Post History
Dyalog APL, 14 bytes {6-+/⍵*⍨6÷⍨⍳5} Not bruteforce! An exact implementation of the formula \[ E_n = 6 - \sum_{i=1}^5 \left(\frac i 6\right)^n \] 6- 6 minus +/ the sum of 6÷⍨⍳5 the list 1/...
Answer
#3: Post edited
by
RubenVerg
·
2023-07-09T11:34:16Z (over 1 year ago)
- # Dyalog APL, 14 bytes
- ```apl
- {6-+/⍵*⍨6÷⍨⍳5}
- ```
- Not bruteforce! An exact implementation of the formula
- \[
- E_n = 6 - \sum_{i=1}^5 \left(\frac i 6\right)^n
- \]
- * `6-` 6 minus
- * `+/` the sum of
- * `6÷⍨⍳5` the list 1/6, 2/6, 3/6, 4/6, 5/6
- * `⍵*⍨` to the power of the argument of the function
- ## Formula explanation
- For simplicity, let's set $n=2$. This process trivially expands to higher dimensions.
- If we roll 2 dice, this table represents all the possible pairs:
- ```text
- ⍳6 6
- ┌───┬───┬───┬───┬───┬───┐
- │1 1│1 2│1 3│1 4│1 5│1 6│
- ├───┼───┼───┼───┼───┼───┤
- │2 1│2 2│2 3│2 4│2 5│2 6│
- ├───┼───┼───┼───┼───┼───┤
- │3 1│3 2│3 3│3 4│3 5│3 6│
- ├───┼───┼───┼───┼───┼───┤
- │4 1│4 2│4 3│4 4│4 5│4 6│
- ├───┼───┼───┼───┼───┼───┤
- │5 1│5 2│5 3│5 4│5 5│5 6│
- ├───┼───┼───┼───┼───┼───┤
- │6 1│6 2│6 3│6 4│6 5│6 6│
- └───┴───┴───┴───┴───┴───┘
- ```
- Let's take the maximum of each pair:
- ```text
- ⌈/¨⍳6 6
- 1 2 3 4 5 6
- 2 2 3 4 5 6
- 3 3 3 4 5 6
- 4 4 4 4 5 6
- 5 5 5 5 5 6
- 6 6 6 6 6 6
- ```
- Notice that if we sum the whole table and divide by $6^n$, we get our desired expected value
- ```text
- 36÷⍨+/,⌈/¨⍳6 6
- 4.472222222
- ```
- Alright, so we don't *really* care about the table, just its sum. I'll print it alongside the table as we build it.
- ```text
- (P←{⍵(+/,⍵)})⌈/¨⍳6 6
- ┌───────────┬───┐
- │1 2 3 4 5 6│161│
- │2 2 3 4 5 6│ │
- │3 3 3 4 5 6│ │
- │4 4 4 4 5 6│ │
- │5 5 5 5 5 6│ │
- │6 6 6 6 6 6│ │
- └───────────┴───┘
- ```
- Let's find another way to build the same table (sum). Start with a table of all 6's.
- ```text
- P 6 6⍴6
- ┌───────────┬───┐
- │6 6 6 6 6 6│216│
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- └───────────┴───┘
- ```
- We want to subtract 1 from all entries except for those in the last row or column. Alternatively, we want to subtract this table from ours:
- ```table
- ~6∊¨⌈/¨⍳6 6
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 0 0 0 0 0 0
- ```
- This is a $5\times 5$ square.
- We now get the following table:
- ```text
- P (6 6⍴6)-(~6∊¨⌈/¨⍳6 6)
- ┌───────────┬───┐
- │5 5 5 5 5 6│191│
- │5 5 5 5 5 6│ │
- │5 5 5 5 5 6│ │
- │5 5 5 5 5 6│ │
- │5 5 5 5 5 6│ │
- │6 6 6 6 6 6│ │
- └───────────┴───┘
- ```
- Repeating the same process for squares of side 4, 3, 2 and 1, we get exactly the original table (and sum!) we wanted.
- Expressed mathematically, the expected value can therefore be written as
- \[
\frac 1 {6^n} \left(6\cdot6^n - 5^n - 4^n - 3^n - 2^n - 1^n ight) = 6 - \sum_{i=0}^5 \left(\frac i 6 ight)^n- \]
- # Dyalog APL, 14 bytes
- ```apl
- {6-+/⍵*⍨6÷⍨⍳5}
- ```
- Not bruteforce! An exact implementation of the formula
- \[
- E_n = 6 - \sum_{i=1}^5 \left(\frac i 6\right)^n
- \]
- * `6-` 6 minus
- * `+/` the sum of
- * `6÷⍨⍳5` the list 1/6, 2/6, 3/6, 4/6, 5/6
- * `⍵*⍨` to the power of the argument of the function
- ## Formula explanation
- For simplicity, let's set $n=2$. This process trivially expands to higher dimensions.
- If we roll 2 dice, this table represents all the possible pairs:
- ```text
- ⍳6 6
- ┌───┬───┬───┬───┬───┬───┐
- │1 1│1 2│1 3│1 4│1 5│1 6│
- ├───┼───┼───┼───┼───┼───┤
- │2 1│2 2│2 3│2 4│2 5│2 6│
- ├───┼───┼───┼───┼───┼───┤
- │3 1│3 2│3 3│3 4│3 5│3 6│
- ├───┼───┼───┼───┼───┼───┤
- │4 1│4 2│4 3│4 4│4 5│4 6│
- ├───┼───┼───┼───┼───┼───┤
- │5 1│5 2│5 3│5 4│5 5│5 6│
- ├───┼───┼───┼───┼───┼───┤
- │6 1│6 2│6 3│6 4│6 5│6 6│
- └───┴───┴───┴───┴───┴───┘
- ```
- Let's take the maximum of each pair:
- ```text
- ⌈/¨⍳6 6
- 1 2 3 4 5 6
- 2 2 3 4 5 6
- 3 3 3 4 5 6
- 4 4 4 4 5 6
- 5 5 5 5 5 6
- 6 6 6 6 6 6
- ```
- Notice that if we sum the whole table and divide by $6^n$, we get our desired expected value
- ```text
- 36÷⍨+/,⌈/¨⍳6 6
- 4.472222222
- ```
- Alright, so we don't *really* care about the table, just its sum. I'll print it alongside the table as we build it.
- ```text
- (P←{⍵(+/,⍵)})⌈/¨⍳6 6
- ┌───────────┬───┐
- │1 2 3 4 5 6│161│
- │2 2 3 4 5 6│ │
- │3 3 3 4 5 6│ │
- │4 4 4 4 5 6│ │
- │5 5 5 5 5 6│ │
- │6 6 6 6 6 6│ │
- └───────────┴───┘
- ```
- Let's find another way to build the same table (sum). Start with a table of all 6's.
- ```text
- P 6 6⍴6
- ┌───────────┬───┐
- │6 6 6 6 6 6│216│
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- └───────────┴───┘
- ```
- We want to subtract 1 from all entries except for those in the last row or column. Alternatively, we want to subtract this table from ours:
- ```table
- ~6∊¨⌈/¨⍳6 6
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 0 0 0 0 0 0
- ```
- This is a $5\times 5$ square.
- We now get the following table:
- ```text
- P (6 6⍴6)-(~6∊¨⌈/¨⍳6 6)
- ┌───────────┬───┐
- │5 5 5 5 5 6│191│
- │5 5 5 5 5 6│ │
- │5 5 5 5 5 6│ │
- │5 5 5 5 5 6│ │
- │5 5 5 5 5 6│ │
- │6 6 6 6 6 6│ │
- └───────────┴───┘
- ```
- Repeating the same process for squares of side 4, 3, 2 and 1, we get exactly the original table (and sum!) we wanted.
- Expressed mathematically, the expected value can therefore be written as
- \[
- \frac 1 {6^n} \left(6\cdot6^n - 5^n - 4^n - 3^n - 2^n - 1^n ight) = 6 - \sum_{i=1}^5 \left(\frac i 6 ight)^n
- \]
#2: Post edited
by
RubenVerg
·
2023-07-08T22:48:40Z (over 1 year ago)
add explanation, change code slightly
- # Dyalog APL, 14 bytes
- ```apl
{7-+/⍵*⍨6÷⍨⍳6}- ```
- Not bruteforce! An exact implementation of the formula
- \[
E_n = 7 - \sum_{i=1}^6 \left(\frac i 6 ight)^n- \]
* `7-` 7 minus- * `+/` the sum of
* `6÷⍨⍳6` the list 1/6, 2/6, 3/6, 4/6, 5/6, 6/6* `⍵*⍨` to the power of the argument of the function
- # Dyalog APL, 14 bytes
- ```apl
- {6-+/⍵*⍨6÷⍨⍳5}
- ```
- Not bruteforce! An exact implementation of the formula
- \[
- E_n = 6 - \sum_{i=1}^5 \left(\frac i 6 ight)^n
- \]
- * `6-` 6 minus
- * `+/` the sum of
- * `6÷⍨⍳5` the list 1/6, 2/6, 3/6, 4/6, 5/6
- * `⍵*⍨` to the power of the argument of the function
- ## Formula explanation
- For simplicity, let's set $n=2$. This process trivially expands to higher dimensions.
- If we roll 2 dice, this table represents all the possible pairs:
- ```text
- ⍳6 6
- ┌───┬───┬───┬───┬───┬───┐
- │1 1│1 2│1 3│1 4│1 5│1 6│
- ├───┼───┼───┼───┼───┼───┤
- │2 1│2 2│2 3│2 4│2 5│2 6│
- ├───┼───┼───┼───┼───┼───┤
- │3 1│3 2│3 3│3 4│3 5│3 6│
- ├───┼───┼───┼───┼───┼───┤
- │4 1│4 2│4 3│4 4│4 5│4 6│
- ├───┼───┼───┼───┼───┼───┤
- │5 1│5 2│5 3│5 4│5 5│5 6│
- ├───┼───┼───┼───┼───┼───┤
- │6 1│6 2│6 3│6 4│6 5│6 6│
- └───┴───┴───┴───┴───┴───┘
- ```
- Let's take the maximum of each pair:
- ```text
- ⌈/¨⍳6 6
- 1 2 3 4 5 6
- 2 2 3 4 5 6
- 3 3 3 4 5 6
- 4 4 4 4 5 6
- 5 5 5 5 5 6
- 6 6 6 6 6 6
- ```
- Notice that if we sum the whole table and divide by $6^n$, we get our desired expected value
- ```text
- 36÷⍨+/,⌈/¨⍳6 6
- 4.472222222
- ```
- Alright, so we don't *really* care about the table, just its sum. I'll print it alongside the table as we build it.
- ```text
- (P←{⍵(+/,⍵)})⌈/¨⍳6 6
- ┌───────────┬───┐
- │1 2 3 4 5 6│161│
- │2 2 3 4 5 6│ │
- │3 3 3 4 5 6│ │
- │4 4 4 4 5 6│ │
- │5 5 5 5 5 6│ │
- │6 6 6 6 6 6│ │
- └───────────┴───┘
- ```
- Let's find another way to build the same table (sum). Start with a table of all 6's.
- ```text
- P 6 6⍴6
- ┌───────────┬───┐
- │6 6 6 6 6 6│216│
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- │6 6 6 6 6 6│ │
- └───────────┴───┘
- ```
- We want to subtract 1 from all entries except for those in the last row or column. Alternatively, we want to subtract this table from ours:
- ```table
- ~6∊¨⌈/¨⍳6 6
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 1 1 1 1 1 0
- 0 0 0 0 0 0
- ```
- This is a $5\times 5$ square.
- We now get the following table:
- ```text
- P (6 6⍴6)-(~6∊¨⌈/¨⍳6 6)
- ┌───────────┬───┐
- │5 5 5 5 5 6│191│
- │5 5 5 5 5 6│ │
- │5 5 5 5 5 6│ │
- │5 5 5 5 5 6│ │
- │5 5 5 5 5 6│ │
- │6 6 6 6 6 6│ │
- └───────────┴───┘
- ```
- Repeating the same process for squares of side 4, 3, 2 and 1, we get exactly the original table (and sum!) we wanted.
- Expressed mathematically, the expected value can therefore be written as
- \[
- \frac 1 {6^n} \left(6\cdot6^n - 5^n - 4^n - 3^n - 2^n - 1^n\right) = 6 - \sum_{i=0}^5 \left(\frac i 6\right)^n
- \]
#1: Initial revision
by
RubenVerg
·
2023-07-08T09:38:19Z (over 1 year ago)
# Dyalog APL, 14 bytes
```apl
{7-+/⍵*⍨6÷⍨⍳6}
```
Not bruteforce! An exact implementation of the formula
\[
E_n = 7 - \sum_{i=1}^6 \left(\frac i 6\right)^n
\]
* `7-` 7 minus
* `+/` the sum of
* `6÷⍨⍳6` the list 1/6, 2/6, 3/6, 4/6, 5/6, 6/6
* `⍵*⍨` to the power of the argument of the function