6 ms·
https://itty.bitty.site/#/?XQAAAALfAAAAAAAAAAA0HSzgdXPvDK5AEi+dXR+Wp+VBBrHPUivt9QmuGfWQLaDzPoK1OINSJH8kY65bBkC0AlA1z6uxBe7XA66svOmlKtUCOjl0NqBjRKWb1K7gATXv4mdfv
by Exuma 8y ago
https://itty.bitty.site/#/?XQAAAALfAAAAAAAAAAA0HSzgdXPvDK5AEi+dXR+Wp+VBBrHPUivt9QmuGfWQLaDzPoK1OINSJH8kY65bBkC0AlA1z6uxBe7XA66svOmlKtUCOjl0NqBjRKWb1K7gATXv4mdfvrbLH5qIBs5OFCEc6mlFqEjzlA+4tzQ2BFx2AqwbP6tWoWy6moAmT2l0YRTSngwnsoOTx/0QOUOV6SAmdcHylijJ+B6JLRuPYqaZVy83XDw61WKkzUCsVYmJ9n0ZxIUlcCjMfvDISU/0KPDsTkpWNaIRmjH4eef//j62gA== https://itty.bitty.site/#/?XQAAAALfAAAAAAAAAAA0HSzgdXPvDK5AE...
- TimTheTinker 8y agoHi to you too :-)
- mchahn 8y agoI just get a page with the URL shown.
- Exuma 8y agoyour first clue
- flyGuyOnTheSly 8y agoOK so you keep following the links (by copying and pasting into your address bar) and after about four links it says "oh hi ". Explain yourself! How did you do this?! Impressive!
- sterlind 8y agoSimple! 0. Write oh hi :D, get link0 1. Paste link0, get link1 2. Paste link1, get link2 ... Encoding goes E(T), E(E(T)), etc. let K be the per-char encoding overhead, so E(T) = KT: complexity is O(K^N * T) where N is number of levels and T is length of text at level 0!
- flyGuyOnTheSly 8y agoD'oh! You're right! That IS simple!
- ReverseCold 8y agooh hi (I will say I'm disappointed that it wasn't a technique to make circular URLs.)
- Exuma 8y agoThat's what it started out as, I was going to look at making a quine but then quickly gave up as I am tired
- TimTheTinker 8y agoI think a quine would be impossible since there's no possibility of indirection or data compression.
- ozzmotik 8y agohoi m8 nicely done o7
- aeleos 8y agoIt seems that due to the extra text from https://itty.bitty.site https://itty.bitty.site each iteration increases the size of the next link. I wonder how the urls are generated and if this is a linear or exponential increase.
- TimTheTinker 8y agoMy guess is 1st-degree polynomial. There’s a constant addition with each iteration, plus the contents appear to be base64(ish)-encoded into the next URL, which multiplies the length by a constant.
- comesee 8y agoIt's original_size * growth_factor ^ iterations I would say it's exponential in iterations. Growth factor is probably <2
- Timpy 8y agoVery clever implementation of this technique