Smal Mathgame inbetween

[AddyBaha]

There are three rules to the game:

1. You have to write an equation using exact four 1s
2. Each post with an equation has to result in a number exactly 1 bigger than the last
3. You can use any operator and combination of brackets you like

I'll start

1+((1-1)*1)=1

edit: later on you can use "." to concate e.g.: 9.8 would be 98

[ryoloth]

1+1+(1-1)=2

[Gilberreke]

(1+1+1)*1=3

[magikeh]

(1+1+1+1)= 4

:D I got the easy one

Edit: After the initial 10 digits have been found does this game not become quite trivial in that any number can be created via concatenations?

[BinoAl]

(1+1+1)!-1

[AddyBaha]

((1+1+1)!)*1

After the initial 10 digits have been found does this game not become quite trivial in that any number can be created via concatenations
No since you are only allowed to use four 1s

[dawnraider]

(1+1+1)!+1 = 7

[jorgebonafe]

(1.(log(1))) -(1+1)

Does this count?

[AddyBaha]

(1.(log(1))) -(1*1)

yeah these ones count as well

[magikeh]

(1-1)*1+(Log(1)) = 10

Right, forgot about that only using four ones. Brain told me that we could somehow use 4 ones in each concat

EDIT: Revised: 1*1*1.log(1)

[jorgebonafe]

(1*1*1).1

(1-1)*1+(Log(1)) = 10

Uh.. I don't think that's right... Your is = 0

[magikeh]

(1*1).(1+1) = 12

(1*1*1).1

(1-1)*1+(Log(1)) = 10

Uh.. I don't think that's right... Your is = 0

(1-1)*1 + (Log(1)) = 10
( 0 )*1 + (10)
0 + 10

Edit: Yepp I am waay wrong. missed the concatenation with one on the log(1) to make 10 from above solutions. I instead somehow reasoned that log(1) was equivalent to 10^1 >.>

[William the tuba]

Log(1) is zero, as 10 ^ 0 = 1

And, we skipped 9, but it's easy to catch up to where we were supposed to be.

1 . (1-1) - 1 = 9
1 . (1-1) * 1 = 10
1 . 1 * 1 * 1 = 11
1 . (1 + 1) * 1 = 12
1 . (1 + 1 + 1) = 13

[EtherealWrath]

1.(1/(1+1))= probably cheating to call that 15 :p
1.(1+1+1)!=16

[William the tuba]

If we're allowed to use functions:

int(sqrt(1 + 1) * (1 . sin(1))) = 14

I'm not sure 1 . 0.5 = 15 or 105, but you can use a similar trick for 15:

int(sqrt(1 + 1) * (1 . 1)) = 15

and, for 17 and 18 (assuming tanh and sin is taking radians)

1 . int((1 . log(1)) * tanh(1)) = 17

1 . int((1 . log(1)) * sin(1)) = 18

[dawnraider]

1.(1.log(1))-1 = 19
(1+1).(1-1) = 20
(1+1).(1*1) = 21
(1+1).(1+1) = 22
EDIT: Couple mistakes plus some new numbers
(1+1).int(arcsin(1) + arcsin(1)) = 23
(1+1+1+1)! = 24

[Wafflewaffle]

1.(1.log(1))-1 = 19
(1+1).(1-1) = 20
(1+1).(1*1) = 21
(1+1).(1.1) = 22

(Not that fast) Quick fix
((1+1) . Log (1)) - 1 = 19
(1+1) . (1+1) = 22

(1+1+1+1)! = 24

[William the tuba]

int((1 . log(1)) ^ sqrt(1 + 1)) = 25

If we're allowed to use recursive and nested functions (and bitshifting)...

(1 . A(1, 1)) << 1 = 26

...where A is the Ackermann-Peter function. There's a table partway down, if you don't want to write it out.

int(cosh(1 + 1 + 1 + 1)) = 27

A(A(1, 1), int(cosh(cosh(1)))) - 1 = 28

A(A(1, 1), 1 + 1) = 29

(1 + 1 + 1) . log(1) = 30

[William the tuba]

Actually, iterated hyperbolic functions start breaking things. They grow slowly at first, but close enough to an additive ramp that you can get

1
int(cosh(cosh(1))) = 2
int(sinh(sinh(sinh(sinh(1))))) = 3
int(cosh(cosh(cosh(1)))) = 5

You can get all the numbers between 1 and 10 using two 1's each this way:
(I'm going to use sinhX(1) as shorthand for sinh(sinh(sinh(..., to save space)

1 = 1
int(cosh2(1)) = 2
int(sinh4(1)) = 3
int(cosh2(1)) + int(cosh2(1)) = 4
int(cosh3(1)) = 5
int(sinh4(1)) + int(sinh4(1)) = 6
int(cosh3(1)) + int(cosh2(1)) = 7
int(cosh3(1)) + int(sinh4(1)) = 8
int(sinh4(1)) * int(sinh4(1)) = 9
1 . log(1) = 10

So, with these in hand, you can make any number between 1 and 139 with concatenation (and probably more). Even if recursive functions aren't allowed (which would take away the successor function), these aren't much more complicated than log, sqrt, or the trigonometric functions, say. After 139, you have to start getting creative.

1 . int(cosh2(1)) + int(cosh2(1)) . log(1) = 140
1 . int(cosh2(1)) + int(cosh2(1)) . 1 = 141
1 . int(cosh2(1)) + int(cosh2(1)) . int(cosh2(1)) = 142
1 . int(cosh2(1)) + int(cosh2(1)) . int(sinh4(1)) = 143

We can't do 144 this way, since we'd run out of 1's, but it is 12 ^ 2, which we can represent fairly compactly.

(1 . int(cosh2(1))) ^ (int(cosh2(1))) + log(1) = 144
(1 . int(cosh2(1))) ^ (int(cosh2(1))) + 1 = 145
(1 . int(cosh2(1))) ^ (int(cosh2(1))) + int(cosh2(1)) = 146
(1 . int(cosh2(1))) ^ (int(cosh2(1))) + int(sinh4(1)) = 147

Again, with 4 costing two 1's, we can't do 148 this way. This certainly isn't the end, though, since you get 50 from int(sinh3(cosh(1))), which nicely gets us to 150 for a mere two 1's, so we can subtract 1 and 2 from it, to get 148, 149 and 150. Since we can represent 5 as a single 1, we can get all of 150 to 160 with the first method. We run out of 1's for 164, 166, 167, and 168. 169 is 13 ^ 2, which we can represent with three and subtract from to get all the ones we're missing for 160-170 (we can represent 5 with a single 1, 169 - 5 = 164).

I think I can get solutions to 200, at least, but a more interesting question might be the upper bound. If we were to find single-1 representations of all numbers less than 10? That puts a limit of 9999 on simple concatenation. You can obviously make giant numbers (Knuth's up-arrow notation, anyone?), but whether we can fill in all the holes past a certain point is something else.

[dawnraider]

4 and 6 should be obtainable using a single 1 using the ceiling function instead of the int function in order to round up instead truncating in the equations for 3 and 5 respectively.

[William the tuba]

Funnily enough, I've never seen ceiling in a textbook, but, yes, if we've got the full math.h in here, we can probably do better than just int/floor.

[Gilberreke]

So no one stopped to consider that maybe the fun was in alternatingly figuring these out with many people? :p

You guys basically ended the thread. The idea was to do one at a time :/

[William the tuba]

So no one stopped to consider that maybe the fun was in alternatingly figuring these out with many people? :p

You guys basically ended the thread. The idea was to do one at a time :/

In my defense, I did point out that I was breaking things with the hyperbolic functions.

I think most of the puzzle is preserved if you disallow solutions that include integer truncation, round, floor, ceiling, and the successor function, which would roll back progress to 14. It'd also be nice if there was a clear explanation on how concatenation works with regards to decimals and significant digits. I think it'd make sense if it started concatenating from the first significant digit, regardless of where the decimal point went, which would mean that 1 . (1 / (1 + 1)) would equal 15.

[Gilberreke]

In my defense, I did point out that I was breaking things with the hyperbolic functions.

Even then, just do one number like that and leave it at that. If the next person wants to use that same method, fine, but what would've happened more probably is that that next person would've chosen another way. Sure, there's a million ways to find solutions that are more general in nature and give you a bunch of sequential solutions, but the fun in the game is to just post one example and let the next person decide for him/herself what to do next.

[dawnraider]

True enough, apologies. Although I'm not sure we can get everything without trig and int, there definitely are other solutions than nested cosh/sinh, so I think we can continue without using those.

Edit: should we disallow the use of nested functions (at least of the same type, I.e. Cosh2())?