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fix number tokens and limits

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README.md
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@@ -1,19 +1,28 @@
# 42_EXT_05_computorv1
## todo
- [ ] arg limit ?
- [x] change order is_number_int and is_number_double
- [x] remove double limit in lexer
- [ ] check why number stored in int or double
- [x] get rid of lib math (check in libft)
- [x] use ft_abs() instead of abs() -> actually abs() is in stdlib
- [x] handling sign '-' : "+ -2"
- [x] max exponent len :
## ressources
- project intra : https://projects.intra.42.fr/projects/42cursus-computorv1
- project luke : https://github.com/LuckyLaszlo/computorv1
- project intra : https://projects.intra.42.fr/projects/42cursus-computorv1
- project luke : https://github.com/LuckyLaszlo/computorv1
## install
this project uses submodules (maybe recursively), so either :
this project uses submodules (maybe recursively), so either :
- `git clone --recurse-submodules <repo-url>`
- or, after cloning : `git submodule update --init --recursive`
---
# sqrt implementation
@@ -24,7 +33,7 @@ finding the square root of x
the dichotomie method, or binary search, consist on dividing the range of research by 2 each time, and choosing the right one for the next iteration.
Ex :
Ex :
```
solution
@@ -57,12 +66,11 @@ Ex :
--------|---------------------- --> solution
```
## Newton–Raphson method
it's like a self-correcting binary search, we get rid of the step "choose range", we use the formulae `x/v` to find the next range, with `x` being the number we are trying to get the sqaure root from, and `v` the value found at the previous step.
Ex :
Ex :
```
solution
@@ -91,87 +99,72 @@ Ex :
```
### mathematical proof that each range is automatically in the right range :
- if the value was higher than the answer, then new value is below old value, and vice versa
- how ? :
- define `x`, solution `s = √x`, and value `v = (old_value + x / old_value) / 2`
- supposition : if `v < s` , then `new_v > v`, else `new_v < v` :
- demonstration :
1. if `v < s` :
v < s
<=> v < √x
<=> v² < x (that's actually how we know that v < s)
<=> v²/v < x/v
<=> v < x/v
-> and is s < x/v ? :
v < s
<=> v < √x
<=> v² < x (as previously)
<=> v² < x²/x
<=> v² * x < x²
<=> (v * √x)² < x²
<=> v * √x < x
<=> v * √x < v * x/v
<=> √x < x/v
-> so indeed : if v < √x, then v < √x < x/v == v < s < x/v
-> conclusion, the new range < v , x/v > contains the solution
2. the same demonstration works for `v > s`
### here is a more intuitive demonstration, with x = 20 :
- if the value was higher than the answer, then new value is below old value, and vice versa
- how ? :
- define `x`, solution `s = √x`, and value `v = (old_value + x / old_value) / 2`
- supposition : if `v < s` , then `new_v > v`, else `new_v < v` :
- demonstration :
1. if `v < s` :
v < s
<=> v < √x
<=> v² < x (that's actually how we know that v < s)
<=> v²/v < x/v
<=> v < x/v
-> and is s < x/v ? :
v < s
<=> v < √x
<=> v² < x (as previously)
<=> v² < x²/x
<=> v² _ x < x²
<=> (v _ √x)² < x²
<=> v _ √x < x
<=> v _ √x < v \* x/v
<=> √x < x/v
-> so indeed : if v < √x, then v < √x < x/v == v < s < x/v
-> conclusion, the new range < v , x/v > contains the solution
2. the same demonstration works for `v > s`
### here is a more intuitive demonstration, with x = 20 :
1. **show that if `v² > x` (== `v > s`) then `v > s > x/v`, and if `v² < x` (== `v < s`) then `v < s < x/v` :**
1.1. **for value too high `v > s` :**
1.1.1 **why `v > x/v` :**
- let's take initial value v = 5 :
- is 5² the solution ? 5² == 25 -> so no, 5 is not the sqrt, it's too high
```
v v²
0 (5) 10 15 20 25
v : |----|----|----|----|----|
x/v: |---|---|---|---|---| <----- squiz it, so the previous 5 portions fit the x = 20 size
0 (4) 8 12 16 20
x/v x
```
- the value of the new portion is 4, and we can visually see that it's lower than the previous portion 5
- so : `v > x/v`
1.1.2 **why `s > x/v` :**
- let's take the value v = 5 :
- we already showed that it's too high, now we will see that x/v == 20/5 is too low :
```
v
0 (5) 10 15 20 25
v : |----|----|----|----|----|
x/v: |---|---|---|---|---| <----- squizz
0 *1 *2 *3 *4 *5 -> number of portions
01234 -> portion size
(x/v)²: |---|---|---|---|
0 4 8 12 16
```
- the portion size is smaller than the number of portions, so it's too small to be the sqrt, indeed we visually see that this portion size `x/v` is a root of a smaller number : `(x/v)² == 16`
- so : `s > x/v`
1.1.3. **conclusion :**
- v > s
- and v > x/v (<- this proof is not essential)
- and s > x/v (<- we actually only need this proof)
- so `v > s > x/v`
1.2. **for value too high `v < s` :**
- this is the same demonstration but in other direction, let's just summarize it :
- let's take initial value v = 4 :
```
v v²
0 (4) 8 12 16
(1) v : |---|---|---|---|
(2) x/v: |----|----|----|----| -----> stretch
(3) 0 (5) 10 15 20
x/v x
(4) 0 *1 *2 *3 *4 -> number of portions
012345 -> portion size
(5) (x/v)²: |----|----|----|----|----|
0 5 10 15 20 25
```
- (1) : 4 is not the sqrt of x == 20, it's too smalle : (4² == 16) < 20
- (2) : stretch it, so the previous 4 portions fit the x = 20 size
- (3) : the new portion x/v == 5, is more than v == 4, so `v < x/v`
- (4) : portion size is bigger than number of portions, so it's too big to be the root
- (5) : indeed, we see that the portion² == (x/v)² is bigger than x, so √x < x/v == `s < x/v`
- so `v < s < x/v`
1.1. **for value too high `v > s` :**
1.1.1 **why `v > x/v` :** - let's take initial value v = 5 : - is 5² the solution ? 5² == 25 -> so no, 5 is not the sqrt, it's too high
` v v²
0 (5) 10 15 20 25
v : |----|----|----|----|----|
x/v: |---|---|---|---|---| <----- squiz it, so the previous 5 portions fit the x = 20 size
0 (4) 8 12 16 20
x/v x` - the value of the new portion is 4, and we can visually see that it's lower than the previous portion 5 - so : `v > x/v`
1.1.2 **why `s > x/v` :** - let's take the value v = 5 : - we already showed that it's too high, now we will see that x/v == 20/5 is too low :
` v
0 (5) 10 15 20 25
v : |----|----|----|----|----|
x/v: |---|---|---|---|---| <----- squizz
0 *1 *2 *3 *4 *5 -> number of portions
01234 -> portion size
(x/v)²: |---|---|---|---|
0 4 8 12 16` - the portion size is smaller than the number of portions, so it's too small to be the sqrt, indeed we visually see that this portion size `x/v` is a root of a smaller number : `(x/v)² == 16` - so : `s > x/v`
1.1.3. **conclusion :** - v > s - and v > x/v (<- this proof is not essential) - and s > x/v (<- we actually only need this proof) - so `v > s > x/v`
1.2. **for value too high `v < s` :**
- this is the same demonstration but in other direction, let's just summarize it :
- let's take initial value v = 4 :
```
v v²
0 (4) 8 12 16
(1) v : |---|---|---|---|
(2) x/v: |----|----|----|----| -----> stretch
(3) 0 (5) 10 15 20
x/v x
(4) 0 *1 *2 *3 *4 -> number of portions
012345 -> portion size
(5) (x/v)²: |----|----|----|----|----|
0 5 10 15 20 25
```
- (1) : 4 is not the sqrt of x == 20, it's too smalle : (4² == 16) < 20
- (2) : stretch it, so the previous 4 portions fit the x = 20 size
- (3) : the new portion x/v == 5, is more than v == 4, so `v < x/v`
- (4) : portion size is bigger than number of portions, so it's too big to be the root
- (5) : indeed, we see that the portion² == (x/v)² is bigger than x, so √x < x/v == `s < x/v`
- so `v < s < x/v`