East meets West

[TimeSpaceLightForce]

The 2 cube dice are glued on 1 of their faces
to make a 1 x 1 x 2 die w/ new configurations.
Any 2 of its sides now can add to:
5, 6, 7, 8, 9, 20,11,12,13,14,15,16
__________ __________
/ / /\
/ / / \
/__________ /_________ / \
\ \ \ /
\ \ \ /
\__________\__________\/

What are the glued faces.?

[CaptainEd]

3 and 5

I'm assuming that our Long Die has four Long Faces (LFs) and two Ends.

This gives 6 faces, four LFs and two Es.

I believe that you are saying for each item S in {5, 6, 7, 8, 9, 20,11,12,13,14,15,16}
there are distinct x and y items from {LF1, LF2, LF3, LF4, E1, E2}, where X+Y=S, and that all X+Y are in that set.

If each individual die obeys the principle that opposite faces total 7, then if a Long Face has total X, its opposite LF has total 14-X.

What LF combinations could total 20?

12,8,6,2--this results in a sum of 10, which is not shown as possible

11,9,5,3

10,10,4,4--this does not permit enough cross-sums to make 12 different sums.

So the LFs must be 11, 9, 5, 3.
What are the Es?
First, which sums are due to LF + LF?

20, 16, 14, 12, 8 (need to add 5, 6, 7, 9, 11, 13, 15)

Then the Es must be
4 (yielding 15, 13, 9, 7) 2 (yielding 13, 11, 7, 5)

The end points are 4 and 2, so the glued faces (7-X) are 3 and 5.
What an interesting puzzle! I thought it would require computer work or tedium!

[CaptainEd]

Just so that I'm sure of what you mean. When you say that any two of the sides can be used to form those sums, are you referring to this die as having 6 sides--4 of them have two faces, and 2 of them have just one face?

[jhawk]

Agree. restate and explain what 2 sides mean

[CaptainEd]

Is one die right-handed (Western) and the other one left-handed (Eastern)?

[TimeSpaceLightForce]

i mean any 2 new configurations adds to those numbers ,yes four 2 faces.. no matter which is east or west?

[TimeSpaceLightForce]

3 and 5

I'm assuming that our Long Die has four Long Faces (LFs) and two Ends.

This gives 6 faces, four LFs and two Es.

I believe that you are saying for each item S in {5, 6, 7, 8, 9, 20,11,12,13,14,15,16}

there are distinct x and y items from {LF1, LF2, LF3, LF4, E1, E2}, where X+Y=S, and that all X+Y are in that set.

If each individual die obeys the principle that opposite faces total 7, then if a Long Face has total X, its opposite LF has total 14-X.

What LF combinations could total 20?

12,8,6,2--this results in a sum of 10, which is not shown as possible

11,9,5,3

10,10,4,4--this does not permit enough cross-sums to make 12 different sums.

So the LFs must be 11, 9, 5, 3.

What are the Es?

First, which sums are due to LF + LF?

20, 16, 14, 12, 8 (need to add 5, 6, 7, 9, 11, 13, 15)

Then the Es must be

4 (yielding 15, 13, 9, 7) 2 (yielding 13, 11, 7, 5)

The end points are 4 and 2, so the glued faces (7-X) are 3 and 5.

What an interesting puzzle! I thought it would require computer work or tedium!

horray CaptainEd.. good logic solve , i learned it. 1up thanks.

I actually just cut a styro and mark the pips for die model ,glued sides , count and theres the puzzle.

[Prime]

The 2 cube dice are glued on 1 of their faces

to make a 1 x 1 x 2 die w/ new configurations.

Any 2 of its sides now can add to:

5, 6, 7, 8, 9, 20,11,12,13,14,15,16

__________ __________

/ / /\

/ / / \

/__________ /_________ / \

\ \ \ /

\ \ \ /

\__________\__________\/

What are the glued faces.?

I don't see how you can achive this set of 2-side totals with 2 standard dice.

If we can add any 2 sides there should be ₆C₂ = 15 totals. Whereas, OP lists just 12. If we only add each 2 adjacent sides, then we get a set of 12 2-side totals.

However, I don't see how that set of 2-side totals is possible with standard dice.

Say, you glued 5 and 3, as CaptainEd suggested. Then you have the end sides of 2 and 4. The only possibility to have a sum of two sides equal to 5, is to make one of the long sides 2+1=3. Only then adding that with the end side of 2, we can get 5. But then no two sides add to 9, 10, 11, 16, 17, 18, 19, and 20. But there are two sides which add to 21.

No other variation of glueing allows the set of given 2-side totals either.

What am missing here?

[CaptainEd]

When I try to lay out the actual dice, the only configuration (except for rotations and reflections) that I find looks like this:

Left End = 2
Right End = 4
Long Faces = 36, 65, 41, 12

Alternative view: two dice, called L and R. We are viewing from the front, we can see

* L's left face is 2, L's top is 3, L's front is 6

* R's right face is 4, R's top is 6, R's front is 5

If you roll these dice towards you two steps, you get the opposite faces:

* L's top is 4, L's front is 1

* R's top is 1, R's front is 2

So, the two End faces are 2 and 4, the four Long faces are 9, 11, 5, 3. Add any two of these six numbers, and you get

6, 11, 13, 7, 5, 13, 15, 9, 7, 20, 14, 12, 16, 14, 8.

Removing duplicates, and sorting, you get 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 20

Notice when you look at L from the vertex that shows 1,2,3, they are located Counterclockwise (the "Western" die), but the R die shows 1,2,3 Clockwise (the "Eastern" die). Very cute, TSLF!

Edited January 24, 2013 by CaptainEd

[Prime]

When I try to lay out the actual dice, the only configuration (except for rotations and reflections) that I find looks like this:

Left End = 2

Right End = 4

Long Faces = 36, 65, 41, 12

Alternative view: two dice, called L and R. We are viewing from the front, we can see

* L's left face is 2, L's top is 3, L's front is 6

* R's right face is 4, R's top is 6, R's front is 5

If you roll these dice towards you two steps, you get the opposite faces:

* L's top is 4, L's front is 1

* R's top is 1, R's front is 2

So, the two End faces are 2 and 4, the four Long faces are 9, 11, 5, 3. Add any two of these six numbers, and you get

6, 11, 13, 7, 5, 13, 15, 9, 7, 20, 14, 12, 16, 14, 8.

Removing duplicates, and sorting, you get 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 20

Notice when you look at L from the vertex that shows 1,2,3, they are located Counterclockwise (the "Western" die), but the R die shows 1,2,3 Clockwise (the "Eastern" die). Very cute, TSLF!

Ah, I see..

This puzzle uses two different dice -- left-handed and right-handed.

If you set a die with 1 on top, 2 on the front, then left-handed die would have 3 on the left, while right-handed would have 3 on the right.

It's a good puzzle. Simple to construct with actual dice, not as simple to solve.

Imo, it would be even better defined if OP gave all 12 totals of the adjacent sides, instead of "any two sides" and did not skip the duplicates. Then each face of the dice would be included 4 times in those totals. The sum of two glued sides could be deduced by subtracting all totals from 168 and dividing that by 4. The rest could be deduced by constructing the smallest 2-side sum of 5.

Thanks, CaptainEd for clearing that up.