I wrote up a gist for last week’s Riddler Express, including the extra credit, but forgot to link it here, probably because I was on holiday! Anyway, this week’s Classic looked very tractable.
Tag: Riddler Classic
Riddler Classic, August 14 2020
I did this one analytically and by brute force simulation, and got the same answer with both methods. It’s in a github gist.
Riddler Classic, July 10 2020
My answer last week seems to be wrong; I thought agreeing with LL would be enough. I will need to look more closely at it.
This week’s Classic is about stacking blocks. At first glance I thought it was Tower of Hanoi, but it’s a little different:
Riddler Classic, July 3 2020
Got last week’s Express just right, with the right explanation.
This week’s Classic is based on the US flag, but the maths question is this:
[W]hen N equals 50, N is twice a square and N+1 is a centered pentagonal number. After 50, what is the next integer N with these properties?
Riddler Classic, May 22 2020
Riddler, May 15 2020
Since I’m still learning my way through Jupyter, I tried the Classic as a notebook.
I’m also not sure what the best way to present a notebook is, if I have a WordPress site? The include I used last time wasn’t that great.
For the Express, the best score I can manage is 6. In this grid, there’s no legal place to put a 1 in the top row.
Riddler Classic, February 28 2020
When I first saw this one, I thought it might be a chance to use some of the Erlang queueing equations, which I’ve known about for ages but never used seriously. Instead, I think it’s a bit easier than that. I missed the cutoff for having my name picked though!
Riddler Classic, January 24 2020
EDIT: Nope! Not far wrong, but definitely wrong.
My strategy of starting with the easy cases and building out to more complex cases was used successfully by other solvers. My analysis of the easy cases was correct, but there was a mistake in the way I built out. My diagram probably didn’t help; the one drawn by the winner didn’t distinguish between the larger and smaller piles, and I think that’s where my error crept in.
For example, when I considered (9, 5), I thought it was a losing position because I didn’t see that taking six coins from the larger pile produced (5, 3), since the larger pile became the smaller pile in that move. So the “simplification” I introduced, of specifying one pile as larger, while not wrong, made it easier for me to make a mistake. It helped to break my mental model by suggesting an identity for the piles (“larger” or “smaller”) that isn’t persistent, but depends on the progress of the game. Will try not to do that again!
EDIT 2: Revised the original spread sheet showing the rows that I had calculated wrongly, and producing a correct game state table.
Another week of being fairly confident about this week’s Riddler Classic. (Bolstered by getting exactly the right answer for the last one I did, if by a slightly inefficient method.)
The puzzle is a coin-picking game, with the usual aim of taking the last coins, but the unusual condition of having two coin stacks:
Riddler Classic, January 3 2020
Initially I thought this week’s Classic puzzle might be too tough to solve without some serious insights, but the more I thought about it, the more it seemed like I might be able to break it down to manageable pieces.
Riddler Classic, November 22 2019
Most of the Classics look too difficult for me to be able to solve, but this week’s looked like I could approach it. No code required, either.
Here’s the question:
Five friends … are playing the … Lottery, in which each must choose exactly five numbers from 1 to 70. After they all picked their numbers, the first friend notices that no number was selected by two or more friends. Unimpressed, the second friend observes that all 25 selected numbers are composite (i.e., not prime). Not to be outdone, the third friend points out that each selected number has at least two distinct prime factors. After some more thinking, the fourth friend excitedly remarks that the product of selected numbers on each ticket is exactly the same. …
What is the product of the selected numbers on each ticket?
There might be a neat, elegant way of solving this, but I chipped away at it bit by bit.