bonanova 76 Report post 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 Share this post Link to post Share on other sites

0 Prime 15 Report post 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. Share this post Link to post Share on other sites

0 k-man 26 Report post Posted March 13, 2013 29 distinct ways. Share this post Link to post Share on other sites

0 bonanova 76 Report post 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. Share this post Link to post Share on other sites

0 k-man 26 Report post 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. Share this post Link to post Share on other sites

0 bonanova 76 Report post 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. Share this post Link to post Share on other sites

0 Prime 15 Report post 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 Share this post Link to post Share on other sites

0 bonanova 76 Report post 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. Share this post Link to post Share on other sites

0 Prime 15 Report post 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. Share this post Link to post Share on other sites

0 k-man 26 Report post 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. Share this post Link to post Share on other sites

0 bonanova 76 Report post 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! Share this post Link to post Share on other sites

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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