BigNate469 wrote:

MonkeyBean2 wrote:

co0lcr34t10ns wrote:

GUYS, REMEMBER THAT ALL CIPHERS THAT ARE SOLVED THAT ARE NOT ADDED ARE NOT COUNTED FOR COINS TO ALLOW PEOPLE IN OTHER TIMEZONES TO ACTUALLY GET A CHANCE PLEASE

Your policy on only allowing them to be solved when the index is updated with them is a bit silly, as if someone does solve it before then, other people's chances will already have been ruined as the response will be spoiled. Also, it gives an unfair advantage to people in a similar timezone as you, or in a timezone where they are available when you normally update the post. Counting solutions even if the cipher has not yet been added to the index yet will make it more fair across timezones.

Yeah, maybe just make it so that if it's clear that someone actually did solve it on their own (without looking at the answer, such as if the way it is solved uses a different programming language or method), they can still get DeCoin.

Or, alternatively, just make it so that you get DeCoin when you create a cipher, and not when you solve it.

MonkeyBean2 wrote:

BigNate469 wrote:

snip

I think only the first person who gets the answer right should get DeCoin as otherwise it is easy to just piggyback on other people's work without solving it yourself which isn't nice.

BigNate469 wrote:

MonkeyBean2 wrote:

BigNate469 wrote:

snip

I think only the first person who gets the answer right should get DeCoin as otherwise it is easy to just piggyback on other people's work without solving it yourself which isn't nice.

But that's the system we have now, which isn't working for people who are asleep while (mostly you) are awake and using Scratch.

blubby4 wrote:

This is unfair! I never get a chance to solve anything because everyone does things while I'm asleep

blubby4 wrote:

Have you gone through and removed coins for people that have previously done this? Otherwise it's kind of ok, not really unfair

co0lcr34t10ns wrote:

BigNate469 wrote:

MonkeyBean2 wrote:

BigNate469 wrote:

snip

I think only the first person who gets the answer right should get DeCoin as otherwise it is easy to just piggyback on other people's work without solving it yourself which isn't nice.

But that's the system we have now, which isn't working for people who are asleep while (mostly you) are awake and using Scratch.

And that's why I am sticking to the new rule. We've already managed to get some negativity from people who are in different timezones:

blubby4 wrote:

This is unfair! I never get a chance to solve anything because everyone does things while I'm asleep

And this also means people who are less skilled actually have a fricking shot at solving some of these. I'm also not adding surveys that someone solved before adding. If you solve a cipher early, you will essentially vapourize the cipher. Please don't do that. I think it's not a problem with thee rule itself. If people follow it, there will be no problem. Also adding any unsolved ciphers.

blubby4 wrote:

Have you gone through and removed coins for people that have previously done this? Otherwise it's kind of ok, not really unfair

Potentially. I will sleep on it and see what happens.

MonkeyBean2 wrote:

8to16 wrote:

Cipher Creators: Me
Cipher:

Ĵ Ǥ  ƈ Lj Ƽ ǜ nj Ɣ Lj  ¼  Ƽ ǀ Ɣ Lj Ƅ ǐ Ƥ Ƹ Ɯ  nj Ǥ nj ǐ Ɣ ƴ è  İ Ƥ Ƹ ǔ Ǡ °  Ę Ƥ Lj Ɣ Ƙ Ƽ Ǡ  Ä Ì Ä ¸ À °  Ę ư Ƅ nj Ơ  Ì È ¸ À    Lj Ɣ ư Ɣ Ƅ nj Ɣ  À ¤ 

Difficulty Level: 5
Short Description: I multiplied the unicode character values and added spaces after each character

Cipher Solvers: Me (@MonkeyBean2)
Solution:

My browser / operating system: Linux, Firefox 131.0, Flash 32.0  release 0)

Code used to solve (python):

x = 'Ĵ Ǥ  ƈ Lj Ƽ ǜ nj Ɣ Lj  ¼  Ƽ ǀ Ɣ Lj Ƅ ǐ Ƥ Ƹ Ɯ  nj Ǥ nj ǐ Ɣ ƴ è  İ Ƥ Ƹ ǔ Ǡ °  Ę Ƥ Lj Ɣ Ƙ Ƽ Ǡ  Ä Ì Ä ¸ À °  Ę ư Ƅ nj Ơ  Ì È ¸ À    Lj Ɣ ư Ɣ Ƅ nj Ɣ  À ¤'
result = ''.join([chr(ord(z)//4) for z in x.replace('  ', '\x80').replace(' ', '')])

Method: I looked at the differences between all the character codes after putting them in a set to remove duplicates and then sorting it. I found the minimum difference be 4, which meant that was most likely the factor used.

p.s. I know @blubby4 already solved this, I was working on it when they posted it :( there's no proof however as to if I cheated or not though (I didn't cheat). I'm just posting this cus why not

co0lcr34t10ns wrote:

I forgot to decoin update, gimme a sec
Also all ciphers added before 8:20 EST

blubby4 wrote:

co0lcr34t10ns wrote:

I forgot to decoin update, gimme a sec
Also all ciphers added before 8:20 EST

Is that 8:20 AM or PM?

ThisIsTemp1 wrote:

(#129)
Cipher Creators:Me
Cipher:B1E2A1C1D2B1A1G3D1
Difficulty level:3
Short Descriptions: This isn't that complicated.

ABCDEFG
HIJKLMN
OPQRSTU
VWXYZ

=CONCATENATE(B1,E2,A1,C1,D2,B1,A1,G3,D1)

       Code: B1 E2 A1 C1 D2 B1 A1 G3 D1

BLACKBAUD

8to16 wrote:

Cipher Creators: me
Cipher:

⁹⁼₄ₑ⁄‸₌₀₁₋‸⁻₁₈₀⁽₊‸₏⁹₋‸⁽⁹₌⁽₆‸⁺ₑ‸⁹₆‸⁽₎₁₄‸₃₍₅₉₍⁹₌⁆‸⁚⁽₌₌⁽₊‸₄₍⁻₃‸₆⁽ₐ₌‸₌₁₅⁽

Difficulty Level: 5
Short Description: Clue is in my username

100th post in this topic

                                                                                                                                                      Sadly, this cipher was eaten by an evil kumquat. Better luck next time

                                                                                                                                                      Treat the input as an array of bytes
                                                                                                                                                      Discard every other byte, starting with the second one (so you end up with byte 0, byte 2, byte 4, etc.)
                                                                                                                                                      Subtract 24 from each byte and convert to ASCII
                                                                                                                                                      Prepend an "S"
                                                                                                                                                      I solved this with @MonkeyBean2's "graph every pair of bytes" method. If your cipher is just some sort of rotation, graphing the ciphertext makes it pretty obvious what you did

davidtheplatform wrote:

8to16 wrote:

Cipher Creators: me
Cipher:

⁹⁼₄ₑ⁄‸₌₀₁₋‸⁻₁₈₀⁽₊‸₏⁹₋‸⁽⁹₌⁽₆‸⁺ₑ‸⁹₆‸⁽₎₁₄‸₃₍₅₉₍⁹₌⁆‸⁚⁽₌₌⁽₊‸₄₍⁻₃‸₆⁽ₐ₌‸₌₁₅⁽

Difficulty Level: 5
Short Description: Clue is in my username

100th post in this topic

Solved (mostly):

                                                                                                                                                      Sadly, this cipher was eaten by an evil kumquat. Better luck next time

The first letter seems to be missing (or i decoded it wrong) but it's pretty clear what it should be.
Method:

                                                                                                                                                      Treat the input as an array of bytes
                                                                                                                                                      Discard every other byte, starting with the second one (so you end up with byte 0, byte 2, byte 4, etc.)
                                                                                                                                                      Subtract 24 from each byte and convert to ASCII
                                                                                                                                                      Prepend an "S"
                                                                                                                                                      I solved this with @MonkeyBean2's "graph every pair of bytes" method. If your cipher is just some sort of rotation, graphing the ciphertext makes it pretty obvious what you did

50_scratch_tabs wrote:

ThisIsTemp1 wrote:

(#129)
Cipher Creators:Me
Cipher:B1E2A1C1D2B1A1G3D1
Difficulty level:3
Short Descriptions: This isn't that complicated.

Solver: me
Solution: Blackbaud
Method:
Create a spreadsheet, make it go like this:

ABCDEFG
HIJKLMN
OPQRSTU
VWXYZ

and do this formula:

=CONCATENATE(B1,E2,A1,C1,D2,B1,A1,G3,D1)

       Code: B1 E2 A1 C1 D2 B1 A1 G3 D1

Result:

BLACKBAUD

(I used Google sheets. You might need to make adjustments if you use a different spreadsheet program.)

ThisIsTemp1 wrote:

(#137)
Correct.

50_scratch_tabs wrote:

ThisIsTemp1 wrote:

(#137)
Correct.

Yay!

co0lcr34t10ns wrote:

50_scratch_tabs wrote:

ThisIsTemp1 wrote:

(#137)
Correct.

Yay!

HOW DID YOU EVEN FIND THAT
going to school rn, will add at 3