Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
From Guy Moore, a puzzling progression:
What is the sequence below, and what are its next elements? (The numbers in parentheses provide a helpful hint.)
From Keith Wynroe, a concise circular conundrum:
If N points are generated at random places on the perimeter of a circle, what is the probability that you can pick a diameter such that all of those points are on only one side of the newly halved circle?
Solution to last week’s Riddler Express
Congratulations to 👏 William Cadegan-Schlieper 👏 of Temecula, California, winner of last week’s Riddler Express!
Last week, you bought a new clock, only to be dismayed when you got home to discover that its two hands were identical — the hour hand and the minute hand looked exactly the same. How would you ever know what time it was? You soon realized, however, that this wouldn’t always be a problem. For example, when it’s 12:30, the minute hand is exactly on the 6 and the hour hand is halfway between the 12 and the 1. It can’t be the other way around because if the hour hand were exactly on 6, the minute hand would have to be exactly on 12, which it’s not. So in that case, you know what time it is. But how many times during the day are you not able to tell the time?
The clock’s hands will be indistinguishable 286 times a day. At 264 of those, you will truly not be able to tell the time — and even then only for a split second (and if your eyesight is good enough). Not bad, really, considering such a seemingly defective design.
I received many widely varying approaches to this problem that led to correct solutions. Let’s start with the tidy approach of solver Doug DeMoss:
Every hour, the true minute hand will sweep around the whole clock and the true hour hand will sweep across a one-twelfth segment of the clock. Twelve times during that hour — once per one-twelfth segment — there will come a time where the hour hand and the minute hand could be reasonably mistaken for one another. That is, where they could be reversed and still display a reasonable time of day. For example, at about 10 minutes and 4 seconds past 12, it would be impossible to tell whether the clock is reading that time or about 50 seconds past 2. One of those 12 times, however, is when the hands are perfectly aligned, and in that case, we do know what time it is. So there are 11 times an hour when we can’t tell the time, and there are 24 hours in a day, so the answer is 24×11, or 264.
And here’s an arithmetic submission from solver Michael Branicky:
If the hour hand has moved a fraction of the circle h from midnight, then the minute hand has moved m = h/12. Thus, h = 12m. We are searching for times that are also valid the other way around, or m = 12h. Combining those equations, this occurs when h = 144h, or 143h = 0, where both equations are mod 1. This happens 143 times every 12 hours, when h = k /143 for integer k = 0, 1, …, 142. In a full day, this happens 286 times. However, we must also remove the cases — such as midnight — when the minute and hour hands overlap. These happen when h = m, or when h = 12h (again, mod 1), or whenever 11h is an integer. This happens 11 times every 12 hours, or 22 times every full day. Therefore, if you carefully measure the hands, you cannot tell the time at just 286-22 = 264 times during the day.
Solver Mike Seifert took a graphical approach. He plotted the position of the big hand and the little hand, with the big hand’s position as the x-coordinate and the little hand’s position in the y-coordinate. He then flipped the x and y coordinates. Anywhere the two paths coincide is an ambiguous time:
Or I could just check my phone.
Solution to last week’s Riddler Classic
Congratulations to 👏 Mark Matthews 👏 of Milton, Georgia, winner of last week’s Riddler Classic!
The Virginia Cavaliers and the Baylor Lady Bears cut down the nets in the men’s and women’s NCAA basketball tournaments this month. But what about the unsung transitive champions — the teams that beat those teams during the season, and the teams that beat those teams, and the teams that beat those teams, and so on? Enter last week’s challenge: How many transitive champions were there this past season in the men’s and women’s games? I provided links to the men’s and women’s season results.
There were 359 transitive national champions on the men’s side, including every single Division I school. We’re gonna need a lot more scissors and nets.
Unless you had a lot of time on your hands — and a lot of paper — this problem was an exercise in data analysis and programming. Many solvers were kind enough to share their code, including Benjamin Phillabaum, Colin W., David Fried and Kyle Tripp. Mathematically, this basketball problem is really an exercise in graph theory — the teams are the graph’s vertices and the games are the graph’s directed edges, pointing from the winning team to the losing team. The challenge, therefore, is to see which vertices ultimately connect to the real NCAA champion. And most, as it turns out, did connect.
Luke Benz plotted the men’s Division I transitive champions by degrees of separation and by conference. Teams in the ACC, for example — Virginia’s conference — tended to have tighter claims to the transitive championship, while teams in a conference such as the Patriot League could lay only more distant claims.
The team with the longest path to a coveted transitive national championship were the Division II Fayetteville State Broncos, who can smile upon a chain of eight games. They beat UNC-Asheville who beat USC Upstate who beat Longwood who beat Southern Miss who beat Old Dominion who beat Syracuse who beat Duke who beat Virginia. Congratulations, Broncos!
There were a whopping 1,775 transitive national champions on the women’s side even though true champion Baylor lost only one game all season. (The women’s results to which I linked included many more teams, for one thing, including Division III teams and teams in Canada.) Every Division I women’s team was a transitive national champ — except for the Eastern Kentucky Colonels, who went 2-27 and were winless against other D-I teams.
Eight different teams on the women’s side tied for the longest path, claiming their transitive title thanks to chains of 25 games! One example: the St. Mary’s Lightning (in Alberta) beat Briercrest who beat Okanagan who beat Capilano who beat Vancouver Island who beat Bellevue College who beat Umpqua who beat North Idaho who beat Chandler-Gilbert who beat Anoka-Ramsey who beat Hibbing who beat Bay College who beat Silver Lake who beat Concordia who beat Rochester College who beat Spring Arbor — deep breath — who beat Keiser who beat Florida Memorial who beat Xavier University of Louisiana who beat Southeastern Louisiana who beat SMU who beat South Florida who beat UCLA who beat Cal who beat Stanford who beat Baylor. Phew. Congratulations, Lightning!
Michael Branicky shared his code and plotted the distribution of all the transitive national champions by their shortest distance from beating the actual champion:
Anyway, great season, everybody.
Want more riddles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email me at firstname.lastname@example.org.