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Louis F Stumpers
Number of games in database: 47
Years covered: 1932 to 1969
Overall record: +13 -27 =7 (35.1%)*
   * Overall winning percentage = (wins+draws/2) / total games
      Based on games in the database; may be incomplete.

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D94 Grunfeld (3 games)
B59 Sicilian, Boleslavsky Variation, 7.Nb3 (2 games)
E60 King's Indian Defense (2 games)

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(born Aug-30-1911, died Sep-27-2003, 92 years old) Netherlands

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 page 1 of 2; games 1-25 of 47  PGN Download
Game  ResultMoves Year Event/LocaleOpening
1. L Stumpers vs J Lehr  1-019 1932 EindhovenD18 Queen's Gambit Declined Slav, Dutch
2. L Stumpers vs E Spanjaard  1-055 1938 Dutch Ch prelimE02 Catalan, Open, 5.Qa4
3. E Sapira vs L Stumpers 0-125 1938 NBSB - FlandersD94 Grunfeld
4. H van Steenis vs L Stumpers  1-025 1939 NED-ch11B02 Alekhine's Defense
5. J van den Bosch vs L Stumpers  ½-½58 1939 NED-ch11A48 King's Indian
6. L Stumpers vs S Landau  0-141 1939 NED-ch11D33 Queen's Gambit Declined, Tarrasch
7. A J van den Hoek vs L Stumpers  1-027 1941 BondswedstrijdenB10 Caro-Kann
8. T van Scheltinga vs L Stumpers 1-035 1942 NED-ch12D94 Grunfeld
9. G Fontein vs L Stumpers  ½-½26 1946 NED-ch prelim ID94 Grunfeld
10. L Stumpers vs Cortlever  ½-½50 1946 NED-ch prelim IE60 King's Indian Defense
11. L Stumpers vs J H Marwitz  1-040 1946 NED-ch prelim ID31 Queen's Gambit Declined
12. L Stumpers vs Euwe 0-130 1946 NED-ch prelim IE60 King's Indian Defense
13. W Wolthuis vs L Stumpers  ½-½52 1946 NED-ch prelim IC58 Two Knights
14. C B van den Berg vs L Stumpers  1-058 1946 NED-ch prelim ID19 Queen's Gambit Declined Slav, Dutch
15. L Stumpers vs H van Steenis 0-124 1946 NED-ch prelim ID28 Queen's Gambit Accepted, Classical
16. L Stumpers vs Grob 1-060 1947 Int BA55 Old Indian, Main line
17. V Soultanbeieff vs L Stumpers  ½-½46 1947 Int BD96 Grunfeld, Russian Variation
18. Tartakower vs L Stumpers 1-024 1947 Int BD74 Neo-Grunfeld, Nxd5, 7.O-O
19. L Stumpers vs H van Steenis  0-133 1947 Int BD23 Queen's Gambit Accepted
20. L Stumpers vs T van Scheltinga  1-047 1948 NED-ch14C97 Ruy Lopez, Closed, Chigorin
21. L Stumpers vs H Kramer  0-140 1948 NED-ch14B92 Sicilian, Najdorf, Opocensky Variation
22. J Baay vs L Stumpers  1-040 1948 NED-ch14E37 Nimzo-Indian, Classical
23. L Stumpers vs F Henneberke 1-043 1948 NED-ch14C92 Ruy Lopez, Closed
24. J T Barendregt vs L Stumpers  0-126 1948 NED-ch14C86 Ruy Lopez, Worrall Attack
25. L Stumpers vs A Vinken  0-133 1948 NED-ch14E21 Nimzo-Indian, Three Knights
 page 1 of 2; games 1-25 of 47  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 254 OF 254 ·  Later Kibitzing>
Feb-04-15  beatgiant: <kellmano>
I see, <one (27) has three factors of three> is what you kindly called the <partial> statement.
Premium Chessgames Member
  kellmano: Yes, full method marks. The team I was refereeing had a girl who explained it wonderfully, saying 'There are twenty two multiples of three, but there is an extra three available from all the multiples of 9, of which there are 6, and one more each from 27 and 54. There are way more twos than this, so the answer is 30'. Unfortunate to spot it quickly, explain very clearly but mess up your times-table and get zero.
Premium Chessgames Member
  kellmano: Same team messed up the other question I posed by carrying out the prime factorisation, then concluding that as there are three prime factors, and each of them paired with another odd number, then it must be possible to write these as (a + b)(a - b) with integers for a and b. I'll admit I thought that this was correct and the answer was 3 until I came to mark it. The Owl of Minerva and all.
Premium Chessgames Member
  al wazir: <kellmano: If I tell you the answer is 31, does that help?> I missed the third one in 54. Drat.
Premium Chessgames Member
  Sneaky: In retrospect, my idea of writing down the numbers from 2 to 66 and checking off the factors of three like a chimp wasn't such a bad idea. (I knew that counting the two's wasn't necessary but thought that would be a nice touch, for thoroughness or something.)
Premium Chessgames Member
  al wazir: <kellmano>: The efficient way to solve that first problem is to proceed as follows:

The quotient of 66 divided by 3 is 22.
The quotient of 66 divided by 9 is 7.
The quotient of 66 divided by 27 is 2.

27 + 7 + 2 = 31.

A more interesting problem would have been to find the highest value of n such that 12^n is a factor of 81!

Premium Chessgames Member
  kellmano: <al wazir> Well yes that is essentially the method you and I employed anyway. For 12^n and 81! I guess we'll have to sum the quotients for powers of 2, then powers of three and take the lower number from the sum from the powers of three and half that of the sum of powers of two.

I guess it is more interesting, but remember that the teams had 40 minutes to solve ten such problems. It's possible I've missed some nice shortcut, and if so please let me know.

Premium Chessgames Member
  kellmano: <Sneaky: In retrospect, my idea of writing down the numbers from 2 to 66 and checking off the factors of three like a chimp wasn't such a bad idea>

oh absolutely it's the right idea, but hopefully you'd notice where the threes were coming from and not bother with, say, 35.

Premium Chessgames Member
  Sneaky: How does a computer programmer boil a pot of water?

<If the pot is empty> execute SUBROUTINE A (1. fill pot with water, 2. place on burner, 3. turn stovetop on, 4. wait until boiling, 5. turn stovetop off.)


<If the pot is already filled with water> (1. empty water into sink, 2. execute SUBROUTINE A.)

Premium Chessgames Member
  Sneaky: I was cruising youtube the other day and was watching a math video in which the speaker (who seems to be a college student, but really knows her stuff) claimed that we aren't sure to this day if Euler's constant is a transcendental number or not.

Is that really true? Along with pi, I always thought that e was the spitting-image of a transcendental number.

To make matters more frustrating, she muttered under her breath that we don't even know if e is *rational* or not. We aren't sure that e is not a simple fraction?! I find that a bit hard to swallow. If true, that's pretty darn embarrassing.

Feb-06-15  beatgiant: <Sneaky>
It was proven over a century ago that e is transcendental. What is not known is whether e is a <normal number> (i.e. its digits are uniformly distrubuted).
Feb-06-15  beatgiant: <Sneaky> Are you sure the video was not about gamma, another Euler constant? I'm not sure that one has been proven to be irrational yet.
Premium Chessgames Member
  zanzibar: This is probably a linguistic problem.

There are two Euler constants, e and gamma.

The natural log Euler's constant (or number), e, is both irrational and transcendental.

It's the other one, gamma, aka the Euler–Mascheroni constant, that was being referred to in the video.


Premium Chessgames Member
  al wazir: <Sneaky: How does a computer programmer boil a pot of water?> That joke used to be "How does a mathematician boil a pot of water?" The punch line was "If the pot is already full, pour out the water. Now the conditions of the previous problem hold, and that problem has already been solved."
Premium Chessgames Member
  zanzibar: Since <Stumpers> has turned into a number theory class of late...

Proving that any number is transcendental is tough, there are actually very few examples:

e, pi, e**pi (but not yet pi**e), 2**sqrt(2), ln(2), sin(1) and i**i

Leave it to Cantor to prove that transcendentals are "denser" than algebraic numbers, yet we know so few of them.

A much better write up is here:

Hermite proved e was back in 1873.

The first transcendental was found by Louiville in 1844:

Sum_n=1_inf 10**(-(n!))

Premium Chessgames Member
  al wazir: If e is transcendental, then so must e/2, e/3, e/4, etc., be. Any rational number multiplied by e must be transcendental. So we already have a (countably) infinite set of examples of transcendental numbers.
Premium Chessgames Member
  Sneaky: <There are two Euler constants, e and gamma.> Ahhhh, thanks Zanzibar. Makes more sense.
Feb-11-15  HeMateMe: Interesting baseball stuff. A-Rod apologizes to the Yankees ownership.


They owe him $61M over the next three years. The Yanks are still suing to negate a home run clause, where A-Rod gets incremental $Ms as he passes career home run hitters. He gets $5M if he passes Willie Mays, who has 660 homers. A-Rod is at 551, I think. He gets a total of $30M if he passes everyone.

Yanks are trying to negate the clause, claiming A-Rod is a proven cheater, so the HR totals are worthless. To be successful in court, don't the Yanks need to get MLB to also kick A-Rod out of the record books? And, what about all of the other cheaters who have as much certified guilt as A-Rod? They all will be forced to give up money and get kicked out of the record books? It will be interesting to see if the Yanks' action will set off a chain reaction, if they are successful in court against Rodriguez.

Premium Chessgames Member
  zanzibar: Haven't been in this neck of the woods for a little while, but a recent <Chessbase> article featuring Feynman inspired me:

The video shows a clip where Feynmen is using the game of chess as an analogy to physics.

It's actually a pretty good analogy, and expresses well some of the features of fundamental physics research. But there is perhaps one weakness in the analogy which got me wondering, and where better than <Stumpers> to raise this question:

<Are there any master level games where a pawn promoted to a bishop instead of a queen?>


Premium Chessgames Member
  Shams: <zanzibar> Seems like maybe not.

(Unless you count the Shirov-Kramnik 2005 game which I think most players would not.)

Premium Chessgames Member
  zanzibar: <shams> Here's a few games I found rummaging through the <CG> database:

5-most recent (reverse chronological):

J Rindlisbacher vs R Rapport, 2014 (move 62)

K H Lien vs J Brorsen, 2014 (move 63 - trivial)

Ponomariov vs D Baramidze, 2014 (move 65)

P Gengler vs P Abrantes, 2014 (move 64)

Akopian vs Gao Rui, 2014 (move 49)

I think with all the monkeying around it will be very difficult to find a legitimate example. Maybe if I work from oldest to newest, let's try a few:

A Von Der Goltz vs Von Der Lasa, 1837 (move 34)

E Schallopp vs A Schottlaender, 1885 (finally! move 84)

Marshall vs J Mason, 1902 (move 40 by Marshall!?!)

Vidmar vs Maroczy, 1932 (moves 124, 125)

L Soluch vs E Paoli, 1952 (move 93)

Let's quit while ahead, eh?

Of course, the statistics are low, and the old players did some monkeying around too. Still, it looks like the more recent players tend to monkey a bit more.

Feb-23-15  tbentley: has quite a few bishop (and also rook) underpromotions.
Premium Chessgames Member
  al wazir: In E Schallopp vs A Schottlaender, 1885, 84. Kf6 Kh6 (forced) 85. Rh8# could have been played. So I don't think that example of under-promotion was "legitimate."

But L Soluch vs E Paoli, 1952 is legitimate. (But a ♘ would have been just as good.)

Premium Chessgames Member
  al wazir: <zanzibar>: Thanks for that video!

I agree that deeper understanding has *usually* simplified our picture of physics -- but not always. Who would claim that General Relativity or the Standard Model is simpler than classical physics?

Feb-24-15  diceman: <Sneaky:
How does a computer programmer boil a pot of water?>

...sits it on top of the microprocessor
as a heat-sink.

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