(EDIT: Added more specific numbers.)
So, let me talk about some math I did while I disappeared from this site for a while. Or, um, attempted to do. You'll see.
Now, as I think I've said before, I've basically closed down my whole ordinal-operations project, seeing as I've gone as far as I think can go with it. While I haven't had a lot of time to get math done recently, my intent was to finally return to integer complexity and addition chains.
...but, one last WPOs problem popped up to distract me. The problem was this: Given ordinals α and γ, what's the largest possible value for o(γ×X), where o(X)=α? (Assuming there is a largest one, anyway; if not, that would be quite notable itself!)
I'll skip discussing for now why I wanted an answer to this particular, odd-looking problem; there were several reasons but I don't really want to take the time to go into them. Point was, I wanted an answer to this question, and it seemed doable.
Now I'd already attempted this twice before and given up -- basically because it was too tedious. Why was it so tedious? Well, so, I thought I actually had most of an answer already, actually! I thought I had a recursion that would yield the correct answer. I hadn't proven it would, but I'd proven it was an upper bound, and I figured I'd save proving that it was also a lower bound for later, as that would likely be tricky. First, I wanted to solve the recursion. Because after all that's what I wanted to do -- I wasn't going to be satisfied with a recursion for the answer, I wanted a more explicit finite rule! That's what's I do!
How do you solve a complicated transfinite recursion? You grind out values -- by hand -- until you see the pattern, that's how. I'd done this before, tedious as it could get, and I could do it again. But it had never been quite this bad.
Consider the previous times I'd done this. When I was computing "right multichoose", I had one transfinite variable (α) and one finite variable (k). This meant I could do it one value of k at a time and get a handle on things. While when I was doing this for D(α1×...×αm), there were multiple transfinite variables, but they were all symmetric with each other, and the resulting answer wasn't that complicated (in the recursive case, anyway, which is what we're concerned with here), so it didn't take a lot of grinding out hand computations to figure it out.
Here, we had two variables, both transfinite, with no symmetry between them, and a nasty, complicated answer. What could I do? This is why I gave up the first two times -- it's way harder to make a transfinite-by-transfinite table, than to merely make a transfinite-by-infinite table! And staring at what tables I did have, the whole thing just looked nasty, with some regularity to it but also full of exceptions.
But for this third attempt, I had a plan: I'd take it one value of γ at a time, going row-by-row, similarly to how I worked with "right multichoose" earlier. (I later transposed the table so it was column-by-column.) Since each row was transfinite rather than merely infinite, this meant every individual row would itself take substantial work to solve, but, well... it worked. The plan of attack worked and before long I was making substantial progress, progressively figuring out the answer for higher and higher values of γ.
Occasionally I'd find I made a mistake and had to go back and redo things. This can easily happen with this sort of thing unfortunately -- you think you've found the pattern when you haven't yet. But I found the mistakes and I fixed them, however much apparent progress it undid.
So I spent weeks grinding out these hand computations. And then, on a flight to Seattle for work, while grinding out a few more, I noticed an answer that looked, just... too big. The parameters here were γ=ωω+12, α=ωω2. The answer I got was ωωω³. Just way too large, right?
I must have made a mistake, I figured. But being on the plane, well, it was a bit hard to go back and recheck all my previous work right there! So I figured, OK, first I can at least check that I have made a mistake -- I can check it against the naïve bound, 2γ×α. If it's larger, clearly I messed up. And indeed, it was larger! The naïve bound was a mere ωωω3+12. So I'd messed up somewhere.
Except... now that I'd bothered to actually compute the naïve bound, I noticed something: Lots of my answers were too big! Even ones way, way earlier! Clearly I couldn't have messed them all up -- the early ones I'd checked and rechecked. (I'd eventually find that the first of the answers that was too big occurred at γ=ω+1, α=ω. (Naïve bound: ωω+1. My answer: ωω2.) Yes, all the way back at ω+1. Hoo boy. And for γ≤ω, my answers were just equal to the naïve bound.)
So what was the problem here? I didn't think I'd messed up that far back. Could I have gotten the recursion wrong? I rechecked how I'd derived it, and, no, I'd gotten it right.
Eventually I came to the only conclusion not eliminated: I'd correctly proved that the recursion yielded an upper bound, but my hypothesis that it would yield the exact answer, which I'd never even attempted to verify? Yeah, that just wasn't true at all. All I'd done was come up with a hard-to-describe bound that was, frequently, worse than the naïve bound. Welp.
So yeah, lacking any other approach to the problem, I chose to give up for a third time. Man, that's a lot of time I could've saved if I'd just done this basic sanity check -- checking against the naïve bound -- earlier! Oh well. Just going to have to be satisfied with the naïve bound here.
Anyway, yeah, time to finally get back to integer complexity and addition chains, like I said; once I have some free time to do math again, anyway...
So, let me talk about some math I did while I disappeared from this site for a while. Or, um, attempted to do. You'll see.
Now, as I think I've said before, I've basically closed down my whole ordinal-operations project, seeing as I've gone as far as I think can go with it. While I haven't had a lot of time to get math done recently, my intent was to finally return to integer complexity and addition chains.
...but, one last WPOs problem popped up to distract me. The problem was this: Given ordinals α and γ, what's the largest possible value for o(γ×X), where o(X)=α? (Assuming there is a largest one, anyway; if not, that would be quite notable itself!)
I'll skip discussing for now why I wanted an answer to this particular, odd-looking problem; there were several reasons but I don't really want to take the time to go into them. Point was, I wanted an answer to this question, and it seemed doable.
Now I'd already attempted this twice before and given up -- basically because it was too tedious. Why was it so tedious? Well, so, I thought I actually had most of an answer already, actually! I thought I had a recursion that would yield the correct answer. I hadn't proven it would, but I'd proven it was an upper bound, and I figured I'd save proving that it was also a lower bound for later, as that would likely be tricky. First, I wanted to solve the recursion. Because after all that's what I wanted to do -- I wasn't going to be satisfied with a recursion for the answer, I wanted a more explicit finite rule! That's what's I do!
How do you solve a complicated transfinite recursion? You grind out values -- by hand -- until you see the pattern, that's how. I'd done this before, tedious as it could get, and I could do it again. But it had never been quite this bad.
Consider the previous times I'd done this. When I was computing "right multichoose", I had one transfinite variable (α) and one finite variable (k). This meant I could do it one value of k at a time and get a handle on things. While when I was doing this for D(α1×...×αm), there were multiple transfinite variables, but they were all symmetric with each other, and the resulting answer wasn't that complicated (in the recursive case, anyway, which is what we're concerned with here), so it didn't take a lot of grinding out hand computations to figure it out.
Here, we had two variables, both transfinite, with no symmetry between them, and a nasty, complicated answer. What could I do? This is why I gave up the first two times -- it's way harder to make a transfinite-by-transfinite table, than to merely make a transfinite-by-infinite table! And staring at what tables I did have, the whole thing just looked nasty, with some regularity to it but also full of exceptions.
But for this third attempt, I had a plan: I'd take it one value of γ at a time, going row-by-row, similarly to how I worked with "right multichoose" earlier. (I later transposed the table so it was column-by-column.) Since each row was transfinite rather than merely infinite, this meant every individual row would itself take substantial work to solve, but, well... it worked. The plan of attack worked and before long I was making substantial progress, progressively figuring out the answer for higher and higher values of γ.
Occasionally I'd find I made a mistake and had to go back and redo things. This can easily happen with this sort of thing unfortunately -- you think you've found the pattern when you haven't yet. But I found the mistakes and I fixed them, however much apparent progress it undid.
So I spent weeks grinding out these hand computations. And then, on a flight to Seattle for work, while grinding out a few more, I noticed an answer that looked, just... too big. The parameters here were γ=ωω+12, α=ωω2. The answer I got was ωωω³. Just way too large, right?
I must have made a mistake, I figured. But being on the plane, well, it was a bit hard to go back and recheck all my previous work right there! So I figured, OK, first I can at least check that I have made a mistake -- I can check it against the naïve bound, 2γ×α. If it's larger, clearly I messed up. And indeed, it was larger! The naïve bound was a mere ωωω3+12. So I'd messed up somewhere.
Except... now that I'd bothered to actually compute the naïve bound, I noticed something: Lots of my answers were too big! Even ones way, way earlier! Clearly I couldn't have messed them all up -- the early ones I'd checked and rechecked. (I'd eventually find that the first of the answers that was too big occurred at γ=ω+1, α=ω. (Naïve bound: ωω+1. My answer: ωω2.) Yes, all the way back at ω+1. Hoo boy. And for γ≤ω, my answers were just equal to the naïve bound.)
So what was the problem here? I didn't think I'd messed up that far back. Could I have gotten the recursion wrong? I rechecked how I'd derived it, and, no, I'd gotten it right.
Eventually I came to the only conclusion not eliminated: I'd correctly proved that the recursion yielded an upper bound, but my hypothesis that it would yield the exact answer, which I'd never even attempted to verify? Yeah, that just wasn't true at all. All I'd done was come up with a hard-to-describe bound that was, frequently, worse than the naïve bound. Welp.
So yeah, lacking any other approach to the problem, I chose to give up for a third time. Man, that's a lot of time I could've saved if I'd just done this basic sanity check -- checking against the naïve bound -- earlier! Oh well. Just going to have to be satisfied with the naïve bound here.
Anyway, yeah, time to finally get back to integer complexity and addition chains, like I said; once I have some free time to do math again, anyway...
