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 Lego query (Posted on 2006-10-05)
Imagine you have a 4 x 2 blue lego, a 4 x 2 red lego and a 2 x 2 yellow lego. How many ways, excluding rotations, can you put the legos together (you must use all the pieces)?

 No Solution Yet Submitted by joshua Rating: 3.2500 (4 votes)

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 Proposed solution | Comment 3 of 7 |
I think the easiest way is to fix the position and orientation of say the blue lego to say, horizontal  (will represent with xs)

There will be exactly one rotational symmetry at that point so divide all possibilities by 2 (we'll see that we always multiply by two so this becomes moot)

Now the possibilites are 6 : red in same plane, next plane, or far plane in same (horizontal) or opposite (vertical) orientation.

same plane means yellow connects and the number of ways yellow can connect is 2x (canceled by rot) (yellow top or bottom) (shared edge lengths + 1)

These values are placed in the position of the lower left corner of the red block for vertical orientation (left) and horizontal orentiation (right):
`1233321   1234543212 xxxx2   2   xxxx2 3 xxxx3   3   xxxx3 3     3   2       2 3     3   1234543212     21233321 `

sum of each corner is (1+2+2+3+3)=11 and  (1+2+2+3+4)=12 4x23=92 centers=6x3+2x5=28 total=120
Call this A

far planes is next.  again, yellow connects and 2 choices for above/below (cancelled).  All of the same choices are available as same plane but now spots inside the squares are available with choices = product of horizontal and veritcal choices.  I'll use hex digits to illustrate assuming the same hidden xs:

`1233321   1234543212466642   2468a86423699963   369cfc9633699963   2468a86423699963   12345432124666421233321`
sum of insides (via foil) = (2+3+4+5+4+3+2)x(2+3+2) +(2+3+3+3+2) squared= 23x7+13x13=161+169=330
Call this B

next plane means yellow does not connect so we have 2x for red above / below (cancelled) 2 big choices, yellow shares a plane with one or all three are on different planes.

yellow different plane is simple because there are simply 3x5=15 ways to put a 2x2 on top of a 4x2  and 5x5 + 7x3 = 25+21=46 ways to put a 4x2 on another 4x2 and two blocks (blue or red) to attach  yellow to so 15x46x2=1380 possibilities
Call this C

yellow shares a plane could be with either blue or red so 2x the following (assume shares with red connecting to blue)  Note that the possibilities are all possibilities for placing on blue (15) - those that intersect with red = 15-value from inside of far planes.  Thus, (15x46 - value from far planes)*2 or C-2B

Total then = A + A+b + C + C -2B = 2A + 2C - B = 240+2760-330=  2670

 Posted by Joel on 2006-10-07 04:42:55

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