Ryan O'Donnell
@booleananalysis.bsky.social
I do not, I'm afraid. I guess it's space efficiency vs time efficiency vs pedagogy; perhaps an expert can chip in...
October 28, 2025 at 4:57 PM
I do not, I'm afraid. I guess it's space efficiency vs time efficiency vs pedagogy; perhaps an expert can chip in...
I do phase estimation without QFT in my undergrad course. youtu.be/CMqPutlG59c?...
It's just Hadamard test plus binary search.
It's just Hadamard test plus binary search.
#60/100: 1-qubit Rotation Estimation: Overview || Quantum Computer Programming in 100 Easy Lessons
YouTube video by Ryan O'Donnell
youtu.be
October 28, 2025 at 7:13 AM
I do phase estimation without QFT in my undergrad course. youtu.be/CMqPutlG59c?...
It's just Hadamard test plus binary search.
It's just Hadamard test plus binary search.
This paper looks incredible, by the way.
October 11, 2025 at 1:19 PM
This paper looks incredible, by the way.
J-Live's Braggin' Writes has my top 'thesis' verse, though. ("I displays my credentials over instrumentals...")
May 20, 2025 at 2:42 AM
J-Live's Braggin' Writes has my top 'thesis' verse, though. ("I displays my credentials over instrumentals...")
This!
I like to say,
"Let p|A denote distribution p conditioned on event A.
Imagine a world where the laws of probability are the same, except (p|A)|B need not equal (p|B)|A.
Except you don't have to imagine, because it's literally our world!
Now explore probabilistic algorithms in this world."
I like to say,
"Let p|A denote distribution p conditioned on event A.
Imagine a world where the laws of probability are the same, except (p|A)|B need not equal (p|B)|A.
Except you don't have to imagine, because it's literally our world!
Now explore probabilistic algorithms in this world."
May 17, 2025 at 11:38 AM
This!
I like to say,
"Let p|A denote distribution p conditioned on event A.
Imagine a world where the laws of probability are the same, except (p|A)|B need not equal (p|B)|A.
Except you don't have to imagine, because it's literally our world!
Now explore probabilistic algorithms in this world."
I like to say,
"Let p|A denote distribution p conditioned on event A.
Imagine a world where the laws of probability are the same, except (p|A)|B need not equal (p|B)|A.
Except you don't have to imagine, because it's literally our world!
Now explore probabilistic algorithms in this world."
There's a reason I called it Quantum Computer Programming in 100 Easy Lessons.
April 4, 2025 at 11:13 AM
There's a reason I called it Quantum Computer Programming in 100 Easy Lessons.
Meming aside, this looks like a cool paper...
April 4, 2025 at 11:11 AM
Meming aside, this looks like a cool paper...
I did not know this!
March 14, 2025 at 9:57 PM
I did not know this!
Also Aayush Jain!
March 10, 2025 at 9:43 PM
Also Aayush Jain!
[...] "x1 += x2" and "x1 += x3". Similarly for "x2 -= x4".
Finally, we posit: Doing "x1 += x2" 256 times in a row is equivalent to doing nothing
(maybe shoulda called it "x1 += x2 mod 256")
and sim. for doing "x2 += x3", "x3 += x4" 256x in a row.
Can you prove "x1 -= x3" commutes with "x2 -= x4"?
Finally, we posit: Doing "x1 += x2" 256 times in a row is equivalent to doing nothing
(maybe shoulda called it "x1 += x2 mod 256")
and sim. for doing "x2 += x3", "x3 += x4" 256x in a row.
Can you prove "x1 -= x3" commutes with "x2 -= x4"?
February 10, 2025 at 9:42 PM
[...] "x1 += x2" and "x1 += x3". Similarly for "x2 -= x4".
Finally, we posit: Doing "x1 += x2" 256 times in a row is equivalent to doing nothing
(maybe shoulda called it "x1 += x2 mod 256")
and sim. for doing "x2 += x3", "x3 += x4" 256x in a row.
Can you prove "x1 -= x3" commutes with "x2 -= x4"?
Finally, we posit: Doing "x1 += x2" 256 times in a row is equivalent to doing nothing
(maybe shoulda called it "x1 += x2 mod 256")
and sim. for doing "x2 += x3", "x3 += x4" 256x in a row.
Can you prove "x1 -= x3" commutes with "x2 -= x4"?