# News aggregator

### Haskell Hackathon in Berlin, Friday 26 September 2014 - Sunday 28 September 2014

### Markov Text Generator & Randomness

### Haskell on JVM (using LLVM backend)

### Locating Modules in Haskell

What is the rationale in locating the standard modules in Haskell?

For instance, why Monoid, Functor, etc. are located in Data, while Applicative, Monad, Category, etc. are located in Control?

submitted by BanX[link] [10 comments]

### GHCi: Behave nicely on `-e`, like `ghc` and other programs

### Enhancement: Default cabal to `-p` profiling enabled

### Retrieving information about type families

### GhcPlugin-writing and "finding things"

### Let me tell you about the types of data

### Neil Mitchell: Applicative vs Monadic build systems

*Summary: Shake is a monadic build system, and monadic build systems are more powerful than applicative ones.*

Several people have wondered if the dependencies in the Shake build system are monadic, and if Make dependencies are applicative. In this post I'll try and figure out what that means, and show that the claim is somewhat true.

Gergo recently wrote a good primer on the concepts of Applicative, Monads and Arrows (it is worth reading the first half if you are unfamiliar with monad or applicative). Using a similar idea, we can model a simple build system as a set of rules:

rules :: [(FilePath, Action String)]rules = [("a+b", do a <- need "a"; b <- need "b"; return (a ++ b))

,("a" , return "Hello ")

,("b" , return "World")

]

Each rule is on a separate line, containing a pair of the file the rule produces (e.g. a for the second rule) and the action that produces the files contents (e.g. return "Hello"). I've used need to allow a rule to use the contents of another file, so the rule for a+b depends on the files a and b, then concatenates their contents. We can run these rules to produce all the files. We've written these rules assuming Action is a Monad, using the do notation for monads. However, for the above build system, we can restrict ourselves to Applicative functions:

rules = [("a+b", (++) <$> need "a" <*> need "b"),("a" , pure "Hello ")

,("b" , pure "World")

]

If Action is applicative but not monadic then we can statically (without running any code operating on file contents) produce a dependency graph. If Action is monadic we can't generate a graph upfront, but there are some build systems that cannot be expressed applicatively. In particular, using a monad we can write a "dereferencing" build system:

rules = [("!a", do a <- need "a"; need a),("a" , pure "b")

,("b" , pure "Goodbye")

]

To build the file !a we first require the file a (which produces the contents b), then we require the file b (which produces the contents Goodbye). Note that the first rule has changed b *the content* into b *the file name*. In general, to move information from the file content to a file name, requires a monad. Alternatively stated, a monad lets you chose future dependencies based on the results of previous dependencies.

One realistic example (from the original Shake paper), is building a .tar file from the list of files contained in a file. Using Shake we can write the Action:

contents <- readFileLines "list.txt"need contents

cmd "tar -cf" [out] contents

The only build systems that I'm aware of that are monadic are redo, SCons and Shake-inspired build systems (including Shake itself, Jenga in OCaml, and several Haskell alternatives).

While it is the case that Shake is monadic, and that monadic build systems are more powerful than applicative ones, it is *not* the case that Make is applicative. In fact, almost no build systems are purely applicative. Looking at the build shootout, every build system tested can implement the !a example (provided the file a is not a build product), despite several systems being based on applicative dependencies.

Looking at Make specifically, it's clear that the output: input1 input2 formulation of dependencies *is* applicative in nature. However, there are at least two aspects I'm aware of that increase the power of Make:

- Using $(shell cat list.txt) I can splice the contents of list.txt into the Makefile, reading the contents of list.txt before the dependencies are parsed.
- Using -include file.d I can include additional rules that are themselves produced by the build system.

It seems every "applicative" build system contains some mechanism for extending its power. I believe some are strictly less powerful than monadic systems, while others may turn out to be an encoding of monadic rules. However, I think that an explicitly monadic definition provides a clearer foundation.

### Looking for list comprehensions use cases

### A Next Generation Smart Contract and Decentralized Application Platform

A Next Generation Smart Contract and Decentralized Application Platform, Vitalik Buterin.

When Satoshi Nakamoto first set the Bitcoin blockchain into motion in January 2009, he was simultaneously introducing two radical and untested concepts. The first is the "bitcoin", a decentralized peer-to-peer online currency that maintains a value without any backing, intrinsic value or central issuer. So far, the "bitcoin" as a currency unit has taken up the bulk of the public attention, both in terms of the political aspects of a currency without a central bank and its extreme upward and downward volatility in price. However, there is also another, equally important, part to Satoshi's grand experiment: the concept of a proof of work-based blockchain to allow for public agreement on the order of transactions. Bitcoin as an application can be described as a first-to-file system: if one entity has 50 BTC, and simultaneously sends the same 50 BTC to A and to B, only the transaction that gets confirmed first will process. There is no intrinsic way of determining from two transactions which came earlier, and for decades this stymied the development of decentralized digital currency. Satoshi's blockchain was the first credible decentralized solution. And now, attention is rapidly starting to shift toward this second part of Bitcoin's technology, and how the blockchain concept can be used for more than just money.

Commonly cited applications include using on-blockchain digital assets to represent custom currencies and financial instruments ("colored coins"), the ownership of an underlying physical device ("smart property"), non-fungible assets such as domain names ("Namecoin") as well as more advanced applications such as decentralized exchange, financial derivatives, peer-to-peer gambling and on-blockchain identity and reputation systems. Another important area of inquiry is "smart contracts" - systems which automatically move digital assets according to arbitrary pre-specified rules. For example, one might have a treasury contract of the form "A can withdraw up to X currency units per day, B can withdraw up to Y per day, A and B together can withdraw anything, and A can shut off B's ability to withdraw". The logical extension of this is decentralized autonomous organizations (DAOs) - long-term smart contracts that contain the assets and encode the bylaws of an entire organization. **What Ethereum intends to provide is a blockchain with a built-in fully fledged Turing-complete programming language that can be used to create "contracts" that can be used to encode arbitrary state transition functions, allowing users to create any of the systems described above, as well as many others that we have not yet imagined, simply by writing up the logic in a few lines of code.**

Includes code samples.

### What does dependent type do that Haskell's type system can't?

I'm pretty new to type theory. What are the major differences between dependent type and Haskell's type system. Is Haskell's strictly a Hindley–Milner type system?

I've heard ATS's support for dependent type isn't complete. What's incomplete about it? How do ATS's and Haskell's type system compare to each other - is ATS's type system truly more "advanced" than Haskell's?

submitted by lebbe[link] [41 comments]

### Mark Jason Dominus: When do n and 2n have the same digits?

[This article was published last month on the math.stackexchange blog, which seems to have died young, despite many earnest-sounding promises beforehand from people who claimed they would contribute material. I am repatriating it here.]

A recent question on math.stackexchange asks for the smallest positive integer for which the number has the same decimal digits in some other order.

Math geeks may immediately realize that has this property, because it is the first 6 digits of the decimal expansion of , and the cyclic behavior of the decimal expansion of is well-known. But is this the *minimal* solution? It is not. Brute-force enumeration of the solutions quickly reveals that there are 12 solutions of 6 digits each, all permutations of , and that larger solutions, such as 1025874 and 1257489 seem to follow a similar pattern. What is happening here?

Stuck in Dallas-Fort Worth airport one weekend, I did some work on the problem, and although I wasn't able to solve it completely, I made significant progress. I found a method that allows one to hand-calculate that there is no solution with fewer than six digits, and to enumerate all the solutions with 6 digits, including the minimal one. I found an explanation for the surprising behavior that solutions tend to be permutations of one another. The short form of the explanation is that there are fairly strict conditions on which *sets* of digits can appear in a solution of the problem. But once the set of digits is chosen, the conditions on that *order* of the digits in the solution are fairly lax.

So one typically sees, not only in base 10 but in other bases, that the solutions to this problem fall into a few classes that are all permutations of one another; this is exactly what happens in base 10 where all the 6-digit solutions are permutations of . As the number of digits is allowed to increase, the strict first set of conditions relaxes a little, and other digit groups appear as solutions.

NotationThe property of interest, , is that the numbers and have exactly the same base- digits. We would like to find numbers having property for various , and we are most interested in . Suppose is an -digit numeral having property ; let the (base-) digits of be and similarly the digits of are . The reader is encouraged to keep in mind the simple example of which we will bring up from time to time.

Since the digits of and are the same, in a different order, we may say that for some permutation . In general might have more than one cycle, but we will suppose that is a single cycle. All the following discussion of will apply to the individual cycles of in the case that is a product of two or more cycles. For our example of , we have in cycle notation. We won't need to worry about the details of , except to note that completely exhaust the indices , and that because is an -cycle.

Conditions on the set of digits in a solutionFor each we have $$a_{P(i)} = b_{i} \equiv 2a_{i} + c_i\pmod R $$ where the ‘carry bit’ is either 0 or 1 and depends on whether there was a carry when doubling . (When we are in the rightmost position and there is never a carry, so .) We can then write:

$$\begin{align} a_{P(P(i))} &= 2a_{P(i)} + c_{P(i)} \\ &= 2(2a_{i} + c_i) + c_{P(i)} &&= 4a_i + 2c_i + c_{P(i)}\\ a_{P(P(P(i)))} &= 2(4a_i + 2c_i + c_{P(P(i)})) + c_{P(i)} &&= 8a_i + 4c_i + 2c_{P(i)} + c_{P(P(i))}\\ &&&\vdots\\ a_{P^n(i)} &&&= 2^na_i + v \end{align} $$

all equations taken . But since is an -cycle, , so we have $$a_i \equiv 2^na_i + v\pmod R$$ or equivalently $$\big(2^n-1\big)a_i + v \equiv 0\pmod R\tag{$\star$}$$ where depends only on the values of the carry bits —the are precisely the binary digits of .

Specifying a particular value of and that satisfy this equation completely determines all the . For example, is a solution when because , and this solution allows us to compute

$$\def\db#1{\color{darkblue}{#1}}\begin{align} a_0&&&=2\\ a_{P(0)} &= 2a_0 &+ \db0 &= 4\\ a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 0 \\ a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 1\\ \hline a_{P^4(0)} &= 2a_{P^3(0)} &+ \db0 &= 2\\ \end{align}$$

where the carry bits are visible in the third column, and all the sums are taken . Note that as promised. This derivation of the entire set of from a single one plus a choice of is crucial, so let's see one more example. Let's consider . Then we want to choose and so that where . One possible solution is . Then we can derive the other as follows:

$$\begin{align} a_0&&&=5\\ a_{P(0)} &= 2a_0 &+ \db1 &= 1\\ a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 2 \\\hline a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 5\\ \end{align}$$

And again we have as required.

Since the bits of are used cyclically, not every pair of will yield a different solution. Rotating the bits of and pairing them with different choices of will yield the same cycle of digits starting from a different place. In the first example above, we had . If we were to take (which also solves ) we would get the same cycle of values of the but starting from instead of from , and similarly if we take or . So we can narrow down the solution set of by considering only the so-called bracelets of rather than all possible values. Two values of are considered equivalent as bracelets if one is a rotation of the other. When a set of -values are equivalent as bracelets, we need only consider one of them; the others will give the same cyclic sequence of digits, but starting in a different place. For , for example, the bracelets are and ; the sequences and being equivalent to , and so on.

ExampleLet us take , so we want to find 3-digit numerals with property . According to we need where . There are 9 possible values for ; for each one there is at most one possible value of that makes the sum zero:

$$\pi \approx 3 $$

$$\begin{array}{rrr} a_i & 7a_i & v \\ \hline 0 & 0 & 0 \\ 1 & 7 & 2 \\ 2 & 14 & 4 \\ 3 & 21 & 6 \\ 4 & 28 & \\ 5 & 35 & 1 \\ 6 & 42 & 3 \\ 7 & 49 & 5 \\ 8 & 56 & 7 \\ \end{array} $$

(For there is no solution.) We may disregard the non-bracelet values of , as these will give us solutions that are the same as those given by bracelet values of . The bracelets are:

$$\begin{array}{rl} 000 & 0 \\ 001 & 1 \\ 011 & 3 \\ 111 & 7 \end{array}$$

so we may disregard the solutions exacpt when . Calculating the digit sequences from these four values of and the corresponding we find:

$$\begin{array}{ccl} a_0 & v & \text{digits} \\ \hline 0 & 0 & 000 \\ 5 & 1 & 512 \\ 6 & 3 & 637 \\ 8 & 7 & 888 \ \end{array} $$

(In the second line, for example, we have , so and .)

Any number of three digits, for which contains exactly the same three digits, in base 9, must therefore consist of exactly the digits or .

A warningAll the foregoing assumes that the permutation is *a single cycle*. In general, it may not be. Suppose we did an analysis like that above for and found that there was no possible digit set, other than the trivial set 00000, that satisfied the governing equation . This would *not* completely rule out a base-10 solution with 5 digits, because the analysis only rules out a *cyclic* set of digits. There could still be a solution where was a product of a and a -cycle, or a product of still smaller cycles.

Something like this occurs, for example, in the case. Solving the governing equation yields only four possible digit cycles, namely , and . But there are several additional solutions: and . These correspond to permutations with more than one cycle. In the case of , for example, exchanges the and the , and leaves the and the fixed.

For this reason we cannot rule out the possibility of an -digit solution without first considering all smaller .

The Large Equals Odd ruleWhen is even there is a simple condition we can use to rule out certain sets of digits from being single-cycle solutions. Recall that and . Let us agree that a digit is *large* if and *small* otherwise. That is, is large if, upon doubling, it causes a carry into the next column to the left.

Since , where the are carry bits, we see that, except for , the digit is odd precisely when there is a carry from the next column to the right, which occurs precisely when is large. Thus the number of odd digits among is equal to the number of large digits among . This leaves the digits and uncounted. But is never odd, since there is never a carry in the rightmost position, and is always small (since otherwise would have digits, which is not allowed). So the number of large digits in is exactly equal to the number of odd digits in . And since and have exactly the same digits, the number of large digits in is equal to the number of odd digits in . Observe that this is the case for our running example : there is one odd digit and one large digit (the 4).

When is odd the analogous condition is somewhat more complicated, but since the main case of interest is , we have the useful rule that:

For even, the number of odd digits in any solution is equal to the number of large digits in . Conditions on the order of digits in a solutionWe have determined, using the above method, that the digits might form a base-9 numeral with property . Now we would like to arrange them into a base-9 numeral that actually does have that property. Again let us write and , with . Note that if , then (if there was a carry from the next column to the right) or (if there was no carry), but since is impossible, we must have and therefore must be small, since there is no carry into position . But since is also one of , and it cannot also be , it must be . This shows that the 1, unless it appears in the rightmost position, must be to the left of the ; it cannot be to the left of the . Similarly, if then , because is impossible, so the must be to the left of a large digit, which must be the . Similar reasoning produces no constraint on the position of the ; it could be to the left of a small digit (in which case it doubles to ) or a large digit (in which case it doubles to ). We can summarize these findings as follows:

$$\begin{array}{cl} \text{digit} & \text{to the left of} \\ \hline 1 & 1, 2, \text{end} \\ 2 & 5 \\ 5 & 1,2,5,\text{end} \end{array}$$

Here “end” means that the indicated digit could be the rightmost.

Furthermore, the left digit of must be small (or else there would be a carry in the leftmost place and would have 4 digits instead of 3) so it must be either 1 or 2. It is not hard to see from this table that the digits must be in the order or , and indeed, both of those numbers have the required property: , and .

This was a simple example, but in more complicated cases it is helpful to draw the order constraints as a graph. Suppose we draw a graph with one vertex for each digit, and one additional vertex to represent the end of the numeral. The graph has an edge from vertex to whenever can appear to the left of . Then the graph drawn for the table above looks like this:

A 3-digit numeral with property corresponds to a path in this graph that starts at one of the nonzero small digits (marked in blue), ends at the red node marked ‘end’, and visits each node exactly once. Such a path is called *hamiltonian*. Obviously, self-loops never occur in a hamiltonian path, so we will omit them from future diagrams.

Now we will consider the digit set , again base 9. An analysis similar to the foregoing allows us to construct the following graph:

Here it is immediately clear that the only hamiltonian path is , and indeed, .

In general there might be multiple instances of a digit, and so multiple nodes labeled with that digit. Analysis of the case produces a graph with no legal start nodes and so no solutions, unless leading zeroes are allowed, in which case is a perfectly valid solution. Analysis of the case produces a graph with no path to the end node and so no solutions. These two trivial patterns appear for all and all , and we will ignore them from now on.

Returning to our ongoing example, in base 8, we see that and must double to and , so must be to the left of small digits, but and can double to either or and so could be to the left of anything. Here the constraints are so lax that the graph doesn't help us narrow them down much:

Observing that the only arrow into the 4 is from 0, so that the 4 must follow the 0, and that the entire number must begin with 1 or 2, we can enumerate the solutions:

1042 1204 2041 2104If leading zeroes are allowed we have also:

0412 0421All of these are solutions in base 8.

The case ofNow we turn to our main problem, solutions in base 10.

To find *all* the solutions of length 6 requires an enumeration of smaller solutions, which, if they existed, might be concatenated into a solution of length 6. This is because our analysis of the digit sets that can appear in a solution assumes that the digits are permuted *cyclically*; that is, the permutations that we considered had only one cycle each.

There are no smaller solutions, but to prove that the length 6 solutions are minimal, we must analyze the cases for smaller and rule them out. We now produce a complete analysis of the base 10 case with and . For there is only the trivial solution of , which we disregard. (The question asked for a positive number anyway.)

For , we want to find solutions of where is a two-bit bracelet number, one of or . Tabulating the values of and that solve this equation we get:

$$\begin{array}{ccc} v& a_i \\ \hline 0 & 0 \\ 1& 3 \\ 3& 9 \\ \end{array}$$

We can disregard the and solutions because the former yields the trivial solution and the latter yields the nonsolution . So the only possibility we need to investigate further is , which corresponds to the digit sequence : Doubling gives us and doubling , plus a carry, gives us again.

But when we tabulate of which digits must be left of which informs us that there is no solution with just and , because the graph we get, once self-loops are eliminated, looks like this:

which obviously has no hamiltonian path. Thus there is no solution for .

For we need to solve the equation where is a bracelet number in , specifically one of or . Since and are relatively prime, for each there is a single that solves the equation. Tabulating the possible values of as before, and this time omitting rows with no solution, we have:

$$\begin{array}{rrl} v & a_i & \text{digits}\\ \hline 0& 0 & 000\\ 1& 7 & 748 \\ 3& 1 & 125\\ 7&9 & 999\\ \end{array}$$

The digit sequences and yield trivial solutions or nonsolutions as usual, and we will omit them in the future. The other two lines suggest the digit sets and , both of which fails the “odd equals large” rule.

This analysis rules out the possibility of a digit set with , but it does not *completely* rule out a 3-digit solution, since one could be obtained by concatenating a one-digit and a two-digit solution, or three one-digit solutions. However, we know by now that no one- or two-digit solutions exist. Therefore there are no 3-digit solutions in base 10.

For the governing equation is where is a 4-bit bracelet number, one of . This is a little more complicated because . Tabulating the possible digit sets, we get:

$$\begin{array}{crrl} a_i & 15a_i& v & \text{digits}\\ \hline 0 & 0 & 0 & 0000\\ 1 & 5 & 5 & 1250\\ 1 & 5 & 15 & 1375\\ 2 & 0 & 0 & 2486\\ 3 & 5 & 5 & 3749\\ 3 & 5 & 15 & 3751\\ 4 & 0 & 0 & 4862\\ 5 & 5 & 5 & 5012\\ 5 & 5 & 5 & 5137\\ 6 & 0 & 0 & 6248\\ 7 & 5 & 5 & 7493\\ 7 & 5 & 5 & 7513\\ 8 & 0 & 0 & 8624 \\ 9 & 5 & 5 & 9874\\ 9 & 5 & 15 & 9999 \\ \end{array}$$

where the second column has been reduced mod . Note that even restricting to bracelet numbers the table still contains duplicate digit sequences; the 15 entries on the right contain only the six basic sequences , and . Of these, only and obey the odd equals large criterion, and we will disregard and as usual, leaving only . We construct the corresponding graph for this digit set as follows: must double to , not , so must be left of a large number or . Similarly must be left of or . must also double to , so must be left of . Finally, must double to , so must be left of or the end of the numeral. The corresponding graph is:

which evidently has no hamiltonian path: whichever of 3 or 4 we start at, we cannot visit the other without passing through 7, and then we cannot reach the end node without passing through 7 a second time. So there is no solution with and .

We leave this case as an exercise. There are 8 solutions to the governing equation, all of which are ruled out by the odd equals large rule.

For the possible solutions are given by the governing equation where is a 6-bit bracelet number, one of . Tabulating the possible digit sets, we get:

$$\begin{array}{crrl} v & a_i & \text{digits}\\ \hline 0 & 0 & 000000\\ 1 & 3 & 362486 \\ 3 & 9 & 986249 \\ 5 & 5 & 500012 \\ 7 & 1 & 124875 \\ 9 & 7 & 748748 \\ 11 & 3 & 362501 \\ 13 & 9 & 986374 \\ 15 & 5 & 500137 \\ 21 & 3 & 363636 \\ 23 & 9 & 989899 \\ 27 & 1 & 125125 \\ 31 & 3 & 363751 \\ 63 & 9 & 999999 \\ \end{array}$$

After ignoring and as usual, the large equals odd rule allows us to ignore all the other sequences except and . The latter fails for the same reason that did when . But , the lone survivor, gives us a complicated derived graph containing many hamiltonian paths, every one of which is a solution to the problem:

It is not hard to pick out from this graph the minimal solution , for which , and also our old friend for which .

We see here the reason why all the small numbers with property contain the digits . The constraints on *which* digits can appear in a solution are quite strict, and rule out all other sequences of six digits and all shorter sequences. But once a set of digits passes these stringent conditions, the constraints on it are much looser, because is only required to have the digits of in *some* order, and there are many possible orders, many of which will satisfy the rather loose conditions involving the distribution of the carry bits. This graph is typical: it has a set of small nodes and a set of large nodes, and each node is connected to either *all* the small nodes or *all* the large nodes, so that the graph has many edges, and, as in this case, a largish clique of small nodes and a largish clique of large nodes, and as a result many hamiltonian paths.

This analysis is tedious but is simple enough to perform by hand in under an hour. As increases further, enumerating the solutions of the governing equation becomes very time-consuming. I wrote a simple computer program to perform the analysis for given and , and to emit the possible digit sets that satisfied the large equals odd criterion. I had wondered if *every* base-10 solution contained equal numbers of the digits and . This is the case for (where the only admissible digit set is ), for (where the only admissible sets are and ), and for (where the only admissible sets are and ). But when we reach the increasing number of bracelets has loosened up the requirements a little and there are 5 admissible digit sets. I picked two of the promising-seeming ones and quickly found by hand the solutions and , both of which wreck any theory that the digits must all appear the same number of times.

Thanks to Karl Kronenfeld for corrections and many helpful suggestions.