Most of the talks were 5 minutes so they could tell you a problem or thought of interest but not much more. This is GOOD in that I UNDERSTOOD most of the talks! There were also some talks on Magic and on science literacy (or perhaps science illiteracy) which were also interests of Martin Gardner.
I posted on one problem and its solution so this is either my first or third post on G4G13.
Withouth further ado, here are my descriptions of some of the talks. OR you could just go here
My descriptions are long- I may have lots of posts on G4G13!
EXTENDING THE 10958 PROBLEM by Bill Ames.
(paper on arxiv here)
In 2014 Taneja proposed the following math problem:
Take the digits 1,2,3,4,5,6,7,8,9.
You can put any of +, -, *, /, and powers you want between the digits, or erase the comma to get (for example) 34. Use parenthesis freely. Which natural numbers can you produce?
Here are some examples:
1 = 1 + (2 - 3) - (4 - 5) - (6 - 7) + (8 - 9)
30 = 1 + 2^3 - 4*5 + 6*7 + (8-9)
Which numbers can you get?
Taneja got all the numbers between 0 and 11,111 EXCEPT 10958.
(Is 10958 possible? The talk didn't say.)
Using other operations- square roots and factorials- you can get 10958. The talk was about adding more operations and getting all numbers between 0 and 111,111
THIRTEEN BOUNCES by Gary Antonick.
Since this was G4G13 there were several talks about the number 13. The hotel did not have a 13th floor, and no room was of the form X13 (I was in room 515 and sometimes overshot my room since I assumed it would go 511-513-515, not 511-515). Martin Gardner would have OBJECTED to the lack of a 13th floor and would have condemned triskaidekaphobia (fear of the number 13 and the longest word I ever spelled correctly) as irrational. And he would be right. But I am preaching to the converted (also known as preaching to the choir).
The talk was about firing a cannoball at a perfect reflector. If the cannon is positioned just right the ball will go back into the cannon. Can you make it do this after taking 13 bounces? What if you an only use a compass and straightedge for your calculations?
A GRAMMATICAL APPROACH TO THE CURLING NUMBER CONJECTURE by Duane Bailey.
(For more on the Curling Conjecture Google "Curling Number Conjecture"-- I was going to give a list of papers but there are just too many. There is no wikipedia entry on it, though I did learn about the sport of Curling!)
Take any finite sequence of integers (we will just use natural numbers) for example
2 3 2 3
Write it as X Y Y Y ... Y with as many Y's as possible.
for example
(2 3)2
The number 2 is the Curling Number of the sequence. Append it to the sequence
2 3 2 3 2
Now take the Curling number of this sequence. It is 2 since the sequence is
2 (3 2)2
Append it to the sequence:
2 3 2 3 2 2
Now the Curlng number of this sequence is 1 and we stop.
The conjecture is that if you start with any sequence you will eventually get to 1.
The talk relates the conjecture to aperiodic grammars.
232322. "the Curling number of this sequence is 1 and we stop".
ReplyDeleteActually, 232322 = XYY with X=2323 and Y=2.
An interesting extension to Taneja's problem is to find the minimal length string that can be parsed to produce each natural while keeping his rules. For example, in his paper he has 5556 = 1 + 2 + 3 × (4 × 56 + 7) × 8 + 9. But 5556 = 12345 - 6789 as well.
ReplyDeletewhy do you block comments ??? What it the correct title of this POST : GAHERING OR GATHERING ?
ReplyDelete1) Fixed- thanks
Delete2) Blocked comments? What are you referring to?
Just an algorithmic trick to quickly scan all combinations (feasible for 9 digits and 5 operators): just use postfix notation and recursion
ReplyDeleteE.g.:
"9 8 7 6 5 4 * - 3 ^ + * 2 / 1 - -"
9 - (((8 * (7 + (6 - (5 * 4)) ^ 3)) / 2) - 1)) = 10958
(nothing new because the author says that 10958 is missing from the increasing sequence, not the decreasing one).
In pseudocode:
procedure rec( String pfix, int currdig, int stack[] ) {
if ( currdig == 10 and stack.size == 1 ) {
print( "Found value: " + stack[0] + " postfix expression: " + pfix);
return;
}
if ( currdig < 10 ) {
pfix += currdig + " ";
stack2[] = stack.push(currdig);
rec( pfix, currdig + 1, stack2 );
}
if ( stack.size >= 2) {
for ( i = 0; i < NUMOPS; i++ ) {
String OP = ops[i];
pfix += OP + " ";
stack2 = stack.push ( calc( stack.pop() OP stack.pop() ) );
rec( pfix, currdig, stack2 );
}
}
}
procedure main() {
rec("",1,[]);
}