Here's yet another puzzle adapted from a puzzle book (in my case, with edits to make some of the specifications of the puzzle more clear because I didn't really understand them the first time I read it).

Five professors are spending three days on a retreat, and every day they eat dinner at the same restaurant. This restaurant has a special "roulette menu" with nine items, in which you choose a menu item from 1 to 9 that corresponds to one of the nine available items, but you don't know which number corresponds to which item. When the items are ordered, they are brought to the table in no particular order, so if a group orders multiple items at a time, they cannot figure out individually which item was ordered by whom.

These professors have tasked themselves with figuring out which number corresponds to which item in three days regardless. Each professor will order one item per day. How should they plan their ordering of the items in order to find out which number corresponds to which item for all items?

  • 1
    $\begingroup$ To make the puzzle work as intended, I think it's also necessary to specify that the orders at each meal are "scrambled" so that no individual dish is associated with a single professor. Otherwise they could identify five dishes per meal: e.g. the first professor orders dish #1 and knows that whatever dish he gets is #1, and so on. $\endgroup$
    – David Z
    Dec 25, 2014 at 15:15
  • 1
    $\begingroup$ Real-world application: Chinese restaurants $\endgroup$
    – Lopsy
    Dec 25, 2014 at 23:50
  • $\begingroup$ @Lopsy: Assuming you don't know Chinese, anyway. $\endgroup$
    – user88
    Dec 26, 2014 at 3:37
  • $\begingroup$ @DavidZ That's what I meant with the "no particular order" bit. $\endgroup$
    – user88
    Dec 26, 2014 at 3:38
  • $\begingroup$ @JoeZ. ah, well I didn't get that from your wording. It's possible - in fact it is exactly how most restaurants work - to bring the dishes to the table in arbitrary order (i.e. it's arbitrary which one comes first, which one comes second) and still have each one placed in front of the professor who ordered it. $\endgroup$
    – David Z
    Dec 26, 2014 at 19:14

3 Answers 3


Okay, 15 orders, 9 dishes, that means 6 dupes. So 3 dishes will only be ordered once, obviously never on the same day. Now to split the six dupes so they can all be detected.

The first 3:

Day 1: 1   3
Day 2: 1 2
Day 3:   2 3

The dish that shows on day 1 and 2 is #1, and so on.

Now add in same-day pairs:

Day 1: 1   3 4 4
Day 2: 1 2   5 5 
Day 3:   2 3 6 6

So the dish that's doubled on day 1 is #4, etc.

And now the last three:

Day 1: 1 7 3 4 4
Day 2: 1 2 8 5 5 
Day 3: 9 2 3 6 6

The dish that only shows up on day 1 is #7, and so on.

And then, secure in their triumph, they can leave five bad Yelp reviews warning people away from this crazy place.

  • 4
    $\begingroup$ Very nicely explained. On the contrary to the Yelp reviews, though: now they can publish and sell a guidebook! $\endgroup$
    – jscs
    Dec 25, 2014 at 3:59
  • 2
    $\begingroup$ Gah beat me by seconds! :) $\endgroup$
    – McMagister
    Dec 25, 2014 at 4:00
  • $\begingroup$ Thanks very much, Josh. And better luck next time, McMagister.:) $\endgroup$
    – Len Pitre
    Dec 25, 2014 at 6:11
  • 7
    $\begingroup$ They could also identify a 10th item that nobody orders. That is assuming the set of dishes is known but not their numbers. $\endgroup$
    – Florian F
    Dec 26, 2014 at 13:25
  • $\begingroup$ @McMagister - For once somebody beat you to it! :-) $\endgroup$ Dec 28, 2014 at 19:30

Assumption: Names of the dishes is known. What's not known if their mapping to numbers 1 to 9

Day 1: 1 2 2 3 3


  • Dish that is served once is 1


  • which of the other two is 2 and 3

Day 2: 2 4 4 5 6


  • Between 2 and 3 of yesterday, dish that is served today as well is 2 and other is 3
  • Dish that is served twice today is 4


  • which of the other two is 5 and 6

Day 3: 5 7 7 8 X


  • Between 5 and 6 of yesterday, dish that is served today as well is 5 and other is 6
  • Dish served twice is 7
  • Dish served once that was not served yesterday is 8
  • Dish that was not served on any of these days is 9
  • $\begingroup$ Could you please specify what you mean by X on Day 3. Keep in mind that the puzzle states, “Each professor will order one item per day.“ $\endgroup$ Jul 3 at 2:50
  • $\begingroup$ By X, I meant 5th professor does not need to order anything on 3rd day and that would still provide desired results. If it is required to order something then 5th professor can order anything that has already been determined on 3rd day (Dish 1, 2, 3, 4) 5th professor should not order Dish 6, 8 or 9 (If they order 6, then we can't determine which is 5 and 6. If 5th prof orders Dish 8 then there are two pairs (7 and 8), so we won't be able to determine which one is which. And if 5th prof orders 9 then we can't determine which is Dish 8 and 9) $\endgroup$
    – Avi Dubey
    Jul 3 at 21:01

One important point to remember is that the same dish can be ordered by more than 1 professor on a particular day. For example, on day 1, two or more professors can order dish number 5. This is a very important point. Here is why.

Let's assume that it is never the case that the same dish is ordered by more than one professor on a particular day.

Now, let's say that the dishes ordered on day 1 are dish no. 1 2 3 4 and 5.

To identify these dishes numbered 1 2 3 4 5 , some of these 5 dishes will need to be ordered again.

Case 1: None of these dishes are ordered on day 2. It won't be possible to figure out the number for all these 5 dishes.

Case 2: One of these dishes was ordered on day 2. Again, it won't be possible to identify all the 5 dishes.

In fact, there is no way to identify all these 5 dishes if the restriction is that no dish can be ordered more than once on the same day.

So, the professors will have to order some dishes more than once on the same day.

Day 1

Dishes ordered : 1 1 2 3 4
This wlll help us identify dish no. 1

Day 2

2 55 6 7 helps us identify 2 and 5

Day 3

3 88 9 6 helps us identify 3 and 4, 6 and 7 and 8 and 9.



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