Diversion

To distract us from worldly worries and woes: two brainteasers from Jeopardy! star Ken Jennings.

Two puzzles: NOT SO HARD and YES VERY HARD.

NOT SO HARD. Can you think of a food that becomes a synonym for “later’ when it’s spelled backwards?

YES VERY HARD. I just wrote down a lower-case, seven-letter word”“something you might consult with a lawyer about. I realized that if I added to the middle letter just the smallest of strokes, I would make a new word”“something you might consult with a doctor about. What are the words?

22 Comments

  1. I have no damn clue, but if I think like some of my goofier students for moment, I’d say:

    nuts/stun (i.e., food / get zapped, wake up later)

    –for the first one. Just for the hell of it.

    As for the second one: I’ll be thinking all night about it. When he says “[add]…just the smallest of strokes,” I’m led to such things as changing an “l” to a “t,” or an “o” to a “d,” etc. Maybe even “n” to “m,” but I’m guessing that that extra leg doesn’t qualify as “the smallest of strokes.”

    Posted February 8, 2012 at 9:53 pm | Permalink
  2. Dom says

    I got NOT NOW right off the bat.

    The second must be l to t. If not the clue is bad.

    Posted February 9, 2012 at 12:50 pm | Permalink
  3. Malcolm says

    Very good – I was trying to think of single words, and NOT NOW didn’t occur to me!

    Posted February 9, 2012 at 12:57 pm | Permalink
  4. Malcolm says

    Following in the tradition of my friend G. Orcalimbo Jones at WOMR (log on for the live stream of his weekly show, Fridays from 9 to midnight), I’ll give out a “hale and hearty congratulations” to the first commenter to solve the second puzzle.

    Perhaps even a dollop of whipped cream, and a blast from the upright can.

    Posted February 9, 2012 at 12:59 pm | Permalink
  5. Congrats, Dom! I’m always stymied by these Mensa-esque puzzles because I tend to make conservative assumptions about a given puzzle’s constraints. So yeah, like Malcolm, I assumed that the answer had to involve single words. (The phrase “a synonym,” in Jennings’s formulation, seemed to reinforce the notion that we needed to find a single word.)

    As for “l” to “t”: I agree that’s possible, but I was also wondering, belatedly, about “o” to “a,” assuming an “a” written as a circle with “the smallest of strokes added.”

    I was also mulling over what 7-letter issue I’d see a lawyer about. All that came to mind was “divorce.” I’m pretty sure I’d have to see the doctor if I started saying “divarce,” a problem for which the diagnosis is, “Yer Oyrish! Say ‘divarce’ all y’want!”

    Posted February 9, 2012 at 1:46 pm | Permalink
  6. I am not a big fan of puzzles, though I occasionally enjoy a bridge “play-of-the-hand”.

    But I remember one that a college roommate of mine challenged me with (50 years ago!):

    Given a scale and 12 identical objects, except that one of the 12 has a different weight (either heavier or lighter). What weighing strategy guarantees identifying the “different” object in at most 3 uses of the balance.

    It took me several hours to figure it out.

    Posted February 9, 2012 at 2:57 pm | Permalink
  7. Dom says

    F. set the coins out in a row
    And chalked on each a letter so
    To form the words

    F AM NOT LICKED

    An idea in his brain clicked
    And now his mother he’ll enjoin

    MA DO LIKE
    ME TO FIND
    FAKE COIN

    No two agree in their effect
    As is by pen and patience checked
    For instance, should the dud be L
    And heavy, here’s the way to tell:
    First weighing, down the right must come;
    the others – equilibrium!

    Posted February 9, 2012 at 6:46 pm | Permalink
  8. Charles says

    Long time no comment!

    I’m still puzzling over the very hard one. I want to put forward an attempt at TheBigHenry’s puzzle, although my answer is based on two assumptions:

    1) “Scale” means “balance scale” where two items or groups of items are weighed against each other.

    2) You have to know if the odd object is lighter or heavier. I don’t know if there is a solution if the odd object could be either lighter or heavier. At least, I haven’t thought of a solution yet.

    Given those two assumptions (the second of which is probably the most questionable), the solution would be as follows:

    Divide the 12 items into groups of four. Put two of those groups on the scales. If one is heavier or lighter than the other (depending on the information we have), that group contains the odd object. If they balance out, then the third group contains the odd object. Take that single group of four and divide it into two groups of two. Weigh those and pick out the group that is either lighter or heavier. Put these two on the scales and you know which is the odd object.

    This wouldn’t work if we didn’t know whether the odd object was lighter or heavier, as we wouldn’t know which group to select in the first round (if the two chosen come out different) or the second round (if the two chosen in the first round come out the same). If there is an answer to this scenario, I haven’t been able to think of it. You would have to establish whether the odd object is lighter or heavier, and that would require one extra use of the scales.

    (If the two first-round groups came out different, for example, you would have to weigh one of those groups against the third group. If you weigh the lighter group against the third group and they come out the same, the odd object is heavier, etc. The same goes if the two first-round groups come out the same. Pick either group and weigh it against the third group. If the third group is lighter, the odd object is lighter, etc. You have to know this before you split up the odd group and start weighing, though, because you have to know if you’re looking for a lighter or heavier object.)

    I am provisionally stumped.

    Posted February 9, 2012 at 8:15 pm | Permalink
  9. Dom says

    Your second assumption is wrong. You do not know, beforehand, if the coin is light or heavy.

    If your assumption were right, then in general, if there were N coins, you would need Log-base-3 weighings to find the odd coin. For example, with 3 coins, you need only 1 weighing, for 9 coins, you need two, for 27 coins you need 3, and so on.

    TheBigHenry’s problem needs three weighings. Letter the coins FAMNOTLICKED, then weigh

    1. MADO against LIKE
    2. METO against FIND
    3. FAKE against COIN

    Say in the first weighing, the LIKE pan goes down. Then it can’t be I, can’t be K, can’t be E. Must be L and L is heavy.

    Posted February 9, 2012 at 10:02 pm | Permalink
  10. @Dom at 6:46 pm:

    Interesting poetry, but I can’t make heads or tails out of it.

    ____________________________________________

    @Charles at 8:15 pm:

    Yes, “scale” means “balance scale”, but no, you are not given whether the coin is heavier or lighter. If you were given that information a priori, the problem becomes trivial.

    ____________________________________________

    @Dom at 10:02 pm:

    If the “LIKE” pan goes down, then either one of the 4 “LIKE” coins is heavy, or one of the 4 “MADO” coins is light, but you can’t know at that point which of the two possibilities is the correct one (since you haven’t been told whether the odd one is light or heavy). This leaves you with two weighings to determine which of the 8 coins [LIKEMADO] is the odd one.

    “Must be L and L is heavy” is not the correct conclusion.

    Posted February 10, 2012 at 1:39 am | Permalink
  11. Spoiler. Click at own risk. This is one of many, many online solutions to the “12 objects in 3 weighings” problem. Upshot: there is no concise way to explain how this is done, but the above-linked page contains a table that comes as close to concision as one is likely to get.

    Posted February 10, 2012 at 3:09 am | Permalink
  12. Dom says

    @TheBigHenry at 1:39

    I mistakenly gave the impression that you knew L was heavy at the first weighing. But
    no, “L” must be heavy because the other two weighings will prove that. M, A, D, O, I, K, E can not be odd (either heavy or light) because they appear in the other two weighings which show equilibrium. So L is odd, and it must be heavy, since the pan went down.

    Posted February 10, 2012 at 8:03 am | Permalink
  13. @Dom at 8:03 am

    Dom, my point was that “L must be heavy” is not the only conclusion that can be determined for the complete solution of the problem, because your analysis only covered part of the ensemble of possibilities.

    I checked Kevin’s linked-solution (Google rules!) and the summary chart appears to be correct.

    After the first weighing, the more complex of the two possible outcomes is ABCD ≠ EFGH, where “≠” is the “not equal” symbol.

    The key to solving the more complex of the two outcomes (after the first weighing) is its corresponding second weighing, in which mixed combinations of potentially light or heavy coins (along with known “good” ones) are compared using the balance scale.

    Sorry to have high-jacked the conversation, Malcolm.

    Posted February 10, 2012 at 1:52 pm | Permalink
  14. Dom says

    The EA ROCHE Group owns such companies as Genetech, although they usually call themselves ROCHE, but don’t be picky. I can easily see suing them, that is, going to a lawyer. Now if we change the fourth letter by the smallest of strokes, and we get EARACHE, something we’d go to a doctor about.

    Okay, I’m not thinking about this anymore.

    Henry, I still think you’re wrong. If the LIKE pan goes down and the others are in equilibrium, then L is heavy.

    Posted February 10, 2012 at 2:23 pm | Permalink
  15. Dom,

    Using your nomenclature, if the LIKE pan goes down (while the MADO pan goes up) on the first weighing, then one can only conclude, at this point, that the 4 FNTC coins are in “equilibrium”. You are then left with two weighings to determine whether one of the 4 LIKE coins is heavy or one of the 4 MADO coins is light. The conclusion you give, that L must be heavy, is just one of 8 possible outcomes, and you have two weighings left with which to identify the correct 1 of the 8 possibilities.

    Posted February 10, 2012 at 2:51 pm | Permalink
  16. Dom says

    Henry, that’s what I meant. The second two weighings — METO vs FIND, and FAKE vs COIN — are in equilibrium so the four MADO coins are neither heavy nor light, and the three IKE coins are neither heavy nor light. L is the odd coin, and its pan went down, so it must be heavy.

    Look at it this way. There are 24 possibilities: F is light, F is heavy, A is light, A is heavy, and so on.

    Take each possibility and get the configurations of pans at the three weighings. For example, if F is light, we get …

    Weigh 1: Equal
    Weigh 2: Down – Up
    Weigh 3: Up – Down

    If F is heavy we get …

    Weigh 1: Equal
    Weigh 2: Up – Down
    Weigh 3: Down – Up

    … and so on. You’ll find that you get 24 DISTINCT configurations. You can make a table. If you do all three weighings and tell me the configurations (eg, Equal, Up-Down, Down-Up), I can look this up in my table and tell you “F is heavy”.

    Posted February 10, 2012 at 3:24 pm | Permalink
  17. Dom,

    I think what you are missing is that, in order to set up your look-up table, it would require more than a total of 3 weighings. But all of this is weighing me down and giving me a migraine.

    So I will concede that I may, in fact, be missing something. After all, I was wrong once (back in ’76, when I thought I had made a mistake). :)

    Peace out, dude. Obama is awesome!

    Posted February 10, 2012 at 3:47 pm | Permalink
  18. Malcolm says

    OK everybody: I’ve been in Wellfleet on my own for few days, and just picked up the lovely Nina at the train in Providence. As we were driving back I told her about this puzzle, and she got it in about ten minutes:

    redress / redness.

    Posted February 10, 2012 at 6:38 pm | Permalink
  19. All hail the Nina!

    Posted February 10, 2012 at 10:38 pm | Permalink
  20. Malcolm says

    Yep, she’s quite a gal!

    Posted February 10, 2012 at 10:45 pm | Permalink
  21. Malcolm says

    Oh — and a hale and hearty congratulations to you, Nina, with all the trimmings.

    Posted February 11, 2012 at 1:38 pm | Permalink
  22. Dom says

    Bah! I still like EA ROCHE / EARACHE.

    Posted February 11, 2012 at 5:05 pm | Permalink

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