More recently, I saw it mentioned again in a shark movie that I happened to sit through, The Reef.  In the movie, a sailboat runs aground and capsizes in shallow water; rather than stay with the boat, the people on board try to swim north toward a distant island.  It doesn’t go well.

Before leaving the boat, one guy looks at his watch, looks at the sun, mumbles a few calculations, then says, “That’s north,” and off they go.  I wondered if those calculations were correct, particularly since the movie was set in Australia– the Southern Hemisphere– and the method didn’t sound quite like I remembered it.  After some additional searching on the web, along with some experiments and calculations of my own, I learned that descriptions of this “Boy Scout” survival trick are frequently incorrect, incomplete, or misleading.  But more interestingly, such descriptions are almost never accompanied by a warning of just how inaccurate the method can be, even when it is used properly.  My suggestion: keep your GPS-equipped smart phone with you at all times.

How it works: If you are north of the tropics, hold your watch face horizontal, and turn so that the hour hand points toward the sun.  Then the ray bisecting the angle between the hour hand and 12 o’clock noon points approximately true south.  (I will deal with the Southern Hemisphere later.)  See the figure below for an example.

Example of using a watch as a compass (Northern Hemisphere, 10:30 am).

The first potential source of confusion is which direction is north and which is south: do you bisect the “small” angle or the “large” angle between the hour hand and 12 o’clock?  Some sources suggest that “north will be the direction further from the sun”– that is, bisecting the smaller angle points south.  Although this is usually the case, it can fail when the sun is up before 6:00 am or after 6:00 pm.  It seems simpler to just remember that south bisects the angle that sweeps in time from the hour hand toward noon (clockwise in the morning, counter-clockwise in the afternoon).

How well it works: This is what I found most interesting about this problem.  One survivalist blog (a somewhat amusing phenomenon, if you think about it) makes the strangely precise and grossly false claim that “the accuracy of this method is within 8 degrees (US and Canada).”  In fact, the error in the estimate of direction quite often exceeds 30 degrees, and can even exceed 80 degrees depending on where– and when– you are.

There are two primary sources of error.  The most obvious source of error is the accuracy of your watch.  Several online sites state that “the direction will be correct if the watch is set for true local time, without adjustments for Daylight Savings Time [sic].”  More precisely, the method is more accurate if your watch reads 12 o’clock when the sun is at its highest point in the sky, due directly south.

The problem is that “standard time” and “apparent solar time” rarely coincide exactly.  Everyone in a particular time zone thinks it is the same time, despite the fact that those on the eastern edge of a time zone will see the sun rise approximately one hour earlier than those on the western edge.  To make matters worse, the time zones in the U.S. are rather erratically shaped, resulting in even larger differences than would be the case if time zone boundaries were simple, straight lines of longitude, 15 degrees apart:

The time zones in the contiguous 48 states.

The second source of error is the latitude of your position: as several online sites suggest, “the further you are from the equator, the more accurate this method will be.”  This is essentially true, and is by far the larger of the two sources of error.  What is often left out, however, is that the method isn’t simply less accurate near the equator, it effectively doesn’t work “at all.”  In the tropics– the band around the earth between about 23.5 degrees south and 23.5 degrees north– not only is the sun higher in the sky, making it more difficult to accurately estimate its direction in the first place, the sun’s motion is also not as well-behaved (why?), so that the error in your estimate of direction can approach 180 degrees, even assuming an accurate estimate of the sun’s direction.

So how well does it work?  As approximately “best case” behavior, following is a plot showing the error in estimated direction over the course of the year 2012, at the Northwest Angle Inlet in Lake of the Woods, Minnesota (the northernmost point in the contiguous 48 states).

Compass error in Lake of the Woods, MN, for the year 2012.

The general shape of this plot is typical; the method works best in the fall and winter months, with the worst case behavior in the middle of the year.  Even this far north, errors can exceed 30 degrees.

At the other extreme, the following plot is for the Florida Keys, which suffer from being just north of the Tropic of Cancer, where errors are the worst.  However, the method still works reasonably well during the “cold” months.

Compass error in the Florida Keys for the year 2012.

Finally, back to the Southern Hemisphere.  Here, the method is only slightly different: if you are south of the tropics, then point 12 o’clock at the sun, and north bisects the angle between the hour hand and 12 o’clock noon (with the same “sweeping in time” sense described above).  I found several different incorrect variations on this online; perhaps the most disappointing was an otherwise very cool interactive Mathematica demonstration on the Wolfram web site.

And the method used in the Australian shark movie?  The description of the method turned out to be correct… but the action takes place at approximately 10 degrees south latitude, where all bets are off.  At any rate, it wasn’t enough to escape a great white shark.

References:

1. Meeus, Jean, Astronomical Algorithms (2nd Ed.). Richmond: Willmann-Bell, Inc., 2009.
2. Muller, Eric, Shapefile of the Time Zones of The United States. [link]

1. If a light bulb is dropped from a given floor and breaks, then any other light bulb will also break if dropped from that or any higher floor.
2. If a light bulb is dropped from a given floor and does not break, then neither it nor any other light bulb will break if subsequently dropped from that or any lower floor.
3. A light bulb may be dropped any number of times until it breaks, after which it is unusable and cannot be dropped again.

A “drop” consists of taking a single light bulb in the elevator up to a particular floor, dropping the light bulb from that floor, taking the elevator back down, and checking whether the dropped light bulb broke.

If the building has floors, and you have light bulbs at your disposal, what is the least number of drops required to guarantee that after at most drops you can determine the highest floor from which a light bulb may be dropped without breaking?

Each variant of the problem from last week simply changed the value of , the number of available light bulbs.  The basic idea is most clearly illustrated with the simplest case of ; with just a single light bulb, we can do no better than dropping from floor 1, then floor 2, 3, 4, etc., until the bulb finally breaks, requiring drops in the worst case.

At the other extreme, suppose that we have “plenty” of light bulbs that we can afford to break (i.e., ).  In this case, intuition correctly suggests that a binary search is the right approach: for a 100-floor building, start by dropping a light bulb from floor 50.  If it breaks, then drop a second bulb from floor 25, otherwise drop the second bulb from floor 75, reducing the number of “candidate” floors by approximately half with each drop.  The maximum number of drops required is , or 7 drops in the case of the original interview question from last week:

Problem 1: “Given 20 ‘destructible’ light bulbs (which breaks at certain height), and a building with 100 floors, how do you determine the height that the light bulb breaks?” – Asked at QUALCOMM

As I said last week, I suspect that this question actually contains a typo, since 20 light bulbs is “more than plenty;” we will only ever actually use 7 of them.  Things get more interesting if we have a more limited supply of light bulbs to break.  In the usual statement of this problem, , and the final variant from last week asked the question for .  However, we can tackle all of these variants at once by solving the general case.

The key to the solution is to turn the problem around, so to speak, and solve for instead of .  That is, given a maximum number of drops , and a supply of light bulbs, what is the tallest building that we can handle?  More precisely, what is the maximum number of floors for which we can guarantee to find the “critical” floor in at most drops?

Consider the first drop from the as-yet-unknown optimal floor .  If the bulb breaks, then by Property 1 (and 3) above, there must be at most floors below floor .  If the bulb does not break, then by Property 2 (and 3), there must be at most floors above floor .  Thus, satisfies the following recurrence relation:

Using the identity

it is not difficult to show that

This gives us a solution to all three original problems.  For any given number of floors and number of light bulbs , we simply find the smallest number of drops for which .  To answer Problem 2, for example, for a 100-floor building with 2 light bulbs, we need 14 drops in the worst case.

Problem 3: Similarly, with 5 or more light bulbs we can do it in 7 drops… but we must be careful, since a binary search will only work with 7 or more light bulbs.  If we only have 5 (or 6) bulbs, the optimal strategy is slightly different.  Fortunately, the counting argument above gives a recipe for determining from which floor to drop the next light bulb in any situation: assuming that we have calculated the optimal number of drops remaining, the next drop should be from floor , counting from the lowest remaining “candidate” floor.  For example, beginning with 5 light bulbs, the first bulb should be dropped from floor 57.  If it breaks, the second bulb is dropped from floor 26.  If it also breaks, the third bulb is dropped from floor 11.  If it breaks, the fourth bulb is dropped from floor 4.  If it breaks, the fifth and final bulb must be dropped from floors 1, 2, and 3 in the worst case, for a total of 7 drops.

Finally, using the identity

we can see how the binary search arises in the case where we have “more than enough” light bulbs.

But subject matter seems to keep dropping in my lap.  In just the last two weeks I learned about some cool new science, a brilliant and elegant algorithm I had not heard of, and a nice twist on an “old” mathematical puzzle.

That is plenty to choose from; but since I love puzzles, and since I am still technically trying to be on vacation, this week I will just briefly present the puzzle, and we’ll see how it goes from there.

A variant of the problem appeared in a year-end compilation of “Top 25 Oddball Interview Questions” (the entire list is at glassdoor.com):

Problem 1: “Given 20 ‘destructible’ light bulbs (which breaks at certain height), and a building with 100 floors, how do you determine the height that the light bulb breaks?” – Asked at QUALCOMM

Before continuing, although I like this problem as a mathematical puzzle, I tend to dismiss this (or similar puzzles) as a useful interview question, for reasons I have discussed before.  But whether asked in an interview or not, note that this is not the possibly-vague “lateral thinking” type of problem as suggested here.  For those who have not heard this one before, we can make the problem precise as follows:

All light bulbs are identical in that they all have the following properties:

1. If a light bulb is dropped from a given floor and breaks, then any other light bulb will also break if dropped from that or any higher floor.
2. If a light bulb is dropped from a given floor and does not break, then neither it nor any other light bulb will break if subsequently dropped from that or any lower floor.
3. A light bulb may be dropped any number of times until it breaks, after which it is unusable and cannot be dropped again.

A “drop” consists of taking a single light bulb in the elevator up to a particular floor, dropping the light bulb from that floor, taking the elevator back down, and checking whether the dropped light bulb broke.  Your objective is to determine the highest floor from which a light bulb may be dropped without breaking, using as few drops (i.e., making as few elevator trips) as possible.  What should your strategy be to minimize the worst case number of drops required?

Problem 2: This problem caught my eye because I suspect that it actually contains a typo– but an interesting typo.  In the usual statement of the problem, instead of having 20 light bulbs at your disposal, you have only 2, and if/once they are both broken, you must have enough information to determine the highest floor from which a light bulb may be safely dropped.  Now what should your strategy be?

Problem 3: The above two variants of the problem are actually somewhat special cases, “endpoints” of a more general problem that still has a very elegant solution.  In this third “twist” variant, suppose that you have 5 light bulbs; now what should your strategy be to minimize the maximum number of drops required?  As usual, can you generalize as a function of the number of floors and light bulbs?

The harder second problem asks essentially the same question as discussed in the Deranged Secret Santa post from last year: suppose that n participants in a Secret Santa gift exchange each put a slip of paper with their name on it into a single hat.  To determine gift-giving assignments, each participant in turn draws a slip from the hat, replacing it with another if he draws his own name.  What is the probability that this procedure “fails,” with the last person to draw finding his own name on the last slip in the hat?

I did some simulation and doodling on this problem last year, with limited success.  After revisiting the problem this year, there are a few more concrete and interesting results to report.

First, as usual, the problem has already been solved; the solution by Brian Parsonnet appears as sequence A136300 in the Online Encyclopedia of Integer Sequences.  (The sequence is parameterized by the number n of participants, where each integer value of the sequence is the numerator of the corresponding probability with denominator .)  However, the given recurrence relation is rather complex, and appears to require time and space to compute.

It turns out that we can solve the problem more simply and efficiently (in quadratic time) by transforming the problem slightly.  Note that the statement of the problem does not explicitly specify any particular order in which participants draw slips from the hat.  That is, it does not really matter who is the last person to draw; all that we want to know is the probability that the last person– whoever that may be– ends up drawing their own name.

At each turn, then, we can randomize the selection of the next person to draw, without affecting the desired probability.  (In the context of last week’s airplane boarding problem, the passengers do not stand in a line; they jostle in a group near the jetway entrance, and each next person to board is selected randomly.)  This helps a lot: given total participants, let be the number still waiting to draw, and let be the number waiting to draw whose names have already been drawn.  Then the probability that the last person to draw is left with his own name is , where is given by the following recurrence (with apologies for WordPress’s limited LaTeX multi-line formatting):

As the following plot shows, the probability of failure varies approximately as ; the probability for in the original problem is approximately 0.0095018.

Probabiilty that the last person draws his own name vs. group size.

Finally, this slight transformation of the problem yields an interesting additional bonus: recall the “warning” given in last year’s post, suggesting that you probably do not want to use this procedure at all for your Secret Santa gift exchange.  The problem is that all possible permutations (specifically, all possible derangements) mapping gifters to recipients are not equally likely.  Even if you handle the small failure possibility of the last person drawing his own name by starting the whole procedure over, some participants are still more likely to draw certain names than others.  (As mentioned last year, the most extreme deviation is near the end of the line, with the last person to draw being nearly twice as likely to draw the next-to-last person as to draw the first person in line.  See the Mathematica source code below to calculate these probabilities.)

We can relax that warning a bit, based on the following observation: as in the puzzle solution above, if we modify the procedure slightly by first randomizing the order in which participants draw names… then the resulting random assignment does at least have the nice property that each person is equally likely to draw any other person’s name!

I verified this numerically for using the code below, but I think this follows generally from a simple symmetry argument.  However, this is a weaker statement than the stronger claim that all derangements are equally likely, which still does not hold ( is a minimal counterexample).

board[seats_, assigned_, prob_] :=
 Module[
  {passenger, s, p},
  If[Length[seats] === 1,
   If[seats === {Length[assigned] + 1}, Return[]];
   MapIndexed[
    (overall[[#2[[1]], #1]] += prob) &,
    Append[assigned, First[seats]]
    ],
   passenger = Length[assigned] + 1;
   s = DeleteCases[seats, passenger];
   p = 1/Length[s];
   Scan[
    board[DeleteCases[seats, #], Append[assigned, #], prob*p] &,
    s
    ]
   ]
  ]

n = 10;
overall = Table[0, {n}, {n}];
board[Range[n], {}, 1];

1. You are standing at the very end of a line of 100 holiday travelers (you and 99 others) waiting to board an airplane.  Each passenger in line has a boarding pass assigning him or her to one of the 100 seats on the airplane.  However, the first person in line is rather confused, and so instead of sitting in her assigned seat, she boards the airplane and simply selects a seat at random (possibly her own).  As each subsequent passenger boards the airplane, he sits in his assigned seat if it is empty, or selects an empty seat at random if it is not.  What is the probability you, the last person to board, will be able to sit in your assigned seat?

2. Preparing to return home after the holiday, you find yourself once again last in a line of 100 travelers waiting to board an airplane.  This time, it is not just the first person in line who is confused; after a stressful holiday with their families, now every person in line is not just confused but deranged.  That is, each passenger in turn boards the airplane and selects a seat at random from all currently empty seats, excluding their own.  What is the probability that you, the last person to board, will be forced to sit in your assigned seat?

The first problem is an oldie but a goodie.  I like it because it can be approached in several different ways.  It is easy to simulate using a computer, which quickly leads to conjecture at the solution.  Given the conjecture, the proof is a nice application of mathematical induction.  But there are also nice symmetry arguments via which the problem may be solved more directly.

The second problem is more difficult, and is the real motivation for this post.  Solutions to both are welcome in the comments– until then, consider in what other way(s) this second problem might arise in relation to the holidays?

Following is Mathematica code that enumerates all possible cycles, unique up to permutations.  It is essentially a lexicographic enumeration of cycles of polygons, where .  (We need at least 3 polygons, since the interior angle of any regular polygon is strictly less than 180 degrees.  There can be at most 6 polygons, since the smallest interior angle of any regular polygon is 60 degrees.)  The idea is to enumerate “candidate prefixes” of polygons, since the number of sides in the -th polygon is determined by the others:

Do[
    n=Table[3,{m-1}]; (* Start with all triangles *)
    While[True,
        s=Total[180-360/n]; (* Compute sum of interior angles *)
        k=Length[n];        (* Usually we increment the last polygon *)
        If[s>180,           (* Remaining polygon contributes < 180 deg *)
            p=360/(s-180);  (* Compute number of sides of remaining polygon *)
            If[p<Last[n],
                t=Split[n]; (* Remaining polygon is too small *)
                If[Length[t]>1,
                    k=k-Length[t//Last], (* Increment at last difference *)
                    Break[]              (* Stop if all polygons are equal *)
                ],
                If[IntegerQ[p],Print[Append[n,p]]] (* Record valid cycles *)
            ]
        ];
        n[[Range[k,Length[n]]]]=n[[k]]+1 (* Move to next candidate cycle *)
    ],
    {m,3,6}
]

This yields the following output, where it turns out that the desired largest polygon is realized in the lexicographically first cycle:

{3,7,42}
{3,8,24}
{3,9,18}
{3,10,15}
{3,12,12}
{4,5,20}
{4,6,12}
{4,8,8}
{5,5,10}
{6,6,6}
{3,3,4,12}
{3,3,6,6}
{3,4,4,6}
{4,4,4,4}
{3,3,3,3,6}
{3,3,3,4,4}
{3,3,3,3,3,3}

Another reason that I like this problem– beyond the fact that its solution is the answer to “Life, the Universe, and Everything”– is that the answer, 42, is… well, large.  That is, my own intuition would not have suggested a cycle might involve a polygon with that many sides.

This phenomenon of “surprisingly large numbers” in mathematics is what reminded me of this problem and motivated the last couple of posts.  The subject of conversation was actually ”surprisingly large minimal counterexamples,” or the dangers of “poor man’s induction:” observing a property for many values of n, and supposing that the property then holds for all n.

There are a lot of interesting examples of this sort of thing, from Polya’s conjecture (the smallest value of n for which the conjecture fails: 906,150,257) to the Chebyshev Bias or “prime race” (23,338,590,792).  And although Graham’s number is not exactly the same thing– it is “merely” a proven upper bound on a minimum solution to an easily stated problem in graph theory– as far as mathematically relevant large numbers go, it’s one of the more mind-blowing ones.

Example using 3 triangles and 2 squares.

Hint (sort of): There is a much better title for this post, but unfortunately it would give away the solution to the problem.

Rules of the game

The rules are simple: the game begins with a deck of n matching pairs of cards (2n cards total), shuffled and arranged face-down on a table.  A player’s turn consists of flipping up two cards, one at a time.  If they match, the player removes the pair and takes another turn.  Otherwise, the cards are returned face-down to the table and the next player takes a turn.  Play continues until all pairs are removed (or until all players agree to end the game), at which point the player with the most collected pairs wins the game.

The references at the end of this post provide extensive analysis of the game for two players.  However, there are some interesting additional subtleties in rules and strategy that seem to have been either missed or omitted.  Following is a summary of my analysis of the game, with a focus on these additional details.

First, how is this a game of strategy?  As its name implies, the game is intended to be a memory exercise, remembering the locations of previously revealed matching cards.  Playing against a person– or a computer– with a perfect memory would not be much fun… unless we leveled the playing field by modifying the rules slightly, so that selected unmatched cards remain face-up on the table even after a player’s turn ends.  In this version of the game, everyone has a perfect memory, since all previously selected but unmatched cards are visible and known to both players.

This new game might seem rather boring: on each turn, simply collect any face-up pairs, flip up a face-down card, collect the new pair if possible, otherwise flip up another face-down card.  Right?  Wrong.  Optimal strategy frequently consists of seemingly strange moves, such as flipping up a card, and if its match is not known, selecting a known, unmatching face-up card, effectively skipping the rest of the turn.  And there are still stranger moves, as we shall see.

Game states

We can represent a game state as a tuple , where is the total number of pairs of cards still on the table, is the number of face-up cards on the table, and is the number of face-up matched pairs on the table.  The game state must satisfy

Finally, is the “score” of the next player to move, or the difference of the numbers of pairs collected by each player.  The initial state of a game is of the form .

Optimal strategy depends on what we are playing for.  That is, what is the value of a completed game ?  The references focus on the case where, for example, each player wins one dollar from the other player for each pair collected.  In this case, we are trying to maximize the expected gain, or expected value of .  However, I think a more common and realistic valuation is simply “win, lose, or draw,” with , which corresponds to maximizing the probability of winning, disregarding the margin of victory.  [Edit: After some discussion, I realized that this correspondence only holds when ties are impossible; i.e., the number of pairs is odd.  For even , the game may end in a draw, and maximizing the expected value of does not necessarily maximize the probability of winning outright.]

Moves

A move by a player is a two-step process, consisting of selecting two cards, with the observation of the first card possibly affecting the choice of the second.  Considering all possible such selections, even “those that might not look clever,” as Gerez puts it, there are 9 different moves:

1. PP: collect a face-up matching pair (and take another turn).
2. OO: pick up two face-up (“old”) but non-matching cards, effectively skipping the turn.  (This is what Zwick refers to as a 0-move.)
3. PN: pick up one of a face-up pair and a face-down (“new”) card, effectively skipping the turn while revealing an additional card.
4. ON: pick up a face-up non-paired card and a face-down card.
5. NPO: pick up a face-down card and its match if known, otherwise a non-matching face-up card.  (This is what Zwick refers to as a 1-move.)
6. NPN: pick up a face-down card and its match if known, otherwise another face-down card.  (This is what Zwick refers to as a 2-move, and is the move that many players might use exclusively.)
7. NNN: pick up two face-down cards.
8. NON: pick up a face-down card and, if its match is known, a non-matching face-up card (leaving a known pair on the table!), otherwise another face-down card.  (This is what Gerez and Zwick refer to as a “sacrifice.”)
9. NNO: pick up a face-down card and, if its match is known, another face-down card (leaving a pair on the table), otherwise a face-up card.

The OO move is particularly interesting, since it effectively amounts to a “pass,” which raises the question of how to ensure that the game terminates.  In the gain-maximizing case focused on in the references (i.e., ), it is not difficult to show that the game is symmetric, in the sense that if the OO move is optimal for one player, then it is optimal for the other as well.  In other words, if one player wants to end the game, the other player will agree to end the game as well.

However, when the goal is to maximize the probability of winning, the game is no longer symmetric, and things get more complicated.  There are several different reasonable rule variations that deal with the OO move, each of which yields slightly different strategy and advantage for each player, in approximately decreasing order of my personal preference:

1. “Agreed end:” A player may make an OO move, and if the other player also makes an immediately following OO move, then the game ends.
2. “Ko rule:” This is similar to the ko rule in Go that prohibits returning to the same game state: an OO move may not be immediately followed by another OO move.
3. “Quit while you’re ahead:” An OO move unilaterally ends the game, with the outcome determined by the current difference of collected pairs.  In the gain-maximizing case only, this is equivalent to an agreed end as in (1).
4. “Withdraw:” An OO move unilaterally ends the game in a draw.

Of course, given all of these complications, another possibility is to simply disallow the OO move altogether.  This variant actually yields the most interesting optimal playing strategy.  In the game where the objective is to maximize the probability of winning, and the OO move is prohibited, the following table shows a portion of the optimal playing strategy, for game states of the form :

Optimal strategy for game states (n, k, 0, 0).

For brevity and clarity, the common NPO and NPN moves are indicated by 1 and 2, respectively.  There are a couple of unexpected weird moves: for game states with , the optimal strategy is the NON “sacrifice.”  Also, for (i.e., the game is tied and six cards remain on the table with one face-up), you can psyche out your opponent with an NNO move: if you happen to flip up the matching card, leave the pair on the table and flip up another face-down card!

A final computer science note: analysis of optimal strategy for the Memory game is essentially a dynamic programming problem.  Mathematica’s automatic memoization once again makes this particularly easy; source code in PDF format is available here.

References:

1. S. H. Gerez, An Analysis of the “Memory” Game, 65-Afternoon Project Report (in handwritten Dutch with English summary).  University of Twente, 1983. [PDF]
2. I. Stewart, Concentration: a winning strategy.  Scientific American, 265(4) (October 1991):103-105.
3. U. Zwick and M. S. Paterson, The memory game.  Theoretical Computer Science, 110 (1993):169-196. [PDF]

Some additional online searching suggests that this “trick” of using your head to increase the key fob’s range also seems to work as well or even better by placing the fob under your chin, sometimes accompanied by opening your mouth.  There are a lot of proposed explanations for why this works, along with repeated experiments confirming that it does work.  One of the more interesting explanations is the following quote from a 2009 New York Times article:

The trick turns your head into an antenna, says Tim Pozar, a Silicon Valley radio engineer.  Mr. Pozar explains, “You are capacitively coupling the fob to your head. With all the fluids in your head it ends up being a nice conductor. Not a great one, but it works.”

At this point I was certainly intrigued, but skeptical.  I think the experimental setup is very important here, with the potential for either unintended misinterpretation of results… or even the intentional misdirection of a magician’s trick.  The experiment is reproduced again in this Unplggd article:

One step back beyond the point of being out of range, I confirmed that the remote didn’t work. Then I raised the remote [my emphasis], pressed it to my chin, and squeezed the button. I heard a “chirp!” in the distance and saw the headlights flash. Huh. So I started walking backwards. All in all I was able to back up 42 steps, roughly 85 feet, before I was no longer able to make contact with the car by holding the remote to my chin.

Sometimes– as in the neutrino timing experiment mentioned in my last post– science is hard, being some combination of expensive, difficult, or complicated to carry out.  But in this case, it was relatively cheap, easy, and straightforward to simply try it myself, and see what happened.

It is nice being married to a woman who not only tolerates but enables these projects and excursions of mine.  Last Saturday morning, she and I drove to the reasonably flat, completely empty parking lot at the nearby high school.  She recorded the maximum distance from which I could consistently lock the car using the remote key fob, while holding the fob in several different positions and orientations.

(Distances were measured in increments of 10 feet, conveniently marked by the intersections of painted lines on the lot.  By “consistently lock the car” I mean that I could lock the car with each of three button presses, but 10 feet farther away I failed on at least two of three attempts.  I was somewhat surprised that the range actually dropped off this sharply.)

Holding the fob in “typical” fashion, away from my body at approximately chest height and pointing it at the car, the fob’s range was 80 feet; i.e., at 90 feet it stopped working.  As in the above experiments, I then held the fob under my chin… and the fob’s range increased to 110 feet!  So far, so good, although opening or closing my mouth did not appear to make any difference.  I also tried holding the fob to my right temple, where the range increased to 140 feet.

See where this is going?  I think the problem is that, at least in the descriptions I have read or videos I have seen, this is where the experiment ends.  But I also tried holding the fob at chin height, in the same orientation relative to the car, but away from my body, and the range remained at 110 feet.  Similarly, holding the fob at the height of my temple, but away from my body, yielded a range of 140 feet.  (We did actually observe one success out of three attempts at 150 feet.)

Finally, the most effective method that I tried was holding (and pointing) the fob straight up over my head, which increased the range to 190 feet.  In other words, the higher the fob, the longer the range.  We observed no effect of the fob’s position relative to my head, only relative to the ground.

Myth busted?  It would seem so… but I wonder if others have tried this experiment, also distinguishing height above the ground from position relative to the head as separate controlled parameters, and observed different results.  Maybe I was doing something wrong?  It would certainly make for a much cooler “did you know” phenomenon if there really was something to the “capacitive coupling to your head.”

This past week, scientists at CERN announced some very interesting results suggesting that neutrinos may be able to travel faster than the speed of light.  This is understandably big news, so you almost certainly didn’t read it here first.  I just have a few additional observations.

First, the initial paper is available on arXiv.org here.  I am no particle physicist, but the basic story is not very complicated, and with my usual educational bent, I think this makes for an interesting and useful exercise even for young students… because it essentially boils down to distance = rate * time.  Scientists measured the time required for neutrinos to travel 731.278 km (about 454 miles) from a particle accelerator in Switzerland to a detector in Italy.  For those here in the U.S., imagine particles traveling underground between Baltimore, Maryland, and Martha’s Vineyard in Massachusetts– or for those reading back home, between Kansas City and Dallas, Texas.

Traveling at the speed of light (exactly 299,792,458 meters per second), the trip should take about 0.0024392808 second.  However, detection equipment on the receiving end observed the arrival of neutrinos about 60 nanoseconds earlier than this, suggesting that the particles were traveling faster than light.  That’s not much less time… but it’s a lot more speed.  It means that neutrinos would beat light to the finish line by just about 60 feet (one light-nanosecond is approximately one foot), but in doing so they exceed the speed of light by more than 16,000 miles per hour, about the speed of the Space Shuttle or the International Space Station in orbit around the Earth.  (This is a small fractional increase in speed, though, relative to the speed of light: just about 25 parts per million.)

As usual, I recommend reading the original source.  Something that the “popular” press frequently seems to miss in this story is that it’s not completely new.  Neutrinos have behaved in funky ways before; the paper mentions and references similar observations back in 2007, although with admittedly less precise measurements.

Finally, it is the improved measurement technique and precision that are under much of the scrutiny in this most recent experiment.  As described in the paper, the scientists are not able to actually measure the times of departure and arrival of any single neutrino.  Instead, they collect timing information for multiple particles over a larger interval, and use statistical techniques to “shift” entire distributions of departure/arrival times until they “line up.”  Even for the non-physicist, there is some interesting mathematics involved.

As I said, I’m not a particle physicist.  But even watching from the sidelines, I find it exciting, and not a little inspiring, to watch science in action like this.  There is an enormous amount of attention, review, and criticism of these experiments, and that scrutiny is what science is all about.  The basic practice of science may not be changing, but I think the speed of science may very well be undergoing a radical increase even as we speak.  Similar to the polymath projects started by Timothy Gowers, today’s technologically small world allows very large numbers of eyes and brains working on a problem, which can have the doubly-beneficial effect of “parallelizing” more exploratory investigation while at the same time more quickly recognizing and abandoning dead ends.

Do I think the result will be fundamental changes to our current understanding of physics?  My uneducated guess is that I doubt it.  It could turn out to be as simple as some unexpected systematic measurement error.  Or maybe there really is a new phenomenon here, but one that merely requires refinement, not complete rejection, of currently prevailing theories.

But I could be wrong.