Search found 8 matches

by lifthrasiir
Sat Aug 27, 2011 10:10 am
Forum: Challenges Solved
Topic: VIndicated
Replies: 15
Views: 1280

Re: VIndicated

megabreit wrote:Funny challenge... especially for Emacs and Windows users :-)
^V in the standard gvim Windows installation is mapped to <paste>, so the challenge is slightly harder for the "Followers of Bill"
You have C-q as an alternative.
by lifthrasiir
Sat Aug 27, 2011 10:06 am
Forum: Challenges Solved
Topic: How is your Grammar?
Replies: 16
Views: 1648

Simple solution: g = { 'A': 'is', 'B': 'mm', 'C': 'oo', 'D': 'rgr', 'E': 'ryg', 'F': 'dth', 'G': 'you', 'H': 'esol', 'I': 'ionA', 'J': 'GDaBarA', 'K': 'veECFHutI', 'L': 'PQ', 'M': 'n', 'N': 'm', 'O': 'oaNcho', 'P': 'MO', 'Q': 'NR', 'R': 'sky', 'S': 'JKL', } a = 'S'; b = '' while a != b: a, b = ''.jo...
by lifthrasiir
Sat Aug 27, 2011 9:47 am
Forum: Challenges Solved
Topic: Frequent These Parts
Replies: 7
Views: 794

I have used Kasiski examination to guess the key length. Four out of five four-character repetitions are found to have the distance of multiples of 8 ("iudz" for 136, "rgai" for 64, "dzbh" for 152, "sdyc" for 184), so the key length is likely 8. Since the word...
by lifthrasiir
Mon Aug 22, 2011 11:02 am
Forum: Challenges Solved
Topic: Deluge
Replies: 3
Views: 935

My solution: 005c! calls the recursive procedure at (0,0) 0^0:1^35*:*1- pushes non-zero iff 0 <= x <= 15 2^0:3^35*:*1- pushes non-zero iff 0 <= y <= 15 *35*? if one or both is zero, then it is out of bound 1^1^44**+0^ pushes a memory location for current position (twice) <4?d if it is empty then ski...
by lifthrasiir
Sun Aug 21, 2011 9:52 pm
Forum: Challenges Solved
Topic: Some Randoms
Replies: 2
Views: 366

Re: Some Randoms

IIMOG wrote:I used GNAT (GPS). Compiling worked fine, but I had to do the gnatbind and gnatlink by hand.
A single "gnatmake testran.adb" should be enough.
by lifthrasiir
Sun Aug 21, 2011 6:51 pm
Forum: Challenges Solved
Topic: Substitute Teacher
Replies: 28
Views: 2667

Another possible approach is to exploit very long word. The ciphertext PWZEHVNWIEDZ would suggest that this word has exactly three distinct duplicate characters at position 2 & 8, 3 & 12 and 4 & 10 and nothing else. The quick search on my laptop gave this: $ grep -E '^.(.)(.)(.)...\1.\3....
by lifthrasiir
Sun Aug 21, 2011 12:37 pm
Forum: Challenges Solved
Topic: calculator
Replies: 20
Views: 1480

I took 2 hours to solve this challenge, due to the lack of assembler or other tools whatsoever. But I think the lack of assembler was actually a pro, not a con, as the code size dramatically changes depending on how the jump/call target is calculated. My code currently weighs 174 bytes (with six no-...
by lifthrasiir
Sat Aug 20, 2011 10:23 pm
Forum: Challenges Solved
Topic: Sum Some Primes
Replies: 4
Views: 986

Sum Some Primes

Thank you Wolfram|Alpha.

Input: sum of Prime[i] for i=49999951..50000000
Output: 49122557320

Of course the "correct" way to solve this problem is finding a fast enough primality testing routine for nine-digit numbers. Plain sieving wouldn't help, so I'd use SPRP-based testing maybe.