My Math Forum Is it possible to create a circle with all Domino tiles?
 User Name Remember Me? Password

 Number Theory Number Theory Math Forum

 June 21st, 2015, 08:23 AM #1 Newbie   Joined: Jun 2015 From: trebek Posts: 3 Thanks: 0 Is it possible to create a circle with all Domino tiles? Is it possible, according to Domino rules, to create a circle with all Domino tiles from the set, when the symbols: a) 1, 2, 3, 4, 5, 6 b) 0, 1, 2, 3, 4, 5, 6 are allowed? What I was trying to do first is brute force it with a computer program, but that kinda got out of hand, I have a working domino set now with no brute forcing yet, but that took me 4 hours. Is there a way to solve this using graph theory and/or combinatorics? (Since those are the topics we are doing right now) I'm thinking it's related to the Euler Circle or something similar.
 June 21st, 2015, 08:45 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra I would go: (0,1) (1,2) (2,3) ... (5,6) (6,0) (steps of 1 round the clock) (0,2) (2,4) (4,6) (6,1) ... (3,5) (5,0) (steps of 2) (0,3) (3,6) (6,2) ... (1,4) (4,0) (steps of 3) That's 21 pieces and the seven doubles can be added anywhere appropriate. This plan works where $n$, the number of different designs is coprime to all of 2, 3, ..., $\tfrac12 (n-1)$. It is this feature that allows each set to return to the zero in exactly $n$ steps. We have $\tfrac12 (n-1)$ sets of $n$ plus the $n$ doubles to get $\tfrac12 n(n+1)$ dominoes in the path.
 June 21st, 2015, 09:26 AM #3 Newbie   Joined: Jun 2015 From: trebek Posts: 3 Thanks: 0 Could I do it like this? I am counting for a) 21 vertices and for b) 28 vertices. Then for the degree I will subtract one from a) 21, 21-1 = 20 and b) 28-1 = 27. Euler says the degree of each vertex has to be even in order for there to be a circle. So the circle should work for a), but nor for b), right?
 June 21st, 2015, 09:40 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 You shouldn't subtract 1. The "circle" is possible (and extremely easy) for (b), but not for (a). For any circle, the numbers occur in pairs, but case (a) has each number occurring 5 times, which is impossible if they occur in pairs.
 June 21st, 2015, 09:51 AM #5 Newbie   Joined: Jun 2015 From: trebek Posts: 3 Thanks: 0 Okay, thank you for the answers!

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post jemes Elementary Math 5 December 4th, 2014 07:51 AM Aska123 Algebra 9 January 29th, 2014 11:06 PM johnsl Real Analysis 0 December 16th, 2010 04:50 AM dervast Advanced Statistics 7 July 2nd, 2010 07:17 AM sovixi Advanced Statistics 2 April 23rd, 2007 12:13 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.