OK I AM AGREE THAT QUBIT IS EXIST MEAN S HEAVY BIT EXIST THEN QUBIT

ANTIQBITS QBUITS ANTIQUANTOM COMPUTING QUANTIOM COMPUTNG QBIT WILL BE MEASURE THE THINGS NOT THE BIT AN DIN HERE WE COULD SAY LOOK LIKE BINARY BUT NOT BINARY QUBIT AND AND CONSUS 10 OPERSTON MAKE QUBUIT AND ANTI QUBIT IF TUS ON DIFFEREMT DIFFERENT QUANTOM BUT SBY SINGLE IT DPEEN ON BINOMISL OR X-A POW N  


NOW STATE  HAS ONE STATE (0,1) OR (1,0)

SECOND STATE IS (1,0)


ANTI STATE (0,0) AND (1,1)


QUBIT HAS STATE AND CONSDIER TO  RESULT TO GET FROM THESE STATE 

QUBIT STATE (1,0) AND QUBIT STATE(0,1)

ANTIQUBITSTATE(0,0) SECODN ANTIQUBITSTATE (1,1)


FIRST STEP ONLY FOR OR NOT FOR AND 

OR DEPEDNET 

QUBIT STATE (0 OR/AND 1) QUBUIT(0 OR/AND 1)

QUBIT STATE(1,OR/AND 0) AND QUBIT(1 OR/AND 0)

1111

11

ALL STATET FROM QUBIT VALUE SOLVE THE LOGICALS

ANTIQUBIT

QUBIT(0 OR/AND 0) QUBUIT(0 OR/AND 0)

QUBIT(1 OR/AND 1) AND QUBIT(1 OR/AND 1)  0011 TIS ANTUIQUBIT COMES FROMS 

FROM STTATE OF QUBT TO SOLVE WE GET TEH NEXT STATE TA CALLLED QUBIT 


WE GOT ONE Q BIT 0011 JUST LIKE THIS MULIPLTE HIZARDS HAS COME TO SOLVE THE STTA OF QUBIT THAT IS UNKNOWS TO REACHED TO THE TOOTAL QUANTOM AN DIT COMES BY OR AND AND WE DO THI STTE SO QUIANTOPM STATE WE FIND BY SINGLY LOGICALL;Y SOLVE THE HIZARDS VERY SIPLY IF MASS OF QUANTOM HAS GVE AND FREQUENCY IOF TAT THE WHAT KLEVLE OF IT TO SOLVE IT AND TO LEVEL OF THAT WE NEEED TO IOOR NEEED ATA FROM QUANTOM W ENEED TO DIO THIS BWE BRING IN THE BIT IN RECIORD OF QUANTOM HERE IN THRERES WE DO ONW TIG WE GOTO QUANTOM NUMBER AND QUNATOM STATE THAT SOMHERE QUANTOPN WILL BE THIS VALUE SUCH FOR SO BY THTA WE CUDL USE (0011)POWN AND ANY THINGS


0011 IS QUNATOM STTAE ACHIVED BY 0AND 1 THTA W E HAVE TO DO IT WE GOT THE QUANTOOM STATE BECAUSE WE VANISH THE 0 AND 1 STATE WE GOT ANTIQBIT STATE 


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