Euler Conjecture and CDC 6600 (opens in new tab)

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40 pointszaikunzhang3mo ago7 comments

7 comments

jmclnx3mo ago

FWIW on my W541, Slackware I get

i^5 = 61917364224 j^5 + k^5 + l^5 + n^5 = 61917364224 27 84 110 133 144

Slackware W541:

real 1m6.703s

user 1m6.658s

sys 0m0.002s

Output is the same, which is aloways good. But times are slower because I am on a ~10 year old system :)

So I gave it a whirl on sdf.org's Debian system Debian 6.1.140-1 (2025-05-22) x86_64 GNU/Linux kenel 6.1.0-37-amd64

real 1m9.615s

user 1m9.595s

sys 0m0.016s

That system is a true multi user system and had 54 users logged in when I ran it. So I think it is better than I expected.

FWIW, I believe my first paid job was on a 6600 while in college, but it could have been a 7600. There were upgrading the system from the 6600 when I was there.

nroets3mo ago

Here's a fairly fast algorithm: Look for a number that appears in both { j⁵+k⁵+l⁵ } and { i⁵-n⁵ } where 0 < j <= k <= l and 0 < n < i

If we only consider l and i under 250, then the sets would contain less than 3 million integers each.

Strength reduction can be used to replace all the multiplications with additions: https://en.wikipedia.org/wiki/Strength_reduction

asplake3mo ago

Re precomputing fifth powers, seems Fortran not only has array comprehensions, but compile-time array comprehensions. It was never exactly my cup of tea, but nearly 40 years out of university it seems Fortran has kinda kept up!

nurettin3mo ago

I like the bald guy in the comments idea golfing and posting his own numbers.

Someone3mo ago

For a limit of 10,000⁵. Reading https://www.ams.org/journals/mcom/1967-21-097/S0025-5718-196... the original search seems to have gone up to 250⁵. That’s a search space that’s 40⁵ ≈ 10⁸ larger.

They also forget to break out of the loops when the sum of, say, the first three fifth powers already is larger than 10,000⁵.

nroets3mo ago

It's likely that the original search also used strength reduction to save a lot of cycles (effectively replacing all multiplications with additions):

https://en.wikipedia.org/wiki/Strength_reduction

zaikunzhangOP3mo ago

7 comments

jmclnx3mo ago

FWIW on my W541, Slackware I get

i^5 = 61917364224 j^5 + k^5 + l^5 + n^5 = 61917364224 27 84 110 133 144

Slackware W541:

real 1m6.703s

user 1m6.658s

sys 0m0.002s

Output is the same, which is aloways good. But times are slower because I am on a ~10 year old system :)

So I gave it a whirl on sdf.org's Debian system Debian 6.1.140-1 (2025-05-22) x86_64 GNU/Linux kenel 6.1.0-37-amd64

real 1m9.615s

user 1m9.595s

sys 0m0.016s

That system is a true multi user system and had 54 users logged in when I ran it. So I think it is better than I expected.

FWIW, I believe my first paid job was on a 6600 while in college, but it could have been a 7600. There were upgrading the system from the 6600 when I was there.

nroets3mo ago

Here's a fairly fast algorithm: Look for a number that appears in both { j⁵+k⁵+l⁵ } and { i⁵-n⁵ } where 0 < j <= k <= l and 0 < n < i

If we only consider l and i under 250, then the sets would contain less than 3 million integers each.

Strength reduction can be used to replace all the multiplications with additions: https://en.wikipedia.org/wiki/Strength_reduction

asplake3mo ago

nurettin3mo ago

I like the bald guy in the comments idea golfing and posting his own numbers.

Someone3mo ago

They also forget to break out of the loops when the sum of, say, the first three fifth powers already is larger than 10,000⁵.

nroets3mo ago

It's likely that the original search also used strength reduction to save a lot of cycles (effectively replacing all multiplications with additions):

https://en.wikipedia.org/wiki/Strength_reduction

zaikunzhangOP3mo ago