Count how many months are having a full 31 days by counting knuckles
Jelly, 21 bytes
ḅ12r/ị7RḂṁ12¤żC$S×⁼Ṣ$
Takes input like [[y, m], [y, m]].
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How it works
ḅ12r/ị7RḂṁ12¤żC$S×⁼Ṣ$ Main link. Argument: [[a, b], [c, d]]
ḅ12 Unbase 12; yield [x, y] := [ 12a + b, 12c + d].
r/ Reduce by range; yield [x, ..., y].
¤ Combine the five links to the left into a niladic chain.
7 Set the return value to 7.
R Range; yield [1, 2, 3, 4, 5, 6, 7].
Ḃ Bit; yield [1, 0, 1, 0, 1, 0, 1].
ṁ12 Mold 12; repeat the Booleans to create an array of length
12. Yields [1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1].
ị At-index; yield the elements of the array to the right at
the indices (1-based and modular) of the array to the left.
$ Combine the two links to the left into a monadic chain.
C Complement; map t -> 1-t over the array.
ż Zip the original array with the complements.
S Take the sum of each column.
$ Combine the two links to the left into a monadic chain.
Ṣ Sort [[a, b], [c, d]].
⁼ Compare the result with [[a, b], [c, d]], yielding 1 if
the input is sorted, 0 if not.
× Multiply the results to both sides.
Python 2, 92 90 86 80 bytes
lambda a,b,c,d:[(bin(2741)[2:]*(c+1-a))[b-1:-12+d or None].count(x)for x in'10']
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6 more by converting to a lambda, with thanks to @math_junkie for the idea. Now outputs a list containing the two numbers.
Previous non-lambda version (86 bytes)
a,b,c,d=input()
for x in'10':print(bin(2741)[2:]*(c+1-a))[b-1:-12+d or None].count(x),
Try it online old!
2 saved with thanks to @ovs for helping me get rid of the len(k). I hadn't thought about using None.
Input is a list of integers in the format y1,m1,y2,m2
Some credit due to @KeerthanaPrabhakaran who got bin(2741)[2:] before I did which saves 1 byte over hard coding the binary string.
JavaScript (ES6), 70 68 67 64 bytes
Takes input as two integers in yyyymm format, in currying syntax (a)(b). Outputs an array of two integers [knuckles, grooves].
a=>g=(b,c=d=0)=>a>b?[c,d-c]:g(--b,c+!((b%=100)>11||b/.87&!!++d))
Formatted and commented
a => // main function: takes start date (a) as input / returns g
g = ( // recursive function g, which takes:
b, // - b = end date
c = d = 0 // - c = number of knuckles
) => // and also keeps track of: d = total number of months
a > b ? // if a is greater than b:
[ c, d - c ] // stop recursion and return the final result
: // else:
g( // do a recursive call to g():
--b, // - decrement the end date
c + // - increment the # of knuckles if
!( // both of these conditions are false:
(b %= 100) // - the end month (now stored in b in 0-based indexing)
> 11 || // is greater than 11
b / 0.87 & !!++d // - the number of days in this month is not 31
) // (at the same time, d is incremented if the first
) // condition is false)
Test cases
NB: The third test case is not included in this snippet, because it won't work unless your browser has Tail Call Optimization enabled.
let f =
a=>g=(b,c=d=0)=>a>b?[c,d-c]:g(--b,c+!((b%=100)>11||b/.87&!!++d))
console.log(f(201703)(201902)); // 14 10
console.log(f(201701)(202008)); // 26 18
console.log(f(201802)(201803)); // 1 1
console.log(f(201601)(201601)); // 1 0
console.log(f(201612)(201611)); // 0 0