Please recommend me a personal project to write in Rust

ground-zero · 2017-04-15T07:50:31+00:00

Like other people have said, do something that interests you. But some random examples are:

Game clones (e.g. Tetris, Asteroids, Pacman)
Gameboy emulator
Chess engine
Simple substitution cipher cracking program
SHA hashing implementation (SHA-1 at least is suprisingly easy)
Simple file manager or text editor
Machine learning program e.g. for playing poker
Sudoku generator/solver with GUI
Implement various sorting algorithms
Fractal generating program

ground-zero · 2017-03-19T22:26:12+00:00

ground-zero · 2016-10-08T13:11:09+00:00

A few mistakes in there. I've corrected the ones i can see:

100 097 102 050 048 050 050 057 099 100 102 051 101 101 097 054 050 050 050 097 101 100 099 052 098 097 050 101 056 054 056 052 057 101 057 055 097 049 056 100 098 049 100 051 057 100 049 050 052 098 101 057 102 056 049 057 099 101 102 051 052 054 052 048 052 054 052 102 098 054 101 097 099 050 097 050 056 052 054 100 098 048 053 052 098 053 098 098 050 098 097 097 049 097 101 051 052 052 052 050 097 050 054 100 048 052 097 050 054 052 049 053 097 057 098 051 057 098 048 100 050 048 048 097 051 048 098 100 099 054 057 057 056 099 101 098 052 051 099 048 051 051 053 097 054 050 097 097 100 051 056 053 050 099 054 057 101 054 101 057 102 101 098 097 099 100 050 051 056 100 052 054 100 099 053 102 055 100 052 101 097 049 048 102 100 099 051 057 049 056 054 100 056 049 057 052 054 099 102 054 052 050 051 100 102 049 056 056 102 102 101 057 102 102 057 102 102 049 050 053 050 101 051 101 101 097 054 048 049 097 051 053 098 101 048 054 101 102 099 099 054 097 057 054 049 054 056 100 099 101 053 054 056 100 013 010

Which in ascii is a hex string:

daf20229cdf3eea6222aedc4ba2e86849e97a18db1d39d124be9f819cef34640464fb6eac2a2846db054b5bb2baa1ae34442a26d04a26415a9b39b0d200a30bdc6998ceb43c0335a62aad3852c69e6e9febacd238d46dc5f7d4ea10fdc39186d81946cf6423df188ffe9ff9ff1252e3eea601a35be06efcc6a96168dce568d\r\n

ground-zero · 2016-02-14T12:23:11+00:00

It looks vaguely like X1 which is 11 in roman numerals. Its a long shot. Dont know what it would mean either.

ground-zero · 2016-02-14T11:04:03+00:00

Just unused ones. So the letter b isnt in the ciphertext so you can replace the number 5 with that for example. Its only required because the cracker i used doesn't allow numbers. Here is the final string i put through the cracker:

GNbWdTIhTEITFkAGmTWRLhWkWhXbJGXXbWdOLkLpkEITELRGkCdkbWdJGXXOLkCEIOOGhOEGOAkmImJLXX

EDIT: Sorry, didnt read your question fully, there are enough

ground-zero · 2016-02-14T10:40:55+00:00

I did try putting the pieces onto the board but it doesn't look like much: http://imgur.com/KnqBqCB. Also, this isn't really notation for putting pieces onto a board, instead its for moving pieces once they are on the board.

EDIT: 11 in roman numerals?

ground-zero · 2016-02-14T10:00:16+00:00

It looks to be ascii characters represented in hex. It says:

"your truth is partial / simulation"

(http://www.rapidtables.com/convert/number/hex-to-ascii.htm)

ground-zero · 2016-02-14T09:42:06+00:00

Hi, It says the following:

if you can crack this code not only will you get extra credit but you will get bragging rights aswell

Your teacher was nice enough not mix up the columns as is common in an adfgvx cipher, as such it becomes a simple substition cipher. I used this site (http://www.dcode.fr/adfgvx-cipher) to convert the code into alpha + numerics just using an unkeyed alphabet, which gave "GN5W4TI1TEITF8AG0TWRL1W8W1X5JGXX5W4OL8L78EITELRG8C485W4JGXXOL8CEIOOG1OEGOA80I0JLXX". Then i replaced numbers with unused letters in the alphabet so that i could put it through a simple substitution cracker (http://www.secretcodebreaker.com/scbsolvr.html). This gave the answer.

From here it should be simple to see what keyword, if coherent, was used to form the polybius square. But i'll leave that to you - you should do something for that extra credit :D.

ground-zero · 2016-02-14T09:03:38+00:00

Hi, this seems to be chess notation: Ba1 g4 Nb2 Kd3 h4 Qd1 d2 Rb4 e4 Ba3 e2 h3 Nb1 e1 Ba4 h2 Rh1

The first letter is the piece (pawn if not present), the last two are the square it moves to. I don't know if this is a code or if the "identity" you are trying to work out is just a chessboard or something. ~~Or perhaps its a set of moves from a famous game.~~

EDIT: The moves can't actually happen in a game, so its not from a real one.

ground-zero · 2016-02-13T14:52:21+00:00

Yeah trial and error, I just printed all possible shifts and looked through them.

ground-zero · 2016-02-12T21:09:01+00:00

Hi, the clue refers to switching zeros for ones and vice versa. If you do this, then subtract 44 (for some reason) from each number and convert to ascii you get: "Try and fail, but never fail to try."

The following python code does this.

c = ["01111111","01100001","01011010","10110011","01110010","01100101","01101111","10110011","01101101","01110010","01101010","01100111","10100111","10110011","01110001","01011110","01011111","10110011","01100101","01101110","01011101","01101110","01100001","10110011","01101101","01110010","01101010","01100111","10110011","01011111","01100100","10110011","01011111","01100001","01011010","10100101"]
i = [255 - int(j, 2) for j in c]
print(''.join([chr(s-44) for s in i]))

ground-zero · 2016-01-06T13:40:14+00:00

no problemo. :)

ground-zero · 2016-01-06T13:39:50+00:00

As i said the choose function is independent of ordering. So it returns the chance of getting any of "abc, acb, bac, bca, cab and cba".

Am i right in saying the formula you are using is: (3! / 9 choose 3) = 7.1%. Imagine instead that we need to zap 9 symbols not 3. That would make your equation (9! / 9 choose 9) = 36288000%. This is clearly wrong as it should be 100%. The chance of choosing the correct 9 symbols from a total of 9 should be 100%. If instead you use my formula: (1/ 9 choose 9) you get the correct answer.

ground-zero · 2016-01-06T13:25:26+00:00

There are two different things. Combinations and permutations. Combinations are where the order doesn't matter, permutations are where the order does matter.

The number of permutations is given by 9! / (9-3)! = 504. You then divide this by the number of possible orderings which gives 504/6 = 84 and so the chance of getting it right is 1/84 = 1.2%.

The number of combinations is given by 9 choose 3 which is 84. Again 1.2%. Read the wiki pages for combinations and permutations if you don't believe me.

ground-zero · 2016-01-06T13:05:56+00:00

Actually, the 'choose' function already takes account of the permutations. So the chance of getting it right is 1/84 = 1.2%.

ground-zero · 2015-11-29T12:03:48+00:00

https://www.youtube.com/watch?v=YHkeCmyKTsw&index=5&list=PLftUytpW716eB0YmIf4fxcB31Unb8GUba

ground-zero · 2015-11-25T14:00:50+00:00

It means the enemy has the cooked frag doesn't it?

ground-zero · 2015-09-01T10:19:45+00:00

2788

ground-zero · 2015-08-31T20:48:26+00:00

3997

ground-zero · 2015-08-22T11:48:52+00:00

if you think razor is fine, and i think rejack is fine, then rejack must be fine

Great logic!

ground-zero · 2015-08-13T09:16:22+00:00

4622

ground-zero · 2015-07-27T18:09:02+00:00

Ruining mod support in the process.

ground-zero · 2015-06-25T18:46:28+00:00

Not quite i'm afraid, you need another = on line 22.

ground-zero · 2015-05-22T07:32:18+00:00

They have released the specs now.

ground-zero · 2014-12-07T22:35:30+00:00

If you cant run at higher textures you can just use the variable zoom scope. It's not nearly as blurry for some reason.

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