Count_2__Wildness

Count 2: Wildness

Programming

Challenge

DESCRIPTION
My friend said that my code is still too long.
Can you help me again?

count2-wildness.cddc19q.ctf.sg 12221

* gcc version 7.4.0 (Ubuntu 7.4.0-1ubuntu1~18.04)
* This challenge is not a pwn challenge.

MD5("count2-wildness.zip"): AACC124AAF1DDB6D1F455AFBA6B898ED

ATTACHED FILES
count2-wildness.zip

Solution

Same as before, we need to do codegolf until 41 chars.

Previously we had this of 45 chars

main(i){for(i=1;i<10000;i++)printf("%d,",i);}

---

If we look at stackoverflow, we get some tips

https://codegolf.stackexchange.com/questions/2203/tips-for-golfing-in-c/2230#2230

Abuse main's argument list to declare one or more integer variables:
main(a){for(;++a<28;)putchar(95+a);}

And we have this important piece of info.

This solution also abuses the fact that a (a.k.a. argc) starts out as 1, provided the program is called with no arguments.

Since the argument starts at 1 by default, we can omit the initialiser.

We can trim some chars by putting the increment together with the comparison statement

main(i){for(;++i<10000;)printf("%d,",i);}

Submit to the server

Your code : main(i){for(;++i<10000;)printf("%d,",i);}
Sorry, my friend doesn't like to go wild! -> '<'

Oops, this time, we have additional characters we can't use

# wild check
wild = '<== w1lD ==>'
for ch in code:
    if ch in wild:
        print "\nSorry, my friend doesn't like to go wild! -> '{}'".format(ch)
        sys.exit(-1)

We have to change the less than < sign.

• Since i is incrementing, we can check if i == 10000 and stop there, but == is also forbidden.
• We can check if 2 numbers are equal to each other by checking its ANDed value

So we have the following which prints from 1 through 9999 inclusive, where i initialises with 0.

main(i){for(;++i^10000;)printf("%d,",i);}

We are almost there... But we can't use the number 1

Your code : main(i){for(;++i^10000;)printf("%d,",i);}
Sorry, my friend doesn't like to go wild! -> '1'

In order to overcome that, we need to replace i < 10000 with i <= 9999.

---

Looking at how other people implement their code golf again, there is a minor statement I missed

https://codegolf.stackexchange.com/a/2230

Use global variables to initialize things to zero:

t[52],i;main(c){for(;i<52;)(c=getchar())<11?i+=26:t[i+c-97]++;
for(i=27;--i&&t[i-1]==t[i+25];);puts(i?"false":"true");}

So by initialising global variable, we can start with 0 instead.

Here's a summary of my changes

[a]  main(i){for(;++i^10000;)printf("%d,",i);} // 1 to 9999 inclusive
[b]  main(i){for(;++i^9999;)printf("%d,",i);}  // 1 to 9998 inclusive
[c]  i;main(){for(;++i^9999;)printf("%d,",i);} // 0 to 9998 inclusive
[d]  i;main(){for(;i++^9999;)printf("%d,",i);} // 1 to 9999 inclusive

Final:  i;main(){for(;i++^9999;)printf("%d,",i);}

The solution is 41 chars and we get the flag.

$ nc count2-wildness.cddc19q.ctf.sg 12221

[Count 2: Wildness]

My friend said that my code is still too long.
Can you help me again?

This is my code:
----------------------------------------|
#include <stdio.h>

int main(int argc, char *argv[])
{
	int i;

	for( i = 1 ; i < 10000 ; i++ )
	{
		printf("%d,", i);
	}

	return 0;
}
----------------------------------------/
    
Your code : i;main(){for(;i++^9999;)printf("%d,",i);}
Flag : $CDDC19${This_really_helps_m3_a_lot!}

Flag

$CDDC19${This_really_helps_m3_a_lot!}