Author Topic: [SPOILER] Varr's Puzzle #3  (Read 4193 times)

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Yip

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• Posts: 4007
[SPOILER] Varr's Puzzle #3
« on: March 02, 2004, 09:47:38 pm »
First, if you haven't seen the puzzle, go try it out before reading this.

 Quote DO NOT LOOK BELOW UNLESS YOU ARE READY FOR THE SOLUTION

The green arrows show the correct path.

So that's 3 right, 4 down, 2 up-left*, 3 right, 4 left, 2 down-right, 3 up, 2 right, 3 down.

*at this point you may go 2 up and 2 left rather then 2 up-left.

Azron

• Full Member
• Posts: 231
[SPOILER] Varr's Puzzle #3
« Reply #1 on: March 03, 2004, 08:22:04 pm »
*glad he got the right solution*

The way I looked at this puzzle was to see if any squares could be 'crossed off' for being useless (i.e. cannot be part of the solution).

If we number the columns from left to right as a, b, c, d, e, and the rows from top to bottom as A, B, C, D, E, I can give references to each square, chess stylie.

Firstly, the 0 in cC is clearly redundant, if we land there we can't go anywhere! Next we can cross off all of the 2s that appear in squares bB, dB, bD and dD, since if we land in any of these we can only move to another, so we effectively have a 'cycle', which cannot be escaped.

It's worth noting at this stage that any squares that can only move to squares we have crossed off are also redundant (since this sets up the cycle again, please ask me to clarify if I have failed to explain this properly).

The 4s in squares bA and bE are also eliminated, since they one can only move from one to the other. And thus the 3 in square eD goes too, since this can only move to bA or bD (already ruled out). This leads to square eA going, and square aD (I'll save the details, check the only valid moves from each and you'll see why).

Then aE goes too. cA and cD is another example of a two square cycle.

We now have all the obviously useless squares elminated, and so look at the options for a path from aA to eE. Starting in aA, there is only one square we can move to (dA), and then only to dE. At this point we can move to either bC, or dC then bC, either way is valid, but the former is quicker.

From here every move is 'forced' since there is then only one possible square that can be moved to legally that has not been eliminated already.

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And that's how you overcomplicate a problem '>

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