# The Big Lockdown Math-Off, Match 2

Welcome to the second match in this year’s Big Math-Off. Take a look at the two interesting bits of maths below, and vote for your favourite.

You can still submit pitches, and anyone can enter: instructions are in the announcement post.

Here are today’s two pitches.

## Sam Shah – Self-graphing functions

Sam Shah is a math teacher who took part in last year’s Big Internet Math-Off and tweets about maths at @SamJShah2.

One of those things that never fails to generate awe when I think about it is how powerful equations are in holding information. Like the equation $y=x^2$. It is a rule, indeed, that says “take the input, square it, and there’s your output.” But if you plot all values of $(x,y)$ that satisfy this rule, you can generate a really cool image.

What I’m trying to say is that I often forget that a graph of an equation is a solution set. You can look at all points on the coordinate plane and see: does this satisfy the equation? Like (4,3). That doesn’t work, so we don’t fill it in. But (4,16) does work, so we do. And beautiful images emerge when we do this.

The reason I bring this up is because I want to share with you an equation I’ve loved since the moment I read about it in some popular math book or another when I was in high school.

$\frac{1}{2} \lt \left \lfloor \bmod{\left(\left\lfloor \frac{y}{17} \right \rfloor 2^{-17\lfloor x \rfloor – \bmod{\left( \lfloor y \rfloor, 17 \right)}}, 2\right)} \right\rfloor$

Before I share what it does, let’s look to see what is going on here. Let’s test the point (5,60).

$\frac{1}{2} \lt \left \lfloor \bmod{\left( \left\lfloor \frac{60}{17} \right\rfloor 2^{-17\lfloor 5 \rfloor – \bmod{\left( \lfloor 60 \rfloor, 17 \right)}},2 \right)} \right\rfloor$

$\frac{1}{2} \lt \left\lfloor \bmod{ \left( 3 \cdot 2^{-94}, 2\right)} \right\rfloor$

Then I looked at what is in the floor function by asking google (since I forgot about modular arithmetic with fractions):

Eeeep. A tiiiiny number. So when we evaluate it, we get:

$\frac{1}{2} \lt 0$

That isn’t true. So we don’t consider (5,60) a solution to the equation.

However if we tried the point (0.5,52), we’d see something different:

$\frac{1}{2} \lt \left \lfloor \bmod{\left( \left\lfloor \frac{52}{17} \right\rfloor 2^{-17\lfloor 0.5 \rfloor – \bmod{\left( \lfloor 52 \rfloor, 17 \right)}},2 \right)} \right\rfloor$

$\frac{1}{2} \lt \left\lfloor \bmod{ \left( 3 \cdot 2^{-1}, 2\right)} \right\rfloor$

Again, asking google what’s in the floor function, we get:

Which results in:

$\frac{1}{2} \lt 1$

A true statement. So we can fill in the point (0.5,52) into our solution set.

But look at the original equation… We could have used any $x$ value between 0 and 1, and any $y$ value between 52 and 52.999999999999 and that wouldn’t have changed our evaluation. So we can fill in the entire square

Because of this, our solution set looks like little pixels.

I graphed it on desmos, and found some pixels, including the one we identified above!

Okay, now for the wow factor! I’m going to scroll up a tiiiiny little bit up on the graph (image grabbed from here):

Yup. The equation encodes itself! It actually writes itself out.

BOOM! MATHEMAGIC!

WHAT IN THE WHAT?!?!?!

My floor dropped when I first saw this. You may now stop reading, if you want to just live in a world where this exists and is beautiful and it feels like magic. Some short equation with involves high school level math can draw itself out!

But if you want a little more, but it might kill off the magic a bit, just keep reading…

Yes, this is a famous formula, known as Tupper’s self-referential formula.

Ignoring the negative $y$-values, you can see only a few solutions, all hugging the $y$-axis… Turns out there are more pixels if you scroll up, and they start moving away from the $y$-axis.

I was using Desmos to create these, and the $y$-values for the last image are from 1099900000 to 1099900100. But even Desmos has trouble going to the numbers I need.

But if you go up on the graph to y-values between

960 939 379 918 958 884 971 672 962 127 852 754 715 004 339 660 129 306 651 505 519 271 702 802 395 266 424 689 642 842 174 350 718 121 267 153 782 770 623 355 993 237 280 874 144 307 891 325 963 941 337 723 487 857 735 749 823 926 629 715 517 173 716 995 165 232 890 538 221 612 403 238 855 866 184 013 235 585 136 048 828 693 337 902 491 454 229 288 667 081 096 184 496 091 705 183 454 067 827 731 551 705 405 381 627 380 967 602 565 625 016 981 482 083 418 783 163 849 115 590 225 610 003 652 351 370 343 874 461 848 378 737 238 198 224 849 863 465 033 159 410 054 974 700 593 138 339 226 497 249 461 751 545 728 366 702 369 745 461 014 655 997 933 798 537 483 143 786 841 806 593 422 227 898 388 722 980 000 748 404 719 and 960 939 379 918 958 884 971 672 962 127 852 754 715 004 339 660 129 306 651 505 519 271 702 802 395 266 424 689 642 842 174 350 718 121 267 153 782 770 623 355 993 237 280 874 144 307 891 325 963 941 337 723 487 857 735 749 823 926 629 715 517 173 716 995 165 232 890 538 221 612 403 238 855 866 184 013 235 585 136 048 828 693 337 902 491 454 229 288 667 081 096 184 496 091 705 183 454 067 827 731 551 705 405 381 627 380 967 602 565 625 016 981 482 083 418 783 163 849 115 590 225 610 003 652 351 370 343 874 461 848 378 737 238 198 224 849 863 465 033 159 410 054 974 700 593 138 339 226 497 249 461 751 545 728 366 702 369 745 461 014 655 997 933 798 537 483 143 786 841 806 593 422 227 898 388 722 980 000 748 404 719

and

960 939 379 918 958 884 971 672 962 127 852 754 715 004 339 660 129 306 651 505 519 271 702 802 395 266 424 689 642 842 174 350 718 121 267 153 782 770 623 355 993 237 280 874 144 307 891 325 963 941 337 723 487 857 735 749 823 926 629 715 517 173 716 995 165 232 890 538 221 612 403 238 855 866 184 013 235 585 136 048 828 693 337 902 491 454 229 288 667 081 096 184 496 091 705 183 454 067 827 731 551 705 405 381 627 380 967 602 565 625 016 981 482 083 418 783 163 849 115 590 225 610 003 652 351 370 343 874 461 848 378 737 238 198 224 849 863 465 033 159 410 054 974 700 593 138 339 226 497 249 461 751 545 728 366 702 369 745 461 014 655 997 933 798 537 483 143 786 841 806 593 422 227 898 388 722 980 000 748 404 736

You’ll see what I showed you in the graph above in blue.

It turns out this formula eventually draws every possible set of pixels in the rectangle bounded by $x=0$ and $x=17$, and $y=k$ and $y=k+106$ if you keep scrolling up far enough.

So you’ll get a smiley face. You’ll get pixelated letters in words. You’ll get anything you want! Here’s one particular image this equation draws, from a Numberphile video I recommend which shows PacMan being chased by a ghost eating the formula!

Here’s the video:

I said above that the magic sort of gets lost once you know this.

That happened to me. But then the wonderment returned as I realized this formula generates worlds…

$\frac{1}{2} \lt \left \lfloor \bmod{\left(\left\lfloor \frac{y}{17} \right \rfloor 2^{-17\lfloor x \rfloor – \bmod{\left( \lfloor y \rfloor, 17 \right)}}, 2\right)} \right\rfloor$

… so yeah, this “simple” formula generates all $17 \times 106$ pixel regions. It’s not magic that it generates itself, anymore. But now — at least for me — it’s magic that it can generate worlds.

I’m left with questions:

1. I see the 17 in the formula, which relates to the height of the $106 \times 17$ rectangle. But where does the 106 come from?
2. How could we prove that this formula will generate all possible $106 \times 17$ rectangles? After a certain $k$-value do the $106 \times 17$ rectangles just start to repeat?
3. Why did I see a lot of filled in values for negative $y$-values when I graphed on Desmos?
4. How do I find the $y$-values I scroll to to see a particular rectangle that I define?
5. What was Jeff Tupper working on that led him to this formula?

Here are two webpages (one and two) that look super interesting for me to investigate later. But right now I need to get some sleep so I can figure out how to teach online in a few days. Le sigh.

## Kyle Evans – A Curious Auction

Kyle is a maths teacher and entertainer. His homepage is kyledevans.com.

How much would you be willing to bid to win a one pound coin? You wouldn’t be tempted to bid more than a pound, right? Perhaps if I explain that this is a special kind of auction…

In a regular auction, of course, only the winner has to pay – everyone who bids but doesn’t end up winning leaves with the money they arrived with. In this special auction, however, all money bid by all bidders goes to the auction house, even from those bidders who do not end up winning.

Perhaps you decide a safe way to win the pound would be to bid 90p – that should see off any rival bidders, right? But then a competitor swoops in and bids one pound, perhaps thinking it worth breaking even (spending a pound to win a pound) for the thrill of being the winning bidder.

What should you do next? Remember, you currently stand to lose 90p – in this special auction your losing bid will not be returned to you. But if you carry on and bid £1.10 for the one pound coin, you will essentially only be 10p in debt (losing £1.10 but gaining the £1 you stand to win). Losing 10p is clearly preferable to losing 90p. However, now your rival is in the same position – accept defeat and lose £1, or bid £1.20 and only lose 20p? It’s a no-brainer! They will bid £1.20 and the cycle continues. Where will this all end though?! At every point in the auction a bidder will put themself into a more profitable position by making one more bet, so the auctioneer stands to make… infinite pounds??

I’ve always loved this problem because it gives a lovely first insight into game theory – clearly if I gave the two bidders twenty seconds to discuss tactics before bidding started they would decide that one of them should make the smallest possible bid, say 10p, then they’d split the pound upon leaving the auction hall. But by acting individually, each bidder can put find themself in a position where it makes logical sense to bid more than a pound for a one pound piece. I remember reading that in real auctions people have regularly paid over £2 for a pound coin in this situation, though I can’t remember where. When I googled ‘one pound coin auction’ all I found was commemorative brexit coins for sale on eBay, and life is too short to trawl through all that.

If you know more about this problem or its origin, please do get in touch. Until then, here’s a poem I wrote about one of my favourite problems:

I SPENT TEN POUNDS ON A ONE POUND COIN

I stepped into the auction hall
And nothing seemed unusual
The auctioneer stood to one side
And, “GOING, GOING, GONE!”, she cried
The people sat – intentions hid –
With fingers poised to make their bids
And now and then the auctioneer
Would shout, “SOLD! TO THIS MAN DOWN HERE”

But here’s where things turned somewhat strange
For suddenly, the rules had changed
“The next lot”, said the auctioneer
“Is different, let me be clear
Each penny bid will then be lost
Regardless of the final cost
So if you bid and do not win
Your money’s gone: let that sink in”

This change of rules seemed quite unfair
But soon, excitement filled the air
The auctioneer knelt to the ground
Then held aloft a shiny pound
(I sensed a bargain coming soon
And shuffled through the auction room)
Then she declared, mischievously,
“I start the bidding at 10p!”

I raced in with a 10p bid
What easier way to win a quid?
But then a man right next to me
Stood up and hollered “20p!”
Well, I bid 30 pence of course
Then he bid 40, in due course
The bidding war remained intense
‘Til I had offered 90 pence

Now, have a think, what would you do
If you were in my rival’s shoes?
To spend a pound to make a pound
Does not seem logically sound
But losing 80 pence is worse
Than having nothing in your purse
So he shouted “One pound!” and then
I pondered… and bid one-pound-ten

One-pound-ten for a one pound piece?!
That’s madness! Foolishness at least!
But ten pence loss – think carefully –
Hurts less than losing 90p
See, one more bid, at any cost
Feels, somehow, like a lesser loss
My rival knew the same, and yet
We chased each other into debt

The gavel pounded! I had won!
It dawned on me what I had done…
There’s no sense in what people buy
Even a careful chap, like I
By making steps quite logical
Had ended up a complete fool
Oh, what a crazy club I’ve joined
I spent ten pounds on a one pound coin

So, which bit of maths made you say “Aha!” the loudest? Vote:

Match 2: Sam Shah vs Kyle D Evans

• Sam with functions that graph themselves (64%, 75 Votes)
• Kyle with a curious auction (36%, 42 Votes)

Total Voters: 117

This poll is closed.

The poll closes at 9am BST on Wednesday 15th, when the next match starts.

If you’ve been inspired to share your own bit of maths, look at the announcement post for how to send it in. The Big Lock-Down Math-Off will keep running until we run out of pitches or we’re allowed outside again, whichever comes first.

## About the author

• #### Christian Lawson-Perfect

Mathematician, koala fan, Aperiodical editor. Usually found paddling in the North Sea, or fiddling with computers.

### 3 Responses to “The Big Lockdown Math-Off, Match 2”

1. Shecky R

To Kyle: Hahhh! I have the original version of your puzzle scheduled to run at Math-Frolic in a couple days where I’ve used it before (and am currently just re-running some old posts over next few weeks). It comes from economist Martin Schubik in its American form, as the “Dollar-auction paradox” (and is one of my favorites as well):
https://en.wikipedia.org/wiki/Dollar_auction