If the game of 24s is played with binary operators only, there is a finite number of different possible ways to combine the cards. Given a modern computer, it is perfectly solvable. The basic technique that I use is called algebraic programming, and amounts to combining every possible combination of four cards and every one of the permitted operators in every possible way, then selecting the combinations that equal 24.

This can be done in any programming language, but it’s natural in Mathematica, which natively supports the set operations on which this method relies.

First, some observations:

1. We have four cards, each of which can be any number from 1 to 10. We therefore have 10*10*10*10 (10,000) possible hands, if order matters.

2. If order doesn’t matter, such that {1,2,3,4} is the same as {4,3,2,1}, there are only 715 possible combinations

In[]:= Length[Union[Sort /@ Tuples[Range[10], 4]]]
Out[]= 715

Now we develop our combinations. This can be made easier using Andrzej Kozlowski’s “ReplaceAllList” function from The Mathematica Journal 9:2 © 2004.

ReplaceAllList[expr_, rules_] :=
Module[{i},
Join[ReplaceList[expr, rules],
If[AtomQ[expr], {},
Join @@ Table[
ReplacePart[expr, #, i] & /@
ReplaceAllList[expr[[i]], rules], {i, Length[expr]}]]]]

We have a choice — make our function set commutative, so that {3,2} is processed as both 3-2 and 2-3, or make our card set comprehensive, so that both arrangements are passed through the function. I elected the latter, but the former should work equally well.

Start by finding all the ways of associating four elements.

ClearAll[f]; SetAttributes[f, {Flat, OneIdentity}]; rawFunctions =
Union[FixedPoint[Union[Flatten[ReplaceAllList[#, f[a_, b_] -> g[a, b]]]] &,
f[a, b, c, d]] /. f -> g];

Off[General::"spell1"]; Off[General::"spell"];
divide[x_, y_] /; y != 0 := Divide[x, y]; divide[x_, 0] := Indeterminate;

Then we remove all duplicates and expressions that involve the application of g to three elements.

uniqueFunctions =
DeleteCases[rawFunctions, _?(Not[FreeQ[#, g[x__ /; Length[{x}] >= 3]]] &),
1];

Finally we replace g by Subscript[A, 2],Subscript[A, 1],Subscript[A, 0], in this order, and swap in the numbers from a hand of cards

functionsWNumbers[hands_List] :=
Map[(i = 0;
MapAll[If[# === g, # /. g -> Subscript[A, Mod[++i, 3]], #] &,
uniqueFunctions, Heads -> True] /. {a -> #[[1]], b -> #[[2]],
c -> #[[3]], d -> #[[4]]}) &, hands]

The following step generates every possible combination of operators. For four binary functions and four cards, we have a maximum of three functions per hand and only 64 possible combinations.

generateRules[binaryFunctions_List, numCards_Integer] :=
Module[{raw, refined, numFuncs},
numFuncs = Length[binaryFunctions];
raw = Map[Thread,
Map[RuleDelayed[Array[Subscript[A, # - 1] &, {numFuncs}], #] &,
Distribute[Array[binaryFunctions &, {numFuncs}], List]]];
(* since an actual hand never involves more than three binary operations,
we remove the rules involving unnecessary As.*)

refined = Union[DeleteCases[raw,Apply[Alternatives,
Map[HoldPattern[#] &,
Table[Subscript[A, i] :> _, {i, numCards - 1, numFuncs - 1}]]],
Infinity]];
refined]

Now we map our selected functions over the generic rules from the previous step.

generateHands[functions_List, rules_List] :=
Table[Union[# /. rules] & /@ functions[[i]], {i, 1, Length[functions]}]

When we get our answers, {1,1,2,2} is treated as distinct from {2,2,1,1}. Let’s fix that. The following assumes the cardSet is a single list of four integers {1,1,2,3}, and that the possibilitySet is a list of {{ordered four cards},{solutions}} like
{{6,6,6,6},{((6+6)+6)+6,(6+(6+6))+6,(6+6)+(6+6),6+((6+6)+6),6+(6+(6+6)),6 6-(6+6),(6 6-6)-6}}

solutionsForSet[cardSet_List, possibilitySet_List] :=
Module[{allVarieties, matches},
allVarieties = Permutations[cardSet];
matches = Select[possibilitySet, MemberQ[allVarieties, #[[1]]] &];
Flatten[matches[[All, 2]], 1]]

So, what do we find? Running every possible hand of cards through our 64 possible rules, and seeing which come out equal to 24, we find 566 of the 715 possible hands (about 79% of all possible cases) are solvable with arithmetic alone.

The unsolvable hands are {{1, 1, 1, 1}, {1, 1, 1, 2}, {1, 1, 1, 3}, {1, 1, 1, 4}, {1, 1, 1, 5}, {1, 1, 1, 6}, {1, 1, 1, 7}, {1, 1, 1, 9}, {1, 1, 1, 10}, {1, 1, 2, 2}, {1, 1, 2, 3}, {1, 1, 2, 4}, {1, 1, 2, 5}, {1, 1, 3, 3}, {1, 1, 5, 9}, {1, 1, 5, 10}, {1, 1, 6, 7}, {1, 1, 6, 10}, {1, 1, 7, 7}, {1, 1, 7, 8}, {1, 1, 7, 9}, {1, 1, 8, 9}, {1, 1, 8, 10}, {1, 1, 9, 9}, {1, 1, 9, 10}, {1, 1, 10, 10}, {1, 2, 2, 2}, {1, 2, 2, 3}, {1, 2, 9, 9}, {1, 2, 9, 10}, {1, 2, 10, 10}, {1, 3, 5, 5}, {1, 4, 7, 10}, {1, 4, 8, 10}, {1, 4, 9, 9}, {1, 5, 5, 7}, {1, 5, 5, 8}, {1, 5, 7, 7}, {1, 6, 6, 7}, {1, 6, 7, 7}, {1, 6, 7, 8}, {1, 6, 10, 10}, {1, 7, 7, 7}, {1, 7, 7, 8}, {1, 7, 10, 10}, {1, 8, 9, 9}, {1, 8, 9, 10}, {1, 8, 10, 10}, {1, 9, 9, 9}, {1, 9, 9, 10}, {1, 9, 10, 10}, {1, 10, 10, 10}, {2, 2, 2, 2}, {2, 2, 2, 6}, {2, 2, 7, 9}, {2, 2, 9, 9}, {2, 3, 3, 4}, {2, 5, 5, 5}, {2, 5, 5, 6}, {2, 5, 9, 9}, {2, 6, 7, 7}, {2, 7, 7, 7}, {2, 7, 7, 9}, {2, 7, 8, 10}, {2, 7, 9, 9}, {2, 9, 9, 9}, {2, 9, 9, 10}, {2, 10, 10, 10}, {3, 3, 4, 10}, {3, 3, 5, 8}, {3, 3, 7, 10}, {3, 3, 10, 10}, {3, 4, 6, 7}, {3, 4, 8, 8}, {3, 4, 9, 10}, {3, 5, 5, 5}, {3, 5, 5, 10}, {3, 5, 7, 7}, {3, 5, 8, 10}, {3, 7, 8, 10}, {3, 10, 10, 10}, {4, 4, 5, 9}, {4, 4, 6, 6}, {4, 4, 6, 7}, {4, 4, 9, 9}, {4, 4, 9, 10}, {4, 7, 7, 9}, {4, 7, 7, 10}, {4, 9, 9, 9}, {4, 9, 10, 10}, {4, 10, 10, 10}, {5, 5, 5, 7}, {5, 5, 5, 8}, {5, 5, 5, 10}, {5, 5, 6, 9}, {5, 5,
6, 10}, {5, 5, 7, 9}, {5, 6, 7, 10}, {5, 7, 7, 7}, {5, 7, 7, 8}, {5, 7, 9, 9}, {5, 8, 9, 9}, {5, 8, 9, 10}, {5, 8, 10, 10}, {5, 9, 9, 9}, {5, 9, 9, 10}, {5, 10, 10, 10}, {6, 6, 6, 7}, {6, 6, 7, 7}, {6, 6, 7, 8}, {6, 6, 9, 9}, {6, 6, 10, 10}, {6, 7, 7, 7}, {6, 7, 7, 8}, {6, 7, 7, 9}, {6, 7, 8, 8}, {6, 7, 9, 10}, {6, 8, 10, 10}, {6, 9, 9, 9}, {6, 9, 10, 10}, {7, 7, 7, 7}, {7, 7, 7, 8}, {7, 7, 7, 9}, {7, 7, 7, 10}, {7, 7, 8, 8}, {7, 7, 8, 9}, {7, 7, 8, 10}, {7, 7, 9, 9}, {7, 7, 10, 10}, {7, 8, 8, 8}, {7, 8, 9, 9}, {7, 9, 9, 9}, {7, 9, 9, 10}, {7, 9, 10, 10}, {7, 10, 10, 10}, {8, 8, 8, 8}, {8, 8, 8, 9}, {8, 8, 9, 9}, {8, 8, 9, 10}, {8, 8, 10, 10}, {8, 9, 9, 9}, {8, 9, 9, 10}, {8, 9, 10, 10}, {8, 10, 10, 10}, {9, 9, 9, 9}, {9, 9, 9, 10}, {9, 9, 10, 10}, {9, 10, 10, 10}, {10, 10, 10, 10}}

These same methods can be used with larger number of functions too, as I will show in coming posts.

I am a fan of the mathematical card game “24s”, in which four cards are turned over simultaneously, and all the players search for ways to make the number 24 by combining the four numbers shown with any mathematical operators they like (I have always played with addition, subtraction, multiplication, division, powers, roots, and logarithms, but the list can be made longer or shorter depending on the preferences and mathematical skill of the players). The number on every card must be used, and no card can be used more than once. Face cards are pulled from the deck before playing and aces == 1. Suits don’t matter for the game; the deck is being used only as a random number generator.

The game is addictively fun, combining elements of math, pattern recognition, and memory. I learned the game in high school, and used to play with expedition-mates when climbing high mountains, to test for loss of mental acuity at altitude. Recently, I’ve been teaching my daughter how to play. She does pretty well!

Solutions are possible for most hands (about 83% of hands, for the normal rule set, by my computation), and in most solvable cases, more than one solution is possible. In case of a tie, cards can be distributed back to the players, or victory can be given to the newest player, or to the person who came up with the most solutions, or the most interesting solution.

Take for example the hand 2,3,10,10.

Two solutions are possible. The “easy” one is

But the more impressive one, remembered decades later who saw it played, is

I mention this game because I recently realized that the game is solvable, and that so long as you restrict yourself to binary operators, it is possible to determine absolutely which hands can be turned into 24 and which can’t, and how many solutions are possible for each hand, and what those solutions are.

I’ll discuss my method in future posts.

I often use Mathematica and MySQL together; it’s typically a lot more efficient than trying to keep megabytes of data in the Mathematica notebook itself, and becomes vastly more powerful if I want to use the same data from multiple machines, all of which can tap into the same database.

Mathematica supports these uses natively, and you can send arbitrary SQL code to the server with the SQLExecute command. Interestingly, Mathematica also provides commands that map to major SQL commands, including SQLSelect[] and SQLInsert[]. I have never used these, viewing it as easier to program SQL commands in SQL itself, tuning them in MySQL Workbench.

Today, it occurred to me that Mathematica’s built-in commands might somehow be quicker than sending raw SQL, and I tested this hypothesis. Surprisingly, it turns out to be the case, with results similar to the following for a number of cases.


In[1]:= Needs["DatabaseLink`"]

Out[4]= {0.047, Null}


(The two queries differ slightly because I did not want caching to give an advantage to the second query).

For many years, the MathGroup mailing list has been the best place to go with Mathematica questions. The new Mathematica board at StackExchange may give it a run for its money.

There is a proposal to create a Mathematica section at the popular programming Q&A site StackExchange. I think it’s probably a good idea; the mathgroup mailing list has been a great resource for years but the web is easier and quicker than e-mail sometimes.

Sign up to support it here:

(Or, if the link above is busted, just visit http://area51.stackexchange.com/proposals/15787/mathematica)