### It's in the papers...

I don't necessarily agree with the New York

"Haskell in the Newsroom" is worth your time.

*Times*'s editorial position, but I applaud this fellow."Haskell in the Newsroom" is worth your time.

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I don't necessarily agree with the New York *Times*'s editorial position, but I applaud this fellow.

"Haskell in the Newsroom" is worth your time.

"Haskell in the Newsroom" is worth your time.

- Get link
- Google+
- Other Apps

fivethirtyeight.com has a weekly column, "The Riddler". The Riddler poses two problems in each column. The first, "Riddler Express", is intended to be an easier problem that one can solve quickly, while the second, "Riddler Classic", is more difficult. Modulo vacations, it appears each Friday, and if you submit a solution by the end of the following Sunday (Eastern Time) , you will be among those who might be given credit for the solution in the next column.

The August 28th column's Classic problem is as follows: what's the longest sequence of integers x[i] for 1 ≤ i ≤ n such that

for all i, 1 ≤ x[i] ≤ 100for all i < n, either x[i+1] is a multiple of x[i] or x[i+1] is a factor of x[i]for all i, j, x[i] = x[j] iff i = j, i.e. the x[i] are all distinct (The last constraint avoids trivial sequences like 2, 4, 2, 4, 2, 4... which can go on forever if repeats are permitted.)

One way to characterize this problem is to look at it as a graph with a hu…

The August 28th column's Classic problem is as follows: what's the longest sequence of integers x[i] for 1 ≤ i ≤ n such that

for all i, 1 ≤ x[i] ≤ 100for all i < n, either x[i+1] is a multiple of x[i] or x[i+1] is a factor of x[i]for all i, j, x[i] = x[j] iff i = j, i.e. the x[i] are all distinct (The last constraint avoids trivial sequences like 2, 4, 2, 4, 2, 4... which can go on forever if repeats are permitted.)

One way to characterize this problem is to look at it as a graph with a hu…

After grinding through much of the "20 Intermediate Exercises" and just about all of the Monad Challenges, Monads make much more sense to me.

Therefore, I will follow the excellent advice of Stephen Diehl, and will not write a monad-analogy tutorial. Instead, I will say: do the 20 Intermediate Exercises, and take the Monad Challenges. To paraphrase Euclid, there is no royal road to Monads; the exercises and challenges take you down that road at whose end you'll see at least part of why Monads are so useful.

If you were trying to learn group theory and people were standing around you saying "groups are like the integers", "groups are like Rubik's Cube", "groups are like m x n matrices", "groups are like baryons and mesons", your situation would be much like the student of Haskell amidst all the monad analogy tutorials. In a sense they're backwards. All those things are groups, just as burritos et al. can at least be thought…

Therefore, I will follow the excellent advice of Stephen Diehl, and will not write a monad-analogy tutorial. Instead, I will say: do the 20 Intermediate Exercises, and take the Monad Challenges. To paraphrase Euclid, there is no royal road to Monads; the exercises and challenges take you down that road at whose end you'll see at least part of why Monads are so useful.

If you were trying to learn group theory and people were standing around you saying "groups are like the integers", "groups are like Rubik's Cube", "groups are like m x n matrices", "groups are like baryons and mesons", your situation would be much like the student of Haskell amidst all the monad analogy tutorials. In a sense they're backwards. All those things are groups, just as burritos et al. can at least be thought…

Darn it, I fell into the trap. An early exercise in Python at exercism.io wants a function that in Haskell would have the type **Int -> Bool** and indicate whether a year is a leap year (under Gregorian rules, ignoring when various locales made the switch to keep it simple).

Some submissions are from people not yet comfortable with assigning/returning Booleans other than the constants**True** or **False**, or maybe not comfortable with the lazy nature of Python's **and** and **or**. Their versions are full of **if** statements. I bashed out

**defis_leap_year(year):returnyear%(400ifyear%100==0else4)==0**
and it worked fine... but then I fell in. How about

**return year % 16 == 0 if year % 25 == 0 else year % 4 == 0**
Nah... no need to make the next guy, who could be me, do the prime factorization of 100 again and figure out how that ties to the leap year rules? OK, here's one way to look at it...

**return (year // 100 if year % 100 == 0 else year) % 4 == 0***i.e.* on century boundaries, the century has to be divis…

Some submissions are from people not yet comfortable with assigning/returning Booleans other than the constants

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