Help With C++ Code to break COD Zombies Game Cipher

[u/Jump_Not_Zero]

I am working on a cipher from call of duty black ops 3 and I have a good lead but was wondering if I can write a program that would speed up decrypting the cipher to the next stage.

The cipher text is base64 and displayed as a game texture file visible in game and has the keyword "TheGiant".

I wrote a program to analyse the base64 and realise you don't need many characters if you modulo subtract as a key.

In fact you only need ten so I used what is assumed to be the keyword and added 1 and 0 base64 character values.

G,T,a,e,h,i,n,t,0,1

6,19,26,30,33,34,39,45,52,53 are the base64 decimal values I am subtracting modulo 64.

The ciphertext seems to be running key and the key starts with some german words and recording elements weights in binary.

This is the first forty characters of ciphertext :

kCmlgFi6GUJNgkNI1Q41fbfyLoCFTCvIqkZiI0KI

This is a possible key for forty characters from the start of the ciphertext :

GeheGehtte00Ge01hantGnaati101te0at0h0th0

This is the output :

zAGjpMgeYhfSSdHgAXxcQyTTCyAPzT

Breakdown of the key

Gehe Geht te00 Ge01 hant Gnaa ti101 te0 at0 te0 at0 h0 th0

What I translate this to is :

German word for Go (imperative) Tellurium '00' Germanium '01' German word for handle German word for Argh Titanium '101' Tellurium'0' Astatine '0' Hydrogen '0' thorium '0'

Now there is a choice of elements like ti and te and binary values this is why I said it's a possible key also the German word hant could be hang which translates to slope.

I think the developers point to the difficulty with using difficult to visually define characters like '1','I' and '0' ,'O' in the texture (Which is solved and where I got the idea to add 1 and 0).

Also the starting German word is a nightmare here is a breakdown.

Gehe and Geht can be joined and google translates it to go and goes but if you join them so there eight letter you get a different meaning.

Gehe Geht == Go (imperative)

Gehe geht == Go Go

GeheGeht == Go/Goes

Gehegeht == Enclosure

What I didn't realise was it was going to decrypt to more base64!

So I have not accounted for the characters '/' and '+' as ascii output in the code I used to test positions and write a report on them.

I will list the c++ code at the bottom of the post here.

What I would like to discuss is do you think this is viable for a partial decrypt and how to break the cipher further.

V sbyybjrq gur ehyrf I have read the rules

// Machine_00.cpp : This file contains the 'main' function. Program execution begins and ends there.
//

#include <array>
#include <bitset>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>


static void testgadget(std::array<std::bitset<6>, 192>& arr_CT, std::vector<unsigned> &vPos) {

    unsigned pos{};

    for (const auto& vx : arr_CT) {


        if (pos % 4 == 0 && vx.test(5) == false && vx.test(4) ) {

            vPos.push_back(pos);

            std::cout << "Q0" << '\t' << pos << '\t' << vx.to_string() << '\n'; }

        if (pos % 4 == 1 && vx.test(3) == false && vx.test(2) == true ) {

            vPos.push_back(pos);

            std::cout << "Q1" << '\t' << pos << '\t' << vx.to_string() << '\n'; }

        if (pos % 4 == 2 && vx.test(1) == false && vx.test(0) == true) {

            vPos.push_back(pos);

            std::cout << "Q2" << '\t' << pos << '\t' << vx.to_string() << '\n'; }



        if (pos % 4 == 3 && vx.test(5) == false) { 

            vPos.push_back(pos);

            std::cout << "Q3" << '\t' << pos << '\t' << vx.to_string() << '\n'; }

        pos++;

    }


}




int main()

{

    std::vector<unsigned> vPos{};

    std::string s_alp = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";

    std::string s_In = "kCmlgFi6GUJNgkNI1Q41fbfyLoCFTCvIqkZiI0KIAXAzP1U1uy1BE4UfPBfpKmmLObjYnQNRBaPtKiVWzc5A4v0w3xle8FOhAGJZ7g4in0wndJxMOvO3dc1M82at2T6935roTqyWDgtGD/hwwRF3oHqFM5Vcw1JtINbsgWRm4o4/quEDkZ7x1B275bX3/Fo1";

    std::array<std::pair<char, std::bitset<6>>, 64> arr_p{};

    for (unsigned char i{};i < arr_p.size();i++) {

        arr_p[i] = std::make_pair(s_alp[i], i);

    }

    std::array<std::bitset<6>, 192> arr_CT{};



    std::array<unsigned, 10> arrs { 6,19,26,30,33,34,39,45,52,53 }; //{ 19,33,30,6,26,34,39,45 }; //{ 45,7,4,32,0,8,13,19 }; // {0,6,8,19,26,30,33,34,39,45};  //{ 0, 2, 6, 19, 20, 26, 28, 32, 45, 46 };

/*    unsigned  c{};

    for (auto &ax_C : arrs) {

        arrs[c] = c;

        c++;
    }
*/
    unsigned i_Count{};

    for (const auto& sx : s_In) {



        for (const auto& ax : arr_p) {

            if (ax.first == sx) {

                arr_CT[i_Count] = ax.second;
            }


        }

        i_Count++;

    }

    std::array<std::bitset<6>, 192> arr_Cc{};

    arr_Cc = arr_CT;

    for (auto ax_s : arrs) {

        std::cout << '\n' << '\n' << ax_s << '\t' << " OFFSET" << '\n' << '\n';



        for ( auto& ax : arr_Cc) {


            auto utemp = ax.to_ulong();

            utemp -= ax_s;

            utemp %=64;



            ax =  utemp ;


        }

        std::string s_Out = "";

        for (auto& ax : arr_Cc) {

            for (auto ax_00 : arr_p) {

                if (ax_00.second == ax) {

                    s_Out.push_back(ax_00.first);

                }

            }


        }



        std::cout << '\n' << s_Out << '\n' << '\n';

        testgadget(arr_Cc,vPos);


        arr_Cc = arr_CT;

    }

    std::cout << '\n' << '\n';



    std::sort(vPos.begin(), vPos.end());

    auto last = std::unique(vPos.begin(), vPos.end());


    vPos.erase(last, vPos.end());

    std::cout << vPos.size() << '\n' << '\n';


    for (auto vx : vPos) {


        std::cout << vx << '\n';

    }



}

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[u/Jump_Not_Zero]

Ciphertext

kCmlgFi6GUJNgkNI1Q41

Shiftkey

GehtGeht0e00Th0Giant

Output

zAxjpMKeY46BOdH

Basicly my theory is there is only ten shift values and the key are german words

The ten shift values are the key "TheGiant" and 1 and 0

All words start capitlised and I am guessing that the '0' and '1' signal wildcards.

The theory is 0 is smallcase wildcard and 1 is uppercase wildcard.

[u/Jump_Not_Zero → u/Jump_Not_Zero]

Sticking to the scheme I have decrypted this

Ciphertext

kCmlgFi6GUJNgkNI1Q41fbfyLoCFTCvIqkZiI0KI

Shift values

GehtGeht1e00Th0GiantG0ei1it01ea0et1t0ht1

German keystream

GehtGehtGehtTheGiantGleiSitzBeametSteht1

German words

Glei Sitz Beamet Steht

Right-away Seat Projecter Stand

If you change the last letter 't' in beamet google translate outputs this.

Glei sitz beamer steht

The projector is set up right here.

Not sure but think I am on the right track not sure why I got the downvotes as I think it is a pretty solid theory.