24

I recently learned about the 24 game. There are several variants of it, but the version I learned goes like this:

Take out the face cards and jokers from a deck and deal four cards (aces or number cards). Aces can be used either as 1 or as 11. Using addition, subtraction, multiplication, and division, with any grouping of operations (paranthesis can be added anywhere), try to come up with an expression that uses all four cards once and equals 24. Division is fractional, so 5 / 2 is 2.5.

For example for A 4 5 8 we have \((1 + 5) * (8 - 4)\).

There is the problem of implementing an algorithm to find a solution given the cards as input.

Brute force

A simple solution is to try all permutations of cards, all possible operations, and all possible groupings.

In the general case, there are 24 ways to arrange the cards - permutations of 4 which is 4!. Each ace doubles this number as we need to consider the case in which we use it as 1 and the case in which we use it as 11.

There are 64 ways to combine operations, since we have 4 operations in 3 possitions, which means \(4^3\) total.

We also have 5 possible groupings:

\[ \begin{align}\begin{aligned}(x_0 \odot x_1) \oplus (x_2 \otimes x_3)\\((x_0 \odot x_1) \oplus x_2) \otimes x_3\\(x_0 \odot (x_1 \oplus x_2)) \otimes x_3\\x_0 \odot ((x_1 \oplus x_2) \otimes x_3)\\x_0 \odot (x_1 \oplus (x_2 \otimes x_3))\end{aligned}\end{align} \]

where \(\odot, \oplus, \otimes\) are placeholders for any operators (they could potentially be the same operator).

A simple solver implementation looks like this:

from itertools import permutations, product
from sys import argv

# Transform arguments to numbers, replace 'A' and 'a' with [1, 11]
args = sum([[1, 11] if arg in 'aA' else
    [int(arg)] for arg in argv[1:5]], [])

# For every permutation of 4 arguments
for xs in permutations(args, 4):
    # If we have more 1s and 11s then aces ignore this permutation
    if sum(n == 1 or n == 11 for n in xs) > len(args) - 4:
        continue

    # For every possible combination of 3 operators
    for ops in product('+-*/', repeat=3):
        # For every possible grouping
        for exp in ['({} {} {}) {} ({} {} {})',
                    '(({} {} {}) {} {}) {} {}',
                    '({} {} ({} {} {})) {} {}',
                    '{} {} ({} {} ({} {} {}))',
                    '{} {} (({} {} {}) {} {})']:

            # Place operands and operators in expression
            exp = exp.format(xs[0], ops[0], xs[1], ops[1], xs[2],
                ops[2], xs[3])

            try:
                # If expression evaluates to 24, we found a solution
                if eval(exp) == 24:
                    print(exp)
                    exit()
            except ZeroDivisionError:
                # Ignore division by zero errors
                pass

# If we get here we tried all combinations and couldn't
# find any solution
print('No solution')

We have ten possible cards (ace and number cards), and taking combination with repetition of 4 cards, we have \(\frac{(n + r - 1)!}{r! * (n - 1)!}\) for \(n = 10, r = 4\), so a total of \(\frac{(10 + 4 - 1)!}{4! * (10 - 1)!} = 715\) possible games.

Feeding all possible games to the code above, we can see that there are 117 games which have no solution. The remaining 598 games are solvable.

We can optimize the above solution further by observing that we only need three of the five groupings to cover all cases. Take, for example, \(x_0 \odot ((x_1 \oplus x_2) \otimes x_3)\). Now if \(\odot\) is a commutative operation (addition or multiplication), we can rewrite this to the equivalent \(((x_1 \oplus x_2) \otimes x_3) \odot x_0\), and since we any way take all permuations of arguments and operators, this ends up getting covered by the \(((x_0 \odot x_1) \oplus x_2) \otimes x_3\) case. For non-commutative operations, for example subtraction, notice that if we do have a solution \(x_0 - ((x_1 \oplus x_2) \otimes x_3) = 24\), since \(x_0\) is at most 11, it means we need to subtract a negative number from it in order to get 24. This implies that at least \(\oplus\) or \(\otimes\) is also a subtraction. If \(\otimes\) is a subtraction, we can rewrite the expression \(x_0 - ((x_1 \oplus x_2) - x_3)\) as \((x_0 + x_3) - (x_1 \oplus x_2)\). If \(\otimes\) is not a subtraction but \(\oplus\) is, we have \(x_0 - ((x_1 - x_2) \otimes x_3)\) which is equivalent with \(x_0 - (- (x_2 - x_1) \otimes x_3)\). If \(\otimes\) is addition, this becomes \(x_0 - (x_3 - (x_2 - x_1))\) = \((x_0 - x_3) + (x_2 - x_1)\). If \(\otimes\) is multiplication or division, this becomes \(x_0 - (- (x_2 - x_1) \otimes x_3)\) = \(x_0 + ((x_2 - x_1) \otimes x_3)\) = \(((x_2 - x_1) \otimes x_3) + x_0\).

Similar rewrites can be done if \(\odot\) is division by observing that we would have to divide with a fractional number in order to get 24, so at least one of \(\oplus\) or \(\otimes\) is also a division. This means we only need the groupings

\[ \begin{align}\begin{aligned}(x_0 \odot x_1) \oplus (x_2 \otimes x_3)\\((x_0 \odot x_1) \oplus x_2) \otimes x_3\\(x_0 \odot (x_1 \oplus x_2)) \otimes x_3\end{aligned}\end{align} \]

to find all possible solutions. Our solution becomes:

from itertools import permutations, product
from sys import argv

args = sum([[1, 11] if arg in 'aA' else
    [int(arg)] for arg in argv[1:5]], [])

for xs in permutations(args, 4):
    if sum(n == 1 or n == 11 for n in xs) > len(args) - 4:
        continue

    for ops in product('+-*/', repeat=3):
        for exp in ['({} {} {}) {} ({} {} {})',
                    '(({} {} {}) {} {}) {} {}',
                    '({} {} ({} {} {})) {} {}']:

            exp = exp.format(xs[0], ops[0], xs[1], ops[1], xs[2],
                ops[2], xs[3])

            try:
                if eval(exp) == 24:
                    print(exp)
                    exit()
            except ZeroDivisionError:
                pass

print('No solution')

This means that for games without aces, we need to check 24 (\(4!\)) permutations of cards, 64 (\(4^3\)) combinations of operators, and 3 groupings. That is \(4! * 4^3 * 3 = 4608\) tests. For games with aces, we double this number for each ace to account for both the 1 and 11 cases.

Minimizing Number of Tests

A more interesting question is what is the minimum number of tests we need to perform in order to correctly find a solution for all solvable games.

It is obvious that there are expressions which can never evaluate to 24 for any game. For example \((x_0 - x_1) - x_2) - x_3\), since \(x_i \in \{1, 2, ... 11\}\).

It is also obvious that we perform a lot of redundant tests, since, for example, all of the below expressions are equivalent for all possible inputs:

\[ \begin{align}\begin{aligned}(x_0 + x_1) + (x_2 + x_3)\\((x_0 + x_1) + x_2) + x_3\\(x_0 + (x_1 + x_2)) + x_3\\(x_0 + x_1) + (x_3 + x_2)\\((x_0 + x_1) + x_3) + x_2\\(x_0 + (x_1 + x_3)) + x_2\\...\end{aligned}\end{align} \]

and so on for all permutations of \(x_0, x_1, x_2, x_3\).

Let’s generate all possible permutations of cards, combinations of operators, and groupings as above:

import itertools

operands = list(itertools.permutations(range(4), 4)) # 24 of these
operators = list(itertools.product('+-*/', repeat=3)) # 64 of these
groupings = ['({} {} {}) {} ({} {} {})',
             '(({} {} {}) {} {}) {} {}',
             '({} {} ({} {} {})) {} {}'] # 3 of these

Note that here we are looking at all possible games so operands are permutations of indexes from 0 to 3, not actual cards. We can also take all possible games as combinations of 4 numbers from 1 to 11. Here we generate 1 and 11 games for each ace, so we end up with 1001 possible games instead of 715:

inputs = list(itertools.combinations_with_replacement(range(1, 12), 4))
# 1001 of these

We can now write a function that, for a given game, generates all possible expressions which evaluate to 24:

def solutions_for(inp):
    for xs in operands:
        for ops in operators:
            for exp in groupings:
                try:
                    if eval(exp.format(inp[xs[0]], ops[0],
                            inp[xs[1]], ops[1], inp[xs[2]],
                            ops[2], inp[xs[3]])) == 24:
                        yield exp.format(f'x{xs[0]}', ops[0],
                            f'x{xs[1]}', ops[1], f'x{xs[2]}',
                            ops[2], f'x{xs[3]}')
                except ZeroDivisionError:
                    pass

This is very similar with the initial solution, except that we don’t have to worry about aces (we assume they are already converted to either 1 or 11), and we use the permutations of indexes to determine the order of terms as input is going to always be in increasing order (as generated by itertools.combinations_with_replacement). So instead of placing inp[0], op[0], inp[1], op[1], inp[2], op[2], inp[3] in the expression to be evaluated as we did in the initial solution, since inp is fixed, we come up with permutations of operands from operands, so we are placing inp[xs[0]], op[0], inp[xs[1]], op[1], inp[xs[2]], op[2], inp[xs[3]] in the expression instead. We also return the expression replacing the operands with x0, x1, x2, x3 since we don’t care about their particular values, rather the expression we are using.

For example calling:

list(solutions_for([2, 4, 7, 8]))

yields

['((x0 * x2) - x3) * x1', '(x1 * x2) - (x3 / x0)',
 '((x2 * x0) - x3) * x1', '((x2 / x0) * x3) - x1',
 '(x2 / (x0 / x3)) - x1', '(x2 * x1) - (x3 / x0)',
 '((x2 * x3) / x0) - x1', '(x2 * (x3 / x0)) - x1',
 '((x3 / x0) * x2) - x1', '(x3 / (x0 / x2)) - x1',
 '((x3 * x2) / x0) - x1', '(x3 * (x2 / x0)) - x1']

These are all possible expression which evaluate to 24 for the game 2 4 7 8.

We can compute the list of all expressions which evaluate to 24 for every possible game:

results = []
for inp in inputs:
    result = set(solutions_for(inp))

    # Only append the set of expressions to the list if
    # non-empty (if game has at least one solution)
    if result:
        results.append(result)

We can take the union of the sets in results and get the set of all expressions that evaluate to 24 for at least one game:

expressions = set()
for result in results:
    expressions = expressions.union(result)

The size of this set is 1809. We are guaranteed that for any possible game, no other expression evaluates to 24 since we generated all possible solutions for all possible games. Which means we can test just these 1809 expression for any game and determine whether it is solvable or not, which is better than our original 4608 (or more for games with aces).

Here we eliminated all expressions which never evaluate to 24, but we still have all the redundant tests in our set of expressions. It is also possible to have an expression \(E_0\) which solves all games some expression \(E_1\) solves, plus some other games. In which case we wouldn’t ever need to test using \(E_1\) since \(E_0\) would still solve all games that \(E_1\) would solve.

More formally, expressions is our universe \(\mathcal{U}\) of tests and results is a set of sets \(R = \{ R_0, R_1 ... R_n \}\) where \(R_i \subset \mathcal{U} \space \forall i \in \{ 0, 1 ... n \}\). We want to find the smallest set \(H \subset \mathcal{U}\) such that \(H \cap R_i \neq \varnothing \space \forall i \in \{ 0, 1 ... n \}\).

The good news is that this is actually a well known problem called the hitting set problem [1]. The bad news is this problem is NP-hard. Even with clever pruning, trying out combinations of expressions to find the smallest \(H\) has factorial complexity and even for small sets it quickly reaches astronomical numbers.

Approximation

Since finding an optimal solution is too computationally expensive, we can at least attempt to find a good enough solution.

The greedy algorithm which solves the hitting set problem works as follows: build up the solution by selecting at each step the element which hits the highest number of sets which were not hit so far.

# Start with an empty set
min_expressions = set()

# While we have unhit sets
while results:
    min_expression, max_hitting = set(), 0

    # For each expression in our universe
    for expr in expressions:
        hitting = sum([1 for result in results if expr in result])
        if hitting > max_hitting:
            min_expression, max_hitting = expr, hitting

    # We found the expression hitting most unhit sets
    min_expressions.add(min_expression)

    # Remove hit sets from results
    results = [result for result in results if
        min_expression not in result]

Interestingly enough, since we are working with sets and hashing is randomized in Python, I got different results across different runs of this algorithm. For cases where there are multiple max hitting sets (sets intersecting the same number of other sets), we non-deterministically select one, since iteration over sets is based on the randomized key order. I got solutions ranging from 110 to 114 expressions. This gives us an upper bound of 110 - we must perform at most 110 tests to find a solution for a game.

We can use the above code to generate a set of expressions and dump it into a source file, together with the code to test input:

from sys import argv

expressions = [
    "(x3 + x1) * (x2 - x0)", "((x2 * x3) - x0) / x1",
    "(x3 + (x1 + x0)) * x2", "(x0 - x3) * (x1 - x2)",
    "(x1 * x0) + (x2 - x3)", "((x2 * x0) * x3) - x1",
    "(x3 + (x1 + x2)) + x0", "(x2 * x0) + (x3 - x1)",
    "((x0 / x1) * x2) * x3", "(x1 * (x3 - x2)) + x0",
    "((x3 + x2) - x0) + x1", "(x0 * x1) * (x3 - x2)",
    "(x1 * x0) - (x2 + x3)", "(x3 / x1) * (x2 + x0)",
    "((x2 + x1) + x3) * x0", "(x3 - x0) / (x1 / x2)",
    "(x3 + x1) * (x2 / x0)", "(x1 * x3) + (x0 * x2)",
    "((x3 * x1) + x2) + x0", "((x0 - x2) + x1) * x3",
    "((x3 / x0) + x2) * x1", "(x3 + (x2 * x1)) - x0",
    "(x1 * x3) - (x2 - x0)", "(x1 * (x3 - x0)) - x2",
    "((x2 + x1) * x3) / x0", "(x0 * x3) - (x1 + x2)",
    "(x1 + (x3 * x0)) * x2", "(x0 * (x2 + x3)) - x1",
    "(x1 * x3) - (x0 + x2)", "(x3 - (x2 / x1)) * x0",
    "(x0 - (x1 / x2)) * x3", "(x3 * x0) + (x2 - x1)",
    "((x3 / x0) + x1) + x2", "(x3 + (x1 * x2)) + x0",
    "(x0 + x2) * (x1 + x3)", "(x0 * (x3 - x1)) + x2",
    "((x0 + x3) * x1) * x2", "(x2 * (x0 + x3)) / x1",
    "(x2 - (x1 + x0)) * x3", "(x3 * x2) - (x0 / x1)",
    "((x0 + x3) * x1) + x2", "((x0 * x3) + x1) - x2",
    "(x2 + (x3 - x1)) * x0", "(x2 * (x0 + x1)) + x3",
    "((x2 + x3) * x0) + x1", "(x0 - (x3 / x2)) * x1",
    "((x0 + x2) - x3) * x1", "((x3 / x2) + x0) * x1",
    "(x1 / x0) + (x2 + x3)", "((x1 * x0) - x2) * x3",
    "((x0 + x1) + x2) * x3", "(x2 * (x3 - x1)) - x0",
    "((x2 * x1) - x0) * x3", "((x3 - x1) * x2) + x0",
    "(x3 / (x0 + x1)) * x2", "((x1 * x0) + x3) + x2",
    "(x3 + x0) * (x2 - x1)", "(x2 - x0) * (x3 - x1)",
    "((x3 / x0) * x2) + x1", "((x2 * x1) - x0) - x3",
    "(x0 + (x3 - x1)) * x2", "(x0 * x2) / (x3 - x1)",
    "((x3 * x2) - x1) / x0", "(x2 - (x0 / x3)) * x1",
    "(x3 - x2) * (x0 + x1)", "(x0 * x2) - (x3 + x1)",
    "((x2 - x0) * x3) + x1", "(x1 * (x3 - x0)) + x2",
    "(x1 * (x0 + x3)) - x2", "((x0 + x1) * x3) + x2",
    "((x1 - x2) + x3) * x0", "((x3 - x1) * x2) * x0",
    "((x2 + x1) - x3) * x0", "(x1 + (x3 / x2)) * x0",
    "(x2 / (x0 / x1)) + x3", "(x2 / (x3 - x0)) * x1",
    "(x3 * x1) - (x0 * x2)", "((x1 + x0) * x2) * x3",
    "(x1 - (x2 / x3)) * x0", "(x2 + (x3 * x0)) + x1",
    "((x2 * x3) + x1) / x0", "(x3 - x0) * (x2 + x1)",
    "(x1 * x3) + (x2 - x0)", "(x3 * (x2 - x0)) - x1",
    "(x0 + (x1 - x3)) * x2", "(x1 * (x3 - x2)) - x0",
    "(x0 + (x2 * x3)) - x1", "(x2 + x3) / (x0 / x1)",
    "(x0 * x3) / (x2 - x1)", "(x2 - (x3 / x0)) * x1",
    "(x0 - (x1 - x2)) * x3", "(x3 + x1) + (x2 * x0)",
    "((x3 - x1) - x0) * x2", "(x1 * (x2 - x0)) - x3",
    "(x2 + x0) * (x3 - x1)", "((x3 - x1) * x2) / x0",
    "((x1 * x3) - x2) * x0", "((x1 + x3) * x0) - x2",
    "((x3 - x2) + x0) * x1", "((x2 * x0) - x3) * x1",
    "(x2 * (x1 + x0)) - x3", "(x0 + (x1 / x2)) * x3",
    "((x2 - x1) * x3) + x0", "((x2 / x1) + x3) * x0",
    "(x1 * x0) - (x2 / x3)", "((x3 * x0) - x1) * x2",
    "((x2 - x0) * x1) + x3", "((x1 + x0) * x3) - x2",
    "(x1 * x0) / (x3 - x2)", "(x3 * (x1 - x0)) - x2"
]

# Since we no longer try all permutations of cards, we
# need to split inputs containing aces by replacing aces
# with both 1 and 11
def get_input(args):
    if 'A' not in args:
        return [args]

    idx = args.index('A')
    return get_input(args[:idx] + ['1'] + args[idx+1:]
        ) + get_input(args[:idx] + ['11'] + args[idx+1:])

for args in get_input(argv[1:5]):
    # We also expect inputs to be in sorted order now
    args = sorted([int(arg) for arg in args])

    for exp in expressions:
        for i in range(4):
            # Replace x0 ... x3 with args[0] ... args[3]
            exp = exp.replace('x' + str(i), str(args[i]))
        try:
            if eval(exp) == 24:
                print(exp)
                return
        except ZeroDivisionError:
            pass

print('No solution')

We can test this by ensuring that we still see 117 games without solution when we try to solve all 715 games, which is indeed the case. We reduced the number of tests we perform on a game from 4608 to 110.

Unique Expressions

There are a couple of other interesting facts we can determine from the set of all solutions for all games: the set of unique expressions and out of them, the subset of expressions which are absolutely required in order to solve all possible games.

By unique expression I mean picking an expression and eliminating all other equivalent expressions from the set (for example keeping only \(((x_0 + x_1) + x_2) + x_3\) while dropping all other permutations and groupings of addition). We can easily do this by defining equivalent expressions as expressions which solve the exact same games. So if \(E_1\) solves some subset \(R_{E_1}\) of \(R\), an expression \(E_2\) is equivalent to it if the set \(R_{E_2}\) of games it solves is equal to \(R_{E_1}\). We can define equivalence as:

def equivalent(exp1, exp2):
    for result in results:
        if exp1 in result and exp2 not in result:
            return False
        if exp1 not in result and exp2 in result:
            return False
    return True

With this function, we can go over our univese \(\mathcal{U}\) of expressions and eliminate all expressions which are the equivalent of another expression:

expressions = list(expressions)
i1, i2 = 0, 0
while i1 < len(expressions):
    i2 = i1 + 1
    while i2 < len(expressions):
        if equivalent(expressions[i1], expressions[i2]):
            expressions = expressions[:i2] + expressions[i2+1:]
        else:
            i2 += 1
    i1 += 1
expressions = set(expressions)

The resulting set of unique expressions has 273 elements. This means these 273 expressions are enough to solve all possible games and, more so, adding any other expression to this set would be redundant. This is a lower bound than our original 1809 expressions which solve games, but higher than our previously found bound of 110 expressions. Note that this data point is not useful in the greedy algorithm shown in the previous section, since once that algorithm picks an expression, it would automatically discard all other equivalent expressions, since it eliminates all games which are solved by the picked expression from the search space. This should put the problem in combinatorial perspective though, as we need to select from 273 candidates to find the smallest hitting set.

Once we have eliminated equivalent expressions, we can update the set of game results accordingly, by intersecting each \(R_i\) with our new \(\mathcal{U}\):

results = [list(expressions.intersection(result)) for result in results]

Now we can search \(R\) for sets with a single element. This will give us expressions which must be part of our solution, otherwise eliminating them would cause a solvable game to appear as unsolvable:

min_expressions, games_with_unique_result = set(), 0
for result in results:
    if len(result) == 1:
        min_expressions = min_expressions.union(result[0])
        games_with_unique_result += 1

Turns out there are 62 expressions which (ignoring equivalences) provide unique solutions to games. They are:

{'(x1 + x2) - (x0 - x3)', '(x1 + x0) * (x2 * x3)',
 '((x1 + x0) - x2) * x3', '((x3 + x2) * x0) + x1',
 '(x0 + x3) * (x2 - x1)', '((x2 * x1) / x0) + x3',
 '((x3 - x1) * x2) - x0', '(x1 - (x3 / x2)) * x0',
 '((x1 * x0) - x2) - x3', '((x2 - x1) - x0) * x3',
 '((x1 * x0) - x3) + x2', '(x2 + (x1 / x0)) + x3',
 '((x1 + x3) - x2) * x0', '(x2 - (x3 - x0)) * x1',
 '(x2 + (x1 + x3)) + x0', '(x1 * x0) - (x2 / x3)',
 '(x3 + (x2 * x1)) - x0', '((x0 * x1) + x3) + x2',
 '(x2 + (x1 - x3)) * x0', '((x0 * x1) - x2) * x3',
 '(x2 * x0) + (x1 + x3)', '(x2 + (x3 * x0)) + x1',
 '((x0 * x3) - x1) + x2', '(x0 + (x2 - x1)) * x3',
 '((x3 * x1) - x2) - x0', '(x1 - x2) * (x0 - x3)',
 '((x3 - x0) - x1) * x2', '(x1 * (x2 - x0)) - x3',
 '(x0 / (x3 - x2)) * x1', '((x0 + x1) - x3) * x2',
 '((x3 - x1) + x0) * x2', '((x1 + x0) * x3) + x2',
 '((x1 + x0) * x3) - x2', '(x3 / x1) * (x2 + x0)',
 '((x3 - x1) * x0) + x2', '(x3 * (x2 - x0)) + x1',
 '((x0 + x2) + x1) * x3', '(x2 * (x0 + x1)) - x3',
 '(x2 * x1) / (x3 - x0)', '(x3 + x0) + (x1 * x2)',
 '(x0 + (x3 * x2)) - x1', '(x0 - x2) * (x1 - x3)',
 '((x2 + x3) * x1) / x0', '(x2 / x0) * (x3 + x1)',
 '(x3 * (x1 - x0)) - x2', '((x3 + x1) * x0) - x2',
 '((x0 + x3) * x1) - x2', '((x3 + x2) * x0) - x1',
 '(x3 + x0) / (x1 / x2)', '(x0 + x2) * (x3 - x1)',
 '(x2 * (x3 - x1)) + x0', '((x0 * x3) - x2) - x1',
 '((x1 * x2) - x3) - x0', '(x3 - (x1 - x2)) * x0',
 '(x2 * x3) / (x1 + x0)', '(x3 - x2) * (x0 + x1)',
 '(x0 / (x3 - x1)) * x2', '((x2 * x3) - x0) / x1',
 '(x2 * (x3 - x0)) / x1', '((x1 + x0) * x2) + x3',
 '(x1 * (x2 - x0)) + x3', '(x3 * (x2 - x1)) + x0'}

This is a lower bound for our problem, since at the very least we need these expressions in order to correcly solve all possible games. We also computed the number of games with a unique solution, which is 122. The remaining games are either unsolvable or admit multiple solutions. Note we considered aces as 1s and aces as 11s as distinct games in the above analysis. We could search for equivalent games (number of 1s and 11s is the same) and see if we can eliminate some expressions from the list above. This is left as an exercise to the reader.

Summary

  • There are 715 distinct games in 24, with 117 of them unsolvable and the remaining 598 having at least one solution.
  • A brute-force solution checks 4608 expressions (double for each ace) to determine that a game is unsolvable.
  • Only 1809 expressions (out of the 4608) solve at least one game.
  • Finding the minimum number of expressions we need to check in order to find whether a game has a solution is equivalent to the hitting set problem which is NP-hard.
  • A greedy algorithm finds a set of 110 expressions which is enough to find a solution for any solvable game.
  • Removing equivalent expressions, we are left with 273 expressions which all solve at least one different game than any other. Out of these, 62 expressions are unique solutions to 122 games, so they must necessarily be checked in order to find solutions for all possible games.

[1]Wikipedia explains the set cover problem which is equivalent to the hitting set problem (one can be converted to the other).