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?
Simple question for this one:
In Riddler City, the city streets follow a grid layout, running north-south and east-west. You’re driving north when you decide to play a little game. Every time you reach an intersection, you randomly turn left or right, each with a 50 percent chance.
After driving through 10 intersections, what is the probability that you are still driving north?
A fairly simple one this week, I think:
[W]hat is the longest word that doesn’t share any letters with exactly one state?
The word list is indicated, and “state” means a US state. It takes much longer to download the word list than to solve the first part of the problem!
Someone asked me recently about managing outsourced piecework, where people outside your company are paid by the number of tasks performed. I did this on various projects for over a year with a previous employer. Our trial-and-error approach had quite a few errors, but we did eventually establish a stable system. This is an outline of that successful system, with a few notes about things we tried that didn’t work.
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.
How do you resolve the tension between falling into a filter bubble on one hand, and having your media be an abyss of stupidity and hostility on the other?
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:
Labour took one hell of a beating, and they’re electing a new leader. Those in favour of continuity sometimes produce figures aiming to show that continuity is fine, and can bring victory.
So here are four graphs of things that might matter, versus the number of seats won. One of these is not like the other three.