Minetest

This was a series of logic circuit challenges in Minetest.

Sidenote: It was a really fun challenge!

Part 1

A simple logic circuit, control the inputs so that the output is True.

UMDCTF{d...}

Part 2

XOR the top and bottom lamps.

10110 XOR 01101 = 11011

UMDCTF-{d1g...}

Part 3

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...}

Part 4

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...}

Part 5

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...)

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