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Comics & Activities 02/17/2025

Rainier Pederson | Retrograde Staff

Mini Meta Crossword by Alana Platt

How to Solve Mini Meta Crosswords

How to Solve Mini Meta Crosswords

Adapted from Pete Muller and Andrew White’s Mini Metta Crosswords: As Seen in The Washington Post

A set of Mini Meta crosswords is made up of at least six puzzles. For each of the first crosswords, one of the entries serves as one part of a multiple-part clue, in order. The answer to that clue is a word or phrase (at least five letters long) hidden within the last mini crossword. The hidden answer may start in any square in that crossword, and will be found snaking through the grid, moving horizontally and vertically (but not diagonally) without revisiting any squares.

For instance, say you’ve just solved the six grids in a Mini Meta set and the first five grids look like this:

Look for words that are related and could flow together or connect other words together. Be flexible and keep trying even if your first idea doesn’t work out! There are two proper nouns in the fourth and fifth grids: PAUL and SIMON. PAUL, SIMON, and MUSIC all go together so the phrase might look like this:

Now that we know the clue is MUSIC GENRE FOR PAUL SIMON we can look at the final crossword to try and solve the puzzle! In this case, the answer is FOLK-ROCK.

For those of you who feel confident in your Mini Meta Entry answers, please submit your answers along with your contact information to tinyurl.com/feb-xword-prize for a chance to win a handmade prize! Winners will be chosen randomly and featured in the next issue of The Retrograde. Only current UTD students and faculty may enter to win, and only correct answers will be considered. Entries are due by Feb. 24 at 11:59 p.m.

KenKen by Rainier Pederson

Your goal is to fill up the 4×4 grid with numbers 1-4. Each column and row must have exactly one of each number, just like in sudoku. Each box (an area outlined with thick black lines) has a target number and a mathematical operation listed in the top-left corner. The numbers within the box must result in the target number when the mathematical operation is employed. For example, a box with 3+ could be satisfied with 1 and 2, as 1+2=3. Similarly, a box with 2÷ could be satisfied with either (1,2) or (2,4), as 2÷1=2 and 4÷2=2.  

The puzzle is currently available as a PNG, with an interactive version coming soon.  

Example puzzle:

Here is a practice KenKen to familiarize yourself with the solving process.

This is an empty puzzle. To start, put a 4 in the box labeled 4 containing only one grid square.
This is an empty puzzle. To start, put a 4 in the box labeled 4 containing only one grid square.
Now, let's look over to the box labeled 5+, which means the numbers within the box must add up to 5.
Now, let's look over to the box labeled 5+, which means the numbers within the box must add up to 5.
There are two ways a box with two numbers could add up to 5 — either 1+4 or 2+3. Because there is already a 4 in the bottom row, it cannot be the (1,4) pair, so the two numbers must be 2 and 3. We do not know the order yet, but that is still information we can use to solve the puzzle.
There are two ways a box with two numbers could add up to 5 — either 1+4 or 2+3. Because there is already a 4 in the bottom row, it cannot be the (1,4) pair, so the two numbers must be 2 and 3. We do not know the order yet, but that is still information we can use to solve the puzzle.
Because there is already a 4 in the bottom row, and because there will be a 2 and a 3 in that row, the last remaining number in the row must be a 1. Keep in mind that every column and row has exactly one of each number. We can use this information to solve the box the 1 is a part of.
Because there is already a 4 in the bottom row, and because there will be a 2 and a 3 in that row, the last remaining number in the row must be a 1. Keep in mind that every column and row has exactly one of each number. We can use this information to solve the box the 1 is a part of.
In a 4x4 grid, there is only one set of numbers that can satisfy 3-, which is 4-1=3. Because the 1 is accounted for, the other number in the box must be 4. Now, let's look at the 6+ box on the top.
In a 4x4 grid, there is only one set of numbers that can satisfy 3-, which is 4-1=3. Because the 1 is accounted for, the other number in the box must be 4. Now, let's look at the 6+ box on the top.
Because all three numbers in the box are in the same row, the numbers in the box cannot repeat. The only way this works is if the numbers are 1+2+3=6. While we do not know which number goes where, this is still helpful information.
Because all three numbers in the box are in the same row, the numbers in the box cannot repeat. The only way this works is if the numbers are 1+2+3=6. While we do not know which number goes where, this is still helpful information.
Because the first three numbers in the top row are 1, 2 and 3, that only leaves 4 for the last number in the top row.
Because the first three numbers in the top row are 1, 2 and 3, that only leaves 4 for the last number in the top row.
While we are looking at the 8x box, note that 1x2x4=8, and 3 is not a factor of 8. While we do not know where the remaining 1 and 2 go, this is still worth taking note of.
While we are looking at the 8x box, note that 1x2x4=8, and 3 is not a factor of 8. While we do not know where the remaining 1 and 2 go, this is still worth taking note of.
Because the 3 in the last column cannot go in the 8x box, it must go in the last square of the last column.
Because the 3 in the last column cannot go in the 8x box, it must go in the last square of the last column.
With the 3 in the 5+ box taken care of, the only remaining number is 2. This also finishes the last row.
With the 3 in the 5+ box taken care of, the only remaining number is 2. This also finishes the last row.
If we look at the box above the 2 we just placed, we have a 4÷. The only way to make that work is 4÷1 = 4, so we can note that the two numbers in the box must be 1 and 4.
If we look at the box above the 2 we just placed, we have a 4÷. The only way to make that work is 4÷1 = 4, so we can note that the two numbers in the box must be 1 and 4.
Because there is already a 4 in the third row, the 4 from the 4÷ must be in the second row.
Because there is already a 4 in the third row, the 4 from the 4÷ must be in the second row.
That puts the 1 from the 4÷ box in the third row.
That puts the 1 from the 4÷ box in the third row.
We can then put a 3 at the top of the third column to complete the column. With that, let's continue looking at the 6+ box.
We can then put a 3 at the top of the third column to complete the column. With that, let's continue looking at the 6+ box.
The 1 in the bottom row means we can't put the 1 in the 6+ box in the second column, so it must go in the first column.
The 1 in the bottom row means we can't put the 1 in the 6+ box in the second column, so it must go in the first column.
This then puts the 2 in the second column, which now has all but one of the squares filled.
This then puts the 2 in the second column, which now has all but one of the squares filled.
This leaves the 3 in the second row of the second column. With that, let's continue looking at the 8+.
This leaves the 3 in the second row of the second column. With that, let's continue looking at the 8+.
The 3 we just placed in the 8+ box means the other two boxes add up to 5. This can be done with either a (1,4) pair for 1+4=5, or a (2,3) pair for 2+3=5.
The 3 we just placed in the 8+ box means the other two boxes add up to 5. This can be done with either a (1,4) pair for 1+4=5, or a (2,3) pair for 2+3=5.
The first column already has 1 and 4, so the remaining 5 can't be built from 1 and 4. Because this box has a bend, we can have repeat numbers — note that the rules say only one of each number can occur in a given row or column, but not that numbers cannot repeat within boxes.
The first column already has 1 and 4, so the remaining 5 can't be built from 1 and 4. Because this box has a bend, we can have repeat numbers — note that the rules say only one of each number can occur in a given row or column, but not that numbers cannot repeat within boxes.
Because the second row already has a 3, the other 3 must go in the third row of the first column.
Because the second row already has a 3, the other 3 must go in the third row of the first column.
We can put the last 2 in the 8+ box, finishing both that box and the first column.
We can put the last 2 in the 8+ box, finishing both that box and the first column.
The second and third rows already have three of their four numbers placed, so we can easily complete both rows, finally solving the puzzle.
The second and third rows already have three of their four numbers placed, so we can easily complete both rows, finally solving the puzzle.

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