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MechCube: Escape – Walkthrough with All Endings

Posted on January 3, 2020

MechCube is not only a set of extraordinary puzzles and tricky riddles. This is a non-linear game in point-and-click adventure genre with a lot of endings, secrets, easter eggs and references to modern pop culture. Here is a guide to go through the game, you can find all complete endings with this guide.

 

Mech Cube: Escape walkthrough Room 1-11

 

MechCube: Escape 10 Items & 5 Endings

Features of the MechCube: Escape

👁 First of all, it’s a set of peculiar, extraordinary and sometimes incredibly tricky riddles and puzzles. To solve them, you need not only logic and attention to details, but also the ability to think outside the box

👁 This is a non-linear point-and-click adventure. You can go in any direction and solve puzzles in any order

👁 Terrible traps and exciting feeling of meeting with the unknown

👁 Try to find all the endings and collectibles hidden in this game

👁 Secret levels, easter eggs and an lots of references to modern pop culture

👁 Stuck? No problem! A built-in hint system will help in a difficult situation

👁 Awesome atmosphere created by hand-drawn graphics and a deep soundtrack

👁 No casual match-3 puzzles, just logical problems and thinking outside the box

👁 No time limits, you can play offline

👁 Amazing graphics optimized for tablets and smartphones

👁 Boost your brain power in this colorful adventure full of secrets and mysteries!

👁 What you’re waiting for? Start solving puzzles right now!

1 thought on “MechCube: Escape – Walkthrough with All Endings”

  1. Maximiliano Antonelli says:
    November 12, 2020 at pm12:35

    %room14MatlabSolver
    % 0 = R, 1 = G, 2 = B.

    A = [1 1 0 1 0 0 0 0 0; …
    1 1 1 0 1 0 0 0 0; …
    0 1 1 0 0 1 0 0 0; …
    1 0 0 1 1 0 1 0 0; …
    0 1 0 1 1 1 0 1 0; …
    0 0 1 0 1 1 0 0 1; …
    0 0 0 1 0 0 1 1 0; …
    0 0 0 0 1 0 1 1 1; …
    0 0 0 0 0 1 0 1 1];

    P = [0; 0; 0; 1; 1; 1; 2; 2; 2];

    %C is the initial condition
    C = [0; 2; 1; 1; 0; 2; 2; 0; 1];
    %is
    % R B G
    % G R B
    % B R G

    %Inicializing N
    N = rand(9,1);

    while sum(mod(N,1)) > 0
    K = randi([0 2],9,1);
    N = (A*K + C – P)/3;
    disp(num2str(K’))
    end

    %Result for this initial condition is
    %k = 0 1 1 2 2 0 2 2 2
    %It meens that we need to taouch
    % 0 1 1
    % 2 2 0
    % 2 2 2

    Reply

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