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Louis Stumpers

Most played openings

D94		Grunfeld (3 games)
B59		Sicilian, Boleslavsky Variation, 7.Nb3 (2 games)
D31		Queen's Gambit Declined (2 games)
D45		Queen's Gambit Declined Semi-Slav (2 games)
E60		King's Indian Defense (2 games)
E21		Nimzo-Indian, Three Knights (2 games)
C65		Ruy Lopez, Berlin Defense (2 games)

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LOUIS STUMPERS (born Aug-30-1911, died Sep-27-2003, 92 years old) Netherlands [what is this?] Frans Louis Henri Marie Stumpers was born in Eindhoven, Netherlands, on 30 August 1911. (1) He was champion of the Eindhoven Chess Club in 1938, 1939, 1946, 1947, 1948, 1949, 1951, 1952, 1953, 1955, 1957, 1958, 1961 and 1963, (2) and champion of the North Brabant Chess Federation (Noord Brabantse Schaak Bond, NBSB) in 1934, 1935, 1936, 1937, 1938, 1939, 1940, 1941, 1942, 1943, 1944, 1946, 1948, 1949, 1950, 1951, 1952, 1953, 1954, 1955, 1959, 1961, 1962, 1963, 1964, 1965, 1966 and 1967. (3) Stumpers participated in five Dutch Chess Championships, with his high-water mark a fourth place finish in 1948, (4) and represented his country at the 1st European Team Championship in Vienna in 1957 (two games, vs Josef Platt and Max Dorn). (5) From 1945 until about 1956, he was first Secretary and then Chairman of the NBSB. (3) Stumpers was a physicist, and worked for the Philips company as an assistant from 1928. During 1934-1937, he studied at the University of Utrecht, where he took the master's degree. (6) In 1938 Stumpers was again employed at Philips, (6) and at a tournament in 1942, he supplied the hungry chess players with food from his employer. (3) After the war, Stumpers made a career in physics, with patents and awards on information ("radio") technology. He received degrees from several universities and colleges, including in Poland and Japan. (1, 3, 6) Stumpers retired from Philips in 1972, but continued teaching, (6) partly as professor at the University of Utrecht (1977-1981). (7) He was also Vice President (1975-1981) and Honorary President (1990-2003) of URSI, the International Union of Radio Science. (8) Louis Stumpers married Mieke Driessen in 1954. They had five children, three girls and two boys. (6) 1) Online Familieberichten 1.0 (2016), http://www.online-familieberichten...., Digitaal Tijdschrift, 5 (255), http://www.geneaservice.nl/ar/2003/... 2) Eindhovense Schaakvereniging (2016), http://www.eindhovenseschaakverenig... 3) Noord Brabantse Schaak Bond (2016), http://www.nbsb.nl/pkalgemeen/pk-er... Their main page: http://www.nbsb.nl. 4) Schaaksite.nl (2016), http://www.schaaksite.nl/2016/01/01... 5) Olimpbase, http://www.olimpbase.org/1957eq/195... 6) K. Teer, Levensbericht F. L. H. M. Stumpers, in: Levensberichten en herdenkingen, 2004, Amsterdam, pp. 90-97, http://www.dwc.knaw.nl/DL/levensber... Also available at http://www.hagenbeuk.nl/wp-content/... 7) Catalogus Professorum Academiæ Rheno-Traiectinæ, https://profs.library.uu.nl/index.p... 8) URSI websites (2016), http://www.ursi.org/en/ursi_structu... and http://www.ursi.org/en/ursi_structu... Suggested reading: Eindhovense Schaakvereniging 100 jaar 1915-2015, by Jules Welling. Stumpers' doctoral thesis Eenige onderzoekingen over trillingen met frequentiemodulatie (Studies on Vibration with Frequency Modulation) is found at http://repository.tudelft.nl/island... This text by User: Tabanus. The photo was taken from http://www.dwc.knaw.nl. Last updated: 2022-04-04 00:17:13

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page 1 of 3; games 1-25 of 63  Game   Result Moves Year Event/Locale Opening 1. L Stumpers vs J Lehr 1-0 19 1932 Eindhoven D18 Queen's Gambit Declined Slav, Dutch 2. L Prins vs L Stumpers   1-0 39 1936 NED-ch prelim B20 Sicilian 3. E Sapira vs L Stumpers 0-1 25 1938 NBSB-Flanders D94 Grunfeld 4. L Stumpers vs E Spanjaard   1-0 55 1938 NED-ch prelim E02 Catalan, Open, 5.Qa4 5. A J Wijnans vs L Stumpers   1-0 36 1939 NED-ch B05 Alekhine's Defense, Modern 6. J van den Bosch vs L Stumpers   ½-½ 58 1939 NED-ch A48 King's Indian 7. L Stumpers vs S Landau 0-1 41 1939 NED-ch D33 Queen's Gambit Declined, Tarrasch 8. H van Steenis vs L Stumpers   1-0 25 1939 NED-ch B02 Alekhine's Defense 9. L Stumpers vs H Kramer   0-1 36 1940 Hilversum E25 Nimzo-Indian, Samisch 10. L Stumpers vs S Landau   ½-½ 34 1940 Hilversum D31 Queen's Gambit Declined 11. A van den Hoek vs L Stumpers   1-0 27 1941 Bondswedstrijden B10 Caro-Kann 12. T van Scheltinga vs L Stumpers 1-0 35 1942 NED-ch12 D94 Grunfeld 13. W Wolthuis vs L Stumpers   ½-½ 52 1946 NED-ch prelim I C58 Two Knights 14. L Stumpers vs J H Marwitz   1-0 40 1946 NED-ch prelim I D31 Queen's Gambit Declined 15. G Fontein vs L Stumpers   ½-½ 26 1946 NED-ch prelim I D94 Grunfeld 16. L Stumpers vs H van Steenis 0-1 24 1946 NED-ch prelim I D28 Queen's Gambit Accepted, Classical 17. C van den Berg vs L Stumpers   1-0 58 1946 NED-ch prelim I D19 Queen's Gambit Declined Slav, Dutch 18. L Stumpers vs Euwe 0-1 30 1946 NED-ch prelim I E60 King's Indian Defense 19. L Stumpers vs N Cortlever   ½-½ 50 1946 NED-ch prelim I E60 King's Indian Defense 20. L Stumpers vs H Grob 1-0 60 1947 Baarn Group B A55 Old Indian, Main line 21. L Stumpers vs H van Steenis   0-1 33 1947 Baarn Group B D23 Queen's Gambit Accepted 22. Tartakower vs L Stumpers 1-0 24 1947 Baarn Group B D74 Neo-Grunfeld, 6.cd Nxd5, 7.O-O 23. V Soultanbeieff vs L Stumpers   ½-½ 46 1947 Baarn Group B D96 Grunfeld, Russian Variation 24. L Stumpers vs A Vinken   0-1 33 1948 NED-ch sf E21 Nimzo-Indian, Three Knights 25. L Prins vs L Stumpers   ½-½ 30 1948 NED-ch sf D02 Queen's Pawn Game  page 1 of 3; games 1-25 of 63

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< Earlier Kibitzing  · PAGE 25 OF 94 ·  Later Kibitzing> Oct-16-17    beatgiant: 306^.2 = 3.14155... finally I manage to get more digits out than I put in. Oct-16-17    al wazir: <john barleycorn: What about e.g. factorials?> They're pretty elementary. I'm willing to include them too. <beatgiant: Did I get that right?> Yes. Oct-16-17    beatgiant: <factorials> That means the new leaderboard is: (97 + 9/22)^(1/4) = 3.141592652... efficiency = 9-7 = 2 355/113 = 3.1415929... efficiency = 7-6 = 1 2 + 4!^(1/4!) = 3.14158... efficiency = 5-4 = 1 306^.2 = 3.14155... efficiency = 5-4 = 1 I've found lots with efficiency 0 by now, so no point in listing all those. Oct-17-17    al wazir: I think I know how Ramanujan found that magical (97 + 9/22)^(1/4) approximation. By raising pi to the fourth power (squaring it twice), he got 97.40909102... . This is almost equal to 97.40909090909..., a repeating decimal, which is a rational. Then he either calculated (or being Ramanujan, already knew) that 9/22 = 0.409090909.... I'm no Ramanujan, but I can use the same trick. For example, if I take the fourth root of pi (i.e., do the opposite of what he did), I get 1.331335364..., which is almost the same as 1.33133133..., which is equal to 1330/999. So we have (133/99.9)^4 = 3.141554592... This doesn't win a prize for efficiency, but that's because my fraction is more unwieldy than R's. Oct-17-17    john barleycorn: now that abbreviations like ^ and ! are allowed how about the use of summation and product symbols (sigma and pi)? Oct-17-17    beatgiant: Beware of geeks bearing grifts. Oct-17-17    alexmagnus: <Yes, same holds true for prime numbers. same sporty approach but practically not that relevant, imo.> The largest primes in recent years had two sources: 1. Search for Mersenne primes 2. Seventeen or Bust project. Both have relevance. Mersenne primes are connected to perfect numbers. The question whether there are infinitely many perfect numbers is the oldest unsolved mathematical problem (and there are exactly as many even perfect numbers as there are Mersenne primes - whether any odd perfect numbers exist is another age-old question). And Seventeen or Bust war set up to answer the question whether 78557 is the smallest Sierpinski number - that is, the smallest k such that k*2^n+1 is composite for every n. Oct-17-17    alexmagnus: No, beware of gifts bearing Greeks Oct-17-17    beatgiant: <al wazir> <because my fraction is more unwieldy> My fractions are almost always more unwieldy. Got a suggestion? Oct-17-17    john barleycorn: Talking about unwieldy. Here is an aesthetically pleasing one just made up of 1,2,+,*,/,(,),and √,√√, etc. a=((√√2)+√(1/√2))/2 b=(√√2)/√2 c= (2+√2)*b*(1+a)/(1+b) Let A= (a+1)/(2√a) B= (√a)*(1+b)/(a+b) Then P= c*B*(1+A)/(1+B)= 3,141592661 Oct-17-17    john barleycorn: the numeric values for a, b, c, A, B if somebody wants to check. The n-th iteration by the way gives at least 2^n correct digits of pi. a=1,0150517651, b=8408964153, c=3,1426067539, A=1,0000278991, B=0,9993269535 (all mistakes are intentional) Oct-17-17    al wazir: Inspired by the insight I reported in my previous post, I began raising pi to rational powers r, testing r = 1/2, 2, 1/3, 3, 2/3, 3/2, etc., systematically. With r = 5/9 this yields 1.888836531... ≈ 17/9, which looks promising. But (17/9)^(9/5) = 3.141749405... is accurate only to four places. However, the result of raising pi to the power 9/5 is 7.850002062..., and 7.85^(5/9) = 3.141592195..., which has an efficiency of +2. Oct-17-17    al wazir: <john barleycorn: now that abbreviations like ^ and ! are allowed how about the use of summation and product symbols (sigma and pi)?> Those symbols aren't on my keyboard, and <chessgames.com> doesn't accept TeX (\Sigma and \Pi). So I am ruling out symbols that can't be typed with a single keystroke, e.g., those requiring unitype or the equivalent. Oct-17-17    beatgiant: It's as if the Trojan horse were blocked because a customs official ruled it contained a commercial quantity of hardwood lumber. Oct-17-17    john barleycorn: <beatgiant> <trojan horse> 355/113 = (3*10^2+5*10+5)/(10^2+10+3) here we go. :-) Oct-17-17    al wazir: <john barleycorn: Are we still comparing to> 3.141592653589...? Yes. Oct-17-17    john barleycorn: ok then Oct-17-17    Sneaky: I happen to know (read: I was told by an angel) that there are over 100 consecutive digits of pi starting with 14159265... buried deep within the digits of e. Unfortunately, it's so deep into the mantissa that it would take over 100 digits to tell you where it is. However, as it just turns out, this 100+ digit number is precisely what you get when you take the product of all the words first verse of the book of Genesis, if you interpret the words as Hebrew numbers. Two questions: (1) Is my solution to your contest invalid as it doesn't conform to the rules? (2) If my tall-tale were actually true, would you dismiss this as a mere coincidence or would your theosophical viewpoint of the world change? Oct-17-17    beatgiant: Using the factorial facility to beat 355/113: 355/(5!-7) = 3.1415929... Oct-17-17    alexmagnus: <Sneaky> Sometimes you don't need to go hundreds of digits to find interesting coincidences. My favorite one is e^pi being very close to pi+20 (it's pi+19.99909997....) Oct-17-17    alexmagnus: <(2) If my tall-tale were actually true, would you dismiss this as a mere coincidence or would your theosophical viewpoint of the world change?> I actually can imagine it <is> true, especially if you don't put an exact condition on your Genesis product (which language, how you convert words to numbers, etc.). Oct-17-17    beatgiant: Or if you want the all 5's and 7's version, (5*7 + 5!)/(5! - 7) = 3.1415929... Oct-17-17    ughaibu: How about updating the leader board? Oct-17-17    beatgiant: (97 + 9/22)^(1/4) = 3.141592652... efficiency = 9-7 = 2 (5*7 + 5!)/(5! - 7) = 3.1415929... efficiency = 7-5 = 2 7.85^(5/9) = 3.1415921... efficiency = 7-5 = 2 2 + 4!^(1/4!) = 3.14158... efficiency = 5-4 = 1 306^.2 = 3.14155... efficiency = 5-4 = 1 Oct-17-17    ughaibu: Thanks. Jump to page #   (enter # from 1 to 94) search thread:    < Earlier Kibitzing  · PAGE 25 OF 94 ·  Later Kibitzing>

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