Minetest
Last updated
Last updated
This was a series of logic circuit challenges in Minetest.
Sidenote: It was a really fun challenge!
A simple logic circuit, control the inputs so that the output is True.
UMDCTF{d...}
XOR the top and bottom lamps.
10110 XOR 01101 = 11011
UMDCTF-{d1g...}
Observe the inputs, and note down which inputs are True for each 'pulse'. The PAUSE
inputs mean that none of the inputs is True.
Translate this to binary:
010 001 100 111 010 101 101 100 011 011 000 100 000 101 100 100
Then, convert the binary to ASCII.
ANSWER: FullAd
UMDCTF-{d1g1...}
Translate this to binary:
001 101 000 111 010 000 110 011 010 000 100 011 000 101 110 100
Then, convert the binary to ASCII.
ANSWER: 4t3b1t
UMDCTF{d1g1L...}
Let the bits be: a b c d e f g h (from MSB to LSB). Analyse the circuit to come up with a boolean expression:
((f XOR h) . ( (a XOR c) . ( ~(e XOR h) . (~a . b) ) )) . (~(g XOR d))
Simplify:
f XOR h . a XOR c . ~(e XOR h) . ~a . b . ~(g XOR d)
Since a=0 and b=1, and a XOR c, then c=1
Since ~(e XOR h) and ~(g XOR d), e = h and g = d.
a, b, c are fixed, so there are only 4 possible combinations (e = h = either 0 or 1, g = d = either 0 or 1, f = ~h)
ANSWER: div
UMDCTF{...3k}
From here we can kinda guess the flag: d1g1L0g3k. It worked!
(I probably didn't get the full solution, but this is good enough I guess...)