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Louis Stumpers
L Stumpers 
 
Number of games in database: 55
Years covered: 1932 to 1969
Overall record: +13 -32 =10 (32.7%)*
   * Overall winning percentage = (wins+draws/2) / total games.

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D94 Grunfeld (3 games)
C65 Ruy Lopez, Berlin Defense (2 games)
D45 Queen's Gambit Declined Semi-Slav (2 games)
E60 King's Indian Defense (2 games)
B59 Sicilian, Boleslavsky Variation, 7.Nb3 (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) He participated in five Dutch Chess Championships, with a 4th place 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 and 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 he 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, he 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) He 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 Academię Rheno-Traiectinę, 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...

Last updated: 2017-06-26 02:43:54

 page 1 of 3; games 1-25 of 55  PGN Download
Game  ResultMoves YearEvent/LocaleOpening
1. L Stumpers vs J Lehr  1-0191932EindhovenD18 Queen's Gambit Declined Slav, Dutch
2. Prins vs L Stumpers  1-0391936NED-ch prelimB20 Sicilian
3. L Stumpers vs E Spanjaard  1-0551938Dutch Ch prelimE02 Catalan, Open, 5.Qa4
4. E Sapira vs L Stumpers 0-1251938NBSB - FlandersD94 Grunfeld
5. A J Wijnans vs L Stumpers  1-0361939NED-chB05 Alekhine's Defense, Modern
6. J van den Bosch vs L Stumpers  ½-½581939NED-ch11A48 King's Indian
7. L Stumpers vs S Landau 0-1411939NED-ch11D33 Queen's Gambit Declined, Tarrasch
8. H van Steenis vs L Stumpers  1-0251939NED-chB02 Alekhine's Defense
9. L Stumpers vs H Kramer  0-1361940HilversumE25 Nimzo-Indian, Samisch
10. A J van den Hoek vs L Stumpers  1-0271941BondswedstrijdenB10 Caro-Kann
11. T van Scheltinga vs L Stumpers 1-0351942NED-ch12D94 Grunfeld
12. W Wolthuis vs L Stumpers  ½-½521946NED-ch prelim IC58 Two Knights
13. L Stumpers vs J H Marwitz  1-0401946NED-ch prelim ID31 Queen's Gambit Declined
14. G Fontein vs L Stumpers  ½-½261946NED-ch prelim ID94 Grunfeld
15. L Stumpers vs H van Steenis 0-1241946NED-ch prelim ID28 Queen's Gambit Accepted, Classical
16. C B van den Berg vs L Stumpers  1-0581946NED-ch prelim ID19 Queen's Gambit Declined Slav, Dutch
17. L Stumpers vs Euwe 0-1301946NED-ch prelim IE60 King's Indian Defense
18. L Stumpers vs Cortlever  ½-½501946NED-ch prelim IE60 King's Indian Defense
19. L Stumpers vs Grob 1-0601947Int BA55 Old Indian, Main line
20. L Stumpers vs H van Steenis  0-1331947Int BD23 Queen's Gambit Accepted
21. Tartakower vs L Stumpers 1-0241947Int BD74 Neo-Grunfeld, 6.cd Nxd5, 7.O-O
22. V Soultanbeieff vs L Stumpers  ½-½461947Int BD96 Grunfeld, Russian Variation
23. J T Barendregt vs L Stumpers  0-1261948NED-ch14C86 Ruy Lopez, Worrall Attack
24. L Stumpers vs C Vlagsma  0-1451948NED-ch14C65 Ruy Lopez, Berlin Defense
25. L Stumpers vs A Vinken  0-1331948NED-ch14E21 Nimzo-Indian, Three Knights
 page 1 of 3; games 1-25 of 55  PGN Download
  REFINE SEARCH:   White wins (1-0) | Black wins (0-1) | Draws (1/2-1/2) | Stumpers wins | Stumpers loses  
 

Kibitzer's Corner
< Earlier Kibitzing  · PAGE 303 OF 303 ·  Later Kibitzing>
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.
Oct-18-17  beatgiant: <alexmagnus>
Funny you should mention e and pi, I was thinking given the factorial function, we might try Stirling's formula.

n! =approx (2*pi*n)^.5 * (n/e)^n

Solving for pi, we have

((n!*(e/n)^n)^2)/2n = pi

I tried this for n=29, the highest n that shows as an exact value on my calculator.

((29!*(2.71828/29)^29)^2)/(2*29) = 3.15...

It must better with higher n, but my calculator itself is probably using some version of Stirling's formula to estimate the factorials. Leading to... you guessed it... a circularity.

Oct-18-17
Premium Chessgames Member
  SwitchingQuylthulg: (97+9/22+11/88062328.3)^(1/4) has an efficiency of 22 - 18 = +4.

Of course, there's a trivial way to beat that - recite pi up to the first spot where five consecutive zeros appear, and stop.

Oct-18-17
Premium Chessgames Member
  al wazir: 5*7 + 5! = 155. Try 3*5! - 5 instead.

So (3*5! - 5)/((5! - 7) = 355/(5! - 7) = 71/(4! - 7/5) = 3.1415929... is a co-leader (efficiency = 7-5 = 2).

Oct-18-17  beatgiant: With our factorial capability, we can generate arbitrarily large numbers with a single digit, like 9!, 9!!, 9!!!, etc. and if Stirling's formula converges, and if one has an arbitrary precision math package and a lot of compute power, the method should become very <al wazir>-efficient. That's the power of composing with looping constructs.

<SwitchingQuylthulg> <recite pi up to the first spot where five consecutive zeros appear> And that's a big weakness in our definition of efficiency. But is it definitely known that there is a spot where five consecutive zeros appears?

Oct-18-17  beatgiant: <SwitchingQuylthulg> Checking on MathOverflow, I learned <There is a sequence of 12 zeroes starting at position 1755524129973> citing Fabrice Bellard https://bellard.org/pi/pi2700e9/pid...

So there's our new champion: The first 1755524129972 digits of pi yeild 1755524129972 + 12 digits of accuracy, for efficiency of 12.

Oct-18-17
Premium Chessgames Member
  al wazir: <beatgiant: But is it definitely known that there is a spot where five consecutive zeros appears?> Although the digits in the decimal expansion of pi are self-evidently not random, they satisfy every test of randomness. So there must be a place where this occurs. Since the chances of this happening as a result of a truly random process are 1 in 10^5, we would expect it to occur within the first 10^5 or so places.

That said, the first occurrence of six consecutive 9s starts in the 762nd place: https://en.wikipedia.org/wiki/Six_n...

Oct-18-17
Premium Chessgames Member
  alexmagnus: <That said, the first occurrence of six consecutive 9s starts in the 762nd place: > What's more interesting, the <next> occurence of six identical digits is again that of 9s.
Oct-18-17
Premium Chessgames Member
  alexmagnus: We all surely remember: last year AlphaGo beat one of the best Go players in the world 4:1.

Well, now <that> version of AlphaGo lost to a newer version, called AlphaGo Zero, by a score of <0 to 100>. With taking up less resources than the original.

https://gizmodo.com/stunning-ai-bre...

Imagine. It's like going from Deep Blue to something stronger than Stockfish within a year. After AphaGo itself took a leap from something comparable to the chess engine Kaissa within a year.

Oct-18-17  beatgiant: <al wazir>
<there must be a place where this occurs>... is actually not <definitely known> but is an open conjecture, unless you've got a major new result up your sleeve.
Oct-18-17
Premium Chessgames Member
  WannaBe: I hope the contestant who wrote down Tibet, realize that it's a land locked country... Dumbazz

http://www.sfgate.com/entertainment...

And a naval officer does not know ocean geography. Ouch.

Oct-18-17
Premium Chessgames Member
  SwitchingQuylthulg: (7.85 + 738/357823923.23114723657983449515923235473) ^ (5/9) has an efficiency of 52 - 46 = 6, unless that giant decimal is huge enough to throw off multiple online calculators.

With a factorial, it gets an efficiency of 7:

(7.85 + 738 / (8!*8874.6+51.23114723657983449515923235473)) ^ (5/9)

Oct-18-17
Premium Chessgames Member
  al wazir: <SwitchingQuylthulg: (7.85 + 738/357823923.23114723657983449515923235473) ^ (5/9) has an efficiency of 52 - 46 = 6, unless that giant decimal is huge enough to throw off multiple online calculators.>

I'm a little skeptical about those online calculators. For example, the one at http://web2.0calc.com/ evaluates (97+9/22+11/88062328.3)^(1/4) as differing from pi in the 16th decimal place.

Which calculators did you use?

Oct-18-17
Premium Chessgames Member
  SwitchingQuylthulg: <al wazir> WolframAlpha's calculator, https://apfloat.appspot.com/ (with lots of extra 0s after the decimal points to encourage precision) and http://keisan.casio.com/calculator. The first two gave the same result, while the last only displayed 50 digits, all identical with pi.
Oct-18-17
Premium Chessgames Member
  al wazir: <beatgiant: <there must be a place where this occurs>... is actually not <definitely known> but is an open conjecture, unless you've got a major new result up your sleeve.> No, it's not a new result.

The first 0 occurs only in the 40th place. If no zeros ever occurred, the sequence would fail the tests for "randomness."

The same thing is true if 00, 000, 0000, 00000, or any other finite sequence you care to choose never occurs. It's just a little less blatant.

Oct-18-17  beatgiant: <al wazir>
We can only ever have finite subsets of the decimal expansion for pi to test for randomness. That is good evidence, but not proof.

It's widely believed, but not as far as I know proven, that <any finite sequence you care to choose> does eventually occur.

Oct-18-17
Premium Chessgames Member
  al wazir: <beatgiant: We can only ever have finite subsets of the decimal expansion for pi to test for randomness. That is good evidence, but not proof.> You're right.

Pi has been calculated out to one trillion digits or so. But that means we can't know anything about the frequency of occurrence of sequences two trillion digits long.

However, I'm sure that the frequency with which short sequences like 00000 occur has been checked -- though not beyond the trillionth digit.

Oct-19-17  beatgiant: Updated leaderboard with the contributions of our world-beater <SwitchingQuylthulg>:

3 + (the next 1755524129972 digits of pi)
Efficiency: 12
Method: Look for a long string of 0's in the digits of pi. This method will scale to arbitrarily high efficiency if the conjecture about the normality of pi is true. That would kill the whole contest or motivate a new definition of efficiency.

Runner-up:
(7.85 + 738 / (8!*8874.6+51.23114723657983449515923235473)) ^ (5/9) Efficiency: 7
Method: fine-tune a previous efficient solution.

(97+9/22+11/88062328.3)^(1/4)
Efficiency: 4
Method: fine-tune a previous efficient solution.

And one up-and-comer:
((n!*(e/n)^n)^2)/2n = pi
Method: Stirling's formula. The initial test with n=29 and using 6 digits of e was not very efficient, but it might scale with very large values of n, which we can generate with a single digit via repeated factorials. But I lack the resources to test the idea.

Oct-19-17
Premium Chessgames Member
  Count Wedgemore: <al wazir: The first 0 occurs only in the 40th place>

Actually, the 32nd..

<beatgiant: It's widely believed, but not as far as I know proven, that <any finite sequence you care to choose> does eventually occur.>

While it is fascinating to ponder the distinct possibilty that pi may contain the answer to every question one could possibly want to ask, it is also not of any practical use, is it?

Ater all, a string that contains everything contains nothing.

Oct-19-17
Premium Chessgames Member
  john barleycorn: < Count Wedgemore: ...

Ater all, a string that contains everything contains nothing.>

Amen to that. However, the irrationality of pi may be contagious to the mind of its researchers.

Anyway, I never gave up upon a search for next week's lotto numbers under the known digits of pi. :-)

Oct-19-17
Premium Chessgames Member
  al wazir: <Count Wedgemore: <al wazir: The first 0 occurs only in the 40th place<<>>> Actually, the 32nd.>

I slipped and wrote that in octal there. Sorry about that.

Oct-19-17  beatgiant: Here's a new spinoff stumper based on <al wazir>'s stumper. As we saw, the <al wazir>-efficiency of representations of pi is increasing without bound but apparently very slowly.

The new stumper is:
Define AW(b) as the best attainable al wazir efficiency for a given budget of b digits. The rules are same as before - we can use +, -, x, /, ^, !, ., (, ), and we have an nth-root oracle. I show the first few values below.

The new stumper is, what's the smallest number b such that AW(b) > 2?

AW(1) = 0, solution 3

AW(2) = 0, solutions 3.1, 3 + .1

AW(3) = 0, solutions 3.14, 3 + .14, 22/7, (4!-2)/7

AW(4) = 1, solution 2 + 4!^(1/4!)

AW(5) = 2, solutions 7.85^(5/9), 355/(4! - 7)

Oct-20-17
Premium Chessgames Member
  WannaBe: Okay, any predictions on when SkyNet will come online? 2030? 2040?

http://www.foxnews.com/tech/2017/10...

Oct-20-17
Premium Chessgames Member
  diceman: <WannaBe:

Okay, any predictions on when SkyNet will come online? 2030? 2040?>

2035
August 13 @ 8:38AM EST.

(all Fibonacci numbers)

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