Song of the Day #189 – Van Morrison and Bob Dylan

Today’s song features the greatest singer ever to emerge from Ireland. Sorry, U2 fans, it’s not Bono. And it’s not Enya or Dolores O’Riordan of the Cranberries. It’s not the Corrs, or Elvis Costello, or the Pogues, or Glen Hansard. It’s a singer with one of the most timeless, soulful voices ever: Van Morrison. Today’s video features Van Morrison collaborating with Bob Dylan on a live duet of Van’s One Irish Rover. I usually avoid static cover art videos but I made an exception today due to the rarity and the raw beauty of the audio. Enjoy this treat from the Youtube rock and roll time capsule.

CAPTCHA The Moment

Do you know what CAPTCHAs are? They’re those ubiquitous word recognition challenges that web services use to make sure you’re a human being. Invented by researchers at Carnegie Mellon University in 2000, CAPTCHA stands for “Completely Automated Public Turing test to tell Computers and Humans Apart”. Ticketmaster, for example, uses CAPTCHAs to prevent automated scalper-bots from buying up all the tickets to popular concerts.

It’s a pretty cool idea – it exploits the fact that computer software, advanced though it may be, has a difficult time doing something we humans take for granted: recognizing messy, ambiguously constructed text. The downside, of course, is that in order to stay a step ahead of the bad guys, over time CAPTCHAs have gotten harder to recognize by humans. Take this one, for example, with which I was just presented when signing up for a new service:

I can tell the second word is “urnice” but am I supposed to recognize that first word? As far as I can tell, it’s an ink stain.

Perhaps now might be a good time to admit that I impose CAPTCHAs on people leaving comments on this blog. I do so reluctantly to thwart a high level of spam comments, however, it annoys me to no end that in order to deter spammers, I have to make life more difficult for legitimate visitors. Recently I added support for integrated Facebook comments, which seems like a nice way to solve both problems (it’s convenient and largely spam-proof), the only downside being that I’m aiding and abetting Facebook’s inexorable march toward world domination.

Saturday Puzzle #19 – Measuring Mystery

This is another one that Microsoft and other companies have used as an interview question but it’s a little easier than some of my recent brain benders. You have two empty containers – one has a capacity of five liters of water and the other can hold at most three liters of water. Both containers are made of clear plastic and have absolutely no markings anywhere. Here’s your challenge: given an unlimited supply of water, I want you to come up with a way to measure exactly four liters of water. Leave me a comment if you find the answer. Good luck!

Solution: I received nine answers to this week’s puzzle and all of them were correct! I have some very smart friends. :) Before I share the solution, I’d like to recognize a few noteworthy submissions:

  • Simon Banks and Morag Livingston are living proof that married couples think alike.
  • Muzaffer Peynirci submitted a brilliant algorithm that works independently of which container holds five liters and which holds three. In essence, Muzaffer solved a much harder problem: measure the four liters of water using the two containers while blindfolded!
  • Demonstrating admirable perseverance, Al Pessot submitted one accurate solution and then followed up with an equally correct but more efficient algorithm. Similarly nice work was submitted by Mudassir Ansari and Ricardo Agudo.
  • Katy Gustafson submitted five (5!) different ways to find the answer, including one approach involving boiling water and another involving sound waves! She concedes that one of her solutions may be in error but that still gave her two more correct answers than anyone else, not to mention the Out of the Box Thinking Prize.

Well done, all! As noted, there are a few ways to solve this one but here’s a three step approach that I find the simplest:

  1. Let’s call the two containers C5 and C3. Fill up C5 and pour its contents into C3 until the latter is filled to the brim. At this point C5 has two liters and C3 has three liters.
  2. Empty C3 and pour the contents of C5 into C3. At this point C5 is empty and C3 has two liters.
  3. Fill C5 and pour its contents into C3 until the latter is filled to the brim. You’ve just added one liter to C3 and removed one liter from C5. Therefore, at this point C3 has three liters and C5 has four liters and you’re done.

Song of the Day #187 – Damien Jurado

Seattle might just be the best place on Earth for finding original acoustic music. Today’s video is another great Black Cab Session. This one features veteran singer/songwriter and local hero Damien Jurado, recorded in 2008 at the Green Man Festival in the back seat of a London cab. The song, Last Rights, performed live with Jena Conrad, features folky fingerpicking and lovely vocal harmonies. Sit back, relax and enjoy four minutes of pure acoustic bliss.

Song of the Day #186 – Pete Townshend

About ten years ago I sold my vinyl record collection but there were about a dozen albums I simply could not part with, even if I were never able to play them again. Three of those records were Who’s Next, Tommy and Quadophenia, which brings me to today’s artist: one of my all-time favorite guitar heroes, prolific songwriter, singer, music legend and composer of the soundtrack of my youth: Pete Townshend. In this video, Pete plays three songs in his dressing room, before a show in 2000: Tattoo, I’m One and an improvised acoustic jam. By the way, does anyone know how Pete does that high speed “stutter strum” heard around 3:50? That seems like guitar magic to me.

Song of the Day #185 – They Might Be Giants

Today’s video is both entertaining and educational. That, in a nutshell, is what I like most about today’s artist. They’re talented musicians who write about the most interesting and unusual subjects imaginable. For a music loving computer nerd like me, that’s a very appealing combination. Enjoy this animated version of Why Does The Sun Shine?, while learning a few things about our solar system and enjoy thinking about the warmth coming our way this spring, courtesy of They Might Be Giants and The Sun. If you like today’s song, also check out Song of the Day #109.

We Are All Inventors

In recognition of Twitter’s fifth anniversary, Robert Scoble published an historic circa-2006 video interview with the three founders of Twitter (see below). This video reveals the founder’s views, while the company was still in its infancy, on what Twitter was intended to be and how it was expected to be used. What I find most interesting about it is that, with hindsight from 2011, it seems apparent that Twitter’s creators really didn’t fully understand what they had created and how it would ultimately be used.

There’s a popular myth that inventors possess a laser-like vision of how their product or service will be used but, especially with disruptive technologies like social networking, creators usually have just a faint glimmer of an invention’s full potential (as is now widely known, Facebook started out as an online directory for college students). Ultimately, the party that decides how a unique technology will be used is the end user, who often creates new and interesting modalities that were never anticipated by inventors.

This is, at least partly, what makes modern technologies so exciting: there’s a grass-roots, crowd sourced, participatory aspect, which gives all of us a stake in the innovation process. Essentially, we are all inventors because we all help extend and enhance the ideas of the creators. I believe the most successful companies will be those who recognize and embrace that reality – companies that involve, engage and, to use Guy Kawasaki’s term, enchant their users by treating them a little less like customers and a little more like collaborators.

Song of the Day #184 – Mark Knopfler

There’s a conventional way to fingerpick a guitar, which you’ve probably learned about if you’ve ever taken lessons. You’re supposed to use your thumb to pluck the bass (top) strings and your fingers to pick the lower strings. Mark Knopfler‘s self-taught style defies those and many other conventions and it’s precisely that innovative approach, that refusal to conform to convention, that characterizes Knopfler’s unique sound. Best known for his days with Dire Straights, over the past 15 years Knopfler has quietly released nine remarkably original solo records. Today’s video features Mark performing Old Pigweed from his excellent 2002 release The Ragpicker’s Dream, live in Norway.

Song of the Day #183 – Mary Gauthier

Today’s song captures an extraordinarily difficult moment in the life of Louisiana singer/songwriter Mary Gauthier. Like many people who are adopted in early childhood, Gauthier spent a great deal of time wondering about her parents. The song March 11, 1962, performed live in the video below, captures the painful conversation she had when she finally reconnected with her birth mother. It comes from Mary’s poignant 2010 release The Foundling, which tells the story of her life. It’s a sad, touching and, ultimately, inspirational tale about the power of music to heal the most broken of hearts.

Saturday Puzzle #18 – Weighing Your Options

Imagine you have nine uniformly sized white balls, eight of which weigh precisely the same amount, and one is decidedly heavier or lighter than the rest. You also have an accurately calibrated balance scale, which you can use to compare the weight of any two sets of objects.

Here’s today’s challenge: with only three weighings, tell me how you can discover which ball is different and whether it’s heavier or lighter than the other eight.

Bonus challenge for the hard core puzzlers: same problem with twelve balls instead of nine. Again, in only three weighings, tell me how you can find the odd ball (so to speak) and whether it’s heavier or lighter than the other eleven.

Leave me a comment below with your answer. I’ll post the solution on Tuesday. Happy puzzling!

Solution: Let’s partition the nine balls into three groups of three. Mark the first group A1/A2/A3, the second group B1/B2/B3 and the third group C1/C2/C3. Now weigh the three A balls against the three B balls. There are two possible outcomes:

  1. The scale balances. In that case, the odd ball is in group C so weigh C1 against C2.
    1. If the scale balances, then C3 is the odd ball. Weigh C3 against any other ball to see if it’s heavier or lighter than the others.
    2. If the scale doesn’t balance, let’s assume C1 is on top and C2 is on the bottom end of the scale (analysis of the reverse case is similar). So either C1 is light or C2 is heavy. Weigh C1 against any ball (other than C2) – if C1 ends up on the high end then C1 is light. If the scale balances then you know C2 is heavy.
  2. The scale doesn’t balance. Let’s assume group A is on top and group B is on the bottom end of the scale (analysis of the reverse case is similar). We therefore know that there is either a light ball in group A or a heavy ball in group B. Next weigh group A against group C. If the scale shows an imbalance, we know that group A contains a light ball. If the scale balances, we know group B contains a heavy ball.
    In either case, we’ve narrowed down the odd ball to one of three possibilities and we also know whether it’s light or heavy. With our last weighing, we take any two balls from the suspect group and weigh them against each other. If the scale balances, we know the odd ball is the remaining ball from that group (and we know whether it’s light or heavy). If the scale does not balance, we know which ball is odd by whether that group is known to contains a heavy or a light ball.

Now for the harder, twelve-ball problem…start in a similar fashion by partitioning the balls into three groups of four (marking them A1-4/B1-4/C1-4) and weigh group A against group B. Regardless of whether the scale balances or not, rotate three balls, i.e. shift A1-3 to the B side, B1-3 off the scale and three C1-3 onto the A side and weigh again, noting the outcome. There are five possible changes in the result between weighings one and two:

  1. The scale changes from balanced to unbalanced. Because the first weighing (between groups A and B) balanced, the odd ball must be in group C. The only new balls introduced to the scale in weighing two are C1-3 so one of them must be the odd one. We can also tell whether the odd ball from group C is light or heavy by whether the group C balls are on the top or bottom side of the scale after weighing two. And from the analysis of the nine ball problem, we know that when we have the odd ball narrowed down to one of three *and* we know whether it’s light or heavy, we can solve the problem with one additional weighing.
  2. The scale changes from unbalanced to balanced. If, after the first weighing, the scale was unbalanced, let’s assume group A was on top and group B was on the bottom end of the scale (analysis of the reverse case is similar). Therefore, we know we have either a light ball in group A or a heavy ball in group B. After the second weighing the scale becomes balanced. The only way that can happen is if the odd ball was one of the three from group B we removed from the lower end of the scale. Thus, we know one of B1-3 is heavy and we can find it in one additional weighing.
  3. The scale changes from unbalanced one way to unbalanced the other way (i.e. it reverses polarity). If, after the first weighing, the scale was unbalanced, let’s assume group A was on top and group B was on the bottom end of the scale (analysis of the reverse case is similar). Therefore, we know we have either a light ball in group A or a heavy ball in group B. After the second weighing the scale becomes unbalanced in the opposite way. The only way that can happen is if we shifted a light ball from group A (i.e. one of A2-4) to the other side of the scale. We therefore have a group of three balls containing one that is known to be light and we can find the odd ball in one additional weighing.
  4. Scale remains balanced. This implies that after the first two weighings, the odd ball has not yet appeared on the scale. But there’s only one ball that has never been on the scale: C4. So in our third weighing, we compare C4 with any other ball to see if it’s light or heavy.
  5. Scale remains unbalanced (in the same way). For both weighings, let’s assume group A was on top and group B was on the bottom end of the scale (analysis of the reverse case is similar). If the scale remains unbalanced in the same way for both weighings, then the odd ball is one that remained in the same position for both weighings. There are only two balls in that category: A4 and B4. So either A4 is light or B4 is heavy. In our third weighing, compare A4 to any ball (other than B4). If it doesn’t balance then A4 is light. If it does balance then B4 is heavy.

Random thoughts about technology, politics and the arts.