How to count the recursion times in Haskell?

Question

I am trying to write a function, given two dates, calculates the distance (in days) between the two dates, using recursion. The function dateDistance takes two dates and uses nextDate to find the next valid date. goodDate makes sure its a valid date.

I am trying to get the result by counting recursions and returning the count. Each time the recursion happens the count variable n should increase and at the end of the recursion when it reaches the end condition (fy == sy && fm == sm && sd == fd) = n it shall return the n.

   leapYear x = if ((x `mod` 4) == 0) && ((x `mod` 100) /= 0) || ((x `mod` 400) == 0) 
                 then True
                 else False

    goodDate (y,m,d)
        | (d<1 || m>12 || m<1 || y==0) = False
        | m `elem` [1,3,5,7,8,10,12] = d<32
        | m `elem` [4,6,9,11] = d<31
        | leapYear y = d<30
        | otherwise= d<29 

    nextDate (y,m,d) =
      if goodDate(y,m,(d+1)) 
      then (y,m,(d+1))
      else if goodDate(y,(m+1),1)
           then (y,(m+1),1)
           else ((y+1),1,1)



    dateDistance (fy,fm,fd) (sy,sm,sd) 
    |(fy == sy && fm == sm && sd == fd) = n
    |otherwise = dateDistance nextDate (fy,fm,fd) (sy,sm,sd)
     where n = 0

sdx23 · Accepted Answer · 2017-01-22 14:23:16Z

2

Actually, you don't need to remember it, even though you could do so, by passing it from call to call as an argument.

Let's answer two questions:

What is the base case of your recursion? Exactly, when both dates are equal; the distance is 0 then.
What is one recursion step? Well, if the dates f and s are not equal, we search for the nextDate (rather next-day!), lets call it f1. f1 is one day distant from f, but one day nearer to s. f2 ( = nextDate f1) is two days from f, but even nearer to s. So: dateDistance f s = 1 + dateDistance f1 s and dateDistance f1 s = 1 + dateDistance f2 s.

answered Jan 22, 2017 at 14:23

sdx23

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2 Comments

ShocKwav3_ Over a year ago

1. fy,fm,fd is the starting year,month,day and sy,sm,sd is the target date. when both are equal then its not zero its the total recursion steps.i put n=0 there just te see if the syntax is okay. but i want to increment n each step until the recursion stops and then show the total value of n as the total distance. 2. f1 is one recursion step.

Paul Johnson Over a year ago

You are thinking in imperative terms, which is not how Haskell works. The answer for two dates the same is zero. The answer for any other pair of dates is one plus the answer for the day after the first date.

Paul Johnson · Accepted Answer · 2017-01-22 14:18:13Z

1

You need to pass the "n" explicitly through the recursive calls. Make it an extra parameter of your dateDistance function and put n+1 in the recursive call. Then call it with an initial value of 0.

You can rename the dateDistance function to something else and rewrite dateDistance to call it with the initial value of 0.

answered Jan 22, 2017 at 14:18

Paul Johnson

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4 Comments

ShocKwav3_ Over a year ago

i tried to do 'dateDistance (fy,fm,fd) (sy,sm,sd) n |(fy == sy && fm == sm && sd == fd) = n |otherwise = dateDistance nextDate (fy,fm,fd) (sy,sm,sd) . n+1 where n = 0' but it did not work. is it wrong how i write? @Paul

sdx23 Over a year ago

In this case you need to leave out the where-clause and just supply 0 from a wrapper. E.g. rename dateDistance to dateDistanceGo and define dateDistance f s = dateDistanceGo f s 0.

ShocKwav3_ Over a year ago

could you elaborate a little bit as a answer . i am very new in haskell and functional programming. @sdx23

Paul Johnson Over a year ago

Actually sdx23's answer below is better.

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