It seems to me that JEV is RETE with more generic pattern matching.

Comments URL: https://news.ycombinator.com/item?id=49786485

Points: 1

# Comments: 2

It seems like this should be possible. You would need to ask questions about what the person brought up in a sentence, e.g. "My MOM calls all the time".

Then you write JEV "rules" to construct a reply based on various probabilities.

Comments URL: https://news.ycombinator.com/item?id=49786471

Points: 1

# Comments: 0

I want to play with the sensitivity. Do you have a circuit diagram for measuring the Hall effect?

Comments URL: https://news.ycombinator.com/item?id=49585473

Points: 1

# Comments: 1

I own about that many. Fortunately a large percentage is PDFs. Never the less I have about 200 linear feet of shelf space to hold my books in my "library" and in my home office.

I developed open source software for 25 years even during the stretch when I was living in my car. I never made money from open source.

So tell me why you think this doesn't work.

The best way to cut through this noise is to visit a Tesla dealer and ask for a test drive.

From this book it is clear that the gas car industry is doomed. Car companies cannot support the change to electric.

From this book it is clear that the energy industry is doomed. Energy from hydrocarbons cannot compete with wind and solar.

It is a worthwhile and enlightening read.

Comments URL: https://news.ycombinator.com/item?id=48107212

Points: 2

# Comments: 2

My professional advice is "Don't learn to program".

Your question is like asking... should I learn to shoe horses?

I bought Starlink Mini in January. I installed it in my Tesla.

I just got a notification that I exceeded the Starlink Mini speed limit.

Apparently FSD "hurry mode" goes too fast for Starlink.That's annoying.

Comments URL: https://news.ycombinator.com/item?id=47297411

Points: 1

# Comments: 0

E1 [Find remained] Divide m by n and r be the remainder. (We will know that 0 <= r < n)

E2 [Is it zero?] If r = 0, the algorithm terminates, n is the answer

E3 [Reduce] Set m = n, then n = r, then go back to step E1 -----------------------------------------------------------------------

[3] Change Algorithm E so that all trivial replacement operations such as m = n are avoided. Write this new algorithm in the style of Algorithm E and call it Algorithm F.

======================================================================== Claude Sonnet 4.6 ========================================================================

Algorithm F

Given two positive integers m and n, find their greatest common divisor.

F1 [Find remainder] Divide m by n and let r be the remainder. (We will know that 0 ≤ r < n.)

F2 [Is it zero?] If r = 0, the algorithm terminates and n is the answer.

F3 [Find remainder] Divide n by r and let m be the remainder. (We will know that 0 ≤ m < r.)

F4 [Is it zero?] If m = 0, the algorithm terminates and r is the answer.

F5 [Find remainder] Divide r by m and let n be the remainder. (We will know that 0 ≤ n < m.) Go back to F2.

Explanation

The key insight is that instead of shuffling values with temporary replacements like:

m = n, n = r

we reuse the existing variables (m, n, r) in rotation, dividing and storing each new remainder directly into the variable that is no longer needed:

Step Divides Remainder stored in F1...m by n...r F3...n by r...m F5...r by m...n

Each variable naturally becomes the "oldest" value at each stage and is safely overwritten. This eliminates all trivial assignments while preserving the logic of Euclid's algorithm. □

Comments URL: https://news.ycombinator.com/item?id=47271921

Points: 3

# Comments: 0

Knuth Art Of Computer Programming V1 p9 ----------------------------------------------------------------------- Algorithm E (Euclid's algorithm). Given two positive integers m and n, find their greatest common divisor, that is, the largest positive integer that evenly divides both m and n.

E1 [Find remained] Divide m by n and r be the remainder. (We will know that 0 <= r < n)

E2 [Is it zero?] If r = 0, the algorithm terminates, n is the answer

E3 [Reduce] Set m = n, then n = r, then go back to step E1 -----------------------------------------------------------------------

Prove that m is always greater than n at the beginning of step E1, except possibly the first time this step occurs.

======================================================================== Claude Sonnet 4.6 ========================================================================

Proof

We want to show that m > n at the start of E1, for every iteration except possibly the first.

First Execution of E1

When E1 is executed the first time, m and n are simply whatever positive integers were given as input. There is no guarantee that m > n, which is why the exception exists. For example, with m = 2166 and n = 6099, we have m < n on the first call.

All Subsequent Executions of E1

After the first execution of E1, we reach E3 only if r ≠ 0 (checked at E2). At E3, the replacements are:

m = n n = r

Then we return to E1. So at the start of the next E1, the new values are:

m_new = n_old and n_new = r_old

From E1, we know r is the remainder when m is divided by n, so by definition:

0 ≤ r < n

Since E3 is only reached when r ≠ 0 (E2 would have terminated otherwise), we tighten this to:

0 < r < n

Therefore:

n_new = r_old < n_old = m_new

Which gives us:

m_new > n_new

Conclusion

After the first iteration, the new m is always the old n, and the new n is always the remainder r, which is strictly less than the old n. Therefore m > n holds at the beginning of every execution of E1 except possibly the first, which is what we wanted to prove. □

Comments URL: https://news.ycombinator.com/item?id=47271831

Points: 1

# Comments: 0

[1] The values of variables m and n, can be interchanged using t = m, then m = n, then n = t. Show how the values of four variables (a,b,c,d) can be rearranged to (b,c,d,a) by a sequence of replacements. In other words, the new value of a is to be the original value of b, etc. Try to use the minimum number of replacements.

======================================================================== Claude Sonnet 4.6 ========================================================================

We want to transform (a, b, c, d) → (b, c, d, a) Using a temporary variable t, here is the sequence:

t = a a = b b = c c = d d = t

Trace through to verify:

Step.: a b c d t Start: a b c d — t = a: a b c d a a = b: b b c d a b = c: b c c d a c = d: b c d d a d = t: b c d a a

Result: (a, b, c, d) → (b, c, d, a)

This uses 5 replacements, which is the minimum needed — one to save the displaced value of a into t, three to shift b→a, c→b, d→c, and one final step to place the saved value into d.

Comments URL: https://news.ycombinator.com/item?id=47271761

Points: 1

# Comments: 0

E1 [Find remained] Divide m by n and r be the remainder. (We will know that 0 <= r < n)

E2 [Is it zero?] If r = 0, the algorithm terminates, n is the answer

E3 [Reduce] Set m = n, then n = r, then go back to step E1

----------------------------------------------------------------------- Knuth Art Of Computer Programming V1 p9 Problem 1.1.4

[4] Using Algorithm E compute the greatest common divisor of 2166 and 6099. Label each step of the algorithm as it is used.

======================================================================== Claude Sonnet 4.6 ========================================================================

Starting with m = 2166, n = 6099: E1: 2166 ÷ 6099 = 0 remainder 2166 → r = 2166 E2: r ≠ 0, continue E3: m = 6099, n = 2166

E1: 6099 ÷ 2166 = 2 remainder 1767 → r = 1767 E2: r ≠ 0, continue E3: m = 2166, n = 1767

E1: 2166 ÷ 1767 = 1 remainder 399 → r = 399 E2: r ≠ 0, continue E3: m = 1767, n = 399

E1: 1767 ÷ 399 = 4 remainder 171 → r = 171 E2: r ≠ 0, continue E3: m = 399, n = 171

E1: 399 ÷ 171 = 2 remainder 57 → r = 57 E2: r ≠ 0, continue E3: m = 171, n = 57

E1: 171 ÷ 57 = 3 remainder 0 → r = 0 E2: r = 0, algorithm terminates GCD(2166, 6099) = 57

Comments URL: https://news.ycombinator.com/item?id=47271665

Points: 1

# Comments: 0

Comments URL: https://news.ycombinator.com/item?id=47257755

Points: 1

# Comments: 0