

Recognizing when two arithmetic expressions are essentially the same

[ Warning: The math formatting in the RSS / Atom feed for this article is badly mutilated. I suggest you read the article on my blog. ] In this article, I discuss “twenty-four puzzles”. The puzzle «4 6 7 9 ⇒ 24» means that one should take the numbers 4, 6, 7, and 9, and combine them with the usual arithmetic operations of addition, subtraction, multiplication, and division, to make the number 24. In this case the unique solution is !!6·\frac{7 + 9}{4}!!. When the target number after the ⇒ is 24, as it often is, we omit it and just write «4 6 7 9». Every example in this article has target number 24. This is a continuation of my previous articles on this topic: A simple but difficult arithmetic puzzle (July 2016)

Solving twenty-four puzzles (March 2017) My first cut at writing a solver for twenty-four puzzles was a straightforward search program. It had a couple of hacks in it to cut down the search space by recognizing that !!a+E!! and !!E+a!! are the same, but other than that there was nothing special about it and I've discussed it before. It would quickly and accurately report whether any particular twenty-four puzzle was solvable, but as it turned out that wasn't quite good enough. The original motivation for the program was this: Toph and I play this game in the car. Pennsylvania license plates have three letters and four digits, and if we see a license plate FBV 2259 we try to solve «2 2 5 9». Sometimes we can't find a solution and then we wonder: it is because there isn't one, or is it because we just didn't get it yet? So the searcher turned into a phone app, which would tell us whether there was solution, so we'd know whether to give up or keep searching. But this wasn't quite good enough either, because after we would find that first solution, say !!2·(5 + 9 - 2)!!, we would wonder: are there any more? And here the program was useless: it would cheerfully report that there were three, so we would rack our brains to find another, fail, ask the program to tell us the answer, and discover to our disgust that the three solutions it had in mind were: $$ 2 \cdot (5 + (9 - 2)) \\ 2 \cdot (9 + (5 - 2)) \\ 2 \cdot ((5 + 9) - 2) $$ The computer thinks these are different, because it uses different data structures to represent them. It represents them with an abstract syntax tree, which means that each expression is either a single constant, or is a structure comprising an operator and its two operand expressions—always exactly two. The computer understands the three expressions above as having these structures: It's not hard to imagine that the computer could be taught to understand that the first two trees are equivalent. Getting it to recognize that the third one is also equivalent seems somewhat more difficult. Commutativity and associativity I would like the computer to understand that these three expressions should be considered “the same”. But what does “the same” mean? This problem is of a kind I particularly like: we want the computer to do something, but we're not exactly sure what that something is. Some questions are easy to ask but hard to answer, but this is the opposite: the real problem is to decide what question we want to ask. Fun! Certainly some of the question should involve commutativity and associativity of addition and multiplication. If the only difference between two expressions is that one has !!a + b!! where the other has !!b + a!!, they should be considered the same; similarly !!a + (b + c)!! is the same expression as !!(a + b) + c!! and as !!(b + a) + c!! and !!b + (a + c)!! and so forth. The «2 2 5 9» example above shows that commutativity and associativity are not limited to addition and multiplication. There are commutative and associative properties of subtraction also! For example, $$a+(b-c) = (a+b)-c$$ and $$(a+b)-c = (a-c)+b.$$ There ought to be names for these laws but as far as I know there aren't. (Sure, it's just commutativity and associativity of addition in disguise, but nobody explaining these laws to school kids ever seems to point out that subtraction can enter into it. They just observe that !!(a-b)-c ≠ a-(b-c)!!, say “subtraction isn't associative”, and leave it at that.) Closely related to these identities are operator inversion identities like !!a-(b+c) = (a-b)-c!!, !!a-(b-c) = (a-b)+c!!, and their multiplicative analogues. I don't know names for these algebraic laws either. One way to deal with all of this would to build a complicated comparison function for abstract syntax trees that tried to transform one tree into another by applying these identities. A better approach is to recognize that the data structure is over-specified. If we want the computer to understand that !!(a + b) + c!! and !!a + (b + c)!! are the same expression, we are swimming upstream by using a data structure that was specifically designed to capture the difference between these expressions. Instead, I invented a data structure, called an Ezpr (“Ez-pur”), that can represent expressions, but in a somewhat more natural way than abstract syntax trees do, and in a way that makes commutativity and associativity transparent. An Ezpr has a simplest form, called its “canonical” or “normal” form. Two Ezprs represent essentially the same mathematical expression if they have the same canonical form. To decide if two abstract syntax trees are the same, the computer converts them to Ezprs, simplifies them, and checks to see if resulting canonical forms are identical. The Ezpr Since associativity doesn't matter, we don't want to represent it. When we (humans) think about adding up a long column of numbers, we don't think about associativity because we don't add them pairwise. Instead we use an addition algorithm that adds them all at once in a big pile. We don't treat addition as a binary operation; we normally treat it as an operator that adds up the numbers in a list. The Ezpr makes this explicit: its addition operator is applied to a list of subexpressions, not to a pair. Both !!a + (b + c)!! and !!(a + b) + c!! are represented as the Ezpr SUM [ a b c - ] which just says that we are adding up !!a!!, !!b!!, and !!c!!. (The - sign is just punctuation; ignore it for now.) Similarly the Ezpr MUL [ a b c ÷ ] represents the product of !!a!!, !!b!!, and !!c!!. (Please ignore the ÷ sign for the time being.) To handle commutativity, we want those [ a b c ] lists to be bags. Perl doesn't have a built-in bag object, so instead I used arrays and required that the array elements be in sorted order. (Exactly which sorted order doesn't really matter.) Subtraction and division This doesn't yet handle subtraction and division, and the way I chose to handle them is the only part of this that I think is at all clever. A SUM object has not one but two bags, one for the positive and one for the negative part of the expression. An expression like !!a - b + c - d!! is represented by the Ezpr: SUM [ a c - b d ] and this is also the representation of !!a + c - b - d!!, of !!c + a - d - b!!, of !!c - d+ a-b!!, and of any other expression of the idea that we are adding up !!a!! and !!c!! and then deducting !!b!! and !!d!!. The - sign separates the terms that are added from those that are subtracted. Either of the two bags may be empty, so for example !!a + b!! is just SUM [ a b - ] . Division is handled similarly. Here conventional mathematical notation does a little bit better than in the sum case: MUL [ a c ÷ b d ] is usually written as !!\frac{ac}{bd}!!. Ezprs handle the associativity and commutativity of subtraction and division quite well. I pointed out earlier that subtraction has an associative law !!(a + b) - c = a + (b - c)!! even though it's not usually called that. No code is required to understand that those two expressions are equal if they are represented as Ezprs, because they are represented by completely identical structures: SUM [ a b - c ] Similarly there is a commutative law for subtraction: !!a + b - c = a - c + b!! and once again that same Ezpr does for both. Ezpr laws Ezprs are more flexible than binary trees. A binary tree can represent the expressions !!(a+b)+c!! and !!a+(b+c)!! but not the expression !!a+b+c!!. Ezprs can represent all three and it's easy to transform between them. Just as there are rules for building expressions out of simpler expressions, there are a few rules for combining and manipulating Ezprs. Lifting and flattening The most important transformation is lifting, which is the Ezpr version of the associative law. In the canonical form of an Ezpr, a SUM node may not have subexpressions that are also SUM nodes. If you have SUM [ a SUM [ b c - ] - … ] you should lift the terms from the inner sum into the outer one: SUM [ a b c - … ] effectively transforming !!a+(b+c)!! into !!a+b+c!!. More generally, in SUM [ a SUM [ b - c ] - d SUM [ e - f ] ] we lift the terms from the inner Ezprs into the outer one: SUM [ a b f - c d e ] This effectively transforms !!a + (b - c) - d - (e - f))!! to !!a + b + f - c - d - e!!. Similarly, when a MUL node contains another MUL , we can flatten the structure. Say we are converting the expression !!7 ÷ (3 ÷ (6 × 4))!! to an Ezpr. The conversion function is recursive and the naïve version computes this Ezpr: MUL [ 7 ÷ MUL [ 3 ÷ MUL [ 6 4 ÷ ] ] ] But then at the bottom level we have a MUL inside a MUL , so the 4 and 6 in the innermost MUL are lifted upward: MUL [ 7 ÷ MUL [ 3 ÷ 6 4 ] ] which represents !!\frac7{\frac{3}{6\cdot 4}}!!. Then again we have a MUL inside a MUL , and again the subexpressions of the innermost MUL can be lifted: MUL [ 7 6 4 ÷ 3 ] which we can imagine as !!\frac{7·6·4}3!!. The lifting only occurs when the sub-node has the same type as its parent; we may not lift terms out of a MUL into a SUM or vice versa. Trivial nodes The Ezpr SUM [ a - ] says we are adding up just one thing, !!a!!, and so it can be eliminated and replaced with just !!a!!. Similarly SUM [ - a ] can be replaced with the constant !!-a!!, if !!a!! is a constant. MUL can be handled similarly. An even simpler case is SUM [ - ] which can be replaced by the constant 0; MUL [ ÷ ] can be replaced with 1. These sometimes arise as a result of cancellation. Cancellation Consider the puzzle «3 3 4 6». My first solver found 49 solutions to this puzzle. One is !!(3 - 3) + (4 × 6)!!. Another is !!(4 + (3 - 3)) × 6!!. A third is !!4 × (6 + (3 - 3))!!. I think these are all the same: the solution is to multiply the 4 by the 6, and to get rid of the threes by subtracting them to make a zero term. The zero term can be added onto the rest of expression or to any of its subexpressions—there are ten ways to do this—and it doesn't really matter where. This is easily explained in terms of Ezprs: If the same subexpression appears in both of a node's bags, we can drop it. For example, the expression !!(4 + (3 -3)) × 6!! starts out as MUL [ 6 SUM [ 3 4 - 3 ] ÷ ] but the duplicate threes in SUM [ 3 4 - 3 ] can be canceled, to leave MUL [ 6 SUM [ 4 - ] ÷ ] The sum is now trivial, as described in the previous section, so can be eliminated and replaced with just 4: MUL [ 6 4 ÷ ] This Ezpr records the essential feature of each of the three solutions to «3 3 4 6» that I mentioned: they all are multiplying the 6 by the 4, and then doing something else unimportant to get rid of the threes. Another solution to the same puzzle is !!(6 ÷ 3) × (4 × 3)!!. Mathematically we would write this as !!\frac63·4·3!! and we can see this is just !!6×4!! again, with the threes gotten rid of by multiplication and division, instead of by addition and subtraction. When converted to an Ezpr, this expression becomes: MUL [ 6 4 3 ÷ 3 ] and the matching threes in the two bags are cancelled, again leaving MUL [ 6 4 ÷ ] In fact there aren't 49 solutions to this puzzle. There is only one, with 49 trivial variations. Identity elements In the preceding example, many of the trivial variations on the !!4×6!! solution involved multiplying some subexpression by !!\frac 33!!. When one of the input numbers in the puzzle is a 1, one can similarly obtain a lot of useless variations by choosing where to multiply the 1. Consider «1 3 3 5»: We can make 24 from !!3 × (3 + 5)!!. We then have to get rid of the 1, but we can do that by multiplying it onto any of the five subexpressions of !!3 × (3 + 5)!!: $$ 1 × (3 × (3 + 5)) \\ (1 × 3) × (3 + 5) \\ 3 × (1 × (3 + 5)) \\ 3 × ((1 × 3) + 5) \\ 3 × (3 + (1×5)) $$ These should not be considered different solutions. Whenever we see any 1's in either of the bags of a MUL node, we should eliminate them. The first expression above, !!1 × (3 × (3 + 5))!!, is converted to the Ezpr MUL [ 1 3 SUM [ 3 5 - ] ÷ ] but then the 1 is eliminated from the MUL node leaving MUL [ 3 SUM [ 3 5 - ] ÷ ] The fourth expression, !!3 × ((1 × 3) + 5)!!, is initially converted to the Ezpr MUL [ 3 SUM [ 5 MUL [ 1 3 ÷ ] - ] ÷ ] When the 1 is eliminated from the inner MUL , this leaves a trivial MUL [ 3 ÷ ] which is then replaced with just 3, leaving: MUL [ 3 SUM [ 5 3 - ] ÷ ] which is the same Ezpr as before. Zero terms in the bags of a SUM node can similarly be dropped. Multiplication by zero One final case is that MUL [ 0 … ÷ … ] can just be simplified to 0. The question about what to do when there is a zero in the denominator is a bit of a puzzle. In the presence of division by zero, some of our simplification rules are questionable. For example, when we have MUL [ a ÷ MUL [ b ÷ c ] ] , the lifting rule says we can simplify this to MUL [ a c ÷ b ] —that is, that !!\frac a{\frac bc} = \frac{ac}b!!. This is correct, except that when !!b=0!! or !!c=0!! it may be nonsense, depending on what else is going on. But since zero denominators never arise in the solution of these puzzles, there is no issue in this application. Results The Ezpr module is around 200 lines of Perl code, including everything: the function that converts abstract syntax trees to Ezprs, functions to convert Ezprs to various notations (both MUL [ 4 ÷ SUM [ 3 - 2 ] ] and 4 ÷ (3 - 2) ), and the two versions of the normalization process described in the previous section. The normalizer itself is about 35 lines. Associativity is taken care of by the Ezpr structure itself, and commutativity is not too difficult; as I mentioned, it would have been trivial if Perl had a built-in bag structure. I find it much easier to reason about transformations of Ezprs than abstract syntax trees. Many operations are much simpler; for example the negation of SUM [ A - B ] is simply SUM [ B - A ] . Pretty-printing is also easier because the Ezpr better captures the way we write and think about expressions. It took me a while to get the normalization tuned properly, but the results have been quite successful, at least for this problem domain. The current puzzle-solving program reports the number of distinct solutions to each puzzle. When it reports two different solutions, they are really different; when it fails to support the exact solution that Toph or I found, it reports one essentially the same. (There are some small exceptions, which I will discuss below.) Since there is no specification for “essentially the same” there is no hope of automated testing. But we have been using the app for several months looking for mistakes, and we have not found any. If the normalizer failed to recognize that two expressions were essentially similar, we would be very likely to notice: we would be solving some puzzle, be unable to find the last of the solutions that the program claimed to exist, and then when we gave up and saw what it was we would realize that it was essentially the same as one of the solutions we had found. I am pretty confident that there are no errors of this type, but see “Arguable points” below. A harder error to detect is whether the computer has erroneously conflated two essentially dissimilar expressions. To detect this we would have to notice that an expression was missing from the computer's solution list. I am less confident that nothing like this has occurred, but as the months have gone by I feel better and better about it. I consider the problem of “how many solutions does this puzzle really have to have?” been satisfactorily solved. There are some edge cases, but I think we have identified them. Code for my solver is on Github. The Ezpr code is in the Ezpr package in the Expr.pm file. This code is all in the public domain. Some examples The original program claims to find 35 different solutions to «4 6 6 6». The revised program recognizes that these are of only two types: !!4 × 6 × 6 ÷ 6!! MUL [ 4 6 - ] !!(6 - 4) × (6 + 6)!! MUL [ SUM [ 6 - 4 ] SUM [ 6 6 - ] ÷ ] Some of the variant forms of the first of those include: $$ 6 × (4 + (6 - 6)) \\ 6 + ((4 × 6) - 6) \\ (6 - 6) + (4 × 6) \\ (6 ÷ 6) × (4 × 6) \\ 6 ÷ ((6 ÷ 4) ÷ 6) \\ 6 ÷ (6 ÷ (4 × 6)) \\ 6 × (6 × (4 ÷ 6)) \\ (6 × 6) ÷ (6 ÷ 4) \\ 6 ÷ ((6 ÷ 6) ÷ 4) \\ 6 × (6 - (6 - 4)) \\ 6 × (6 ÷ (6 ÷ 4)) \\ \ldots

$$ In an even more extreme case, the original program finds 80 distinct expressions that solve «1 1 4 6», all of which are trivial variations on !!4·6!!. Of the 715 puzzles, 466 (65%) have solutions; for 175 of these the solution is unique. There are 3 puzzles with 8 solutions each («2 2 4 8», «2 3 6 9», and «2 4 6 8»), one with 9 solutions («2 3 4 6»), and one with 10 solutions («2 4 4 8»). The 10 solutions for «2 4 4 8» are as follows: !!4 × 8 - 2 × 4 !! SUM [ MUL [ 4 8 ÷ ] - MUL [ 2 4 ÷ ] ] !!4 × (2 + 8 - 4) !! MUL [ 4 SUM [ 2 8 - 4 ] ÷ ] !!(8 - 4) × (2 + 4) !! MUL [ SUM [ 8 - 4 ] SUM [ 2 4 - ] ÷ ] !!4 × (4 + 8) ÷ 2 !! MUL [ 4 SUM [ 4 8 - ] ÷ 2 ] !!(4 - 2) × (4 + 8) !! MUL [ SUM [ 4 - 2 ] SUM [ 4 8 - ] ÷ ] !!8 × (2 + 4/4) !! MUL [ 8 SUM [ 1 2 - ] ÷ ] !!2 × 4 × 4 - 8 !! SUM [ MUL [ 2 4 4 ÷ ] - 8 ] !!8 + 2 × (4 + 4) !! SUM [ 8 MUL [ 2 SUM [ 4 4 - ] ÷ ] - ] !!4 + 4 + 2 × 8 !! SUM [ 4 4 MUL [ 2 8 ÷ ] - ] !!4 × (8 - 4/2) !! MUL [ 4 SUM [ 8 - MUL [ 4 ÷ 2 ] ] ÷ ] A complete listing of every essentially different solution to every «a b c d» puzzle is available here. There are 1,063 solutions in all. Arguable points ​ There are a few places where we have not completely pinned down what it means for two solutions to be essentially the same; I think there is room for genuine disagreement. Any solution involving !!2×2!! can be changed into a slightly different solution involving !!2+2!! instead. These expressions are arithmetically different but numerically equal. For example, I mentioned earlier that «2 2 4 8» has 8 solutions. But two of these are !! 8 + 4 × (2 + 2)!! and !! 8 + 4 × 2 × 2!!. I am willing to accept these as essentially different. Toph, however, disagrees. A similar but more complex situation arises in connection with «1 2 3 7». Consider !!3×7+3!!, which equals 24. To get a solution to «1 2 3 7», we can replace either of the threes in !!3×7+3!! with !!(1+2)!!, obtaining !!((1 + 2) × 7) + 3!! or !! (3×7)+(1 +2)!!. My program considers these to be different solutions. Toph is unsure. It would be pretty easy to adjust the normalization process to handle these the other way if the user wanted that. Some interesting puzzles «1 2 7 7» has only one solution, quite unusual. (Spoiler) «2 2 6 7» has two solutions, both somewhat unusual. (Spoiler) Somewhat similar to «1 2 7 7» is «3 9 9 9» which also has an unusual solution. But it has two other solutions that are less surprising. (Spoiler) «1 3 8 9» has an easy solution but also a quite tricky solution. (Spoiler) One of my neighbors has the license plate JJZ 4631 . «4 6 3 1» is one of the more difficult puzzles. What took so long? Back in March, I wrote: I have enough material for at least three or four more articles about this that I hope to publish here in the coming weeks. But the previous article on this subject ended similarly, saying I hope to write a longer article about solvers in the next week or so. and that was in July 2016, so don't hold your breath. And here we are, five months later! This article was a huge pain to write. Sometimes I sit down to write something and all that comes out is dreck. I sat down to write this one at least three or four times and it never worked. The tortured Git history bears witness. In the end I had to abandon all my earlier drafts and start over from scratch, writing a fresh outline in an empty file. But perseverance paid off! WOOOOO. [ Addendum 20170825: I completely forgot that Shreevatsa R. wrote a very interesting article on the same topic as this one, in July of last year soon after I published my first article in this series. ] [ Addendum 20170829: A previous version of this article used the notations SUM [ … # … ] and MUL [ … # … ] , which I said I didn't like. Zellyn Hunter has persuaded me to replace these with SUM [ … - … ] and MUL [ … ÷ … ] . Thank you M. Hunter! ] [ Yet more on this topic! ]

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