Palindromic Numbers Consisting of the Product of Two 3-digit Numbers

[ben64]

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@jcarmody wrote:

I'm in the high 30's, now...

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That's the best I was able to get. 3 ms is sure hard to beat 🙂

Ben64

[jcarmody]

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@jcarmody wrote:

I'm in the high 30's, now...

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I cut it in half. 🙂  I wonder if the string conversion is the bottleneck?

[altenbach]

I was using string conversion too. The rest of the code just needs to be a little smarter. 🙂

[jcarmody]

I made it. 

[altenbach]

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@jcarmody wrote:

I made it. 

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I guess you should go back to the drawing board. On a more modern computer mine is well below 0.8ms (now on a 3.1GHz CPU, most likley even faster on a current generation CPU). Only using one core.

Remember, the 3ms was on an ancient laptop. 😄

[jcarmody]

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altenbach wrote:

I guess you should go back to the drawing board. On a more modern computer mine is well below 0.8ms (now on a 3.1GHz CPU, most likley even faster on a current generation CPU). Only using one core.

Remember, the 3ms was on an ancient laptop. 😄

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I was happy to do better than brute-force.

[Oligarlicky]

I'm hitting 0.22 ms (single core) by observing that all 6 digit palindromes are multiples of 11, by not using string compare, not bothing to do a palindrome check for any products smaller than my currently largest product, and by starting from 999 and working backwards.

[altenbach]

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@Oligarlicky wrote:

I'm hitting 0.22 ms (single core) by observing that all 6 digit palindromes are multiples of 11,....

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Ah, did not know that. OK, if I include that test, mine drops to 0.12ms (Still using string compare, though)  😄

[jcarmody]

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@Oligarlicky wrote:

[...] by observing that all 6 digit palindromes are multiples of 11 [...]

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That's what I was waiting for.

[Oligarlicky]

I'm noticing some weird stuff depending on how I benchmark. If I stick it all in a for loop I can hit ~2 ns/ itteration. To be fair, I'm only doing 256 palindrome checks in my search (everything else gets filtered out).