It was inevitable, really... Each fragment of shell has exactly three sharp points, joined by smooth curves. While the King's horses can count reasonably well, his men have been known to confuse donuts and coffee cups, and thus can hardly be trusted with subtle geometric distinctions like adding angles. Between them, they'll eventually manage to label each corner of each fragment with the number of fragments that originally surrounded that point, but if there are any two fragments with identical labels, it's guaranteed that they'll get switched and the reassembly will go poorly. (Even XYZ and XZY will get swapped!) Things could certainly have been worse...all pieces could have an identical set of labels! But is it possible that all fragments' label-sets are distinct, giving HD at least a shot at an open-carton funeral?

Lest this otherwise seem to defy fate, it so happens that the King's llamas and women are able to distinguish concavity from convexity (or at least stickiness from not) well enough to orient a piece labeled XXX or XXY correctly...but is their help sufficient, or for that matter necessary?

(For purposes of this problem, despite occasionally being drawn with what appears to be the beginnings of a digestive tract, you may safely assume that Humpty Dumpty is topologically a sphere)

Extension: Now that it's known to be possible, how few triangles can be used while satisfying the criteria (and how many such solutions exist?) How do these answers change when we eschew the extra help and insist that all three labels are distinct on every triangle?

  • $\begingroup$ I'd mention how I discovered this and an interesting extension, but will save that (and perhaps another tag or two) for later to avoid spoilers. $\endgroup$ Apr 9 '18 at 18:20
  • $\begingroup$ The horses/men cannot distinguish (a,b,c) from (c,b,a). Humpty Dumpty had a great reputation, but was was not the most uniformly curved of eggs, so the 'extra help' only allows correct orientation of (a,a,b), which would otherwise be problematic. Perhaps the llamas have already checked the concavity of the fragments with three distinct labels a few too many times for the stickiness to be a reliable indicator... $\endgroup$ Apr 9 '18 at 18:45
  • $\begingroup$ If I understand this correctly - all the fragments are 'triangular' pieces of a sphere, but we don't know how many there are. The Question is: Can they be put together by these people with their talents and limitations as described? $\endgroup$
    – tom
    Apr 9 '18 at 19:02
  • $\begingroup$ @tom, I think there was a now-deleted comment that correctly clarified this. It is very easy to find a pattern of cracks which cannot be reassembled within these limitations. The question is: does any possible pattern exist which could be put back together again? $\endgroup$ Apr 9 '18 at 20:09
  • 1
    $\begingroup$ Interestingly, nowhere in the nursery rhyme does it suggest that Humpty Dumpty is an egg. $\endgroup$
    – Astralbee
    Apr 19 '18 at 9:15

Yes, there are many solutions that meet these criteria. There is a simple mathematical proof for this, but it's slightly too large to fit in this answer recursive programming is much easier than graph theory.

One such solution is:

enter image description here

This picture takes some interpreting. Triangle ABC forms the "back" side of the sphere. All the other triangles have no other points inside.

The list of triangles in this solution, as well as their "number of fragments around each point" labels are shown below.

[A, B, C]   [4, 8, 8] 
[A, B, D]   [4, 8, 6] 
[C, D, E]   [7, 6, 5] 
[B, D, E]   [8, 6, 5] 
[A, D, F]   [4, 6, 4] 
[A, C, F]   [4, 7, 4] 
[B, C, G]   [8, 7, 5] 
[C, E, G]   [7, 5, 5] 
[C, D, H]   [7, 6, 3] 
[D, F, H]   [6, 4, 3] 
[C, F, H]   [7, 4, 3] 
[B, E, I]   [8, 5, 5] 
[E, G, I]   [5, 5, 5] 
[B, G, J]   [8, 5, 4] 
[G, I, J]   [5, 5, 4] 
[B, I, K]   [8, 5, 3] 
[I, J, K]   [5, 4, 3] 
[B, J, K]   [8, 4, 3]

If you sort the list of labels, it becomes obvious that all are unique:

[3, 4, 5]
[3, 4, 6]
[3, 4, 7]
[3, 4, 8]
[3, 5, 8]
[3, 6, 7]
[4, 4, 6]
[4, 4, 7]
[4, 5, 5]
[4, 5, 8]
[4, 6, 8]
[4, 7, 8]
[5, 5, 5]
[5, 5, 7]
[5, 5, 8]
[5, 6, 7]
[5, 6, 8]
[5, 7, 8]

However, it is also obvious that:

I'm using triangles of the form XXY (meaning the llama skills are required), as well as triangles of the form XXX (which are neither clearly allowed nor clearly prohibited).

  • $\begingroup$ Huzzah! However, either my program or yours contains an error...will elaborate once I am substantially more confident which. ^_^ $\endgroup$ Apr 10 '18 at 6:45
  • $\begingroup$ Just wrote a new program from scratch to double check (much easier than it used to be, hooray for FOSS!). The example you give fulfills the requirements, but another statement in your answer is false as presented (I wouldn't nitpick if the corrected version weren't my motivation for sharing in the first place.) $\endgroup$ Apr 10 '18 at 13:24
  • $\begingroup$ @AndyJuell Hmm perhaps a hint for what's wrong? Is there a solution with fewer faces that I missed? $\endgroup$
    – Selvek
    Apr 11 '18 at 13:42
  • $\begingroup$ Precisely...... $\endgroup$ Apr 11 '18 at 14:11
  • $\begingroup$ Hmm you are correct, my method of adding faces does not account for every possibility. I will need to reexamine... $\endgroup$
    – Selvek
    Apr 11 '18 at 14:46

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