Here are three problems, all original with me, that you can solve in your head. They're perfect for keeping from being bored when you're out walking or otherwise unable to safely read or do other tasks. Inspired by Sloane's strange-looking A258107 (2, 3, 4, 82000), I thought about the smallest number (other than "1") that's represented by nothing but odd digits in the first N bases (binary, ternary, etc.). I quickly proved to myself that no number has this property for bases 2, 3, and 4. What about just in odd bases? I quickly proved to myself that no number (except "1") has this property in bases 3, 5, 7, and 9. I had been saving the dead president's problem for just after the next president died. But I've decided that would be in bad taste. So here goes: There are currently 38 dead US presidents. Their dates of birth and death are all known. If I sort them in order of birth, they're in a certain order, starting with Washington and ending with Kennedy. If I sort them in order of death, they're in a slightly different order, starting with Washington and ending with Ford. Suppose I sort them by the date on which they had been (or will have been) dead longer than they were alive? I get yet a third slightly different order. (Once again it begins with Washington and ends with Ford. So does sorting them by when they will have been dead twice as long as they were alive.) The question is: How many different sort orders are there, as you vary the parameter X for "dead X times longer than alive" over all real numbers, including negatives. Yes, you can solve it in your head, and no, the answer isn't 38 factorial. All of these puzzles of course give rise to more challenging puzzles.
If done w/ sufficient resolution, and barring a measure-zero set of coincidences, your procedure w/ N dead presidents should result in N(N-1)/2 different orderings. And nice question -- I solved it while driving in the car to get dinner. :-) On 2016-03-26 12:06, Keith F. Lynch wrote:
Here are three problems, all original with me, that you can solve in your head. They're perfect for keeping from being bored when you're out walking or otherwise unable to safely read or do other tasks.
[...]
I had been saving the dead president's problem for just after the next president died. But I've decided that would be in bad taste. So here goes:
There are currently 38 dead US presidents. Their dates of birth and death are all known. If I sort them in order of birth, they're in a certain order, starting with Washington and ending with Kennedy. If I sort them in order of death, they're in a slightly different order, starting with Washington and ending with Ford. Suppose I sort them by the date on which they had been (or will have been) dead longer than they were alive? I get yet a third slightly different order. (Once again it begins with Washington and ends with Ford. So does sorting them by when they will have been dead twice as long as they were alive.)
The question is: How many different sort orders are there, as you vary the parameter X for "dead X times longer than alive" over all real numbers, including negatives. Yes, you can solve it in your head, and no, the answer isn't 38 factorial.
All of these puzzles of course give rise to more challenging puzzles.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 27/03/2016 03:38, William R. Somsky wrote:
If done w/ sufficient resolution, and barring a measure-zero set of coincidences, your procedure w/ N dead presidents should result in N(N-1)/2 different orderings. And nice question -- I solved it while driving in the car to get dinner. :-)
I think you have an off-by-one error. N=1: N(N-1)/2 = 0, but obviously there's one ordering. N=2: N(N-1)/2 = 1. Suppose we have A (born 0, died 1) and B (born -1, died 2). Then just "born" and "died" already give us both possible orderings. N=3: N(N-1)/2 = 3. Suppose we have A(0..0), B (1..2) and C(0..3). Then, if h is very small and positive, we have the following orderings: - b + h(d-b): A < C < B - b - h(d-b): C < A < B - b + 1(d-b): A < B < C - b - 9(d-b): C < B < A (You get one *change* in ordering per "line joining two of the points", not one *ordering*.) -- g
Um.. Yeah.. Off by one. I gave how many changes of orderings there are, so add one for the total number of orderings. DOH! ;-) On Mar 27, 2016 04:01, "Gareth McCaughan" <gareth.mccaughan@pobox.com> wrote:
On 27/03/2016 03:38, William R. Somsky wrote:
If done w/ sufficient resolution, and barring a measure-zero set of
coincidences, your procedure w/ N dead presidents should result in N(N-1)/2 different orderings. And nice question -- I solved it while driving in the car to get dinner. :-)
I think you have an off-by-one error.
N=1: N(N-1)/2 = 0, but obviously there's one ordering.
N=2: N(N-1)/2 = 1. Suppose we have A (born 0, died 1) and B (born -1, died 2). Then just "born" and "died" already give us both possible orderings.
N=3: N(N-1)/2 = 3. Suppose we have A(0..0), B (1..2) and C(0..3). Then, if h is very small and positive, we have the following orderings: - b + h(d-b): A < C < B - b - h(d-b): C < A < B - b + 1(d-b): A < B < C - b - 9(d-b): C < B < A
(You get one *change* in ordering per "line joining two of the points", not one *ordering*.)
-- g
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Very cute puzzle. Believe it or not, this is related to the clever algorithm of Nimrod Megiddo for linear programming in 2 dimensions. The idea is that if you have N linear inequalities in 2 dimensions (which is what this problem had) the vertices of the polytope given by their intersection come from intersections of of pairs of the lines (there might be some degeneracy) which are N(N-1)/2. Megiddo's idea for minimizing a linear form over the polytope was a divide and conquer, which done cleverly, only took time *linear* in N. Victor On Sun, Mar 27, 2016 at 8:33 PM, William R Somsky <wrsomsky@gmail.com> wrote:
Um.. Yeah.. Off by one. I gave how many changes of orderings there are, so add one for the total number of orderings. DOH! ;-) On Mar 27, 2016 04:01, "Gareth McCaughan" <gareth.mccaughan@pobox.com> wrote:
On 27/03/2016 03:38, William R. Somsky wrote:
If done w/ sufficient resolution, and barring a measure-zero set of
coincidences, your procedure w/ N dead presidents should result in N(N-1)/2 different orderings. And nice question -- I solved it while driving in the car to get dinner. :-)
I think you have an off-by-one error.
N=1: N(N-1)/2 = 0, but obviously there's one ordering.
N=2: N(N-1)/2 = 1. Suppose we have A (born 0, died 1) and B (born -1, died 2). Then just "born" and "died" already give us both possible orderings.
N=3: N(N-1)/2 = 3. Suppose we have A(0..0), B (1..2) and C(0..3). Then, if h is very small and positive, we have the following orderings: - b + h(d-b): A < C < B - b - h(d-b): C < A < B - b + 1(d-b): A < B < C - b - 9(d-b): C < B < A
(You get one *change* in ordering per "line joining two of the points", not one *ordering*.)
-- g
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (5)
-
Gareth McCaughan -
Keith F. Lynch -
Victor Miller -
William R Somsky -
William R. Somsky