hacker.org

maybe you like puzzler.sourceforge.net same as i did ... the enthusiasm of that guy is amazing ...

send me a pn if you want to get the puzzle def python script ... (ps: python-solver took about 4 min using algorithm x)

here comes my simple solution: first: load all numbers onto stack (while setting mem completly to zero) 882** 1-0^<1v01^>0^7?037*-g then: create hashtable { mod 127 } to reduce overall compare-operations ... d882**1- 1^1^/1^*2^1v-1^* 0^<0^54*0+?3^-2?4gddp!1+056*-g d2v1v> 088*1--g happy coding!

hi there, tried this one with the hvm java tool provided here somewhere, much easier to manage the relative jumps there + comments possible ... thumbs up! my solution: ... [edit: shouldn't been provided here, after-solve link took me here] ... took 30335 cycles to solve the example figure.. happy c ...

i can tell u a little more about the result that i examined by hand (+golly): there are 2 periodical activities in this world: *154 .. the upper thing i called "glider machine"(A), the reflector (B), creating new gliders +(G1)/(G2) *174 .. the moving "pathbuilder" (C), creating n...

here comes my (straightforward 8) ) receipt (>0 numbers only, but passed the test) ... 0<1</2+0^1-/1-<p which means: load the 2 nums, divide .. result is 0 (if S0<S1) or any n >0. then calc ( (n+2) / (n+1) ) which gives "2" for n=0 only, "1" else. looking forward for more broken ...

hi all,

here comes my solution to this hvm-hack:

(formatted with spaces, in the form
jumpToStart!compareRoutine! initSize outerLoop innerLoop+call(compare) checkInnerLoop )

9g!-1?$<pd! 5 1-0^2?2gd! 0^<1^1-0^<2^3c 0^7?054*-g!dd085*-7-g

happy coding!

max.

look, here comes my quick'n-dirty solution for this one... it uses a !subroutine! for comparing the s0/s1 values in place .. 44*g!1^1^:1:2?1vd$! 0<1<5c 2<5c 3<5c 4<5c 5<5c 6<5c 7<5c 8<5c 9<5c 91+<5c 92+<5c 93+<5c 94+<5c 95+<5c 96+<5c 97+<5c 98+<5c 99+<5c 99+1+<5c p lots of bits wasted, but why not, ...