Challenge Coins
Challenge problem again?
I posted this challenge yesterday and got two attempts both suggesting that it was easier than it really is. I repeat it for the benefit of those who like this sort of thing.
Coins are arranged on a table in a rectangular array of r ranks top to bottom and f files left to right with f > = r and r >= 2.
Coins are identical except for the top right hand corner one which is different. Also the opposite corner one is missing.
The object is to get the special coin into the missing position. Each move you slide a coin up or down or left or right (but not diagonally) into an empty space. Coins must stay in the array.
Find a formula for m, the minimum possible number of moves, in terms of f and r. It is worthwhile double checking that you have minimum figures before looking for a pattern. If correct, formula should give m(f = 6,r=3)=29 and m(f=6,r=6) = 37.
I will leave a couple of hints here in 24 hours and the answer in 48 hours.
Initially the coin is at position (f,r) and the hole at position (1,1).
1) We need to get the hole adjacent to the coin in order for it to move.
This requires (f-1) moves to get it to the bottom right corner, then (r-2) moves to get it just below the special coin.
1 more move gets to coin to position (f,r-1) with the hole above it.
So that's (f+r-2) moves so far.
2) Now let's try to move the coin in a zigzag fashion to the first row.
To get it from (f,r-1) to (f-1,r-1) requires 3 moves.
Now the hole is to the right of hte coin.
Another 3 moves gets it to (f-1,r-2) with the hole above it.
Continuing in similar vein, it will reach the first row after
6(r-2) moves and it will be in position (f-r+2, 1) with the hole above it.
3) It now requires 5 moves to move the coin one position to the left with the hole above it.
So 5*(f-r) moves will take the coin to position (2,1) with the hole above it.
4) It needs only 3 more moves to get to position (1,1).
So the total is m(f,r) = (f+r-2) + 6(r-2) + 5(f-r) + 3
= 6f + 2r - 11
This matches m(6,6)=37,
but it's giving m(6,3) = 31
I'll see if I can improve on this.
***EDIT***
OK I think I got it.
My earlier analysis was not optimal because I was initially moving the hole to (f, r-1) such that the initial movement of the coin was down. It is better if the hole is moved to (f-1, r) such that the initial movement of the coin is to the left.
Comparing these two setups, the coin can move to a given position on the first row in the same number of moves. The difference is that with my original setup, the hole was to the right of the coin as it hit the first row. With the new setup, the hole is at the top. So it saves 2 moves when f>r. When f=r, the coin reaches the first row and first column at the same time so it gives the same answer in both cases.
That means,
m = 6f + 2r - 11 when f = r
m = 6f + 2r - 13 when f > r
Or in computer terminology
m(f,r) = 6f + 2r - 11 - 2(f > r)
where (f > r) = 1 if true, 0 otherwise
So the optimum path is a zigzag with the initial movement of the special coin to the left, followed by a straight line movement along the first row if necessary.
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Challenge Coin Collecting 101
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