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2024-12-22
I showed off and part-solved a prototype version of this puzzle with Katie Steckles in the fifteenth Finite Group livestream.
You can watch a recording of this stream, and watch our future streams if you sign up to
our Patreon.
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...
What is a regular expression
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:^(not inside square brackets) means the start of the text.$means the end of the text.- A set of characters are written between
[and]means any of these characters.
For example, the regular expression - A set of characters are written between
[^and]means any character except one of these. For example, the regular expression/^[^df]$/matches any character except d or f. ?means the previous item may or may not appear. For example, the regular expression/^ab?c$/matches "ac" or "abc".+means the previous item can appear one or more times. For example, the regular expression/^ab+c$/matches "abc", "abbc", "abbbc", and so on.*means the previous item can appear zero or more times. For example, the regular expression/^ab*c$/matches "ac", "abc", "abbc", and so on.{and}with a number between means the previous item can appear this number of times. For example, the regular expression/^ab{4}c$/matches "abbbbc".|means either the stuff to the left or the stuff to the right. For example, the regular expression/^ab|c$/matches "ab" or "c"..means any character. For example, the regular expression/^ab|.$/matches "ab" or any single character.(and)can be used to group together other symbols. Groups in brackets can be later referred to by writing\1,\2, etc to refer to the first, second, etc bracketed group. For example, the regular expression/^(a|b)\1$/matches "aa" or "bb".
/^[abc]$/ matches "a", "b", or "c".
The puzzle
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:
- each row (reading left to right) satisfies the regular expression to the right of the row;
- each column (reading top to bottom) satisfies the regular expression below the column.
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.
/^w+yw+$/
/^([kw]+)[^kw]\1$/
/^(g|wwwg|gww)+.$/
/^wy?g*y+w+$/
/^((w|gg)(ww|g)){3}$/
/^[wg](w|g)[gw](.)\2+\1{2}$/
/^.g*[^y]$/
/^([gk][gk][gk])\1\1$/
/^yw+kw+y$/
/^w*b(bb)+w*$/
/^(w+)w?(bb?)\2\2\1$/
/^(www|bbb)+$/
/^w+gyw+$/
/^[wg]*y[wg]*$/
/^.*gwg.*gwb.*$/
/^[^g]+g+[^g]+$/
/^y?g+y?g+k?b+$/
/^[w]+g*w[^w]+$/
/^w+g+wg+[^g]+$/
/^w*yw*g+w*$/
/^w*y?g?y?w*$/
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2024-12-04
As usual, I spent some time this November,
designing this year's Chalkdust puzzle Christmas card
(with some help from TD).
![](javascript:showlimage("card2024.jpg"))
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...
| 1. | What is the largest number you can make by using the digits 1 to 4 to make two 2-digit numbers, then mutiplying the two numbers together? | Answer |
| 2. | What is the largest number you can make by using the digits 0 to 9 to make a 2-digit number and a 8-digit number, then mutiplying the two numbers together? | Answer |
| 3. | The expansion of \((2x+3)^2\) is \(4x^2+12x+9\). The sum of the coefficients of \(4x^2+12x+9\) is 25. What is the sum of the coefficients of the expansion of \((30x+5)^2\)? | Answer |
| 4. | What is the sum of the coefficients of the expansion of \((2x+1)^{11}\)? | Answer |
| 5. | What is the geometric mean of all the factors of 41306329? | Answer |
| 6. | What is the largest number for which the geometric mean of all its factors is 92? | Answer |
| 7. | What is the sum of all the factors of \(7^4\)? | Answer |
| 8. | How many numbers between 1 and 28988500000 have an odd number of factors? | Answer |
| 9. | Eve found the total of the 365 consecutive integers starting at 500 and the total of the next 365 consecutive integers, then subtracted the smaller total from the larger total. What was her result? | Answer |
| 10. | Eve found the total of the \(n\) consecutive integers starting at a number and the total of the next \(n\) consecutive integers, then subtracted the smaller total from the larger total. Her result was 22344529. What is the largest possible value of \(n\) that she could have used? | Answer |
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Matt, great card this year! Problems 1 and 2 are slightly ambiguous though in that you did not specify that each digit could only be used once.
I initially thought the answers were simply 44×44 = 1936 and 99×99999999 = 9899999901, respectively ????×1 ×1
I initially thought the answers were simply 44×44 = 1936 and 99×99999999 = 9899999901, respectively ????
Dan Whitman
×1 ×1 I find that I can enter seven correct answers without issue. however, an eighth answer causes the entire tree to vanish.
I'm using Firefox on Windows 11.
I'm using Firefox on Windows 11.
hakon
@HJ: I can't reproduce that error on Firefox or Chrome on Ubuntu - although I did notice I'd left some debug outputting on, which I've now removed. Perhaps that was causing the issue.
If anyone else hits this issue, please let me know.
If anyone else hits this issue, please let me know.
Matthew
On my machine (Mac, using either Firefox or Chrome, including private mode so no plugins) the puzzle disappears when I complete the answers for 1, 3 and 9. I'm presuming my answers are correct -- the pattern they create is pretty clear and looks reasonable.
HJ
I fond this card quite amusing. If I were clever enogh, I cold solve more of the problems! - Cheers from the USA
mitch
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2024-11-21
The mscroggs.co.uk Advent Calendar is back for its tenth year!
Behind each door, there will be a puzzle with a three digit solution. The solution to each day's puzzle forms part of a logic puzzle:
It's nearly Christmas and something terrible has happened: there's been a major malfunction in multiple machines in Santa's toy factory, and
not enough presents have been made. Santa has a backup warehouse full of wrapped presents that can be used in the case of severe emergency, but the warehouse is locked.
You need to help Santa work out the code to unlock the warehouse so that he can deliver the presents before Christmas is ruined for everyone.
The information needed to work out the code to the warehouse is known by Santa and his three most trusted elves: Santa is remembering a three-digit number,
and each elf is remembering a one-digit and a three-digit number. If Santa and the elves all agree that the emergency warehouse should be opened, they can work out the code for the door as follows:
- Santa tells his three-digit number to the first elf.
- The first elf subtracts her three-digit number then multiplies by her one-digit number. She tells her result to the second elf.
- The second elf subtracts his three-digit number then multiplies by his one-digit number. He tells his result to the third elf.
- The third elf subtracts their three-digit number then multiplies by their one-digit number. Their result is a five-digit number that is the code to unlock the warehouse.
But this year, there is a complication: the three elves are on a diplomatic mission to Mars to visit Martian Santa and cannot be contacted, so you need to piece together their
numbers from the clues they have left behind.
Behind each day (except Christmas Day), there is a puzzle with a three-digit answer. Each of these answers forms part of a clue about Santa's and the elves' numbers.
You must use these clues to work out the code for the warehouse.
You can use this page to try opening the door. If you enter an incorrect code three times, the door mechanism locks until the following day.
Ten randomly selected people who solve all the puzzles, open the warehouse, and fill in the entry form behind the door on the 25th will win prizes!
![](javascript:showlimage("prizes2018.jpg"))
The prizes will include an mscroggs.co.uk Advent 2024 T-shirt. If you'd like one of the T-shirts from a previous Advent, they are available to order at merch.mscroggs.co.uk.
The winners will be randomly chosen from all those who submit the entry form before the end of 2024. Each day's puzzle (and the entry form on Christmas Day) will be available from 5:00am GMT. But as the winners will be selected randomly,
there's no need to get up at 5am on Christmas Day to enter!
As you solve the puzzles, your answers will be stored. To share your stored answers between multiple devices, enter your email address below the calendar and you will be emailed a magic link to visit on your other devices.
To win a prize, you must submit your entry before the end of 2024. Only one entry will be accepted per person. If you have any questions, ask them in the comments below,
on Bluesky,
or on Mastodon.
If you'd like to chat with other solvers, we'll be discussing the Advent Calendar in the #scroggs-advent-calendar channel in the Finite Group Discord: you
can join the Discord by following the link in this post on Patreon (you'll need to become a free member
on Patreon to unlock the post).
So once December is here, get solving! Good luck and have a very merry Christmas!
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I am so appreciative that you continue to make these puzzles every year. I think I started doing them in 2017 and always enjoy them. Thank you!×1
Jessica
×1 Thanks Scroggs - first time I've done this and very much enjoyed the days and also the meta-puzzling. Brilliant!!. If you run in future years I have one request for a tiny tweak - I find the numbers on the advent calendar for the days very small for my ageing eye-sight - any chance of a bigger font? And last suggestion - when it gets to the end, provide a link to your "buy me a cup of tea" page as this is more than worth a few £s :-). Thanks again :-)
×1
Justin
×1 @Seth Cohen: Thanks Seth, I solved it by looking at the 5×5 picture and thinking harder. Thanks to Matthew for another enjoyable set of puzzles. I look forward to reading the proper solution.
Reza
Had a great time doing this puzzles again this year! 23 was particularly fun :) Thanks for taking the time to make this!
Bill V.
I enjoyed working the advent puzzles. Thank you for providing such fun entertainment and math challenges! Attempted most without any programming help but some begged for a programming solution. Refreshed some former Python skills to happily solve a few puzzles. Looking forward to math solutions. Merry Christmas and Happy Holidays!×1
Tony
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2024-02-20
Back in November, I wrote about making 2n-page zines.
Thanks to some conversations I had at Big MathsJam
in later November, I've been able to work out how many 128-page zines there are: 315434.
The insight
At Big MathsJam, Colin Beveridge pointed out
something he'd noticed about the possible zines: when drawing the line connecting the pages
in order, there were some line segments that were always included. For example, here are all of the possible
64-page zines:
![](javascript:showlimage("64zines.png"))
Every single one of these includes these line segments:
![](javascript:showlimage("64zines-always.png"))
Colin conjectured that for a zine of any size, a pattern like this of alternative horizontal segments
must always be included. He was close to justifying this, and since MathsJam I've been able to fill
in the full justificication.
The justificiation
First, consider the left-most column of pages. They must be connected like this:
![](javascript:showlimage("64zines-leftcol.png"))
If they were connected in any other way, there would be two vertical connections in a row,
which would create a page that is impossible to open (as every other connection must be a horizontal
that ends up in the spine). Additionally, the horizontal lines in this diagram must all be in the
spine (as otherwise we again get pages that cannot be opened).
Next, consider a horizontal line that's in the spine (shown in red below), and we can look
at all the possible ways to draw the line through the highlighted page, paying particular
attention to the dashed blue line:
![](javascript:showlimage("64zines-highlighted.png"))
The six possible ways in which the line could travel through the highlighted page are:
![](javascript:showlimage("64zines-options.png"))
The three options in the top row do not give a valid zine: the leftmost diagram has two vertical
connections in a row (leading to pages that do not open). The other two diagrams in the top row
have the horizontal line that we know is in the spine, followed by a horizontal line not in the
spine, then a vertial line: this vertical line should be in the spine, but as it is vertical
it cannot be (without making a page that doesn't open).
In each of the diagrams in the bottom row, the connection shown in dashed blue
is included and must be in the spine: in the leftmost diagram, the horizontal line that we know is in the spine
is followed by a horizontal not in the spine, then the horizinal in the dashed blue position
that must therefore be in the spine. The othe other two diagrams in the bottom row,
the dashed blue position is connected to a vertical line: this means that the dashed blue connection
must be in the spine (as otherwise the vertical would cause a page that doesn't open).
Overall, we've now shown that the leftmost column of lines must always be included and
must all be in the spine; and for each horizontal line in the spine, the line to the right of it
after a single gap must also be included and in the spine. From this, it follows that all the horizontal
lines in Colin's pattern must always be included.
Calculating the number of 128-page zines
Now that I knew that all these horizonal lines are always included, I was able to update
the code I was using to find all the possible zines
to use this. After a few hours, it had found all 315434 possibilites. I was very happy to get this
total, as it was the same as the number that
Luna (another attendee of Big MathsJam) had calculated but wasn't certain was correct.
The sequence of the number of 2n-page zines,
including the newly calculated number,
is now published on the OEIS.
I think calculating number of 256-page zines is still beyond my code though...
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2024-01-07
Welcome to 2024 everyone! Now that the Advent calendar has disappeared, it's time to reveal the answers and announce the winners.
But first, some good news: with your help, the machine was fixed in time for Santa to deliver presents and Christmas was saved!
Now that the competition is over, the questions and all the answers can be found here.
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.
Highlights
My first highlight is the puzzle from 4 December. I like this puzzle, because at first it looks really difficult, and the size of the factorial involved is impossibly large,
but the way of solving it that I used essentially just ignores the factorial leading to a much easier question.
4 December
If \(n\) is 1, 2, 4, or 6 then \((n!-3)/(n-3)\) is an integer. The largest of these numbers is 6.
What is the largest possible value of \(n\) for which \((n!-123)/(n-123)\) is an integer?
My next pair of highlights are the puzzles from 6 and 7 December. I always enjoy a surprise appearance of the Fibonacci sequence, and a double enjoyed a
double appearance in two contexts that at first look completely different.
6 December
There are 5 ways to tile a 4×2 rectangle with 2×1 pieces:
![](javascript:showlimage("advent2023-6dec.png"))
How many ways are there to tile a 12×2 rectangle with 2×1 pieces?
7 December
There are 8 sets (including the empty set) that contain numbers from 1 to 4 that don't include any consecutive integers:
How many sets (including the empty set) are there that contain numbers from 1 to 14 that don't include any consecutive integers?
My next highlight is the puzzle from 13 December. I love a good crossnumber, and had a lot of fun making this small one up. (If you enjoyed this one, you should check out the
crossnumbers I write for Chalkdust.)
13 December
Today's number is given in this crossnumber. No number in the completed grid starts with 0.
|
|
| |||||||||||||||||||||||||||||||||
My final highlight is the puzzle from 22 December. I enjoy that you can use one of the circle theorems to solve this, despite there being no circles directly involved in the question.
22 December
There are 4 ways to pick three vertices of a regular quadrilateral so that they form a right-angled triangle:
![](javascript:showlimage("advent2023-22dec.png"))
In another regular polygon with \(n\) sides, there are 14620 ways to pick three vertices so that they form a right-angled triangle. What is \(n\)?
Hardest and easiest puzzles
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.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 |
| Day | |||||||||||||||||||||||
It looks like the hardest puzzles were on
23 and
12 December;
and the easiest puzzles were on
1,
3,
5, and
11 December.
Fixing the machine
To finish the Advent calendar, you were tasked with fixing the machine. The answers to all the puzzles were required to
be certain of which combination of parts were needed to fix the machine, but it was possible to reduce the number of options
to a small number and get lucky when trying these options. This graph shows how many people fixed the machine on each day:
| 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
| Day | |||||||||||||||
The winners
And finally (and maybe most importantly), on to the winners: 180 people managed to fix the machine. That's slightly fewer than last year:
| 2015 | 2016 | 2017 | 2018 | 2019 | 2020 | 2021 | 2022 | 2023 |
| Year | ||||||||
From the correct answers, the following 10 winners were selected:
- Matt Thomson
- Matthieu
- Steve Paget
- Millie
- Eleanor
- Alex Bolton
- Brennan Dolson
- UsrBinPRL
- Daniel Low
- Erik
Congratulations! Your prizes will be on their way shortly.
The prizes this year include 2023 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, Adam NH, Aidan Dodgson, AirWrek, Alan Buck, Alejandro Villarreal, Alek2ander, Alex, Alex Hartz, Allan Taylor, Andrew Roy, Andrew Thomson, Andrew Turner, Andy Ennaco, Ashley Jarvis, Austin Antoniou, Becky Russell, Ben, Ben Boxall, Ben Reiniger, Ben Tozer, Ben Weiss, Bill Russ, Bill Varcho, Blake, Bogdan, Brian Wellington, Carl Westerlund, Carmen, Carnes Family, Cathy Hooper, Chris Eagle, Chris Hellings, Colin Brockley, Connors of York, Corbin Groothuis, Dan Colestock, Dan May, Dan Rubery, Dan Swenson, Dan Whitman, Daphne, David and Ivy Walbert, David Ault, David Berardo, David Fox, David Kendel, David Mitchell, Deborah Tayler, Diane, Donald Anderson, Duncan S, Dylan Madisetti, Ean, Elise Raphael, Emelie, Emily Troyer, Emma, Eric, Eric Kolbusz, Ewan, Frank Kasell, Fred Verheul, Gabriella Pinter, Gareth McCaughan, Gary M, Gary M. Gerken, George Witty, Gert-Jan, Grant Mullins, Gregory Wheeler, Guillermo Heras Prieto, Heerpal Sahota, Helen, Herschel, Iris Lasthofer, Ivan Molotkov, Jack, Jack H, Jacob Y, James Chapman, Jan Z, Jay N, Jean-Sébastien Turcotte, Jen Sparks, Jenny Forsythe, Jessica Marsh, Jim Ashworth, Joe Gage, Johan, Jon Palin, Jonathan Chaffer, Jonathan Thiele, Jorge del Castillo Tierz, K Brooks, Kai, Karen Climis, Kevin Docherty, Kevin Fray, Kirsty Fish, Kristen Koenigs, lacop, Lazar Ilic, Lewis Dyer, Lisa Stambaugh, Lise Andreasen, Lizzie McLean, Louis, Magnus Eklund, Marco van der Park, Mark Fisher, Mark Stambaugh, Martijn O., Martin Harris, Martin Holtham, Mary Cave, Matthew Schulz, Max, Merrilyn, Mihai Zsisku, Mike Hands, Miles Lunger, Mr J Winfield, Nadine Chaurand, Naomi Bowler, Nathan Whiteoak, Nick C, Nick Keith, Niji Ranger, Pamela Docherty, Pierce R, Qaysed, Rashi, Ray Arndorfer, rea, Reuben Cheung, Riccardo Lani, Richard O, Rob Reynolds, Robby Brady, Roger Lipsett, Roni, Rosie Paterson, RunOnFoot, Ruth Franklin, Ryan Wise, Sage Robinson, Sam Dreilinger, Sarah, Scott, Sean Henderson, Seth Cohen, Shivanshi, Shreevatsa, Stephen Cappella, Steve Blay, TAS, Tehnuka, The Johnston Family, Tina, Tony Mann, Trent Marsh, tripleboleo, Valentin VĂLCIU, Vinny R, William Huang, Yasha, and Yuliya Nesterova,
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.
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In your solution for the 12th, I think there's still a little work to do: to check that the answer is the smallest integer that works. For that, because 241 is prime, you only have a handful of values to check.×4 ×4 ×4 ×4 ×4
Ben Reiniger
×4 ×4 ×4 ×4 ×4 (you've left the "drones" in at the beginning of the Winners section)×2 ×2 ×3 ×1 ×2
Ben Reiniger
×2 ×2 ×3 ×1 ×2 On the 6th and 7th, there's also a direct bijection: in the tiling, horizontal tiles must occur in aligned pairs (else they split left/right into odd number of 1x1 blocks). Encode a tiling with the set of horizontal locations of the left ends of the horizontal-tile-pairs.×2 ×2 ×2 ×2 ×2
Ben Reiniger
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