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I stopped at a street corner and asked for directions to Burger Jack. Unfortunately, the person I wasked was Larry Longway, whose directions are guaranteed to be too complicated. He said,"You are now facing north:

Go straight for two blocks.

$\ \ \ $Turn left. Go straight for one block.

$\ \ \ $Turn right. Go stright for three blocks.

$\ \ \ $Turn right. Go straight for five blocks.

$\ \ \ $Turn right. Go straight for three blocks.

$\ \ \ $Turn left. Go straight for one block.

$\ \ \ $Turn right. Go straight for four blocks.

$\ \ \ $Turn left. Go straight for two blocks.

$\ \ \ $Turn left. Go straight for one block.

$\ \ \ $Turn left. Go straight for five blocks, and you are there.

By the time I arrived I was out of breath and Burger Jack was closed. Please give me the directions for the shortest path from my original spot to Burger Jack (assume no street have dead ends.)

Below I included a picture of my work as a rough sketch. Here is the points I have labeled on my graph: (0,0) as the starting point (-1,2) (4,5) (4,2) (5,2) (5,-2) (7,-2) (7,-1) (2,-1) Ending point

So, he is two blocks east and one block south from his original point. I am looking for a confirmation on this answer if anyone else has time to work the problem, it would be greatly appreciated.

$\hskip{2.3in}$enter image description here

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Thanks for embedding my picture and placing another tag for the question, Kannappan Sampath. I tried putting the needed syntax for it originally, but the picture nor the link showed up the first time around. – Joe Feb 13 '12 at 22:09
That looks correct. – David Mitra Feb 13 '12 at 22:09
Thanks David. If any others could check the solution, that would be nice. :) – Joe Feb 13 '12 at 22:10
Thanks for cleaning up the formatting of the directions David. I had clicked shift enter for a line break for each one, but it appears it did not format correctly (it just brought all of them together). Again, thank you. – Joe Feb 13 '12 at 22:14
Besides being within my wheelhouse, so to speak, I especially appreciate this question for all the work that you demonstrated. – The Chaz 2.0 Feb 13 '12 at 22:15
up vote 1 down vote accepted

I combined the "turn and go straight..." commands to just

(Cardinal direction: N/S/E/W):(# of blocks)











Now clearly many of these "cancel" with others (e.g. E:5 cancels with W:5, N:3 cancels with S:3). We are left with a net change of E:2 and S:1, corresponding to the point $(2, -1)$ on the plane if you started from the origin.

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Ah, clever thinking The Chaz. Thanks for your input. – Joe Feb 13 '12 at 22:12
Well I was trying to write this out on a pizza box, and kept getting lost in the paragraph, so I decided to simplify things so I could return to eating :) – The Chaz 2.0 Feb 13 '12 at 22:14
Haha StackExchange and eating - quite the multitasker. – Joe Feb 13 '12 at 22:15

Another (but basically the same) way to think about it: Suppose the grid of streets is a lattice, and your starting point is the origin. I will treat each step of Larry's directions as a vector (which can be identified with a point in $\mathbb{R}^2$).

The location you are searching for is:

$$ (0,2) + (-1,0) + (0,3) + (5,0) + (0,-3) + (1,0) + (0,-4) + (2,0) + (0,1) + (-5,0) $$

Add all the components you get: $$ (0 -1 + 0 + 5 + 0 + 1 + 0 + 2 + 0 -5,2 + 0 + 3 + 0 -3 + 0 -4 + 1 + 0) = (2,-1). $$

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Ah, another way to look at - similar to The Chaz. Nice work. – Joe Feb 13 '12 at 23:27

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