Friday, 29 December 2017

Watch along with MATM: Flatland

It's been a long time coming, but we're finally back on track. Next Friday will be our 8th episode where we discuss and watch Flatland: The Movie.


Once again we have a star-studded cast of voice actors including Martin Sheen, Kristen Bell and Michael York.

You can buy a DVD copy from the company behind the film, or Amazon does stock imported US versions. However, they are quite expensive and will take a little while to arrive.

For those of you who do not want to seek an illegal online streaming copy (which we do not condone here at Maths At) there is always the possibility of reading the book by Edwin A. Abbott, which sticks pretty close to the movie, except for one or two plot points. The book also goes into much more detail about the Flatland universe.

Be there or be square.

Monday, 25 December 2017

Maths at: Christmas

It's the most wonderful time... of the year.

Merry Christmas and happy holidays to all of our listeners. 

Join us as we discuss (amongst other things):
  • our favourite Christmas films (not really mathematical);
  • the differences between a mathematicians office Christmas party and an actual Christmas party (slightly mathematical);
  • the number of presents your true is giving and how many legs they'll have (quite mathematical).
So fill your boots with Turkey and in your after dinner stupor enjoy our CHRISTMATHS! We're better than the Queen's speech.

GUEST INTRO AND OUTRO: Ma Woolley.


As festive as Ben can get!
Further reading links:
Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Friday, 15 December 2017

A very merry christmas to all our listeners!


A very merry Christmas  to all our listeners. We're hoping to record one more seasonal themed podcast before Santa comes, so let us know your Christmas mathematical questions in the comments below.

Christmas card with apologies to Hilbert, all major churches, and humour in general.

MATM Appendix: Boy born on a Tuesday

During Maths at the Movies: The Imitation Game Ben introduced the following puzzle:

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

What is the probability that both children are boys? 

Although counter-intuitive, it can be shown that the solution is 1/3.

He also posed a similar, but different question:

I have two children. One of them is a boy who was born on a Tuesday.

What is the probability that both children are boys?

Although very similar, the answer turns out to be 13/27, which is quite different. However, Ben got confused when giving the answer during our pi podcast and paid penance by making a video
 

However, there is still some debate about the answer.

Join us this week as we record all from the same location and try and tease apart the complexity of weird conundrum.


Yup, we really are altogether!


Friday, 1 December 2017

Maths at the Movies: Moneyball

In this (Hallowe'en themed?) episode we watch the movie Moneyball.

The story of a young mathematician, played by Jonah Hill, struggling to be heard amongst the angry men that make up baseball. And I think Brad Pitt was in there somewhere as well.
  • Can Jonah produce a winning team by redefining how statistics are used?
  • Why does Thomas keep saying the phrase "dick swinging"?
  • Where did Ben get a hold of those baseball sounds?
All of these questions and more are not answered in our Moneyball podcast.

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



Further reading links:

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

Monday, 27 November 2017

Puzzle from pi: Crossing a river

Although there was no Dev Patel, or tigers in this pi, we still managed to squeeze a boat into this week's puzzle.

This puzzle is a variant on an old favourite. We don't have chickens, corn, or foxes, but we still want to cross a river. Here's the question:
A husband and wife come to the edge of a river, where they find two children with a small boat. The boat can hold either one child, two children, or one adult. How do you get everyone across the river in the minimum possible number of crossings?

Post your answers below, and we'll give you the answer in the next podcast!

Sunday, 26 November 2017

Watch along with MATM: Moneyball

Brad Pitt is a sexy man. He's done action, comedy, drama and more. A true acting force to be reckoned with.

Join us as we watch him sit around in chairs and talk to men:
  • Young men;
  • Old men;
  • Sporty men;
  • Angry men;
  • Mathematical men.
All the men you could ever want in a film are right here for your viewing pleasure.

Yup, it's Moneyball: baseball, men and apparently some maths? We've really got to stop Ben choosing the films!

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.

Friday, 17 November 2017

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:

Subscribe via iTunes.

Thursday, 16 November 2017

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!

Wednesday, 15 November 2017

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

Tuesday, 14 November 2017

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

Saturday, 11 November 2017

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.
 

Friday, 10 November 2017

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.

Thursday, 9 November 2017

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.



Wednesday, 8 November 2017

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.

Friday, 3 November 2017

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:

Subscribe via iTunes.

Monday, 30 October 2017

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 Cumberbatches 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, 27 October 2017

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

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