Broken Keys

[klavierspieler21]

I "solved" this by using the fact that max(a,b) = (|a-b|+a+b)/2, and Newton's method to calculate the square root of (a-b)^2. However, it shouldn't have passed when a-b = 0 because I'm dividing by 0 in newton's method.

I got error messages but I still "passed"...

Here's my code:
0<1<1^1^-0^*11^1^/+2/1^1^/+2/1^1^/+2/1^1^/+2/1^1^/+2/1^1^/+2/1^1^/+2/1^1^/+2/1^1^/+2/1^1^/+2/1vd++2/p

The repeating 1^1^/+2/ is the recursive newton's method applied enough times.

[ftfish]

I used the fact: when a<b, a/b equals 0.
max(a,b)=(a/b *a + b/a *b)/(a/b + b/a)

here's my code:

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0<0<1</*1<1<0</*+0<1</1<0</+/p

[livinskull]

eat this:

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1<0</0^/<p

getting divide-by-zero warnings if second is larger, but who cares as long as it works

[coderT]

eat this:

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1<0</0^/<p

getting divide-by-zero warnings if second is larger, but who cares as long as it works

damn i thought that's not allowed.
Here is how I solved it:
1. except memory location 0, fill the whole memory with the content of memory location 1.
2. location = Mem[0] / Mem[1]
3. the greater number is: Mem[location]

this is stupid but worked.

[jonik555]

Oh no, I found other solution, but it is solution for next challenge Very nice one!!!!

[OKOB]

Another conventionally correct solution (All tests passed!)
111111111111111111111101<0</v<p!
Longer (1111111111111111111111101<0</v<p!) - not all tests passed
Shorter ( 11111111111111111111101<0</v<p!) - not all tests passed


Bad tests set or error in 'v' comand implementation (Longer don't work!)

[arthur]

I used the fact: when a<b, a/b equals 0.
max(a,b)=(a/b *a + b/a *b)/(a/b + b/a)

here's my code:

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0<0<1</*1<1<0</*+0<1</1<0</+/p

It seems not to work for negative integer. if a<0<b and |a|>|b|, a/b != 0.
my solution is

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0<1<-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-0^2/-1+4*g    1<p!0<p!

works for |a-b|<2^27

[whattheh@ck]

I sort of cheated

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0<1</9*0^52*/3>3<25**-3<+g1<p!000000000000000000000000000000000<p!

all of those 0's seem to be a good enough buffer to pass all the tests...

I just fixed it to be more exact:

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1<0<1</9*0^52*/3>3<25**-3<+0^52*/3>3<25**-3<+3/gp!00<p!

it works due to an interesting fact that the sum of the digits in a multiple of 9 is 9.

[zjorzzzey]

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0<1<1^1^10>01>/<vdp

Works because memory is initialized with zeroes by default.

Seen more beautifull solutions though...

[kalunos]

here comes my (straightforward ) 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 keys ....

happy coding!

[AMindForeverVoyaging]

0<1<-088*8*8*8*8*8*8*8*8*-2*/1+4*g0<p!1<p!

Dividing the difference of the two numbers by -(2^31) gives either -1 or 0, from there it's easy.

Now to do it without the division instruction... I really have no clue how to

[compudemon]

i cheated with divide
if smallnumber > 0 and < bignumber then ( smallnumber + bignumber - 1) / bignumber = 1
else = 0

0<1</88888****1-+8/8/8/8/8/4*g1<p!0<p

yeah this wont work if the test has a big number in it but who cares

[Chnty]

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 1<0</0^2*1v1+/<p 

Math rules!
Because 2x/(x+1) is 0 for 0 and 1 for anything else

[sascha27]

I kind of used a shift operation, 32 times so it was big enough for integer values

0<1<-2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2/01-*<p

:P

[rhian]

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1<0<1</>0<p

i just save the value of M1 into the result of M0/M1. if M0 < M1 then the result is 0 and higher M1 value is stored into M0 which I print in any case.