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Task/Prime-triangle/Visual-Basic-.NET/prime-triangle.vb
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116
Task/Prime-triangle/Visual-Basic-.NET/prime-triangle.vb
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Option Strict On
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Option Explicit On
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Imports System.IO
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''' <summary>Find solutions to the "Prime Triangle" - a triangle of numbers that sum to primes.</summary>
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Module vMain
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Public Const maxNumber As Integer = 20 ' Largest number we will consider.
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Dim prime(2 * maxNumber) As Boolean ' prime sieve.
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''' <returns>The number of possible arrangements of the integers for a row in the prime triangle.</returns>
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Public Function countArrangements(ByVal n As Integer) As Integer
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If n < 2 Then ' No solutions for n < 2.
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Return 0
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ElseIf n < 4 Then
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' For 2 and 3. there is only 1 solution: 1, 2 and 1, 2, 3.
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For i As Integer = 1 To n
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Console.Out.Write(i.ToString.PadLeft(3))
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Next i
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Console.Out.WriteLine()
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Return 1
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Else
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' 4 or more - must find the solutions.
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Dim printSolution As Boolean = true
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Dim used(n) As Boolean
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Dim number(n) As Integer
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' The triangle row must have 1 in the leftmost and n in the rightmost elements.
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' The numbers must alternate between even and odd in order for the sums to be prime.
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For i As Integer = 0 To n - 1
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number(i) = i Mod 2
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Next i
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used(1) = True
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number(n) = n
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used(n) = True
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' Find the intervening numbers and count the solutions.
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Dim count As Integer = 0
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Dim p As Integer = 2
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Do While p > 0
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Dim p1 As Integer = number(p - 1)
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Dim current As Integer = number(p)
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Dim [next] As Integer = current + 2
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Do While [next] < n AndAlso (Not prime(p1 + [next]) Or used([next]))
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[next] += 2
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Loop
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If [next] >= n Then
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[next] = 0
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End If
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If p = n - 1 Then
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' We are at the final number before n.
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' It must be the final even/odd number preceded by the final odd/even number.
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If [next] <> 0 Then
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' Possible solution.
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If prime([next] + n) Then
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' Found a solution.
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count += 1
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If printSolution Then
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For i As Integer = 1 To n - 2
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Console.Out.Write(number(i).ToString.PadLeft(3))
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Next i
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Console.Out.WriteLine([next].ToString.PadLeft(3) & n.ToString.PadLeft(3))
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printSolution = False
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End If
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End If
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[next] = 0
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End If
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' Backtrack for more solutions.
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p -= 1
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' There will be a further backtrack as next is 0 ( there could only be one possible number at p - 1 ).
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End If
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If [next] <> 0 Then
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' have a/another number that can appear at p.
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used(current) = False
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used([next]) = True
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number(p) = [next]
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' Haven't found all the intervening digits yet.
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p += 1
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ElseIf p <= 2 Then
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' No more solutions.
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p = 0
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Else
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' Can't find a number for this position, backtrack.
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used(number(p)) = False
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number(p) = p Mod 2
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p -= 1
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End If
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Loop
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Return count
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End If
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End Function
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Public Sub Main
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prime(2) = True
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For i As Integer = 3 To UBound(prime) Step 2
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prime(i) = True
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Next i
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For i As Integer = 3 To Convert.ToInt32(Math.Floor(Math.Sqrt(Ubound(prime)))) Step 2
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If prime(i) Then
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For s As Integer = i * i To Ubound(prime) Step i + i
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prime(s) = False
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Next s
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End If
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Next i
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Dim arrangements(maxNumber) As Integer
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For n As Integer = 2 To UBound(arrangements)
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arrangements(n) = countArrangements(n)
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Next n
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For n As Integer = 2 To UBound(arrangements)
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Console.Out.Write(" " & arrangements(n))
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Next n
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Console.Out.WriteLine()
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End Sub
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End Module
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