Linear Chinese Checkers
You have a Chinese Checkers board that consists of an extremely long but single row. On the left side of the board there are a googolplex ( ) of blue marbles all lined up in a row, and on the right side there are a googolplex plus one ( ) of red marbles all lined up in a row. Initially, there are no spaces between any two blue marbles and no spaces between any two red marbles, but there is one space between the two marble sets.
Just like in Chinese Checkers, a turn consists of either moving a marble one step in any direction to an adjacent empty space, or by jumping a marble over another single marble into an empty space.
What is the digit sum of the minimum number of turns needed to switch the red and blue marbles (so that the googolplex plus one of red marbles are all on the left side and the googolplex of blue marbles are all on the right side, with no spaces between any two blue marbles and no spaces between any two red marbles, but one space between the two marble sets)?
The answer is 5.
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1 solution
For 1 blue marble and 2 red marbles, the minimum number of turns using S slides and J jumps to switch the marble sets is S , J , S , J , S , for 2 J ’s, 3 S ’s, and a total of 5 turns.
For 2 blue marbles and 3 red marbles, the minimum number of turns using S slides and J jumps to switch the marble sets is S , J , S , 2 J , S , 2 J , S , J , S , for 6 J ’s, 5 S ’s, and a total of 1 1 turns.
Continuing this pattern for n blue marbles and n + 1 red marbles, the minimum number of turns for S slides and J jumps to switch the marble sets is S , J , S , 2 J , S , 3 J , … S , n J , S , n J , S , … 3 J , S , 2 J , S , J , S , for n 2 + n J ’s, 2 n + 1 S ’s, and a total of n 2 + 3 n + 1 turns.
When n is a power of 1 0 greater than 1 , (like a googolplex), the total number of turns is in the form of 1 0 0 0 … 0 0 0 3 0 0 0 … 0 0 0 1 , which has a digit sum of 1 + 3 + 1 = 5 .