Maths at: Tidying & toilets

It is a truth universally acknowledged, that noone likes tidying up. So, if you want a reason to not bother then give us a listen.

Once again your regulars, Ben, Liz and Thomas, seek answers to such questions as:

• Why are people in London Freezing?
• Do butterflies fart? 
• How many toilets do you need to make everyone happy?
  

 All questions and no answers in today's Maths at:

 

Further reading: 

• House always messy? Just blame it on the second law of thermodynamics
• No more queuing at the ladies' room
• Do butterflies fart?
  

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Travelling Salesman

The government didn't want you to hear this podcast. They tried to pay us off! Just remember as you listen to it: we are not responsible for any repercussions if you decide to watch Travelling Salesman.

Today's discussion points:

• How many tickets do you have to buy to win the lottery?
• Squashed research!
• How many actors can you have playing the same character?

All this and more in this weeks Maths at

In you're interested in watching Travelling Salesman you don't need to buy it off Amazon, you can just watch it on YouTube! They can't even give it away!

Further reading:

• Should you feel guilty about watching the film on YouTube, you can give the creators your money, through their official website.
• An easy description of the P vs NP problem.
• Sorting algorithms through dance.
• Color visualization of sorting algorithms.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Math at the Movies: x+y

Well, this was a pain in the backside to edit. The film is so tawdry and dull that we kept getting lost on tangents. Fear not though faithful listener, Thomas has edited the two hours of guff down to a single hour of solid... bronze.

Today's discussion points include: 

• How should you flip a mattress?
• Does the culture you grow up in influence how you learn maths?
• BUMFIT!

From our mouths to your ears, enjoy!

 

If you want to watch x+y, you can follow the link below.

Further reading links:

• As per usual some artistic license was taken with this true story, this website provides the fact behind the fiction;
• Fancy testing yourself with real IMO problems?
• What is the difference between IMO problems and research mathematics?

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Valentine's day

Although we've finished our first series, we couldn't miss St Valentine's day.

What could be more romantic than scientifically analysing love and romance using mathematics?

Amongst other things we discuss:

• Why should you dump the first 37% of your partners?
• How many snogs away from Prince Charles are you?
• Are you a validator or an avoider?

Join your hosts as they show you their romantic side and be surprised that any of them actually have partners!

 

 
Further reading links:

• Ben's bunkum length of love formula;
• Drawing hearts graphically;
• Jim Murray's Maths of Marriage book or online lecture;
• Mobius strip earrings, for you significant other.

Subscribe via iTunes.
 
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley, @benmparker.

Maths at the Movies: Interstellar plus SPECIAL GUEST

This week we watch Interstellar.

We are very excited to be joined by the expertise of

Cat Harris

Whereas Thomas and Ben are just about good enough understand their own mathematical fields Cat has a wide range of disciplines under her belt spanning both physics and movie effects. We couldn't ask for a better guest!

If you're interested in watching Interstellar you can follow the Amazon below.

Further reading links:

• The hour long Interstellar documentary;
• The Interstellar book;
• All you could ever want to know about Sex in Space;
• Dissolving the dead with alkaline hydrolysis.

Subscribe via iTunes.
 
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley, @benmparker and @cat_harris_fx

Watch along with MATM: Interstellar and SPECIAL GUEST!

Next Friday we will be critiquing the maths and physics of Interstellar.

What's more, because this is the 10th episode (and because Ben and Thomas aren't physicists) we will be joined by a SPECIAL GUEST.

Not only is she a Cambridge physics graduate, but she now works in film digital effects. Who could ask for more?

Why not watch along with us? You can buy a digital version of the film through Amazon by clicking on the image below.

Maths at the Movies: The Oxford Murders

This week we discuss we discuss The Oxford Murders:

• Thomas laughs at the word bra;
• Liz wants a prime number named after her;
• Ben just wants his coffee bringing to him;
• And we all think about mathematically defining pasta shapes.

Yup, it's a case of a bad movie, with very little to talk about. At least it made for a fun recording!

If you're interested in watch The Oxford Murders, then you're weird, but you can buy the DVD from Amazon below.

Further reading links:

• Self descriptive numbers
• Online integer sequences database
• Fermat's last theorem
• Magic square and noughts and crosses

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Watch along with MATM: The Oxford Murders

Don't do it.

The Oxford Murders is terrible.

(Thomas is very sorry for choosing it).

However, should you want to waste 104 minutes of your life you can buy the DVD from Amazon by following the link below.

 

Such a waste of talented people!
Such a waste of mathematical logic!

We didn't read it... but the book might be better?

Maths at the Movies: Flatland

This week we take a walk in the lower dimensions as we talk about Flatland: The Movie.

• How much is your life worth?
• Could 2D animals exist?
• Did George Orwell rip this story off by adding a third dimension?

All of these questions and more are interrupted in our Flatland: The Movie podcast.

Join us in our search for the third dimension and beyond!

If you're interested in watching Flatland: The Movie  you can follow the Amazon link below.

Further reading links:

• galleries of beautiful fractal images;
• learn more about maths in the courtroom;
• the the company behind the film have made a number of educational resources and other films;
• why not read the book the film is based on?
• This links takes you to one of Martin Gardener's review of Planiverse, Alexander Dewdney's fully thought out and scientifically accurate version of Flatland.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker. 

Maths at the Movies: pi

This week we watched pi.

Sadly there were no tigers, boats or Dev Patel in this movie.

Nope, this film was an avant garde, mood piece, seething with questioning the meaning of truth and its place in the lives of humans and the universe.

In other words pretentious.

However, none Darren Aronofsky's nonsense matters. What you should be listening for is Ben's rendition of "Euclid's people" a song of his own creation sung to the tune of "Common people" by pulp.

You've got to hear it to believe it.

If you're interested in watching pi you can follow the Amazon link below.

 

Further reading links:

• play go online;
• find your birthday in pi;
• more pi facts than you can shake a stick at;
• weird ways of calculating pi;
• known digits of pi over time.

Subscribe via iTunes.

Answer to The Imitation Game

As you'll hear in the next podcast Ben cocks up the explanation of his original questions from Maths at: the Movies, The Imitation Game:

I have two children. One of them is a boy and they were born on a Tuesday.

What is the probability that both children are boys?

This is a hard question, and Ben ****ed up the explanation when he tried to do it live. So, as penance, we made him sit down and explain it as a video.

Here's a simpler question written out much nicer:

I have two children. One of them is a boy.
What is the probability that both children are boys?

Now you may think the probability is 50%, but that is not so (note that we are assuming that boy and girl births are equally likely). The reason is because we have more information about the children.

Suppose we denote a boy by "b" and a girl by "g". Further, we capitalise the letter to denote the elder child. In this way we could have the following combinations of children:

Bb

Gb

Bg

Gg

However, we know we have at least one boy, so we can't have Gg. Out of the possibilities that are left, namely Bb, Gb and Bg, there is only one way to get two boys, the chance is 1/3! Counter-intuitive no?

Note that if we had posed the problem as I have two children and my eldest is a boy then (using the above argument) the probability of have a second boy is then 1/2.

Probability can be a tricksy animal. Even for a Cambridge educated lecturer!

Answer to The Man Who Knew Infinity

At the end of Maths at: the Movies, The Man Who Knew Infinity Thomas posed two teasers to you.

A simple one to start you off.
I buy a bottle and a cork for £1.10. The bottle costs £1 more than the cork.

How much does the cork cost?

A moments thought should show that the bottle costs £1.05, whilst the cork costs only 5p. If you go it right first time well done! The answer most people tend to give if they don't pause for a second is 10p.

Now, for the more difficult question:

I live on a street with more than one house. All the houses on this street are numbered consecutively, 1, 2, 3,..., etc. Amazingly, I live in the house such that if you add up all the house numbers below me and all the house numbers above me then they come to exactly the same answer.

What is the minimum number of houses on this street and what is my house number?

The smallest answer, excluding the one house case is 8 houses on the street and I live at number 6, thus, 1+2+3+4+5=15=7+8.

There are actually an infinite number of increasing solutions to this problem. Although the solution can be found using basic algebra and a knowledge of continued fractions the details can get a bit hairy. Thus, I direct the interested reader to the following two wonderful expositions on the matter:

http://www.johnderbyshire.com/Opinions/Diaries/Puzzles/2009-06.html
http://www.angelfire.com/ak/ashoksandhya/winners2.html#PUZZANS4

Answer to Donald Duck in Mathmagic Land

During Maths at: the Movies, Donald Duck in Mathmagic Land Ben presented the following conundrum.

Ben's greengrocer uses a balance scale, like the one seen above, and only has a 40kg weight. However, the greengrocer fortuitously broke the weight into four pieces of integer weight that will allow them to  measure out every integer of kilograms from 1kg to 40kg. What are the four weights?

Normally, for a problem like this, you'd think of the binary sequence 1, 2, 4, ..., because, as shown in the gold chain problem, you can construct any number using combinations of these numbers. However, with only four weights, we would have 1, 2, 4, 8, from which we could produce a maximum of 15, falling far short of the 40kg total.

The crux of the problem is that in binary we can only add or not add a weight. In this problem, because we are using a set of balance scales we have three possibilities:
(a) Not add the weight, denoted 0;
(b) Add the weight to the left side, denoted L;
(c) Add the weight to the right side, dented R.

Because we have three possibilities, instead of two, we might think about using the numbers based around powers of 3 (the trinary system), rather than those based around powers of 2 (the binary system). Thus, our weights would be 1, 3, 9, 27. Adding these together does indeed give 40kg, but how would we use them to weigh out 2kg?

Put the 1kg on the left and the 3kg on the right. This produces a deficit of 2kg in the left pan, so we add apples to the left pan until it balances and, voila, we know we have two kilograms of apples.
Thus, 2 is represented as LR00 in our system.

What about 5kg? Similar to the above put the 1 and  3kg weights in the left and the 9 in the right. This produces a deficit of 5kgs in the left pan. Thus, 5kg is represented by LLR0.

Using this ideas we can produce the following table

Weight to be measured	Trinary encoding	Effective calculation
1	R000	1=1
2	LR00	-1+3=2
3	0R00	3=3
4	RR00	1+3=4
5	LLR0	-1-3+9=5
6	0LR0	-3+9=6
7	RLR0	1-3+9=7
8	L0R0	-1+9=8
9	00R0	9=9
10	R0R0	1+9=10
11	LRR0	-1+3+9=11
12	0RR0	3+9=12
13	RRR0	1+3+9=13
14	LLLR	-1-3-9+27=14
15	0LLR	-3-9+27=15
16	RLLR	1-3-9+27=16
17	L0LR	-1-9+27=17
18	00LR	-9+27=18
19	R0LR	1-9+27=19
20	LRLR	-1+3-9+27=20
21	0RLR	3-9+27=21
22	RRLR	1+3-9+27=22
23	LL0R	-1-3+27=23
24	0L0R	-3+27=24
25	RL0R	1-3+27=25
26	L00R	-1+27=26
27	000R	27=27
28	R00R	1+27=28
29	LR0R	-1+3+27=29
30	0R0L	3+27=30
31	RR0R	1+3+27=31
32	LLRR	-1-3+9+27=32
33	0LRR	-3+9+27=33
34	RLRR	1-3+9+27
35	L0RR	-1+9+27=35
36	00RR	9+27=36
37	R0RR	1+9+27=37
38	LRRR	-1+3+9+27=38
39	0RRR	3+9+27=39
40	RRRR	1+3+9+29=40

Watch along with MATM: pi

Apple?

Cherry?

Raspberry?

Unfortunately, none of these.

This Friday we'll be releasing our pi podcast. Darren Aronofsky's surreal, disturbing, art house, pretentious, feature debut about the descent into insanity of the Jewish Mathematician Max Cohen.

One of the reasons Thomas started this podcast was to talk about this film and to ensure that everyone hated it as much as him.

We don't recommend it, but if you want to watch along with us then can buy a digital version of the film through Amazon by clicking on the image below.

 

Puzzle from The Imitation Game

At the end of Maths at: the Movies, The Imitation Game Ben  presented the following question

I have two children. One of them is a boy and they were born on a Tuesday.

What is the probability that both children are boys?

Now, unfortunately, Ben royally screws up the explanation of this answer in the next podcast. However, we are recording an appendix episode to put the world to rights. In the mean time though we can solve the slightly simpler puzzle:

I have two children. One of them is a boy.

What is the probability that both children are boys?

The answer will appear in the next podcast.

Puzzle from The Man Who Knew Infinity

At the end of Maths at: the Movies, The Man Who Knew Infinity Thomas posed two teasers to you.

A simple one to start you off.
I buy a bottle and a cork for £1.10. The bottle costs £1 more than the cork.

How much does the cork cost?

And a more difficult one for you to chew on that was apparently given to Ramanujan himself.

I live on a street with more than one house. All the houses on this street are numbered consecutively, 1, 2, 3,..., etc. Amazingly, I live in the house such that if you add up all the house numbers below me and all the house numbers above me then they come to exactly the same answer.

What is the minimum number of houses on this street and what is my house number?

Post your comments or answers below.

If you want to know the answer you can either wait until next week or listen to the next podcast, Maths at: the Movies, The Imitation Game.

Puzzle from Donald Duck in Mathmagic Land

During Maths at: the Movies, Donald Duck in Mathmagic Land Ben presented the following conundrum.

Until recently my greengrocer sold apples in multiples of 40kg. The apples are weighed out using an old set of balance scales and a 40kg weight (see the above picture). However, the greengrocer dropped their weight and it broke into four pieces weighing four different integer values. Just before the greengrocer threw the pieces away I stopped them and showed that the set of four smaller weights could be used to measure out every integer of kilograms from 1kg to 40kg.

What were the values of the four smaller weights?

Post your comments or answers below.

If you want to know the answer you can either wait until next week or listen to the next podcast, Maths at: the Movies, The Man Who Knew Infinity.

Maths at the Movies: The Imitation Game plus SPECIAL GUEST

In this episode we watch the movie The Imitation Game. 

Alongside your regular team of Thomas, Ben and Liz there was only one mathematician with the expertise who could take us through this movie with grace, wit and wisdom. And that mathematician wasn't available so we got 

Dr James Grimes 

instead.

Join us for episode five of Maths at: The Movies as we separate fact from fiction about the life of Alan Turing.

If you're interested in watching The Imitation Game you can follow the Amazon link below.

 

Further reading links:

• more than you would ever want to know about Alan Turing;
• Turing's biography by Andrew Hodges;
• more than you would ever want to know about James Grime;
• finding the truth behind the movies;
• comparing actors and their real-life counter parts.
• play with an online enigma machine;
• play with an online Turing machine.

Subscribe via iTunes.

Watch along with MATM: The Imitation Game and SPECIAL GUEST!

This Friday, we will be watching Banister Crumblebench starring in The Imitation Game. The film follows the life of monumental mathematician Alan Turing and his work on cracking the Enigma code during World War II.

The mathematical genius Alan Turing and marvellous actor Bumblesnuff Crimpysnitch.

Joining your regular team of Liz, Ben and Thomas will be a SPECIAL GUEST mathematician.

Who could it be?

• Is it Bodysnatch Cummerbund?
• Is it Buffalo Custardbath?
• Is it Bundleup Catchyoudeath?

Will Thomas ever get tired of mispronouncing Benedict Cumberbatch's name?

None of these answers and more will be provided on Friday.

Why not watch along with us? You can buy a digital version of the film through Amazon by clicking on the image below.

 

Friday Factoid: There are 451 unique ways to make up 50p.

During The Man who knew Infinity I mentioned an interesting problem:

How many ways were there of making coins add up to the value of 50p?

In the UK, we have coins of value 1,2,5,10,20,50,100, and 200. To get them to add up to 50, we can have three 10p pieces plus one 20p.  Or we could have 25 tuppences (2p pieces). Or many other combinations.

I got this problem via a phone call from a friend of mine who said that their grandchild had been sent home from school with a challenge. I was in the pub with two statistics professors, and we all scratched our heads.

This problem turned out to be quite fun. I thought the answer would be quite large, but I was surprised that there are 451 ways. No wonder running the economy is so difficult!

I eventually came up with a graph which manages to show how we generalise this to any monetary amount.

See how quickly the graph rises with the amount of money to be made up in change? This seems a very difficult problem for a small person, so I suspect the teacher set it as a way to keep a bright child quiet for a few hours!

(Edited to add US Coins: Look how much difference there is for UK and US coins, mostly because USA does not have a 2c coin.)

I looked for a sensible way to do it, but the only real way I can explain it is to come up with some logical way to write these out. I cannot see a clever mathematical trick, so I used a recursive algorithm, which means that we keep using the algorithm within itself until we get the answer.

Either you use one 50p piece (count 1 way)

or you don't use any 50p piece (Call this X).

Now we need X:

Either you use two 20p pieces (call this X1)

or you use one 20p piece (Call this X2)

or you use no 20p pieces (Call this X3)

So X=X1+X2+X3

Now we find X1- we have to find ways to make 10p with no coin bigger than a 10p.

Either we use one 10p piece (count 1)

or we use none (Call this Y1)...

All this algebra produces a complicated sum. Although you can reuse bits of it, it is laborious, and probably very hard to get right.

I got bored in the end, so I cheated. I wrote this short piece of code in MatLab which will take any amount of money, any collection of change, and print out all possible combinations of change to get the required sum. The code can be found below.

The function takes in three arguments, "amount", "possCoins" and "strIn".

• amount is the amount of money we are partitioning;
• possCoins is the set of coins we are partitioning the amount into. For example, in Britain we use coins of denomination of 1p, 2p, 5p, 10p, 20p, 50p, £1 and £2. In America, they have a 25c coin, rather than a 20c coin.
• strIn is a string which is the list of all the coin combinations we have already found if we are part way through the recursion.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function numWays = coins(amount,possCoins,strIn)
    if ((isempty(possCoins)) | (amount<=0))
            numWays=0;
    else
        if (amount==possCoins(1))
           strOut=[strIn,' ',num2str(possCoins(1))];
           disp(strOut)
           numWays=1+coins(amount,possCoins(2:length(possCoins)),strIn);
         else
           strOut=[strIn,' ',num2str(possCoins(1))];

           numWays=coins(amount-possCoins(1),possCoins,strOut)+...

           coins(amount,possCoins(2:length(possCoins)),strIn);
        end
    end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The above code does essentially the same process as described. Using it for the 50p case, we first define the possible coins we can use:

>> possCoins=[50,20,10,5,2,1]
possCoins =

    50    20    10     5     2     1

Then we define strIn to be empty (we start at the beginning so we haven't started the recursion) :

>> strIn=' '

Finally, we call the function above, which gives the following output, in a logical order:

>> coins(amount,possCoins,strIn)

  50
  20 20 10
  20 20 5 5
  20 20 5 2 2 1
  20 20 5 2 1 1 1
  20 20 5 1 1 1 1 1
  20 20 2 2 2 2 2
  20 20 2 2 2 2 1 1
  20 20 2 2 2 1 1 1 1
  20 20 2 2 1 1 1 1 1 1
  20 20 2 1 1 1 1 1 1 1 1
  20 20 1 1 1 1 1 1 1 1 1 1
  20 10 10 10
  20 10 10 5 5
  20 10 10 5 2 2 1
  20 10 10 5 2 1 1 1
  20 10 10 5 1 1 1 1 1
  20 10 10 2 2 2 2 2
  20 10 10 2 2 2 2 1 1
  20 10 10 2 2 2 1 1 1 1
  20 10 10 2 2 1 1 1 1 1 1
  20 10 10 2 1 1 1 1 1 1 1 1
  20 10 10 1 1 1 1 1 1 1 1 1 1
  20 10 5 5 5 5
  20 10 5 5 5 2 2 1
  20 10 5 5 5 2 1 1 1
  20 10 5 5 5 1 1 1 1 1
  20 10 5 5 2 2 2 2 2
  20 10 5 5 2 2 2 2 1 1
  20 10 5 5 2 2 2 1 1 1 1
  20 10 5 5 2 2 1 1 1 1 1 1
  20 10 5 5 2 1 1 1 1 1 1 1 1
  20 10 5 5 1 1 1 1 1 1 1 1 1 1
  20 10 5 2 2 2 2 2 2 2 1
  20 10 5 2 2 2 2 2 2 1 1 1
  20 10 5 2 2 2 2 2 1 1 1 1 1
  20 10 5 2 2 2 2 1 1 1 1 1 1 1
  20 10 5 2 2 2 1 1 1 1 1 1 1 1 1
  20 10 5 2 2 1 1 1 1 1 1 1 1 1 1 1
  20 10 5 2 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 10 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 10 2 2 2 2 2 2 2 2 2 2
  20 10 2 2 2 2 2 2 2 2 2 1 1
  20 10 2 2 2 2 2 2 2 2 1 1 1 1
  20 10 2 2 2 2 2 2 2 1 1 1 1 1 1
  20 10 2 2 2 2 2 2 1 1 1 1 1 1 1 1
  20 10 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1
  20 10 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1
  20 10 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 10 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 10 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 5 5 5 5 5
  20 5 5 5 5 5 2 2 1
  20 5 5 5 5 5 2 1 1 1
  20 5 5 5 5 5 1 1 1 1 1
  20 5 5 5 5 2 2 2 2 2
  20 5 5 5 5 2 2 2 2 1 1
  20 5 5 5 5 2 2 2 1 1 1 1
  20 5 5 5 5 2 2 1 1 1 1 1 1
  20 5 5 5 5 2 1 1 1 1 1 1 1 1
  20 5 5 5 5 1 1 1 1 1 1 1 1 1 1
  20 5 5 5 2 2 2 2 2 2 2 1
  20 5 5 5 2 2 2 2 2 2 1 1 1
  20 5 5 5 2 2 2 2 2 1 1 1 1 1
  20 5 5 5 2 2 2 2 1 1 1 1 1 1 1
  20 5 5 5 2 2 2 1 1 1 1 1 1 1 1 1
  20 5 5 5 2 2 1 1 1 1 1 1 1 1 1 1 1
  20 5 5 5 2 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 5 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 5 2 2 2 2 2 2 2 2 2 2
  20 5 5 2 2 2 2 2 2 2 2 2 1 1
  20 5 5 2 2 2 2 2 2 2 2 1 1 1 1
  20 5 5 2 2 2 2 2 2 2 1 1 1 1 1 1
  20 5 5 2 2 2 2 2 2 1 1 1 1 1 1 1 1
  20 5 5 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1
  20 5 5 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 5 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 5 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 5 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 2 2 2 2 2 2 2 2 2 2 2 2 1
  20 5 2 2 2 2 2 2 2 2 2 2 2 1 1 1
  20 5 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1
  20 5 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1
  20 5 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1
  20 5 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1
  20 5 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
  20 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1
  20 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1
  20 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1
  20 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1
  20 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1
  20 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1
  20 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  20 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
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  5 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  5 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  5 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

ans =

   451