Monday, June 22, 2009

Three Kids Meet Up

Three kids meet up at the park, and out of site of their parents, decide they are best friends forever.  And in accordance with their new resolve, they decide to exchange shoes.  One kid has white shoes, another has black and the third has red shoes.  Each kid had on a matching pair of socks (white shoes-white socks, black shoes-black socks, red shoes-red socks).

After exchanging shoes, each kid has on one shoe from each of the other two kids (They did not exchange socks, to their mother's relief).  At that point, one of the mom's discovers their hiding spot.

How many shoes will the mom have to look at to know which color of shoe each kid is wearing on each foot?  That is, which color shoe each kid wars on his right foot and which color each kids wears on his left foot?  Note that when you look at a shoe, you can see the color sock the kid is wearing.


  1. If the already parents know that the kids aren't wearing their own shoes, they should be able to only look at one foot to figure everything out.

    If you see the white socks has a black left shoe, you already know the right foot has to be red.

    The red sock's left foot has to be white or black, but black is taken, so it must be white.

    The black sock's left foot has to be white. For the right shoe, it has to be red.

    And you can keep going.

  2. I'm sorry, Mike, I just don't get this riddle... I really don't. Sorry! Can you explain later, please?

  3. Jeff got it right.
    There's only 3 colors on each kid. The sock is not changeable so that will determine the other 2 colors.

    Each color shoe is labeled W1, W2, R1, R2, B1 and B2 to identify the left and right ones.

    When the mom sees the sock with one shoe of any color, then the other shoe had to be the third color on the other foot.
    White sock, Red Right Shoe (R1), then
    - W1, W2 (Nope, same color with the sock)
    - R2 (Nope, same color with the first shoe)
    - B1 (Nope, you can't wear the same side of shoe on both feet)
    - B2 (No choice...)

    Then the rest will fall into place automatically...

  4. She probably already knows what color shoes her kid has on and if she knew previously what the other two kids had on then she doesn't even have to see anyone's shoes to know what each one is wearing!

  5. The previous solutions are great if the mother knows that no kid is wearing his own shoes and that no kid is wearing matching shoes.

    If she didn't know that none of the kids wore matching shoes, she'd have to look at two shoes. Basically, she'd have to look at another shoe to eliminate R2 (following Jitzu's example).

    If she didn't have any previous information, she'd have to look at a minimum of four shoes, because the only way she could work out which shoe a kid is wearing is by elimination. If she looked at the wrong four shoes - eg. all three left shoes and one right shoe - she'd have to look at a fifth shoe, because each of the two remaining right shoes could be worn by either kid.

    I'm pretty sure the original answers are right, but the variations make it more interesting!

  6. Sarah>
    She had to look at at least one shoe. If she's not looking at it, then there are two possible answers for her. Let's say her son wore red socks then the possibilities are:

    R: W1 R2
    W: R1 B2
    B: B1 W2
    R: R1 W2
    W: B1 R2
    B: W1 B2

    So, in this case, if not looking but knowing that her son is not wearing his own shoe, how can she know it's RI&W2 or W1&R2?

    At the end, she still had to look at at least one shoe just to determine which side that particular colored shoe her son is putting on.

  7. I say it's 4 shoes. If you say one, you're assuming that (a) the mom is looking at her own kid's shoes first (which she probably would), and that (b) she knows that each kid swapped both shoes, and (c) that the other two kids were wearing matching socks to their shoes (if she were to look at their feet first).
    She wouldn't get the entire combination from looking at just any one shoe. If she looked at any four shoes though, the remaining two would have to be able to be determined by process of elimination. Unless of course, the last two shoes never got put back on...

  8. Bonnie>
    Ok, based on your 3 assumptions
    (a) the mom is looking at her own kid's shoes first (which she probably would), and that
    (b) she knows that each kid swapped both shoes, and
    (c) that the other two kids were wearing matching socks to their shoes (if she were to look at their feet first).

    The number of shoes that she had to look at, is 3 instead of 4. Here's the explanation:

    [W,R,B - Socks
    Wr,Rr,Br - Right Shoes
    Wl,Rl,Bl - Left Shoes]

    Mum arrived, look at her son's shoes (White socks)

    W: Br, Rl

    Then she had to look at another shoe, any shoe, let's say Rr (which is with black socks kid)

    B: Rr, ?

    Then she would know where is Bl, because of assumption (b). The black socks kid had swapped his left shoe and he won't be wearing Br on his left foot. So, Bl is with Red socks kid.

    R: ?, Bl

    And now there are a left shoe and a right shoe remaining, which will fall into place without any prediction. :) So, all she need is just 3 shoes, if she doesn't know their agreement on how to swap their shoes.

  9. Ops, didn't see that your assumptions are on mine.
    Anyway, my answer just need to based on 2 simple assumptions instead of 3:
    (a) The mom knows that there are 3 sets of socks and shoes (Each set has a matching color for socks and shoes but different color from other sets)
    (b) They swap both of their shoes.

    She doesn't have to look at his son first, any one shoe will do.

  10. borin sorry mike!

  11. This has been great. Lots of variations on the answer that have been well thought out.

    I don't know how familiar the rest of you are with the blogging system, but I get an e-mail sent to me whenever someone posts an answer here. It's been great reading over the back and forth throughout the day.

    I was going for just one sock, but that depends on the assumptions I was making (as you have all pointed out!).


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