# More Puzzles

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 Very Difficult Logic Problems Dotted Dwarfs Goto page 1, 2  Next
 Fri Apr 23, 2010 7:35 pm  by tartle The leader of a group of dwarfs wants to test his group. So at night he paints on the back of each dwarf a dot, either red of blue. The next morning he let's the dwarfs gather at a meeting point and tell's them the rules of the game: You may never know what the color of your own dot is. (This is accomplished by saying they may not communicate in any way or use mirrors and such) He will give them a few minutes to discuss a tactic to solve the puzzle. The goal is that the number of dwarfs with a red dot, assume this number is X, on their back gather at the same meetingpoint the number of days, X, after the start of the game. for example if there are 10 red dots, after 10 days they must gather Any other entry upon the meetingpoint of any dwarf results in failure. Now, how must every dwarf think to solve the puzzle?
 Thu May 20, 2010 2:25 pm  by sinb well, the group selects a single person as a leader. The leader can see, let's say Y dots. either X = Y (if his own dot is green) or X = Y+1 (if his own dot is red) The plan is that the group meets at a place every day at a pre-decided point, but the leader is supposed to show up only on (Y-1)'th day. So everyone except for the leader know X, that is they know Y and depending on the leader's dot they can decide on X. When they decide on X (the total number of red dots), then they can decide whether their own dot is red or blue. Now on X'th day, every red dot dwarf meets at the same place if the leader's dot was red, otherwise they all meet at some other pre-decided place to which leader wont come. I dont know if the guy is allowed to indicate X value to others by going to the meetingplace, (it would be indirectly telling everyone which dots they have), but this is the solution I came up with. Let me know if you have some other solution.
 Mon Jun 28, 2010 2:58 pm  by shrek619 I assume there are at least 2 red dots. Consider one dwarf with either red or blue dot, let X be total no. of dots out there. If he is red dotted he sees X-1 dots let it be Y, if he is blue dotted he sees X dots let it be Y. Y is no. of red dots as he sees on other dwarfs. All he need to do is go on to the location on Y+1 day. Each one of them should follow this rule. If he is red dotted he goes to the location on correct day, if he is blue dotted the game is already over on Y th day so there is no need for going to the meeting place. am i right ? is there any other solution
 Mon Nov 29, 2010 9:12 pm  by sudheerkuttiyat Every dwarf must think he has a red dot on his back. Then every one with red dots in the back will meet X days from now, where X-1 is the number of red dots a dwarf with a red dot behind him can see. Anyone with blue dots on the back will see X red dots and will come to meet his mates X+1 days from now.
 Thu Aug 15, 2013 10:57 pm  by ramagurung They are not supposed to communicate or look in the mirror to see the dot, so just don't turn up. I am sure i got this wrong.
 Tue Sep 24, 2013 1:07 pm  by hasz921 Agree with sinb. That should be the solution.
 Tue Sep 24, 2013 1:07 pm  by hasz921 Agree with sinb. That should be the solution.
 Tue Jul 08, 2014 4:52 pm  by sanchit this is indeed a very difficult one . since there should be no communication therefore no chances of appointing a leader . i would xperince difficulty in explaining it . one can search it on google as the blue eye islander. it goes lyk this suppose only two persons had red dots you and me . firstly you will be sure that i have a red dot therefore i must go at first day but i will be thinking same as all others are blue . but not meeting at first day you will come to know that you also have a red dot otherwise seeing all blue dots i would have gone there at once(as there is atleast one dot for sure) , this will give u idea that i also thinks the same therefore there must be one more and which obviously would be you as all others except me are blue same will occur in every case , every one with red dot will know that they too possess a red dot just a day before gathering hope u understood
 Wed Jul 23, 2014 4:51 am  by nicejojo12 If there is only 1 red dot, then he goes around and sees nobody else with one, meaning it's only him. If there are 2 red dots, Imagine that you are one of them. You see that there is another person with a red dot. Thinking that he is the only one, you wait. But that's what he thinks as well. Then seeing as nobody else has a red dot, you realize that you must be the person with the red dot. If there are 3 red dots, then imagine you are one them. You see 2 others also with red dots. You think that those 2 are the only people with red dots, but after the second night you realize that they still need one more person with a red dot. Seeing as there are no other people with red dots, you realize that you must be the one with a red dot. etc.
 Tue Aug 19, 2014 12:01 am  by FredMo I assume all reds must show up at the prescribed meeting place on the right day and blues should show up there never. A simple strategy - All dwarfs agree to meet at a pre-meeting rendezvous point one day before the number of red dots they count. The red folks will rendezvous two days ahead of the correct day, see only reds, and know to go to the meeting place two days hence. The blues will rendezvous the next day, see only blues, and know they should stay away.
 Fri Sep 12, 2014 10:33 am  by MK The DWARFS select a leader. The leader counts the RED dots on others and tells that there are X dots. As the leader is not telling anybody individually and nobody is telling him or to anybody, it is all fair. Now, the DWARFS can see the dot on leader. If it is blue, they go to assemble after X days. If it is RED, they go to assemble after X+1 days. But, the scene is spoiled if leader comes on day X and also if leader does not appear on day X+1. But, as the leader is intelligent, he checks that others are there on day X or not. If they are not to be seen, it means he is BLUE and he need not go. If they are there on day X, it means he has also to go on dayX+1.[/list][/quote]
 Fri Oct 10, 2014 12:49 am  by Decius Every dwarf comes to the arranged location at day X, where X is one greater than the number of red dots that dwarf sees. Every dwarf with a red dot will see X-1 red dots, and arrive on day X. Every dwarf with a blue dot will see X red dots, and not arrive on day X.
 Fri Oct 10, 2014 2:05 am  by elcanrebat This is actually a very simple question. The key to winning this is [color=red]for all dwarves to automatically assume their own dot is red in colour.[/color] for eg, imagine there are 10 dwarves, 5 are blue and 5 are red. a red dot dwarf will see 5 blues and 4 reds. if he automatically assume his colour is red, he will gather after 5 days all 5 red dwarves will gather after 5 days. a blue dwarf will see 4 blues and 5 reds, if he automatically assume his colour is red, he will gather after 6 days. hence 5 blue dwarves will gather after 6 days. ( but they will not appear on the 5th day and will not affect the win) there is no need to elect a leader.
 Thu Jan 08, 2015 6:07 am  by pvedha Solution can be achieved with less than 3 days of activity. Conditions: 1. No of red dots not equal to blue dots. 2. The dwarfs dont have any other meeting place apart from the one instructed. At the meeting point they agree upon this, The dwarfs should gather in pairs. The pair will see each others colors. Consider the dwarfs are named A1, A2, B1, B2..... If there are odd number of dwarfs, let X be the name of the one that cant be paired. Day 1: All the dwarfs who seen red color on their partner should turn up. So, if A1 and A2 both present, they both know they got red color. if only B2 is present, B2 understand he got blue color. Dwarf X (if there is one) should approach a pair with both partners present and show his color. This pair should turn up on day 2 if Dwarf X is red. (WCS-Worst case scenario, no pair got both red, X will try this on Day 2). Day 2: All the dwarfs who were absent on day 1 turn up. (those who seen blue on their partner) Now, if C1 and C2 both present they both know they got blue color. if only D2 is present, D2 understand he got red color. (D1 would have been there on day1). Dwarf X understands he is red if his agred pair turns up. In case of WCS, X tries to find another dwarf and agree for the Day 3. On the day X all the dwarfs with red shall go to the meeting point.
 Sat Feb 28, 2015 7:08 am  by hemant.bisht24 elcanrebat is right
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