Re: You will be sorry you asked: was Re: Counting
Trent Shipley wrote: ... I did not recognize this as a simple pigeonhole problem of 2n choose n, so I cross posted to lists where I thought I might get help and would not be accused of an off topic post. David Hobby, do you live in the Phoenix area? No, upstate New York. This is a design issue. Since EEs do not want metastability screwing up their designs, they minimize the window where metastability can occur. Working from scratch, one could MAXIMIZE a flip-flop's metastable window. If you take it too far, it won't BE metastable any more. It's a matter of definitions. It shouldn't be hard to design a device with three states, where the middle one has any desired degree of stability. Now with an ordinary latch we have an input control lead (it sets the latch to remember current state or receive a new state) an data input lead (if the control is T, then the data lead sets the latch to either T or F), and an output lead that just tells you whether the latch is remembering T or F. More literally the latch has TWO output leads, but if one is T the other must be F, so we gain no data from the second one. So one is the complement of the other. O.K.. I had my Digital Design textbook stolen partway through the semester, but as i remember it a latch was mathematically: A[out] - B[in], Output B[out] - A[in] Sorry, I can't decipher this. What is - supposed to mean, and while we are at it, what are A and B? So, if we have a latch with two output leads we should be able to get it to act as the brain of a trivial qpu: it will perform the not so interesting operation of ADD 1. Unfortunately if we have two latches we do not get either a 2 or a 4 q-bit computer. We just have two weak little 1 q-bit qpu's. They cannot be quantum entangled. I conjecture that a three component latch (3-latch) will not be stable, but have a tendency to ring. (Note this is a K-3 graph.) A[o] - B[i],C[i], Out B[o] - A[i],C[i], Out C[o] - A[i],C[i], Out A 4-latch, however, should be stable. Like a 2-latch I conjecture it should only output bits so that the number of Fs equals the number of Ts. A[o] - B[i], C[i], D[i], Out B[o] - A[i], C[i], D[i], Out C[o] - A[i], B[i], D[i], Out D[o] - A[i], B[i], C[i], Out That is 4 pick 2. The conjecture is that THIS 4-latch can be built and used in a QPU (designed by someone more clever than my self). It will have 6 states. Unfortunately, mapping these states to numbers is not as straightforward as in a determinate ALU, but counting to 3 should be straight forward. I don't understand exactly how the internal parts of your 4-latch are hooked up. You have some sort of mix of excitatory and inhibitory connections? But it should be possible to design something that always has two output leads high, where all six possible pairs of leads might be high. Indeed, if one can build a 4-latch QPU then one should, But it's quantum, too? ... Imagine we have a device that must produce output in pairs so that for every F there is a T. It cannot produce an odd number of outputs. How many symbols can be represented by such a machine's output (how big is the output's alphabet) for any output block size 2N that implicitly includes N F's and N T's. Now aren't you sorry you asked? ... Well, maybe a little... I don't understand the question completely. But if these are just C(2n,n), you can use Sterling's formula to approximate the factorials, and get a very accurate estimate. By the way, the estimate for C(2n,n) is approximately (pi * n)^(-.5) * 4^n ---David ___ http://www.mccmedia.com/mailman/listinfo/brin-l
You will be sorry you asked: was Re: Counting
] - A[i], B[i], C[i], Out That is 4 pick 2. The conjecture is that THIS 4-latch can be built and used in a QPU (designed by someone more clever than my self). It will have 6 states. Unfortunately, mapping these states to numbers is not as straightforward as in a determinate ALU, but counting to 3 should be straight forward. Indeed, if one can build a 4-latch QPU then one should, in theory, be able to use the same solid-state technology to build a QPU of arbitrary size with an even number of output leads. == Abstraction: == For the moment, forget about stupid real-world things like latches and quantum computers. Those are toys for experimental physicists at best and dirty engineers at worst. Let's do some mathematical computer science, albeit inspired by the forgoing speculation. Imagine we have a device that must produce output in pairs so that for every F there is a T. It cannot produce an odd number of outputs. How many symbols can be represented by such a machine's output (how big is the output's alphabet) for any output block size 2N that implicitly includes N F's and N T's. Now aren't you sorry you asked? ... 2th item: 4 places: 0011 0101 0110 1001 1010 1100 6 parity combinations, sum of each row is 2 2^4 possible (16) 4: 6 :8:16 This is C(4,2) = 4!/(2! * 2!) = 6. You are picking 2 places out of 4 possible ones to place the 1s. 3th item: 6 places: 000111 001011 ... 110100 111000 20 parity combinations (if my spreadsheet is right) And this is C(6,3) = 6!/(3! * 3!). 4th item: 8 places: 00010111 . . 11101000 70 parity combinations (if spreadsheet right) And C(8,4), etc. ... NOTE each binary number (or representation of state) has an equal number of 'zeroes' and 'ones'. Because of the parity of zeroes and ones we are only interested in binary numbers with an even number of places. The idea that the sum of symbols for each number equals the rank of the item in the sequence is not proven, but is evident. --- Question: For a given item n in the sequence N, how many parity number combinations will occur in the resulting set of all modulo 2^(2n) binary representations? (How many parity combinations for a given item N?) Is number of binary parities bounded underneath by 2^(2[n-1]) for all N? I don't understand the question completely. But if these are just C(2n,n), you can use Sterling's formula to approximate the factorials, and get a very accurate estimate. ___ http://www.mccmedia.com/mailman/listinfo/brin-l ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Counting
(Replies to plug-discuss preferred) best viewed monospaced 0th item: 0 places: 0 parity combinations, 2^0 possible nil:nil: 1 :1 1th item: 2 places: 01 10 2 parity combinations, sum of each row is 1, 2^2 possible 1:1: 2 :4 2th item: 4 places: 0011 0101 0110 1001 1010 1100 6 parity combinations, sum of each row is 2 2^4 possible (16) 4: 6 :8:16 3th item: 6 places: 000111 001011 001101 001011 001110 010011 010101 010110 011001 011010 011100 100011 100101 100110 101001 101010 101100 110001 110010 110100 111000 20 parity combinations (if my spreadsheet is right) sum of each row is 3 2^6 possible 16: 20 :32:64 4th item: 8 places: 00010111 . . 11101000 70 parity combinations (if spreadsheet right) sum of each row is 4 2^8 possible 64: 70 :128:256 This is beyond my initially weak, and now rusted, skill at counting and combinatorics. NOTE each binary number (or representation of state) has an equal number of 'zeroes' and 'ones'. Because of the parity of zeroes and ones we are only interested in binary numbers with an even number of places. The idea that the sum of symbols for each number equals the rank of the item in the sequence is not proven, but is evident. --- Question: For a given item n in the sequence N, how many parity number combinations will occur in the resulting set of all modulo 2^(2n) binary representations? (How many parity combinations for a given item N?) Is number of binary parities bounded underneath by 2^(2[n-1]) for all N? ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Counting
Trent Shipley wrote: What the I can't find any context for this. But see comments, below. ---David ... 2th item: 4 places: 0011 0101 0110 1001 1010 1100 6 parity combinations, sum of each row is 2 2^4 possible (16) 4: 6 :8:16 This is C(4,2) = 4!/(2! * 2!) = 6. You are picking 2 places out of 4 possible ones to place the 1s. 3th item: 6 places: 000111 001011 ... 110100 111000 20 parity combinations (if my spreadsheet is right) And this is C(6,3) = 6!/(3! * 3!). 4th item: 8 places: 00010111 . . 11101000 70 parity combinations (if spreadsheet right) And C(8,4), etc. ... NOTE each binary number (or representation of state) has an equal number of 'zeroes' and 'ones'. Because of the parity of zeroes and ones we are only interested in binary numbers with an even number of places. The idea that the sum of symbols for each number equals the rank of the item in the sequence is not proven, but is evident. --- Question: For a given item n in the sequence N, how many parity number combinations will occur in the resulting set of all modulo 2^(2n) binary representations? (How many parity combinations for a given item N?) Is number of binary parities bounded underneath by 2^(2[n-1]) for all N? I don't understand the question completely. But if these are just C(2n,n), you can use Sterling's formula to approximate the factorials, and get a very accurate estimate. ___ http://www.mccmedia.com/mailman/listinfo/brin-l
[aztechlist] Counting
(Replies to plug-discuss preferred) best viewed monospaced 0th item: 0 places: 0 parity combinations, 2^0 possible nil:nil: 1 :1 1th item: 2 places: 01 10 2 parity combinations, sum of each row is 1, 2^2 possible 1:1: 2 :4 2th item: 4 places: 0011 0101 0110 1001 1010 1100 6 parity combinations, sum of each row is 2 2^4 possible (16) 4: 6 :8:16 3th item: 6 places: 000111 001011 001101 001011 001110 010011 010101 010110 011001 011010 011100 100011 100101 100110 101001 101010 101100 110001 110010 110100 111000 20 parity combinations (if my spreadsheet is right) sum of each row is 3 2^6 possible 16: 20 :32:64 4th item: 8 places: 00010111 . . 11101000 70 parity combinations (if spreadsheet right) sum of each row is 4 2^8 possible 64: 70 :128:256 This is beyond my initially weak, and now rusted, skill at counting and combinatorics. NOTE each binary number (or representation of state) has an equal number of 'zeroes' and 'ones'. Because of the parity of zeroes and ones we are only interested in binary numbers with an even number of places. The idea that the sum of symbols for each number equals the rank of the item in the sequence is not proven, but is evident. --- Question: For a given item n in the sequence N, how many parity number combinations will occur in the resulting set of all modulo 2^(2n) binary representations? (How many parity combinations for a given item N?) Is number of binary parities bounded underneath by 2^(2[n-1]) for all N? ==AzTechList AzTechList Discussion List http://www.aztechlist.org Subscribe by sending an email to: [EMAIL PROTECTED] Report list related problems/concerns: [EMAIL PROTECTED] Become a Member and/or Sponsor: http://paypal.azipa.org * Co-sourcing with Inflow is the future - http://www.inflow.com * PublicOpinion.com. Share with the world - http://www.publicopinion.com * Contactlink Keeps You Connected! - http://www.contactlink.com * $7.85 Domain Name Registrations/Transfers - http://www.domaindo.com AzTechList== Yahoo! Groups Links * To visit your group on the web, go to: http://groups.yahoo.com/group/aztechlist/ * To unsubscribe from this group, send an email to: [EMAIL PROTECTED] * Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/ ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Counting
There have been 301 conversations since this first message I received on gmail. I participated in 63 of them. I was curious about the number of emails but gmail counts conversations based on subject line instead. So 60 conversations a week with from 1 to 69 messages. Gary Big Mouth Denton #1 on Google for easter lemming notebook 1st! On Mon, 26 Apr 2004 16:47:47 -0500, The Fool [EMAIL PROTECTED] wrote: A terrorist targets liberals ___ http://www.mccmedia.com/mailman/listinfo/brin-l
