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 2018-01-05 

Origins of World War I

In 1969, Sid Sackson published his magnum opus: A Gamut of Games, a collection of 38 games that can all be played with pen and paper or a pack of cards.
One of the best games I've tried so far from the book is James Dunnigan's Origins of World War I. The original version from the book gets you to start by drawing a large table to play on, and during play requires a fair bit of flicking backwards and forwards to check the rules. To make playing easier, I made this handy pdf that contains all the information you need to play the game.

The Rules

Starting

Origins of World War I is a game for five players (although it can be played with 3 or 4 people; details at the end). To play you will need a printed copy of the pdf, a pen or pencil, and a 6-sided dice.
Each player picks one of the five nations along the top of the board: Britain, France, Germany, Russia, or Austria–Hungary. Once you've picked your countries, you are ready to begin.

Taking a Turn

The countries take turns in the order Britain, France, Germany, Russia, or Austria–Hungary: the same order the countries are written across the top and down the side of the board. A player's turn involves two things: (1) adding "political factors"; (2) carrying out a "diplomatic attack".
First the player adds political factors (PFs). On their turn, Britain adds 14 PFs, France adds 12 PFs, Germany adds 16 PFs, Russia adds 10 PFs, and Austria–Hungary adds 10 PFs. These numbers are shown under the names of the countries on the left hand side of the board. A player can add at most 5 PFs to each country per turn, although they may add as many PFs as they like to their own country. The number of PFs a player adds to each country should be written in the boxes in the players column.
For example, Britain may choose to add 5 PFs in Italy, 2 in the Far East and 12 in Britain. This would be added to the board by writing the relevant numbers in the Italy, Far East and Britain rows of the Britain column.
After adding PFs, a player may choose to carry out a diplomatic attack. If so the player chooses one of the other four players to attack, and a country in which this attack takes place. Both players must have some PFs in the country where the attack takes place. The dice is rolled. The outcome of the attack depends on how much the attacker outnumbers the defender and the value rolled: these are shown to the right of the board. The three possible outcomes are: Attacker Eliminated (AE), which causes the attacker's PFs in this country to be reduced to 0; Exchange (EX), which causes both players' PFs in this country to be reduced by the same amount so that one player is left with 0; and Defender Eliminated (DE), wich causes the defender's PFs in this country to be reduced to 0.
For example, Britain may choose to attack Germany in Africa. If Britain and Germany have 10 and 4 PFs in Africa (respectively), then Britain outnumbers Germany 2 to 1. The dice is rolled. If a 1 is rolled, the attacker (Britain) is eliminated, leaving Britain on 0 and Germany on 4. If one of 2-5 is rolled, the players exchange, leaving Britain on 6 and Germany on 0. If a 6 is rolled, the defender (Germany) is eliminated, leaving Britain on 10 and Germany on 0.
The game ends after each play has played 10 turns. The number of turns may be kept track of by crossing out a number in the Turn Counter to the right of the board after each round of 5 turns.

Scoring

If a player has 10 or more PFs in a country, then they have Treaty Rights (TR) with that country. Each player scores point by achieving TR with other countries. TR are not symmetric: if Russia has TR with Germany, then this does not mean that Germany automatically has TR with Russia.
The number of points scored by a player for obtaining TR with other countries are printed in the boxes on the board. The numbers in brackets are only scored if the TR are exclusive: ie if no other country also has TR with that country. Additionally, points are awarded to Britain, France and Germany if the objectives in the boxes at the foot of their columns are satisfied.
For example, Britain scores 3 points if they have TR with Italy, 1 point if they have TR with Greece, 2 points if they have TR with Turkey. and 4 points if they have exclusive TR with the Far East. Britain also scores 10 points if no other nation has more than 12 points.

Alliances

During the game, players are encouraged to make deals with other players: for example, Britain may agree to not add PFs in Serbia if Russia agrees to carry diplomatic attacks against Germany in Bulgaria. Deals can of course be broken by either player later in the game.
Two players may also enter into a more formal alliance, leading to their two nations working together for the rest of the game. These alliances may not be broken. If two players are allied, then at the end of the game, their scores are added: if this total is higher than the scores of the other three players combined, then the allies win; if not, then the highest score among the other three wins.
During a game, it is possible for two different alliances to form (these must be between two different pairs of nations: a country cannot form two alliances, and three countries cannot form a three-way alliance). In this case, a pair of allies wins if their combined score is larger than the combined score of the other three players. If neither pair of allies scores this high, the unallied player wins.

Playing with 3 or 4 Players

Alliances can be used to play Origins of World War I with fewer than 5 players. To play with four players, an alliance can be formed at the start of the game, with one player playing both nations in the alliance. To play with three players, the game can be started with two alliances already in place.

If you've ready this far, then you're now fully prepared to play Origins of World War I, so print the pdf, invite 4 friends over, and have a game...

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 2018-01-02 

Christmas (2017) Is Over

It's 2018, and the advent calendar has disappeared, so it's time to reveal the answers and annouce the winners. But first, some good news: with your help, Santa was able to work out which present each child wanted, and get their presents to them just in time:
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.

4 December

Pick a three digit number whose digits are all different.
Sort the digits into ascending and descending order to form two new numbers. Find the difference between these numbers.
Repeat this process until the number stops changing. The final result is today's number.
This puzzle revealed the surprising fact that repeatedly sorting the digits of a three digit number into ascending and descending order then finding the difference will always give the same answer (as long as the digits of the starting number are all different). This process is known as the Kaprekar mapping.
If four digit starting numbers are chosen, then all starting numbers that do not have three equal digits will eventually lead to 6174. It's not as simple for five digit numbers, but I'll leave you to investigate this...

11 December

Two more than today's number is the reverse of two times today's number.
Ruben pointed something interesting out to me about this question: if you remove the constraint that the answer must be a three digit number, then you see that the numbers 47, 497, 4997, 49997, and in fact any number of the form 49...97 will have this property.

20 December

What is the largest number that cannot be written in the form \(10a+27b\), where \(a\) and \(b\) are nonnegative integers (ie \(a\) and \(b\) can be 0, 1, 2, 3, ...)?

Show Answer & Extension

If you didn't manage to solve this one, I recommend trying replacing the 10 and 27 with smaller numbers (eg 3 and 4) and solving the easier puzzle you get first, then trying to generalise the problem. You can find my write up of this solution here.
Pedro Freitas (@pj_freitas) sent me a different way to approach this problem (related to solving the same question with different numbers on this year's Christmas card). To see his method, click "Show Answer & Extension" in the puzzle box above.

24 December

Today's number is the smallest number with exactly 28 factors (including 1 and the number itself as factors).

Show Answer

I really like the method I used to solve this one. To see it, click "Show Answer" above.
Solving all 24 puzzles lead to the following final logic puzzle:

Advent 2017 Logic Puzzle

2016's Advent calendar ended with a logic puzzle: It's nearly Christmas and something terrible has happened: Santa and his two elves have been cursed! The curse has led Santa to forget which present three children—Alex, Ben and Carol—want and where they live.
The elves can still remember everything about Alex, Ben and Carol, but the curse is causing them to lie. One of the elves will lie on even numbered days and tell the truth on odd numbered days; the other elf will lie on odd numbered days and tell the truth on even numbered days. As is common in elf culture, each elf wears the same coloured clothes every day.
Each child lives in a different place and wants a different present. (But a present may be equal to a home.) The homes and presents are each represented by a number from 1 to 9.
Here are the clues:
21
White shirt says: "Yesterday's elf lied: Carol wants 4, 9 or 6."
10
Orange hat says: "249 is my favourite number."
5
Red shoes says: "Alex lives at 1, 9 or 6."
16
Blue shoes says: "I'm the same elf as yesterday. Ben wants 5, 7 or 0."
23
Red shoes says: "Carol wants a factor of 120. I am yesterday's elf."
4
Blue shoes says: "495 is my favourite number."
15
Blue shoes says: "Carol lives at 9, 6 or 8."
22
Purple trousers says: "Carol wants a factor of 294."
11
White shirt says: "497 is my favourite number."
6
Pink shirt says: "Ben does not live at the last digit of 106."
9
Blue shoes says: "Ben lives at 5, 1 or 2."
20
Orange hat says: "Carol wants the first digit of 233."
1
Red shoes says: "Alex wants 1, 2 or 3."
24
Green hat says: "The product of the six final presents and homes is 960."
17
Grey trousers says: "Alex wants the first digit of 194."
14
Pink shirt says: "One child lives at the first digit of 819."
3
White shirt says: "Alex lives at 2, 1 or 6."
18
Green hat says: "Ben wants 1, 5 or 4."
7
Green hat says: "Ben lives at 3, 4 or 3."
12
Grey trousers says: "Alex lives at 3, 1 or 5."
19
Purple trousers says: "Carol lives at 2, 6 or 8."
8
Red shoes says: "The digits of 529 are the toys the children want."
13
Green hat says: "One child lives at the first digit of 755."
2
Red shoes says: "Alex wants 1, 4 or 2."

Show Answer

Together the clues reveal what each elf was wearing:
Drawn by Alison Clarke
Drawn by Adam Townsend
and allow you to work out where each child lives and what they wanted. Thanks Adam and Alison for drawing the elves for me.
I had a lot of fun finding place names with numbers in them to use as answers in the final puzzle. For the presents, I used the items from The 12 Days of Christmas:
#LocationPresent
1Maidstone, Kenta partridge
2Burcot, Worcestershireturtle doves
3Three Holes, NorfolkFrench hens
4Balfour, Orkneycalling birds
5Fivehead, Somersetgold rings
6Sixpenny Handley, Dorsetgeese
7Sevenhampton, Glosswans
8Leighton Buzzard, Bedsmaids
9Nine Elms, Wiltshireladies
I also snuck a small Easter egg into the calendar: the doors were arranged in a knight's tour, as shown below.
And finally (and maybe most importantly), on to the winners: 84 people submitted answers to the final logic puzzle. Their (very) approximate locations are shown on this map:
From the correct answers, the following 10 winners were selected:
1M Oostrom
2Rosie Paterson
3Jonathan Winfield
4Lewis Dyer
5Merrilyn
6Sam Hartburn
7Hannah Charman
8David
9Thomas Smith
10Jessica Marsh
Congratulations! Your prizes will be on their way shortly. Additionally, well done to Alan Buck, Alessandra Zhang, Alex Burlton, Alex Hartz, Alex Lam, Alexander, Alexander Bolton, Alexandra Seceleanu, Arturo, Brennan Dolson, Carmen Günther, Connie, Dan Whitman, David Fox, David Kendel, Edison Yifeng He, Elijah Kuhn, Eva, Evan Louis Robinson, Felix Breton, Fred Verheul, Henry Hung, Joakim Cronvall, Joe Gage, Jon Palin, Kai Lam, Keith Sutherland, Kelsey, Kenson Li, Koo Zhengqun, Kristen Koenigs, Lance Nathan, Louis de Mendonca, Mark Stambaugh, Martin Harris, Martine Vijn Nome, Matt Hutton, Matthew Schulz, Max Nilsson, Michael DeLyser, Michael Smith, Michael Ye, Mihai Zsisku, Mike Walters, Mikko, Naomi Bowler, Pattanun Wattana, Pietro Alessandro Murru, Raj, Rick, Roni, Ross Milne, Ruben, Ryan Howerter, Samantha Duong, Sarah Brook, Shivanshi, Steve Paget, Steven Peplow, Steven Spence, Tony Mann, Valentin Vălciu, Virgile Andreani, and Yasha Asley, who all also submitted the correct answer but were too unlucky to win prizes this time.
See you all next December, when the advent calendar will return.

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Comments

Comments in green were written by me. Comments in blue were not written by me.
 2018-01-02 
Thanks! The advent calendar was great fun to take part in - and winning something in the process is the cherry on top. The riddles themselves were interesting and varied, they fitted well together in the overall puzzle, and I learned some interesting new bits of maths in the process. And now, to try my hand at the other advent calendars...

I particularly liked the riddle on the 5th of December (with walking 13 units) - it was quite tricky at first, but then I solved it by seeing that you can't end up on a square an even distance away from the centre, so the possible areas are in "circles" from the center with odd side lengths . It was quite reminiscent of showing you can't cover a chessboard with dominoes when two opposite corners are removed .
Lewis
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 2017-12-18 

Christmas Card 2017

Just like last year, I spent some time in November this year designing a puzzle Christmas card for Chalkdust.
The card looks boring at first glance, but contains 10 puzzles. Converting the answers to base 3, writing them in the boxes on the front, then colouring the 1s black and 2s orange will reveal a Christmassy picture.
If you want to try the card yourself, you can download this pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will be automatically converted to base 3 and coloured...
#Answer (base 10)Answer (base 3)
10000000
20000000
30000000
40000000
50000000
60000000
70000000
80000000
90000000
100000000
  1. In a book with 116 pages, what do the page numbers of the middle two pages add up to?
  2. What is the largest number that cannot be written in the form \(14n+29m\), where \(n\) and \(m\) are non-negative integers?
  3. How many factors does the number \(2^6\times3^{12}\times5^2\) have?
  4. How many squares (of any size) are there in a \(15\times14\) grid of squares?
  5. You take a number and make a second number by removing the units digit. The sum of these two numbers is 1103. What was your first number?
  6. What is the only three-digit number that is equal to a square number multiplied by the reverse of the same square number? (The reverse cannot start with 0.)
  7. What is the largest three-digit number that is equal to a number multiplied by the reverse of the same number? (The reverse cannot start with 0.)
  8. What is the mean of the answers to questions 6, 7 and 8?
  9. How many numbers are there between 0 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6?
  10. What is the lowest common multiple of 52 and 1066?

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Comments

Comments in green were written by me. Comments in blue were not written by me.
 2018-01-01 
C - look up something called Frobenius numbers. This problem's equivalent to finding the Frobenius number for 14 and 29.
Lewis
 2017-12-28 
I can solve #2 with code, but is there a tidy maths way to solve it directly?
C
 2017-12-23 
My efforts were flightless.
NHH
 2017-12-20 
What a fun diversion! I have to admit I can't remember how to solve #2 - I finally had to write some code. (So primitive.)
Heather
 2017-12-19 
Thank you for the clarification!

I don't want to spoil the answer for anyone, but my Christmassy picture looks like a well dressed bird.
Claudio
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 2017-11-28 

Christmas (2017) is Coming!

This year, the front page of mscroggs.co.uk will once again feature an advent calendar, just like last year and the year before. 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: Santa and his two elves have been cursed! The curse has led Santa to forget which present three children—Alex, Ben and Carol—want and where they live.
The elves can still remember everything about Alex, Ben and Carol, but the curse is causing them to lie. One of the elves will lie on even numbered days and tell the truth on odd numbered days; the other elf will lie on odd numbered days and tell the truth on even numbered days. As is common in elf culture, each elf wears the same coloured clothes every day.
Each child lives in a different place and wants a different present. (But a present may be equal to a home.) The homes and presents are each represented by a number from 1 to 9.
Santa has called on you to help him work out the details he has forgotten. Behind each day (except Christmas Day), there is a puzzle with a three-digit answer. Each of these answers forms part of a fact that one of the elves tells you. You must work out which combination of clothes each elf wears, which one lies on each day, then put all the clues together to work out which presents need delivering to Alex, Ben and Carol, and where to deliver them.
Ten randomly selected people who solve all the puzzles and submit their answers to the logic puzzle using the form behind the door on the 25th will win prizes! A selection of the prizes are shown below, and will be added to throughout December.
The ten winners will also will one of these winners' medals:
As you solve the puzzles, your answers will be stored. This year, there is a new feature allowing you to synchronise your answers between multiple computers: simply enter your email address below the calendar, and you will be emailed a magic link to visit on your other devices.
Behind the door on Christmas Day, there will be a form allowing you to submit your answers. The winner will be randomly chosen from all those who submit the correct answer before the end of 2017. 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!
To win a prize, you must submit your entry before the end of 2017. Only one entry will be accepted per person. If you have any questions, ask them in the comments below or on Twitter.
So once December is here, get solving! Good luck and have a very merry Christmas!

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Comments

Comments in green were written by me. Comments in blue were not written by me.
 2017-12-27 
Thanks, I've added a clarification to 22
Matthew
 2017-12-24 
Me again

Just for info (clarification?): I read question on 22nd as
22 is two times an odd number. Today's number is the mean of all the answers, on days (including today), that are two times an odd number."

Note my added commas. I was averaging the answers, not the dates. Certainly ambiguous as far as I am concerned.
Only fixed it by 'cheating'. Trying best guessses of averages until I got the correct one.
neal (@zbvif)
 2017-12-24 
Wow. Just discovered I meisread 15th Dec puzzle.

I can tell you that the number of combinations of n As and Bs which contain at at least one uninterrupted sequence of 3 As is 2^n - F3(n+3) where F3 is the fibonaccia variant adding 3 numbers (1,1,2,4,7,13,24 etc.).
Only took me about 8 hours (with some small help form OEIS for the 2 As problem)
Neal (@zbvif)
 2017-12-18 
Assume the pancake is 2D
Matthew
 2017-12-18 
With todays puzzle does the pancake have any thickness i.e can we slice the pancake into 2 circular pancakes each with half the thickness or are we to assume its 2D
Alex
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 2017-11-14 

MENACE at Manchester Science Festival

A few weeks ago, I took the copy of MENACE that I built to Manchester Science Festival, where it played around 300 games against the public while learning to play Noughts and Crosses. The group of us operating MENACE for the weekend included Matt Parker, who made two videos about it. Special thanks go to Matt, plus Katie Steckles, Alison Clarke, Andrew Taylor, Ashley Frankland, David Williams, Paul Taylor, Sam Headleand, Trent Burton, and Zoe Griffiths for helping to operate MENACE for the weekend.
As my original post about MENACE explains in more detail, MENACE is a machine built from 304 matchboxes that learns to play Noughts and Crosses. Each box displays a possible position that the machine can face and contains coloured beads that correspond to the moves it could make. At the end of each game, beads are added or removed depending on the outcome to teach MENACE to play better.

Saturday

On Saturday, MENACE was set up with 8 beads of each colour in the first move box; 3 of each colour in the second move boxes; 2 of each colour in third move boxes; and 1 of each colour in the fourth move boxes. I had only included one copy of moves that are the same due to symmetry.
The plot below shows the number of beads in MENACE's first box as the day progressed.

Sunday

Originally, we were planning to let MENACE learn over the course of both days, but it learned more quickly than we had expected on Saturday, so we reset is on Sunday, but set it up slightly differently. On Sunday, MENACE was set up with 4 beads of each colour in the first move box; 3 of each colour in the second move boxes; 2 of each colour in third move boxes; and 1 of each colour in the fourth move boxes. This time, we left all the beads in the boxes and didn't remove any due to symmetry.
The plot below shows the number of beads in MENACE's first box as the day progressed.

The Data

You can download the full set of data that we collected over the weekend here. This includes the first two moves and outcomes of all the games over the two days, plus the number of beads in each box at the end of each day. If you do something interesting (or non-interesting) with the data, let me know!

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Comments

Comments in green were written by me. Comments in blue were not written by me.
 2018-02-14 
On the JavaScript version, MENACE2 (a second version of MENACE which learns in the same way, to play against the original) keeps setting the 6th move as NaN, meaning it cannot function. Is there a fix for this?
Lambert
 2017-11-22 
what would happen if you loaded the boxes slightly differently. if you started with one bead corresponding to each move in each box. if the bead caused the machine to lose you remove only that bead. if the game draws you leave the bead in play if the bead causes a win you put an extra bead in each of the boxes that led to the win. if the box becomes empty you remove the bead that lead to that result from the box before
Ian
 2017-11-17 
Hi, I was playing with MENACE, and after a while the page redrew with a Dragon Curves design over the top. MENACE was still working alright but it was difficult to see what I was doing due to the overlay. I did a screen capture of it if you want to see it.
Russ
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© Matthew Scroggs 2018