Kominers’s Conundrums: 14 Faces, One Main Event
(Bloomberg Opinion) -- Some puzzles require a potentially stupefying range of knowledge. That’s the case with this week’s Conundrum, in which we’ve boxed up fourteen famous figures.
Some are present-day luminaries, while others are from as far back as the Stone Age. That makes this an ideal puzzle to solve with the help of your full family.
The first goal is to identify each person. But you’re not done when you can just name them – from there, solving the puzzle will still require a couple more logical leaps.
You’re on a quest for an answer that’s a single person. Once you put together the clues, you’ll realize that he or she is hardly hidden.
So what are you waiting for? Call up a few relatives on the phone or Zoom, share this with them, and start solving!
Last Week’s Conundrum
As I mentioned in the column, one of the challenges in this puzzle was that the setup seemed almost under-determined. Nothing was said about how quickly cars moved, or how many were in line – I implied the answer would be the same no matter what.
Nevertheless, a good way to start solving would be to try out some simple examples with a more detailed assumptions: For example, what if all cars move one “space” per minute, meaning that a two-hour line must contain 120 cars? Once Knievel jumps in front, the first car is effectively pushed back a space, resulting in a one minute delay. That, in turn, pushes the second car back one space, and so forth, creating the same one minute delay for each car. Adding those delays up, we get a total of 120 minutes – which is suspiciously the same as the two-hour length specified in the puzzle.
One other quick thought experiment is to imagine that once Knievel jumps the line, the first car gets out of the way and waits for the others to go by. Then, the other cars all cross the bridge exactly when they would have otherwise – no delays. But that takes two hours, which means that the first car ends up crossing two hours late.
And indeed, if we make other assumptions about the cars in line, we seem to always get back to that same two-hour number. So we might conjecture that is indeed the answer – but how to prove it?
The trick, as it turns out, is to imagine a much more laid-back Knievel: If he were to just get in line behind all the other cars, that would be just adding one more driver to the line – going from N drivers, say, to N + 1. The total time it takes for all N + 1 drivers (including Knievel) to get across the bridge is independent of the order they cross in. So then what happens if Knievel skips ahead to the front? He crosses the bridge two hours earlier than he would have otherwise; since the total crossing time is unchanged, that means he must have slowed the other N drivers down by a total of two hours.
Charles Medrano observed if Knievel could actually jump that many cars, he would have to be moving at "screaming speed," which is probably worthy of a traffic ticket. Ross Rheingans-Yoo proposed that we compute the optimal fine with a complicated mechanism involving surveying all the drivers about their cost of lost time. Jennifer Walsh also wanted to tack on a penalty for jumping a motorcycle on the freeway.
The Bonus Round
Alice in Typhoidland (featuring the Mad Hatter’s pipe challenge); "Merlin's Mathematical Mysteries"; "My Word," archived (hat tip: Ellen Kominers); revisiting the first Millionaire winner. Trick pool shots featuring dominoes; triathlon, but with marbles. 5D chess, time travel included (hat tip: Paul Goldsmith-Pinkham); plus the upcoming Co-Puzzle Hunt. And inquiring minds want to know: What happens to all that money on unused gift cards?
We did assume in the Conundrum statement that Knievel drives across the bridge like any other car – if his bridge crossings involved theatrics, that would be a different matter.
Another way to see this, thanks to Jeremy Hurwitz, is as follows: if we let T(k) be the wait time for the k-th car in line, then bumping a car from position k to k+1 creates a delay of T(k+1) - T(k). Adding these up across all the cars, most of the terms cancel, and we see that the total delay is T(LAST CAR) - T(FIRST CAR), which is (2 hours) - (0).That said, Ross Berger and Zoz pointed out that if the cars really could move at the same pace before and after Knievel's jump, then Knievel would have to have landed in the spaces between the cars -- which means he must have great aim, and might not cause much delay at all.
And once again, I learned the puzzle fromAlex Tabarrok.
This column does not necessarily reflect the opinion of the editorial board or Bloomberg LP and its owners.
Scott Duke Kominers is the MBA Class of 1960 Associate Professor of Business Administration at Harvard Business School, and a faculty affiliate of the Harvard Department of Economics. Previously, he was a junior fellow at the Harvard Society of Fellows and the inaugural research scholar at the Becker Friedman Institute for Research in Economics at the University of Chicago.
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