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

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

[what is this?]

 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. E Sapira vs L Stumpers 0-125 1938 NBSB - FlandersD94 Grunfeld
3. L Stumpers vs E Spanjaard  1-055 1938 Dutch Ch prelimE02 Catalan, Open, 5.Qa4
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. L Stumpers vs J H Marwitz  1-040 1946 NED-ch prelim ID31 Queen's Gambit Declined
10. L Stumpers vs Euwe 0-130 1946 NED-ch prelim IE60 King's Indian Defense
11. W Wolthuis vs L Stumpers  ½-½52 1946 NED-ch prelim IC58 Two Knights
12. C B van den Berg vs L Stumpers  1-058 1946 NED-ch prelim ID19 Queen's Gambit Declined Slav, Dutch
13. L Stumpers vs H van Steenis 0-124 1946 NED-ch prelim ID28 Queen's Gambit Accepted, Classical
14. G Fontein vs L Stumpers  ½-½26 1946 NED-ch prelim ID94 Grunfeld
15. L Stumpers vs Cortlever  ½-½50 1946 NED-ch prelim IE60 King's Indian Defense
16. V Soultanbeieff vs L Stumpers  ½-½46 1947 Int BD96 Grunfeld, Russian Variation
17. Tartakower vs L Stumpers 1-024 1947 Int BD74 Neo-Grunfeld, 6.cd Nxd5, 7.O-O
18. L Stumpers vs H van Steenis  0-133 1947 Int BD23 Queen's Gambit Accepted
19. L Stumpers vs Grob 1-060 1947 Int BA55 Old Indian, Main line
20. L Stumpers vs F Henneberke 1-043 1948 NED-ch14C92 Ruy Lopez, Closed
21. J T Barendregt vs L Stumpers  0-126 1948 NED-ch14C86 Ruy Lopez, Worrall Attack
22. L Stumpers vs A Vinken  0-133 1948 NED-ch14E21 Nimzo-Indian, Three Knights
23. L Stumpers vs C Vlagsma  0-145 1948 NED-ch14C65 Ruy Lopez, Berlin Defense
24. L Stumpers vs T van Scheltinga  1-047 1948 NED-ch14C97 Ruy Lopez, Closed, Chigorin
25. L Stumpers vs H Kramer  0-140 1948 NED-ch14B92 Sicilian, Najdorf, Opocensky Variation
 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 249 OF 249 ·  Later Kibitzing>
Nov-23-14
Premium Chessgames Member
  kellmano: If you're the kind of person that likes a challenge try this:

https://www.youtube.com/watch?v=Yaj...

What's wrong with this argument? I'm big enough to admit I didn't get it, but then neither did most of the people in the comments section. Will tag in <micartouse>, <sneaky> and <al wazir> as I think this is a nice puzzle that you'll enjoy if you haven't seen it before.

Nov-23-14  micartouse: <kellmano: What's wrong with this argument?>

I'm almost sure I figured out what's wrong with the argument, but when I scrolled down, I didn't see my answer in the comments! I was very surprised as it only took a few minutes and I don't even remember much geometry.

I actually think it's a lot simpler in nature than most of the commenters are making it, but I can wait to post my solution so I don't spoil the fun.

Nov-23-14
Premium Chessgames Member
  kellmano: <micartouse> I agree it doesn't seem to be in the comments. I could see that the objections to congruence were incorrect.
Nov-23-14
Premium Chessgames Member
  Sneaky: I've seen that proof about equilateral triangles a long time ago, and I don't remember anything about it, other than the fact it's a bogus proof.

I think an easy way to reveal the flaw without much brainwork would be to draw a triangle that is clearly not equilateral and then follow his proof along with that drawing at hand. It's no coincidence he drew a triangle that was very nearly equilateral to prove his point. If you drew a triangle with three very different sides, I think the drawing itself should betray the tomfoolery.

Nov-23-14  micartouse: <It's no coincidence he drew a triangle that was very nearly equilateral to prove his point. If you drew a triangle with three very different sides, I think the drawing itself should betray the tomfoolery.>

Agree, it's a nice optical illusion. I almost have the sense someone discovered the fake proof when trying to work out a real one in the context of a different problem and then tried to retrace their steps to see what happened.

Nov-23-14
Premium Chessgames Member
  Sneaky: <On a lighter note, and not too challenging, how many 3 digit numbers cannot be written as the sum of a string of consecutive integers?>

That's very interesting. I never thought about it until now.

I believe it's more interesting to ask the generalized question: "Which integers cannot be written as a sum of a string of consecutive positive integers?"

At first I was tempted to say that all integers can be expressed as such a sum but then I realized that the number 8 can't be done.

(BTW, it's important to put the word "positive" in there, because otherwise you can sum to any number by starting in the negatives.)

Nov-23-14  micartouse: <On a lighter note, and not too challenging, how many 3 digit numbers cannot be written as the sum of a string of consecutive integers?>

I found this difficult to chew on. I believe the answer is 897.

I had to think of the answer visually: All numbers with an odd factor can be expressed as a rectangle with an odd number of rows. Then you can make one edge diagonal just by pulling blocks off the top and putting them on the bottom.

Example: Your example of 120 can be expressed as 5 rows of 24 blocks. Then rearrange the rows so they have 22, 23, 24, 25, and 26 blocks (it appears your example had a typo).

This procedure can't be followed with any numbers without odd factors. The only 3 digit numbers with this property are powers of 2: 128, 256, and 512. This leaves 897.

Nov-24-14
Premium Chessgames Member
  PhilFeeley: < kellmano: hmmmmm c +21 = m and 5c + 30 = m + 6. Just do a bit of algebra and ....>

I didn't think my algebra was that rusty, but solving your equations gives 4c = -3, which seems quite meaningless. I'm not sure where I went wrong.

My system for the solution was:

c+21=m and 5c+6=m+6, which gives
c=5.25 years, which fits.

Nov-24-14
Premium Chessgames Member
  al wazir: <PhilFeeley: I didn't think my algebra was that rusty, but solving your equations gives 4c = -3, which seems quite meaningless.> Think again.
Nov-24-14
Premium Chessgames Member
  kellmano: yes that is right micartouse. All odd numbers or numbers with an odd factor can be written in this way. The only numbers that are neither odd nor have an odd factor are powers of 2 (which can be shown using prime factorisation rather easily).
Nov-24-14  micartouse: I noticed my visual solution to the consecutive integer problem has flaws in it so it's inadequate as a proof, but it at least gets one in the ballpark of how one could go about proving it algebraically.

For the equilateral triangle problem, I'm pretty sure the flaw is that the angle bisector crosses through the midpoint of the opposite segment. Therefore M and X are the same point, and all his subsequent shapes would collapse.

Nov-24-14  micartouse: <Burning Ropes> You are given two ropes and a lighter. Both ropes take exactly one hour to burn. Some parts of the ropes are thicker than other parts of the rope, therefore not all parts of the rope will burn at the same speed (burning half the rope will not necessarily take 30 minutes).

Find a way to measure 45 minutes of time.

Nov-24-14  micartouse: I couldn't figure this one out, liked the answer.
Nov-24-14
Premium Chessgames Member
  Sneaky: <I'm pretty sure the flaw is that the angle bisector crosses through the midpoint of the opposite segment.> if the triangle is in fact equilateral, yes. If it's not, then his proof seems to hold up--which is disturbing. That's like saying you can't prove that an equilateral triangle is equilateral but you can prove that any other type of triangle is.

I think that the real problem lies in this point "X". He never demonstrated that this point "X" exists, and it's not obvious that the angle bisector will intersect with the perpendicular. Why can't it be parallel, or just veer off to the right or left? But he just drew it that way and everything stems from that.

Nov-24-14  micartouse: <Sneaky> Dang you're right - it has to be isosceles to bisect the opposite side. :( I'm back to square one then - no wonder I was surprised at how quickly I got the answer; it was crap.
Nov-25-14
Premium Chessgames Member
  kellmano: <sneaky> I think it would be fairly easy to prove that unless the triangle is isosceles the angle bisector and perpendicular bisector will meet at one point outside the triangle. You are right to be suspicious about the diagram but for the wrong reason.
Nov-25-14  Schwartz: Nice <Burning Ropes> puzzle, micartouse.

Here's a quick riddle.. not in the usual spirit of this page though:

A man is pushing his car along the road when he comes to a hotel. He shouts, "I'm bankrupt!" Why?

Nov-25-14
Premium Chessgames Member
  OhioChessFan: Lovely, <Schwarz> and I am sure I'm not the only person who likes the non-math questions asked on this page(I think "road" is a bit of an unfair red herring, although it would be kind of hard to ask the question any other way). The math questions might as well be like trying to do Sunday puzzles blindfold and I don't even bother.
Nov-25-14
Premium Chessgames Member
  kellmano: <micartouse> doesn't involve making a figure of 8 from one of them and lighting the middle does it? That's what I thought first but it doesn't work and I can't think of another way to do it.
Nov-25-14
Premium Chessgames Member
  al wazir: <micartouse>: Light one end of of one rope and both ends of the other. When the second rope is completely consumed, that is, when the flames from the two ends meet, exactly half an hour has elapsed. Now light the intact end of the first rope. When the flame from that end meets the one from the end originally lit, 45 minutes have elapsed.
Nov-25-14  micartouse: <al wazir> Yes, well done. I was picturing things like kellmano's post and setting up various contraptions with weights or water but just couldn't make progress.

<OCF> Agree, I enjoy reading a diverse puzzle set from many posters, even if I only dip my toes in the water with the math ones.

Nov-25-14
Premium Chessgames Member
  al wazir: <Schwartz: Here's a quick riddle.> I've run out of gas on this one. My standard riddle-solving technique ("focus on the word or words used in an unusual fashion") fails, despite what seems to be a hint from <OhioChessFan>.
Nov-25-14  Schwartz: He was playing a game of Monopoly.
Nov-26-14
Premium Chessgames Member
  al wazir: <Schwartz>: It's a good puzzle, but I was at least three hints away from getting it. That was good preparation for the two weeks of embarrassment and humiliation I expect to experience during the forthcoming Holiday Present Hunt.
Nov-26-14
Premium Chessgames Member
  Shams: Good news for cryptophiles:
http://www.slate.com/articles/techn...
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