A superconducting chip developed at IBM demonstrates an important step needed for the creation of computer processors that crunch numbers by exploiting the weirdness of quantum physics. If successfully developed, quantum computers could effectively take shortcuts through many calculations that are difficult for today's computers.
IBM's new chip is the first to integrate the basic devices needed to build a quantum computer, known as qubits, into a 2-D grid. Researchers think one of the best routes to making a practical quantum computer would involve creating grids of hundreds or thousands of qubits working together. The circuits of IBM's chip are made from metals that become superconducting when cooled to extremely low temperatures. The chip operates at only a fraction of a degree above absolute zero.
IBM's chip contains only the simplest grid possible, four qubits in a two-by-two array. But previously researchers had only shown they could operate qubits together when arranged in a line. Unlike conventional binary bits, a qubit can enter a "superposition state" where it is effectively both 0 and 1 at the same time. When qubits in this state work together, they can cut through complex calculations in ways impossible for conventional hardware. Google, NASA, Microsoft, IBM, and the U.S. government are all working on the technology.
Nature Communications: Demonstration of a quantum error detection code using a square lattice of four superconducting qubits
(Score: 0) by Anonymous Coward on Friday May 01 2015, @03:30PM
It is more like massively parallel. You can try all the combinations to the lock instantly...
If anyone suggests they've found a shortcut in math they are usually full of shit.
(Score: 3, Informative) by takyon on Friday May 01 2015, @03:56PM
https://en.wikipedia.org/wiki/Quantum_computing [wikipedia.org]
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(Score: 2) by mtrycz on Friday May 01 2015, @04:54PM
Thanks.
One thing I can't decipher is "polynomial speedup". Does this mean that O(2^n) becomes something like O(n^x) for some x, or does that mean that O(n^x) becomes O(n^(x-c)), hence possibly O(n^2) -> O(n).
Or something else entirely?
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(Score: 2) by takyon on Friday May 01 2015, @05:07PM
I think it's like this: https://jsfiddle.net/g4q08m4a/ [jsfiddle.net]
Exponentials will outpace any polynomials eventually.
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(Score: 2) by takyon on Friday May 01 2015, @05:14PM
More readable: https://jsfiddle.net/g4q08m4a/1/ [jsfiddle.net]
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(Score: 2) by maxwell demon on Friday May 01 2015, @05:25PM
The first one is more than polynomial speedup (e.g. seen for Shor's algorithm), the second one is polynomial speedup (e.g. seen for Grover search).
In the case of Grover search, it's in particular O(n) -> O(n1/2).
The Tao of math: The numbers you can count are not the real numbers.
(Score: 1) by dak664 on Friday May 01 2015, @06:32PM
Well that's the idea. But it seems to me the discussion of computation time is lacking; adjacent qubits have to superimpose over some time scale before settling into the answer, and that answer is statistical to boot.
An excited nucleus is basically a qubit(?) and it can take years or centuries to beta decay when coupling to the continuum.