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Louis 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
      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)
C65 Ruy Lopez, Berlin Defense (2 games)
D45 Queen's Gambit Declined Semi-Slav (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 3; games 1-25 of 55  PGN Download
Game  ResultMoves Year Event/LocaleOpening
1. L Stumpers vs J Lehr  1-019 1932 EindhovenD18 Queen's Gambit Declined Slav, Dutch
2. Prins vs L Stumpers  1-039 1936 NED-ch prelimB20 Sicilian
3. E Sapira vs L Stumpers 0-125 1938 NBSB - FlandersD94 Grunfeld
4. L Stumpers vs E Spanjaard  1-055 1938 Dutch Ch prelimE02 Catalan, Open, 5.Qa4
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 Wijnans vs L Stumpers  1-036 1939 NED-chB05 Alekhine's Defense, Modern
8. H van Steenis vs L Stumpers  1-025 1939 NED-ch11B02 Alekhine's Defense
9. L Stumpers vs H Kramer  0-136 1940 HilversumE25 Nimzo-Indian, Samisch
10. A J van den Hoek vs L Stumpers  1-027 1941 BondswedstrijdenB10 Caro-Kann
11. T van Scheltinga vs L Stumpers 1-035 1942 NED-ch12D94 Grunfeld
12. W Wolthuis vs L Stumpers  ½-½52 1946 NED-ch prelim IC58 Two Knights
13. C B van den Berg vs L Stumpers  1-058 1946 NED-ch prelim ID19 Queen's Gambit Declined Slav, Dutch
14. L Stumpers vs H van Steenis 0-124 1946 NED-ch prelim ID28 Queen's Gambit Accepted, Classical
15. G Fontein vs L Stumpers  ½-½26 1946 NED-ch prelim ID94 Grunfeld
16. L Stumpers vs Cortlever  ½-½50 1946 NED-ch prelim IE60 King's Indian Defense
17. L Stumpers vs J H Marwitz  1-040 1946 NED-ch prelim ID31 Queen's Gambit Declined
18. L Stumpers vs Euwe 0-130 1946 NED-ch prelim IE60 King's Indian Defense
19. Tartakower vs L Stumpers 1-024 1947 Int BD74 Neo-Grunfeld, Nxd5, 7.O-O
20. L Stumpers vs H van Steenis  0-133 1947 Int BD23 Queen's Gambit Accepted
21. L Stumpers vs Grob 1-060 1947 Int BA55 Old Indian, Main line
22. V Soultanbeieff vs L Stumpers  ½-½46 1947 Int BD96 Grunfeld, Russian Variation
23. J T Barendregt vs L Stumpers  0-126 1948 NED-ch14C86 Ruy Lopez, Worrall Attack
24. L Stumpers vs A Vinken  0-133 1948 NED-ch14E21 Nimzo-Indian, Three Knights
25. L Stumpers vs C Vlagsma  0-145 1948 NED-ch14C65 Ruy Lopez, Berlin Defense
 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 287 OF 287 ·  Later Kibitzing>
Premium Chessgames Member
  john barleycorn: Still, my question: "When does the player lose?" There is no misreading possible as it is not mentioned in the game description.

the question:

<If the casino is to break even playing this game, how much should they charge — what should be the value of X?>

is absolutely meaningless in that case.

Jul-03-16  beatgiant: <john barleycorn>
The player loses money if the charge X he paid is higher than the amount he won.
Jul-03-16  beatgiant: The gambling game in question is the well-known "St. Petersburg Paradox." Wikipedia has a good article on it.
Premium Chessgames Member
  WannaBe: Here's the link, I'll save y'all from typing it in google. =)

Premium Chessgames Member
  john barleycorn: <beatgiant: <john barleycorn> The player loses money if the charge X he paid is higher than the amount he won.>

Ah, thank you. so the player pays an amount X at the beginning and it is automatically collected by the casino irrelevant of the outcome of the coin flips.

Premium Chessgames Member
  al wazir: It's convention time!

As is traditional, a banquet is held at the conclusion of the Republicratic convention. Two hundred senior partisans, selected by invitation from the party's most principled princes and dignified dignitaries, stream into the ballroom where tables have been set up.

When they arrive they find that seats has been preassigned. Every setting has a name tag, and the invitees are required to sit at the place with their own tag.


Two hundred mostly elderly, increasingly hungry, slightly tipsy (from pre-banquet cocktails) guests mill around, bumping into chairs and tables, stepping on each other's toes, as they try to locate their places. Some give up the search and sit at random, only to be displaced when the rightful occupant finds his name and claims the place. A few headstrong mavericks -- most of them long-serving legislators accustomed to their own safe electoral seats -- redistribute the name tags. As a result, some tables have more tags than chairs.

Is there a better way to do it? (Cf.

Jul-14-16  Schwartz: Just a small amount should search at the same time, unless there's a lot of room.
Premium Chessgames Member
  john barleycorn: get them seated by trained stuff.
Premium Chessgames Member
  Tiggler: More cocktails! Then bring them in on wheelbarrows.
Premium Chessgames Member
  Tiggler: Open seating is the reason Southwest Airlines boards twice as fast and saves billions.
Premium Chessgames Member
  al wazir: <Tiggler: Open seating is the reason Southwest Airlines boards twice as fast and saves billions.> Good point. But in this scenario the seating isn't open.

Since you mention airline seating, I should point out that some airlines *platoon* boarding passengers. Their tickets are labeled "Group 1," "Group 2," and so on, and the groups are assigned seats in such a way that Group 1 goes to the back of the plane where it won't interfere with Group 2. Then Group 2 takes the seats forward of them, and so on.

So one solution to the problem would be to give the banqueters tickets with table numbers. Alternatively, they could simply be told ahead of time what table they are assigned to, and at appropriate intervals the Maître d' could announce over the PA system "Table 1 now seating ... Table 2 now seating ...," etc. But I'm not convinced that there isn't a better way.

I recently attended an outdoor luncheon where they must have been around 200 guests, if not more. Owing to a conflict (it was a chess conflict, as a matter of fact), I arrived an hour after everyone else had been seated, and even knowing the table I was supposed to sit at I had a hell of a time finding my seat. So if everyone arrived at the same time it must have been a mess.

Premium Chessgames Member
  WannaBe: Collatz Conjecture:

Premium Chessgames Member
  john barleycorn: I know Collatz from his work on linear optimization but never knew he has his own problem . ha the easier the problem is to formulate the harder to solve.

Thanks for pointing it out

Premium Chessgames Member
  al wazir: Collatz conjecture: See Louis Stumpers (kibitz #4488) et seq.
Premium Chessgames Member
  WannaBe: Since we like to discuss math here, thought I'd post this:

Saw "The Man Who Knew Infinity" last night, solid movie. Very enjoyable.

I give it an "A".

Premium Chessgames Member
  alexmagnus: The Collatz conjecture mention here, huh. But as R.K. Guy said: <"Don't try to solve these problems!"> (so was the title of a paper including several unsolved "elementary" math problems, among them the Collatz)

Another problem from the same paper is similar, but involving division by 3 instead of 2:

Start with a natural number n. If it is divisible by 3, multiply it by 2/3. If the remainder from division by 3 is 1, take 4n/3-1/3. And if the remainder is 2, take 4n/3+1/3. The "unsolvable" problem: does the sequence beginning with n=8 ever go into a cycle or does it diverge? Also, what if we trace the same sequence from n=8 <backwards>?! It appears to go from large numbers, through n=8 and then back to large numbers.

Why n=8? Because for n<8 the sequence does go into a cycle (1 just repeats endlessly, 2 forms a (2,3)-cycle, 4 forms a (4,5,7,9,6) cycle). There is only one other cycle known - (44, 59, 79, 105, 70, 93, 62, 83, 111, 74, 99, 66). Whether any more exist, is another open question.

Premium Chessgames Member
  al wazir: Here's a puzzle inspired by the Collatz Conjecture. Take any positive integer, double it and add 1. If the result is a prime, stop. If not, double it again and add 1. If this number is a prime, stop. If not, keep repeating the process. Will the sequence of numbers resulting from this procedure always end in a prime?

Note that the sequences created this way increase as powers of 2, i.e., exponentially. The distribution of primes is known to be logarithmic. (This is the content of the Prime Number Theorem, Primes are rather abundant for small integers; they become less abundant the higher you go. But the '2x+1' procedure yields numbers that are even farther apart. So what we have here is a situation analogous to Arctic foxes hunting lemmings: the farther north you go, the rarer members of both species become, and by the time you reach the North Pole it is doubtful whether any kills take place.

I conjecture that the answer is no. Future mathematicians will refer to this as the Al Wazir Conjecture.

Here are some examples.

1, 3.
2, 5.
3, 7.
4, 9, 19.
5, 11.
6, 13.
7, 15, 31.
8, 17.
9, 19.
10, 21, 43.
11, 23, 47.
12, 25, 51, 103.
13, 27, 55, 111, 223.
14, 29.
15, 31.
16, 33, 67.
17, 35, 71.
18, 37.
19, 39, 79.
20, 41.

Premium Chessgames Member
  al wazir: Oops. The eleventh line of that list should read 11, 23. Sorry.
Premium Chessgames Member
  alexmagnus: <Al wazir> The answer is no. Your numbers have a form 2n+1, 4n+3, 8n+7 and so on - that is n*2^k+2^k-1 = (n+1)2^k-1. So, we need a number x for which x*2^k-1 is always composite - then x-1 is your counterexample.

The numbers x from above are known as Riesel numbers. It is proven that there are infinitely many of them. The smalles number <proven> to be Riesel is 509203 (for it, x*2^k-1 will always be divisible by one of 3, 5, 7, 13, 17, 241). There are some smaller numbers for which it is unknown whether they are Riesel or not (smallest of them being 2293).

So, if you start your sequence at 509202, you'll never end up with a prime.

Premium Chessgames Member
  alexmagnus: <for it [509203], x*2^k-1 will always be divisible by one of 3, 5, 7, 13, 17, 241>

In case you need the proof, here the remainders of 2x+1 recursion as a cycle for 509202:

3: (1, 0)
5: (0, 1, 3, 2)
7: (3, 0, 1)
13: (11, 10, 8, 4, 9, 6, 0, 1, 3, 7, 2, 5)
17: (3, 7, 15, 14, 12, 8, 0, 1)
241: (180, 120, 0, 1, 3, 7, 15, 31, 63, 127, 14, 29, 59, 119, 239, 238, 236, 232, 224, 208, 176, 112, 225, 210)

So, the numbers divisible by those have the k index which satisfies one of:

0 mod 2
1 mod 4
2 mod 3
7 mod 12
7 mod 8
3 mod 24

Now search for numbers that fit neither congruence.

First line eliminates all even numbers. Second line leaves only number 3 mod 4 behind. Third line: only 3 mod 4 AND 0 or 1 mod 3 - that is (by the Chinese remainder theorem), either 3 mod 12 or 7 mod 12.

The next line eliminates the numbers 7 mod 12 - leaving only with 3 mod 12.

The next line tells that our 3 mod 12 numbers may not be 7 mod 8 - that is, they are 0, 1, 2, 3, 4, 5 or 6 mod 8. 0, 2, 4, and 6 fall out because they are not 1 mod 2. 1 and 5 fall out because they are not 3 mod 4. So, our number is 3 mod both 8 and 12 - that is, 3 mod 24.

But the last line excludes numbers 3 mod 24.

That means, no counterexamples exist.

Premium Chessgames Member
  Domdaniel: <alexmagnus> That's beautiful, thank you. I knew that recursive loops existed, but not where to find them.
Premium Chessgames Member
  al wazir: <alexmagnus>: Outstanding!

That shoots downs my bid to become an eponym. But it looks as if Riesel was ahead of me in line.

Premium Chessgames Member
  Domdaniel: <al wazir> -- < it looks as if Riesel was ahead of me > That's nothing - I've 'discovered' large chunks of 18th century maths. I once even 'invented' Bernoulli numbers, about 300 years after Bernoulli. And don't get me started on Euler, who repeatedly pipped me to the post... with 250 years to spare.
Premium Chessgames Member
  john barleycorn: <Domdaniel: ... I once even 'invented' Bernoulli numbers, about 300 years after Bernoulli. ...>

We all know how hard it is to invent something that is already invented.

Premium Chessgames Member
  alexmagnus: When I was 10, I "invented" logarithms.

Actually I often come up with some mathematical ideas - but Googling quickly reveals those ideas are well known. This is actually why I knew about the Riesel numbers - I once had ideas along similar lines as <al wazir> (slightly differnt - but eventually also arriving at the question of existence of what is known as Riesel numbers). The proof I posted if from me though - I'm pretty sure if you google you'll find some quicker proof (after all, my proof comes from <knowing> that 509203 is a Riesel number, while whoever made the original proof arrived at it first somehow!)

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