INSTANT INSANITY PUZZLE

By Patrick K. Asaba

INSTANT INSANITY PUZZLE

 The Instant Insanity puzzle is played with four cubes. Each face of the cube is colored with one of the four colors: red ( R ), white ( W ), blue ( B ) and green ( G ).

  Note: All front faces are WHITE The objective of the puzzle is to stack the cubes one on top of another so that on each long face of the resulting shape, all four colors show up exactly once. Determining the Instant Insanity solution is a very difficult task.

INSTANT INSANITY PUZZLE

 Determining how to arrange these four cubes in a stack that is an Instant Insanity solution is a very difficult task. The problem is that all the cubes are not the same.

 Observe that there are 24 symmetries of the cube, thus in doing this puzzle with four blocks which can each be placed in 24 different positions; that is: 24 4 = 331,776 possible stacks.

Wow!!

Solving the Puzzle

  One way of solving puzzle is to play around with these four cubes until you can figure out the solution, however I promise you that this will drive you insane.

The other way to look at this problem is by the decomposition principle:   Pick one pair of opposite faces on each cube for left and right sides of the stack so that these two sides of the stack will have one face of each color.

Then pick a different pair of opposite faces on each cube for the front and back sides of the stack so that these two sides will have one face of each.

Looking back

 In modeling the graph we use a Multigraph; – A multigraph is a generalized graph in which multiple edges are allowed, that is two or more edges can join the same two vertices; and loops are allowed, that is edges of the form (

a , a

).

} A loop Multiple edges { a

GRAPH MODEL (cube 1)

 Make one vertex for each of the four colors.

  For each pair of the opposite faces on cube

i,

create an edge with label

i

joining the two vertices representing the colors of these two opposite faces.

For opposite faces:

l

1 =

blue , r

1 =

white

between

blue

and

white

; for

f

1 =

red , b

1 on cube 1, draw edge labeled 1 =

red ,

draw a loop labeled 1 at

red ;

and

t

1 =

green , u

1 =

blue ,

drew edge labeled 1 between

green

and

blue .

1 1 1

 This graph only presents cube 1.

GRAPH MODEL (cube 2)

 For opposite faces:

l

2 =

white, r

2 =

blue

on cube 2, draw an edge labeled 2 between

white

and

blue

; for

f

2 =

white, b

2 =

white,

draw a loop labeled 2 at

white;

and for

t

2 =

white, u

2 =

red ,

drew an edge labeled 2 between white and

red .

2 2 2

GRAPH MODEL (cube 3)

 For opposite faces:

l

3 =

blue , r

3 =

green

on cube 3, draw an edge labeled 3 between

blue

and

green

; for

f

3 =

white, b

3 =

red ,

draw an edge labeled 3 between

white

and

red ;

and for

t

3 =

red , u

3 =

red ,

drew a loop labeled 3 at

red .

3 3 3

GRAPH MODEL (cube 4)

 For opposite faces:

l

4 =

green , r

4 =

red

on cube 4, draw an edge labeled 4 between

green

and

red

; for

f

4 =

white, b

4 =

green ,

draw an edge labeled 4 between

white

and

green ;

and for

t

4 =

blue , u

4 =

green ,

drew an edge labeled 4 between

blue and green .

4 4 4

Graph Model of instant Insanity puzzle

 Now we can construct a single multiple graph with 4 vertices and all the 12 edges associated with the 4 cubes.

1 3 3 4 1 2 4 4 3 2 1 2

Characterize the Solution

 Now you restate decomposition principle in terms of subgraphs.

 One subgraph will represent the left-right sides; and  The second Subgraph will represent the front-back sides.

 The subgraphs must:  use all four vertices.

   contain four edges, one from each cube.

use each edge only once.

have each vertex at degree 2.

SUBGRAPHS

 There are three subgraphs associated with our graph model. We are using edge-disjoint to find the subgraphs.

3 4 1 4 2 3 1 3 2 2 1 4

A B C

SUBGRAPHS

  Only two subgraphs meet the required standards. One of the subgraphs will represent the left/right sides and the second one will represent the front/back sides.

3 4 4 2 1 3 1

Left/right sides

2

Front/back sides

Building the solution

 Now you can stack the cubes as labeled.

l 3 r r l 4 l r l 2 r 1

Left/right sides

3 b f 1 b b f 4 b f 2

Front/back sides

f

Solution!!!

Try this!!!

   Find a multigraph for the set of cubes presented below.

If there is a solution, find two subgraphs that solve the puzzle.

Is there another solution? If so show it; and if not explain why not?

Cube 1 Cube 2 Cube 3 Cube 4

Solution

 This is a multigraph the for a different set of cubes.

1 4 3 2 1 3 1 2 4 3 2 4

Solution

 Now you restate decomposition principle in terms of subgraphs.

   One subgraph will represent the left-right sides; and The second Subgraph will represent the front-back sides.

Then label left/right and front/back to stack the cubes.

r l 1 3 r l f b 4 3 b 1 f 2 l r 4

Left/right side

l r b f 2

Front/back side

b f

Solution

 There is no other solution, since the third subgragh does not meet the required standards.

1 4 2 3