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For ten years now, I've been setting the Chalkdust crossnumber. Over this time, I've developed a lot of tricks and tools for setting good puzzles.

Although puzzles involving words in grids of squares have been around since at least the 1800s, the first crossword (or word-cross as the author called it) puzzle was published in the New York World newspaper in 1913. Since this, crosswords have become a staple in newspapers and magazines around the world.

In the UK, cryptic crosswords were invented and grew in popularity in the 1900s and now appear in most newspapers and magazines. The cryptic crossword in the Listener magazine became infamous as the hardest such puzzle. the Listener ceased publication in 1991, but the Listener crossword still lives on and is now published on Saturdays in the Times newspaper.

In early 2015, we were discussing ideas for regular content in our brand new maths magazine, and wanted to include a more mathematical crossword-style puzzle. Like Venus emerging from the shell, the crossnumber was born.

Of course, we didn't invent the crossnumber puzzle: the first known crossnumber puzzle was written by Henry Dudeney and published by Strand Magazine in 1926; four times a year, the Listener features a numerical puzzle; and the UKMT (United Kingdom Mathematics Trust) have included a crossnumber as part of their team challenge for many years. There are also plenty of other publications that include crossnumbers, including the very enjoyable Crossnumbers Quarterly, that's been publishing collections of the puzzles four times a year since 2016. But perhaps we'll be mentioned in a footnote in a book about the history of puzzles.

But anyway, we'd decided we needed a crossnumber, so I needed to write one...

The first step when creating a puzzle is to create the grid. Like many publications, we restrict ourselves to using grids of squares (for the main crossnumber at least---we allow more freedom in other puzzles that we feature).

Typically, a publication will impose some restrictions on the placement of the black squares in its puzzle. The most popular restrictions are:

There are two forms of symmetry that are used in crossword grids: rotational symmetry and reflectional symmetry. Both order 2 (rotate the grid 180° and it looks the same) and order 4 (rotate the grid 90° and it looks the same) rotational symmetry are commonly used in crossword grids, with order 2 rotational symmetry often being the minimal allowable amount of symmetry. You'll want to be careful when making a grid with order 4 rotational symmetry, as it's very easy to make your grid into an accidental swastika.

With a few notable exceptions, the grids for the Chalkdust crossnumber have some form of symmetry, and also follow the other two restrictions. We do, however, allow some flexibility on these restrictions if it allows us to use an interesting grid. For crossnumber #14 (shown on the left), we experimented with translational symmetry, leading to a grid that was not simply connected and with a greater than usual proportion of black squares. For crossnumbers #11 and #6, the grids had a nice property: rotating the grid leads to the same pattern of squares with the colours inverted. Once again we had to break the requirements on connectedness and the proportion of black squares to do this.

For crossnumber #17, we skipped imposing symmetry entirely and instead formed all 18 pentominoes from the black squares in the grid. This really pleased me, as there were clues in the puzzle related to the number of pentominoes and the number of black squares in the grid.

For American style crosswords, there's an additional restriction that is imposed: all white squares must be checked. A white square is called 'checked' if it is part of an across entry and a down entry---and so you can fill that square in by solving one of two different clues. Due to this, American crosswords will have large rectangles composed entirely of white squares and never have lines of alternating black and white squares as commonly seen in British puzzles.

When writing a crossword, it is common to pick the words to include while making the grid, as trying to find valid words or phrases to fill a predetermined grid is a challenging task. (Thankfully, there's software out there that can help you make grids from a list of words.) For crossnumbers, filling the grid is a much easier task as any string of digits not starting with a zero is a valid entry. This ease of filling the grid is what allowed us to use the interesting restriction-breaking grids mentioned in this section.

For crosswords, it is also common to disallow the use of two-letter words. For the crossnumber, removing two-digit numbers would remove the potential for a lot of fun number puzzles, so we don't impose this restriction here. We allow two-letter words in the Chalkdust cryptic too, as they can be really useful when trying to make a grid with our additional restriction that the majority of the included words should be related to maths.

Once I've made the grid for a crossnumber, the next task is to write the clues.

Often, I start this task by picking a fun mathematical or logic puzzle to include in the clues. Sometimes, this is a single clue, such as this one from crossnumber #1:

Or this clue from crossnumber #5:

Other times, this could be a set of clues that refer to each other and reveal enough information to work out what one of the entries should be, such as these clues from crossnumber #10:

Once I've included a few sets of clues like this, it's time to write the rest of the clues. As any crossnumber solvers will have noticed, my favourite type of clue to add from this point on is a clue that refers to another entry.

More recently, I've begun adding an additional mechanic to each crossnumber. This started in crossnumber #13, when all the clues involved two conditions which were joined by an and, or, xor, nand, nor or xnor connective. Mechanics in later puzzles have included the clues being given in a random order without clue numbers (#14), some clues being false (#16), and each clue being satisfied by both the entry and the entry reversed (#19). I really hope that you enjoy the 'fun' mechanic I used in this issue's puzzle.

Around the same time as I started playing with additional mechanics, my taste in puzzles shifted. Older crossnumbers had been quite computational, and often needed some programming for a few of the clues, but more recently I have become a greater fan of logic puzzles and number puzzles that can be solved by hand. To reflect the change in the type of puzzle I was setting we added the phrase 'but no programming should be necessary to solve the puzzle' to the instructions, starting with crossnumber #14.

One of the most important things to watch out from when writing and checking clues is accidental ambiguity due to writing maths in words.

For example, the clue 'A factor of 6 more than 2D' could be read in two ways: this could be asking the solver to add 6 to 2D, then find a factor of the result; or it could be asking the solver to add 1, 2, 3, or 6 (ie a factor of 6) to 2D.

As long as I spot clues like this, it can usually be fixed with some rewording. In this example, I'd rewrite the clues as either '2D plus a factor of 6.' or 'This number is a factor of the sum of 6 and 2D.'

In my time setting the crossnumber, I've got a lot better at spotting ambiguity in clues, and do this by reading through the clues and trying to be a mega-pedant and intentionally misinterpret them. It can be really helpful to get someone else to help with this check though, as remembering what you intended to mean when writing a clue can make it hard to read them critically.

Perhaps the most difficult part of setting a crossnumber is checking that there is exactly one solution to the completed puzzle.

To help with this task, I've written a load of Python code to help me find all the solutions to the puzzle. I run this a lot while writing clues to make sure there's no area of the puzzle where I've left multiple options for a digit. I intentionally use a lot of brute force in this code so that it's really good at catching situations where there are multiple answers to a puzzle where I only found one solution by hand.

Once I've got all the clues and my code says the solution is unique, I do a full solve of the puzzle by hand. This is both to confirm that the code's conclusion was not due to a bug, and to check that the difficulty of the puzzle is reasonable.

Following this checking, and a little proofreading, the puzzle is ready for publishing. Then the fun part begins, as I get to chill with a nice cup of tea and wait for people to submit their answers.

Welcome to 2025 everyone! It's time to reveal the answers to the Advent Calendar puzzles and announce the winners. But first, some good news: with your help, Santa was able to open the warehouse to get the presents and Christmas was saved!

Now that the competition is over, the questions and all the answers can be found at mscroggs.co.uk/puzzles/advent2024. Before announcing the winners, I'm going to go through some of my favourite puzzles from the calendar and a couple of other interesting bits and pieces.

My first highlight is the puzzle from 2 December. I like this puzzle, because at first it looks really difficult, as there are so many different combinations that could be rolles with 100 dice. But after some thought, it turns out that the answer isn't any different to if there were only 2 dice.

My next highlight is the puzzle on 6 December, as it can be solved by spotting a really satisfying pattern.

My final highlight is the puzzle from 23 December. I like this puzzle so much that I've written a whole blog post about it, and am in the middle of writing a second blog post about it.

Once you've entered 24 answers, the calendar checks these and tells you how many are correct. I logged the answers that were sent for checking and have looked at these to see which puzzles were the most and least commonly incorrect. The bar chart below shows the total number of incorrect attempts at each question.

It looks like the hardest puzzles were on 23 and 8 December; and the easiest puzzles were on 1 and 15 December.

To finish the Advent calendar, you were tasked with opening the warehouse full of presents. The answers to all the puzzles were required to be certain of which combination of parts were needed to open the warehouse, but it was possible to reduce the number of options to a small number and get lucky when trying these options. This year only one person got lucky warly. This graph shows how many people opened the warehouse on each day:

And finally (and maybe most importantly), on to the winners: 237 people managed to open the warehouse. That's a record high number!

From the correct answers, the following 10 winners were selected:

Congratulations! Your prizes will be on their way shortly.

The prizes this year include 2024 Advent calendar T-shirts. If you didn't win one, but would like one of these, I've made them available to buy at merch.mscroggs.co.uk alongside the T-shirts from previous years.

Additionally, well done to 100118220919, Aaron Stiff, Aglahir, Aidan Dodgson, Alan Buck, Alan Shotkin, alek2ander, Alex, Alex Hartz, Aline M. Lee, Allan Taylor, Anastasiia, Andrea Chlebikova, Andrew Brady, Andrew Roy, Andrew T, Andrew Thomson, Andy Ennaco, Ashley Jarvis, Austin Antoniou, Becky Russell, Ben, Ben Reiniger, Ben Tozer, Ben Weiss, Berl Steiner, Bill Varcho, Blake, BlueStorminator, Bogdan, Brian Wellington, Carmen, Carol Cooper, Cathy Hooper, Celer, Chris Eagle, Colin Beveridge, Colin Brockley, Corbin Groothuis, Dahé, Dan May, Dan Swenson, Dan Whitman, Daniel A, Daniel Low, Daphne, Dave, Dave Cowen, David, David Ault, David Banham, David Berardo, David Fox, David Kendel, Dhruv Pisharody, Dimitri Cacouris, Donald Anderson, Dylan Madisetti, Ean, Elizabeth Ann Madisetti, Ellie Winters, Elytre, Emily Troyer, Eoin Davey, Eric, Eric Kolbusz, Erick Lee, Erik Eklund, Evil Loanword, Ewan, Ewan Leeming, Fionn Woodcock, Frank Kasell, Fred Verheul, Gabriella Pinter, Gareth McCaughan, Gary M, Gary M. Gerken, George Witty, Gert-Jan, glocq, Green, Greg W., Guillermo Heras Prieto, Hannah Harris, Heerpal Sahota, Helen, Helen, Henri, Herschel, Holly Carnes, Håkon Balteskard, Iris Lasthofer, Ivan Andrus, Ivan Molotkov, Izzi, Jack, Jake, Jamas Enright, James Dolengewicz, Jamie van Otter, Jan Zemba, Jay Winter, JDev, Jean-Noël Monette, Jen Sparks, Jessica Marsh, Jim Ashworth, Joe Gage, Johan, Johannes C. Huber, John Drake, John Warwicker, Jon Palin, Jonathan Chaffer, Jonathan Thiele, Jorge del Castillo Tierz, Judith Werner, Justin Roughley, Kai, Kat Yates, Katerina Stergiopoulou, Katharine Velleman, Kelly Spoon, Kenneth McComas, Kevin Docherty, kil k, Kirsty Fish, Kristen Koenigs, Larry Goddard, Laura Kaplan, Lauren W., Lazar Ilic, Lincoln Ford, Lise Andreasen, Lizzie McLean, Logan Davis, Louis, Lucas Bowman, Lyra, Magnus Eklund, Marco van der Park, Mark Fisher, Mark Lydon, Martijn, Martin Harris, Martin Holtham, Martine Nome, Matthieu, Michael DeLyser, Mihai Zsisku, Mike Graczyk, Mike Hands, Miles Lunger, Millie, Mr J Winfield, Nadim Shawky, Nadine Chaurand, Naomi, Nazneen Molu, Neil Bastian, Neil M., Nick C, Nick Keith, Nikos Iakovidis, Noah Molder, Oliver Mills, olligobber, Pamela, Pete Chare, Peter Rowlett, Philip Corradi, Pierce R, Qaysed, Quinnifer, Rachel Sheridan, Rashi Gaba, Ray Arndorfer, Reuben, Riccardo Lani, Rob Reynolds, Roger Lipsett, Roni, Rosie P, Ross Ogilvie, Ruth Franklin, Sadie Sage Robinson, Sam Dreilinger, Sami Wannell, Sarah, Sarah B, Scott, Sean Henderson, Seth Cohen, Shannon, Shivanshi, stephen, Stephen Cappella, stephen kirkham, Steve Blay, Steven Richardson, Stuart MacDonald, Sumaya Felic, Tehnuka, The Connors of York, The Secret Kite, Tim Dufton, Tina, Todd, Tony Mann, Tony Palmer, Trevor Casey, tripleboleo, tt1991, Tyaka, Tyler St Clare, UsrBinPRL, Valentin VĂLCIU, Victor Miller, Vinayak, Vinny R, wilberforce314, William Huang, Yasha, Yuliya Nesterova, and Zoran Morrissey-Ralevic who all also completed the Advent calendar but were too unlucky to win prizes this time or chose to not enter the prize draw.

See you all next December, when the Advent calendar will return.

The puzzle on day 23 of this year's Advent calendar is an interesting puzzle. After some hasty editing once it was pointed out to me that the original wording was nonsense, the puzzle was this:

As the squares that a square is friendly with are the same as those that would be attacked by a chess king on the square, the puzzle could be re-expressed as: What is the fewest number of kings that need to be placed on a 23×23 chess board so that every square is attacked or occupied by at least two kings (with no more than one king placed on each square)?

I came up with this puzzle while playing around with square grids in early November last year. I started by working out how many squares need to be coloured for some smaller grids:

This sequence was not on the OEIS, so I set about trying to work out the general formula for a term in this sequence. Let \(a(n)\) be the minimum number of squared coloured for an \(n\) by \(n\) grid. To work out the formula for \(a(n)\), we'll split the problem into 3 cases: when \(n\) is a multiple of 3, when \(n\) is 1 less than a multiple of 3, when \(n\) is 2 less than a multiple of 3.

When \(n\) is 1 less than a multiple of 3, it can be written as \(3k-1\) for some integer \(k\).

We can get an upper bound for the number of squares by finding a pattern of coloured squares that works, as the minimum number of squares will then be at most this. For a 2×2 grid, we can colour 2 squares like this:

For a 5×5 grid, we can colour 8 squares like this:

For an 8×8 grid, we can colour 18 squares like this:

We can continue this pattern: in general for a \((3k-1)\)×\((3k-1)\) grid, we colour in pairs of squares with a gap of one between them in every third row. This would involve colouring \(k\) rows of \(2k\) squares, so a total of \(2k^2\) squares.

Therefore \(a(3k-1)\leqslant2k^2\).

To obtain a lower bound, consider these squares in 5×5 grid:

or these squares in 8×8 grid:

The squares that neighbour these highlighted squares do not overlap, so two different squares must be coloured for each of these. This means that at least 2×4 (for a 5×5 grid) and 2×9 (for a 8×8 grid) squares must be coloured.

Continuing the pattern of highlighting the squares in every third row and every third column, we see that for a \((2k-1)\)×\((2k-1)\) grid, we must colour at least \(2k^2\) squares, and so \(a(3k-1)\geqslant2k^2\).

Combining the lower and upper bounds, we can see that \(2k^2\leqslant a(3k-1)\leqslant2k^2\), and so \(a(3k-1)=2k^2\). As 23 is one less than a multiple of 3, this first case is enough to solve the puzzle in the Advent calendar.

When \(n\) is 2 less than a multiple of 3, it can be written as \(3k-2\) for some integer \(k\). We can obtain upper and lower bounds in a very similar way.

To get an upper bound, consider this pattern of coloured squares:

For a \((3k-2)\)×\((3k-2)\) grid, this pattern involves colouring \(k\) rows of \(2k\) squares, and so \(a(3k-2)\leqslant2k^2\).

To get a lower bound, consider this pattern of highlighted squares:

In exactly the same was as for \(n=3k-1\), we can see from this pattern that \(a(3k-2)\geqslant2k^2\).

Once again, our upper and lower bounds are equal, and so \(a(3k-2)=2k^2\).

Here's where this question gets trickier/more interesting. If \(n\) is a multiple of 3, then it can be written at \(3k\) for some integer \(k\).

We can obtain an upper bound from this pattern:

For a \(3k\)×\(3k\) grid, this pattern involved colouring \(k\) rows of \(2k+1\) squares, and so \(a(3k)\leqslant2k^2+k\).

As in the previous sections, we can get an lower bound using this pattern:

For a \(3k\)×\(3k\) grid, the lower bound requires at least \(2k^2\) squares to be coloured.

We can increase this lower bound by 1 by looking at the bottom right corner: the previous lower bound can colour at most one square adjacent to this, so there must be at least one more square next to that corner coloured.

So we have arrived at the lower bound \(a(3k)\geqslant2k^2+1\).

In this case, the lower and upper bounds are not the same, so we are left with \(2k^2+1\leqslant a(3k)\leqslant 2k^2+k\). I've written a Python script to compute terms of this sequence and have used it to find that:

For \(n=12\) and larger than this, the script takes a very long time, so I haven't checked any more terms.

Overall, we've shown that:

This can be written slightly more succinctly as (where \(\lfloor x\rfloor\) means \(x\) rounded down to an integer):

I've submitted the sequence I have so far to the OEIS: A379726.

Based on the terms computed above, I conjecture that \(a(3k)=2k^2 + k\). If you have any ideas how I could increase my lower bound, or for any other way I could prove this, let me know!

I clearly haven't already made enough Christmas puzzles this year, so I've made another one. If you've used regular expressions before, head straight to mscroggs.co.uk/regexmas to try the puzzle. If you've not, read on...

Regular expressions are strings of characters that can be used in multiple programming languages to validate text. Regular expressions are usually written between two / characters. Between the slashes, characters have the following meaning:

My regular expression Christmas puzzle is shown below. You can either solve it on this page or at mscroggs.co.uk/regexmas using the buttons or your keyboard, or you can download this PDF of the puzzle.

In the grid below, write r, g, b, c, m, y, k, or w in every square so that:

The squares containing an r will be coloured red, those containing a g will be coloured green, those containing a b will be coloured blue, those containing a c will be coloured cyan, those containing an m will be coloured magenta, those containing a y will be coloured yellow, those containing a k will be coloured black, and those containing a w will be left white.

As usual, I spent some time this November, designing this year's Chalkdust puzzle Christmas card (with some help from TD).

The card contains 10 puzzles. By splitting the answers into pairs of digits, then drawing lines between the dots on the cover for each pair of digits (eg if an answer is 201304, draw a line from dot 20 to dot 13 and another line from dot 13 to dot 4), you will reveal a Christmas themed picture. Colouring any region containing an even number of unused dots green and colour any region containing an odd number of unused dots red or blue will make the picture even nicer.

If you're in the UK and want some copies of the card to send to your maths-loving friends, you can order them at mscroggs.co.uk/cards.

If you want to try the card yourself, you can download this printable A4 pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will automatically be used to join the dots and the appropriate regions coloured in...