Re: Why is Bitcoin safe against a quantum computer?

As we all know elliptic curve cryptography is vulnerable to a quantum computer. For a conventional computer difficulty of breaking 256-bit key equals 256/2=128 bits. For a quantum computer it's just sqrt(256)=16 bits.
Bitcoin address is a hashed public key of 256-bit EC. Hashes are resistant to quantum algos, so while someone keeps his public key unknown it's OK. But when he wants to transfer his money he must reveal the key.
Let's assume that an attacker with a quantum computer monitors all transactions. The attacker can pick any key while a transaction awaits to be included into a block. Now imagine that miners choose transactions with higher fees. The attacker can issue other transaction (when he picks the private key) that transfer coins to his address and set a higher fee. Or he could switch his mining rig on and try to find a block himself. With 0.1% of all hashpower he needs only 5 days to solve a block with 50% chance.

Seems Bitcoin is NOT safe. Or am I wrong?

very wrong as the power of mathematics used increases the level of the craking cpu rises exponetially meaning that soon the step required from the cpu will be unfathomable

i call this type of cpu that requires the next cpu to be almost infinite in computing power, a number powered cpu

Obviously u are talking about some other Bitcoin implementation. What about the current one?

unless you have studied mathematics for years on end id expect this concept to be a little difficult to graph

write yourself an oct counter then examine this true oct counter

#include
#define P printf
#define F for

int main(void){

int dec,oct=0,x;

F(dec=1;dec<=1000000;dec++){

oct++;

F(x=10; (oct/(x/10))>0; x*=10){

if((oct%x)==(8*(x/10))){

oct+=(2*(x/10)); } }

P("%5d%5d\n",dec,oct);
}
return 0;
}