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Introducing Probability into Chip Design

prostoalex writes "The August issue of Intel Developer Update has an interview with Shekhar Borkar, Intel Fellow and Director of Circuit Research at Intel Corp. talking about the future of microprocessor design and what goes on inside Intel Labs. Borkar tells why we need even faster processors and how probability will make its way into future chip designs - "It's like the shift from Newtonian mechanics to quantum mechanics. We will shift from the deterministic designs of today to probabilistic and statistical designs of the future.""

197 of 271 comments (clear)

  1. 1 + 1 by rastos1 · · Score: 5, Funny
    1 + 1 = 2. However there is a 0.0009% probability of it being 1.999999999.

    Sorry could not resist.

    1. Re:1 + 1 by SpaghettiPattern · · Score: 5, Funny

      That was done before in the first batches of Pentium 0.99999999.

      --

      I hadn't the slightest objection to his spending his time planning massacres for the bourgeoisie... (P.G. Wodehouse)
    2. Re:1 + 1 by file-exists-p · · Score: 1

      If they were distributed with a null mean value, I would have no problem summing computations with that sort of error. The law of large number ensures the result would be close to the good answer with high proba.

    3. Re:1 + 1 by CatWrangler · · Score: 1

      There is a 90% chance that you are correct. But there is only a 50% chance of that.

      --

      ---
      When you come to a fork in the road, take it! --Yogi Berra--

    4. Re:1 + 1 by El+Cubano · · Score: 1

      That was done before in the first batches of Pentium 0.99999999.

      So that would be why the first batches of Pentium chips are actualy 486.99999-class processors.

    5. Re:1 + 1 by roman_mir · · Score: 1

      Well, if you want to build a real improbability engine you need to be able to do that kind of math. One step at a time...

  2. so does that mean improbability drives too? by Wameku · · Score: 5, Funny

    UM, Ford. theres an infinite number of monkeys outside that want to talk to us about a script for hamlet they've hammered out. PROBABILITY FACTOR OF 1 to 1: any other problems are your own lookout.

  3. Is this new? by Jugalator · · Score: 4, Insightful

    "We will shift from the deterministic designs of today to probabilistic and statistical designs of the future"

    Doesn't branch predictions in current processors use probabilities already?

    --
    Beware: In C++, your friends can see your privates!
    1. Re:Is this new? by bentini · · Score: 4, Interesting
      Probabilities that will always be the same if you run the exact same sequence of commands.

      What he appears to be suggesting is transistors that we acknowledge to be based in an analog world -- their state depends not only on the data you feed them, but also on the temperature they are immersed in, etc.

    2. Re:Is this new? by mawi · · Score: 1

      Yes but this is about a lower level, AFAICS.

    3. Re:Is this new? by mgessner · · Score: 3, Informative

      I can't speak for all branch prediction models, but in the PowerPC, the answer is yes, but it's static.

      In the PowerPC, unless a hint is given by the programmer/compiler, forward branches (positive offset) are predicted as NOT taken, and backward branches are predicted as BEING taken.

      This is simply because lots of branching (aside from function calls) takes place in for, while and do-while loops (or for, while, and repeat-until for you Pascal geeks :)

      --
      "Sometimes the truth is stupid." - Lawrence, creator of Prime Intellect
    4. Re:Is this new? by Jugalator · · Score: 4, Informative

      Oh, I see... The page now loaded for me, and it seems they're simply considering the fact that previously, hardware performance didn't vary that much, but now when we've got down to real small components -- down to atom level -- that are packed closely together, the probability that the chips will behave differently from environment changes becomes greater. And as the probability of chips "misbehaving" increases, there will also be an increasing need of chips that can take this probability of fluctations into account.

      I first thought the article was about speeding up stuff by probabilities and statistics, but it's mostly about solving a currently theoretical problem that might soon become an actual, real world problem. And to solve that problem, we might even have to move away from some of the computer architecture as we know it today.

      --
      Beware: In C++, your friends can see your privates!
    5. Re:Is this new? by AdEbh · · Score: 1

      Branch prediction is an algorithm that is used to guess where the execution flow will go next based on the past.

      What this guy is talking about is having a processor that has a certain percentage of it's internal components (transistors) faulty, and yet still give the correct result.

      I remember about a guy that was doing a similar thing only his basic unit was a CPU. This is at a lot smaller level.

      If you think that's old hat you're a smarter man than me. 'Cause I'll be damned if I can even begin to think of how they are going to tackle that one.

      - Alex

  4. Less or more by mawi · · Score: 1, Funny

    Software always had it. User: "It'll *probably* work" "It *should* work" If hardware gets more, software needs less...

  5. sounds good for AMD by Anonymous Coward · · Score: 1, Funny

    'cause if intel is gonna make a processor that's in a way kinda right most of the time...i'm going for the green stuff instead :-)

  6. Is that 1.999 repeating? by TrekkieGod · · Score: 4, Funny
    Is that .9 repeating? If so, there's a 100% possibility of 1 + 1 = 1.9999...

    .999... is exactly equal to 1. To the non-believers out there, consider that 1/3 = .333..., and that 1/3 + 1/3 + 1/3 = 1.

    --

    Warning: Opinions known to be heavily biased.

    1. Re:Is that 1.999 repeating? by blackcoot · · Score: 1, Insightful

      1 is the canonical representation of 0.9999... in the decimal system. incidentally, it is also the canonical representation of 1.00000...1. just to throw a spanner in the works, what's the largest real number smaller than 1? most folks want to say 0.99999... any mathheads out there who've done calculus or analysis more recently than i want to take a crack at it? (fwiw, i think the supremum of the set of all numbers less than 1 is 1, but the set itself doesn't contain a maximum)

    2. Re:Is that 1.999 repeating? by TrekkieGod · · Score: 2, Insightful
      Uh...

      Ummm, but .333 repeating + .333 repeating + .333 repeating does not equal 1

      .333... + .333... + .333... = .999...

      That's why .999... exactly equals 1. That was the argument. 1/3 * 3 in fraction equals 1, 1/3 * 3 in decimal form must also equal one.

      There was a remainder when you divided 1/3 that got thrown out.

      What in the name of Gauss are you talking about? Remainder thrown out? No remainders at all...decimals...0.333..., 3 repeats forever. If you kept doing the division, you keep getting 3. Trust me, pal 1/3 exactly equals .333...

      It's incredible how much resistance there is by people whenever this is mentioned in a math class...it's a solid argument.

      --

      Warning: Opinions known to be heavily biased.

    3. Re:Is that 1.999 repeating? by TrekkieGod · · Score: 1
      the geek and correct way to do it...yes. Although I'd argue 1/3 = .333... to be a statement that can also be proven in on of itself (and a lot easier to visualize than asking non-calculus people to take that sum).

      That said...there probably aren't many non-calculus people in slashdot :)

      --

      Warning: Opinions known to be heavily biased.

    4. Re:Is that 1.999 repeating? by csloetjes · · Score: 1

      You are wrong. Welcome to the world of limits.

    5. Re:Is that 1.999 repeating? by sco08y · · Score: 1

      Trust me, pal 1/3 exactly equals .333...

      It's incredible how much resistance there is by people whenever this is mentioned in a math class...it's a solid argument.


      It's the same reason why the square root of two was called an "irrational" number.

    6. Re:Is that 1.999 repeating? by blancolioni · · Score: 4, Insightful


      The supremum of all reals less than one is one. The set itself, as you said, doesn't have a maximum element.

      In a not-at-all-patronising way, I'm surprised that this is even up for discussion on /. but that's probably my bad. Anyway, say X was the maximum real number less than one. Let Y = 1 - (1 - X) / 2. Now clearly Y is less than 1, but also Y - X = (1 - X) / 2 which is > 0 since 1 - X > 0, so Y > X, and therefore X is not the maximum.

    7. Re:Is that 1.999 repeating? by sco08y · · Score: 2, Insightful

      Didn't mean to post right away...

      Anyway, the reason is that people have conceptual issues with infinities.

    8. Re:Is that 1.999 repeating? by csloetjes · · Score: 1

      the sum of (9*(10^-x)) for x from 1 to y will always be 1-(10^-y). Even for very large values of y, (10^-y) will never *equal* zero.

    9. Re:Is that 1.999 repeating? by 56ker · · Score: 1

      OK, try it this way then:-

      1/3 + 1/3 = 2/3

      2/3 = 0.67 (to 2 s.f.)
      1/3 = 0.33 (to 2 s.f.)

      0.67 + 0.33 = 1.00

      1/3 can't be expressed (without an error) in decimal form as it's recurring. It's that error that causes a problem in your calculation.

      I agree that .3 recurring * 3 does not equal 1. However .3 is an approximation of a third (and it always will be as you don't have an infinite number of decimal points). The difference between .3 recurring * 3 and 1 is due to a round off error - the size of which is determined by how inaccurate your value for 1/3 is. .3 recurring is always going to be a little lower than a third.

    10. Re:Is that 1.999 repeating? by Anonymous Coward · · Score: 1, Insightful
      Even for very large values of y, (10^-y) will never *equal* zero.

      Except when y = infinite (more precisely aleph-0). Who said the sum is a finite sum?

    11. Re:Is that 1.999 repeating? by hanssprudel · · Score: 2, Interesting

      i've been using a computer for so long that i'm half convinced that the reals are a hoax invented by physicists to make their sums easier ;-)

      Well, at least you are, or rather were, in good company. When Lindemann proved the transcendence of pi, Kronecker asked:

      "Of what use is your beautiful investigation of pi? Why study such problems when irrational numbers do not exist?"

    12. Re:Is that 1.999 repeating? by Shamashmuddamiq · · Score: 1, Funny
      It's true! ...and the same goes for any numeric base:

      In base 7, 0.66666666... is equal to 1
      In base 2, 0.11111111... is equal to 1
      etc.

      Notice that, in binary (base 2):
      1/3 = 0.010101010101...
      2/3 = 0.101010101010...
      3/3 = 0.111111111111...

      --
      ...just my 2 gil.
    13. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 4, Informative

      An easier form of this proof that I used back in 1979 or thereabouts is when arguing with those who don't understand that 0.9999... is equal to 1. I had learned this proof even earlier from my excellent math teacher in high school.

      Let x be 0.99999....

      x = 0.9999....
      10x = 9.9999...
      10x - x = 9.9999... - 0.9999... = 9
      10x - x = 9 (the infinite trail of nines drop off in the subtraction)
      but also, 10x - x = 9x
      So, 9x = 9
      Therefore x = 1

      With this form of the proof, it is easier to see how the trailing nines just drop off in the subtraction. After thinking about this, the key seems to be that after multiplying 0.9999... by ten, to get 9.9999..., you still have the same number (infinity) of nines behind the decimal point.

      --
      The price of freedom is eternal litigation.
    14. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 2, Insightful

      My excellent high school math teacher had explained to me, when I had asked, exactly what is an irrational number?

      Numbers like 1/3, such as 0.3333... are rational, because even though they repeat, such numbers are exactly representable by the ratio of two integers.

      In fact, any decimal number that can be expressed as the ratio of two integers, is a rational number. Even a number such as...

      0.939287357853918724781923498753298235789

      is a rational number. It is just

      939287357853918724781923498753298235789 divided by 1000000000000000000000000000000000000000.

      Similarly, even a long decimal number, with some trailing set of digits repeating, can be expressed as the ratio of two integers.

      An irrational number, such as PI, or the square root of 2, can NOT be expressed as the ratio of two integers.

      --
      The price of freedom is eternal litigation.
    15. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 2, Insightful

      .3 recurring is always going to be a little lower than a third.

      No. 0.3333... with some finite number of 3's is going to be less than 1/3.

      .3 recurring is exactly equal to 1/3. Exactly. Because the number of trailing 3's is infinite. See the proof about 0.9999... recurring being equal to 1.

      --
      The price of freedom is eternal litigation.
    16. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      There was a remainder when you divided 1/3 that got thrown out. So your argument is bullshiat.

      See this proof .

      --
      The price of freedom is eternal litigation.
    17. Re:Is that 1.999 repeating? by LittleBigLui · · Score: 1

      well... i would say that

      _n_
      \ -a
      lim /__ 10 * 0.3
      n->infinity a=1

      is 1.

      but actually writing out 0.333.. ad infinitum is major work, and writing it out three times and summing it up is exactly as much work (ha, infinity rules!) and you better hire lots of monkeys to do that work for you ;)

      but IANAM

      --
      Free as in mason.
    18. Re:Is that 1.999 repeating? by Ompaloskeptic · · Score: 1

      How about this: .999... + .000...1 = 1 Yes, I know you can't have a new digit after the repeating digit, but it describes it so nicely.

      --
      Good health is merely the slowest possible rate at which one can die.
    19. Re:Is that 1.999 repeating? by eluusive · · Score: 1, Interesting

      There is a problem with your "proof."

      Let x be 0.99999.... x = 0.9999....
      10x = 9.9999...
      10x - x = 9.9999... - 0.9999... = 9
      10x - x = 9 (the infinite trail of nines drop off in the subtraction)
      ^-- this step is artificially inflates the value of 10x, in reality this would be 8.99999999999999...1
      So the question herein lies, what happens when you shift an infinitely repeating decimal 1 place to the left? Does it magically gain a new decimal place on the right? This may happen, but that violates the basic laws of normal numbers. These mechanics of an infintely repeating decimal cannot be used as the basis of a algebraic proof.

      Because you're not saying .9999.. = 1, you're saying:
      9.999... - .999... = 9
      but 9.9999... is not 10(.99999...)

      10(.99999...) has one less decimal place on the right than 9.9999... But still an infinite amount of decimal places. Just because they're both infinite doesn't mean it's not a smaller infinity.

    20. Re:Is that 1.999 repeating? by TuringTest · · Score: 1

      Glad to see I'm not the only one believing that! Indeed I would extend that feeling to natural numbers too, and in fact to all mathematical objects.

      But then, I've been using computers since I was 10... Would you say that is the reason? 8-)

      --
      Singularity: a belief in the "God" idea with the "demiurge" relation inverted.
    21. Re:Is that 1.999 repeating? by bytesmythe · · Score: 1

      Just for another perspective on this:

      (I'm going to use the tilde (~) to represent the repeating digits.)

      z = 0.999~
      10z = 9.999~
      10z - z = 9.999~ - z
      9z = 9
      z = 1

      Thus, 0.999~ = 1, QED.

      --
      bytesmythe
      Hypocrisy is the resin that holds the plywood of society together.
      -- Scott Meyer
    22. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      Is an infinite number of nines minus an infinite number of nines zero?

      Is 0.99999...., minus 0.99999.... equal to zero?

      If so, then 9.99999...,
      minus 0.99999....
      is exactly 9.

      You are thinking of a finite number of nines. I'm thinking of an infinite number of nines.

      Here is another question for you. What is infinity minue one? An infinite number of nines minus one nine.

      In fact, infinity minus one has practical consequence. The Constitution gives Congress the power to grant copyright holders an exclusive right to copy for a limited time. Just some months ago when the whole copyright extension business was in the news, and the Supreme Court was asked to decide whether Congress could extend copyrights this long, our good buddy Jack Valenti (from the Motion Picture Ass. of America) was suggesting that Congress should change the limited time of copyrights to be "Forever minus a day". Maybe we need to get a mathematician to testify before Congress as to what the value of infinity minus one is. Obviously, there are plenty of people, including you, who do not know that it is infinity.

      --
      The price of freedom is eternal litigation.
    23. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      I think we've all known those things since elementary school as the very definition of rational and irrational.

      Good for you. I didn't learn it until high school. In the 70's.

      (That is NINTEEN-seventies, in case you were wondering which century.)

      --
      The price of freedom is eternal litigation.
    24. Re:Is that 1.999 repeating? by Ramze · · Score: 1
      actually, you are wrong. This isn't a function with limits. It's simple math.

      3/3 = 1, but if you instead put in 0.9 in long-division, you get a remainder of 3 which means the next slot would be another 9 giving a remainder of 3 and repeating forever. Try it on paper with long division...

      1/3 is .3333333333333333333333333333333333333333333333333 33333333333333333333333333333333333333333333333333 33333333333333333333333333333333333333333333333333 3333333333333333333333333 etc.

      3/3 if you put in .9 instead of 1 at the top in long division, you get a remainder of 3 (because 3*0.9 is 2.7, leaving .3 which when divided by .09 gives .27 leaving a remainder of .03 thus being divided by .009 gives .027 w/ a remainder of .003... infinity)

      Also, logically, the difference between .9 repeating and 1 is that at some point past infinity, there is a unit subtracted from the last digit leaving a 9... since there is no such thing as infinity +1, there is no "last" digit from which to subtract a unit from and at infinity, .9 repeating is exactly = to 1 -- not approaching it as in a function.

      The idea of numbers repeating to infinity is really a math trick. if .9 repeating did not exactly = 1, then 1/3 + 1/3 + 1/3 could not exactly = 1, which must be true for 3/3 =1.

      The same way 1/3 can be expressed as .33333333 in long-division where 3 goes into 1 three times leaving remainder 1 over and over all the way to infinity, 3 can go into 3 nine times leaving remainder 3 for infinity giving .9999999

    25. Re:Is that 1.999 repeating? by ebuck · · Score: 1, Informative

      How does a mathematically incorrect assumption become insightful here on Slashdot?

      Infinitely repeating digits (aka 0.9...) don't share the same meaning as those that terminate (aka 0.9) treating them the same is inappropriate, and only useful in demonstrating the flaw in logic that they can be treated the same.

      x = 0.9999...
      10 = 10 * 0.9999...
      10x - x = 10 * 0.9999... - 0.9999
      10x + (-1)x = 10 * 0.9999... + (-1) * 0.9999...
      (10 - 1)x = (10 - 1) * 0.9999...
      9x = 9 * 0.9999...

      Unless somewhere along the line you make the error of assuming that 0.9999... = 1, you cannot arrive at 9x = 9.

    26. Re:Is that 1.999 repeating? by CycleMan · · Score: 1

      An irrational number is any number of hours my manager suggests I will need to rewrite a section of code.
      Or the number of dollars my wife wishes to spend on shoes.

    27. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 2, Informative

      Unless somewhere along the line you make the error of assuming that 0.9999... = 1, you cannot arrive at 9x = 9.

      I never made such an assumption. Let's review my proof.

      Let x = 0.99999.....

      Now, don't you agree that 10x would be 9.999999..... ?

      So far I am not assuming what you have said.

      Now is it true that 9.9999.... minus 0.9999.... would be exactly 9? An infinite number of nines minus an infinite number of nines is zero.

      But there is an even simpler proof that someone else mentioned. If there is a difference between 1 and 0.9999... then there must be some number in between these two. Would you be so kind as to tell me what what number is?

      --
      The price of freedom is eternal litigation.
    28. Re:Is that 1.999 repeating? by 56ker · · Score: 1

      It is always going to be less than a 1/3 as there is always going to be a remainder - even with an infinite number of decimals.

      It's always going to be:-

      10/3 = 3 r 1

      10/3 = 3 r 1 etc etc etc

    29. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      How does a mathematically incorrect assumption become insightful here on Slashdot?

      This is not a mathematically incorrect assumption.

      --
      The price of freedom is eternal litigation.
    30. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      there is always going to be a remainder - even with an infinite number of decimals.

      Um, no.

      When you think you have a remainder, then there is simply NOT an infinite number of digits. You have stopped at less than infinity. You have a 0.3333 that is NOT recurring. It is 0.333 recurring that we are talking about here. You are talking about 0.3333 with some fixed finite number of 3's.

      When you have some remainder, you have not finished yet. Add an additional infinite number of 3's and keep dividing. Then you have 1/3.

      This is like looking at an expansion of some function (cos or sin) or value such as pi. There are various expansions that will give you the value of pi. But only when you add an infinite number of terms to the expansion. Now, in implementing a calculator, you can do, say the first fifty terms of the expansion and have a good enough result. But not the real thing. No calculator does an infinite number of steps when you calculate, say, the cosine of some value. But at the same time, you don't see it stated that "the following series is approximately equal to pi", it is stated that "the following series is equal to pi". You just need to do the infinite number of steps. Sorry, I don't have an example of one handy.

      --
      The price of freedom is eternal litigation.
    31. Re:Is that 1.999 repeating? by Eskarel · · Score: 1

      This is in fact not true, however, the proof which I have witnessed on at least two occaisions requires calculus which I have done my best to remove all traces of from my mind. You will just have to trust me, as well as take into account the fact that all proofs given so far to the contrary are flawed.

    32. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      Here is another good proof after some Googling . It goes like this....

      Do we agree that 0.99999..... is equal to 0.9 + 0.099999...?
      If so, then do we also agree that 0.99999.... is equal to 0.99 + 0.00999999....?

      Then we find that no matter small a fraction we add to 0.99999... that we get a number greater than 1.

      0.9 + 0.1 + 0.099999... > 1
      0.99 + 0.01 + 0.009999... > 1
      0.999 + 0.001 + 0.0009999.... > 1


      No matter how small a fraction 0.0000.......001 you add, you get a number greater than 1.

      As I have already stated here. If 0.99999... is less than 1, then please state a number in between the two. For any number A < B, then there exists a number Z such that A < Z < B. Please state Z for me.

      --
      The price of freedom is eternal litigation.
    33. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1
      --
      The price of freedom is eternal litigation.
    34. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1
      --
      The price of freedom is eternal litigation.
    35. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1
      --
      The price of freedom is eternal litigation.
    36. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      Please show me these mathematical texts.

      Please see this .

      After googling, there seems to be plenty of argument that 0.9999... is equal to 1. Therefore, 0.33333... is equal to 1/3.

      I'll use the same proofs...

      If there is a difference between 0.33333... and 1/3, then there must be a number in between the two. Please identify this number.

      Would you agree that 0.333333... is equal to 0.3 + 0.033333...?
      Is it also equal to 0.33 + 0.0033333...?
      Is it also equal to 0.333 + 0.00033333...?

      But no matter how small a fraction I add, I get a number larger than 1/3.
      0.3 + 0.1 + 0.033333.... > 1/3
      0.33 + 0.01 + 0.00333333.... > 1/3
      0.333 + 0.001 + 0.00033333..... > 1/3

      So no matter how small a fraction I add to 0.33333... I get a number larger than 1/3.

      --
      The price of freedom is eternal litigation.
    37. Re:Is that 1.999 repeating? by cgibbard · · Score: 1

      The parent is true, but the reasoning given is somewhat weak.

      I think that all of the confusion that appears to crop up all the time around this consideration is that highschools rarely, if at all, give a rigourous definition of the reals. Often I see them defined as simply the set of infinite decimal expansions. While it is possible to define the reals in terms of infinite decimal expansions, it is necessary to actually define every number with trailing 9's to be equal to a number with trailing 0's and otherwise each digit being equal, but the one right before the trailing starts incremented by one. (i.e. 5.7499999... = 5.75000...) Thus, every number of the form n*(10^m) where n, m are integers, gets two decimal representations by definition.

      This however, is not considered by many to be the most elegant definition of the reals available. The simplest way to express what the real numbers are, is as the unique ordered field satisfying the least upper bound property, but this is not very constructive. To actually talk about real numbers, you need some construction that meets these requirements. Most mathematicians use either Cauchy sequences of rationals or Dedekind cuts of rationals to define the real numbers.

      Please see PlanetMath's entry for Real Number for a definition in terms of Cauchy sequences, and their entry on Dedekind Cuts has a very nice, thorough explanation and definition in terms of Dedekind cuts. Both of these are more elegant and easier to reason about than digit strings if you actually want to prove some property about the reals.

      If you have any questions about this topic, or would like me to clarify or expand on something, reply to this, and I'll try to respond. Alternately, I'm available on irc.freenode.org or irc.slashnet.org as Cale. On FreeNode, I'm usually idling in #math.

    38. Re:Is that 1.999 repeating? by Breakfast+Pants · · Score: 1

      Sounds like you are using some sort of axiom based logic there buddy, Godel's proof casts some uncertainty on your conclusions.

      --

      --

      WHO ATE MY BREAKFAST PANTS?
    39. Re:Is that 1.999 repeating? by Stalyn · · Score: 2, Informative

      ahhh jeeze.

      1. first you apply rational analysis to an irrational number but whatever.

      x = 0.9999....

      lim x = 1
      x -> 1

      lim 10x = 10

      lets supposed x = 0.999
      thefore 10x = 9.99
      10x - x = 8.991

      now again x = 0.9999...
      10x = 9.9999....

      10x - x = 8.999....1

      the infinite series of 9s is reduced by one when you multiply it by 10. it is a common misconception that all infinites are equal, this is not true.

      let x = 0.9999... where n is the number of decimal places.

      x to the n places * 10 = x to the n - 1 places

      proof

      let n = 3

      0.999 * 10 = 9.99
      n = 2

      let n = inf

      0.9999.... * 10 = 9.9999....

      n = inf - 1

      lim n = inf
      n -> inf

      lim n - 1 = inf
      n -> inf

      --
      The best education consists in immunizing people against systematic attempts at education. - Paul Feyerabend
    40. Re:Is that 1.999 repeating? by jcast · · Score: 1

      the sum of (9*(10^-x)) for x from 1 to y will always be 1-(10^-y). Even for very large values of y, (10^-y) will never *equal* zero.

      Repeat after me: the limit of a sequence needn't be an element of that sequence.
      --
      There are reasons why democracy does not work nearly as well as capitalism.
      -- David D. Friedman
    41. Re:Is that 1.999 repeating? by Ramze · · Score: 1

      uh... no. This is the difference between displaying a value as fraction or a whole number in base 10 mathematics. 1/3 is .3 repeating 2/3 is .6 repeating and 3/3 is .9 repeating or 1. They are equivilants -- not approximations.

    42. Re:Is that 1.999 repeating? by jcast · · Score: 1

      What? Goedel himself knew full well there were (are) consistent axiom systems---he frickin' proved the consistency of arithmetic!

      --
      There are reasons why democracy does not work nearly as well as capitalism.
      -- David D. Friedman
    43. Re:Is that 1.999 repeating? by Sivaram_Velauthapill · · Score: 1

      Not exactly what you were looking for but you DID help me out although it wasn't what you were expecting lol... Thanks for linking to that math site. I never knew such a site existed. It is similar to Wikipedia and I like it (people working together to help everyone :) ). I'm bookmarking it for sure :)

      If you know of other free specialized websites, can you e-mail me or respond to this message. Website for things like a chemistry, physics, astronomy, sociology, etc would help. I personally find these online "encyclopedias" very helpful. Quickly checking out a specialized site to see the description of say a quasar or a black hole is better than trying to search the web.

      Sivaram Velauthapillai

      --
      Sivaram Velauthapillai
      Seeking the meaning of life... @slashdot of all places ;)
    44. Re:Is that 1.999 repeating? by Ramze · · Score: 1
      Actually, to clarify...

      .9 repeating as a result is an error in division logic

      It exists only because division is done "improperly" in a sense. 1/3 + 1/3 + 1/3 = 3/3 = 1... however if you use the infinitely repeating decimal places, you get .3 repeating + .3 repeating + .3 repeating = .9 repeating or 1.

      The remainders from the 3 division problems addd up to a total which should not have left a remainder to divide by, thus giving 1 insead of the .9 repeating.

      ex: 1/1 = 1 , but you can also say 1/1 is .9 repeating b/c of "incorrect" long division which leaves a remainder it should not have left -- which is in the case of every instance of .9 repeating. 1 goes into 1 .9 times leaving .1 which 1 goes into .09 times, leaving .01, and so on to infinity b/c the proper answer should have simply been 1 -- but .9 repeating to infinity is also a solution.

      This number does not "approach" 1, but is 1 in a different notation. The CAUSE of the infinite repeating 9's is due to the remainder which should not be there in the first place. To think of it as a number which is 1 minus infinite decimal places of zeros with a 1 somewhere at the end of infinity is not only rediculous, but impossible as nothing ever reaches infinity, much less goes past it.

    45. Re:Is that 1.999 repeating? by EtherealSys · · Score: 1

      consider that 1/3 = .333..., and that 1/3 + 1/3 + 1/3 = 1.

      Sir, I think you've just breached the limit of human understanding. This kind of computation is only as accurate as your calculator. Sadly, my abacus doesn't do fractions.

      "Why is a penny for your thoughts, but you have to put you're two cents in? Somebody's making a penny." - Some comedian whos name i forgot

      --

    46. Re:Is that 1.999 repeating? by zCyl · · Score: 1

      0.99...9 + 0.00...1 = 1, yes, but this is not the question at hand. The flaw is in your usage of infinite. Infinite does not mean, "Everytime we think of it, we add another 9." It means that the listing of 9's is boundless. It is then incorrect to say, "I will add 1 to the last 9," because there by definition is no last 9.

    47. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      1. first you apply rational analysis to an irrational number but whatever.

      0.9999... is not an irrational number. It is rational. That is, it can be expressed as the ratio of two integers.

      For instance, 0.333333... is rational, because it is expressed as 1/3.

      0.99999.... is rational because it is expressed as 1/1.

      1/3 = 0.33333.....

      Multiply both sides by 3, you get...

      3/3 = 0.999999......

      --
      The price of freedom is eternal litigation.
    48. Re:Is that 1.999 repeating? by csloetjes · · Score: 1

      Im glad I learned mathematics in a proper institution, not slashdot.

    49. Re:Is that 1.999 repeating? by Stalyn · · Score: 1

      if p and q are rational numbers then p/q is rational.

      1/3 is rational but 0.333.... is not

      decimals are different then integers.

      --
      The best education consists in immunizing people against systematic attempts at education. - Paul Feyerabend
    50. Re:Is that 1.999 repeating? by Pharmboy · · Score: 2, Funny

      But then, I've been using computers since I was 10...

      Was that 10 as in ten, or 10 as in two?

      I bought my first computer when I was 00010010, but that was almost 00010101 years ago...:)

      --
      Tequila: It's not just for breakfast anymore!
    51. Re:Is that 1.999 repeating? by Stalyn · · Score: 1

      my bad you are right 0.333... is rational and so is 0.999....

      however i still think 0.999... does not equal 1. yet this is still a matter of debate.

      --
      The best education consists in immunizing people against systematic attempts at education. - Paul Feyerabend
    52. Re:Is that 1.999 repeating? by Angry+Black+Man · · Score: 1

      maybe you shouldnt be. IT is NOT A LIMIT. A limit would be something like X in which X is the smallest possible positive number:

      F(x)=1-x, x>0 LIMIT as X approaches 0, f(x) approaches 1.

      That would be a limit. (1-x) != 1 ever. It would merely approach 1. However, 1-x != .9 repeating ever either. It would terminate somewhere before infiniti. .9 repeating DOES NOT TERMINATE. It is NOT merely an estimation, but it is the actual number. You can never subtract anything from 1 and get .9 repeating. Thus, .9 ...= 1

      heres another quick article and paradox abotu it

      I love how when people either a) just made tehmselves look stupid or b) dont have the mental capacity to counter a claim, they just say something like "Im glad i learned math in a proper institution." Well, "learned" must be the wrong word here.

      --
      the byproduct of years of oppression by the white man
    53. Re:Is that 1.999 repeating? by blackcoot · · Score: 1

      if you've been using a computer since you were ten, you should have a firm belief in groups, rings, and fields, just of the finite variety ;-P

    54. Re:Is that 1.999 repeating? by csloetjes · · Score: 1

      I'm glad Im not an asshole

    55. Re:Is that 1.999 repeating? by ebuck · · Score: 1

      No I do not agree that 10x would be 9.9999...

      The whole problem lies between 1 and 0.999...
      The first number is definatly a unit, whereas the last number is constantly being defined. To understand the last number would take an infinite number of expansions of "the next digit"
      so if you multiply 10 * 0.99999..., you would have to first consider 10 * 0.9 + 10 * 0.09 + 10 * 0.009 + ... which you could perceive as 9.999...

      However for you to make that perception, you infer that your result is correct based on induction.

      When you subtract 0.999... from 9.999... you can only imply that the result is 9 if you expand one of your infinite series of 9's one repetition more than the other, so in fact you are arguing that one of the numbers is artificially inflated "just a bit more".

      The correct way to subtract the two and maintain the same degree of percision (if such a thing really made sense at all) would be similar to
      9 - 0.9 + 0.9 - 0.09 + 0.09 - 0.009 + ...
      or
      8.1 + 0.81 + 0.081 + ...
      which would compact down to
      8.999999.....1
      assuming that you could carry on the expansion forever, then the result could be considered
      8.99999....
      but not 9. To achieve a result of 9, you have to opt to expand one of the infinite series preferentially (instead of in lockstep).

      Preferential expansion is not permitted, otherwise I could argue that 9.999999 - 0.9999... equals 9.000000999999 simply because I opted to expand one side just a bit more than the other.

      Happy expanding. Cheers!

    56. Re:Is that 1.999 repeating? by ebuck · · Score: 1

      The assumption that you can preferentially expand one repeating series one digit by multiplying it with 10, without expanding the same repeating series one digit in the non-multiplied context is mathematically incorrect.

      If it were not then the value of 0.3333... - 0.3333... would not equal zero. But it is a basic tenet of mathematics that for all x, x-x = 0 If I can expand the precision of one series preferentially, then I could claim that x-x = 0.000033333..... which would mean that x is not equal to x (an impossiblity unless you change all of the rules, in which case, you don't need to prove it using math :P )

      Which is why it simply boils down to this: you cannot treat infintely repeating series of digits like you can fractions, since fractions do not suffer from errors of percision whereas infinitely repeating digits do.

      3(1/3) = 1 but 3 (0.3333...) = 0.9999...
      because 1/3 is not equal to 0.3333... but is only
      capabable of being represented as 0.3333... due to the percision of the base 10 (decimal) number system.

    57. Re:Is that 1.999 repeating? by Tom7 · · Score: 1

      No, it's true. 0.9999999... = 1.

    58. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      my bad you are right 0.333... is rational and so is 0.999.... however i still think 0.999... does not equal 1. yet this is still a matter of debate.

      0.999... is indeed rational. Expressable as the ratio of two integers. 1/1.

      --
      The price of freedom is eternal litigation.
    59. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      So you do not agree that 10 times 0.99999... is 9.999....?

      Then there is not much else to say.


      The whole problem lies between 1 and 0.999... The first number is definatly a unit, whereas the last number is constantly being defined.

      Not true. The last number has an infinite number of nines. Not some continually expanding number. Quit thinking about getting closer and closer to infinity, and just get there. Sort of like the man who will never crash into the wall because he will always just get closer and closer to it. Just get to infinity. The purpose of the elipsis is to indicate an infinite number of repititions, not an ever expanding number. But an Infinite number. Complete.


      The correct way to subtract the two...

      Do you really expect me to buy that? The correct subtraction is...
      Term 1: 9 + 0.9 + 0.09 + 0.009 .....
      Term 2: 0 + 0.9 + 0.09 + 0.009 .....


      Now take term 1 minus term 2. The infinite nines drop off. What is so difficult to grasp about that? Is that your pre-formed conclusion that 0.9999... is not 1 getting in your way? Refusal to face a simple conclusion.

      If you don't agree that 10 times 0.999... is 9.9999... then there really isn't any more to say.

      --
      The price of freedom is eternal litigation.
    60. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      because 1/3 is not equal to 0.3333

      But 1/3 is 0.3333... exactly. (Notice I put on elipsis after the last 3.) I could even correctly restate it that 1/3 is exactly equal to 0.3...

      Too bad there isn't also a way to write a horizontal bar over the 3 on Slashdot.

      --
      The price of freedom is eternal litigation.
    61. Re:Is that 1.999 repeating? by Anonym0us+Cow+Herd · · Score: 1

      because 1/3 is not equal to 0.3333...

      Let me add one more thing to my other reply.

      The whole purpose of the elipsis notation is so that you can express any rational number in decimal form.

      (Rational number being any number which can be expressed as the ration of two integers.)

      A rational number such as 1/3 could not be expressed exactly in decimal form without the invention of the elipsis notation. That is the whole point of the special notation. (Or writing a horizontal bar over the trailing repeating digits.)

      How can you write the statement I quoted above, inclucing the elipsis, when the whole purpose of that notation is to provide a way so that any rational number could be written conveniently in decimal form?

      0.999999.... is also a rational number. Expressable as the ratio of two integers. 1/1. See all of the proofs I have given in this topic. See the references I have linked. Google for it yourself (although I included a Google set of results in one of my replies in this thread).

      If 1/3 is exactly 0.33333.... (the whole purpose of the invention of the special notation) then it stands to reason that 3/3 is exactly expressable as 0.99999....

      And note that I did not write some long string of 9's and stop. I used a recognized mathematical notation to express that the nines are infinite. Not part way to infinity. Not somewhere along the way, in your mind where you are thinking of some "last 9". But all the way to infinity. Stop thinking of "one the way" to infinity, and actually get there. Stop thinking of what a function approaches, and think of what it is when it finally gets there at infintiy. (Sort of like the guy who will never run into the wall because he will always have to get halfway there first, and it will take an infinite number of halfways, therefore he will never hit the wall. Therefore the repeating nines will never equal 1.)

      --
      The price of freedom is eternal litigation.
    62. Re:Is that 1.999 repeating? by ZigMonty · · Score: 1
      Another proof.

      0.999... = 0.9+0.09+0.009+...

      This is a geometric series with initial term 0.9 and common ratio 0.1.

      The infinite sum is:
      initial term/(1-common ratio)
      = 0.9/(1-0.1)
      = 0.9/0.9
      = 1

      See MathWorld.

    63. Re:Is that 1.999 repeating? by sco08y · · Score: 1

      Yeah, that explains it pretty well.

      Now, suppose we have three sets: Integers, Rationals and Reals.

      Which of these is true, and why:

      a. All three have the same cardinality.

      b. Reals has a greater cardinality than Rationals which has a greater cardinality than Integers.

      c. Integers and Rationals have the same cardinality, but Reals has a greater cardinality.

      d. Rationals and Reals have the same cardinality, which is greater than that of Integers.

      Hint: by "A has the same cardinality as B" I mean that if I step through every element in set A, there is a corresponding element in set B.

      Think about it before you Google "Cantor Diagonal Method".

  7. Douglas by howman · · Score: 1

    how are we going to fit a restaurant into my computer... I barely have room for my tower in my apartment as it is...

    --
    flinging poop since 1969
    1. Re:Douglas by Nighttime · · Score: 1

      how are we going to fit a restaurant into my computer...

      If it was a bistro, it would be Somebody Else's Problem.

      --
      I've got a fever and the only prescription is more COBOL.
  8. I remember... by Muad'Dave · · Score: 5, Interesting

    ...back in the heady days of Concurrent Computer their top-of-the-line 3280 processor has "usual branch" instructions. The compiler could use the usual branch instructions to provide hints about the probability of the branch being taken to the processor. In a loop, for instance, you'd use a "usual branch not equal" (UBNE) instruction to send execution back to the top. This would indicate to the processor that it should preemptively invalidaate the cache and pipeline.

    I'm sure many mainstream processors have this now, but it's funny to think that CCUR had this technology in the late 1980's.

    --
    Tiller's Rule: Never use a word in written form that you've only heard and never read. You will end up looking foolish.
    1. Re:I remember... by hey · · Score: 1

      I always figured this was used for stuff like this:

      for (i = 0; i 10; i++)
      { // do something
      }

      The branch to jump outside the loop
      is the unlikely one. This is probably
      a good assumption to make for loops.

    2. Re:I remember... by PenguiN42 · · Score: 1

      Why are you invalidating the cache and pipeline on a branch? wow.

      Anyway, static branch prediction has been around for a long long time. That's not what intel is talking about.

      --
      The following sentence is true. The preceding sentence was false.
    3. Re:I remember... by bsharma · · Score: 1

      >>but it's funny to think that CCUR had this technology in the late 1980's Analog Computers are even older than that; and they used true quantum mechanics since they computed based on voltage, current etc., What he is saying is that eventually, all computation will become 'Analog' - in ways much more sophisticated than in 1950's.

  9. The more things change by Peldor · · Score: 2, Funny

    Hey, this is nothing new as anyone who owned an original Pentium can tell you. It probably gave you the right answer, except for the occassional FDIV.

  10. it is an old story by wannasleep · · Score: 5, Informative

    In the interview, a lot of things have been left out. The topic is first and foremost old. It goes back to the 80s. Statistical variations have always been taken into account by using worst cases. Problem is that the worst case approach sucks in the latest technologies, so more sophisticated methods have to be used. There has been a lot of research in the last 10 years (Check american, german, and italian universities, just to name few).
    Also, the problem is old, meaning that analog designers had to deal with these problems since the early stages (example: the offset in the operational amplifiers is caused by transistor performance mismatch). Now, digital designs are affected too. First on the clocking network and now all the rest. Furthermore, it is widely known (in the community) that interconnect variations are of the same order of magnitude of the device (i.e. transistor)performance variations, and on the top of that dynamic effects (like cross talk) may severely affect the performance.
    I don't agree with him on the fact that all the variations are gaussian, there is plenty of literature that states the contrary, and major chip makers know it very well.
    Last but not least, there are already tools that deal with statistical variations, although none of them can handle a microprocessor, as they are mostly circtuit simulation-based. All in all, the good news is that awareness is spreading thru the designers.

    1. Re:it is an old story by fupeg · · Score: 2, Informative

      I think there is a slight misunderstanding here. The "variance" of analogy systems is really just manufacturing fault tolerance. The manufacturing process is imperfect and so there is a standard error associated with each quality of a given component. This is a little different than the kind of variance that is related to quantum mechanics.

      Imagine an AND gate that is a single silicon atom. For such a gate to be "open" a single electron would have to be "flow" through it. This requires the electron to bond to the atom and it's ability to do that is a function of the position of the other electrons in the silicon atom. Quantum mechanics tells us that the positions of these electrons is not determined UNTIL something (such as the electron in this case) interacts with the atom. Until the time of interaction, the electrons have no discrete positions but are instead described by probability distributions. So this is an AND gate that is non-deterministic. Not only will "not always work", but you have no way of knowing if will or will not until you fire an electron at it, and just because it "fails" once has no bearing on the probability of it "failing" again. This is similar to the analog analogy, but it is even less deterministic.

      I think Brokar caused some of the confusion in his interview by bringing up the analogy of variance in speeds of P4 chips. For that to be an accurate analogy, each P4 chip's speed would also vary each time you measured it. Anyways, a chip designed around this kind of non-deterministic behavior would be impressive. And chips will have to be designed this way for the size of components to get much smaller (an atomic diameter is around 0.1 nm.)

  11. Old news... by qtp · · Score: 2, Funny

    Didn't intel already do this whith the original Pentium?

    --
    Read, L
  12. Re:Corporate Feces by Peldor · · Score: 4, Funny

    At least it didn't say Pentium(R) 4(R). Not for lack of trying, I'm sure.

  13. I fail to see by Capt'n+Hector · · Score: 1

    How is this like the transition from newton to quantum mechanics? Hasn't computing always been quantized? I for one would love to see a change form quantum type computing to newton type computing using smooth instead of discrete. Ready for infinite FPS? How about ultra-realistic physics models? It seems to me like this intel fellow is merely riding the quantum-this quantum-that hype.

    --
    Quid festinatio swallonis est aetherfuga inonusti?
    Africus aut Europaeus?
    1. Re:I fail to see by SilentTristero · · Score: 1
      Because QM is inherently probabilistic. It's not the quantization he's talking about, it's the "probabilization".

      -- Tristero

    2. Re:I fail to see by alenm · · Score: 1

      He is probably reffering to the fact that the designers have to take the effects of quantum mecanics into account when they design chips. The didn't need to before. The newton mecanics describe the world pretty well for many purposes.

  14. About time! by sco08y · · Score: 2, Funny

    Kinda like we've been releasing software that "probably" works for the past 40 years?

    It's good to see computer engineering is finally catching up with computer science!

  15. Celeron chips by Hoplite3 · · Score: 1

    Does this mean that the new bargain line for intel will have a higher stardard deviation to its error? Imagine buging a 5 p/m 3 ghz chip. This could be big for overclockers. Perhaps that three gig chip was mislabeled and you have an eight ghz model on your hands ...

    --
    Use the Firehose to mod down Second Life stories!
    1. Re:Celeron chips by wannasleep · · Score: 1

      yes and no..
      first, the variability is caused by the variability of the process used to produce the chips because most of the physical phenomena that are used to produce a chip are stochastic in nature (e.g. diffusion, implant, heating etc.).
      Then, there is the economic factor. When you by a 2GHz pentium and a 1.8 GHz pentium, you are buying the same design and process. One was just luckier than the other. Then the manufacturer tests them and discovers that one meets the 1.8 specs and the other one the 2 specs as well. One is sold as 1.8 and the other for a higher price. Kinda like if a baker were tasting what he sells and put a higher price tag on the best tasting one.
      So, why can you overclock? Manufacturers might make errors in the tests, or simply a chip might fail a test in a condition you never use the microprocessor. Or, even better, they might simply be labelling say 2.4 GHz processors as 2GHz so that the 2.4 price doesn't drop...

  16. HAL by Anonymous Coward · · Score: 2, Funny

    This will make HAL even worse:

    OPEN the DOOR HAL!

    Proberbly not, Dave

    1. Re:HAL by mwood · · Score: 1

      Or:

      "Open the pod bay doors, HAL."

      "I don't think so, Dave."

      "Awright, it's time for MORE POWER!!!"

      Kaboom!

  17. Random by t_allardyce · · Score: 1

    Great, for 50 years people have been saying "computers can never make mistakes" and while allot of hardware and software has been designed like shit and does make mistakes (FDIV Pentium) atleast it makes consistent mistakes. Does this mean that a=b will sometimes be a=c? Will computers sometimes make mistakes just like humans? Or am i talking total crap.. probably.

    --
    This comment does not represent the views or opinions of the user.
  18. Good! by MagicMerlin · · Score: 1

    Maybe now they can probabalisticly determine themselves a 64 bit chip with good x86 32 bit compatiblity :)

  19. I was expecting SOMEone to do this... by ChuyMatt · · Score: 1

    Soon we will have improbability chips powering our processes! It will take us where ever we need to with out even a graphics card or ram! That is until some odd white robots come and take our chips away. Damn Kricket bots!

  20. I can now relax... by KingRamsis · · Score: 1

    from the interview...
    Look at the whole proactive computing model, where computers will anticipate our needs and sometimes take action on our behalf. That's one.

    so I can relax and count on my Opera v40 to post the first post for me?

  21. Already happens by PsiPsiStar · · Score: 2, Funny

    Isn't probability already a part of chip design.
    "Our new P4 has a 40% probability of being out in May, a 20% chance of being out in June..."

    --

    ___
    It's the end of my comment as I know it and I feel fine.
    1. Re:Already happens by Alsee · · Score: 1

      Our new P4 has a 40% probability of being out in May, a 20% chance of being out in June

      Exactly, and a 95% chance of being delayed until at least September.

      -

      --
      - - You can't take something off the Internet! That's like trying to take pee out of a swimming pool.
    2. Re:Already happens by Atticu5 · · Score: 1

      Well, I'm happy now, because now I know that the new P4 has a 40 + 20 + 95 = 155% chance of coming out before September. It's a sure-fire thing; thanks guys!

  22. Re:Corporate Feces by AdEbh · · Score: 1

    It's not 4(R) as you can't register a pure digit number like that. That's why Intel started using the name Pentium instead of 586.

    -Alex

  23. Think error correction by elwinc · · Score: 2, Insightful

    I believe the kind of probabalistic computing Intel's talking about is analagous to error correction. On your average data CD about 15% of the bits are redundant and devoted to error correction. This reduces the probability of erroneously reading the CD, although the probability of error is still non-zero. Same deal with ECC memory. I'm guessing Intel is looking at ways to apply that kind of trick to the computational logic.

    --
    --- Often in error; never in doubt!
  24. That is a very smart man. by mr_luc · · Score: 3, Insightful

    Listen to that guy. He just GETS it.

    I am actually, to some extent, inspired by that article. Corporate BS policies aside, whatever you think of Intel or AMD or any other company as a company, as a political entity, or as a producer or consumer goods, you still have to feel good that there are people like that, people that just GET the overriding vision of advancing technology, and are actively working to advance it.

    I don't have time advance technology much in my current job. I don't have the mind or the skills or the time for boundary-pushing endeavors. Some at /. do, and contribute all their mind and skills and time to furthering open-source and other efforts, and that is very commendable.

    But as we often lament, it sometimes seems like the Big Boys don't have the same spark. Let's not forget that somewhere within the pudge of even the fattest multinational technology company, there are brilliant, passionate minds working to further everything we hold dear. These are people who aren't just brilliant scientists or passionate geeks -- they're both. And they're on our side. :)

    1. Re:That is a very smart man. by mr_luc · · Score: 1

      Well, he got a degree from Notre Dame in 1981. Granted, he was born in India, but he's been here long enough that by any reasonable standards he's an American.

      Not that it would matter if he came over from India yesterday. He's a smart man, he's very qualified, and he holds a position of great authority that commands great respect. This isn't a case of a mildly trained person who will work for peanuts being brought in to lay off a highly trained technician that wants an honest wage. Positions like the one he holds are earned purely on the basis of merit.

    2. Re:That is a very smart man. by thebigmacd · · Score: 1

      No, he took an India's job. If it were an American's job, an American would have it. If Americans quit being nationalistic dumbasses and solved the problem by being more adept than foreigners, there might actually be some improvement in employment of American nationals.

    3. Re:That is a very smart man. by JDBrechtel · · Score: 1

      You're a troll....Right?

    4. Re:That is a very smart man. by evilviper · · Score: 1
      Let's not forget that somewhere within the pudge of even the fattest multinational technology company, there are brilliant, passionate minds

      And odds are, they are all working as janitors at Microsoft hq.
      --
      Slashdot gets worse every day... Pipedot: News for nerds, without the corporate slant
  25. Now all I need is a nice, hot cup of tea.. by ElVee · · Score: 1

    It's all in the Brownian motion, baby.

    --
    - Pithy comment goes here.
  26. Chip Name? by jasondlee · · Score: 1

    I hope they call their next chip the Beeblebrox...

    jason

    --
    jason
    Have a good day?! Impossible! I'm at work!
  27. That's peice of cake by MC68040 · · Score: 1

    Just get a muffin and a cup of tea, that seemed to work fine in the hitchikers guide to the galaxy as for probabilistic designs. Although don't try it out at too massive starships and such, as we've learned that it could be an bad idea (ref. starship titanic).

  28. The future is now! by supergerwalk · · Score: 2, Funny

    And we'll all be traveling in flying cars while eating meals in pill form!!

  29. Faster processors? by gr8_phk · · Score: 1

    No wonder we'll need faster processors. The more times you run the same calculation, the more certain you can be of the result. For critical applications that need 10-20 sigma reliablility, we'll need very fast processors indeed.

  30. Chaos Theory by Quietdemon · · Score: 1
    Well probability theories being what they are, I guess that acceptable by the majority. Of course seeing as the majority might take some time to react to any problem known by the few who understand what problems probability theories pose...

    There's another theory that's called the Chaos Theory and given that anything left unguarded will most likely go haywire or be vulnerable in it's unguarded state, I'm not sure I'm a fond believer of the 0.00009 error margin.

    What about compounding the problem though time? Isn't it remotely possible that we could be slowly adding to the problem instead of just remaining status quo? The news that companies like Intel release to the general public have to do with current technology and not compounded problems with technology. Isn't it possible that the flaw can accumulate through time and with the changing of the guard (The older scientists leaving and the newer scientists picking up the slack) that there could be some loss of information concerning the acceptable flaws of the chip designs?

    I know that when my computer crashes it isn't because the probability error factor kicked in, but more likely that my computer did what it could do given the circumstances of it's invention. The human body isn't perfect either and nothing we could possibly invent could ever be perfect. How can we expect our computer chips to be? It's nothing more than human after all.

    QD

    1. Re:Chaos Theory by geekoid · · Score: 1

      Do you get all your chaos theroy from Jerassic park, or did you just pick the worse explanation of Chaos Theory by accident?

      --
      The Kruger Dunning explains most post on /. http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
    2. Re:Chaos Theory by Quietdemon · · Score: 1
      Actually I got it from reading the geekoid wannabe manual, which subsequently was used as poop paper for my dog.

      I skipped the comebacks chapter...cause nothing in there was satisfactory.

  31. I'm a non-believer. by Amorpheus_MMS · · Score: 1

    I've heard that 0,99999.. is 1 before, and it just doesn't make sense - theoretically those nines go on forever, so it's always an infinitely small (not zero!) part away from the next full number. Fractals and rounding errors don't convince me.

    1. Re:I'm a non-believer. by Charbal · · Score: 2, Informative

      Alright. So, say that x is 0.99999... with an infinite number of nines. You claim that x is a number distinct from 1 (since you claim 1 - x is nonzero). Since the the rationals and irrationals are both dense in the reals, we know that if we pick any two distinct reals, we'll always have numbers between them. That is, it's impossible to pick a number and the "next" real number without skipping over any. Yet that is essentially what you've done by positing that the gap between these numbers is infinitely small.

      Put in other terms,

      1 - x = limit(n->infinity, 1 - sum(9*10^(-i), i=1..n)) = limit(n->infinity, 10^(-n))

      So let y equal 1 - x. We can show that if y != 0, then 1 - x < y even with a finite number of digits. Just pick n to be the least integer which is at least -log(y)+1 and then 10^n < y just considering a finite number of digits, n.

      That is, if you tell me what the nonzero gap between the numbers that you expect to see, one can show that the gap is less than that. Since the gap is a nonnegative number that is less than all nonzero (ie, positive) choices, it is zero.

      Therefore, 0.9999999..... = 1.

      On an even more off-topic note, has anyone else started to think that trolls that bring up controversial mathematical statements are probably the best way to get responses nowadays? Not to say that the OP was a troll, but if it was, congratulations on all the responses. Also, congratulations to the Monty Hall problem poster some days back that was a troll.

      --
      Prudence forbids me to explain myself further. - Isaac Barre, 1765
    2. Re:I'm a non-believer. by schmink182 · · Score: 1

      1/3 = .33333...
      3 * (1/3) = 3 * (.33333...)
      1 = .99999...
      Q.E.D.

      How 'bout that?

    3. Re:I'm a non-believer. by Amorpheus_MMS · · Score: 1

      The problem is that 1/3 isn't .3333 - the latter only approaches the former more with every digit. Thanks to our base 10 system, 1/3 is somewhere in between .3333~33 and .3333~34 but doesn't have an exact match. (Add the neglected .000~01 to every third third and it works out fine...)

    4. Re:I'm a non-believer. by Amorpheus_MMS · · Score: 1

      >That is, it's impossible to pick a number and the
      >"next" real number without skipping over any.
      >Yet that is essentially what you've done by
      >positing that the gap between these numbers is
      >infinitely small.

      It's not skipping over any numbers if the difference is infinitely small, which is the key here. I think this is more of a logic than a math problem, is it not obvious that since an infinitely small 1 seperates the two numbers they can't be the same? The smallest theoretically possible number is your nonzero gap. Put all the fancy math aside and think about it. (The biggest problem with this theoretical matter is that real math will break it, and since math is never wrong people believe it. :))

      Look at it graphically. On one side you have 1. On the other: 0, ., and a whole lot of 9. Not the same! ;)

    5. Re:I'm a non-believer. by luckyguesser · · Score: 1

      It's not skipping over any numbers if the difference is infinitely small

      A gap that is infinitely small is not a gap. However, using "fancy math" ;), there will always be another number between any 2 set numbers. If you insist that there is a point close to infinity where the gap is not an issue, come back to me with the 2 numbers that you measured from =). Look at it this way: half-way between 0 and 1 is 0.5. Halfway between 0.5 and 1 is 0.75. When are you going to decide that the gap is "inifnitely small" enough?

      --


      The power of Christ compiles you.
      A Random Blog
    6. Re:I'm a non-believer. by Breakfast+Pants · · Score: 1

      "Therefore, 0.9999999..... = 1."

      Hmm, that doesn't stand up to Godel's proof.

      --

      --

      WHO ATE MY BREAKFAST PANTS?
    7. Re:I'm a non-believer. by Amorpheus_MMS · · Score: 1

      A gap as big as the smallest possible gap is still a gap. Obviously there can be two numbers with nothing in between when we're talking about infinite decimals. 0.999999~ will get closer to 1 with every digit, but won't equal it ever. There isn't anything else to say... Free your mind! ;)

    8. Re:I'm a non-believer. by luckyguesser · · Score: 1

      By admitting that .9999_ is never = to 1, you admit that they are distinct numbers, which, by definition, must have a gap between them in which can be found another real number. Taking that information back into Charbal's post, .999- is = 1.

      --


      The power of Christ compiles you.
      A Random Blog
    9. Re:I'm a non-believer. by Amorpheus_MMS · · Score: 1

      >which, by definition, must have a gap between them
      >in which can be found another real number

      That definition must be flawed when dealing with infinite decimals. I'm not going to spend any more time on this, but it seems people don't think this theoretical thing through but rely on fallacies like 1/3 = 0,333~ to decide.

    10. Re:I'm a non-believer. by Angry+Black+Man · · Score: 1

      Dude your the only one with the fallacy. Its counter-intuitive. The reason most people dont get calculus and advacned level math/physics is because it's coutner-intuitive and they are so set in their ideas of "common sense (that they cant prove)". Face it, your common sense is wrong. The primate in you is still insisting that .99999~ only approaches 1, but repeated mathematical proofs have shown you otehrwise. You are just too stubborn to forget your .999~ does not "get closer to 1 with every digit" you add. YOU CANNOT ADD A DIGIT TO .999~... IT IS IMPOSSIBLE. THERE IS NO WAY TO ADD A DIGIT WHEN THE DIGITS REPEAT FOREVER. You are attempting to visualize .9 on a graph in a real world visualization, and in your head with every .9 you add, your little imaginary graph is getting closer to 1. But thats not how it works. STOP TRYING TO VISUALIZE IT. Get over your intuition, or you will never succeed in any higher level math/science studies.

      X is the smallest possible positive number. Already I know you are probably trying to visualise this by thinking of it as (.00000~1). Dont. Dont try to assign a number to X. Just know that X is the smallest possible postive number. Period.

      f(x)=1-x (x>0) . As X approaches 0, f(x) approaches 1. ALSO, as X approaches 0, f(x) approahces .999 repeating.

      There is no number that you can subtract from 1 to give you .9 repeating. Because 1 minus any number, no matter how small, would terminate somewhere. .999~ DOES NOT TERMINATE. Thus, .9~=1

      --
      the byproduct of years of oppression by the white man
    11. Re:I'm a non-believer. by Tom7 · · Score: 1

      No!! 1/3 is exactly .33333..., repeating infinitely. Are you saying that the rationals aren't embedded in the reals? Did you ever take a discrete math course?

  32. This idea has been around for ages by Theovon · · Score: 2, Informative

    Random number generators are used in ASIC and FPGA logic placement and interconnect routing.

    The goal is to place and route logic in a way that meets the designer's timing and area constraints. The problem is that a deterministic algorithm for that is NP-complete. Instead of considering all possibilities, a number of randomly-generated possibilities are considered, with some ability to make adjustments when one is chosen.

    The randomly-generated possibilities, of course, are not completely random -- it's a matter of multiple gates competing for the same fixed-location logic cell, etc. Who gets the one closest to where they all want to be? Where do you place the rest? What about others competing for THOSE locations? It's complicated. :)

  33. A first step by PeterChenoweth · · Score: 1, Funny

    Well, we've got to have probability processors before we can ever hope to make an Infinite Improbability Drive !

  34. Probability by smatt-man · · Score: 1

    What's the probability that I can afford one? 0.00000000000000000

    --

    ---
    Lousy rotten karmic retribution.
    1. Re:Probability by evilviper · · Score: 1
      Amd:
      It will probably melt through your desk.

      I've said it before (just yesterday, in fact) and I'm sure I'll have to say it again, but here goes...

      AMD XP processors have a lower heat output than Intel P4s, and P4s, while putting out more heat, have a SIGNIFICANTLY lower maximum operating temperature.
      --
      Slashdot gets worse every day... Pipedot: News for nerds, without the corporate slant
  35. Re:Apples? by mwood · · Score: 1

    Nah, the Newtonian Mechanics' Union just sued them instead.

  36. This is a big mistake... by drgroove · · Score: 1

    They should just paint the chip pink and put an SEP (Somebody Else's Problem) field around it - saves mucking about with all of that improbability stuff.

  37. A first step? by djeaux · · Score: 2, Interesting
    Assuming that the overarching goal of computer (and software) design is to emulate the human brain -- or even the brain of a flatworm -- hardware is going to have to break free of the confines of binary true-false logic, tight tolerances, etc., and embrace variation.

    Since physical science (and by extrapolation, engineering) is built on a "reductionist" paradigm where every problem is broken to its simplest components & solved piecemeal at that level, it makes sense for a "probabilistic" approach to chip design to happen some time. Might as well be now.

    But when we operate under the reductionist model, we forget emergent properties at the system level. In developing a "learning" system -- which again, I assume to be the overarching goal -- we have to learn to deal with variation. Situations are almost never exactly the same. In the beginning of a "learning" system, things probably (pun intended) do look random. But as special cases, exceptions, subtle cues, etc. are encountered by the system & incorporated into the decision-making process, things appear to become increasingly deterministic.

    So, if a "probabilistic" chip design is implemented properly, it likely will look pretty "deterministic" to the end-user, who expects certain kinds of results.

    The problem now is that the hardware is "deterministic" & any attempt to create a "probabilistic" learning system has to happen in software. Right now, the limit to AI, IMO, is simply that chips aren't even in the same league with neurons. "Learning" software built on "learning" hardware ought to be a pretty powerful concept.

    Of course, this may just be a way to get around the fact that manufacturing may be pushing the limits for tight tolerances & probabilistic chip design is the only out. Whatever it takes to force a paradigm shift.

    "Most places a paradigms won't buy you a cup of coffee..."

    --
    "Obviously, I'm not an IBM computer any more than I'm an ashtray" (Bob Dylan)
    1. Re:A first step? by dvdeug · · Score: 1

      Assuming that the overarching goal of computer (and software) design is to emulate the human brain -- or even the brain of a flatworm

      It's one goal, but if the human brain could accurately, reliably and quickly make ballistics tables, or spell check, or run nuclear simulations, we would never have made computers. There are a lot of things that a computer is very good at that humans suck at, and breaking that to make computers better at doing stuff that humans can do already is no win. Even if computers never think like humans, they will still be useful tools.

  38. Re:Trinary Logic... a nonserious answer by mwood · · Score: 1

    Oh, great. See Gerrold's _When HARLIE Was One_ to see what happens when you allow ten-state logic into your designs. :-)

  39. Alternate proof by FiloEleven · · Score: 1

    x = 0.999...
    10x = 9.999...

    10x - x = 9x

    9.999... - 0.999... = 9

    9x = 9

    x = 1

  40. Probability by geekoid · · Score: 2, Funny

    Intel:
    The addition is probably right.

    Amd:
    It will probably melt through your desk.

    Me:
    I will probably be modded to Hell.

    --
    The Kruger Dunning explains most post on /. http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
  41. Laws of Probability by lmahan · · Score: 1

    Its interesting that Intel is looking at moving into the laws of probability with chip design. Microsoft has been doing it for years with their operating system.

  42. statistical? by sdjunky · · Score: 1

    "We will shift from the deterministic designs of today to probabilistic and statistical designs of the future"

    Statistics? There are lies, damn lies, and 824633702441(WARNING! STAT FLOATING POINT ERROR)

  43. Your failure by geekoid · · Score: 1

    is that you don't relize that the editors don't read the article.probably.

    --
    The Kruger Dunning explains most post on /. http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
  44. Quality of life by gilesjuk · · Score: 1

    The article mentions the increasing productivity and quality of life that increasing speeds will bring. Yet computers are becoming noisier all the time, for some this is a reduction in their productivity and quality of life.

  45. Infinity != Infinity, for all values of Infinity by IncohereD · · Score: 1

    You are thinking of a finite number of nines. I'm thinking of an infinite number of nines.

    You missed the parent's point that just because two things are infinite, doesn't make them equal. Some of them approach infinity at different rates, or are clearly larger or smaller than another. Infinity's just a handy notation, it is NOT an actual specific value.

    There's a whole wonderful, nasty math of infinities out there for you to do a graduate degree in mathematics on. Saddle up and ride.

  46. Imagine the new games... by RedlumF · · Score: 1

    "You are in an abstract state. Both dead and alive. Press any key to open the door and really find out!"

    Bah.
    /Alex

  47. Re:Infinity != Infinity, for all values of Infinit by Anonym0us+Cow+Herd · · Score: 2, Informative

    I'm not talking about different functions approaching infinity at different rates.

    I'm talking about the fact that 9.99999.... minus nine is exactly 0.99999.....

    Therefore 9.99999..... minus 0.99999.... must be exactly 9.

    --
    The price of freedom is eternal litigation.
  48. Sign in Bed-n-Breakfast at the End of the Universe by The+Monster · · Score: 1

    "Heisenberg may have slept here"

    --

    [100% ISO 646 Compliant]
    SVM, ERGO MONSTRO.

  49. Re:Trinary Logic... a serious issue... by fireboy1919 · · Score: 1

    There's already a field devoted to what you descibe, and it's not called trinary logic because there are more than three states.

    It's fuzzy logic, and the idea is that along with true and false there are also degrees of truth. If true is represented as 1, and false as 0, then an operand can have any value between zero and one. Here are a few fuzzy operators:
    Not: 1-operand
    And: min(operand1, operand2)
    Or: max(operand1, operand2)

    Notice that this is a superset of the original operators; with normal true and false values they still behave the same way.

    These operators give us a nice way to quantify "possibly," so that we can actually do math using it.

    --
    Mod me down and I will become more powerful than you can possibly imagine!
  50. Probability of defect to be detected by the user by acozer · · Score: 1

    Probably he means that, in the future, parts not approved by the quality control could be sold for users that doesn't use their computer too much. So, the probability of a malfuncion will be small enough to justify the high profits selling garbage.

  51. Re:Infinity != Infinity, for all values of Infinit by Anonym0us+Cow+Herd · · Score: 1

    just because two things are infinite, doesn't make them equal. Some of them approach infinity at different rates, or are clearly larger or smaller than another

    I would like to see some references to different values of infinity please. That is like saying that one is not equal to one for different values of one.


    There's a whole wonderful, nasty math of infinities out there for you to do a graduate degree in mathematics on

    And I'd really like to hear what the authors of such works would have to say about the whole 1 = 0.999... discussion. After all, this is not new. This discussion was very old and settled when my high school math teacher taught it to me 25 years ago.

    --
    The price of freedom is eternal litigation.
  52. Faster Processors... by MojoRilla · · Score: 2, Funny

    Q4: What are some other applications that need more power?

    Look at the whole proactive computing model, where computers will anticipate our needs and sometimes take action on our behalf. That's one.


    When he said this, all I could think of was, yeah, computers need more power to run the heavy virus workload and still make them usable.

    1. Re:Faster Processors... by AllTheGoodNamesWereT · · Score: 1

      Microsoft Research has a number of projects looking at user interfaces that anticipate our needs etc. -- see for example http://research.microsoft.com/adapt.

      I'm not sure whether I find this encouraging or frightening. When you describe it the way Borkar did ("anticipate our needs and sometimes take action on our behalf") it sounds positive. However, if you describe it as "computer systems that don't always display consistent behavior," maybe it's not such a good thing.

      I remember seeing a presentation at the Hot Chips conference (www.hotchips.org) a number of years ago. At the beginning of the session, the presenter (an IBM research scientist) was having problems getting Powerpoint to work right on his notebook computer. His comment: "I love the non-determinism of Windows."

  53. An even simpler proof by lipi · · Score: 1

    If two numbers are not equal, then you can pick a number between them.

    Now try to do that with 0.99999... and 1.

  54. ROTL by Leroy_Brown242 · · Score: 1

    Now if only we could get ROTL back, my RC5-72 speeds on my P4 would keep up with every other cpu out there.

  55. Re:Infinity != Infinity, for all values of Infinit by Mac+Degger · · Score: 1

    Infinity is not a number; it's not a value...it's a concept. Go look on kuroshin and do a search for infinity...they had a article which explained it pretty decently.

    --
    -- Waht? Tehr's a preveiw buottn?
  56. Re:Infinity != Infinity, for all values of Infinit by pkhuong · · Score: 1

    lim 1/x x-> 0 = [infinity]
    lim 2/x x-> 0 = [infinity]
    lim (2/x)/(1/x) x -> 0 = [infinity]/[infinity] = 2

    Clearly, two infinites are not always equal, just like it is with any other number when we are considering a limit. In fact, I believe infinity itself is not a number, but a concept(the biggest number you can think of), and as such, we can only have numbers or values that approach it.

    Similarly, for 1 = 0.999...
    It seems that this is not the case. When the difference approaches 1/[infinity], it approaches 0+. However, 0+ != 0, and there still is a difference, an infinitely small one, but one none the less. To say that an infinitely small value is equal to zero would mean calculus can not exist. Integration and differentiation all rely on the fact that values that approach zero are not actually equal to 0, and can therefore be used as dividers(sp?) and multipliers.

    It is incredibly easy for High School teachers to create misunderstanding regarding infinity since it is such a complex concept, and that given what they teach, such complexity is often not useful. Thankfully, my HS math teachers from 9th grade up were college teachers (eh... private HS+College), and forced us to twist our brains around it as much as they could.

    --
    Try Corewar @ www.koth.org - rec.games.corewar
  57. The future of computing... by GameGod0 · · Score: 1

    Probabilistic/Quantum Computing: You buy a dual processor computer where a single CPU exists in both sockets at the same time.

  58. Hardly non-deterministic computing by Roxton · · Score: 2, Interesting

    He's not talking about non-deterministic computing. He's talking about ways to salvage the chip if one or more subcircuits don't function correctly. The article isn't very technical, but this probably alludes to having redundant circuits, possibly even taking the answer that the most redundant circuits produce.

    I'm not a smart enough man to know whether or not this is feasible. Keep in mind that introducing these redundancy checks actually increases the "length" of the circuit, increasing propogation delays. If this system works at all, you can be certain that it will be very rigidly subjected to the law of diminishing returns.

    -Roxton

    1. Re:Hardly non-deterministic computing by SuiteSisterMary · · Score: 1

      In many computer systems designed for orbit, that use, say, 486s, they'll put in three of them. Any given calc gets run through all three. Which ever answer is most popular between all three, that's the answer used.

      Why? Because 1+1 = 2, unless a stray bit of radiation or cosmic energy has flipped a bit in your processor.

      --
      Vintage computer games and RPG books available. Email me if you're interested.
  59. Finally... by HTH+NE1 · · Score: 1

    ...a processor that can has the probabilty necessary to support a DWIM (Do What I Mean) instruction!

    --
    Oh, say does that Star-Spangled Banner entwine / The myrtle of Venus with Bacchus's vine?
  60. Re:Infinity != Infinity, for all values of Infinit by Anonym0us+Cow+Herd · · Score: 1

    If A
    If 0.99999... is truly less than 1, then please name me a number that falls in between these two.

    --
    The price of freedom is eternal litigation.
  61. Re:Infinity != Infinity, for all values of Infinit by Anonym0us+Cow+Herd · · Score: 1

    Let me try that again, this time, previewing first :_)

    If A < B, then there must be some other number Z such that A < Z < B.

    If 0.9999... is less than 1, then please name me a number that falls in between the two.

    --
    The price of freedom is eternal litigation.
  62. Haven't they already done this? by El · · Score: 1

    ... and wasn't it named "The Pentium floating point bug"??? Damn it, I don't want a chip that probably gives the right answer, I want a chip that deterministically gives the right answer!

    --

    "Freedom means freedom for everybody" -- Dick Cheney

  63. Re:Infinity != Infinity, for all values of Infinit by Tyler+Eaves · · Score: 1

    Not exactly.

    The average of a and b (assuming b is the larger of the two, IF they are different at all), is a = avg = b. Note those equal to.

    --
    TODO: Something witty here...
  64. Re:Infinity != Infinity, for all values of Infinit by Anonym0us+Cow+Herd · · Score: 1

    Easy as can be. The average of those two numbers falls between them.

    So what is that number?

    From this web page, I got the following text....

    Here's another enlightening argument from Burger . I never met anybody who thought 0.999... greater than 1. So, if it's not equal to 1, it is less than 1. Let's think of the average of 0.999... and 1. As an average of any two numbers, it's greater than 0.999... but is less than 1. Can we determine its decimal expansion? Say, what is its integer part. Since it's less than 1 but greater than 0 < 0.999..., its integer part is bound to be 0. What about its first decimal digit. Since 0.9 < 0.999..., that digit must be 9. And the second one? Since 0.99 < 0.999..., the second digit must also be 9. And so on. It appears like the average of 0.999... and 1 is 0.999... If the latter is denoted as X, (X + 1)/2 = X. X + 1 = 2X. X = 1. The conclusion can't be helped.

    --
    The price of freedom is eternal litigation.
  65. 0.999 = 1 ?! NO! by ericandrade · · Score: 1

    Does 1/infinity = 0? no. It's allmost non-existant, but there's something happening. something keeps getting divided. the same with 0.9999.. it doesn't equal 1 because there's a little something missing, the same value as 1/infinity.

  66. Re:Infinity != Infinity, for all values of Infinit by Anonym0us+Cow+Herd · · Score: 1
    --
    The price of freedom is eternal litigation.
  67. Re:Infinity != Infinity, for all values of Infinit by Anonym0us+Cow+Herd · · Score: 1
    --
    The price of freedom is eternal litigation.
  68. You are wrong... by rmdyer · · Score: 1

    You err'ed with your first line...

    "Let x be 0.999..."

    This is most definitely NOT possible.
    A non-complete can never be assigned.

    That is like saying..."I'll go to infinity and bring you back some ice cream." I'm sorry, by logic the conditional expression will never be evaluated when dealing with infinity.

    +1-1

    1. Re:You are wrong... by Anonym0us+Cow+Herd · · Score: 1

      Well, if I can't assign a rational number like 1 (which is 0.99999....) or 1/3 (which is 0.33333...) then I how I can even assign an irrational, such as...

      Let x = pi

      or... Let x = sqrt( 2 )

      At least rational numbers can be expressed as the ratio of two integers (1/1) or (1/3). Irrational numbers cannot.

      See these other proofs .

      --
      The price of freedom is eternal litigation.
  69. It will work but I do not know why by ratfynk · · Score: 1
    "It's like the shift from Newtonian mechanics to quantum mechanics. We will shift from the deterministic designs of today to probabilistic and statistical designs of the future."

    So these processors will work but as quantum physics states we cannot know why, or will they just work in an alternate universe with threads to this? Very interesting but I smell some very unexpected results. Dissappearing sales revenues is first that comes to mind, rediculous developement costs is on the plate as well! I wish success but getting lead down blind vectors of developement will make it very expensive to execute, and produce product at reasonable costs.

    --
    OH THE SHAME I fell off the wagon and use sigs again!
  70. then the improbability computer will appear... by kipple · · Score: 1

    ...making windows completely reliable. and the only blue screen will display a huge "42".

    --
    -- There are two kind of sysadmins: Paranoids and Losers. (adapted from D. Bach)
  71. Re:NO! NO? NO?! by ericandrade · · Score: 1
    It's a question about survival of values. They can't dissapear just because they are drowned in a large string of numbers which don't seem to have an end.

    Yes, the imaging with infinity is a bit flawed, but here's a logical run thru...

    1/10 = 0? no

    1/90909 = 0? no

    1//(10^infinity) = 0? no

    but it's pretty close...

    why would the inverse, 0.99999 = 1 if it just isn't. there's that really smally thingy that isn't there.

  72. 0.999...?1 by ericandrade · · Score: 1
    1/10 = 0? no

    1/90909 = 0? no

    1/(10^infinity) = 0? no

    but it's pretty close...

    why would the inverse, 0.99999 = 1 if it just isn't. there's that really smally thingy that isn't there.

    1. Re:0.999...?1 by ericandrade · · Score: 1
      this:

      Is that .9 repeating? If so, there's a 100% possibility of 1 + 1 = 1.9999...

      should read this:

      Is that .9 repeating? If so, there's a 99.999...% possibility of 1 + 1 = 1.9999...

  73. Illusion by nitpick1 · · Score: 1

    1/3 is practically unrepresentable as a decimal in base 10. Therefore we've all agreed to the illusion that .33333... is the same thing as 1/3. We stop calculating because we all have much better things to do. Otherwise actually slcing a pizza would take forever, which it demonstrably does not. Unlike this discussion...

  74. Re:NO! NO? NO?! by jcast · · Score: 1

    1/10 = 0? no

    1/90909 = 0? no

    1//(10^infinity) = 0? no

    BEEP! Logic error! The behavior of a limit cannot in general be determined from the behavior of a limiting sequence. (After all, 1/3 is the limit of a sequence of numbers expressable as finite decimals and sqrt 2 is the limit of a sequence of rational numbers.) Don't feel bad though, I swear to God Hawking makes the same mistake in A Short History of Time
    --
    There are reasons why democracy does not work nearly as well as capitalism.
    -- David D. Friedman
  75. Re:Infinity != Infinity, for all values of Infinit by SirLestat · · Score: 1

    About the:
    lim 1/x x-> 0 = [infinity]
    lim 2/x x-> 0 = [infinity]
    lim (2/x)/(1/x) x -> 0 = [infinity]/[infinity] = 2

    First 1/x x-> 0 isn't equal to infinity. Its undefined because x -> 0+ = infinity and x -> 0- = -infinity.

    Lets say we cange both with 0+
    But (2/x)/(1/x) x -> 0 is equal to [infinity]/[infinity] but that gets you nowhere. That doesn't mean that because the equation can't be resolved in that state that the real solution (2) is equal to that unsolvable state.
    It is just plain illogical to compare infinite with infinite. It's not smaller, not larger, not even equal.

  76. Funny.... by voxel · · Score: 1

    I introduced probability into my Intel CPU very easily. I overclocked it... Now I have a good probability of it doing funny things and executing code in an order no one can predict! Hmm, talk about true random numbers!

    - Voxel

    --
    Modesty is one of life's greatest attributes
  77. Limiting sequence... by ericandrade · · Score: 1

    The fact that the behaviour of a limiting sequence doesn't determine a particular behaviour of a limit is really interesting! But does it apply to limiting eveything over 0?

    1. Re:Limiting sequence... by jcast · · Score: 1

      The fact that the behaviour of a limiting sequence doesn't determine a particular behaviour of a limit is really interesting! But does it apply to limiting eveything over 0?

      What do you mean by ``limiting everything over 0''? What sequence or class of sequences are you limiting? Clearly, a decreasing sequence of strictly positive values can limit to 0.
      --
      There are reasons why democracy does not work nearly as well as capitalism.
      -- David D. Friedman
  78. Infinite Hotel by zCyl · · Score: 1

    So the question herein lies, what happens when you shift an infinitely repeating decimal 1 place to the left? Does it magically gain a new decimal place on the right? This may happen, but that violates the basic laws of normal numbers.

    Yes, it does. The strangeness of this principle of infinities is more clearly seen when you consider the problem of the infinite hotel.

    Suppose there is a hotel with a countably (each room is numbered, 1 through infinity) infinite number of rooms, and every room is full. In the middle of the night a bus stops by with an infinite number of passengers. The bus driver asks the hotel manager if he has room for the new guests even though the hotel is full. The hotel manager says, "Sure, there's plenty of room." How does he fit the new guests?

    There are a lot of solutions, but the simplest is just that every guest already in the hotel moves from room n to room 2*n. Then all the new guests fill in the odd numbered rooms (of which there are an infinite number).

    This probably violates your common sense intuition, but that's because common sense intuition deals with finite numbers of things. If you think about it for a while, it makes sense for this to work in a hotel with an infinite number of rooms.

    The idea of shifting all the 9's left one spot is the same thing. There are an infinite number of 9's to the right of them to fill in all the previous spots. If you go to the infinite hotel, kick out the guest in room 1, and ask all the other guests to move from room n to room n-1, then all your rooms are full again, even though you have one less guest.

  79. Article summary by geekee · · Score: 1

    Basically this guy is saying that the variance of transistor parameters will increase in the futute, so more attention must be paid to circuit design to avoid producing a lot of chip that perform poorly. People already use monte carlo sims to determine the effects of process variation in analog designs, and to bound digital design performance. Digital circuit designers must look at new ways to overcome the issue of transistor performance variations, rather than accepting them as inevitable is the message, I believe.

    --
    Vote for Pedro
  80. QUICK HIDE YOUR PROCESSOR by StormyWeather · · Score: 1

    Before Zaphod Beeblebrox steals it!

  81. Can I Connect To The Internet Today? by Master+of+Transhuman · · Score: 1


    Probably.

    Maybe.

    Will I be /.'d today?

    Oh, that's a certainty.

    --
    Richard Steven Hack - This sig is TOO GODDAMN SHORT TO DO ANYTHING USEFUL WITH! MORONS!
  82. Re:Infinity != Infinity, for all values of Infinit by IncohereD · · Score: 1

    If 0.99999... is truly less than 1, then please name me a number that falls in between these two.

    I'm not saying you're wrong - I in fact agree with you - I'm just saying your proof was a little sketchy.

  83. Re:Still stuck on middle school math eh? by floW+enoL · · Score: 1

    My god. Have *any* of you who are posting your "proofs" and "arguments" *ever* taken a college math class? Listen up, because I'll only explain this once.

    A rational number is defined as a pair (p, q), where p is an integer and q is a non-zero natural number. Notice there is *nothing* about decimal expansions or repeading decimals mentioned in this definition.

    However, for the sake of convenience, given a rational number (p, q) and a base b (a natural number > 1), one could define a unique b-ary expansion e = {bn...b3b2b1.d1d2d3...dn, r} where bi and di are natural numbers in {0, 1,...,b - 1}, bn != 0, dn != 0, and r = 0 or 1. (r represents whether the last digit dn repeats or not).

    The question is, what other conditions do we need to impose on e to make it unique? Guess what: the only time when two decimal expansions arise for the same number is when (p / q) is a whole number w, where the two expansions are {(w - 1).9, 1} (repeating nines) and {w, 0}. So to ensure uniqueness, we *define* the decimal expansion to be (w, 0}.

    Simple. Silly-ass questions like HURR DOES .9999... EQUAL 1? can be simply replied to with "That is a stupid and irrelevant question." Equality on rational numbers is defined solely in terms of p and q and it just so happens that for convenience, equality can be considered in terms of the decimal expansions *unless* you try to fuck things up by trying to calculate and "prove" things about meaningless strings of symbols like .99999.... By the way, note that all of the above "proofs" that .9999... = 1 are fallacious, simply because you cannot assume that you can operate on infinite decimal expansions like finite decimal expansions.

    So in summary, asking: "Does .9999... = 1?" is just as silly as asking "Does 1 / 0 = infinity?" or "Is the sentence 'All your base are belong to us' passive voice or acive voice?" Why? Because the symbol ".9999..." is meaningless in terms of the analysis of real numbers, just as the symbol "=" has no meaning when combined with the concepts of "1 / 0" (which is also meaningless) and "infinity" (which is also meaningless in this context).

  84. Good for Bill Gates by jigyasubalak · · Score: 1

    Now he can say that your windows crash has 50% to do with your CPU choosing not to compute.

    --
    The best planning can be done after the project completes.
  85. C'est La Poo by NiTRiX · · Score: 1

    Theorems are crap. It's just a word used to creatively disguise people as having some sort of grasp on the way things work. The fact that this world survives ignorance by believing in a defined and unchallenged form of mathematics is astounding. Who, or what, is to determine the essence of mathematics? Mathematics itself is a theorem. It's a concept, a religious belief even. It can be twisted and distorted in so many fashionable ways until finally you're left to question the very subsistence of whatever you happen to be putting into equation.

    Math, my fellow slashdotters, is a tool much like the Swiss Army Knife. While highly functional and infinitely flexible, it can't save you from choking on those delightful honey-roasted peanuts.

    --


    on the sixth day God created man.
    on the seventh day, man returned the favor.
  86. Re:0.999 = 1 ?! NO! by Tom7 · · Score: 1

    Wrong. There are several proofs of this here. It's counter-intuitive, but it's what you get when you accept things like infinite decimal expansions.

  87. Re:Trinary Logic... a serious issue... by Hentai · · Score: 1

    Actually, though, that won't work probabilistically. The actual functions should be:

    #define NOT(x) (1-x)
    #define AND(x, y) (x * y)
    #define OR(x, y) NOT(AND(NOT(x), NOT(y)))

    This is how statistics works, anyways.

    --
    -Hentai [in vita non pacem est]
  88. Re:Trinary Logic... a serious issue... by fireboy1919 · · Score: 1

    Do you mean "this is how probability works?"

    So not means "if the statistical likelihood is 40%, then the likelihood of the reverse is 60%," and "if the statistical likelihood of A is 50% and B is 50%, then the chance of getting in both is 25%."

    Fuzzy logic isn't the same as likelihood. It's about sets, and it turns out to be useful more often that likelihood - enough so to make machines specifically to do fuzzy operations. To put it into perspective, fuzzy logic answers the question, what the amount that an element is in this set, if the boundary of the set is not clearly defined. This is especially useful when talking about quantum particles, for instance.

    So if you've got an item that is 50% in set A, and 70% in set B, then it there it's 50% in A and B, and 70% in A or B.

    --
    Mod me down and I will become more powerful than you can possibly imagine!
  89. Re:Still stuck on middle school math eh? by RobertB-DC · · Score: 1

    Simple. Silly-ass questions like HURR DOES .9999... EQUAL 1? can be simply replied to with "That is a stupid and irrelevant question.

    Finally, an answer I can agree with!

    (But deep down inside, I feel like 0.999... is still != 1.0, though it feels very, very close)

    --
    Stressed? Me? Of course not. Stress is what a rubber band feels before it breaks, silly.