24 Game Solver: gives you all dissimilar solutions. Try it!
For example, for the card with the numbers 2, 3, 6, Q(12), possible solutions are:
2 + 4 + 6 + 12 = 24
4 × 6 ÷ 2 + 12 = 24
12 ÷ 4 × (6 + 2) = 24
...
Here similar solutions are filtered out, e.g.
2 + 4 + 6 + 12 = 24
(12 + 6) + (4 + 2) = 24 // associativity
4 × 6 ÷ 2 + 12 = 24
6 ÷ 2 × 4 + 12 = 24 // commutativity
We have other similar situations, e.g.
4 × 6 + 5 - 5 = 24
4 × 6 × 5 / 5 = 24 // useless duplicated numbers
12 × (13 × 1 - 11) = 24
12 × (13 ÷ 1 - 11) = 24 // useless 1
(2 + 2) × (3 + 3) = 24
2 × 2 × (3 + 3) = 24 // 2 + 2 = 2 × 2
(2 × 6 - 6) × 4 = 24
(6 + 6) ÷ 2 × 4 = 24 // 2 × a - a = (a + a) ÷ 2
We need a set of all inequivalent expressions involving 4 operands. E.g. a × b ÷ c + d = d + a ÷ c × b, so we have a × b ÷ c + d in this set but not d + a ÷ c × b.
We listed all expressions by combination of parenthesis skeletons, operand positions and operators. Then we tested each expression with 4 random numbers (and run several times to avoid collision). Equivalent expressions always have same result, and vice versa. Finally, we found out 1,170 (matches with sequence A140606) inequivalent expressions.
For 4 different numbers (a, b, c, d), we can test each expression to check whether it equals to 24. E.g. 2, 3, 6, 12, there are a + b + c + d, b × c ÷ a + d and other 8 expressions match to 24.
But this does not work for 4 numbers with 2 same (a, b, c, c), or with 1 (1, a, b, c), or with 2, ... So we must have other sets of inequivalent expressions:
a, a, b, c // two a's may be commuted, or may be useless
a, a, b, b
a, a, a, b
a, a, a, a
1, a, b, c // 1 may be useless
1, a, a, b
1, a, a, a
1, 1, a, b
1, 1, a, a
1, 1, 1, a
2, 2, a, b // 2 + 2 = 2 × 2
2, 2, a, a
2, 2, 2, a
1, 2, 2, a
2, a, a, b // 2 × a - a = (a + a) ÷ 2
In our JavaScript programs, pregen.js generated these sets of inequivalent expressions and saved as 24-expressions.js. 24.js just decides which set is used and find out all matched expressions.
For situation a, a', b, c where a' = a + 1, we must consider that a' - a may be useless. So we must filter out all expressions that contain (a' - a) ×, × (a' - a) and ÷ (a' - a), then try b + c and b × c in addition.
1, 1, 5, 5 has only one dissimilar solution, because we regard (a + 1) × (a - 1) and a × a - 1 × 1 as the same.
2, 4, 6, 6 has a pair of solutions that seems similar: (6 - (4 - 2)) × 6 (A) and (6 - 4 ÷ 2) × 6 (B), but we have not considered 4 - 2 similar to 4 ÷ 2.
2, 4, 6, 6 has another pair of solutions that seems similar: (6 - 4 + 2) × 6 (C) and (6 - 4) × 2 × 6 (D), but we have not considered (a" - a) + 2 similar to (a" - a) × 2 where a" = a + 2.
It is hard to implement the above two considerations, because A is similar to C obviously but B is not similar to D at all.
2, 4, 8, 10 has 11 solutions:
10 + 8 + 4 + 2 = 24
(10 - 4) × 8 ÷ 2 = 24
(10 × 4 + 8) ÷ 2 = 24
((10 + 2) × 8 ÷ 4 = 24
10 × 2 + 8 - 4 = 24
(10 - 2) × 4 - 8 = 24
8 × 4 - 10 + 2 = 24
(8 ÷ 4 + 10) × 2 = 24
(8 × 2 - 10) × 4 = 24
(10 - 8 ÷ 2)) × 4 = 24
10 × 4 - 8 × 2 = 24
1, 3, 4, 6
1, 4, 5, 6 (2 solutions)
1, 5, 5, 5
1, 6, 6, 8
1, 8, 12, 12
2, 2, 11, 11
2, 2, 13, 13
2, 3, 5, 12
2, 4, 10, 10
2, 5, 5, 10
2, 7, 7, 10
3, 3, 7, 7
3, 3, 8, 8
4, 4, 7, 7
5, 5, 7, 11
5, 7, 7, 11
1, 7, 13, 13
6, 12, 12, 13
1, 6, 11, 13
6, 11, 12, 12
5, 10, 10, 13
1, 5, 11, 11
5, 10, 10, 11
4, 8, 8, 13
4, 4, 10, 10
4, 8, 8, 11
6, 9, 9, 10
3, 8, 8, 10
3, 5, 7, 13
3, 6, 6, 11
1, 2, 7, 7
5, 8, 9, 13
5, 9, 10, 11
4, 7, 11, 13
4, 9, 11, 11
4, 10, 10, 11
6, 7, 7, 11
3, 5, 8, 13
5, 5, 8, 11
2, 3, 13, 13
Combinations with solutions like a × b - a × c = 24 are not listed, e.g. 10, 12, 12, 12
3, 7, 9, 13 has 2 solutions:
(7 - 13 ÷ 3) × 9 = 24 // fractional solution
9 × 7 - 13 × 3 = 24 // large number as intermediate step
