Problem 776. TRON strategy toggle

Created by Christopher

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59 Solutions
18 Solvers

Last Solution submitted on Nov 19, 2022

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Richard Zapor on 20 Jun 2012

Suggestion on implementation. Usage of persistent with a flag that bikes are isolated will save time.

rifat on 26 Jan 2015

Anyone knows why I cannot change recursion limit by set(0,'RecursionLimit',1200)? As far as I recall, I had done it before in Cody. Has Cody blocked access to this option?

Jan Orwat on 26 Jan 2015

Hello Rifat. I've made some tests and had no problem to change 'RecursionLimit' to 1200 although for a very simple function I've been able to do up to 993 recursions, but for any higher number cody server returned 'temporary unavailability of MATLAB Service' error. Please note that this problem occured when a function which was using recursion were changeing 'RecursionLimit' by itself (even if 'set(...' is called only once). When limit was changed from main function and a subfunction was recursed, or when limit was changed from testsuite everything works normal.

rifat on 26 Jan 2015

yes.. I changed recursion limit from the subfunction. thanks for the tips.. :)

Solution Comments

Solution 8629475

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Meg Noah on 15 Oct 2022

:-)

Solution 8629370

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1 Comment

GeeTwo on 4 Jul 2022

As I commented elsewhere, only 1198 loops are necessary. The longest path in a square array avoiding adjacent squares is a spiral from a corner to a point near the center. (For odd, a deep slalom gives the same answer.) The number of steps is:
odd: (n^2+2n-3)/2
even:(n^2+2n-4)/2
For 48, this is 1198, not 2500.

Solution 8628650

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1 Comment

GeeTwo on 4 Jul 2022

Lessons from other solutions:
My convolution should have been with:
[ 0 1 0; 1 1 1; 0 1 0].
filter2 works better than conv2 here; no need to trim edges.

Solution 591183

1 Comment

1 Comment

GeeTwo on 4 Jul 2022

Great solution, and proper!
I doubt you're saving anything with t, as you only use it once and don't need to enclose it in parenthesis.
2500 is overkill. Should be based on 48^2, and then only a bit over half. The longest possible path separated by "tracks" is a spiral in from one corner. For odd boards (where a slalom can be as long), the formula is (n^2+2n-3)/2. For even, it's (n^2+2n-4)/2. For 48, this is 1198.

Solution 100904

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1 Comment

Richard Zapor on 20 Jun 2012

Recursive Grow Flood Fill

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Problem 776. TRON strategy toggle

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Problem 776. TRON strategy toggle

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