The Genome Game

Legos - Solution

The puzzle is solved by building a 5 x 4 x 4 lego brick (as seen in the background of the provided piece of paper) from the given pieces.

The provided paper identifies six different lego brick types, followed by a list of triples. Each triple is three digits (or question marks) of the same color. Each triple corresponds to a single provided lego brick. For example, the blue ?23 in the fourth row corresponds to the blue 1x4 brick.

Each triple is to be interpreted as three coordinates (x,y,z) in the 3-space implied by the gray axes in the background. Where a question mark appears the coordinate's value is unknown. Each brick is placed such that it occupies the cell corresponding to that coordinate. Orientation is not provided; it must be inferred.

For example, consider the blue 2x3 at 301. The third digits indicates that it is definitely in layer 1, but could occupy any of the following locations within that layer:

Bricks whose triples contain question marks have much more flexibility. For example, the 4x1 brick at ?23 could be anywhere on level 3, as long as it intersects row 2 (the third row from the top).

The first insight is to realize that almost all the bricks have their z-coordinate specified. Separate the bricks with known z-coordinates into their respective layers. Sort those bricks into groups by z coordinate:

Each layer in the finished brick is 4x5, meaning that the total number of pips on each layer must add up to 20. The column titled "Total Pips" identifies the number of pips in the bricks known to be in a given layer. The column titled "Pips Needed" is 20 minus this number. We need to distribute the bottom row's pieces among the top four so that all rows add up to 20.

We start with the z=3 row: it requires only one additional pip, which can only be provided by the blue 1x1 brick. Now that the blue 1x1 brick is gone, row z=1 requires the black 1x3 brick. The blue 2x2 brick must go to row z=2, with the remaining two bricks going to the z=0 level:

You can now look at this problem as four simpler subproblems of filling a 4x5 grid paying attention only to the x and y coordinates. Level z=3 could be solved through the following reasoning:

		The 1x3 yellow brick (033) must touch (0,3). There are only two possible positions.
		Consider the horizontal orientation. There are now only four possible placements for the 1x4 blue brick (?23). Every one of these positions leaves at least one exactly-1-pip-wide empty space to be filled outside of column 0. Since the only remaining 1-pip-wide piece (the blue 0??) must be in column 0, none of these positions can lead to a solution. Therefore the yellow brick must not be horizontal.
		There is now only one place where the blue 1x1 brick (0??) can go. There are now five possible placements for the 1x4 blue brick (?23).
		Three of these positions leave exactly-1-pip-wide empty spaces. Neither of the remaining two pieces can fill such gaps, so we eliminate those possible placements of the 1x4 blue brick.
		We know the black 2x3 brick (1?3) appears somewhere in column 1. Therefore the blue 1x4 brick must go in column 4, otherwise there would be no spaces left for the black brick.
		The remaining two bricks can only fill the remaining space in two ways (both horizontal, either one on top). We know the red 2x3 brick (?23) touches row 2, so it must be the bottom brick.

Techniques similar to the above may be applied to the other levels. The solutions to the various levels are as follows:

Z=0

Z=1

Z=2

Z=3

Put the layers together with z=0 on the bottom, z=1 on top of that, etc. and you get the following brick:

Front ("FT")

Back ("PT")

The dashed border around the puzzle suggests the color sequence blue-red-yellow-black. Reading the letters on the sides of the structure in that order, you get FTPT, or "Ft. Pt.", which is an abbreviation for Fort Point.