What are the mathematical/computational principles behind this game?
My kids have this fun game called Spot It! The game constraints (as best I can describe) are:
The principle of the game is: flip over 2 cards and whoever first picks the matching picture gets a point.
Here's a picture for clarification:
(Example: you can see from the bottom 2 cards above that the matching picture is the green dinosaur. Between the bottom-right and middle-right picture, it's a clown's head.)
I'm trying to understand the following:
What are the minimum number of different pictures required to meet these criteria and how would you determine this?
Using pseudocode (or Ruby), how would you generate 55 game cards from an array of N pictures (where N is the minimum number from question 1)?
Update:
Pictures do occur more than twice per deck (contrary to what some have surmised). See this picture of 3 cards, each with a lightning bolt: 
Finite Projective Geometries
The axioms of projective (plane) geometry are slightly different than the Euclidean geometry:
Now, add "finite" into the soup and you have the question:
Can we have a geometry with just 2 points? With 3 points? With 4? With 7?
There are still open questions regarding this problem but we do know this:
Q points, then Q = n^2 + n + 1 and n is called the order of the geometry. n+1 points in every line. n+1 lines. Total number of lines is also Q .
And finally, if n is prime, then there does exists a geometry of order n .
What does that have to do with the puzzle, one may ask.
Put card instead of point and picture instead of line and the axioms become:
Now, lets take n=7 and we have the order-7 finite geometry with Q = 7^2 + 7 + 1 . That makes Q=57 lines (pictures) and Q=57 points (cards). I guess the puzzle makers decided that 55 is more round number than 57 and left 2 cards out.
We also get n+1 = 8 , so from every point (card), 8 lines pass (8 pictures appear) and every line (picture) has 8 points (appears in 8 cards).
Here's a representation of the most famous finite projective (order-2) plane (geometry) with 7 points, known as Fano Plane , copied from Noelle Evans - Finite Geometry Problem Page
I was thinking of creating an image that explain how the above order-2 plane could be made a similar puzzle with 7 cards and 7 pictures, but then a link from the math.exchange twin question has exactly such a diagram: Dobble-et-la-geometrie-finie
So there are k=55 cards containing m=8 pictures each from a pool of n pictures total. We can restate the question 'How many pictures n do we need, so that we can construct a set of k cards with only one shared picture between any pair of cards?' equivalently by asking:
Given an n-dimensional vector space and the set of all vectors, which contain exactly m elements equal to one and all other zero, how big has n to be, so that we can find a set of k vectors, whose pairwise dot products are all equal to 1?
There are exactly (n choose m) possible vectors to build pairs from. So we at least need a big enough n so that (n choose m) >= k. This is just a lower bound, so for fulfilling the pairwise compatibility constraint we possibly need a much higher n.
Just for experimenting a bit i wrote a small Haskell program to calculate valid card sets:
Edit: I just realized after seeing Neil's and Gajet's solution, that the algorithm i use doesn't always find the best possible solution, so everything below isn't necessarily valid. I'll try to update my code soon.
module Main where
cardCandidates n m = cardCandidates' [] (n-m) m
cardCandidates' buildup 0 0 = [buildup]
cardCandidates' buildup zc oc
| zc>0 && oc>0 = zerorec ++ onerec
| zc>0 = zerorec
| otherwise = onerec
where zerorec = cardCandidates' (0:buildup) (zc-1) oc
onerec = cardCandidates' (1:buildup) zc (oc-1)
dot x y = sum $ zipWith (*) x y
compatible x y = dot x y == 1
compatibleCards = compatibleCards' []
compatibleCards' valid [] = valid
compatibleCards' valid (c:cs)
| all (compatible c) valid = compatibleCards' (c:valid) cs
| otherwise = compatibleCards' valid cs
legalCardSet n m = compatibleCards $ cardCandidates n m
main = mapM_ print [(n, length $ legalCardSet n m) | n<-[m..]]
where m = 8
The resulting maximum number of compatible cards for m=8 pictures per card for different number of pictures to choose from n for the first few n looks like this:
This brute force method doesn't get very far though because of combinatorial explosion. But i thought it might still be interesting.
Interestingly, it seems that for given m, k increases with n only up to a certain n, after which it stays constant.
This means, that for every number of pictures per card there is a certain number of pictures to choose from, that results in maximum possible number of legal cards. Adding more pictures to choose from past that optimal number doesn't increase the number of legal cards any further.
The first few optimal k's are:
I just found a way to do it with 57 or 58 pictures but now I have a very bad headache, I'll post the ruby code in 8-10 hours after I slept well! just a hint my my solution every 7 cards share same mark and total 56 cards can be constructed using my solution.
here is the code that generates all 57 cards that ypercube was talking about. it uses exactly 57 pictures, and sorry guy's I've written actual C++ code but knowing that vector <something> is an array containing values of type something it's easy to understand what this code does. and this code generates P^2+P+1 cards using P^2+P+1 pictures each containing P+1 picture and sharing only 1 picture in common, for every prime P value. which means we can have 7 cards using 7 pictures each having 3 pictures(for p=2), 13 cards using 13 pictures(for p=3), 31 cards using 31 pictures(for p=5), 57 cards for 57 pictures(for p=7) and so on...
#include <iostream>
#include <vector>
using namespace std;
vector <vector<int> > cards;
void createcards(int p)
{
cards.resize(0);
for (int i=0;i<p;i++)
{
cards.resize(cards.size()+1);
for(int j=0;j<p;j++)
{
cards.back().push_back(i*p+j);
}
cards.back().push_back(p*p+1);
}
for (int i=0;i<p;i++)
{
for(int j=0;j<p;j++)
{
cards.resize(cards.size()+1);
for(int k=0;k<p;k++)
{
cards.back().push_back(k*p+(j+i*k)%p);
}
cards.back().push_back(p*p+2+i);
}
}
cards.resize(cards.size()+1);
for (int i=0;i<p+1;i++)
cards.back().push_back(p*p+1+i);
}
void checkCards()
{
cout << "---------------------n";
for(unsigned i=0;i<cards.size();i++)
{
for(unsigned j=0;j<cards[i].size();j++)
{
printf("%3d",cards[i][j]);
}
cout << "n";
}
cout << "---------------------n";
for(unsigned i=0;i<cards.size();i++)
{
for(unsigned j=i+1;j<cards.size();j++)
{
int sim = 0;
for(unsigned k=0;k<cards[i].size();k++)
for(unsigned l=0;l<cards[j].size();l++)
if (cards[i][k] == cards[j][l])
sim ++;
if (sim != 1)
cout << "there is a problem between cards : " << i << " " << j << "n";
}
}
}
int main()
{
int p;
for(cin >> p; p!=0;cin>> p)
{
createcards(p);
checkCards();
}
}
again sorry for the delayed code.
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