Enter a couple of letters in each pair of brackets such that they serve as an end of the word on the left and begin another valid word with the letters on the right:
CO (XX) ST
PA (XX) AY
MO (XX) EN
NO (XX) ON
KI (XX) RE
BU (XX) OT
NI (XX) NT
sample solution: CO(RE)ST
Off the top of my head:
COLA LAST
PAST STAY
MOTH THEN, or MOTE TEEN
NONE NEON
BUST STOT
NICE CENT
From computer search:
coca cast
coco cost
coho host
cola last
cole lest
coma mast
cone nest
cope pest
cops psst
core rest
cote test
cove vest
pash shay
past stay
moke keen
momi mien
mope peen
mote teen
moth then
noir iron
nolo loon
none neon
nope peon
kilo lore
kiwi wire
bubo boot
bury ryot
bush shot
nice cent
nide dent
nidi dint
nipa pant
nite tent
Curiously STOT wasn't in my word list. It's another word for PRONK, defined as "(of a springbok or other antelope) leap in the air with an arched back and stiff legs, typically as a form of display or when threatened."
DefDbl A-Z
Dim crlf$, word1$(7), word2$(7)
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
word1(1) = "co": word2(1) = "st"
word1(2) = "pa": word2(2) = "ay"
word1(3) = "mo": word2(3) = "en"
word1(4) = "no": word2(4) = "on"
word1(5) = "ki": word2(5) = "re"
word1(6) = "bu": word2(6) = "ot"
word1(7) = "ni": word2(7) = "nt"
w1$ = Space(4)
For i = 1 To 7
Open "c:\words\words4.txt" For Binary As #1
Do
Get #1, , w1
If EOF(1) Or Left(w1, 2) > word1(i) Then Exit Do
If Left(w1, 2) = word1(i) Then
If isWord(Right(w1, 2) + word2(i)) Then
Text1.Text = Text1.Text & w1 & " " & Right(w1, 2) + word2(i) & crlf
End If
End If
Loop
Text1.Text = Text1.Text & crlf
Close 1
Next
Text1.Text = Text1.Text & crlf & " done"
End Sub
Function isWord(w$)
n = Len(w$)
w1$ = Space$(n)
Open "c:\words\words" + LTrim$(Str$(n)) + ".txt" For Binary As #2
l = LOF(2) / n
low = 1: high = l
Do
middle = Int((low + high) / 2)
Get #2, (middle - 1) * n + 1, w1$
If w1$ = w$ Then isWord = 1: Close 2: Exit Function
If w1$ < w$ Then low = middle + 1 Else high = middle - 1
Loop Until low > high
isWord = 0
Close 2
End Function
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Posted by Charlie
on 2017-07-06 09:53:55 |