The Seven Bridges of Königsberg/Section 18
Since the sum of the numbers opposite A, B, C, etc., is double the number of bridges, it is clear that if this sum is increased by two and then divided by 2 the result will be the number written at the top. When all the numbers in the second column are even, and the half of each is written down in the third column, the total of this column will be one less than the number at the top. In that case it will always be possible to cross all the bridges. For in whatever region the journey begins, there will be an even number of bridges leading to it, which is the requirement. In the Königsberg problem we could, for instance, arrange matters so that each bridge is crossed twice, which is equivalent to dividing each bridge into two, whence the number of bridges leading to each region would be even.