The Universe of Discourse

Sun, 02 Apr 2017

/dev/null Follies

A Unix system administrator of my acquaintance once got curious about what people were putting into /dev/null. I think he also may have had some notion that it would contain secrets or other interesting material that people wanted thrown away. Both of these ideas are stupid, but what he did next was even more stupid: he decided to replace /dev/null with a plain file so that he could examine its contents.

The root filesystem quickly filled up and the admin had to be called back from dinner to fix it. But he found that he couldn't fix it: to create a Unix device file you use the mknod command, and its arguments are the major and minor device numbers of the device to create. Our friend didn't remember the correct minor device number. The ls -l command will tell you the numbers of a device file but he had removed /dev/null so he couldn't use that.

Having no other system of the same type with an intact device file to check, he was forced to restore /dev/null from the tape backups.

[Other articles in category /Unix] permanent link

Sun, 05 Mar 2017

Solving twenty-four puzzles

Back in July, I wrote:

Lately my kids have been interested in puzzles of this type: You are given a sequence of four digits, say 1,2,3,4, and your job is to combine them with ordinary arithmetic operations (+, -, ×, and ÷) in any order to make a target number, typically 24. For example, with 1,2,3,4, you can go with $$((1+2)+3)×4 = 24$$ or with $$4×((2×3)×1) = 24.$$

I said I had found an unusually difficult puzzle of this type, which is to make 2,5,6,6 total to 17. This is rather difficult. (I will reveal the solution later in this article.) Several people independently wrote to advise me that it is even more difficult to make 3,3,8,8 total to 24. They were right; it is amazingly difficult. After a couple of weeks I finally gave up and asked the computer, and when I saw the answer I didn't feel bad that I hadn't gotten it myself. (The solution is here if you want to give up without writing a program.)

From now on I will abbreviate the two puzzles of the previous paragraph as «2 5 6 6 ⇒ 17» and «3 3 8 8 ⇒ 24», and others similarly.

The article also inspired a number of people to write their own solvers and send them to me, and comparing them was interesting. My solver followed the tree search technique that I described in chapter 5 of Higher-Order Perl, and which has become so familiar to me that by now I can implement it without thinking about it very hard:

  1. Invent a data structure that represents the state of a possibly-incomplete search. This is just a list of the stuff one needs to keep track of while searching. (Let's call this a node.)

  2. Build a function which recognizes when a node represents a successful search.

  3. Build a function which takes a node, computes all the ways the search could proceed from that point, and returns a list of nodes for those slightly-more-advanced searches.

  4. Initialize a queue with a node representing a search that has just begun.

  5. Do this:

      until ( queue.is_empty() ) {
        current_node = queue.get_next()
        if ( is_successful( current_node ) ) { print the solution }
        queue.push( slightly_more_complete_searches( current_node ) )

This is precisely a breadth-first search. To make it into depth-first search, replace the queue with a stack. To make a heuristically directed search, replace get_next with a function that looks at the queue and chooses the best-looking node from which to proceed. Many other variations are possible, which is the advantage of this synthetic approach over letting the search arise organically from a recursive searcher. (Higher-Order Perl says “Recursive functions naturally perform depth-first searches.” (page 203)) In Python or Ruby one would be able to use yield and would not have to manage the queue explicitly, but in this case the queue management is trivial.

In my solver, each node contains a list of available expressions, annotated with its numerical value. Initially, the expressions are single numbers and the values are the same, say

    [ [ "2" => 2 ], [ "3" => 3 ], [ "4" => 4 ], [ "6" => 6 ] ]

Whether you represent expressions as strings or as something more structured depends on what you need to do with them at the end. If you just need to print them out, strings are good enough and are easy to handle.

A node represents a successful search if it contains only a single expression and if the expression's value is the target sum, say 24:

    [ [ "(((6÷2)+3)×4)" => 24 ] ]

From a node, the search should proceed by selecting two of the expressions, removing them from the node, selecting a legal operation, combining the two expressions into a single expression, and inserting the result back into the node. For example, from the initial node shown above, the search might continue by subtracting the fourth expression from the second:

    [ [ "2" => 2 ], [ "4" => 4 ], [ "(3-6)" => -3 ] ]

or by multiplying the second and the third:

    [ [ "2" => 2 ], [ "(3×4)" => 12 ], [ "6" => 6 ] ]

When the program encounters that first node it will construct both of these, and many others, and put them all into the queue to be investigated later.


    [ [ "2" => 2 ], [ "(3×4)" => 12 ], [ "6" => 6 ] ]

the search might proceed by dividing the first expression by the third:

    [ [ "(3×4)" => 12 ], [ "(2÷6)" => 1/3 ] ]

Then perhaps by subtracting the first from the second:

    [ [ "((2÷6)-(3×4))" => -35/3 ] ]

From here there is no way to proceed, so when this node is removed from the queue, nothing is added to replace it. Had it been a winner, it would have been printed out, but since !!-\frac{35}3!! is not the target value of 24, it is silently discarded.

To solve a puzzle of the «a b c d ⇒ t» sort requires examining a few thousand nodes. On modern hardware this takes approximately zero seconds.

The actual code for my solver is a lot of Perl gobbledygook that may not be of general interest so I will provide a link for people who are interested in deciphering it. It also represents my second attempt: I lost the code that I described in the earlier article and had to rewrite it. It is rather bigger than I would have liked.

My puzzle solver in Perl.

Stuff goes wrong

People showed me a lot of programs to solve this, and many didn't work. There are a few hard cases that several of them get wrong.


Some puzzles require that some subexpressions have fractional values. Many of the programs people showed me used integer arithmetic (sometimes implicitly and unintentionally) and failed to solve those puzzles. We can detect this by asking for a solution to «2 5 6 6 ⇒ 17», which requires a fraction. The solution is !!6×(2+(5÷6))!!. A program using integer arithmetic will calculate !!5÷6 = 0!! and fail to recognize the solution.

Several people on Twitter made this mistake and then mistakenly claimed that there was no solution at all. Usually it was possible to correct their programs by changing

        inputs = [ 2, 2, 5, 6 ]


        inputs = [ 2.0, 2.0, 5.0, 6.0 ]

or something like that.

Some people also surprised me by claiming that I had lied when I stated that the puzzle could be solved without any “underhanded tricks”, and that the use of intermediate fractions was itself an underhanded trick. Your Honor, I plead not guilty. I originally described the puzzle this way:

You are given a sequence of four digits, say 1,2,3,4, and your job is to combine them with ordinary arithmetic operations (+, -, ×, and ÷) in any order to make a target number, typically 24.

The objectors are implicitly claiming that when you combine 5 and 6 with the “ordinary arithmetic operation” of division, you get something other than !!\frac56!!. This is an indefensible claim.

I wasn't even trying to be tricky! It never occurred to me that fractions were something that some people would consider underhanded, and now that it has been suggested, I reject the suggestion. Folks, the result of division can be a fraction. Fractions are not some sort of obscure mathematical pettifoggery. They have been with us for at least 3,500 years now, so it is time everyone got used to them.

Floating-point error

Some programs used floating-point arithmetic to deal with the fractions and then fell foul of floating-point error. I will defer discussion of this to a future article.

I've complained about floating-point numbers on this blog before. ( 1 2 3 4 5 ) God, how I loathe them.

Expression construction

A more subtle error that several programs made was to assume that all expressions can be constructed by combining a previous expression with a single input number. For example, to solve «2 3 5 7 ⇒ 24», you multiply 3 by 7 to get 21, then add 5 to get 26, then subtract 2 to get 24.

But not every puzzle can be solved this way. Consider «2 3 5 7 ⇒ 41». You start by multiplying 2 by 3 to get 6, but if you try to combine the 6 with either 5 or 7 at this point you will lose. The only solution is to put the 6 aside and multiply 5 by 7 to get 35. Then add the 6 and the 35 to get 41.

Another way to put this is that an unordered binary tree with 4 leaves can take two different shapes. (Imagine filling the green circles with numbers and the pink squares with operators.)

The right-hand type of structure is sometimes necessary, as with «2 3 5 7 ⇒ 41». But several of the proposed solutions produced only expressions with structures like that on the left.

Here's Sebastian Fischer's otherwise very elegant Haskell solution, in its entirety:

    import Data.List ( permutations )

    solution = head
      [ (a,x,(b,y,(c,z,d)))
        | [a,b,c,d] <- permutations [2,5,6,6],
           ops <- permutations [((+),'+'),((-),'-'),((*),'*'),((/),'/')],
           let [u,v,w] = map fst $ take 3 ops,
           let [x,y,z] = map snd $ take 3 ops,
           (a `u` (b `v` (c `w` d))) == 17

You can see the problem in the last line. a, b, c, and d are numbers, and u, v, and w are operators. The program evaluates an expression to see if it has the value 17, but the expression always has the left-hand shape. (The program has another limitation: it never uses the same operator twice in the expression. That second permutations should be (sequence . take 3 . repeat) or something. It can still solve «2 5 6 6 ⇒ 17», however.)

Often the way these programs worked was to generate every possible permutation of the inputs and then apply the operators to the input lists stackwise: pop the first two values, combine them, push the result, and repeat. Here's a relevant excerpt from a program by Tim Dierks, this time in Python:

  for ordered_values in permutations(values):
    for operations in product(ops, repeat=len(values)-1):
      result, formula = calc_result(ordered_values, operations)

Here the expression structure is implicit, but the current result is always made by combining one of the input numbers with the old result.

I have seen many people get caught by this and similar traps in the past. I once posed the problem of enumerating all the strings of balanced parentheses of a given length, and several people assumed that all such strings have the form ()S, S(), or (S), where S is a shorter string of the same type. This seems plausible, and it works up to length 6, but (())(()) does not have that form.

Division by zero

A less common error exhibited by some programs was a failure to properly deal with division by zero. «2 5 6 6 ⇒ 17» has a solution, and if a program dies while checking !!2+(5÷(6-6))!! and doesn't find the solution, that's a bug.

Programs that worked

Ingo Blechschmidt (Haskell)

Ingo Blechschmidt showed me a solution in Haskell. The code is quite short. M. Blechschmidt's program defines a synthetic expression type and an evaluator for it. It defines a function arb which transforms an ordered list of numbers into a list of all possible expressions over those numbers. Reordering the list is taken care of earlier, by Data.List.permutations.

By “synthetic expression type” I mean this:

    data Exp a
        = Lit  a
        | Sum  (Exp a) (Exp a)
        | Diff (Exp a) (Exp a)
        | Prod (Exp a) (Exp a)
        | Quot (Exp a) (Exp a)
        deriving (Eq, Show)

Probably 80% of the Haskell programs ever written have something like this in them somewhere. This approach has a lot of boilerplate. For example, M. Blechschmidt's program then continues:

    eval :: (Fractional a) => Exp a -> a
    eval (Lit x) = x
    eval (Sum  a b) = eval a + eval b
    eval (Diff a b) = eval a - eval b
    eval (Prod a b) = eval a * eval b
    eval (Quot a b) = eval a / eval b

Having made up our own synonyms for the arithmetic operators (Sum for !!+!!, etc.) we now have to explain to Haskell what they mean. (“Not expressions, but an incredible simulation!”)

I spent a while trying to shorten the code by using a less artificial expression type:

    data Exp a
        = Lit  a
        | Op ((a -> a -> a), String) (Exp a) (Exp a)

but I was disappointed; I was only able to cut it down by 18%, from 34 lines to 28. I hope to discuss this in a future article. By the way, “Blechschmidt” is German for “tinsmith”.

Shreevatsa R. (Python)

Shreevatsa R. showed me a solution in Python. It generates every possible expression and prints it out with its value. If you want to filter the voluminous output for a particular target value, you do that later. Shreevatsa wrote up an extensive blog article about this which also includes a discussion about eliminating duplicate expressions from the output. This is a very interesting topic, and I have a lot to say about it, so I will discuss it in a future article.

Jeff Fowler (Ruby)

Jeff Fowler of the Recurse Center wrote a compact solution in Ruby that he described as “hot garbage”. Did I say something earlier about Perl gobbledygook? It's nice that Ruby is able to match Perl's level of gobbledygookitude. This one seems to get everything right, but it fails mysteriously if I replace the floating-point constants with integer constants. He did provide a version that was not “egregiously minified” but I don't have it handy.

Lindsey Kuper (Scheme)

Lindsey Kuper wrote a series of solutions in the Racket dialect of Scheme, and discussed them on her blog along with some other people’s work.

M. Kuper's first draft was 92 lines long (counting whitespace) and when I saw it I said “Gosh, that is way too much code” and tried writing my own in Scheme. It was about the same size. (My Perl solution is also not significantly smaller.)

Martin Janecke (PHP)

I saved the best for last. Martin Janecke showed me an almost flawless solution in PHP that uses a completely different approach than anyone else's program. Instead of writing a lot of code for generating permutations of the input, M. Janecke just hardcoded them:

    $zahlen = [
      [2, 5, 6, 6],
      [2, 6, 5, 6],
      [2, 6, 6, 5],
      [5, 2, 6, 6],
      [5, 6, 2, 6],
      [5, 6, 6, 2],
      [6, 2, 5, 6],
      [6, 2, 6, 5],
      [6, 5, 2, 6],
      [6, 5, 6, 2],
      [6, 6, 2, 5],
      [6, 6, 5, 2]

Then three nested loops generate the selections of operators:

 $operatoren = [];
 foreach (['+', '-', '*', '/'] as $x) {
   foreach (['+', '-', '*', '/'] as $y) {
     foreach (['+', '-', '*', '/'] as $z) {
       $operatoren[] = [$x, $y, $z];

Expressions are constructed from templates:

        $klammern = [
          '%d %s %d %s %d %s %d',
          '(%d %s %d) %s %d %s %d',
          '%d %s (%d %s %d) %s %d',
          '%d %s %d %s (%d %s %d)',
          '(%d %s %d) %s (%d %s %d)',
          '(%d %s %d %s %d) %s %d',
          '%d %s (%d %s %d %s %d)',
          '((%d %s %d) %s %d) %s %d',
          '(%d %s (%d %s %d)) %s %d',
          '%d %s ((%d %s %d) %s %d)',
          '%d %s (%d %s (%d %s %d))'

(I don't think those templates are all necessary, but hey, whatever.) Finally, another set of nested loops matches each ordering of the input numbers with each selection of operators, uses sprintf to plug the numbers and operators into each possible expression template, and uses @eval to evaluate the resulting expression to see if it has the right value:

   foreach ($zahlen as list ($a, $b, $c, $d)) {
     foreach ($operatoren as list ($x, $y, $z)) {
       foreach ($klammern as $vorlage) {
         $term = sprintf ($vorlage, $a, $x, $b, $y, $c, $z, $d);
         if (17 == @eval ("return $term;")) {
           print ("$term = 17\n");

If loving this is wrong, I don't want to be right. It certainly satisfies Larry Wall's criterion of solving the problem before your boss fires you. The same approach is possible in most reasonable languages, and some unreasonable ones, but not in Haskell, which was specifically constructed to make this approach as difficult as possible.

M. Janecke wrote up a blog article about this, in German. He says “It's not an elegant program and PHP is probably not an obvious choice for arithmetic puzzles, but I think it works.” Indeed it does. Note that the use of @eval traps the division-by-zero exceptions, but unfortunately falls foul of floating-point roundoff errors.


Thanks to everyone who discussed this with me. In addition to the people above, thanks to Stephen Tu, Smylers, Michael Malis, Kyle Littler, Jesse Chen, Darius Bacon, Michael Robert Arntzenius, and anyone else I forgot. (If I forgot you and you want me to add you to this list, please drop me a note.)

Coming up

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.

[Other articles in category /math] permanent link

Thu, 23 Feb 2017

Miscellaneous notes on anagram scoring

My article on finding the best anagram in English was well-received, and I got a number of interesting comments about it.

  • A couple of people pointed out that this does nothing to address the issue of multiple-word anagrams. For example it will not discover “I, rearrangement servant / Internet anagram server” True, that is a different problem entirely.

  • Markian Gooley informed me that “megachiropteran / cinematographer” has been long known to Scrabble players, and Ben Zimmer pointed out that A. Ross Eckler, unimpressed by “cholecystoduodenostomy / duodenocholecystostomy”, proposed a method almost identical to mine for scoring anagrams in an article in Word Ways in 1976. M. Eckler also mentioned that the “remarkable” “megachiropteran / cinematographer” had been published in 1927 and that “enumeration / mountaineer” (which I also selected as a good example) appeared in the Saturday Evening Post in 1879!

  • The Hacker News comments were unusually pleasant and interesting. Several people asked “why didn't you just use the Levenshtein distance”? I don't remember that it ever occured to me, but if it had I would have rejected it right away as being obviously the wrong thing. Remember that my original chunking idea was motivated by the observation that “cholecystoduodenostomy / duodenocholecystostomy” was long but of low quality. Levenshtein distance measures how far every letter has to travel to get to its new place and it seems clear that this would give “cholecystoduodenostomy / duodenocholecystostomy” a high score because most of the letters move a long way.

    Hacker News user tyingq tried it anyway, and reported that it produced a poor outcome. The top-scoring pair by Levenshtein distance is “anatomicophysiologic physiologicoanatomic”, which under the chunking method gets a score of 3. Repeat offender “cholecystoduodenostomy / duodenocholecystostomy” only drops to fourth place.

    A better idea seems to be Levenshtein score per unit of length, suggested by user cooler_ranch.

  • A couple of people complained about my “notaries / senorita” example, rightly observing that “senorita” is properly spelled “señorita”. This bothered me also while I was writing the article. I eventually decided although “notaries” and “señorita” are certainly not anagrams in Spanish (even supposing that “notaries” was a Spanish word, which it isn't) that the spelling of “senorita” without the tilde is a correct alternative in English. (Although I found out later that both the Big Dictionary and American Heritage seem to require the tilde.)

    Hacker News user ggambetta observed that while ‘é’ and ‘e’, and ‘ó’ and ‘o’ feel interchangeable in Spanish, ‘ñ’ and ‘n’ do not. I think this is right. The ‘é’ is an ‘e’, but with a mark on it to show you where the stress is in the word. An ‘ñ’ is not like this. It was originally an abbreviation for ‘nn’, introduced in the 18th century. So I thought it might make sense to allow ‘ñ’ to be exchanged for ‘nn’, at least in some cases.

    (An analogous situation in German, which may be more familiar, is that it might be reasonable to treat ‘ö’ and ‘ü’ as if they were ‘oe’ and ‘ue’. Also note that in former times, “w” and “uu” were considered interchangeable in English anagrams.)

    Unfortunately my Spanish dictionary is small (7,000 words) and of poor quality and I did not find any anagrams of “señorita”. I wish I had something better for you. Also, “señorita” is not one of the cases where it is appropriate to replace “ñ” with “nn”, since it was never spelled “sennorita”.

    I wonder why sometimes this sort of complaint seems to me like useless nitpicking, and other times it seems like a serious problem worthy of serious consideration. I will try to think about this.

  • Mike Morton, who goes by the anagrammatic nickname of “Mr. Machine Tool”, referred me to his Higgledy-piggledy about megachiropteran / cinematographer, which is worth reading.

  • Regarding the maximal independent set algorithm I described yesterday, Shreevatsa R. suggested that it might be conceptually simpler to find the maximal clique in the complement graph. I'm not sure this helps, because the complement graph has a lot more edges than the original. Below right is the complement graph for “acrididae / cidaridae”. I don't think I can pick out the 4-cliques in that graph any more than the independent sets in the graph on the lower-left, and this is an unusually favorable example case for the clique version, because the original graph has an unusually large number of edges.

    But perhaps the cliques might be easier to see if you know what to look for: in the right-hand diagram the four nodes on the left are one clique, and the four on the right are the other, whereas in the left-hand diagram the two independent sets are all mixed together.

  • An earlier version of the original article mentioned the putative 11-pointer “endometritria / intermediator”. The word “endometritria” seemed pretty strange, and I did look into it before I published the article, but not carefully enough. When Philip Cohen wrote to me to question it, I investigated more carefully, and discovered that it had been an error in an early WordNet release, corrected (to “endometria”) in version 1.6. I didn't remember that I had used WordNet's word lists, but I am not surprised to discover that I did.

    A rare printing of Webster's 2¾th American International Lexican includes the word “endometritriostomoscopiotomous” but I suspect that it may be a misprint.

  • Philippe Bruhat wrote to inform me of Alain Chevrier’s book notes / sténo, a collection of thematically related anagrams in French. The full text is available online.

  • Alexandre Muñiz, who has a really delightful blog, and who makes and sells attractive and clever puzzles of his own invention. pointed out that soapstone teaspoons are available. The perfect gift for the anagram-lover in your life! They are not even expensive.

  • Thanks also to Clinton Weir, Simon Tatham, Jon Reeves, Wei-Hwa Huang, and Philip Cohen for their emails about this.

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Wed, 22 Feb 2017

Moore's law beats a better algorithm

Yesterday I wrote about the project I did in the early 1990s to find the best anagrams. The idea is to give pair of anagram words a score, which is the number of chunks into which you have to divide one word in order to rearrange the chunks to form the other word. This was motivated by the observation that while “cholecysto-duodeno-stomy” and “duodeno-cholecysto-stomy” are very long words that are anagrams of one another, they are not interesting because they require so few chunks that the anagram is obvious. A shorter but much more interesting example is “aspired / diapers”, where the letters get all mixed up.

I wrote:

One could do this with a clever algorithm, if one were available. There is a clever algorithm, based on finding maximal independent sets in a certain graph. I did not find this algorithm at the time; nor did I try. Instead, I used a brute-force search.

I wrote about the brute-force search yesterday. Today I am going to discuss the clever algorithm.

The plan is to convert a pair of anagrams into a graph that expresses the constraints on how the letters can move around when one turns into the other. Shown below is the graph for comparing acrididae (grasshoppers) with cidaridae (sea urchins):

The “2,4” node at the top means that the letters ri at position 2 in acrididae match the letters ri at position 4 in cidaridae; the “3,1” node is for the match between the first id and the first id. The two nodes are connected by an edge to show that the two matchings are incompatible: if you map the ri to the ri, you cannot also map the first id to the first id; instead you have to map the first id to the second one, represented by the node “3,5”, which is not connected to “2,4”. A maximal independent set in this graph is a maximal selection of compatible matchings in the words, which corresponds to a division into the minimum number of chunks.

Usually the graph is much less complicated than this. For simple cases it is empty and the maximal independent set is trivial. This one has two maximal independent sets, one (3,1; 5,5; 6,6; 7,7) corresponding to the obvious minimal splitting:

and the other (2,4; 3,5; 5,1; 6,2) to this other equally-good splitting:

In an earlier draft of yesterday's post, I wrote:

I should probably do this over again, because my listing seems to be incomplete. For example, it omits “spectrum / crumpets” which would have scored 5, because the Webster's Second list contains crumpet but not crumpets.

I was going to leave it at that, but then I did do it over again, and this time around I implemented the “good” algorithm. It was not that hard. The code is on GitHub if you would like to see it.

To solve the maximal independent set instances, I used a guided brute-force search. Maximal independent set is NP-complete, and so the best known algorithm for it runs in exponential time. But the instances in which we are interested here are small enough that this doesn't matter. The example graph above has 8 nodes, so one needs to check at most 256 possible sets to see which is the maximal independent set.

I collated together all the dictionaries I had handy. (I didn't know yet about SCOWL.) These totaled 275,954 words, which is somewhat more than Webster's Second by itself. One of the new dictionaries did contain crumpets so the result does include “spectrum / crumpets”.

The old scored anagram list that I made in the 1990s contained 23,521 pairs. The new one contains 38,333. Unfortunately most of the new stuff is of poor quality, as one would expect. Most of the new words that were missing from my dictionary the first time around are obscure. Perhaps some people would enjoy discovering that that “basiparachromatin” and “Marsipobranchiata” are anagrams, but I find it of very limited appeal.

But the new stuff is not all junk. It includes:

10 antiparticles paternalistic
10 nectarines transience
10 obscurantist subtractions

11 colonialists oscillations
11 derailments streamlined

which I think are pretty good.

I wasn't sure how long the old program had taken to run back in the early nineties, but I was sure it had been at least a couple of hours. The new program processes the 275,954 inputs in about 3.5 seconds. I wished I knew how much of this was due to Moore's law and how much to the improved algorithm, but as I said, the old code was long lost.

But then just as I was finishing up the article, I found the old brute-force code that I thought I had lost! I ran it on the same input, and instead of 3.5 seconds it took just over 4 seconds. So almost all of the gain since the 1990s was from Moore's law, and hardly any was from the “improved” algorithm.

I had written in the earlier article:

In 2016 [ the brute force algorithm ] would probably still [ run ] quicker than implementing the maximal independent set algorithm.

which turned out to be completely true, since implementing the maximal independent set algorithm took me a couple of hours. (Although most of that was building out a graph library because I didn't want to look for one on CPAN.)

But hey, at least the new program is only twice as much code!

38333 anagrams, scored

[ Addendum: The program had a minor bug: it would disregard capitalization when deciding if two words were anagrams, but then compute the scores with capitals and lowercase letters distinct. So for example Chaenolobus was considered an anagram of unchoosable, but then the Ch in Chaenolobus would not be matched to the ch in unchoosable, resulting in a score of 11 instead of 10. I have corrected the program and the output. Thanks to Philip Cohen for pointing this out. ]

[ Addendum 20170223: More about this ]

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Tue, 21 Feb 2017

I found the best anagram in English

I planned to publish this last week sometime but then I wrote a line of code with three errors and that took over the blog.

A few years ago I mentioned in passing that in the 1990s I had constructed a listing of all the anagrams in Webster's Second International dictionary. (The Webster's headword list was available online.)

This was easy to do, even at the time, when the word list itself, at 2.5 megabytes, was a file of significant size. Perl and its cousins were not yet common; in those days I used Awk. But the task is not very different in any reasonable language:

  # Process word list
  while (my $word = <>) {
    chomp $word;
    my $sorted = join "", sort split //, $word;  # normal form
    push @{$anagrams{$sorted}}, $word;

  for my $words (values %anagrams) {
      print "@$words\n" if @$words > 1;

The key technique is to reduce each word to a normal form so that two words have the same normal form if and only if they are anagrams of one another. In this case we do this by sorting the letters into alphabetical order, so that both megalodon and moonglade become adeglmnoo.

Then we insert the words into a (hash | associative array | dictionary), keyed by their normal forms, and two or more words are anagrams if they fall into the same hash bucket. (There is some discussion of this technique in Higher-Order Perl pages 218–219 and elsewhere.)

(The thing you do not want to do is to compute every permutation of the letters of each word, looking for permutations that appear in the word list. That is akin to sorting a list by computing every permutation of the list and looking for the one that is sorted. I wouldn't have mentioned this, but someone on StackExchange actually asked this question.)

Anyway, I digress. This article is about how I was unhappy with the results of the simple procedure above. From the Webster's Second list, which contains about 234,000 words, it finds about 14,000 anagram sets (some with more than two words), consisting of 46,351 pairs of anagrams. The list starts with

aal ala

and ends with

zolotink zolotnik

which exemplify the problems with this simple approach: many of the 46,351 anagrams are obvious, uninteresting or even trivial. There must be good ones in the list, but how to find them?

I looked in the list to find the longest anagrams, but they were also disappointing:

cholecystoduodenostomy duodenocholecystostomy

(Webster's Second contains a large amount of scientific and medical jargon. A cholecystoduodenostomy is a surgical operation to create a channel between the gall bladder (cholecysto-) and the duodenum (duodeno-). A duodenocholecystostomy is the same thing.)

This example made clear at least one of the problems with boring anagrams: it's not that they are too short, it's that they are too simple. Cholecystoduodenostomy and duodenocholecystostomy are 22 letters long, but the anagrammatic relation between them is obvious: chop cholecystoduodenostomy into three parts:

cholecysto duodeno stomy

and rearrange the first two:

duodeno cholecysto stomy

and there you have it.

This gave me the idea to score a pair of anagrams according to how many chunks one had to be cut into in order to rearrange it to make the other one. On this plan, the “cholecystoduodenostomy / duodenocholecystostomy” pair would score 3, just barely above the minimum possible score of 2. Something even a tiny bit more interesting, say “abler / blare” would score higher, in this case 4. Even if this strategy didn't lead me directly to the most interesting anagrams, it would be a big step in the right direction, allowing me to eliminate the least interesting.

This rule would judge both “aal / ala” and “zolotink / zolotnik” as being uninteresting (scores 2 and 4 respectively), which is a good outcome. Note that some other boring-anagram problems can be seen as special cases of this one. For example, short anagrams never need to be cut into many parts: no four-letter anagrams can score higher than 4. The trivial anagramming of a word to itself always scores 1, and nontrivial anagrams always score more than this.

So what we need to do is: for each anagram pair, say acrididae (grasshoppers) and cidaridae (sea urchins), find the smallest number of chunks into which we can chop acrididae so that the chunks can be rearranged into cidaridae.

One could do this with a clever algorithm, if one were available. There is a clever algorithm, based on finding maximal independent sets in a certain graph. (More about this tomorrow.) I did not find this algorithm at the time; nor did I try. Instead, I used a brute-force search. Or rather, I used a very small amount of cleverness to reduce the search space, and then used brute-force search to search the reduced space.

Let's consider a example, scoring the anagram “abscise / scabies”. You do not have to consider every possible permutation of abscise. Rather, there are only two possible mappings from the letters of abscise to the letters of scabies. You know that the C must map to the C, the A must map to the A, and so forth. The only question is whether the first S of abscise maps to the first or to the second S of scabies. The first mapping gives us:

and the second gives us

because the S and the C no longer go to adjoining positions. So the minimum number of chunks is 5, and this anagram pair gets a score of 5.

To fully analyze cholecystoduodenostomy by this method required considering 7680 mappings. (120 ways to map the five O's × 2 ways to map the two C's × 2 ways to map the two D's, etc.) In the 1990s this took a while, but not prohibitively long, and it worked well enough that I did not bother to try to find a better algorithm. In 2016 it would probably still run quicker than implementing the maximal independent set algorithm. Unfortunately I have lost the code that I wrote then so I can't compare.

Assigning scores in this way produced a scored anagram list which began

2 aal ala

and ended

4 zolotink zolotnik

and somewhere in the middle was

3 cholecystoduodenostomy duodenocholecystostomy

all poor scores. But sorted by score, there were treasures at the end, and the clear winner was

14 cinematographer megachiropteran

I declare this the single best anagram in English. It is 15 letters long, and the only letters that stay together are the E and the R. “Cinematographer” is as familiar as a 15-letter word can be, and “megachiropteran” means a giant bat. GIANT BAT! DEATH FROM ABOVE!!!

And there is no serious competition. There was another 14-pointer, but both its words are Webster's Second jargon that nobody knows:

14 rotundifoliate titanofluoride

There are no score 13 pairs, and the score 12 pairs are all obscure. So this is the winner, and a deserving winner it is.

I think there is something in the list to make everyone happy. If you are the type of person who enjoys anagrams, the list rewards casual browsing. A few examples:

7 admirer married
7 admires sidearm

8 negativism timesaving
8 peripatetic precipitate
8 scepters respects
8 shortened threnodes
8 soapstone teaspoons

9 earringed grenadier
9 excitation intoxicate
9 integrals triangles
9 ivoriness revisions
9 masculine calumnies

10 coprophagist topographics
10 chuprassie haruspices
10 citronella interlocal

11 clitoridean directional
11 dispensable piebaldness

“Clitoridean / directional” has been one of my favorites for years. But my favorite of all, although it scores only 6, is

6 yttrious touristy

I think I might love it just because the word yttrious is so delightful. (What a debt we owe to Ytterby, Sweden!)

I also rather like

5 notaries senorita

which shows that even some of the low-scorers can be worth looking at. Clearly my chunk score is not the end of the story, because “notaries / senorita” should score better than “abets / baste” (which is boring) or “Acephali / Phacelia” (whatever those are), also 5-pointers. The length of the words should be worth something, and the familiarity of the words should be worth even more.

Here are the results:

38333 anagrams, scored

In former times there was a restaurant in Philadelphia named “Soupmaster”. My best unassisted anagram discovery was noticing that this is an anagram of “mousetraps”.

[ Addendum 20170222: There is a followup article comparing the two algorithms I wrote for computing scores. ]

[ Addendum 20170222: An earlier version of this article mentioned the putative 11-pointer “endometritria / intermediator”. The word “endometritria” seemed pretty strange, and I did look into it before I published the article, but not carefully enough. When Philip Cohen wrote to me to question it, I investigated more carefully, and discovered that it had been an error in an early WordNet release, corrected (to “endometria”) in version 1.6. I didn't remember that I had used WordNet's word lists, but I am not surprised to discover that I did. ]

[ Addendum 20170223: More about this ]

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Thu, 16 Feb 2017

Automatically checking for syntax errors with Git's pre-commit hook

Previous related article
Earlier related article

Over the past couple of days I've written about how I committed a syntax error on a cron script, and a co-worker had to fix it on Saturday morning. I observed that I should have remembered to check the script for syntax errors before committing it, and several people wrote to point out to me that this is the sort of thing one should automate.

(By the way, please don't try to contact me on Twitter. It won't work. I have been on Twitter Vacation for months and have no current plans to return.)

Git has a “pre-commit hook” feature, which means that you can set up a program that will be run every time you attempt a commit, and which can abort the commit if it doesn't like what it sees. This is the natural place to put an automatic syntax check. Some people suggested that it should be part of the CI system, or even the deployment system, but I don't control those, and anyway it is much better to catch this sort of thing as early as possible. I decided to try to implement a pre-commit hook to check syntax.

Unlike some of the git hooks, the pre-commit hook is very simple to use. It gets run when you try to make a commit, and the commit is aborted if the hook exits with a nonzero status.

I made one mistake right off the bat: I wrote the hook in Bourne shell, even though I swore years ago to stop writing shell scripts. Everything that I want to write in shell should be written in Perl instead or in some equivalently good language like Python. But the sample pre-commit hook was written in shell and when I saw it I went into automatic shell scripting mode and now I have yet another shell script that will have to be replaced with Perl when it gets bigger. I wish I would stop doing this.

Here is the hook, which, I should say up front, I have not yet tried in day-to-day use. The complete and current version is on github.


    function typeof () {
        case $filename in
            *.pl | *.pm) echo perl; exit ;;

        line1=$(head -1 $1)
        case $line1 in '#!'*perl )
            echo perl; exit ;;

Some of the sample programs people showed me decided which files needed to be checked based only on the filename. This is not good enough. My most important Perl programs have filenames with no extension. This typeof function decides which set of checks to apply to each file, and the minimal demonstration version here can do that based on filename or by looking for the #!...perl line in the first line of the file contents. I expect that this function will expand to include other file types; for example

               *.py ) echo python; exit ;;

is an obvious next step.

    if [ ! -z $COMMIT_OK ]; then
        exit 0;

This block is an escape hatch. One day I will want to bypass the hook and make a commit without performing the checks, and then I can COMMIT_OK=1 git commit …. There is actually a --no-verify flag to git-commit that will skip the hook entirely, but I am unlikely to remember it.

(I am also unlikely to remember COMMIT_OK=1. But I know from experience that I will guess that I might have put an escape hatch into the hook. I will also guess that there might be a flag to git-commit that does what I want, but that will seem less likely to be true, so I will look in the hook program first. This will be a good move because my hook is much shorter than the git-commit man page. So I will want the escape hatch, I will look for it in the best place, and I will find it. That is worth two lines of code. Sometimes I feel like the guy in Memento. I have not yet resorted to tattooing COMMIT_OK=1 on my chest.)

    exec 1>&2

This redirects the standard output of all subsequent commands to go to standard error instead. It makes it more convenient to issue error messages with echo and such like. All the output this hook produces is diagnostic, so it is appropriate for it to go to standard error.

    for file in $(git diff --cached --name-only | sort) ; do

allOK is true if every file so far has passed its checks. badFiles is a list of files that failed their checks. the git diff --cached --name-only function interrogates the Git index for a list of the files that have been staged for commit.

        type=$(typeof "$file")

This invokes the typeof function from above to decide the type of the current file.


When a check discovers that the current file is bad, it will signal this by setting BAD to true.

        echo "##  Checking file $file (type $type)"
        case $type in
            perl )
                perl -cw $file || BAD=true
                [ -x $file ] || { echo "File is not executable"; BAD=true; }
            * )
                echo "Unknown file type: $file; no checks"

This is the actual checking. To check Python files, we would add a python) … ;; block here. The * ) case is a catchall. The perl checks run perl -cw, which does syntax checking without executing the program. It then checks to make sure the file is executable, which I am sure is a mistake, because these checks are run for .pm files, which are not normally supposed to be executable. But I wanted to test it with more than one kind of check.

        if $BAD; then

If the current file was bad, the allOK flag is set false, and the commit will be aborted. The current filename is appended to badFiles for a later report. Bash has array variables but I don't remember how they work and the manual made it sound gross. Already I regret not writing this in a real language.

After the modified files have been checked, the hook exits successfully if they were all okay, and prints a summary if not:

    if $allOK; then
        exit 0;
        echo ''
        echo '## Aborting commit.  Failed checks:'
        for file in $(echo $badFiles | tr ';' ' '); do
            echo "    $file"
        exit 1;

This hook might be useful, but I don't know yet; as I said, I haven't really tried it. But I can see ahead of time that it has a couple of drawbacks. Of course it needs to be built out with more checks. A minor bug is that I'd like to apply that is-executable check to Perl files that do not end in .pm, but that will be an easy fix.

But it does have one serious problem I don't know how to fix yet. The hook checks the versions of the files that are in the working tree, but not the versions that are actually staged for the commit!

The most obvious problem this might cause is that I might try to commit some files, and then the hook properly fails because the files are broken. Then I fix the files, but forget to add the fixes to the index. But because the hook is looking at the fixed versions in the working tree, the checks pass, and the broken files are committed!

A similar sort of problem, but going the other way, is that I might make several changes to some file, use git add -p to add the part I am ready to commit, but then the commit hook fails, even though the commit would be correct, because the incomplete changes are still in the working tree.

I did a little tinkering with git stash save -k to try to stash the unstaged changes before running the checks, something like this:

        git stash save -k "pre-commit stash" || exit 2
        trap "git stash pop" EXIT

but I wasn't able to get anything to work reliably. Stashing a modified index has never worked properly for me, perhaps because there is something I don't understand. Maybe I will get it to work in the future. Or maybe I will try a different method; I can think of several offhand:

  • The hook could copy each file to a temporary file and then run the check on the temporary file. But then the diagnostics emitted by the checks would contain the wrong filenames.

  • It could move each file out of the way, check out the currently-staged version of the file, check that, and then restore the working tree version. (It can skip this process for files where the staged and working versions are identical.) This is not too complicated, but if it messes up it could catastrophically destroy the unstaged changes in the working tree.

  • Check out the entire repository and modified index into a fresh working tree and check that, then discard the temporary working tree. This is probably too expensive.

  • This one is kind of weird. It could temporarily commit the current index (using --no-verify), stash the working tree changes, and check the files. When the checks are finished, it would unstash the working tree changes, use git-reset --soft to undo the temporary commit, and proceed with the real commit if appropriate.

  • Come to think of it, this last one suggests a much better version of the same thing: instead of a pre-commit hook, use a post-commit hook. The post-commit hook will stash any leftover working tree changes, check the committed versions of the files, unstash the changes, and, if the checks failed, undo the commit with git-reset --soft.

Right now the last one looks much the best but perhaps there's something straightforward that I didn't think of yet.

[ Thanks to Adam Sjøgren, Jeffrey McClelland, and Jack Vickeridge for discussing this with me. Jeffrey McClelland also suggested that syntax checks could be profitably incorporated as a post-receive hook, which is run on the remote side when new commits are pushed to a remote. I said above that running the checks in the CI process seems too late, but the post-receive hook is earlier and might be just the thing. ]

[ Addendum: Daniel Holz wrote to tell me that the Yelp pre-commit frameworkhandles the worrisome case of unstaged working tree changes. The strategy is different from the ones I suggested above. If I'm reading this correctly, it records the unstaged changes in a patch file, which it sticks somewhere, and then checks out the index. If all the checks succeed, it completes the commit and then tries to apply the patch to restore the working tree changes. The checks in Yelp's framework might modify the staged files, and if they do, the patch might not apply; in this case it rolls back the whole commit. Thank you M. Holtz! ]

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Wed, 15 Feb 2017

More thoughts on a line of code with three errors

Yesterday I wrote, in great irritation, about a line of code I had written that contained three errors.

I said:

What can I learn from this? Most obviously, that I should have tested my code before I checked it in.

Afterward, I felt that this was inane, and that the matter required a little more reflection. We do not test every single line of every program we write; in most applications that would be prohibitively expensive, and in this case it would have been excessive.

The change I was making was in the format of the diagnostic that the program emitted as it finished to report how long it had taken to run. This is not an essential feature. If the program does its job properly, it is of no real concern if it incorrectly reports how long it took to run. Two of my errors were in the construction of the message. The third, however, was a syntax error that prevented the program from running at all.

Having reflected on it a little more, I have decided that I am only really upset about the last one, which necessitated an emergency Saturday-morning repair by a co-worker. It was quite acceptable not to notice ahead of time that the report would be wrong, to notice it the following day, and to fix it then. I would have said “oops” and quietly corrected the code without feeling like an ass.

The third problem, however, was serious. And I could have prevented it with a truly minimal amount of effort, just by running:

    perl -cw the-script

This would have diagnosed the syntax error, and avoided the main problem at hardly any cost. I think I usually remember to do something like this. Had I done it this time, the modified script would have gone into production, would have run correctly, and then I could have fixed the broken timing calculation on Monday.

In the previous article I showed the test program that I wrote to test the time calculation after the program produced the wrong output. I think it was reasonable to postpone writing this until after program ran and produced the wrong output. (The program's behavior in all other respects was correct and unmodified; it was only its report about its running time that was incorrect.) To have written the test ahead of time might be an excess of caution.

There has to be a tradeoff between cautious preparation and risk. Here I put everything on the side of risk, even though a tiny amount of caution would have eliminated most of the risk. In my haste, I made a bad trade.

[ Addendum 20170216: I am looking into automating the perl -cw check. ]

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Tue, 14 Feb 2017

How I got three errors into one line of code

At work we had this script that was trying to report how long it had taken to run, and it was using DateTime::Duration:

    my $duration = $end_time->subtract_datetime($start_time);
    my ( $hours, $minutes, $seconds ) =
    $duration->in_units( 'hours', 'minutes', 'seconds' );

    log_info "it took $hours hours $minutes minutes and $seconds seconds to run"

This looks plausible, but because DateTime::Duration is shit, it didn't work. Typical output:

    it took 0 hours 263 minutes and 19 seconds to run

I could explain to you why it does this, but it's not worth your time.

I got tired of seeing 0 hours 263 minutes show up in my cron email every morning, so I went to fix it. Here's what I changed it to:

    my $duration = $end_time->subtract_datetime_absolute($start_time)->seconds;
    my ( $hours, $minutes, $minutes ) = (int(duration/3600), int($duration/60)%60, $duration%3600);

I was at some pains to get that first line right, because getting DateTime to produce a useful time interval value is a tricky proposition. I did get the first line right. But the second line is just simple arithmetic, I have written it several times before, so I dashed it off, and it contains a syntax error, that duration/3600 is missing its dollar sign, which caused the cron job to crash the next day.

A co-worker got there before I did and fixed it for me. While he was there he also fixed the $hours, $minutes, $minutes that should have been $hours, $minutes, $seconds.

I came in this morning and looked at the cron mail and it said

    it took 4 hours 23 minutes and 1399 seconds to run

so I went back to fix the third error, which is that $duration%3600 should have been $duration%60. The thrice-corrected line has

    my ( $hours, $minutes, $seconds ) = (int($duration/3600), int($duration/60)%60, $duration%60);

What can I learn from this? Most obviously, that I should have tested my code before I checked it in. Back in 2013 I wrote:

Usually I like to draw some larger lesson from this sort of thing. … “Just write the tests, fool!”

This was a “just write the tests, fool!” moment if ever there was one. Madame Experience runs an expensive school, but fools will learn in no other.

I am not completely incorrigible. I did at least test the fixed code before I checked that in. The test program looks like this:

    sub dur {
      my $duration = shift;
      my ($hours, $minutes, $seconds ) = (int($duration/3600), int($duration/60)%60, $duration%60);
      sprintf  "%d:%02d:%02d", $hours, $minutes, $seconds;

    use Test::More;
    is(dur(0),  "0:00:00");
    is(dur(1),  "0:00:01");
    is(dur(59), "0:00:59");
    is(dur(60), "0:01:00");
    is(dur(62), "0:01:02");
    is(dur(122), "0:02:02");
    is(dur(3599), "0:59:59");
    is(dur(3600), "1:00:00");
    is(dur(10000), "2:46:40");

It was not necessary to commit the test program, but it was necessary to write it and to run it. By the way, the test program failed the first two times I ran it.

Three errors in one line isn't even a personal worst. In 2012 I posted here about getting four errors into a one-line program.

[ Addendum 20170215: I have some further thoughts on this. ]

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Tue, 07 Feb 2017

How many 24 puzzles are there?

[ Note: The tables in this article are important, and look unusually crappy if you read this blog through an aggregator. The properly-formatted version on my blog may be easier to follow. ]

A few months ago I wrote about puzzles of the following type: take four digits, say 1, 2, 7, 7, and, using only +, -, ×, and ÷, combine them to make the number 24. Since then I have been accumulating more and more material about these puzzles, which will eventually appear here. But meantime here is a delightful tangent.

In the course of investigating this I wrote programs to enumerate the solutions of all possible puzzles, and these programs were always much faster than I expected at first. It appears as if there are 10,000 possible puzzles, from «0,0,0,0» through «9,9,9,9». But a moment's thought shows that there are considerably fewer, because, for example, the puzzles «7,2,7,1», «1,2,7,7», «7,7,2,1», and «2,7,7,1» are all the same puzzle. How many puzzles are there really?

A back-of-the-envelope estimate is that only about 1 in 24 puzzles is really distinct (because there are typically 24 ways to rearrange the elements of a puzzle) and so there ought to be around !!\frac{10000}{24} \approx 417!! puzzles. This is an undercount, because there are fewer duplicates of many puzzles; for example there are not 24 variations of «1,2,7,7», but only 12. The actual number of puzzles turns out to be 715, which I think is not an obvious thing to guess.

Let's write !!S(d,n)!! for the set of sequences of length !!n!! containing up to !!d!! different symbols, with the duplicates removed: when two sequences are the same except for the order of their symbols, we will consider them the same sequence.

Or more concretely, we may imagine that the symbols are sorted into nondecreasing order, so that !!S(d,n)!! is the set of nondecreasing sequences of length !!n!! of !!d!! different symbols.

Let's also write !!C(d,n)!! for the number of elements of !!S(d,n)!!.

Then !!S(10, 4)!! is the set of puzzles where input is four digits. The claim that there are !!715!! such puzzles is just that !!C(10,4) = 715!!. A tabulation of !!C(\cdot,\cdot)!! reveals that it is closely related to binomial coefficients, and indeed that $$C(d,n)=\binom{n+d-1}{d-1}.\tag{$\heartsuit$}$$

so that the surprising !!715!! is actually !!\binom{13}{9}!!. This is not hard to prove by induction, because !!C(\cdot,\cdot)!! is easily shown to obey the same recurrence as !!\binom\cdot\cdot!!: $$C(d,n) = C(d-1,n) + C(d,n-1).\tag{$\spadesuit$}$$

To see this, observe that an element of !!C(d,n)!! either begins with a zero or with some other symbol. If it begins with a zero, there are !!C(d,n-1)!! ways to choose the remaining !!n-1!! symbols in the sequence. But if it begins with one of the other !!d-1!! symbols it cannot contain any zeroes, and what we really have is a length-!!n!! sequence of the symbols !!1\ldots (d-1)!!, of which there are !!C(d-1, n)!!.

0 0 0 0 1 1 1
0 0 0 1 1 1 2
0 0 0 2 1 1 3
0 0 0 3 1 1 4
0 0 1 1 1 2 2
0 0 1 2 1 2 3
0 0 1 3 1 2 4
0 0 2 2 1 3 3
0 0 2 3 1 3 4
0 0 3 3 1 4 4
0 1 1 1 2 2 2
0 1 1 2 2 2 3
0 1 1 3 2 2 4
0 1 2 2 2 3 3
0 1 2 3 2 3 4
0 1 3 3 2 4 4
0 2 2 2 3 3 3
0 2 2 3 3 3 4
0 2 3 3 3 4 4
0 3 3 3 4 4 4

Now we can observe that !!\binom74=\binom73!! (they are both 35) so that !!C(5,3) = C(4,4)!!. We might ask if there is a combinatorial proof of this fact, consisting of a natural bijection between !!S(5,3)!! and !!S(4,4)!!. Using the relation !!(\spadesuit)!! we have:

$$ \begin{eqnarray} C(4,4) & = & C(3, 4) + & C(4,3) \\ C(5,3) & = & & C(4,3) + C(5,2) \\ \end{eqnarray}$$

so part of the bijection, at least, is clear: There are !!C(4,3)!! elements of !!S(4,4)!! that begin with a zero, and also !!C(4,3)!! elements of !!S(5, 3)!! that do not begin with a zero, so whatever the bijection is, it ought to match up these two subsets of size 20. This is perfectly straightforward; simply match up !!«0, a, b, c»!! (blue) with !!«a+1, b+1, c+1»!! (pink), as shown at right.

But finding the other half of the bijection, between !!S(3,4)!! and !!S(5,2)!!, is not so straightforward. (Both have 15 elements, but we are looking for not just any bijection but for one that respects the structure of the elements.) We could apply the recurrence again, to obtain:

$$ \begin{eqnarray} C(3,4) & = \color{darkred}{C(2, 4)} + \color{darkblue}{C(3,3)} \\ C(5,2) & = \color{darkblue}{C(4,2)} + \color{darkred}{C(5,1)} \end{eqnarray}$$

and since $$ \begin{eqnarray} \color{darkred}{C(2, 4)} & = \color{darkred}{C(5,1)} \\ \color{darkblue}{C(3,3)} & = \color{darkblue}{C(4,2)} \end{eqnarray}$$

we might expect the bijection to continue in that way, mapping !!\color{darkred}{S(2,4) \leftrightarrow S(5,1)}!! and !!\color{darkblue}{S(3,3) \leftrightarrow S(4,2)}!!. Indeed there is such a bijection, and it is very nice.

To find the bijection we will take a detour through bitstrings. There is a natural bijection between !!S(d, n)!! and the bit strings that contain !!d-1!! zeroes and !!n!! ones. Rather than explain it with pseudocode, I will give some examples, which I think will make the point clear. Consider the sequence !!«1, 1, 3, 4»!!. Suppose you are trying to communicate this sequence to a computer. It will ask you the following questions, and you should give the corresponding answers:

  • “Is the first symbol 0?” (“No”)
  • “Is the first symbol 1?” (“Yes”)
  • “Is the second symbol 1?” (“Yes”)
  • “Is the third symbol 1?” (“No”)
  • “Is the third symbol 2?” (“No”)
  • “Is the third symbol 3?” (“Yes”)
  • “Is the fourth symbol 3?” (“No”)
  • “Is the fourth symbol 4?” (“Yes”)

At each stage the computer asks about the identity of the next symbol. If the answer is “yes” the computer has learned another symbol and moves on to the next element of the sequence. If it is “no” the computer tries guessing a different symbol. The “yes” answers become ones and “no” answers become zeroes, so that the resulting bit string is 0 1 1 0 0 1 0 1.

It sometimes happens that the computer figures out all the elements of the sequence before using up its !!n+d-1!! questions; in this case we pad out the bit string with zeroes, or we can imagine that the computer asks some pointless questions to which the answer is “no”. For example, suppose the sequence is !!«0, 1, 1, 1»!!:

  • “Is the first symbol 0?” (“Yes”)
  • “Is the second symbol 0?” (“No”)
  • “Is the second symbol 1?” (“Yes”)
  • “Is the third symbol 1?” (“Yes”)
  • “Is the fourth symbol 1?” (“Yes”)

The bit string is 1 0 1 1 1 0 0 0, where the final three 0 bits are the padding.

We can reverse the process, simply taking over the role of the computer. To find the sequence that corresponds to the bit string 0 1 1 0 1 0 0 1, we ask the questions ourselves and use the bits as the answers:

  • “Is the first symbol 0?” (“No”)
  • “Is the first symbol 1?” (“Yes”)
  • “Is the second symbol 1?” (“Yes”)
  • “Is the third symbol 1?” (“No”)
  • “Is the third symbol 2?” (“Yes”)
  • “Is the fourth symbol 2?” (“No”)
  • “Is the fourth symbol 3?” (“No”)
  • “Is the fourth symbol 4?” (“Yes”)

We have recovered the sequence !!«1, 1, 2, 4»!! from the bit string 0 1 1 0 1 0 0 1.

This correspondence establishes relation !!(\heartsuit)!! in a different way from before: since there is a natural bijection between !!S(d, n)!! and the bit strings with !!d-1!! zeroes and !!n!! ones, there are certainly !!\binom{n+d-1}{d-1}!! of them as !!(\heartsuit)!! says because there are !!n+d-1!! bits and we may choose any !!d-1!! to be the zeroes.

We wanted to see why !!C(5,3) = C(4,4)!!. The detour above shows that there is a simple bijection between

!!S(5,3)!! and the bit strings with 4 zeroes and 3 ones

on one hand, and between

!!S(4,4)!! and the bit strings with 3 zeroes and 4 ones

on the other hand. And of course the bijection between the two sets of bit strings is completely obvious: just exchange the zeroes and the ones.

The table below shows the complete bijection between !!S(4,4)!! and its descriptive bit strings (on the left in blue) and between !!S(5, 3)!! and its descriptive bit strings (on the right in pink) and that the two sets of bit strings are complementary. Furthermore the top portion of the table shows that the !!S(4,3)!! subsets of the two families correspond, as they should—although the correct correspondence is the reverse of the one that was displayed earlier in the article, not the suggested !!«0, a, b, c» \leftrightarrow «a+1, b+1, c+1»!! at all. Instead, in the correct table, the initial digit of the !!S(4,4)!! entry says how many zeroes appear in the !!S(5,3)!! entry, and vice versa; then the increment to the next digit says how many ones, and so forth.

!!S(4,4)!!(bits)(complement bits)!!S(5,3)!!
0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 1 4 4 4
0 0 0 1 1 1 1 0 1 0 0 0 0 0 1 0 1 1 3 4 4
0 0 0 2 1 1 1 0 0 1 0 0 0 0 1 1 0 1 3 3 4
0 0 0 3 1 1 1 0 0 0 1 0 0 0 1 1 1 0 3 3 3
0 0 1 1 1 1 0 1 1 0 0 0 0 1 0 0 1 1 2 4 4
0 0 1 2 1 1 0 1 0 1 0 0 0 1 0 1 0 1 2 3 4
0 0 1 3 1 1 0 1 0 0 1 0 0 1 0 1 1 0 2 3 3
0 0 2 2 1 1 0 0 1 1 0 0 0 1 1 0 0 1 2 2 4
0 0 2 3 1 1 0 0 1 0 1 0 0 1 1 0 1 0 2 2 3
0 0 3 3 1 1 0 0 0 1 1 0 0 1 1 1 0 0 2 2 2
0 1 1 1 1 0 1 1 1 0 0 0 1 0 0 0 1 1 1 4 4
0 1 1 2 1 0 1 1 0 1 0 0 1 0 0 1 0 1 1 3 4
0 1 1 3 1 0 1 1 0 0 1 0 1 0 0 1 1 0 1 3 3
0 1 2 2 1 0 1 0 1 1 0 0 1 0 1 0 0 1 1 2 4
0 1 2 3 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 2 3
0 1 3 3 1 0 1 0 0 1 1 0 1 0 1 1 0 0 1 2 2
0 2 2 2 1 0 0 1 1 1 0 0 1 1 0 0 0 1 1 1 4
0 2 2 3 1 0 0 1 1 0 1 0 1 1 0 0 1 0 1 1 3
0 2 3 3 1 0 0 1 0 1 1 0 1 1 0 1 0 0 1 1 2
0 3 3 3 1 0 0 0 1 1 1 0 1 1 1 0 0 0 1 1 1
1 1 1 1 0 1 1 1 1 0 0 1 0 0 0 0 1 1 0 4 4
1 1 1 2 0 1 1 1 0 1 0 1 0 0 0 1 0 1 0 3 4
1 1 1 3 0 1 1 1 0 0 1 1 0 0 0 1 1 0 0 3 3
1 1 2 2 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 2 4
1 1 2 3 0 1 1 0 1 0 1 1 0 0 1 0 1 0 0 2 3
1 1 3 3 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 2 2
1 2 2 2 0 1 0 1 1 1 0 1 0 1 0 0 0 1 0 1 4
1 2 2 3 0 1 0 1 1 0 1 1 0 1 0 0 1 0 0 1 3
1 2 3 3 0 1 0 1 0 1 1 1 0 1 0 1 0 0 0 1 2
1 3 3 3 0 1 0 0 1 1 1 1 0 1 1 0 0 0 0 1 1
2 2 2 2 0 0 1 1 1 1 0 1 1 0 0 0 0 1 0 0 4
2 2 2 3 0 0 1 1 1 0 1 1 1 0 0 0 1 0 0 0 3
2 2 3 3 0 0 1 1 0 1 1 1 1 0 0 1 0 0 0 0 2
2 3 3 3 0 0 1 0 1 1 1 1 1 0 1 0 0 0 0 0 1
3 3 3 3 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0

Observe that since !!C(d,n) = \binom{n+d-1}{d-1} = \binom{n+d-1}{n} = C(n+1, d-1)!! we have in general that !!C(d,n) = C(n+1, d-1)!!, which may be surprising. One might have guessed that since !!C(5,3) = C(4,4)!!, the relation was !!C(d,n) = C(d+1, n-1)!! and that !!S(d,n)!! would have the same structure as !!S(d+1, n-1)!!, but it isn't so. The two arguments exchange roles. Following the same path, we can identify many similar ‘coincidences’. For example, there is a simple bijection between the original set of 715 puzzles, which was !!S(10,4)!!, and !!S(5,9)!!, the set of nondecreasing sequences of !!0\ldots 4!! of length !!9!!.

[ Thanks to Bence Kodaj for a correction. ]

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Tue, 31 Jan 2017

Strangest Asian knockoff yet

Below, One Liberty Place, the second-tallest building in my home city of Philadelphia. (Completed 1987, height 288 meters.)

Below, Zhongtian International Mansion at Fortune Plaza, the tallest building in Ürümqi, capital city of Xinjiang in northwest China.

(Completed 2007, height 230 meters.)


[ Addendum: Perhaps I should mention that One Liberty Place is itself widely seen as a knockoff of the much more graceful and elegant Chrysler Building in New York City. (Completed 1930, height 319 meters.) ]

[ Addendum: I brought this to the attention of GroJLart, the foulmouthed architecture blogger who knows everything, absolutely everything, about Philadelphia buildings, and he said “Thanks. I wrote an article on the same subject in 2011”. Of course. ]

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Mon, 30 Jan 2017

Digit symbols in the Parshvanatha magic square

In last month's article about the magic square at the Parshvanatha temple, shown at right, I said:

It has come to my attention that the digit symbols in the magic square are not too different from the current forms of the digit symbols in the Gujarati script. The temple is not very close to Gujarat or to the area in which Gujarati is common, so I guess that the digit symbols in Indian languages have evolved in the past thousand years, with the Gujarati versions remaining closest to the ancient forms, or else perhaps Gujarati was spoken more widely a thousand years ago. I would be interested to hear about this from someone who knows.

Shreevatsa R. replied in detail, and his reply was so excellent that, finding no way to improve it by adding or taking away, I begged his permission to republish it without change, which he generously granted.

Am sending this email to say:

  1. Why it shouldn't be surprising if the temple had Gujarati numerals
  2. Why the numerals aren't Gujarati numerals :-)

The Parshvanatha temple is located in the current state of Madhya Pradesh. Here is the location of the temple within a map of the state:

And here you can see that the above state of Madhya Pradesh (14 in the image below) is adjacent to the state of Gujarat (7):

The states of India are (sort of) organized along linguistic lines, and neighbouring states often have overlap or similarities in their languages. So a priori it shouldn't be too surprising if the language is that of a neighbouring state.

But, as you rightly say, the location of the Parshvanatha temple is actually quite far from the state (7) where Gujarat is spoken; it's closer to 27 in the above map (state named Uttar Pradesh).

Well, the Parshvanatha temple is believed to have been built "during the reign of the Chandela king Dhanga", and the Chandela kings were feudatories (though just beginning to assert sovereignty at the time) of the Gurjara-Pratihara kings, and "Gurjara" is where the name of the language of "Gujarati" comes from. So it's possible that they used the "official" language of the reigning kings, as with colonies. In fact the green area of the Gurjara-Pratihara kings in this map covers the location of the Parshvanatha temple:

But actually this is not a very convincing argument, because the link between Gurjara-Pratiharas and modern Gujarati is not too strong (at least I couldn't find it in a few minutes on Wikipedia :P)

So moving on...

Are the numerals really similar to Gujarati numerals? These are the numbers 1 to 16 from your blog post, ordered according to the usual order:

These are the numerals in a few current Indic scripts (as linked from your blog post):

Look at the first two rows above. Perhaps because of my familiarity with Devanagari, I cannot really see any big difference between the Devanagari and Gujarati symbols except for the 9: the differences are as minor as variation between fonts. (To see how much the symbols can change because of font variation, one can go to Google Fonts' Devanagari page and Google Fonts' Gujarati page and click on one of the sample texts and enter "० १ २ ३ ४ ५ ६ ७ ८ ९" and "૦ ૧ ૨ ૩ ૪ ૫ ૬ ૭ ૮ ૯" respectively, then "Apply to all fonts". Some fonts are bad, though.)

(In fact, even the Gurmukhi and Tibetan are somewhat recognizable, for someone who can read Devanagari.)

So if we decide that the Parshvanatha temple's symbols are actually closer not to modern Gujarati but to modern Devanagari (e.g. the "3" has a tail in the temple symbols which is present in Devanagari but missing in Gujarati), then the mystery disappears: Devanagari is still the script used in the state of Madhya Pradesh (and Uttar Pradesh, etc: it's the script used for Hindi, Marathi, Nepali, Sanskrit, and many other languages).

Finally, for the complete answer, we can turn to history.

The Parshvanatha temple was built during 950 to 970 CE. Languages: Modern Gujarati dates from 1800, Middle Gujarati from ~1500 to 1800, Old Gujarati from ~1100 to 1500. So the temple is older than the earliest language called "Gujarati". (Similarly, modern Hindi is even more recent.) Turning to scripts instead: see under Brahmic scripts.

So at the time the temple was built, neither Gujarati script nor Devanagari proper existed. The article on the Gujarati script traces its origin to the Devanagari script, which itself is a descendant of Nagari script.

At right are the symbols from the Nagari script, which I think are closer in many respects to the temple symbols.

So overall, if we trace the numerals in (a subset of) the family tree of scripts:

Brahmi > Gupta > Nagari > Devanagari > Gujarati

we'll find that the symbols of the temple are somewhere between the "Nagari" and "Devanagari" forms. (Most of the temple digits are the same as in the "Nagari" example above, except for the 5 which is closer to the Devanagari form.)

BTW, your post was about the numerals, but from being able to read modern Devanagari, I can also read some of the words above the square: the first line ends with ".. putra śrī devasarmma" (...पुत्र श्री देव‍सर्म्म) (Devasharma, son of...), and these words have the top bar which is missing in Gujarati script.

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Thu, 15 Dec 2016

Let's decipher a thousand-year-old magic square

The Parshvanatha temple in Madhya Pradesh, India was built around 1,050 years ago. Carved at its entrance is this magic square:

The digit signs have changed in the past thousand years, but it's a quick and fun puzzle to figure out what they mean using only the information that this is, in fact, a magic square.

A solution follows. No peeking until you've tried it yourself!

There are 9 one-digit entries
and 7 two-digit entries
so we can guess that the entries are the numbers 1 through 16, as is usual, and the magic sum is 34. The appears in the same position in all the two-digit numbers, so it's the digit 1. The other digit of the numeral is , and this must be zero. If it were otherwise, it would appear on its own, as does for example the from or the from .

It is tempting to imagine that is 4. But we can see it's not so. Adding up the rightmost column, we get

+ + + =
+ 11 + + =
(10 + ) + 11 + + = 34,

so that must be an odd number. We know it isn't 1 (because is 1), and it can't be 7 or 9 because appears in the bottom row and there is no 17 or 19. So must be 3 or 5.

Now if were 3, then would be 13, and the third column would be

+ + + =
1 + + 10 + 13 = 34,

and then would be 10, which is too big. So must be 5, and this means that is 4 and is 8. ( appears only a as a single-digit numeral, which is consistent with it being 8.)

The top row has

+ + + =
+ + 1 + 14 =
+ (10 + ) + 1 + 14 = 34

so that + = 9. only appears as a single digit and we already used 8 so must be 7 or 9. But 9 is too big, so it must be 7, and then is 2.

is the only remaining unknown single-digit numeral, and we already know 7 and 8, so is 9. The leftmost column tells us that is 16, and the last two entries, and are easily discovered to be 13 and 3. The decoded square is:


I like that people look at the right-hand column and immediately see 18 + 11 + 4 + 8 but it's actually 14 + 11 + 5 + 4.

This is an extra-special magic square: not only do the ten rows, columns, and diagonals all add up to 34, so do all the four-cell subsquares, so do any four squares arranged symmetrically about the center, and so do all the broken diagonals that you get by wrapping around at the edges.

[ Addendum: It has come to my attention that the digit symbols in the magic square are not too different from the current forms of the digit symbols in the Gujarati script. ]

[ Addendum 20161217: The temple is not very close to Gujarat or to the area in which Gujarati is common, so I guess that the digit symbols in Indian languages have evolved in the past thousand years, with the Gujarati versions remaining closest to the ancient forms, or else perhaps Gujarati was spoken more widely a thousand years ago. I would be interested to hear about this from someone who knows. ]

[ Addendum 20170130: Shreevatsa R. has contributed a detailed discussion of the history of the digit symbols. ]

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