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[2014-9-28. : 2:33 am]  Moose -- I feel as though the increasing power of hardware has made many optimizations less useful.[2014-9-28. : 2:32 am]  Jack -- That's not true for all CS majors to be fair, majority are like that though with a bachelors[2014-9-28. : 2:31 am] Jack -- We're like engineers, maths is a means to an end, not an end in and of itself[2014-9-28. : 2:30 am] Jack -- maybe CS phd/masters, or those who took a maths minor and did a lot of 3D[2014-9-28. : 2:27 am] Moose -- Or, if you're just programming and want to see how to solve problems, you could just get a Schaum's outline and do problems, lel[2014-9-28. : 2:24 am] Moose -- http://www.math.brown.edu/~treil/papers/LADW/LADW-2014-09.pdf Chapters 1, 2, 3 should more than cover an undergrad Linear Algebra course[2014-9-28. : 2:21 am] O)FaRTy1billion[MM] -- All this time spent talking about it, I could've learned it by now.  [2014-9-28. : 2:20 am] O)FaRTy1billion[MM] -- lil-Infernolil-Inferno shouted: Also I forget everything from linear algebra. Instead have transferred it to me, since I need linear algebra for my programming  [2014-9-28. : 2:00 am] Moose -- I would be very surprised if they got into finding an optimal general solution for that in an intro class.[2014-9-28. : 2:00 am] Moose -- In most introductory linear algebra classes, they aren't going to teach you more than finding the inverse for general cases[2014-9-28. : 1:51 am] Moose -- Though you said most efficient way, I just gave you "a way". I can't really prove that's the most efficient.[2014-9-28. : 1:51 am] Moose -- It's non-singular, so you can find A-1 and left-multiply by it k times[2014-9-28. : 12:48 am] MasterJohnny -- what is the most efficient way to solve A^k x=b where k is some positive integer[2014-9-28. : 12:01 am] Moose -- Yeah, I didn't like abstract algebra. Started to get me turned off to pure math.[2014-9-27. : 11:58 pm] Moose -- http://youtu.be/UTby_e4-Rhg I can tell you about the Finite Simple Group (of Order Two)[2014-9-27. : 11:21 pm]  LoveLess -- Time to ostrasize MZ, he gave up on anime because he has grown up.[2014-9-27. : 11:10 pm] O)FaRTy1billion[MM] -- But I think that's the point of simplifying the expression[2014-9-27. : 11:10 pm] O)FaRTy1billion[MM] -- I guess input first helps avoid redundancy ... like if you already had a P AND Q you'd never accidentally make a second[2014-9-27. : 11:08 pm] O)FaRTy1billion[MM] -- That's also how I'd build it when actually putting the chips on the breadboard[2014-9-27. : 11:07 pm] O)FaRTy1billion[MM] -- like the last operation is or, so I'd put an or on the right. Next, one input is P AND Q, so I'd put an AND on one of the OR inputs, and connect the AND inputs to P and Q, etc. ...[2014-9-27. : 11:07 pm] O)FaRTy1billion[MM] -- I always just sort of did it starting from the output ;o[2014-9-27. : 11:05 pm] Jack -- To make the circuit, simplest way is to make each variable a "column" wire going down your page. Then you branch off each wire a NOTted column wire, and from there you just use logic gates to create the rest of the circuit.[2014-9-27. : 11:04 pm] O)FaRTy1billion[MM] -- also you can get rid of several ANDs if you just make it PQ + (~P and Q and R), but I'm too lazy to try to go further than that.[2014-9-27. : 11:03 pm] O)FaRTy1billion[MM] -- dunno, you actually explained it. I just sort of said some stuff and then gave the answer[2014-9-27. : 11:02 pm] Jack -- You don't have to but you can perform boolean algebra optimizations to make that a smaller expression now.[2014-9-27. : 11:02 pm] Jack -- First row, (P . Q . R). Second row, (P . Q . ~R). Fifth row, (~P . Q . R). Then you just OR those together, which gives you (P . Q . R) + (P . Q . ~R) + (~P . Q . R) = Value[2014-9-27. : 11:01 pm] Jack -- In your truth table, the true outputs are fewer than the falses, so we use those.[2014-9-27. : 10:53 pm] Jack -- Anyway demon you only have to worry about the smallest of (number of true outputs) or (number of false outputs)[2014-9-27. : 10:53 pm] O)FaRTy1billion[MM] -- Oh. No, he didn't mean outputs. He meant combinations of inputs[2014-9-27. : 10:53 pm] Jack -- Dem0nDem0n shouted: there are 3 variables, so there are 8 different outputs, but i don't know how to make all of those into a circuit Dem0n did[2014-9-27. : 10:52 pm] O)FaRTy1billion[MM] -- JackJack shouted: There's only one output there Who said there wasn't?[2014-9-27. : 10:31 pm] O)FaRTy1billion[MM] -- Value= (P and Q and R) or (P and Q and not-R) or (not-P and Q and R), then you just simplify that[2014-9-27. : 10:29 pm] O)FaRTy1billion[MM] -- lol thats funny, it's like the inverse of one I randomly typed in my example[2014-9-27. : 10:29 pm] O)FaRTy1billion[MM] -- Both are random examples. You can use either method for any truth table, it's just fewer expressions this way so it should be simpler to simplify[2014-9-27. : 10:27 pm] O)FaRTy1billion[MM] -- here is what I mean ... And then you just simplify the expression 'out = ' for either one |
I like algebra
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