Data update

Task/Prime-triangle/FreeBASIC/prime-triangle.basic

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+Dim Shared As Uinteger maxNumber = 20 ' Largest number we will consider.
+Dim Shared As Uinteger prime(2 * maxNumber) ' prime sieve.
+
+Function countArrangements(Byval n As Uinteger) As Uinteger
+    Dim As Uinteger i
+    If n < 2 Then ' No solutions for n < 2.
+        Return 0
+    Elseif n < 4 Then
+        ' For 2 and 3. there is only 1 solution: 1, 2 and 1, 2, 3.
+        For i = 1 To n
+            Print Using "###"; i;
+        Next i
+        Print
+        Return 1
+    Else
+        ' 4 or more - must find the solutions.
+        Dim As Boolean printSolution = True
+        Dim As Boolean used(n)
+        Dim As Uinteger number(n)
+        ' The triangle row must have 1 in the leftmost and n in the rightmost elements.
+        ' The numbers must alternate between even and odd in order for the sums to be prime.
+        For i = 0 To n - 1
+            number(i) = i Mod 2
+        Next i
+        used(1) = True
+        number(n) = n
+        used(n) = True
+        ' Find the intervening numbers and count the solutions.
+        Dim As Uinteger count = 0
+        Dim As Uinteger p = 2
+        Do While p > 0
+            Dim As Uinteger p1 = number(p - 1)
+            Dim As Uinteger current = number(p)
+            Dim As Uinteger sgte = current + 2
+            Do While sgte < n Andalso (Not prime(p1 + sgte) Or used(sgte))
+                sgte += 2
+            Loop
+            If sgte >= n Then
+                sgte = 0
+            End If
+            If p = n - 1 Then
+                ' We are at the final number before n.
+                ' It must be the final even/odd number preceded by the final odd/even number.
+                If sgte <> 0 Then
+                    ' Possible solution.
+                    If prime(sgte + n) Then
+                        ' Found a solution.
+                        count += 1
+                        If printSolution Then
+                            For i = 1 To n - 2
+                                Print Using "###"; number(i);
+                            Next i
+                            Print Using "###"; sgte; n
+                            printSolution = False
+                        End If
+                    End If
+                    sgte = 0
+                End If
+                ' Backtrack for more solutions.
+                p -= 1
+                ' There will be a further backtrack as next is 0 ( there could only be one possible number at p - 1 ).
+            End If
+            If sgte <> 0 Then
+                ' have a/another number that can appear at p.
+                used(current) = False
+                used(sgte) = True
+                number(p) = sgte
+                ' Haven't found all the intervening digits yet.
+                p += 1
+            Elseif p <= 2 Then
+                ' No more solutions.
+                p = 0
+            Else
+                ' Can't find a number for this position, backtrack.
+                used(number(p)) = False
+                number(p) = p Mod 2
+                p -= 1
+            End If
+        Loop
+        Return count
+    End If
+End Function
+
+Dim As Integer i, s, n
+prime(2) = True
+For i = 3 To Ubound(prime) Step  2
+    prime(i) = True
+Next i
+For i = 3 To Cint(Sqr(Ubound(prime))) Step 2
+    If prime(i) Then
+        For s = i * i To Ubound(prime) Step i + i
+            prime(s) = False
+        Next s
+    End If
+Next i
+
+Dim As Integer arrangements(maxNumber)
+For n = 2 To Ubound(arrangements)
+    arrangements(n) = countArrangements(n)
+Next n
+For n = 2 To Ubound(arrangements)
+    Print arrangements(n);
+Next n
+Print
+
+Sleep