Solving Calcudoku

In the left column of the puzzle we see a region of two squares with the sum 4. Since Calcudoku rules don’t allow the same number to appear more than once in a row or a column the only combination to satisfy this requirement is 1+3 though we don’t yet know in which order the numbers are placed. We can therefore place 4 in the bottom square of the left column, and 3, the block remainder, in the adjacent square.

We can use the same method for regions with multiplication. In the bottom row of the example is a region of two squares with the multiplication product 2. This means the only combination can be 1x2 though we don’t yet know in which order the numbers are placed. Since we also have 4 in the bottom row, the only candidate for the square on the right side of the bottom row is 3. After placing 3 we can also place 1, the region remainder, in the square above it.

In the bottom row of the example we see a two-square region with the sum 4. Since Calcudoku rules don’t allow the same number to appear more than once in a row or a column the only combination can be 1+3. However, the top-left square already contains 3 which means there is only one way to place 1 and 3 in the bottom row as shown in the diagram.

This example is similar to the ones above except there are two empty squares at the top of the left column and the candidates for those squares are 4 and 5. However, 5 is too large to place in the top-left region so 4 remains the only candidate as shown. We can now place 1, the region remainder, in the adjacent square.

We can use the same method for regions with multiplication. The top row of the example contains a three-square region with the multiplication product 6. Since the only combination allowed is 1x2x3 we can deduce that 4 and 5 are in the remaining two squares in the top row. However, we can’t place 5 in the right-most square because 8 can only be the result of 2x4. Therefore we must place 4 in the top-right square and 2, the region remainder, in the square under it.

The example Calcudoku puzzle which uses multiplication and division operations. Similar to the previous examples, the only candidate for the left square in the second row of the puzzle below is 3. After placing 3 we can also place 1 in the square below since it is the only way to get the division result of 3.

Here is another example of using the unique region method in a Calcudoku puzzle with multiplication and division. Let’s examine the region with 2÷ in the left column. There are only two combinations possible, 2÷1 and 4÷2 which make 1, 2 and 4 the only possible candidates for this region. However, the fourth row already contains 1 and 4 leaving 2 as the only possibility for the left square in the fourth row. Unlike previous examples, we cannot yet place the region remainder since both 1 and 4 are possible.

Sometimes its harder to identify unique region methods as shown in the example. The only combination possible for the top region in the right column is 1+2. However, when we examine the three-square region in the first row we see only 2+4+5 is possible which excludes 2 from the top square in the right column and leaves 1 as its only candidate. After placing 1 in the top square of the right column we can also place 2, the region remainder, in the square underneath.

The top row of the puzzle has two fully contained regions shaded in gray, one with the sum 2 (which is actually a given) and the other with the sum 8. Together these two regions sum up to 2+8=10, but we know that any row or any column in a 5x5 Calcudoku puzzle must sum up to 1+2+3+4+5=15. This means the difference 15-10=5 is caused by the un-shaded square in the top row and we can therefore place in it 5.

Here is another example with a twist. The left column contains two regions shaded in gray, one with the sum 10 which is fully contained in the left column, and one with the sum 9 which has only two squares contained in the left column. Together these two regions sum up to 10+9=19, but we know that any row or any column in a 5x5 Calcudoku puzzle must sum up to 1+2+3+4+5=15. This means the difference 19-15=4 is caused by the square in the bottom region which isn’t contained in the left column and we can therefore place in it 4.

The grid remainder method is also useful for rows and columns containing regions with multiplication. The right column of the puzzle contains two regions shaded in gray, one with the product 8 which is fully contained in the right column, and one with the product 60 which has only two squares contained in the right column. Together these two regions have a product of 8x60=480, but we know that the product of any row or any column in a 5x5 Calcudoku puzzle must be 1x2x3x4x5=120. This means the quotient 480÷120=4 is caused by the square which isn’t contained in the right column and we can therefore place in it 4.

In the example, the only combination for the sum 14 in the shaded region is 4+5+5. This situation is true for all three-square L-shaped regions of a 5x5 puzzle. In addition to knowing which numbers should be placed we also know where they should be since the 5’s must be diagonal to avoid a number appearing in the same row or column more than once.

Here is another example where the intra region method can help find some of the numbers in a region. In the puzzle three out of the four squares of the shaded region are contained in one column. This means the only possible combination for these three squares is 1+2+3=6 because any other combination will reach or exceed 7, the sum of this region, thus not allowing any number in its fourth square. We can therefore place 1, the region remainder calculated by 7-6, in the fourth square.

Intra region methods can be used equally well for regions with multiplication. In the example, the only combination for the product 80 in the shaded region is 4x4x5. This situation is true for all three-square L-shaped regions of a 5x5 puzzle. In addition to knowing which numbers should be placed we also know where they should be since the 4’s must be diagonal to avoid a number appearing in the same row or column more than once.

Larger regions present more challenging methods of using the intra region method. Let’s look at the shaded region with the sum 8. Using the minimum sum approach, the smallest possible sum in the top row of this region is 1+2=3 and the smallest possible sum in the second row is also 1+2=3. This means that 3+3=6 out of the sum 8 in this region is already accounted for leaving only 1 and 2 as candidates for the last square. However, since 2 is already used in the same column, 1 must be placed in the bottom square of this region.

In 5x5 Calcudoku puzzles, the only combinations for the product 32 in four squares are 1x2x4x4 and 2x2x2x4. If we look at the shaded L-shaped region it is obvious that the second combination is not possible because no matter how we place the numbers there will always be 2 in the same row or column. This means only 1x2x4x4 is allowed and 4 must be placed in the left square to avoid conflict with the other 4.

We know the left column must contain 5 so let’s assume it is placed in one of the red dotted squares. This means the minimum possible sum of the shaded region will be 1+2+5=8 in the left column, 1+2+3=6 in the second column and 1+2=3 in the third column, which altogether add up to 8+6+3=17. Since 17 is more than the sum of this region, we know our initial assumption is incorrect and 5 must be placed in the bottom row of the left column.

This method is similar to the one above except it uses an opposite assumption to prove a number must be used in a certain region. Let’s assume 5 in the second row is not placed in one of the red dotted squares. This means the maximum possible sum of the shaded region will be 2+5=7 in the top row, 2+3+4=9 in the second row and 5 in the third row, which altogether add up to 7+9+5=21. Since 21 is less than the sum of this region, we know our initial assumption is incorrect and 5 must be placed in one of the squares with the red dots. Although we don’t know in which red dotted square 5 should be placed, we now know exactly where 1 and 5 are placed in the left column.

As with any 5x5 Calcudoku puzzle, the sum of all squares in the left column of the shaded region below must be 1+2+3+4+5=15. Since the total sum of this region is 20, we still need to place numbers in the two red dotted squares which together sum up to 5. Now let’s examine the second column from the left to see where 5 can be placed. We can’t place 5 in a red dotted square because we will exceed the sum 20 in the shaded region. We also can’t place 5 in the L-shaped region above the red dotted squares because only 1x1x4 or 1x2x2 can produce 4. Therefore 5 can only be placed in the top square of the second column.