From dlitwin@geoworks.com Thu Sep 14 23:45:03 1995 Return-Path: Received: from quark.geoworks.com ([198.211.201.100]) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA20748; Thu, 14 Sep 95 23:45:03 EDT Received: from radium.geoworks.com by quark.geoworks.com (4.1/SMI-4.0) id AA15388; Thu, 14 Sep 95 20:40:52 PDT Date: Thu, 14 Sep 95 20:40:52 PDT From: dlitwin@geoworks.com (David Litwin) Message-Id: <9509150340.AA15388@quark.geoworks.com> To: cube-lovers@life.ai.mit.edu Subject: Alexander's Star While we are on the subject of the Alexander's star, I have never been entirely satisfied with the arrangment of its solution. I've noticed that most people who look at it don't even know it is solved and I have to explain that the solution is that all the stickers of pieces laying in a common plane around the points are the same color. Hard to say, but I can point it out to people. To this end I have spent some time trying to find a more satisfying solution, one more visually clear and simple. I've only come up with one alternative that I consider reasonable, and it isn't as pure as I would like. This solution involves grouping colors in the depressions of the star. The main problem lies in the fact that the edges come together in groups of three, but there are 10 stickers of each color so at some point having all the depressions of the star a single color breaks down. For this reason I choose one color (White is my preference) to be an exception and have it remain in the original configuration, i.e. all in two parallel planes. With the rest of the colors, I group them in groups of five: three in one depression, and two in an adjacent depression with the third color of this second depression being one of a white. The result of this is a set of interlocking "diamonds" that group visually because the are of the same color. Unrolled, the star would look like this (this should just fit on a display of 80 columns): /|\ /|\ /|\ /|\ /|\ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / Y_|_Y \ / B_|_B \ / O_|_O \ / R_|_R \ / G_|_G \ / _-' W `-_ \ / _-' W `-_ \ / _-' W `-_ \ / _-' W `-_ \ / _-' W `-_ \ _/-'_________`-\_/-'_________`-\_/-'_________`-\_/-'_________`-\_/-'_________`-\ |\-,_ _,-/|\-,_ _,-/|\-,_ _,-/|\-,_ _,-/|\-,_ _,-/ | \ `-_Y_-' / | \ `-_B_-' / | \ `-_O_-' / | \ `-_R_-' / | \ `-_G_-' / | \ Y " Y / | \ B " B / | \ O " O / | \ R " R / | \ G " G / | \ | / | \ | / | \ | / | \ | / | \ | / |_O \ | / R_|_R \ | / G_|_G \ | / Y_|_Y \ | / B_|_B \ | / O_ O `-_ \ | / _-' R `-_ \ | / _-' G `-_ \ | / _-' Y `-_ \ | / _-' B `-_ \ | / _-' _____`-\|/-'_________`-\|/-'_________`-\|/-'_________`-\|/-'_________`-\|/-'____ _,-/ \-,_ _,-/ \-,_ _,-/ \-,_ _,-/ \-,_ _,-/ \-,_ W_-' / \ `-_W_-' / \ `-_W_-' / \ `-_W_-' / \ `-_W_-' / \ `-_ " O / \ R " R / \ G " G / \ Y " Y / \ B " B / \ O | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ |/ \|/ \|/ \|/ \|/ \ I haven't found a nice way of having more solid depressions than this. Has anyone else found any nice solutions? Dave Litwin