Posted March 13, 2013 I have 12 sticks and I have painted 8 of them red and 4 of them blue. I then glued the sticks together to form the frame of a cube. So now I'm wondering how many total ways could I have assembled the sticks into cubes that could be distinguished from each other. Assume I can pick up the cube and move it around any way I like, but I cannot deform or re-assemble it 0 Share this post Link to post Share on other sites

0 Posted March 14, 2013 Is there a consensus? Can you construct the possible configurations in groups, using words or a sketch? My first try to solve this was to take 12 things 4 at a time, then remove the symmetrically equivalent solutions. But I kept removing too many cases. Next try was to group the edges and distribute 4 blues among the groups. I'm not convinced yet of my answer. 0 Share this post Link to post Share on other sites

0 Posted March 14, 2013 Here is how I approached the problem: 1. There is 1 configuration where 4 blue sticks are all on one side of the cube. Let's call it "O" configuration. 2. Now I removed one of them, so 3 remaining blue sticks form a "C" on one side. Since "C" is assymetrical, every position of the 4th blue stick produces a distinct configuration - 8 distinct "C" configurations. 3. The only 1 remaining configuration when all 4 blue sticks are connected is "M" configuration (blue sticks form a letter "M" in isometric projection of the cube). Next, let's review configurations where 3 blue sticks are connected and the 4th stick is separate excluding any "C" configurations (3+1): 4. There is 1 "Y" configuration where 3 blue sticks connect in one corner (similarly to "M" named after the letter "Y" formed in isometric projection) 5. There are 6 "Z" configurations where 3 blue sticks are connected in a 3D zigzag - it can be left or right oriented and each has 3 possible positions for the 4th blue stick. Next, configurations where no 3 sticks are connected - "2+2", "2+1+1" and "1+1+1+1": 6. There are 6 "2+2" configurations. To eliminate duplicates I just pick a pair of 2 connected sticks and lock them in one place. Then find all other possible locations for 2 other sticks without being connected to the first 2. 7. There are 4 "2+1+1" configurations. Same idea - lock the 2 connected sticks in place and find possible locations for the remaining 2 without them being connected to the 2 and each other. 8. There are 2 "1+1+1+1" configurations. So, my total is 1 "O" + 8 "C" + 1 "M" + 1 "Y" + 6 "Z" + 6 "2+2" + 4 "2+1+1" + 2 "1+1+1+1" = 29 configurations. 0 Share this post Link to post Share on other sites

0 Posted March 15, 2013 Here is how I approached the problem: 1. There is 1 configuration where 4 blue sticks are all on one side of the cube. Let's call it "O" configuration. 2. Now I removed one of them, so 3 remaining blue sticks form a "C" on one side. Since "C" is assymetrical, every position of the 4th blue stick produces a distinct configuration - 8 distinct "C" configurations. 3. The only 1 remaining configuration when all 4 blue sticks are connected is "M" configuration (blue sticks form a letter "M" in isometric projection of the cube). Next, let's review configurations where 3 blue sticks are connected and the 4th stick is separate excluding any "C" configurations (3+1): 4. There is 1 "Y" configuration where 3 blue sticks connect in one corner (similarly to "M" named after the letter "Y" formed in isometric projection) 5. There are 6 "Z" configurations where 3 blue sticks are connected in a 3D zigzag - it can be left or right oriented and each has 3 possible positions for the 4th blue stick. Next, configurations where no 3 sticks are connected - "2+2", "2+1+1" and "1+1+1+1": 6. There are 6 "2+2" configurations. To eliminate duplicates I just pick a pair of 2 connected sticks and lock them in one place. Then find all other possible locations for 2 other sticks without being connected to the first 2. 7. There are 4 "2+1+1" configurations. Same idea - lock the 2 connected sticks in place and find possible locations for the remaining 2 without them being connected to the 2 and each other. 8. There are 2 "1+1+1+1" configurations. So, my total is 1 "O" + 8 "C" + 1 "M" + 1 "Y" + 6 "Z" + 6 "2+2" + 4 "2+1+1" + 2 "1+1+1+1" = 29 configurations. Very nice approach. Nice categories and elimination of symmetric cases. In the final 1 + 1 + 1 + 1 case. Looks like front face has two V [H] edges, and back face has two H [V] edges. Which [rotation] would be one case, not two? Maybe I'm missing something. 0 Share this post Link to post Share on other sites

0 Posted March 15, 2013 (edited) Good spacial vision exercise. Without looking at an actual model of a cube, I am not sure I got it right. Next, let's try it in 4-D. There are 7 ways in which blue-colored edges of the cube can be conected. And I see the arrangements, which cannot be obtained one from another for each variation as following: So it is 25, unless I missed or duplicated some arrangements. Edited March 15, 2013 by Prime 0 Share this post Link to post Share on other sites

0 Posted March 15, 2013 Here is how I approached the problem: 1. There is 1 configuration where 4 blue sticks are all on one side of the cube. Let's call it "O" configuration. 2. Now I removed one of them, so 3 remaining blue sticks form a "C" on one side. Since "C" is assymetrical, every position of the 4th blue stick produces a distinct configuration - 8 distinct "C" configurations. 3. The only 1 remaining configuration when all 4 blue sticks are connected is "M" configuration (blue sticks form a letter "M" in isometric projection of the cube). Next, let's review configurations where 3 blue sticks are connected and the 4th stick is separate excluding any "C" configurations (3+1): 4. There is 1 "Y" configuration where 3 blue sticks connect in one corner (similarly to "M" named after the letter "Y" formed in isometric projection) 5. There are 6 "Z" configurations where 3 blue sticks are connected in a 3D zigzag - it can be left or right oriented and each has 3 possible positions for the 4th blue stick. Next, configurations where no 3 sticks are connected - "2+2", "2+1+1" and "1+1+1+1": 6. There are 6 "2+2" configurations. To eliminate duplicates I just pick a pair of 2 connected sticks and lock them in one place. Then find all other possible locations for 2 other sticks without being connected to the first 2. 7. There are 4 "2+1+1" configurations. Same idea - lock the 2 connected sticks in place and find possible locations for the remaining 2 without them being connected to the 2 and each other. 8. There are 2 "1+1+1+1" configurations. So, my total is 1 "O" + 8 "C" + 1 "M" + 1 "Y" + 6 "Z" + 6 "2+2" + 4 "2+1+1" + 2 "1+1+1+1" = 29 configurations. Very nice approach. Nice categories and elimination of symmetric cases. In the final 1 + 1 + 1 + 1 case. Looks like front face has two V [H] edges, and back face has two H [V] edges. Which [rotation] would be one case, not two? Maybe I'm missing something. Never mind, I think I see two cases after all. I think that's it. 0 Share this post Link to post Share on other sites

0 Posted March 15, 2013 Good spacial vision exercise. Without looking at an actual model of a cube, I am not sure I got it right. Next, let's try it in 4-D. There are 7 ways in which blue-colored edges of the cube can be conected. And I see the arrangements, which cannot be obtained one from another for each variation as following:colorcube.gif So it is 25, unless I missed or duplicated some arrangements. Correction: It is 10 arrangements for the variation 4. Then the answer is the same as K-man has found. 0 Share this post Link to post Share on other sites

0 Posted March 15, 2013 Actually, we should split the difference and go with 27. The variation 4 on the diagram in my previous post should have 8 arrangements. K-man's zig-zag (5) should have 4, instead of 6 variations. There are 2 different 3-side zig-zags, and each has 2 (not 3) arrangements with the 4th disjointed side. 0 Share this post Link to post Share on other sites

0 Posted March 15, 2013 Actually, we should split the difference and go with 27. The variation 4 on the diagram in my previous post should have 8 arrangements. K-man's zig-zag (5) should have 4, instead of 6 variations. There are 2 different 3-side zig-zags, and each has 2 (not 3) arrangements with the 4th disjointed side. colorcube2.gif Good one, Prime! Those two are in fact the same. The drawing certainly helps to see it. 0 Share this post Link to post Share on other sites

0 Posted March 18, 2013 Someone posed this problem to me, and I tried several ways to count cases and finally gave up. This prize is a cooperative sharing of k-man and Prime, but I can mark only one post. So I marked the post that got the final count. Good job both! 0 Share this post Link to post Share on other sites

Posted

I have 12 sticks and I have painted 8 of them red and 4 of them blue.

I then glued the sticks together to form the frame of a cube.

So now I'm wondering how many total ways could I have assembled the sticks into cubes that could be distinguished from each other.

Assume I can pick up the cube and move it around any way I like, but I cannot deform or re-assemble it

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