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Nintendo - Hire Me - Challenge

Solution:

First i separated the algorithem in 3 parts.

Part A:

for(u8 j=0;j<32;j++) {
	output[j]=confusion[input[j]];
	input[j]=0;
}

Part B:

for(u8 j=0;j<32;j++)
	for(u8 k=0;k<32;k++)
		input[j]^=output[k]*((diffusion[j]>>k)&1);

Part C:

for(u8 i=0;i<16;i++)
		output[i]=confusion[input[i*2]]^confusion[input[i*2+1]+256];

Part A was the easiest part. This is just a basic list to convert the value. Because the list contains some values multiple times we can determain if reverse Part B will give a valid value.

Part B needs first need some rewriting. Watch the file : /Solution/Rewrite/HireMe.cpp line 125. These xor pattern can be reverse with Gaussian elimination. See /Solution/Part B/

Part C has a lot of different possebilities to check. Basicly in the /Solution/Reverse/main.cpp I bruteforce all possebilities and check if he can reverse Part A and B 256 times. If he pass he found a valid key.

In the end he cracked the challenge in like 15 seconds.

Nintendo - HireMe/Solution/Reverse$ ./main
Crack Part C
Found final input = 0x5d, 0xc5, 0xda, 0x80, 0x81, 0xad, 0xe, 0xe6, 0xda, 0x88, 0xd9, 0x67, 0x43, 0x63, 0x1c, 0x87, 0x8b, 0x72, 0x60, 0x24, 0x4d, 0xbc, 0xd9, 0xa7, 0xfa, 0x86, 0x19, 0x26, 0x1e, 0x69, 0x3, 0x6

Compile and Crack:

cd Solution/Reverse
./compile
./main

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