[AdamSpeight2008]

Code Golf (Hole 1, Code Love)

It February, the month of Love.
For those with partners, remember it Valentines Day on the 14^th
(WTF, A coder with a partner?)

Challenge
Fill in the code to calculate what percentage one member loves another.

Example
Locke loves supersloth how much?

LOVES
22022 <-- Count of each letter.
 4224 <-- (2+2), (2+0), (0+2), (2+2)
  646 <-- (4+2), (2+2), (2+4)
   00 <-- (5+4), (4+6)          ' Last one 

Note how the 4+6 is 0 and not 10, this so the calculation converges.
Extra Clarification The values in each must be 0<= x < 10 at all stages, this includes the count of each letter.
So Locke loves supersloth 00% of the time.

Letter casing is iqnored so Aa are the same.

Function Loves(ByVal PersonA As String, ByVal PersonB As String) As Integer)
' Your code goes here '
End Function

Scoring Rules
You can assume the inputs will be non null or nothing.
Must compile with

Option Strict On
Option Infer On

Is the same code golf standard, the count of non-whitespace character inside the function method.
For this hole an allowance of -6 (to account for the required keyword Return.

Edit: Update and corrected calculation

Example Submission

Spoiler

 Function Love(ByVal PersonA As String, ByVal PersonB As String) As Integer
    Dim l As Integer = 0
    Dim o As Integer = 0
    Dim v As Integer = 0
    Dim e As Integer = 0
    Dim s As Integer = 0
    For Each c In PersonA
      If Char.ToUpper(c) = "L"c Then l += 1
      If Char.ToUpper(c) = "O"c Then o += 1
      If Char.ToUpper(c) = "V"c Then v += 1
      If Char.ToUpper(c) = "E"c Then e += 1
      If Char.ToUpper(c) = "S"c Then s += 1
    Next
    For Each c In PersonB
      If Char.ToUpper(c) = "L"c Then l += 1
      If Char.ToUpper(c) = "O"c Then o += 1
      If Char.ToUpper(c) = "V"c Then v += 1
      If Char.ToUpper(c) = "E"c Then e += 1
      If Char.ToUpper(c) = "S"c Then s += 1
    Next
    Dim n() As Integer = {l, o, v, e, s}
    Dim r2(3) As Integer
    r2(0) = (n(0) + n(1)) Mod 10
    r2(1) = (n(1) + n(2)) Mod 10
    r2(2) = (n(2) + n(3)) Mod 10
    r2(3) = (n(3) + n(4)) Mod 10
    Dim r3(2) As Integer
    r3(0) = (r2(0) + r2(1)) Mod 10
    r3(1) = (r2(1) + r2(2)) Mod 10
    r3(2) = (r2(2) + r2(3)) Mod 10
    Dim r4(1) As Integer
    r4(0) = (r3(0) + r3(1)) Mod 10
    r4(1) = (r3(1) + r3(2)) Mod 10
    Return (r4(0) * 10) + r4(1)
  End Function

Score: 725 - (6) = 719

This post has been edited by AdamSpeight2008: 03 February 2011 - 07:02 PM

[TMKCodes]

I'm sorry this is not VB.NET, but I noticed it after I had written this in C++.

Spoiler

int loves(std::string pA, std::string pB) {
	int l = 0, o = 0, v = 0, e = 0, s = 0;
	int n[5] = {l,o,v,e,s}, r2[4], r4[2], r3[3];
	for(unsigned int i = 0; i < pA.size(); i++) {
		if(toupper(pA.at(i)) == 'L') l += 1;
		if(toupper(pA.at(i)) == 'O') o += 1;
		if(toupper(pA.at(i)) == 'V') v += 1;
		if(toupper(pA.at(i)) == 'E') e += 1;
		if(toupper(pA.at(i)) == 'S') s += 1;
	}
	for(unsigned int i = 0; i < pB.size(); i++) {
		if(toupper(pB.at(i)) == 'L') l += 1;
		if(toupper(pB.at(i)) == 'O') o += 1;
		if(toupper(pB.at(i)) == 'V') v += 1;
		if(toupper(pB.at(i)) == 'E') e += 1;
		if(toupper(pB.at(i)) == 'S') s += 1;
	}
	for(int i = 0; i < 4; i++) { r2[i] = (n[i] + n[i+1]) % 10; }
	for(int i = 0; i < 3; i++) { r3[i] = (r2[i] + r2[i+1]) % 10; }
	for(int i = 0; i < 2; i++) { r4[i] = (r3[i] + r3[i+1]) % 10; }
	return (r4[0] * 10) + r4[1];
}

This post has been edited by TMKCodes: 03 February 2011 - 05:38 PM

[AdamSpeight2008]

TMKCodes: That's Ok but only VB.net ones will officially be scored.

As as target the best I can do so a far is 171c 148c.

This post has been edited by AdamSpeight2008: 03 February 2011 - 07:32 PM
Reason for edit:: Updating Score

[Curtis Rutland]

Here's my go. It's certainly no 148, but I think it's pretty good for a non-regular VB.NET programmer.

Spoiler

Function Love(ByVal PersonA As String, ByVal PersonB As String) As Integer
    Dim d As New Dictionary(Of Char, Integer) From {{"l"c, 0}, {"o"c, 0}, {"v"c, 0}, {"e"c, 0}, {"s"c, 0}}
    Dim g = Sub(s As String) s.ToLower.ToList.ForEach(Sub(x) If (d.ContainsKey(x)) Then d(x) += 1)
    g(PersonA)
    g(PersonB)
    Dim f = Function(a As List(Of Integer))
                Dim r As New List(Of Integer)
                For i = 0 To a.Count() - 2
                    r.Add((a(i) + a(i + 1)) Mod 10)
                Next
                Return r
            End Function
    Dim r4 = f(f(f(d.Select(Function(x) x.Value).ToList)))
    Return (r4(0) * 10) + r4(1)
End Function

Edit: Ignore the last one. Here's a slightly smaller version:

Spoiler

Function Love(ByVal PersonA As String, ByVal PersonB As String) As Integer
    Dim d As New Dictionary(Of Char, Integer) From {{"l"c, 0}, {"o"c, 0}, {"v"c, 0}, {"e"c, 0}, {"s"c, 0}}
    Dim g = Sub(s As String)
                For Each c In s.ToLower
                    If (d.ContainsKey(c)) Then d(c) += 1
                Next
            End Sub
    g(PersonA)
    g(PersonB)
    Dim f = Function(a As List(Of Integer))
                Dim q As New List(Of Integer)
                For i = 0 To a.Count - 2
                    q.Add((a(i) + a(i + 1)) Mod 10)
                Next
                Return q
            End Function
    Dim r = f(f(f(d.Select(Function(x) x.Value).ToList)))
    Return (r(0) * 10) + r(1)
End Function

Edit 2: Saved myself 90 characters doing this:

Spoiler

Function Love3(ByVal PersonA As String, ByVal PersonB As String) As Integer
    Dim g = Function(c As Char, s As String) s.ToLower.Count(Function(x) x = c)
    Dim d = "loves".Select(Function(c) g(c, PersonA) + g(c, PersonB))
    Dim f = Function(a As List(Of Integer))
                Dim q As New List(Of Integer)
                For i = 0 To a.Count - 2
                    q.Add((a(i) + a(i + 1)) Mod 10)
                Next
                Return q
            End Function
    Dim r = f(f(f(d.ToList)))
    Return (r(0) * 10) + r(1)
End Function

Edit 3: And I just saved 27 more characters over the last one. This is an iterative process

Spoiler

Function Love3(ByVal PersonA As String, ByVal PersonB As String) As Integer
    Dim g = Function(c As Char, s As String) s.ToLower.Count(Function(x) x = c)
    Dim f = Function(a As IEnumerable(Of Integer)) Enumerable.Range(0, a.Count - 1).Select(Function(i) (a(i) + a(i + 1)) Mod 10)
    Dim r = f(f(f("loves".Select(Function(c) g(c, PersonA) + g(c, PersonB)))))
    Return (r(0) * 10) + r(1)
End Function

This post has been edited by insertAlias: 04 February 2011 - 11:02 AM

[modi123_1]

452-6 = 446 using some pretty ugly looking coding practices..
460-6 = 454

Spoiler

  Function Loves(ByVal PersonA As String, ByVal PersonB As String) As Integer
        Dim t As Int32 = 0
        PersonA = PersonA.ToUpper

        Dim l As New List(Of Int32)({0, 0, 0, 0, 0})
        Dim m As New List(Of Int32)

        While 0 <> 1
            For Each i As Char In PersonA
                If i = "L" Then l(0) += 1
                If i = "O" Then l(1) += 1
                If i = "V" Then l(2) += 1
                If i = "E" Then l(3) += 1
                If i = "S" Then l(4) += 1
            Next
            If PersonA = PersonB.ToUpper Then Exit While
            PersonA = PersonB.ToUpper
        End While

        While l.Count > 1
            For i As Int32 = 0 To l.Count - 2
                t = l(i) + l(i + 1)
                If t > 9 Then t = Convert.ToInt32(CStr(CStr(t)(CStr(t).Length - 1)))
                m.Add(t)
            Next
            l = m
            m = New List(Of Int32)
        End While

        Return l(0)
    End Function

This post has been edited by modi123_1: 04 February 2011 - 11:34 AM

[Curtis Rutland]

@modi, when I ran yours it just infinitely looped.

[modi123_1]

What was your input?
Nevermind.. I fixed it... stupid paint fumes in the office today.

This post has been edited by modi123_1: 04 February 2011 - 11:34 AM

[AdamSpeight2008]

Remember you need to do a mod 10 on initial count the count of the letters.
So

 L O V E S
13 1 0 4 4 becomes 3 1 0 4 4

I think it a good challenge.
As you consider the char cost of using Function() and Sub() when you use lambda functions.

This post has been edited by AdamSpeight2008: 04 February 2011 - 11:57 AM

[Curtis Rutland]

Updated to fix that issue:

Spoiler

Function Love3(ByVal PersonA As String, ByVal PersonB As String) As Integer
    Dim g = Function(c As Char, s As String) s.ToLower.Count(Function(x) x = c) Mod 10
    Dim f = Function(a As IEnumerable(Of Integer)) Enumerable.Range(0, a.Count - 1).Select(Function(i) (a(i) + a(i + 1)) Mod 10)
    Dim r = f(f(f("loves".Select(Function(c) g(c, PersonA) + g(c, PersonB)))))
    Return (r(0) * 10) + r(1)
End Function

261 - 6 = 255

I have considered the character cost, but I'm not sure of a way that would be shorter. I wish I could use some of my extension methods. But to be fair, the code in those would have to be counted too.

This post has been edited by insertAlias: 04 February 2011 - 11:59 AM

[AdamSpeight2008]

Another validation check for you

Gorian LOVES BenignDesign = 58%

Score: 145c

Spoiler

Function Love(ByVal PersonA As String,
              ByVal PersonB As String) As Integer
  Dim n = (From c In "SEVOL"
           Select (From x In (PersonA & PersonB).ToUpper Where c = x).Count()).ToList
  For z = 1 To 3
    For i = 0 To 3
      n(i) = (n(i) + n(i + 1)) Mod 10
    Next
  Next
  Return 10 * n(1) + n(0)
End Function

I'd be very surprised if could be coded in less chars.

This post has been edited by AdamSpeight2008: 04 February 2011 - 04:10 PM

[Curtis Rutland]

Pretty damn slick, Adam.

[leoJava]

OK, next try (while logged in...) 138c.

Spoiler

Both the previous version and this one unwrap the adding algorithm in the return method.
This version gets rid of .toUpperCase() via ASCII Table Masking. This also allows reducing the X array length and makes the character offsets within X array a trivial matter.

int Loves(String A, String B)/> {
    int[] X = new int[32];
    for (int a : (A+B)/>.getBytes()) ++X[a/64*a%32];
    int L=X[12],O=X[15],V=X[22],E=X[5],S=X[19];
    return (L+3*O+3*V+E)%10*10+(O+3*V+3*E+S)%10;
}

This post has been edited by leoJava: 05 February 2011 - 09:33 PM