The words
again, event and
title have the same structural pattern i.e. 12134.
The corresponding pattern of the word
bedressed is 123425523, same as a well known location in the USA.
What location?
Unexpected correct answers welcomed!
Having lived there 8 years, many years ago, I should have known this without the computer, but:
The program I wrote for this has a bug that didn't affect the outcome. The requirement that the first letter not match the rest of the letters was actually met by the first letters all being capital; I should have converted to consistent case before doing the pattern matching.
I used a data base from the census bureau that had only cities, villages and towns, but not boroughs. As entity names were delimited by space I searched for a space to terminate the name, but of course some locations have a space within the name, so these parts of the output need be ignored.
The output:
CA 0645400 02411020 Manhattan Beach city 25 A 10197388 10360 3.937 0.004 33.889632 -118.39737
IL 1746357 02399240 Manhattan village 47 A 17064204 0 6.589 0. 41.420153 -87.98072
KS 2044250 00485618 Manhattan city 25 A 49985444 113130 19.299 0.044 39.188193 -96.605746
MN 2739806 02395829 Manhattan Beach city 25 A 3963008 2549 1.53 0.001 46.7356 -94.143626
MT 3047575 02412941 Manhattan town 43 A 4526462 0 1.748 0. 45.861756 -111.333536
Eliminating the ones with Beach in their names, that leaves
Manhattan, Illinois
Manhattan, Kansas
Manhattan, Montana
Of course there's the borough of Manhattan in New York City, coterminous with New York county.
And most of that borough/county is on Manhattan Island, to which the puzzle might refer.
DefDbl A-Z
Dim crlf$
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
Open "2017_Gaz_place_national.txt" For Input As #1
Do
Line Input #1, l$
ls$ = l
l$ = Mid(l, 21)
ix = InStr(l, " ")
n$ = Left(l, ix - 1)
good = 1
If Len(n) = 9 Then
For i = 1 To Len(n) - 1
ix = InStr(Mid(n, i + 1), Mid(n, i, 1))
Select Case i
Case 1: If ix <> 0 Then good = 0: Exit For
Case 2: If ix <> 3 Then good = 0: Exit For
Case 3: If ix <> 6 Then good = 0: Exit For
Case 4: If ix <> 0 Then good = 0: Exit For
Case 5: If ix <> 3 Then good = 0: Exit For
Case 6: If ix <> 1 Then good = 0: Exit For
Case 7: If ix <> 0 Then good = 0: Exit For
Case 8: If ix <> 0 Then good = 0: Exit For
End Select
Next
If good Then
Text1.Text = Text1.Text & ls & crlf
End If
End If
DoEvents
Loop Until EOF(1)
Close 1
Text1.Text = Text1.Text & crlf & " done"
End Sub
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Posted by Charlie
on 2017-10-10 13:35:40 |