what to do with the data?
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clm
Posted on: Fri Mar 02, 2012 12:37 pm
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

Re: what to do with the data?
giulio wrote: Patrick, concerning the calculation of difficulty of a puzzle, as quoted by clm from your paper, you say: Quote: “The difficulty of a puzzle is estimated by averaging the natural logarithm of the number of possible permutations for each row and each column (after applying the restrictions imposed by each cage), divided by the size of the puzzle (for example, the puzzle in Figure 1 has a rating of 0, the puzzle in Figure 2 a rating of 62).” ... 288x is obtained by multiplying the prime factors 2222233. The resulting combinations are 9841, 9822, 9442, 8661, 8632, 8433, and 6642. This means that the 2 cells in the top row can contain one of the following 17 combinations: 98, 94, 92, 91, 86, 84, 83, 82, 81, 64, 63, 62, 61, 43, 42, 41, and 32, which generate 17 x 2 = 34 permutations. ... Interesting concept for evaluating the "rating", pse add "288x" = [3,4,4,6], the combination 6443, making a total of 8 combinations, though it does not modify the total of 17 combinations you show.




giulio
Posted on: Fri Mar 02, 2012 12:51 pm
Posts: 54 Joined: Thu Nov 03, 2011 8:52 am

Re: what to do with the data?
Oops! I overlooked 6443... Almost always one can replace 62 with 43... clm, I was just describing what I think Patrick uses to rate the puzzle difficulty. Let see whether he confirms it.




clm
Posted on: Fri Mar 02, 2012 4:09 pm
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

Re: what to do with the data?
pnm wrote: starling wrote: clm, on the solver rating, I was assuming there was some factor it was multiplied by so that you didn't just end up with a bunch of meaninglessly close decimals. True, I may have forgotten to mention a factor of 10 in there, I'll doublecheck. Patrick Yes, starling, the rating is an integer, so some additional rounding function must be included in the formula. The Patrick’s paper is neat, clearly presented, bright and interesting (apart of the big effort), the little precision I made (perhaps with the hope that Patrick includes a more extended formula in the next edition of the paper) is a consequence of some “deformation” I have since child, some tendency to always question the numerical conclusions, very reluctant to accept the results (by the way, I identify myself with you when you revisit your past loving numbers, I suppose like most calcudokers here, when I was 6 years old I loved chess, I used to calculate the successive powers of 2 to perhaps 2 ^ 25 or so, the Rubik’s cube … and one day I lost myself from my parents in a big city because I was studying the shape of the stones but I could return to the hotel, one kilometer away, just remembering those shapes rather than asking for the name of the hotel, my parents had gone to the police … ). Let me have some digressions just to relax the debate. Actually I am an extremely relativistic person, I do not believe in almost nothing, for instance, in the big bang, etc. (I am quite sure Hawking and Penrose do not believe either); around 500 years ago very bright minds were calculating how many angels could occupy the equivalent of a squared millimeter, ridiculous right? unless the intention was to measure “the size of a pure spirit”. And more or less in 1890 some very bright journalist wrote that with the actual curve of increase of car horses in Madrid’s city, in 1950, the debris of the animals will reach the second floor of the full town (and he made consistent calculations). And, in 1900, more or less, the duration of the Sun was going to be 10,000 years considering its mass, the emitting energy, the fact of being a “middle life” star and its nature, “a good anthracite” to burn. When you make tourism they always tell you about numbers … the number of stones in the Egypt’s pyramids, the weight of the Vatican, how tall is the Eiffel’s Tower, the Empire State, the weight of the whole Manhattan’s Island, the weight of the buildings, how long is a river, how long is the Chinese Wall, the number of islands around a bay, etc., other times the average of beer per capita in a determined country … One day, visiting Nuremberg, the guide commented to a group of tourists that the city was producing every year “one billion” of those very small typical Nuremberg’s sausages, I inmediately said “you mean one billion … european … or one american billion?” (as we know in Europe one billion is 10 ^ 12 so one thousand times the american billion), “one billion … european, of course”, then I said “that’s impossible”, the answer was “it’s in the books”, I said “suppose a sausage weights 10 grams, make the calculations please”, everybody was laughing but after some time everybody was calculating and the conclusion was that it was impossible; but this a very typical misunderstanding when translating and retranslating info. The natural logarithm reduces drastically the size of the operand, i.e., ln(1 googol) = 230.26 … (being the googol = 10 to the power of 100, so higher than the total number of particles in the Universe, estimated to be in the order of 10 ^ 80), which divided by 9 equals around 25, and we have “solver ratings” > 100 !. That took my attention. The total number os sudoku grids is (generously in the comparison) in the order of the Avogadro number (with the very big numbers we use to say that 10 ^ 30 approx equals 10 ^ 31). In the raw calculation I made at the end of my previous post, (9!) ^ 9 = 1.09 … x 10 ^ 50 so more or less the squared root of a googol. This number decreases quickly when you start introducing restrictions (the column condition first) and the sudoku condition (digits 1 to 9 all different in every one of the 9 3x3 boxes). In the thread about the “strange cages” we saw the big amount of possibilities (permutations) that the big cages produce. Logically, when we apply more restrictions (many cages, 34 in today’s 9x9) the number of combinations and permutations should decrease clearly going to a number much lower than the “Avogadro number”. I apologize for the long post.




pnm
Posted on: Fri Mar 02, 2012 6:44 pm
Posts: 2193 Joined: Thu May 12, 2011 11:58 pm

Re: what to do with the data?
clm wrote: I will comment first on the “solving rate”. I am a little confused. When you say (page 2):
I looked at the code again, and made a mistake: the division by the size (dimension, e.g. 6 for 6x6) doesn't happen. Patrick




giulio
Posted on: Sat Mar 03, 2012 12:53 am
Posts: 54 Joined: Thu Nov 03, 2011 8:52 am

Re: what to do with the data?
pnm wrote: clm wrote: I will comment first on the “solving rate”. I am a little confused. When you say (page 2):
I looked at the code again, and made a mistake: the division by the size (dimension, e.g. 6 for 6x6) doesn't happen. Patrick So Patrick, can you confirm that you calculate the difficulty as I explained in my recent post (minus the division by the size)? clm, No need to apologise for your interesting post. I welcome your skepticism.




starling
Posted on: Sat Mar 03, 2012 1:54 am
Posts: 175 Joined: Fri May 13, 2011 2:11 am

Re: what to do with the data?
clm wrote: (as we know in Europe one billion is 10 ^ 12 so one thousand times the american billion)
Wait, what? What do you call 10 ^ 9, then?




sneaklyfox
Posted on: Sat Mar 03, 2012 3:42 am
Posts: 422 Location: Canada Joined: Fri May 13, 2011 2:43 am

Re: what to do with the data?
starling wrote: clm wrote: (as we know in Europe one billion is 10 ^ 12 so one thousand times the american billion)
Wait, what? What do you call 10 ^ 9, then? Really??? 10 ^ 12 is a billion in Europe??? 10 ^ 3 thousand 10 ^ 6 ? 10 ^ 9 ? 10 ^ 12 billion 10 ^ 15 trillion???




giulio
Posted on: Sat Mar 03, 2012 3:59 am
Posts: 54 Joined: Thu Nov 03, 2011 8:52 am

Re: what to do with the data?
sneaklyfox wrote: starling wrote: clm wrote: (as we know in Europe one billion is 10 ^ 12 so one thousand times the american billion)
Wait, what? What do you call 10 ^ 9, then? Really??? 10 ^ 12 is a billion in Europe??? 10 ^ 3 thousand 10 ^ 6 ? 10 ^ 9 ? 10 ^ 12 billion 10 ^ 15 trillion??? You have to use languages other than English: 10 ^ 3: Italian=mille; German=Tausend; French=mille 10 ^ 6: milione; Million; million 10 ^ 9: miliardo; Miliarde; milliard 10 ^ 12: bilione; Billion; billion 10 ^ 15: trilione; Trillion; trillion As you can see, miliardo/Miliarde/milliard replace the English billion, and billion and the rest are shifted 'up' one place.




sheldolina
Posted on: Sat Mar 03, 2012 9:10 am
Posts: 73 Location: Delft, The Netherlands Joined: Fri May 13, 2011 7:15 am

Re: what to do with the data?
giulio wrote: sneaklyfox wrote: Really??? 10 ^ 12 is a billion in Europe??? 10 ^ 3 thousand 10 ^ 6 ? 10 ^ 9 ? 10 ^ 12 billion 10 ^ 15 trillion??? You have to use languages other than English: 10 ^ 3: Italian=mille; German=Tausend; French=mille 10 ^ 6: milione; Million; million 10 ^ 9: miliardo; Miliarde; milliard 10 ^ 12: bilione; Billion; billion 10 ^ 15: trilione; Trillion; trillion As you can see, miliardo/Miliarde/milliard replace the English billion, and billion and the rest are shifted 'up' one place. And that's still not completely correct  on this Wikipedia page and on this one a full list of respectively French and Dutch names can be found.




clm
Posted on: Sat Mar 03, 2012 9:31 am
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

Re: what to do with the data?
starling wrote: clm wrote: (as we know in Europe one billion is 10 ^ 12 so one thousand times the american billion)
Wait, what? What do you call 10 ^ 9, then? We usually said "one thousand million ... ", for instance, "five thousand million dollars ..." "one billion three hundred and 25 thousand million dollars ... ", actually the more new term "miliardo" (in Spain "millardo" accepted in 1995 by the "RAE", Real Academia Española of the language) (this word, being new, sounds strange to many people and many people is not accustomed to its use yet or simply do not know it). In other cases, in Europe, the use of Giga, for other purposes, like the particles accelerator, "Gigatron" (10 ^ 9 eV of energy) where you say "Bevatron" (one billion eV), but this sounds more logical using the greek symbols, though in Europe nobody would have used "Bevatron" in the sense 10 ^ 9. After that we have (sneaklyfox, 10 ^ 6 million): billion = 10 ^ 12 (we say "billón" in Spain) biliardo ("billardo") (in Spanish, French, German, Russian, Dutch, ... ) = 10 ^ 15 trillion = 10 ^ 18 ("trillón") (you use trillion for 10 ^ 12) and later triliardo ("trillardo") = 10 ^ 21 ... though it seems that the idea is to unify in the future all this numbering system, a little under the influence of the English speaking countries (in some way inversely to what happens with the "measuring system" where the european influence is higher). And a little more about this. A typical "mistake" is to translate wrongly, like they do many times in TV (specially with money) when they say "the world's population has reached 7 billion persons ... " (translated from English) because for us that is 7.000.000.000.000 and not 7.000.000.000 (here we use the dot instead of the colon) when the total population of the planet (since the beginning of the evolution is estimaded to be 250,000,000,000 souls (Assimov I think).






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