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March 19[edit]

Amdahl's law variant, for when sequential portion excludes only itself?[edit]

Does a variant of Amdahl's law exist for the case where the parallel-before-sequential or sequential-before-parallel dependency only applies to each work unit, so that one instance of the sequential portion can run concurrently with any number of instances of the parallel portion (and, unlike with Gustafson's law, the amount of sequential work is asymptotically proportional to the total amount of work)? An example would be a Java program like this:

public static void example(int i) {
  final int sequentialOutput;
  synchronized {
    sequentialOutput = sequentialPortion(i);
  }
  parallelPortion(sequentialOutput);
}
public static void main(String[] args) {
  for (int i=0; i < SCALING_PARAMETER; i++) {
    final int thisI = i; // necessary in Java
    ForkJoinPool.commonPool().submit(() -> example(thisI));
  }
  ForkJoinPool.commonPool().awaitQuiescence(); // wait until all tasks are done
}

I believe that in the above case, the maximum throughput will be achieved (neglecting overheads and memory constraints) if the number of threads in ForkJoinPool.commonPool() is at least the total running time of example divided by the running time of sequentialPortion, and each thread has a processor core. But under Amdahl's law, maximum throughput could only be approached asymptotically. Neon Merlin 04:09, 19 March 2018 (UTC)[reply]

How does space complexity affect empirical time complexity?[edit]

As a program's memory usage increases, so will its memory access times, as it begins to run out of registers, then to miss the L1 cache, then miss the L2 cache, then miss the L3 cache, then hard-fault to disk, then hard-fault to a "disk" that's an abstraction of tertiary storage. As well, AFAICT, RAM stick specifications involve a tradeoff between access speed and capacity, even on an unlimited budget. Are any models available of how an algorithm's space complexity affects the difference between its theoretical and empirical big-O time complexity? Neon Merlin 04:47, 19 March 2018 (UTC)[reply]

In modern usage, all programs are swapped all the time. Therefore, the rule is that you decrease time in an algorithm by using more space. You don't increase time by using more space. 209.149.113.5 (talk) 12:53, 19 March 2018 (UTC)[reply]

@NeonMerlin: Yes there is a lot of work on this. Cache-oblivious algorithm has some references that can get you started. 173.228.123.121 (talk) 04:33, 21 March 2018 (UTC)[reply]

Freenode banning me and the possibility I am a zombie[edit]

IRC banned me a few hours ago. A report dated 2013 or something said something like code 17 and that their automated open proxy scanner was able to exploit software running on my machine. I just got back onto IRC with no problem. Odd. Am I a zombie? Is this some sort of false alarm? Anyone? Anna Frodesiak (talk) 08:13, 19 March 2018 (UTC)[reply]

This is fairly common if you're on a dynamic IP range, in certain locations. Freenode uses a blacklist to ban IPs, so it probably doesn't refer to what's currently happening on your computer. It just means you've been assigned an IP address that was once assigned to a computer that was compromised in the past. Just wait to be reassigned a new IP and you'll be good. -- zzuuzz ^(talk) 08:36, 19 March 2018 (UTC)[reply]

Phew. Thanks, zzuuzz. I just paid my annual net bill. Maybe they gave me a new IP. Is there a way to tell what my old one was and if it was static and the same about my new one? Anna Frodesiak (talk) 08:40, 19 March 2018 (UTC)[reply]

Probably the easiest thing to do is visit https://whatismyipaddress.com/ and note your address over a period of time. I'll tentatively wager that it changes on its own within a day. If you restart your router it may change again. -- zzuuzz ^(talk) 08:48, 19 March 2018 (UTC)[reply]

Wonderful. I will check and keep a record. I just have a computer and am not sure what a router is. I don't think I have one. Search engine pics show a box with rabbit ears and wires. They look fancy and important. Maybe I should get one. Then again, these things always give me trouble. I'd probably end up putting balls of tinfoil on the rabbit ears to make it work properly, and then it still wouldn't. Thank you kindly, zzuuzz. Anna Frodesiak (talk) 08:59, 19 March 2018 (UTC)[reply]

Anna, your computer is connected to the internet either by wifi or through a cable. Either way the network connection or wifi access point probably comes from a box that you got from your ISP, typically sitting on the floor somewhere. That's your router. The box will be connected to your phone line or cable TV connector, and it will also be plugged into a wall socket to get power. To change addresses, try unplugging the power plug, waiting several seconds, then plugging it back in. The router will spend some time (as much as a minute or so) booting itself back up and reconnecting to the net. You'll probably see a bunch of flashing lights on the front of the router while this is happening, and you won't be able to use the internet during the process. Eventually the lights will stop flashing and settle down, and that means you're back online. At that point try looking at whatsmyipaddress.com again and see if your address has changed. 173.228.123.121 (talk) 04:30, 21 March 2018 (UTC)[reply]

What kind of card is this?[edit]

I bought this card on ebay and it was stated to be an IBM Solid Logic Technology card. The card is very similar to the card pictured in that article, but the components are different. Reading the article section "Later developments", it sounded like a Solid Logic Dense card. But it has no resistors (or any components) on the other side. So what kind of card is it? Bubba73 ^{You talkin' to me?} 14:30, 19 March 2018 (UTC)[reply]

It looks like a transistor array used as a digital buffer. The top transistor is the controller or clock. On a complete circuit, you'd expect a ground with a capacitor and regulator power, but this is a card. So, that could be on the main board and removed from the card. To get a real idea of what is going on, you need to look at the opposite side to see how the transistors are connected. 209.149.113.5 (talk) 19:49, 19 March 2018 (UTC)[reply]

The other side of the card has solder spots like the other side. You can see the three leads from each transistor soldered there, but there are no connections between the transistors. Perhaps this is memory or registers? Bubba73 ^{You talkin' to me?} 20:03, 19 March 2018 (UTC)[reply]

It'll be a multilayer printed circuit board, there are layers with connections inside the board and the thru vias which are where those solder spots are connect to copper layers that are beside the holes. Dmcq (talk) 17:55, 20 March 2018 (UTC)[reply]

OK, I can't really see that. But there are two sets of 12 pins each on the top and two sets on the bottom, and they have connectors going into the board. But do you know what kind of card it is? Last night I ordered a big book on the IBM 360 and 370 that might tell. Bubba73 ^{You talkin' to me?} 18:02, 20 March 2018 (UTC)[reply]

Best guess from me: "Plug-in card (detail) with 4 x 8 transistors 291 (plus 4 x 4 transistors COSEM GV -6924-), presumably magnetic core memory readout" (Google translate from German) from IBM-291_TI at RadioMuseum.org Found it by looking for the Texas Instruments (TI) 291 transistor. You can confirm it by finding that type of card, whatever term IBM used. The Computer History Museum collection might have a picture. It could predate the IBM 360. StrayBolt (talk) 19:18, 20 March 2018 (UTC)[reply]

Yes, the picture at that website matches the card (with T.I. 291 transistors). Bubba73 ^{You talkin' to me?} 19:23, 20 March 2018 (UTC)[reply]

The book on the 360 and 370 came in today. So far I haven't found that card. I also have the predecessor Early IBM Computers and it isn't in there either. Bubba73 ^{You talkin' to me?} 01:52, 23 March 2018 (UTC)[reply]