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August 2017 Puzzle Periodical - Solar Scramble

By James M., NSA Operations Researcher

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Can you reattach the sun's rays in this month's Puzzle Periodical?

Problem:

In anticipation of next week's solar eclipse on Monday, 21 August 2017, this month's NSA Puzzle Periodical asks: Can you solve this logic challenge to help reattach the sun's rays? 

Due to bizarre and unexplained cosmic fluctuations, the sun's rays have fallen off. In order to save summer you have to put them back on in the right way. Here is what you know:

  1. Each of the 10 rays is uniquely labeled with a number in the range 1-10.
  2. The spots between the rays are marked with the absolute difference of the ray labels that are next to them.
  3. There are four consecutive rays that have even number labels.
Diagram of a circle with 10 numerical ray labels around the perimeter, and 10 unlabeled spots between each number. The numbers going clockwise and beginning at the 12:30 position and ending at the 11:30 position are 7, 4, 1, 7, 8, 4, 3, 4, 2, 6.

Following the above rules, label each spot on the sun with the number of the ray that goes there. 

Once the rays are correctly placed, add up the label values of rays that are directly opposite each other on the sun. Convert these values to letters using the substitution 1=A, 2=B, …, 26=Z and read them clockwise starting at the top to find out what the sun can do now that it has its rays again.

Illustration of a total solar eclipse

Click to see the answer!

Solution:

Diagram of a circle with 10 numerical ray labels around the perimeter, and 10 spots between each number that now contain numbered labels. The original ray numbers going clockwise and beginning at the 12:30 position and ending at the 11:30 position are 7, 4, 1, 7, 8, 4, 3, 4, 2, 6. The new labeled spots beginning at the 12:00 position (prior to the first ray number 7) and working clockwise are: 10, 3, 7, 8, 1, 9, 5, 2, 6, and 4.

Adding opposite ray labels:

The same circle diagram as the prior diagram, but now featuring the opposing ray labels above the spot labels. Starting with the spot labeled 10 at the 12:00 position and working clockwise, the numbers shown are 19, 8, 9, 14, and 5.

Converting to letters:
19 = S, 8 = H, 9 = I, 14 = N, 5 = E

Final answer is SHINE

Getting to the Solution:

There are many ways individual spots can be labeled, so we want to narrow our search space as much as possible. To do this, we look at the given differences and consider how many pairs of numbers could generate them. Smaller differences can be made by many pairs of numbers, while larger differences are limited to a few pairs.

 

Start with the biggest difference shown, which is the 8 on the bottom. There are only two pairs of numbers in the range 1-10 that have a difference of 8: 2 and 10, and 1 and 9. There are also two ways that each of those pairs could be placed, depending on which number is placed on the bottom of the sun. We'll try each possibility to see where it leads.

  • 10 on the bottom - Moving counterclockwise from 10, the following labels are forced: 2, 9, 8, 4. Once we reach 4, however, we are stuck, since neither 4+7 nor 4-7 are in the range 1-10. So this arrangement will not work.
  • 2 on the bottom - Going counterclockwise from 2, we are forced to place the 10 and the 3. Clockwise from 2, 6 is forced. Next must be 9, since 3 is already placed. We now have three even numbers surrounded by odd numbers. Since there are only two even numbers left to place, there is no way we can get the four even numbers in a row that the third condition demands. So this arrangement will not work.
  • 1 on the bottom - Moving counterclockwise from 1, 9 and then 2 are forced. Next must be 3, since 1 is already placed. 7 is then forced. From here we are stuck, however, since neither 7+7 nor 7-7 are in the range 1-10. So this arrangement will not work. 
  • 9 on the bottom – Going counterclockwise from 9, the following labels are forced: 1, 8, 7, 3, 10, 4. We can't yet say if 2 or 6 should go next, so we switch to going clockwise from 9. Here 5 is forced, then 2, since 7 is already used. This leaves 6 for the last spot. So this is the only arrangement that will work. Finally, note that this arrangement also satisfies the third condition.