On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
This topic was previously discussed in another thread, but I don't think
it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
and this point seems unresolved in that thread. I aim to resolve this
point here, completely rigorously.
Let us call a position "crowded" if it contains strictly more than 15 checkers
on the bar.
Suppose, by contradiction, that we have reached the first occurrence
of a crowded position in a game.
Let k be the number of checkers that were brought in from the bar on the
previous roll.
If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
hit myself so none of my checkers get sent to the bar, leaving a max of 15. >> Suppose k is 1. If I bring that checker in, we again get a max of 15, as
before, because I can't hit myself.
If the previous roll is a dance, then the position repeats contradicting the >> claim of first occurrence.
Suppose k >= 2. If I dance completely, the previous argument applies.
If I dance with one man, we either end up with fewer checkers on the
bar or the same number of checkers on the bar depending on whether
the entering number hits. If I don't dance at all, then I take at least two >> checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
Paul
Sorry, k needs to be the number of the last-roller's checkers that were on the bar before the
most recent shake.
This topic was previously discussed in another thread, but I don't think
it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
and this point seems unresolved in that thread. I aim to resolve this
point here, completely rigorously.
Let us call a position "crowded" if it contains strictly more than 15 checkers
on the bar.
Suppose, by contradiction, that we have reached the first occurrence
of a crowded position in a game.
Let k be the number of checkers that were brought in from the bar on the previous roll.
If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
hit myself so none of my checkers get sent to the bar, leaving a max of 15. Suppose k is 1. If I bring that checker in, we again get a max of 15, as before, because I can't hit myself.
If the previous roll is a dance, then the position repeats contradicting the claim of first occurrence.
Suppose k >= 2. If I dance completely, the previous argument applies.
If I dance with one man, we either end up with fewer checkers on the
bar or the same number of checkers on the bar depending on whether
the entering number hits. If I don't dance at all, then I take at least two checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
Paul
On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
This topic was previously discussed in another thread, but I don't think >> it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
and this point seems unresolved in that thread. I aim to resolve this
point here, completely rigorously.
Let us call a position "crowded" if it contains strictly more than 15 checkers
on the bar.
Suppose, by contradiction, that we have reached the first occurrence
of a crowded position in a game.
Let k be the number of checkers that were brought in from the bar on the >> previous roll.
If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
hit myself so none of my checkers get sent to the bar, leaving a max of 15.
Suppose k is 1. If I bring that checker in, we again get a max of 15, as >> before, because I can't hit myself.
If the previous roll is a dance, then the position repeats contradicting the
claim of first occurrence.
Suppose k >= 2. If I dance completely, the previous argument applies.
If I dance with one man, we either end up with fewer checkers on the
bar or the same number of checkers on the bar depending on whether
the entering number hits. If I don't dance at all, then I take at least two
checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
Paul
Sorry, k needs to be the number of the last-roller's checkers that were on the bar before theThanks for clarifying...I was having trouble following your argument.
most recent shake.
So when you say, "we have reached the first occurrence of a crowded position," you mean for example that White has just completed her
checker play, leaving a crowded position with Black on roll?
And k is the number of White's checkers on the bar? And in your narrative, is "I" White or Black?
Tim Chow
On Thursday, March 30, 2023 at 1:49:02 PM UTC+1, Timothy Chow wrote:
On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
This topic was previously discussed in another thread, but I don't think
it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
and this point seems unresolved in that thread. I aim to resolve this >> point here, completely rigorously.
Let us call a position "crowded" if it contains strictly more than 15 checkers
on the bar.
Suppose, by contradiction, that we have reached the first occurrence
of a crowded position in a game.
Let k be the number of checkers that were brought in from the bar on the
previous roll.
If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
hit myself so none of my checkers get sent to the bar, leaving a max of 15.
Suppose k is 1. If I bring that checker in, we again get a max of 15, as
before, because I can't hit myself.
If the previous roll is a dance, then the position repeats contradicting the
claim of first occurrence.
Suppose k >= 2. If I dance completely, the previous argument applies. >> If I dance with one man, we either end up with fewer checkers on the
bar or the same number of checkers on the bar depending on whether
the entering number hits. If I don't dance at all, then I take at least two
checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
Paul
Sorry, k needs to be the number of the last-roller's checkers that were on the bar before theThanks for clarifying...I was having trouble following your argument.
most recent shake.
So when you say, "we have reached the first occurrence of a crowded position," you mean for example that White has just completed herWell, yes. But how could I possibly have meant anything else?
checker play, leaving a crowded position with Black on roll?
Please could you explain the ambiguity here?
And k is the number of White's checkers on the bar? And in your narrative, is "I" White or Black?White.
Tim Chow
On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
This topic was previously discussed in another thread, but I don't think >> it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
and this point seems unresolved in that thread. I aim to resolve this
point here, completely rigorously.
Let us call a position "crowded" if it contains strictly more than 15 checkers
on the bar.
Suppose, by contradiction, that we have reached the first occurrence
of a crowded position in a game.
Let k be the number of checkers that were brought in from the bar on the >> previous roll.
If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
hit myself so none of my checkers get sent to the bar, leaving a max of 15.
Suppose k is 1. If I bring that checker in, we again get a max of 15, as >> before, because I can't hit myself.
If the previous roll is a dance, then the position repeats contradicting the
claim of first occurrence.
Suppose k >= 2. If I dance completely, the previous argument applies.
If I dance with one man, we either end up with fewer checkers on the
bar or the same number of checkers on the bar depending on whether
the entering number hits. If I don't dance at all, then I take at least two
checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
Paul
Sorry, k needs to be the number of the last-roller's checkers that were on the bar before theThanks for clarifying...I was having trouble following your argument.
most recent shake.
So when you say, "we have reached the first occurrence of a crowded position," you mean for example that White has just completed her
checker play, leaving a crowded position with Black on roll? And
k is the number of White's checkers on the bar? And in your narrative,
is "I" White or Black?
---
Tim Chow
On Thursday, March 30, 2023 at 1:49:02 PM UTC+1, Timothy Chow wrote:These rather tortuous explanations are dealt with more succinctly in Backgammon Funfair: "It is possible to have fifteen White checkers on the bar. Then a Red checker can come to the bar only by being hit by a White checker entering from the bar.
On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
This topic was previously discussed in another thread, but I don't think
it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
and this point seems unresolved in that thread. I aim to resolve this >> point here, completely rigorously.
Let us call a position "crowded" if it contains strictly more than 15 checkers
on the bar.
Suppose, by contradiction, that we have reached the first occurrence
of a crowded position in a game.
Let k be the number of checkers that were brought in from the bar on the
previous roll.
If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
hit myself so none of my checkers get sent to the bar, leaving a max of 15.
Suppose k is 1. If I bring that checker in, we again get a max of 15, as
before, because I can't hit myself.
If the previous roll is a dance, then the position repeats contradicting the
claim of first occurrence.
Suppose k >= 2. If I dance completely, the previous argument applies. >> If I dance with one man, we either end up with fewer checkers on the
bar or the same number of checkers on the bar depending on whether
the entering number hits. If I don't dance at all, then I take at least two
checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
Paul
Sorry, k needs to be the number of the last-roller's checkers that were on the bar before theThanks for clarifying...I was having trouble following your argument.
most recent shake.
So when you say, "we have reached the first occurrence of a crowded position," you mean for example that White has just completed her
checker play, leaving a crowded position with Black on roll? And
k is the number of White's checkers on the bar? And in your narrative,
is "I" White or Black?
---I think we can be a lot simpler as follows:
Tim Chow
Clearly the 15 figure can be obtained by making an inner board point and then strolling around,
picking up all 15 blots.
But can this be beaten?
To get more than 16 checkers on the bar, we need checkers on the bar of both colours.
Consider the position immediately before the maximum was first obtained, and consider the
next roll.
If all that roll did was bring checkers in from the bar, with no time to do extra non-entering plays.
Then clearly there were at least as many checkers entered as blots hit, and we didn't increase
the on-bar number.
Alternatively, suppose that this prior roll did make non-entering plays. That means that
the prior roller had all their checkers off the bar and we don't beat 15.
Paul
On Friday, March 31, 2023 at 2:07:45 PM UTC+1, peps...@gmail.com wrote:the total on the bar cannot exceed fifteen."
On Thursday, March 30, 2023 at 1:49:02 PM UTC+1, Timothy Chow wrote:
On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
This topic was previously discussed in another thread, but I don't think
it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
and this point seems unresolved in that thread. I aim to resolve this >> point here, completely rigorously.
Let us call a position "crowded" if it contains strictly more than 15 checkers
on the bar.
Suppose, by contradiction, that we have reached the first occurrence >> of a crowded position in a game.
Let k be the number of checkers that were brought in from the bar on the
previous roll.
If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
hit myself so none of my checkers get sent to the bar, leaving a max of 15.
Suppose k is 1. If I bring that checker in, we again get a max of 15, as
before, because I can't hit myself.
If the previous roll is a dance, then the position repeats contradicting the
claim of first occurrence.
Suppose k >= 2. If I dance completely, the previous argument applies. >> If I dance with one man, we either end up with fewer checkers on the >> bar or the same number of checkers on the bar depending on whether
the entering number hits. If I don't dance at all, then I take at least two
checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
Paul
Sorry, k needs to be the number of the last-roller's checkers that were on the bar before theThanks for clarifying...I was having trouble following your argument.
most recent shake.
So when you say, "we have reached the first occurrence of a crowded position," you mean for example that White has just completed her checker play, leaving a crowded position with Black on roll? And
k is the number of White's checkers on the bar? And in your narrative, is "I" White or Black?
---I think we can be a lot simpler as follows:
Tim Chow
Clearly the 15 figure can be obtained by making an inner board point and then strolling around,
picking up all 15 blots.
But can this be beaten?
To get more than 16 checkers on the bar, we need checkers on the bar of both colours.
Consider the position immediately before the maximum was first obtained, and consider the
next roll.
If all that roll did was bring checkers in from the bar, with no time to do extra non-entering plays.
Then clearly there were at least as many checkers entered as blots hit, and we didn't increase
the on-bar number.
Alternatively, suppose that this prior roll did make non-entering plays. That means that
the prior roller had all their checkers off the bar and we don't beat 15.
PaulThese rather tortuous explanations are dealt with more succinctly in Backgammon Funfair: "It is possible to have fifteen White checkers on the bar. Then a Red checker can come to the bar only by being hit by a White checker entering from the bar. Hence
I don't see this argument as being in any way sketchy.
These rather tortuous explanations are dealt with more succinctly in Backgammon Funfair: "It is possible to have fifteen White checkers on the bar. Then a Red checker can come to the bar only by being hit by a White checker entering from the bar.Hence the total on the bar cannot exceed fifteen."
I don't see this argument as being in any way sletchy.
On 3/31/2023 1:29 PM, Raymond Kershaw wrote:Hence the total on the bar cannot exceed fifteen."
These rather tortuous explanations are dealt with more succinctly in Backgammon Funfair: "It is possible to have fifteen White checkers on the bar. Then a Red checker can come to the bar only by being hit by a White checker entering from the bar.
I don't see this argument as being in any way sletchy.The trouble is that this argument only says what can happen if you
start with 15 White checkers on the bar. It doesn't prove that
there isn't some other way of getting to (for example) 8 White
checkers and 8 Red checkers on the bar without first getting 15
checkers of one color on the bar.
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