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Week 3 Puzzle: Search Logic

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Matt Wieteska - Code Master Matt Wieteska - Code Master | 15:29 UK time, Tuesday, 9 August 2011

In this week's third and final episode of The Code, Marcus will be demonstrating the ability of maths to predict seemingly unpredictable events. From the identity of Jack the Ripper to the outbreak of the next flu epidemic, spotting the patterns that underpin events can allow us to produce accurate models of what will happen in the future. 

On his trip to their London offices, Marcus learns about how Google can correlate a rise in certain search terms with an oncoming outbreak of the flu virus, or the rise of a pop sensation. By working through these relationships logically, Google are able to produce startlingly accurate predictions of future global trends. 

It's this trip that has inspired our puzzle for week three. The puzzle below will test your ability to use logic and reasoning in order to deduce a currently unknown fact. We present: Search Logic. Get your pens and paper ready!

The grid below describes three people's internet searches: Ada, Bill and Chris. As you can see, each search took a certain amount of time. It's your job, using the information we've provided, to fill the grid in and answer the question at the bottom. 


Search Logic Grid



There were nine searches made in total. Each search was for one of the clues from one of the episodes of The Code. The nine searches were: 

Episode One: Pi, Stars, Cicadas

Episode Two: Bubbles, Hex, Dice

Episode Three: Search, Hand, Flock

Each person searched for one clue per episode. 

No search was repeated, either by the same person, or by a different person. 

Everyone's first searches were from different episodes. 

Episode Three's clues all took as many minutes to search for as there are letters in the search term. For example, "Three" would have taken five minutes. 

Bill searched for his Episode Three clip first.

Ada did not search for "search".

The person who searched for "hand" also searched for "bubbles".

The "stars" search took 15 minutes and did not occur at the same time as the "hexes" search. 

The person who searched for "cicadas" and "dice" spent 18 minutes in total searching. 

If all the above is true, who searched for "Pi"? 


  • Comment number 1.

    Great puzzle, I love these! Unfortunately a bit of a clue in the Codebreaker, but luckily I didn't need it.

  • Comment number 2.

    Not very clever to have only one option on Codebreaker!!

  • Comment number 3.

    Ada did

  • Comment number 4.

    It is a shame about the divergence from reality that 'three' takes 5 minutes and 'seventeen'takes 9 minutes. This makes it seem too academic.

  • Comment number 5.

    re Martin
    oh horror a maths/logic question that's academic, though personally i don't think counting the letters in a word is particularly challenging

  • Comment number 6.

    I believe that one and a half clues:
    - "did not occur at the same time as the "hexes" search"; and also
    - "The person who searched for "hand" also searched for "bubbles""
    are unnecessary to solve the problem. Anyone want to prove me wrong?

  • Comment number 7.

    Ada - person 1. Because Pi has 2 letters. None of the clues were necessary surely?

  • Comment number 8.

    Your all boffins arguing or scoring points about details, it's supposed to be fun!

    I've enjoyed solving the clues, or trying to, even I worked out Ada in 30secs. So what. I won't get the 3 words for the Ultimate challenge, missing some clues, but I have had fun trying. I remember the Golden Hare, yes I'm that old, which was fun too. Good luck to those who get to the final 3. I hope the book will be unlocked for those who just enjoy puzzles. Great trophy, love the way it is constructed. My daughter was inspired by the fun and interesting way maths is in our world. Anything to spark her interest in a subject I found really hard, mystery facts about the world we live in and making Maths just a tad more interesting. Thanks to the BBC perhaps, something in the future more directed at the average child, book series......

  • Comment number 9.

    Hi anyone who may be reading this, I'm after a pointer!

    In episode 3, the formula for calculating a population from one year to the next w given as:

    P next = R [growth rate] * P [population this year] - R * P * P [to allow for deaths]


    Pnext = R * P (1-P)

    Now by my calculation, no matter what the starting population, from year 2 onwards you start getting negative populations - if I'm in error could someone show me how please?

    Many thanks in advance!

  • Comment number 10.

    @jkk398 - yes, also you don't need to know who searched for 'dice'

    i guess if you want to fill in the whole grid that's all useful though, so one for the completists ;)

  • Comment number 11.

    @welshpaul23 That looks right to me, and suggests Marcus has got a bit of a simplification going on here. The second P is supposed to be some function (let's call it K) such that KPR gives the number of deaths in the population in this iteration. Now if K is itself linearly increasing in P then one can write it as K = kP, so that what the equation really needs to say is:

    P next = RP(1 - kP)

    The k could practically be ignored in terms of giving a rough model of what the population will do qua fluctuation, but certain conditions relating k, P and R need to be in place to get chaotic behaviour. You'll get fairly roughly what Marcus was talking about if you set k = 0.0001, your population size between 1 and 10,000 and R = 4 for the above equation.

  • Comment number 12.

    Well thank goodness this was an easy one to do from the circle after that ridiculously dice one in episode 2.

    However I am stuck with catapult, I have got to level 9 and cant see anything on level 6 that crops up on the wheel, any pointers please?

  • Comment number 13.

    Um, I hate to break it to you guys, but the clue says:

    Episode Three's clues all took as many minutes to search for as there are letters in the search term. For example, "Three" would have taken five minutes.

    the key word being EPISODE THREE'S CLUES.

    This means that only search, hand and flock relate to the number of letters, and therefore pi isn't ada, otherwise ther is two 4 letter words: dice and hand, and only one 4 minute time slot.

    remeber to read the question :D

  • Comment number 14.

    Posicoln, "Pi" is still Ada. Here's one full solution:

    The episode 3 clues are such that Person 1 searched for "Hand" second, Person 2 searched for "Search" second, and Person 3 searched for "Flock" first.

    Bill is therefore Person 3, since "Flock" was the only item searched first in episode 3. Ada did not search for "Search", which only leaves Chris as Person 2 and Ada as Person 1.

    Chris searched for 18 minutes, which therefore means his three search items were "Cicadas", "Dice" and "Search

    Since Bill spent 15 minutes on one of his searches, he therefore searched "Flock" first and "Stars" third.

    With Bill searching "Stars" and Chris searching "Cicadas", this leaves Ada searching for "Pi".

    In relation to #6, I've done this without appealing to the one-and-a-half clues as you thought. The "Hexes" half-clue is a very weak statement relative to the rest of the clues, only adding enough to rule out the possibility that Ada searched Hexes third and still leaving open the possibility of searching it first or bubbles third. Ada's "hand/Bubbles" link is stronger but still not enough to fully complete the grid, since you establish enough to determine all of Bill's searches and when they took place, leaving open two options as to the order of Ada's and Chris's searches.

    I've also done this without appeal to the "Different episodes for first searches" clue which is only necessary for full completion, so it's probably even better than you thought at the time. It's certainly possible to weaken the "Dice/Ciceda in 18" and the "Ep 3 number of letters" clues as well without losing the provability of Ada = pi, but you can't get rid of them altogether like you can the other clues.

  • Comment number 15.


    Really appreciate your post and you taking the time to explain what was going on, makes sense now!

    To be honest though I have to call into question the effectiveness of this type of simplification, I'm all for helping people into a subject (in this case maths) but not to the extent that what's said on-screen is just plain wrong, I know most people wouldn't go away and test what was said (althought that's more the pity from my point of view!) but it does leave a nasty taste in my mouth

  • Comment number 16.


    It's not really "Just plain wrong", as such. The form Marcus gave is completely appropriate when the various P values are thought of in terms of population values that are fractions, and the constant k can be trivially 1 (though it needn't be). k, P and R need to be in some specific relationships to model the relevant behaviour - discussing why that is isn't obviously beneficial to the point Marcus is trying to make.

    (Also, apparently the chaos involved when R=4 has quite a lot of exception cases; thanks to B_Rizla for pointing out the Wikipedia page for the Logistic map!)


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