"Have a look at this" said Ernie when he dropped in unannounced yesterday, pulling what looked like an old-fashioned photographic negative from an envelope. I held the small plastic square up to the light and peered at it, but could see nothing except the grey glossy surface. But after a few seconds later it appeared to turn completely transparent so I could see the light fitting through it. "Try the other side", he suggested. I did, and almost dropped it in surprise as I saw an eye staring back at me. A few seconds the eye image faded to a view of the other side of the room. "It's a new ultra-high refractive index plastic from Acme Industries" said Ernie. "They call it slow glass". He went on to explain that the speed of light through a transparent material was inversely proportional to its refractive index. "...and this material has a refractive index of approximately 10^18". I must have looked confused so, putting on his 'explain-this-to-a-small-child' face he continued "So light hitting one side of a surface coating only a manometer thick would take almost 4 seconds to get out the other side".

"Interesting", I said, "but I can't imagine it being very useful. Just consider the risk of accident if you used it in a car windscreen". But Ernie assured me that while previously it had been a 'solution waiting for a problem', he had now identified the perfect problem. "As you know", he announced (I didn't), "there is a lunar eclipse visible from here in just a few days. Two weeks later there will be a solar eclipse visible in the Kzijekistan hinterlands. I would love to see both but it will take a 16 day trek to get to the site of the solar eclipse and set up my equipment." I murmured some commiserations, but Ernie wasn't finished. "Don't you see - the slow glass is perfect. If someone" (he looked pointedly at me) "exposes a 1.0 mm thick sheet to the lunar eclipse the light will take 40 days to get through so I will have plenty of time to get back and make all the required observations".

He assured me I would need no astronomical abilities (just nimble fingers). "My attic sky-light is perfectly oriented for a great view of the eclipse. All you need to do is tape slow glass sheets up to the sky-light during the eclipse, then store them safely for my return". He went on to explain that as slow glass was a 'very new thing' he didn't quite know the best size or exposure times for optimal eclipse viewing. So he had come up with a cleverly designed compromise involving multiple sheets of slow glass and a range of different exposure times. "It's all very simple", he announced passing an envelope to me, "I have writen down explicit instructions for you". And then he looked at his watch, told me he had to rush and finish packing for the solar eclipse trip, and left with a cheerful farewell.

enter image description here

As soon as he left I opened the envelope to read his instructions. The contents consisted 16 statements, each typed on a separate scrap of paper, plus a scrawled note. In the order I removed them from the envelope, the statements read:

  1. The area of the largest sheet of slow glass should be precisely 1/4 the area of the sky-light.
  2. The eclipse will last for precisely 210 minutes.
  3. The smallest sheet of slow glass should be as small as possible consistent with the rest of these instructions.
  4. The exposure time for the largest sheet of slow glass should be precisely half that of the eclipse.
  5. No sheet of slow glass may overlap the edges of the window during its exposure.
  6. The sky-light is precisely 1000 mm square.
  7. Each sheet may be exposed at any time, and at any position and orientation on the window that is consistent with the rest of these instructions.
  8. The exposure time for the smallest sheet of slow glass should be precisely half that of the second to smallest sheet.
  9. Each sheet of slow glass, except for the smallest and the largest should have exposure time exactly equal to the average of the exposure times of the next smaller and next larger sheets.
  10. Each slow glass sheet should be exactly square.
  11. Every part of the eclipse should be recorded by the exposure of at least one sheet of slow glass..
  12. Each sheet of slow glass, except for the smallest and the largest should have side length exactly equal to the average of the side lengths of the next smaller and next larger sheets.
  13. Each sheet's exposure must begin at or after the start of the eclipse and end at or before the end of the eclipse.
  14. Each slow glass sheet will be exposed by taping it flat to the window, without movement, for its entire exposure time.
  15. The area of the smallest sheet of slow glass should be precisely 1/4 that of the second to smallest sheet.
  16. No sheet of slow glass may over-lap or under-lap any other sheet during its exposure.

The note read: Sorry, but I haven't had time to buy the slow glass. If you order it for me from Acme Industries I will compensate you upon my return. Note that Acme will only supply square sheets of slow glass - no rectangles supplied. While their price per square meter is reasonable, they have an exorbitant cutting and handling charge for multi-piece orders. So please just buy a single square piece that is just big enough to cut out all the bits you will need - you can cut it up into the required squares yourself using a heated knife.

I am sure you can see my problem. I don't want to disappoint Ernie by buying too much, or too little, or cutting too many or too few pieces, or exposing them for the wrong time. How big a square piece of slow glass should I order? A cutting diagram and some sort of diagram or schedule for exposing the sheets would also be of great help to me.

Clue 1. If you think of time as a third dimension, what shape does a square sheet of slow glass plus its exposure time (or the sky-light plus the time of the eclipse) make?

Clue 2. (Taking Xoff's comments into account) Just consider where you would put the 9-th largest cube...

  • 2
    $\begingroup$ There should be a Ernie tag or something. I love these Ernie questions :) $\endgroup$
    – Wen1now
    Jan 21, 2017 at 2:35

2 Answers 2


This problem is only about packing space-time cube of size $1,2,...,n$ in a cube of size $2n$. $$\sum_1^n k^3=\left(\frac{n(n+1)}{2}\right)^2$$

As the combined volumes of the $n$ cubes is a polynomial of degree 4 compared to the size of the large cube (of side $2n$) that is $8n^3$ (polynomial of degree 3 ) then there are only a finite number of possible solutions where the volume of the main large cube can contains all the small cubes. So we need $$(2n)^3-\left(\frac{n(n+1)}{2}\right)^2\ge0$$ Hence $n<30$. Is there a valid packing for 29 or even 28 ? I don't know, it seems to be a difficult problem.

EDIT : So, as Tim explained, this is not difficult as you are bound by the 8 biggest cube to be at most 21. And it's easy to see that there is a trivial solution for 21 (as there are a lot of empty spaces between the largest cube placed in the corners) : as you fit 21,20,19,18,17,16,14,15 in 8 consecutive corners, then between 14 and 15 you can place 13, between 14 and 16, you place 12 and 11 (it's ok as 15 and 17 don't use all their half), between 17 and 15 place 10 and 9 and all remaining cubes are easy to place.

So We have to buy a square large enough to fit 21 squares : this is a 58x58 square (as explained in A005842).

Here a solution, that I found, I hope it's ok :


This is a very nice problem !

Comment added by Penguino. An example of a cube-packing exposure schedule is shown in [a] below (the second number on each cube is the height of its top surface). To more clearly demonstrate that there is no solution for N=22, the 14-cube cannot fit in an orthogonal orientation with the 8 larger cubes (as noted by Tim). It can fit if rotated (about the time axis), but then it intersects with the 13-cube and 12-cube (see [b] below). There are other options (for example you can put 14-cube square on the side and rotate 15-cube), but a little exploration shows they all have a similar problem.

enter image description here

  • $\begingroup$ To elaborate on the second hint: there is a trivial way to fit in the biggest 8 cubes, by putting each in a corner. The 9th biggest will have to be able to at least squeeze in between the 7th and 8th biggest. Assuming we have n cubes, the 7th and 8th biggest cubes will be 'n-6' and 'n-7'. Within a '2n' max side value of the cubes, if all cubes were stacked ialigned withx-, y- and time-axis, we can at most fit in a cube with sides 2n - (n-6) - (n-7) = 13. That would put the upper limit to n = 21, and still not prove how. We could however rotate in the x-y-plane... $\endgroup$ Jan 11, 2017 at 12:35
  • $\begingroup$ Solution is still missing one thing - the 'placement chart' of all the squares in the '42x42' window (including overlaps) - in essence it's a summary of what you decribed the cube packing to be like. I'll go ahead and assume issues as far as 'timing of placing/removing' and 'can we actually fit the tape' as I looked at in my solution, are probably overworking the problem. $\endgroup$ Jan 13, 2017 at 7:05
  • $\begingroup$ I'm wondering if the given solution (which is only one possible solution) can be improved upon by finding ways that don't include taking off one piece and putting on another in the exact same time slot (thus accounting for the time of editing the setup). It looks fairly doable to only need a maximum height of 40 in the stacking, thus allowing 2/42th of 210 minutes = (10 minutes) to be divided and act as the required time buffer for executing the actions. $\endgroup$ Jan 16, 2017 at 13:45
  • $\begingroup$ Where this solution did surprise me however (in a good way), was not seeing the 'big 8' as the 'corner cubes', but putting 5th to 8th in the middle layer. This indeed simplies the packing problem alot, which I failed to realise. $\endgroup$ Jan 16, 2017 at 13:48
  • $\begingroup$ @Tim, I am pretty sure you can stagger the cubes to avoid needing to do instant replacement. In my diagram A) you can raise the set of ten cubes a smidgen up from the underlying yellow cubes, without any time gaps appearing, (For example, the 'gap' between 19 and 15 is covered by the remaining exposure time of 21). Then swap cubes 5 and 6, raise the green cubes a smidgen off the yellow ones, and raise 5 so it ends exactly at the end of the eclipse. Note you could also stagger initial placement of the yellows similarly so no need for simultaneous placement either. $\endgroup$
    – Penguino
    Jan 16, 2017 at 20:39

We need to figure out:

  1. Number and area of the squares needed
  2. How to place the squares
  3. When to place/remove each of the squares with regards to the exposure time.

For the time being, I'll first try and solve it for cases where we do NOT put up a piece on a single spot, remove it later and have another piece cover the same spot afterwards. (although point (16) technically allows for it). Whether or not correct, it might serve as inspiration for others attempting to tackle this.

Solution for point 1:

Assume the smallest square has a side x

From (15) we know the second smallest has side 2x

From (9) we know the third smallest has side 3x, and the next few will have sides 5x, 6x, 7x, .. up to the largest piece.

From (1) we know that largest piece is 250000 mm².

From the OEIS (source: https://oeis.org/A005842 ) we know that for squares with sides from 1x to 37x, there is a known upper bound for the size of the surrounding square. (note: we do need series that start from ONE upwards, as the lower bound is '1*x'

We know that the largest piece is exactly 1/4 of the total surface area, so it's sides are half. We are therefor looking for an entry in the sequence where 'n' is exactly half of 'm'. If we list the (n,m) pairs (referencing as in the OEIS link), we find (9,18) the only matching packing where n = m/2.

So assumming 9x = 2500mm, we know x = 2500/9mm, so the smallest square has a side of 2500/9mm, the next 5000/9mm, etc.. untill the largest of 17500/9 = 2500mm.

Solution for point 2:

Unfortunately the OEIS only lists the relation, but not the packing itself. If we know how to pack the smaller squares in the big square, we also know how to cut them from it.

A possible packing is as follows: enter image description here

Red lines are where tape goes, one square = 2500/9mm of side.

Solution for point 3:

We know we have 210 minutes to cover, and we know, in analogy to the sides of the squares, that the exposure times range from y over 2y, 3y, .. up to 9y seconds. They can overlap in time, but at any time we need to have at least one square up. In total we get 45*y exposure time we could possibly cover.

We know however that 9y (exposure for the largest piece) = 105 minutes = 6300 seconds. This means y = 700 seconds, and in total we can cover 45y = 31500 seconds = 525 minutes, so that's plenty.It'll also imply that at times we'll have multiple pieces up at once, as we can only start using our exposure time at t = 0, and have to end it at t = 210 min.

Building off point 2, here's a suggested time-progress of the squares, which fortunately does not imply having to place and remove any squares at the same time (because face it, we are not THAT handy - I've left 30 second intervals between two actions):


Note that the combinations 3-8 and 4-8 are never on at the same time, so there's no issues in placing or removing the holding tape. Or for those who want timed instructions:

enter image description here

  • $\begingroup$ Your longest exposure is supposed to equal half the time of the eclipse (i.e., 105 minutes). $\endgroup$
    – Gareth McCaughan
    Jan 5, 2017 at 13:43
  • $\begingroup$ @GarethMcCaughan Piece #9 (well, the piece with side 2500mm) is on for 6300 seconds, which is 105 minutes), from 6210 to 12510 seconds. $\endgroup$ Jan 5, 2017 at 14:04
  • $\begingroup$ D'oh, so it is. Sorry! $\endgroup$
    – Gareth McCaughan
    Jan 5, 2017 at 14:37
  • $\begingroup$ You are moving in the right direction, but your comment regarding point 16 (and point 7) is relevant. I am sure Ernie expects a smaller smallest piece of slow glass (and hence a larger N). $\endgroup$
    – Penguino
    Jan 5, 2017 at 21:14
  • $\begingroup$ @Penguino - glad to hear there's some merit to the effort, but for now I'm stumped on how to tackle it in an analytic way. I've only managed figure out an N due to a packing relying on them all being there at the same time. Besides that, I'm also struggling with defining what an 'optimal' solution consists off: Minimizing smallest piece, minimising waste after cutting, to what extent does practically matter (ie where the holding tape goes, how fast it can be done, ...) $\endgroup$ Jan 6, 2017 at 10:30

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