# Around the World in Many Days

Deаr Puzzling,

How have you been? I am excited to tell you about my new trip. It’s my most ambitious one yet – a full trip around the world! I will definitely keep you posted about my whereabouts. I hope you stick along, this will be a long ride!

I started my trip with some culture. I visited a beautiful theatre with a long history to see a play by a local author. They had more famous titles on offer as well, but having come this far it would kind of be a shame to watch something I can just as well see at home. I have to say I was not disappointed!

Today’s puzzle is a killer sudoku. The rules are as follows:
そ) Each of the eight letters in the grid represents a different number 1–8.
こ) Fill the grid so that each row, column and 2x4 box contains each letter exactly once.
ば) The values of letters in cages marked with dashed lines must sum to the total indicated in the top-left corner.
ん) Solve the grey background puzzle to extract the final answer.

I hope you enjoy today’s puzzle. There are many more to come this time around, so be sure to stay tuned! Can you guess where I am?

Gladys will return in Colourful Candy Cane Carousel.

Here is the solved killer sudoku:

The key is: 1-K, 2-F, 3-G, 4-E, 5-A, 6-R, 7-I, 8-H

The solution path can be seen below.

The puzzle in the background is a

sokoban, as indicated by the hirigana in the steps: そ, こ, ば, and ん represent the syllables so, ko, ba, and n, respectively. It has a relatively simple solution path:

The numbers traced along the solution path are 8-5-3-4-6-2-7-1-7-6, or HAGERFIKIR, which happens to be the name of a theater.

This means that Gladys must have started out her journey in

Addis Ababa, Ethiopia, or more specifically, the Hager Fikir Theatre!

Full solution path for the sudoku (disclaimer - this is a slightly more logical version of my actual path when solving for the first time so as to make myself look a bit less like an idiot):

First, there are two letters which are each used in three cages within the same box: H and I. These must be 7 and 8 in some order, since only 7 and 8 have three different sums with distinct digits (within the scope of this puzzle, of course). Furthermore, if H was 7, the digits outside of the H cages would have to be 7 and 8, meaning A would also have to be at least 8, a contradiction. Therefore H is 8 and I is 7. We can quickly fill in R as 6 as well due to the cage of size 3:

Let's continue looking at the box with the 3 8-cages. We know that these cages must include 1-7, 2-6, and 3-5, in some order, meaning 4 and 8 must be outside of those three cages. However, one of these is also in a cage, and it obviously can't be 8 or else it wouldn't add up to A. Thus, that number must be 4, and with 6 and up already off the table, A must be 5, K must be 1, and we can use the A in the bottom left to complete a cage:

The top cell of the 3-cell 6-cage must be a 1, due to the 1 in the lower box. That (and the 1-4 cage below it) means that the 5-cage in the upper right box must be a 2-3, in some order, and the 6-cage directly above it must be a 1 and a 5. We can now place the 5 and 6 in the top-left box, and the 8 in the top-right box due to normal sudoku rules:

With a 1 and a 5 already in the column, the rightmost 7-cage in the bottom-left box must be a 3-4, leaving the other as a 1-6. The 6-cage immediately above that one must be a 2-4 (which is resolvable). Now, the minimum sum for the E-cage is 4, which is also our highest remaining unassigned value, so that cage must be a 1-3. With so many possibilities stacked in columns, we can now fill in the top-left box:

With six of the eight letter-digit pairs found, that leaves F and G to be 2 and 3 in some order. There's already a 2 in the seventh row, so F must be 2 and G must be 3. Also, the only possible remaining combination for the 8-cage in between them is 1 and 7. This places the 4 and the 7 in the top-right box, and with the five in the bottom row, we can now complete that column and disambiguate several cages:

Using regular old sudoku rules, we can find the 6 in r4c4 and the 1 and then the 5 in the bottom-right box, as well as the 7 in between them because of killer sudoku. That 5 is particularly helpful, as with a 1 already in that column, the cage directly above it must be a 2-6, leaving a 3-5 in the rest of that box, both of which we can immediately disambiguate. We can also fill in the 7-8 pair in the bottom-left box, the 1-3 pair in the leftmost row, the 2-3 pair in the top-right box, and the 1-7 pair in the fifth column:

From here, you could figure out that the only remaining cage must be a 3-4 and use the 3 in the third column to disambiguate it, but you don't technically need to - the rest of the puzzle can be solved using nothing but columns, rows, and boxes:

• Suggestion: Why not convert them to letters again? (Being pedantic, you have used your own variables to answer the question :) )
– ACB
Commented Dec 3, 2022 at 3:47
• @ACB I suppose I could, but with sums playing a part in the solution, it made more sense (at least to me) to present the sudoku solution with digits and then convert only the relevant ones into letters after the fact (also since you're being pedantic, I will be too: they're not my variables, they're jafe's ;) ) Commented Dec 3, 2022 at 3:54