves would return on the day represented by the number of people in the group on the left. If that counts as "communication," then I'd just stick to the one offered by others.
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