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Merry Christmaths! (card update) (Patreon)

Published:
2024-12-24 22:58:50
Imported:
2024-12

Content

Hello everyone! Thanks again for another year of helping me make a wide variety of silly maths videos.

This is an update on your festive cards. Normally I'd have emailed them to everyone by now but I've just lost the week before christmas to the flu, which left me completely out of action. I'm fine, I was just holed up in an impromptu sick bay at home, feeling sorry for myself and trying to avoid Lucie catching it as well. Skylab is mainly just annoyed I've not taken her to the pub in over a week.

But I've not had the mental capacity to wrangle my terrible python code together to send the christmas cards out. I did try to do one other bit of work, and if you've seen the Secret Santa video which I ended with a last-minute addition of a new system with an obviously fatal flaw, you can deduce exactly how well that went.

You will get your cards before the month is out, I'm just sorry it's not before the 5^2th of December. I appreciate your understanding!

And if you have any Secret Santa improvements: share them below. I'll go through them all when I'm back on my brain.

Comments

SomeBelgianDude

Health comes first, so I'll be here patiently waiting ❤️

David Collison

Best wishes for a speedy recovery and please don't worry about missing the 5^2th. Speaking of squaring, I'm really excited that 2025 ... 45^2 ... will be the only square year of my lifetime unless there are amazing medical advances. I don't know whether you have any late-octogenarian or nonagenarian (or higher) supporters who were alive last time in 1936, but you may possibly have very young supporters who make it to the next time in 2116.

Frans Wilbrink

Get well soon Matt and have agreat Christmas!

Rick Rubenstein

If you don’t have the energy to write terrible Python code, you could just half-ass it and write decent code instead. Everyone would understand under the circumstances.

Noel Markham

Best wishes Matt, get well soon, Merry Christmas

Lawry McCarthy

My daughter Shona McCarthy is in Scotland sharing the same problem as you. Apparently the bug is pretty bad, I hope you still have some sort of good time somehow for Christ-mass.

Nyx Furry

You can still make it for the 2^5 and blame a typo in the script. Don't rush you recovery, take the time needed

Jaap

For the Secret Santa; let's say you have a list of numbers. The amount of numbers matches the amount of people that are participating in the Secret Santa and all numbers need to be unique. 1 2 3 4 Place this list somewhere online where everyone can access it anonymously (google doc) and let people pick one number and delete it from the list. Wait until there are no more numbers left. (No one knows who picked what and it does not matter if you are first or last). Then use the same list of numbers and do that again, but you must pick a different number than you picked the first time. Then everyone announces which number they picked first. The second number you picked will be the person you are buying for. I must say I got the feeling I am missing something obvious cause it seems too simple, if that's the case I'm blaming Christmas.

Bartosz Błaszkiewicz

The last person has no choice of number. For the second peeking they may get themselves.

Jonathan Raisin

Happy Christmas to you Matt. Thank you for the update - more importantly I hope you get well soon.

Niels Olie

Best wishes and Merry Christmas

Ron

Sorry to hear about your illness. I can wait. But what about your daily cycling streak?

Peter Freese

This would require the concept of a central authority (tracking numbers are available, and presumably making sure no one picked multiple numbers). I think the idea is to solve the problem in a distributed fashion, perhaps by simply passing a sheet of paper around the participants. If we used your idea in such a fashion, the second person would immediately know the picked number of the first person.

Peter Freese

It seems like the requirements you created for the shunt are unnecessary. All that's required is that a shunt takes place, and by the time the last person on the list is done, the shunt doesn't create 1-cycle loops (i.e. the last person needs to shift the right column by a random amount ensuring that they don't have both their numbers in the same row).