Comments on: Euler Problem 52

By: fzz

fzz — Fri, 09 Jan 2009 19:23:00 +0000

Actually, using VBA with numbers converted to strings is pretty efficient (for VBA). The following took less than a second runtime.

[FORTRAN-like comparison operators]

Sub euler52()
Dim i As Long, j As Long, k As Long, n As Long, p As Long
Dim s As Double, ns As String, ts As String, dt As Double

dt = Timer
s = Log(1# + 2# / 3#) / Log(10#)
n = 100
k = 3

Do
n = n + 1
If Log(n) / Log(10#) Mod 1# .GT. s Then
k = k + 1
n = 10 ^ k + 1
End If
ns = CStr(n)
For i = 2 To 6
ts = CStr(i * n)
For j = 1 To k
p = InStr(1, ts, Mid$(ns, j, 1))
If p .GT. 0 Then Mid$(ts, p, 1) = " " Else Exit For
Next j
If Trim(ts) .NE. "" Then Exit For
Next i
If i .GT. 6 Then
Debug.Print n, Timer - dt
Exit Do
End If
Loop
End Sub

By: Hans Schraven

Hans Schraven — Fri, 09 Jan 2009 14:43:00 +0000

For the length-test suffices

If Len(i) = Len(6 * i) Then

By: Hans Schraven

Hans Schraven — Fri, 09 Jan 2009 13:17:00 +0000

I used this one

Sub Euler52()
T = Timer
i = 0
Do
i = i + 1
If Len(i) = Len(2 * i) And Len(i) = Len(3 * i) And Len(i) = Len(4 * i) And Len(i) = Len(5 * i) And Len(i) = Len(6 * i) Then
For j = 1 To Len(i)
If InStr(i, Mid(i * 2, j, 1)) * InStr(i * 2, Mid(i, j, 1)) * InStr(i, Mid(i * 3, j, 1)) * InStr(i * 3, Mid(i, j, 1)) * InStr(i, Mid(i * 4, j, 1)) * InStr(i * 4, Mid(i, j, 1)) * InStr(i, Mid(i * 5, j, 1)) * InStr(i * 5, Mid(i, j, 1)) * InStr(i, Mid(i * 6, j, 1)) * InStr(i * 6, Mid(i, j, 1)) = 0 Then Exit For
Next
If j .gt. Len(i) Then Exit Do
End If
Loop
Debug.Print "answer: " & i & " time: " & Timer - T
End Sub

By: Doug Jenkins

Doug Jenkins — Fri, 09 Jan 2009 09:43:00 +0000

Tushar – I used strings because when I used longs I went over the maximum size. I could have just switched to doubles, or with a little more thought have changed my approach so I didn’t get such big numbers, but comparing strings just seemed the obvious way to do it. Also I wasn’t aware that it would be much slow than comparing numbers.

By: Tushar Mehta

Tushar Mehta — Fri, 09 Jan 2009 08:41:00 +0000

Why switch to strings to carry out various comparisons? It is probably so much faster to work with numbers. That's what I did for this problem. The VBA code ran in under 0.2 seconds.
http://www.tushar-mehta.com/misc_tutorials/project_euler/euler052.html

By: Doug Jenkins

Doug Jenkins — Fri, 09 Jan 2009 00:56:00 +0000

Quicker if you start at the top factor(6) and work down to 1. Comes down to 1.1 seconds for me.

By: Doug Jenkins

Doug Jenkins — Fri, 09 Jan 2009 00:51:00 +0000

I did it as a function that will check for any number of factors (but it didn't find a solution going up to 7). Rather than sorting I fed the digits into an array, then formed the array into a string and compared those. You could probably save a millisecond or two there.

1.9 seconds on my machine:

Function p_52(level As Long) As Variant
Dim i As Long, j As Long, Same As Boolean, IntA() As Long, NumDig As Long
Dim Fact As Long, k As Long, l As Long, CheckDig1 As String, CheckDig2 As String
Dim Time As Double

Time = Timer
i = 100
On Error GoTo Err
Do
i = i + 1
CheckDig1 = ""
ReDim IntA(0 To 9, 1 To level)
NumDig = Len(Trim(i))
For Fact = 1 To level
j = i * Fact
If Len(Trim(j)) .GT. NumDig Then
i = 10 ^ NumDig
Exit For
End If
For k = 1 To NumDig
l = Mid(j, k, 1)
IntA(l, Fact) = IntA(l, Fact) + 1
Next k
For k = 0 To 9
If Fact = 1 Then
CheckDig1 = CheckDig1 & IntA(k, Fact)
Else
CheckDig2 = CheckDig2 & IntA(k, Fact)
End If
Next k
If Fact .GT. 1 Then
If CheckDig2 .NE. CheckDig1 Then
CheckDig2 = ""
Exit For
End If
CheckDig1 = CheckDig2
CheckDig2 = ""
End If
Next Fact
If Fact = level + 1 Then
p_52 = Array(i, Timer - Time)
Exit Function
End If
Loop
Err:
p_52 = "stopped at i = " & i

End Function

By: Michael

Michael — Fri, 09 Jan 2009 00:32:00 +0000

Hi fzz -

The answer makes sense…it can’t be over 166666 or i*6 will roll over to more decimal places. I’ve been impressed with the Parallels emulation. Code runs faster than my on my stock PC at work (and that’s a moderately fast HP). I named that tune in 4.96875 notes errr seconds on my Mac with Parallels and XL2002 on the PC side. I’ve just got to figure out how to cleanse my cut and pastes.

The first time I wrote IsPanDigital I didn’t specify the first occurrence only, so all digits got swapped out, and everything was TRUE.

Live and learn…
Thanks for your help and insights. I know that’s what Dick wanted to come from this.

…mrt (that answer’s out there now…interesting that Dick’s challenge was answered by a number 10x higher. Euler forgive us )

By: fzz

fzz — Thu, 08 Jan 2009 23:28:00 +0000

Me, I used awk. With FORTRAN-like comparison operators.

BEGIN {
n = 101
s = log(1 + 2/3)/log(10)
while (check(n)) {
if (log(++n)/log(10) % 1 .GT. s) n = 10 ^ int(1 + log(n)/log(10)) + 1
}
print "answer", n
}

function check(n , i, j, k, t) { # also c as an array
k = length(n)
for (j = 1; j .LE. k; ++j) c[j] = substr(n, j, 1)
for (i = 2; i .LE. 6; ++i) {
t = i * n
for (j = 1; j .LE. k; ++j) sub(c[j], "", t)
if (t != "") return i
}
return 0
}

answer 142857 in 6.234 seconds

By: fzz

fzz — Thu, 08 Jan 2009 23:27:00 +0000

Me, I used awk.

BEGIN {
n = 101
s = log(1 + 2/3)/log(10)
while (check(n)) {
if (log(++n)/log(10) % 1 .GT. s) n = 10 ^ int(1 + log(n)/log(10)) + 1
}
print "answer", n
}

function check(n , i, j, k, t) { # also c as an array
k = length(n)
for (j = 1; j