# The Indisputable Santa Mathematical Advent CalendarDay 8

Happy 8th of December!

Throughout the month, to accompany the release of our book on the Mathematics of ChristmasHannah Fry & I are tweeting out Christmathsy bits and pieces, one a day, advent calendar style. Assuming we don’t run out of ideas, that is…

There’s a tougher-than-usual puzzle tomorrow, so today we’re keeping it light, with a joke:

SOLUTION TO YESTERDAY’S PUZZLE

Scroll down for the solution…

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Solution:

So, this is the ‘official’ solution, with the shortest possible path to the goal, taking in 10 baubles:

Some people noticed you could do actually better than this if you were allowed to perform U-turns, allowing you to complete the maze while touching only 8 baubles. To be honest, I think the word “passing” in the instructions implies that this isn’t really valid, but well done for lateral thinking anyway:

You might also have wanted to pass by all the red baubles on the way to the present (You can’t pass by all the green ones. Why not?), like this:

Personally though, none of these is my favourite solution. Since Santa is wider than the paths that he is supposed to be following, the diagram clearly isn’t to scale. Santa must be much smaller than he appears to be, so perhaps if he squeezes himself really tightly against the wall, he might be able to manage this path, taking in just 2 baubles:

Well done to Elliot Cossum, Remy, Alex Draghici, abra328, Guillaume Laurent, Brett Kreis, Olaf Doschke and Ebexanna Scrooge, who all submitted solutions of one form or another. Merry Christmas to you all!

## 4 thoughts on “The Indisputable Santa Mathematical Advent CalendarDay 8”

1. thomas Post author

You could, certainly. If you can find the shortest path to do so, I’ll stick that up as well. 🙂

1. thomas Post author

Interesting… Still interested to know whether it’s the shortest. Reckon I have a method to work it out. Might muse on it later.