Problem Statement
King Arthur was the ruler in Camelot who had all of those knights over for parties around his round table. He loved inviting the knights over for parties around his round table, he would have them play a game to determine which lucky knight would get something pleasant from the king. First, King Arthur put numbers on the chairs beginning with 1 and continuing around the table, with one chair for each knight. He had the knights sit down so that every chair was occupied. King Arthur himself remained standing. He them went to the first chair and said "you're in" and to the second chair "you're out" and to the third "you're in", he went around the entire table going you're in and you're out until there was only one knight siting. Once, he had gone around the table completely he would say either in or out depending on what he had said to the last person before restarting. The number of knights always varied so one night there could've been 5 knights or 100 knights at his table.
BIG QUESTION
BIG QUESTION
If you knew how many knights were going to be at the table, how
could you quickly determine which chair to sit in
so that you would win?
could you quickly determine which chair to sit in
so that you would win?
Process
When beginning the problem we worked individually to see what each one us could come up with to build a formula and then we would come together as a group to show each other our ideas. What we decided to do first was build an x and y table. Where x was the amount of knights in the table and y being the winning seat/knight.
The pattern we found was that the knight in seat #1 wins if there are 4, 8, 16, 32, 64, etc. knights at the table. The common thing we found with all these numbers were that they all came from 2 to the power of anything. So if the knights equalled 2 to the power of anything the winning seat would be #1. We then noticed that the number after 2 to the power of anything, for example 5, 9, 17, 33, 65 the winning seat would be #3, and the one after that #5 and so on until the winning chair reset to #1 and the whole pattern would start over again. We also noticed that an even numbered seat would never win. It would always be odd. After noticing this pattern we knew for sure what number of knights would cause seat #1 to win.
The pattern we found was that the knight in seat #1 wins if there are 4, 8, 16, 32, 64, etc. knights at the table. The common thing we found with all these numbers were that they all came from 2 to the power of anything. So if the knights equalled 2 to the power of anything the winning seat would be #1. We then noticed that the number after 2 to the power of anything, for example 5, 9, 17, 33, 65 the winning seat would be #3, and the one after that #5 and so on until the winning chair reset to #1 and the whole pattern would start over again. We also noticed that an even numbered seat would never win. It would always be odd. After noticing this pattern we knew for sure what number of knights would cause seat #1 to win.
Solution
For the solution we ended coming up with formula which was 2(x-2 [{ log2^]}+1).Our teacher thought this would be an easier way to solve the problem because we are using log which is something that we used in the past. We were also showed us the integer function and it would then be combined with log which would replace the variable X in the formula.
Evaluation/Reflection