Given any quadruple that is solvable, we always want to know *all* the solutions.
Some solutions are obviously different, some are obviously equivalent, some are less so.

We discuss our "theory" on how we present solutions in a less redudant way. i.e. we only present one solution out of all equivalent solutions. Furthermore, we'd like
to present the most *concise* solution among all equivalent solutions.

## Notations:

a, b, c, d are numbers (1,2,3 etc), x, y, z, w are expressions eg. 1, 1 + 2, (1 + 2 × 6) /5, etc.

Two expressions x and y are equivalent then we say x ↔ y, eg. a + b ↔ b + a, a - (b - c) ↔ a + c - b .

If two expressions x and y are equivalent, and we prefer y over x for its conciseness and/or easy human computer interaction, then we say x → y . For example: a - (b - c) ↔ a + c - b , but we prefer the latter, so
we have a - (b - c) → a + c - b

± is + or -, and ×/ is × or /.

## Part 1: elementary arithmetic equivalences

This part should be rather obvious. 1 × 2 × 3 × 4 and (4 × 3 × ( 2 × 1)) make no difference but obviously a lot of people prefer the first expression for its simplicity.

The rules in Part 1 can be summarized in one line: replace numbers with variables, and if two expressions have the same value for *any * choice of integer values, then these two expressions are equivalent.
For example, for the quadruple {1,2,3,4} and two expressions: (3 + 2 + 1) × 4 and 4 × (1 + 2 + 3) : , we first replace 1 with a, 2 with b, 3 with c and 4 with d, now the
two expressions become (c + b + a) × d and d × ( a + b + c). These two expressions have the same evaluation for any integer quadruple {a,b,c,d}. So we know that these two expressions are equivalent.
Among (3 + 2 + 1) × 4 and 4 × (1 + 2 + 3), we prefer the former one. The reason is that for some human computer interface (think of calculators), brackets are not needed for (3 + 2 + 1) × 4.
aka 4 × (1 + 2 + 3) → (3 + 2 + 1) × 4 . A general rule of thumb in determining our preference over equivalent expressions is that we prefer those with the more complicated part on the left most side.

Before giving the details of the Part 1 rules, we need one more footnote. For two expressions x and y (recall that x could be a + b), when we write x ×/ y, we add brackets around x or y when necessary.

**(1) Commutativity:** x ± y ↔ ± y + x x × y ↔ y × x x ×/ y ×/ z ↔ x ×/ z ×/ y

Preference 1: - and / on the right most side, eg. - a + b → b - a, a / b × c → a × c / b

Preference 2: those with more complicated expression(s) on the left, eg. c + a × b → a × b + c, a × (b + c) → (b + c) × a ...

**(2) Associativity:** x + y ± z ↔ ( x + y ) ± z ↔ x + (y ± z ) x × y ×/ z ↔ ( x × y ) ×/ z ↔ x × (y ×/ z ) ( x / y ) / z ↔ x / y / z

Preference: those with fewer brackets, e.g. (a + b) + c → a + b + c, (a / b) / c → a / b / c .

**(3) Exchangeability between + and -, and, × and / :** x - (y - z) ↔ x + z - y x × z / y ↔ x/(y/z) x /(y × z) ↔ x/y/z

Preference: those with fewer brackets, eg. a - (b - c) → a + c - b, a/(c/b) → a × b / c

**(4) Less negativity is preferred:** a - (b - c) ×/ d ↔ a + (c - b) ×/ d, note: this rule is not covered by rule (3).

Preferences: by preference in rule (3) we have: a - (b - c) ×/ d → (c - b) ×/ d + a

**(5)Tie breakers: ** a × b × c × d or d × c × b × a, (a + b) × (c + d) or (d + c) × (a + b)?

Preferences: doesn't matter. I prefer my bigger numbers on the left most side, i.e. 1 × 2 × 3 × 4 → 4 × 3 × 2 × 1, (1 + 2) × (5 + 3) → (5 + 3) × (2 + 1)

## Part 2: special rules for expressions with value 0 or 1

**(6) Multiplication/Division with 1: ** x × y ↔ x / y, if the evaluation of y is 1.

Preference: x × y, eg. (13 + 11) / (5 - 4) → (13 + 11) × (5 - 4). Similarly (12 + 6 × 2) / 1 → (12 + 6 × 2) × 1.

**(7) Placement of × z, when the value of z is 1: ** if z = 1, then we have: x + y × z ↔ x × z + y ↔ (x + y ) × z

Preferences: (x + y ) × z for symmetry

** (8) The interesting case of {a, b, c, c} **: a × b × c /c ↔ a × b + c - c, and ( a + b ) × c /c ↔ a + b + c - c .

Preference: a × b + c - c or a + b + c - c

(8.b) The final case of {a, b, 1, 1} : a × b × 1 × 1 ↔ a × b + 1 - 1, and ( a + b ) × 1 × 1 ↔ a + b + 1 - 1 .

Preference: a × b + 1 - 1 or a + b + 1 - 1

(8.c) The final final case of 24 : this case is only relevant when we play the bigger game where 24 is one of the 4 numbers.

24 × x / y → 24 + x - y, here x and y are two expressions with the same value

24 + a - b - c → 24 + b + c - a

24 + c - x is preferred over x + 24 - c where x is an expression composed of two numbers, obviously c and x has the same value.

The above three seemingly random rules cut a bunch of "gimmick" different solutions where 24 is envolved.

(8.d) The final final final case of 23 and 25 : this case is only relevant when we play the bigger game where 23 or 25 is one of the 4 numbers.

If the 4 numbers are 23 a b c=a+b then we have the following two solutions:

c/(a+b) + 23 ↔ (a+b)/c + 23

(a+b)/c + 23 is preferred over c/(a+b) + 23.

Similarly for 25 where 25 - (a+b)/c is preferred.

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