Write approaches for Yacht

exercises/practice/yacht/.approaches/config.json

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+{
+  "introduction": {
+    "authors": ["safwansamsudeen"]
+  },
+  "approaches": [
+    {
+      "uuid": "3593cfe3-5cab-4141-b0a2-329148a66bb6",
+      "slug": "functions",
+      "title": "Lambdas with Functions",
+      "blurb": "Use lambdas with functions",
+      "authors": ["safwansamsudeen"]
+    },
+    {
+      "uuid": "eccd1e1e-6c88-4823-9b25-944eccaa92e7",
+      "slug": "if-structure",
+      "title": "If structure",
+      "blurb": "Use an if structure",
+      "authors": ["safwansamsudeen"]
+    },
+    {
+      "uuid": "72079791-e51f-4825-ad94-3b7516c631cc",
+      "slug": "structural-pattern-matching",
+      "title": "Structural Pattern Matching",
+      "blurb": "Use structural pattern matching",
+      "authors": ["safwansamsudeen"]
+    }
+  ]
+}

exercises/practice/yacht/.approaches/functions/content.md

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+## Approach: Using Lambdas with Functions
+Each bit of functionality for each category can be encoded in an anonymous function (otherwise known as a [`lambda` expression][lambda] or lambda form), and the constant name set to that function.
+
+In `score`, we call the category (as it now points to a function) passing in `dice` as an argument.
+
+```python
+def digits(num):
+    return lambda dice: dice.count(num) * num
+
+YACHT = lambda dice: 50 if len(set(dice)) ==  1 else 0
+ONES = digits(1)
+TWOS = digits(2)
+THREES = digits(3)
+FOURS = digits(4)
+FIVES = digits(5)
+SIXES = digits(6)
+FULL_HOUSE = lambda dice: sum(dice) if len(set(dice)) == 2 and dice.count(dice[0]) in [2, 3] else 0
+FOUR_OF_A_KIND = lambda dice: 4 * dice[1] if dice[0] == dice[3] or dice[1] == dice[4] else 0
+LITTLE_STRAIGHT = lambda dice: 30 if sorted(dice) == [1, 2, 3, 4, 5] else 0
+BIG_STRAIGHT = lambda dice: 30 if sorted(dice) == [2, 3, 4, 5, 6] else 0
+CHOICE = sum
+
+def score(dice, category):
+    return category(dice)
+``` 
+
+
+Instead of setting each constant in `ONES` through `SIXES` to a separate function, we create a function `digits` that returns a function, using [closures][closures] transparently.
+
+For `LITTLE_STRAIGHT` and `BIG_STRAIGHT`, we first sort the dice and then check it against the hard-coded value. 
+Another way to solve this would be to check if `sum(dice) == 20 and len(set(dice)) == 5` (15 in `LITTLE_STRAIGHT`).
+In `CHOICE`, `lambda number : sum(number)` is shortened to just `sum`.
+
+In `FULL_HOUSE`, we create a `set` to remove the duplicates and check the set's length along with the individual counts. 
+For `FOUR_OF_A_KIND`, we check if the first and the fourth element are the same or the second and the last element are the same - if so, there are (at least) four of the same number in the array. 
+
+This solution is a succinct way to solve the exercise, although some of the one-liners can get a little long and hard to read. 
+Additionally, [PEP8][pep8] does not recommend assigning constant or variable names to `lambda` expressions, so it is a better practice to use `def`:
+```python
+def digits(num):
+    return lambda dice: dice.count(num) * num
+
+def YACHT(dice): return 50 if len(set(dice)) ==  1 else 0
+ONES = digits(1)
+TWOS = digits(2)
+THREES = digits(3)
+FOURS = digits(4)
+FIVES = digits(5)
+SIXES = digits(6)
+def FULL_HOUSE(dice): return sum(dice) if len(set(dice)) == 2 and dice.count(dice[0]) in [2, 3] else 0
+def FOUR_OF_A_KIND(dice): return 4 * sorted(dice)[1] if len(set(dice)) < 3 and dice.count(dice[0]) in (1, 4, 5) else 0
+def LITTLE_STRAIGHT(dice): return 30 if sorted(dice) == [1, 2, 3, 4, 5] else 0
+def BIG_STRAIGHT(dice): return 30 if sorted(dice) == [2, 3, 4, 5, 6] else 0
+CHOICE = sum
+
+def score(dice, category):
+    return category(dice)
+```
+
+As you can see from the examples, the [ternary operator][ternary-operator] (_or ternary form_) is crucial in solving the exercise using one liners.  
+As functions are being used, it might be a better strategy to spread the code over multiple lines to improve readability.
+```python
+def YACHT(dice):
+    if dice.count(dice[0]) == len(dice):
+        return 50
+    return 0
+```
+
+[closures]: https://www.programiz.com/python-programming/closure
+[ternary-operator]: https://www.tutorialspoint.com/ternary-operator-in-python
+[lambda]: https://docs.python.org/3/howto/functional.html?highlight=lambda#small-functions-and-the-lambda-expression    
+[pep8]: https://peps.python.org/pep-0008/

exercises/practice/yacht/.approaches/functions/snippet.txt

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+def digits(num):
+    return lambda dice: dice.count(num) * num
+YACHT = lambda x: 50 if x.count(x[0]) == len(x) else 0
+ONES = digits(1)
+FULL_HOUSE = lambda x: sum(x) if len(set(x)) == 2 and x.count(x[0]) in [2, 3] else 0
+LITTLE_STRAIGHT = lambda x: 30 if sorted(x) == [1, 2, 3, 4, 5] else 0
+def score(dice, category):
+    return category(dice)

exercises/practice/yacht/.approaches/if-structure/content.md

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+# If structure
+
+The constants here can be set to random, null, or numeric values, and an `if` structure inside the `score` function can determine the code to be executed. 
+
+As one-liners aren't necessary here, we can spread out the code to make it look neater:
+```python
+ONES = 1
+TWOS = 2
+THREES = 3
+FOURS = 4
+FIVES = 5
+SIXES = 6
+FULL_HOUSE = 'FULL_HOUSE'
+FOUR_OF_A_KIND = 'FOUR_OF_A_KIND'
+LITTLE_STRAIGHT = 'LITTLE_STRAIGHT'
+BIG_STRAIGHT = 'BIG_STRAIGHT'
+CHOICE = 'CHOICE'
+YACHT = 'YACHT'
+
+def score(dice, category):
+    if category in (1,2,3,4,5,6):
+         return dice.count(category) * category
+    elif category == 'FULL_HOUSE':
+        if len(set(dice)) == 2 and dice.count(dice[0]) in [2, 3]:
+            return sum(dice) or 0
+    elif category == 'FOUR_OF_A_KIND':
+        if dice[0] == dice[3] or dice[1] == dice[4]:
+            return dice[1] * 4 or 0
+    elif category == 'LITTLE_STRAIGHT':
+        if sorted(dice) == [1, 2, 3, 4, 5]: 
+            return 30 or 0
+    elif category == 'BIG_STRAIGHT':
+        if sorted(dice) == [2, 3, 4, 5, 6]:
+            return 30 or 0
+    elif category == 'YACHT':
+        if all(num == dice[0] for num in dice):
+            return 50
+    elif category == 'CHOICE':
+        return sum(dice)
+    return 0
+```
+Note that the code inside the `if` statements themselves can differ, but the key idea here is to use `if` and `elif` to branch out the code, and return `0` at the end if nothing else has been returned. 
+The `if` condition itself can be different, with people commonly checking if `category == ONES` as opposed to `category == 'ONES'` (or whatever the dummy value is).
+
+This may not be an ideal way to solve the exercise, as the code is rather long and convoluted.
+However, it is a valid (_and fast_) solution.
+Using [structural pattern matching][structural pattern matching], introduced in Python 3.10, could shorten and clarify the code in this situation.
+Pulling some logic out of the `score` function and into additional "helper" functions could also help.
+
+[structural pattern matching]: https://peps.python.org/pep-0636/

exercises/practice/yacht/.approaches/if-structure/snippet.txt

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+ONES = 1
+YACHT = 'YACHT'
+def score(dice, category):
+    if category == 'ONES':
+        ...
+    elif category == 'FULL_HOUSE':
+        ...
+    return 0

exercises/practice/yacht/.approaches/introduction.md

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+# Introduction
+Yacht in Python can be solved in many ways. The most intuitive approach is to use an `if` structure. 
+Alternatively, you can create functions and set their names to the constant names.
+
+## General guidance
+The main thing in this exercise is to map a category (_here defined as constants in the stub file_) to a function or a standalone piece of code. 
+While mapping generally reminds us of dictionaries, here the constants are global.
+This indicates that the most idiomatic approach is not using a `dict`.
+Adhering to the principles of DRY is important - don't repeat yourself if you can help it, especially in the `ONES` through `SIXES` categories!
+
+## Approach: functions
+Each bit of functionality for each category can be encoded in a function, and the constant name set to that function. 
+This can be done by assigning the constant name to a `lambda` or creating a one-line function using the constant as a function name. 
+```python
+```python
+def digits(num):
+    return lambda dice: dice.count(num) * num
+YACHT = lambda dice: 50 if dice.count(dice[0]) == len(dice) else 0
+ONES = digits(1)
+TWOS = digits(2)
+THREES = digits(3)
+FOURS = digits(4)
+FIVES = digits(5)
+SIXES = digits(6)
+FULL_HOUSE = lambda dice: sum(dice) if len(set(dice)) == 2 and dice.count(dice[0]) in [2, 3] else 0
+FOUR_OF_A_KIND = lambda dice: 4 * dice[1] if dice[0] == dice[3] or dice[1] == dice[4] else 0
+LITTLE_STRAIGHT = lambda dice: 30 if sorted(dice) == [1, 2, 3, 4, 5] else 0
+BIG_STRAIGHT = lambda dice: 30 if sorted(dice) == [2, 3, 4, 5, 6] else 0
+CHOICE = sum
+def score(dice, category):
+    return category(dice)
+```
+This is a very succinct way to solve the exercise, although some one-liners get a little long. 
+For more information on this approach, read [this document][approach-functions].
+
+## Approach: if structure
+The constants can be set to random, null, or numeric values, and an `if` structure inside `score` determines the code to be executed. 
+As one-liners aren't necessary here, we can spread out the code to make it look neater:
+```python
+ONES = 1
+TWOS = 2
+THREES = 3
+FOURS = 4
+FIVES = 5
+SIXES = 6
+FULL_HOUSE = 'FULL_HOUSE'
+FOUR_OF_A_KIND = 'FOUR_OF_A_KIND'
+LITTLE_STRAIGHT = 'LITTLE_STRAIGHT'
+BIG_STRAIGHT = 'BIG_STRAIGHT'
+CHOICE = 'CHOICE'
+YACHT = 'YACHT'
+def score(dice, category):
+    if category in (1,2,3,4,5,6):
+         return dice.count(category) * category
+    elif category == 'FULL_HOUSE':
+        if len(set(dice)) == 2 and dice.count(dice[0]) in [2, 3]:
+            return sum(dice) or 0
+    elif category == 'FOUR_OF_A_KIND':
+        if dice[0] == dice[3] or dice[1] == dice[4]:
+            return dice[1] * 4 or 0
+    elif category == 'LITTLE_STRAIGHT':
+        if sorted(dice) == [1, 2, 3, 4, 5]: 
+            return 30 or 0
+    elif category == 'BIG_STRAIGHT':
+        if sorted(dice) == [2, 3, 4, 5, 6]:
+            return 30 or 0
+    elif category == 'YACHT':
+        if all(num == dice[0] for num in dice):
+            return 50
+    elif category == 'CHOICE':
+        return sum(dice)
+    return 0
+```
+Read more on this approach [here][approach-if-structure].
+
+[approach-functions]: https://exercism.org/tracks/python/exercises/yacht/approaches/functions
+[approach-if-structure]: https://exercism.org/tracks/python/exercises/yacht/approaches/if-structure

exercises/practice/yacht/.approaches/structural-pattern-matching/content.md

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+# Structural Pattern Matching
+
+Another very interesting approach is to use [structural pattern matching][structural pattern matching]. 
+Existing in Python since 3.10, this feature allows for neater code than traditional if structures.
+
+By and large, we reuse the code from the [if structure approach][approach-if-structure]. 
+We set the constants to random values and check for them in the `match` structure. 
+`category` is the "subject", and in every other line, we check it against a "pattern". 
+```python
+ONES = 1
+TWOS = 2
+THREES = 3
+FOURS = 4
+FIVES = 5
+SIXES = 6
+FULL_HOUSE = 'FULL_HOUSE'
+FOUR_OF_A_KIND = 'FOUR_OF_A_KIND'
+LITTLE_STRAIGHT = 'LITTLE_STRAIGHT'
+BIG_STRAIGHT = 'BIG_STRAIGHT'
+CHOICE = 'CHOICE'
+YACHT = 'YACHT'
+
+def score(dice, category):
+    match category:
+        case 1 | 2 | 3 | 4 | 5 | 6:
+            return dice.count(category) * category
+        case 'FULL_HOUSE' if len(set(dice)) == 2 and dice.count(dice[0]) in [2, 3]:
+            return sum(dice)
+        case 'FOUR_OF_A_KIND' if dice[0] == dice[3] or dice[1] == dice[4]:
+            return dice[1] * 4
+        case 'LITTLE_STRAIGHT' if sorted(dice) == [1, 2, 3, 4, 5]:
+            return 30
+        case 'BIG_STRAIGHT' if sorted(dice) == [2, 3, 4, 5, 6]:
+            return 30 
+        case 'YACHT' if all(num == dice[0] for num in dice):
+            return 50
+        case 'CHOICE':
+            return sum(dice)
+        case _:
+            return 0
+```
+For the first pattern, we utilize "or patterns", using the `|` operator. 
+This checks whether the subject is any of the provided patterns.
+
+In the next five patterns, we check an additional condition along with the pattern matching. 
+Finally, we use the wildcard operator `_` to match anything. 
+As the compiler checks the patterns (`case`s) in order, `return 0` will be executed if none of the other patterns match. 
+
+Note that the conditions might differ, but the patterns must have hard coded values - that is, you can't say `case ONES ...` instead of `case 1 ...`. 
+This will capture the category and lead to unexpected behavior. 
+
+This code is much clenaer than the corresponding `if` structure code.
+
+[structural pattern matching]: https://peps.python.org/pep-0636/
+[approach-if-structure]: https://exercism.org/tracks/python/exercises/yacht/approaches/if-structure

exercises/practice/yacht/.approaches/structural-pattern-matching/snippet.txt

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+ONES = 1
+YACHT = 'YACHT'
+def score(dice, category):
+    match category:
+        case 1 | 2 | 3 | 4 | 5 | 6:
+            return dice.count(category) * category
+        case _:
+            return 0