Question

pick an integer from 0 to 100 ... can i guess it in 7 tries?

Answer 1

OK, let's go.

Answer 2

rules, of too low, say higher

if too high, say lower

Answer 3

is it 50?

Answer 4

go higher

Answer 5

50+25 ... is it 75?

Answer 6

isn't this some algorithm?

Answer 7

go lower

Answer 8

It's witchcraft.

Answer 9

50+25-12, is it 63

Answer 10

go higher

Answer 11

50+25-12+6, is it 69?

Answer 12

go lower

Answer 13

50 + 25 - 12 + 6 - 3, is it 66

Answer 14

go higher

Answer 15

so, its either 67 or 68 and i have 2 guess left :)

Answer 16

bazinga. 68.

Answer 17

2^7 = 128 so you are bound to limit in on the interval

Answer 18

I was just typing that... haha

Answer 19

Oh!

Answer 20

Cool trick for shrinking a domain

Answer 21

yep, i programmed a rabbit to follow a predefined curve with it :) well, a binary rabbit

Answer 22

i used 1/2^100 chances to get within an acceptable error to move along the curve

Answer 23

so you can do this with any number b/w 1 to 128 in theory?

Answer 24

[0,128]
[0, 64]
[0, 32]
[0,16]
[0,8]
[0,4]
[0,2] .. 0,1,2 is not a good bet for picking 1 of 3

Answer 25

i spose we start with 0,128 bing a 0 guess lol

Answer 26

[0,128]
[0, 64]
[0, 32]
[0,16]
[0,8]
[0,4]
[0,2]
[0,1] .. looks good to me

Answer 27

highest interval for 7 guesses, I meant.

Answer 28

by some version of the nested interval thrm, you can reduce the range of options

Answer 29

well for a safe bet, 128 seems to be the largest

Answer 30

0,130

0,65,130

0,32,65,97,130

0,16,32,48,65,81,97,113,130

0,8,16,24,32,40,48,56,65,73,81,89,97,105,113,121,130

0,4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,65,69,73,
77,81,85,89,93,97,100,105,109,113,117,121,125,130

largest interval is 5 and 2 guess left

Answer 31

largest interval if not guessed is between 0 and 5:

1234 elements
g
3 or 4 for a final guess is not a sure thing

Answer 32

it seems to me in order to generalize it, it'd have to be \(x<2^{n+1}\) where n is the number of guesses

Answer 33

seems like a fair generalization

Answer 34

\[x\le2^n\]

Answer 35

Sorry, 2^n with n+1 being the number of guesses

Answer 36

2^7 = 128, and in 7 guess we can get to x

130 < 2^8 and we would need 8 guess to be sure

Answer 37

10 < 2^4 .... 4 guess

0,5,10

0,2,5,7,10

1 34 6 89 in 2 guesses

Answer 38

with 128 it took 8 guesses I thought

Answer 39

nah, the initial state is not a guess, it was just stating the interval

Answer 40

whoops

Answer 41

I see now

Answer 42

guess 0 was [0,n] :) which aint a guess its just initializing the setup

Answer 43

true, but I count it as a guess

Answer 44

first guess, is it 128? lol

Answer 45

id accept that

Answer 46

so then we get this relation \(2^{n-1}<x\le 2^n\)

Answer 47

lets assume the solution is 128

64
32
16
8
4
2
1

yeah, we would need 1 more guess for 0 or 128

Answer 48

but more often than not, people will not choose an endpoint :)

Answer 49

true

Answer 50

but x is the number of options not the number

Answer 51

0-128 has 129 options

Answer 52

im ready for a thrm :)

if \(k\le2^n,~k\not\in Z^-,\) then the number of guess needed is \(\small n+1\)

Answer 53

I think you need to define k

Answer 54

or possibly k in Z+ or 0

Answer 55

and at maximum n+1

Answer 56

maximum yes, or safest number of guesses

Answer 57

the least upper bound lol

Answer 58

true lol So what did we develop this grand theory for?

Answer 59

annd also, if you publish this, I want my name on it! haha

Answer 60

posterity?

Answer 61

is that 1 c or 2?

Answer 62

haha, and I'll let you cite my real name if you do, but it's 2 ;P

Answer 63

oh im sure someone else many hundreds of years ago ironed this out and got shot afterwards :) right Cardano?

Answer 64

3 total

Answer 65

lmao

Answer 66

one of them was a gambler ... prolly not Cardano, or was it?

Answer 67

idk, but gtg class I'll check back later

Answer 68

have fun ;)