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From: Peter T. Breuer on 24 Sep 2005 17:52 Unruh <unruh-spam(a)physics.ubc.ca> wrote: > "Peter T. Breuer" <ptb(a)oboe.it.uc3m.es> writes: >>Michael Heiming <michael+USENET(a)www.heiming.de> wrote: >>> In comp.os.linux.setup Peter T. Breuer <ptb(a)oboe.it.uc3m.es>: >>>> Jean-David Beyer <jdbeyer(a)exit109.com> wrote: >>>>> Peter T. Breuer wrote (in part): >>>>>> Here. Here's a chance for you. Next number in this series, please: >>>>>> >>>>>> 1 2 4 6 10 12 .. >>>>>> >>>>>> Answers by next friday. > There are an infinite number of answers. Any number from minus infinity to > infinity can be justified as the next number. I'll even tell you that the series is not algebraic, but is arithmetic (I don't mean "linear" by "arithmetic", but rather that its definition falls in the realm of arithmetic). Peter
From: Unruh on 24 Sep 2005 18:20 "Peter T. Breuer" <ptb(a)oboe.it.uc3m.es> writes: >Unruh <unruh-spam(a)physics.ubc.ca> wrote: >> Michael Heiming <michael+USENET(a)www.heiming.de> writes: >>>In comp.os.linux.setup Peter T. Breuer <ptb(a)oboe.it.uc3m.es>: >>>> Michael Heiming <michael+USENET(a)www.heiming.de> wrote: >>>>> In comp.os.linux.setup Peter T. Breuer <ptb(a)oboe.it.uc3m.es>: >>>>>> Jean-David Beyer <jdbeyer(a)exit109.com> wrote: >>>>>>> Peter T. Breuer wrote (in part): >>>>>>>> Here. Here's a chance for you. Next number in this series, please: >>>>>>>> >>>>>>>> 1 2 4 6 10 12 .. >>>>>>>> >>>>>>>> Answers by next friday. >>>>> 22 >>>> No, but that's not a bad attempt. It's a later member of the series all >>>> right! >>>So it's "18" (euler phi function or so), you used? >18 is also a later member of the series, but it is not the next member. >> Or 14 (base 8) >?? I.e 12? We already had 12. Or do you mean 16? No I mean that the sequence is written in base 8 and the next number is 14 in base 8. In base 10 your sequence is 1 2 4 6 8 10 with the next number being 12 in base 10.
From: Unruh on 24 Sep 2005 18:30 "Peter T. Breuer" <ptb(a)oboe.it.uc3m.es> writes: >Unruh <unruh-spam(a)physics.ubc.ca> wrote: >> "Peter T. Breuer" <ptb(a)oboe.it.uc3m.es> writes: >>>Michael Heiming <michael+USENET(a)www.heiming.de> wrote: >>>> In comp.os.linux.setup Peter T. Breuer <ptb(a)oboe.it.uc3m.es>: >>>>> Jean-David Beyer <jdbeyer(a)exit109.com> wrote: >>>>>> Peter T. Breuer wrote (in part): >>>>>>> Here. Here's a chance for you. Next number in this series, please: >>>>>>> >>>>>>> 1 2 4 6 10 12 .. >>>>>>> >>>>>>> Answers by next friday. >> There are an infinite number of answers. Any number from minus infinity to >> infinity can be justified as the next number. >Well, justify "42" to me then! 1-211/20*x+4459/180*x^2-905/48*x^3+959/144*x^4-263/240*x^5+49/720*x^6 >(the series is actually perfectly well defined). Of course it is. You wrote it down up there. >Peter
From: Peter T. Breuer on 24 Sep 2005 18:31 Unruh <unruh-spam(a)physics.ubc.ca> wrote: > "Peter T. Breuer" <ptb(a)oboe.it.uc3m.es> writes: >>Unruh <unruh-spam(a)physics.ubc.ca> wrote: >>> Michael Heiming <michael+USENET(a)www.heiming.de> writes: >>>>In comp.os.linux.setup Peter T. Breuer <ptb(a)oboe.it.uc3m.es>: >>>>> Michael Heiming <michael+USENET(a)www.heiming.de> wrote: >>>>>> In comp.os.linux.setup Peter T. Breuer <ptb(a)oboe.it.uc3m.es>: >>>>>>> Jean-David Beyer <jdbeyer(a)exit109.com> wrote: >>>>>>>> Peter T. Breuer wrote (in part): >>>>>>>>> Here. Here's a chance for you. Next number in this series, please: >>>>>>>>> >>>>>>>>> 1 2 4 6 10 12 .. >>>>>>>>> >>>>>>>>> Answers by next friday. >>>>>> 22 >>>>> No, but that's not a bad attempt. It's a later member of the series all >>>>> right! >>>>So it's "18" (euler phi function or so), you used? >>18 is also a later member of the series, but it is not the next member. >>> Or 14 (base 8) >>?? I.e 12? We already had 12. Or do you mean 16? > No I mean that the sequence is written in base 8 and the next number is 14 Oh, I see. You think it's base 8 because there is no digit beyond 7 (6) so far. Reading it as base 8, it indeed goes 1 2 4 6 8 10 decimal ...! > in base 8. In base 10 your sequence is 1 2 4 6 8 10 with the next number > being 12 in base 10. Why is the first number "1", then, and not "0"? No, it's not a base change trick. The numbers given are in decimal. I can give the next humber if it helps. I suppose the sequence might be described as "computable", as well as "not algebraic" and "in the arithmetic realm". Peter
From: Enrique Perez-Terron on 24 Sep 2005 19:09
On Sat, 24 Sep 2005 22:55:39 +0200, Peter T. Breuer <ptb(a)oboe.it.uc3m.es> wrote: > Peter T. Breuer <ptb(a)oboe.it.uc3m.es> wrote: >> Jean-David Beyer <jdbeyer(a)exit109.com> wrote: >>> Peter T. Breuer wrote: >>>> Michael Heiming <michael+USENET(a)www.heiming.de> wrote: >>>>> In comp.os.linux.setup Peter T. Breuer <ptb(a)oboe.it.uc3m.es>: >>>>>> Jean-David Beyer <jdbeyer(a)exit109.com> wrote: >>>>>>> Peter T. Breuer wrote (in part): >>>>>>>> >>>>>>>> 1 2 4 6 10 12 .. >>>>>>>> >>>>>>>> Answers by next friday. > >>> I liked the puzzel like this where a bunch of small integers were >>> given, and >>> they seemed to have a pattern, but unless you were a resident of >>> Manhattan >>> Island, you would not recognize they were the stops of the 8th Avenue >>> Subway. > >> Don't worry, this puzzle is purely numeric/arithmetic and no tricks >> (there, I shouldn't have given you the clue - now you have so much less >> to consider!). > > I'll add by way of background ... this puzzle was set me by the then > world (math) Olympiad champion when I walked into Trinity bar as an > eighteen year old and was introduced to him as "the scholar at Clare". > > He wasn't much good at conversation. Or walking down the street come to > that. But the pressure of having to think of the answer to keep face then > has meant that the problem has stayed embossed in my head all these > years. > > Peter Thinking aloud.... We are seeking a function f defined over the positive integers N. For each 'n' in N, f(n) is expressible as an arithmetic formula using the operations of addition, substraction, multiplication, exponentiation, and possibly division and remainder, and functional composition; where the operands are n, fixed members of N, and f(i) for i < n. (In the latter case 'i' would have to be subject to similar restricions, i must be expressible in a similar way from n, constants and f(j), where j is recursively expressible in a similar way.) The goal is really to find an f that is expressible with the least number of primitive operations (+ - * ^ / mod f()), such that (f(n))n=1,...,6 is the sequence (1 2 4 6 10 12). Perhaps another operation has to be included: choice. That is, e.g "if x is odd, y, else z". Then the condition should be expressible similarly as the other subexpressions of f(n), and with the relational operators <, =. (Other relational operators >, >= etc are expressible with <, = by inverting the order of the operands, and by exchanging the if and else branches. The simplest possible formula is that which names a constant. E.g., 0) f(n) = 3 (for all n). The next simplest formula involes one operation: eg. (never mind if some of the examples do not make sense for the problem) 1a) f(n) = if n=1, 1, else 2 1b) f(n) = n + 1 1c) f(n) = n - 1 1d) f(n) = n * 1 1e) f(n) = n / 1 1f) f(n) = n mod 1 1g) f(n) = n ^ 1 1h) f(n) = f(1) The next simplest formula involves two operations, eg. 2aa f(n) = if n=1 1, else if n=2 2, else 4 2ab f(n) = if n=1 1, else n+1 2ac f(n) = if n=1 1, else n-1 etc, ad nauseam. There is an upper limit to the complexity: f(n) = if n=1, 1, else if n=2, 2, else if n=3, 4, else if n=4, 6, else if n=5, 10, else 12 in five choice operations. (According to this solution, f(7) is 12.) f(n) = if n = 1, 1 else if n < 5, n*2 - 2 else n*2 has two choices, two multiplications, and one substraction, that's five operations. Rearranging, f(n) = if n=1, 1 else 2 * if n < 5, n - 1, else n has two choices, one multiplication, and one substraction, in all four. Any simpler? If the choice involves any more than evaluating a relatinal operator then those operations also count. To make it more stringent: When we want to compute f(n), we have access to all constants in Z (the integers), and to f(i) for each fixed i=1, ..., n-1. Say these are the memebers of a set S(n; 0). In these sets we must distinguish between a fixed number k in Z and the same number k if k=f(i) for allowable i. We also distinguish k=n. This boils down to making S(n;0) a set of formulas. S(n; 0) = { the names of the members of Z, the formulas "f(i)" where i names a positive integer less than n, and "n", taken as another name of n} Now we define the sets S(n;k) for k>0. The members of S(n;k) are either - members of S(n;k-1), or - of the form "(a op b)" where "a", "b" are in S(n;k-1) and "op" is one of {+ - * / (i.e. integer division) mod ^}, or - "f(a)" where a is in S(n;k-1) AND a has its value in [1..n-1], or - "if a relop b, c, else d", where a, b, c, d are all members of S(n; k-1), and relop is one of {<, =} Now the problem becomes to select a member X of S(7;k) for the smallest possible value of k, such that by replacing the numbers 1 through 6 for "n" in the formula X, the values 1, 2, 4, 6, 10, 12 result. Then the value of X with 7 for "n" is the answer. But at this time, we have another restraint, it was revealed in another post that the number 22 must occur as a value of X when some value higher than 7 is substituted for "n" in X. Notice that the suggested 4-step solution above yields 22 for n=11. It is also of course a question if the set of "primitive" operations should be different. Perhaps this IQ (or IQ + math) test here takes into consideration such knowledge as "7 is a prime", so that "if 'a' is a prime, 'b', else 'c'" should be a a member of S(n; k) whenever 'a', 'b', and 'c' are members of S(n; k-1). Perhaps even "The next prime after x" should be available in S(n; k) whenever 'x' is a member of S(n; k-1) (In the previous text I have been a bit inconsistent with the use of single quotes. The motivation has been to avoid reading 'a' as the indefinite article in English.) One could also argue that in IQ test context, a choice like if n < 5, the A, else B should count as two steps, because you need to figure out both the choice per se, and the condition. Yet another way of evaluating the simplicity of the solution is to count the number of computations done in computing all the numbers f(1) through f(7). This means that the case f(1) only counts as one step since the else branch of the first choice does not influence the result. With this metric, the Fibonacci sequence if n=1, 1, else if n=2, 1, else f(n-1) + f(n-2) has a cheap start, with one point for f(1) and two points for f(2), while all other values are penalized with the to initial choices. It is perhaps reasonable to use a primitive form "Start with the sequence 1, 1, and then for n>2, f(n) = X" together with a metric that costs just one point for this form, and then adds the number of points in the formula X. (This metric does not add points for each value of f computed). But then it would be a cheap solution to Peters challenge: "Start with the sequence 1, 2, 4, 6, 10, 12, and then for n > 6, f(n) = 22" with a cost of one! That is not desirable. We should require that the members of the fixed start are referenced by the formula X for at least some values of n <= 7. -Enrique |