phil1882

come on brainden, just need one more person!

by that I'm guessing you mean most unusual, but every number has some interesting/unusual property about it.

many gears spinning 'round, numbers spill too unbound. sapping energy of a spring, adding adding everything. whether weather is need to know, or complex equations bust from flow. memory, shift, register, digest; come forth solution, progamming guest!

by the way: i have a bunch of semi humorous achievements lined up that you can get as you play. they won't affect the game of course, but they will add an element of levity to the game, and I'll give the person that gets the most my coveted Gold Eagle General award :-).

by the way, there is one with 8 if you allow division.

I'm pretty sure, I've tried several approaches, including powers of 3, and 5.

sure!

edit: subtraction is clearly nessicary in some edge cases, such as 1023.

question: is division ever faster? the opening post mentioned that it affects the problem, but i think multiplication, addition and subtraction are enough to get any minimum number of steps. I'm not sure if even subtraction is necessary.

3*3 = 9, you're allowed to use any combination of previous values to get the next one.

how about this tie sequence...

in this game, each player is given a fleet of 10 star-ships to conquer his opponents. for each planet you conquer, you get more production per turn. specifically: (1 2 3 4 5 6 7 8) A 10 2 3 5 5 3 2 10 B 2 5 4 6 6 4 5 2 C 3 4 6 8 8 6 4 3 D 5 6 8 9 9 8 6 5 E 5 6 8 9 9 8 6 5 F 3 4 6 8 8 6 4 3 G 2 5 4 6 6 4 5 2 H 10 2 3 5 5 3 2 10 each player starts off in their corner. you can move as many ships as you desire, but to conquer a world, the amount you send must be higher than the number currently there. note that in the course of battle the number of ships you lose will be equivalent to the amount you fight. to make things interesting, you can only move like a chess knight. after everyone has submitted their move, the game state will be updated. for every planet you control each one will get +2 ships. you can have multiple planets attack the same target. an example first round... red attacks B3 and C2 with 5 each. blue attacks B6 and C7 with 6 and 4 respectively. green attacks F2 with all 10. purple attacks F7 with 7, and saves the remaining 3 for the next turn. after production the game would look like this: (1 2 3 4 5 6 7 8) A 2 2 3 5 5 3 2 2 B 2 5 3 6 6 4 5 2 C 3 3 6 8 8 6 2 3 D 5 6 8 9 9 8 6 5 E 5 6 8 9 9 8 6 5 F 3 8 6 8 8 6 5 3 G 2 5 4 6 6 4 5 2 H 2 2 3 5 5 3 2 5 anyone interested?

dead?

you have the following allowable operations, addition, multiplication, and parenthesis. further more the only number you are allowed to use is 1. for example, the number 15 could be (1+1+1)*((1+1)*(1+1)+1) express 137 with the fewest number of 1's in a single expression.

hmm my above logic doesn't seem quite right, i'm sure something is off there; you should need to go at least 2 layers to find a contradiction.

so i guess the next logical question is this, if every irrational number is countably infinite in it representation, can you represent all irrational numbers in a countably infinite way? that is, we know we can form any individual irrational number with a series of 1's and 0's. there is no individual irrational number we can't form this way. now imagine we have a countably infinite number of people flipping coins. which irrational number won't be hit?

I'm very interested in knowing how you get 16 flips for the value pi, since pi is transcendental.

okay maybe some clarification is in order. clearly no finite number of flips will be enough, assuming you're not allowed to apply a function to it. but can you mathematically flip an infinite number of coins to represent an irrational number? i guess what I'm asking is, is the digits of a single irrational number countably infinite?

let heads be 1 and tails be 0. can you flip a coin enough times to form an irrational number?