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Beginner’s Guide to Code Algorithms

46

STEP 9

The principle here is to compare the cell with each of the nine numbers, one grid

at a time. If the number is found in cantbelist for a cell, or the cell is occupied by

a number that is not the same as the number being searched, then a count is kept

to see how many cells are there of this kind in that row. If there are 3, then it is a

candidate, less than 3 does not qualify hence count is reset to zero for that grid. If

the count is 6 for that row, this means there are two out of the three adjacent grids

that cannot have this number. Hence the third grid included in this row must have

this number. Therefore, the other rows in this third grid cannot have this number—​

these are then added to the cantbelist.

For putnumber =​ 1 to 9

:

:

  ‘ *** algorithm to refine cantbelist based on one row of 3 by 3 grid

For i =​ 1 To 9

  middlerowcount(i) =​ 0

Next i

  For i =​ 1 To 9 Step 3

    For j =​ 1 To 9 Step 3

      For k =​ 1 To 3

      n =​ Int((i -​ 1) /​ 3) * 3 +​ k

      triocount =​ 0

      For l =​ 1 To 3

        p =​ Int((j -​ 1) /​ 3) * 3 +​ l

          If (sbox(n, p) <> putnumber And sbox(n, p) <> ““) Or

          findincantbelist(putnumber, n, p) =​ 1 Then

          middlerowcount(n) =​ middlerowcount(n) +​ 1

          triocount =​ triocount +​ 1

        End If

      Next l

      If triocount < 3 Then

        middlerowcount(n) =​ middlerowcount(n) -​ triocount

        savedcolumn(n) =​ p

      End If

    Next k

  Next j

Next i

For i =​ 1 To 9

  If middlerowcount(i) =​ 6 Then

    a =​ i Mod 3

    If a =​ 0 Then

      savedrow1 =​ i -​ 2

      savedrow2 =​ i -​ 1

    Else

    If a =​ 1 Then

      savedrow1 =​ i +​ 1

      savedrow2 =​ i +​ 2

    Else