meetforever/GRPO_CL huggingface dataset:
user: There are 32 letters in the Polish alphabet: 9 vowels and 23 consonants. 
 
Your task is to change the letters with diacritics:

```
ą -> a,
ć -> c,
ę -> e,
ł -> l,
ń -> n,
ó -> o,
ś -> s,
ź -> z,
ż -> z
```
and print out the string without the use of the Polish letters.


For example:
```
"Jędrzej Błądziński"  -->  "Jedrzej Bladzinski"
```

Answer: {"fn_name": "correct_polish_letters", "inputs": [["J\u0119drzej B\u0142\u0105dzi\u0144ski"], ["Lech Wa\u0142\u0119sa"], ["Maria Sk\u0142odowska-Curie"], ["W\u0142adys\u0142aw Reymont"], ["Miko\u0142aj Kopernik"], ["J\u00f3zef Pi\u0142sudski"], ["Czes\u0142aw Mi\u0142osz"], ["Agnieszka Radwa\u0144ska"], ["Wojciech Szcz\u0119sny"], ["Za\u017c\u00f3\u0142\u0107 g\u0119\u015bl\u0105 ja\u017a\u0144"], ["W\u00f3\u0142 go pyta: 'Panie chrz\u0105szczu,Po co pan tak brz\u0119czy w g\u0105szczu?'"]], "outputs": [["Jedrzej Bladzinski"], ["Lech Walesa"], ["Maria Sklodowska-Curie"], ["Wladyslaw Reymont"], ["Mikolaj Kopernik"], ["Jozef Pilsudski"], ["Czeslaw Milosz"], ["Agnieszka Radwanska"], ["Wojciech Szczesny"], ["Zazolc gesla jazn"], ["Wol go pyta: 'Panie chrzaszczu,Po co pan tak brzeczy w gaszczu?'"]]}
user: You've been collecting change all day, and it's starting to pile up in your pocket, but you're too lazy to see how much you've found.

Good thing you can code!

Create ```change_count()``` to return a dollar amount of how much change you have!

Valid types of change include:
```
penny: 0.01
nickel: 0.05
dime: 0.10
quarter: 0.25
dollar: 1.00
```

```if:python
These amounts are already preloaded as floats into the `CHANGE` dictionary for you to use!
```
```if:ruby
These amounts are already preloaded as floats into the `CHANGE` hash for you to use!
```
```if:javascript
These amounts are already preloaded as floats into the `CHANGE` object for you to use!
```
```if:php
These amounts are already preloaded as floats into the `CHANGE` (a constant) associative array for you to use!
```

You should return the total in the format ```$x.xx```.

Examples:

```python
change_count('nickel penny dime dollar') == '$1.16'
change_count('dollar dollar quarter dime dime') == '$2.45'
change_count('penny') == '$0.01'
change_count('dime') == '$0.10'
```

Warning, some change may amount to over ```$10.00```!

Answer: {"fn_name": "change_count", "inputs": [["dime penny dollar"], ["dime penny nickel"], ["quarter quarter"], ["dollar penny dollar"], ["dollar dollar dollar dollar dollar dollar dollar dollar dollar dollar penny"]], "outputs": [["$1.11"], ["$0.16"], ["$0.50"], ["$2.01"], ["$10.01"]]}
user: Move the first letter of each word to the end of it, then add "ay" to the end of the word. Leave punctuation marks untouched.

## Examples

```python
pig_it('Pig latin is cool') # igPay atinlay siay oolcay
pig_it('Hello world !')     # elloHay orldway !
```
```C++
pig_it("Pig latin is cool");   // igPay atinlay siay oolcay
pig_it("Hello world !");       // elloHay orldway
```
```Java
PigLatin.pigIt('Pig latin is cool'); // igPay atinlay siay oolcay
PigLatin.pigIt('Hello world !');     // elloHay orldway !
```

Answer: {"fn_name": "pig_it", "inputs": [["Pig latin is cool"], ["This is my string"]], "outputs": [["igPay atinlay siay oolcay"], ["hisTay siay ymay tringsay"]]}
user: In this Kata, we are going to see how a Hash (or Map or dict) can be used to keep track of characters in a string. 

Consider two strings `"aabcdefg"` and `"fbd"`. How many characters do we have to remove from the first string to get the second string?  Although not the only way to solve this, we could create a Hash of counts for each string and see which character counts are different. That should get us close to the answer. I will leave the rest to you. 

For this example, `solve("aabcdefg","fbd") = 5`. Also, `solve("xyz","yxxz") = 0`, because we cannot get second string from the first since the second string is longer.

More examples in the test cases.

Good luck!

Answer: {"fn_name": "solve", "inputs": [["xyz", "yxz"], ["abcxyz", "ayxz"], ["abcdexyz", "yxz"], ["xyz", "yxxz"], ["abdegfg", "ffdb"], ["aabcdefg", "fbd"]], "outputs": [[0], [2], [5], [0], [0], [5]]}
user: In this kata you will be given a sequence of the dimensions of rectangles ( sequence with width and length ) and circles ( radius - just a number ).  
Your task is to return a new sequence of dimensions, sorted ascending by area.

For example,

```python
seq = [ (4.23, 6.43), 1.23, 3.444, (1.342, 3.212) ] # [ rectangle, circle, circle, rectangle ]
sort_by_area(seq) => [ ( 1.342, 3.212 ), 1.23, ( 4.23, 6.43 ), 3.444 ]
```

This kata inspired by [Sort rectangles and circles by area](https://www.codewars.com/kata/sort-rectangles-and-circles-by-area/).

Answer: {"fn_name": "sort_by_area", "inputs": [[[]]], "outputs": [[[]]]}
user: A [perfect power](https://en.wikipedia.org/wiki/Perfect_power) is a classification of positive integers:

> In mathematics, a **perfect power** is a positive integer that can be expressed as an integer power of another positive integer. More formally, n is a perfect power if there exist natural numbers m > 1, and k > 1 such that m^(k) = n.

Your task is to check wheter a given integer is a perfect power. If it is a perfect power, return a pair `m` and `k` with m^(k) = n as a proof. Otherwise return `Nothing`, `Nil`, `null`, `NULL`, `None` or your language's equivalent.

**Note:** For a perfect power, there might be several pairs. For example `81 = 3^4 = 9^2`, so `(3,4)` and `(9,2)` are valid solutions. However, the tests take care of this, so if a number is a perfect power, return any pair that proves it.

### Examples
```python
isPP(4) => [2,2]
isPP(9) => [3,2]
isPP(5) => None
```

Answer: {"fn_name": "isPP", "inputs": [[4], [9], [5]], "outputs": [[[2, 2]], [[3, 2]], [null]]}
user: Write function RemoveExclamationMarks which removes all exclamation marks from a given string.

Answer: {"fn_name": "remove_exclamation_marks", "inputs": [["Hello World!"], ["Hello World!!!"], ["Hi! Hello!"], [""], ["Oh, no!!!"]], "outputs": [["Hello World"], ["Hello World"], ["Hi Hello"], [""], ["Oh, no"]]}
user: You have a grid with `$m$` rows and `$n$` columns. Return the number of unique ways that start from the top-left corner and go to the bottom-right corner. You are only allowed to move right and down.

For example, in the below grid of `$2$` rows and `$3$` columns, there are `$10$` unique paths:

```
o----o----o----o
|    |    |    |
o----o----o----o
|    |    |    |
o----o----o----o
```

**Note:** there are random tests for grids up to 1000 x 1000 in most languages, so a naive solution will not work.

---
*Hint: use mathematical permutation and combination*

Answer: {"fn_name": "number_of_routes", "inputs": [[1, 1], [5, 1], [3, 4], [5, 6], [10, 10], [100, 3], [123, 456]], "outputs": [[2], [6], [35], [462], [184756], [176851], [448843261729071620474858205566477025894357385375655014634306680560209909590802545266425906052279365647506075241055256064119806400]]}
user: Complete the solution so that it returns true if it contains any duplicate argument values. Any number of arguments may be passed into the function.

The array values passed in will only be strings or numbers. The only valid return values are `true` and `false`.

Examples:
```
solution(1, 2, 3)             -->  false
solution(1, 2, 3, 2)          -->  true
solution('1', '2', '3', '2')  -->  true
```

Answer: {"fn_name": "solution", "inputs": [[1, 2, 3, 1, 2], [1, 1], [1, 0], ["a", "b"], ["a", "b", "a"], [1, 2, 42, 3, 4, 5, 42], ["a", "b", "c", "d"], ["a", "b", "c", "c"], ["a", "b", "c", "d", "e", "f", "f", "b"]], "outputs": [[true], [true], [false], [false], [true], [true], [false], [true], [true]]}
user: The number ```89``` is the first integer with more than one digit that fulfills the property partially introduced in the title of this kata. 
What's the use of saying "Eureka"? Because this sum gives the same number.

In effect: ```89 = 8^1 + 9^2``` 

The next number in having this property is ```135```.

See this property again: ```135 = 1^1 + 3^2 + 5^3```

We need a function to collect these numbers, that may receive two integers ```a```, ```b``` that defines the range ```[a, b]``` (inclusive) and outputs a list of the sorted numbers in the range that fulfills the property described above.

Let's see some cases:
```python
sum_dig_pow(1, 10) == [1, 2, 3, 4, 5, 6, 7, 8, 9]

sum_dig_pow(1, 100) == [1, 2, 3, 4, 5, 6, 7, 8, 9, 89]
```
If there are no numbers of this kind in the range [a, b] the function should output an empty list.
```python
sum_dig_pow(90, 100) == []
```
Enjoy it!!

Answer: {"fn_name": "sum_dig_pow", "inputs": [[1, 100], [10, 89], [10, 100], [90, 100], [90, 150], [50, 150], [10, 150], [89, 135]], "outputs": [[[1, 2, 3, 4, 5, 6, 7, 8, 9, 89]], [[89]], [[89]], [[]], [[135]], [[89, 135]], [[89, 135]], [[89, 135]]]}
user: Your task is to create a new implementation of `modpow` so that it computes `(x^y)%n` for large `y`. The problem with the current implementation is that the output of `Math.pow` is so large on our inputs that it won't fit in a 64-bit float.

You're also going to need to be efficient, because we'll be testing some pretty big numbers.

Answer: {"fn_name": "power_mod", "inputs": [[11, 10, 300], [11, 100000, 49], [5, 100000000, 19], [2, 3, 5], [4, 12, 3], [200, 3000, 10], [8132, 21302, 5], [9, 193125, 37], [13, 81230123, 22], [29, 1013293125, 492], [31, 902938423012, 1023]], "outputs": [[1], [32], [5], [3], [1], [0], [4], [26], [19], [161], [961]]}
user: In the wake of the npm's `left-pad` debacle, you decide to write a new super padding method that superceds the functionality of `left-pad`. Your version will provide the same functionality, but will additionally add right, and justified padding of string -- the `super_pad`.

Your function `super_pad` should take three arguments: the string `string`, the width of the final string `width`, and a fill character `fill`. However, the fill character can be enriched with a format string resulting in different padding strategies. If `fill` begins with `'<'` the string is padded on the left with the remaining fill string and if `fill` begins with `'>'` the string is padded on the right. Finally, if `fill` begins with `'^'` the string is padded on the left and the right, where the left padding is always greater or equal to the right padding. The `fill` string can contain more than a single char, of course. 

Some examples to clarify the inner workings:

- `super_pad("test", 10)` returns "      test"
- `super_pad("test", 10, "x")` returns `"xxxxxxtest"`
- `super_pad("test", 10, "xO")` returns `"xOxOxOtest"`
- `super_pad("test", 10, "xO-")` returns `"xO-xO-test"`
- `super_pad("some other test", 10, "nope")` returns `"other test"`
- `super_pad("some other test", 10, "> ")` returns `"some other"`
- `super_pad("test", 7, ">nope")` returns `"testnop"`
- `super_pad("test", 7, "^more complex")` returns `"motestm"`
- `super_pad("test", 7, "")` returns `"test"`

The `super_pad` method always returns a string of length `width` if possible. We expect the `width` to be positive (including 0) and the fill could be also an empty string.

Answer: {"fn_name": "super_pad", "inputs": [["test", 10], ["test", 10, "x"], ["test", 10, "xO"], ["test", 10, "xO-"], ["some other test", 10, "nope"], ["some other test", 10, "> "], ["test", 7, ">nope"], ["test", 7, "^more complex"], ["test", 7, ""], ["test", 10, "< "], ["test", 3], ["test", 3, "> "], ["test", 10, ""], ["test", 10, ">"], ["test", 10, "^"], ["test", 0, "> "], ["test", 0, "^ "], ["test", 0, "< "]], "outputs": [["      test"], ["xxxxxxtest"], ["xOxOxOtest"], ["xO-xO-test"], ["other test"], ["some other"], ["testnop"], ["motestm"], ["test"], ["      test"], ["est"], ["tes"], ["test"], ["test"], ["test"], [""], [""], [""]]}
user: We want to know the index of the vowels in a given word, for example, there are two vowels in the word super (the second and fourth letters). 

So given a string "super", we should return a list of [2, 4].

        Some examples:
        Mmmm  => []
        Super => [2,4]
        Apple => [1,5]
        YoMama -> [1,2,4,6]

**NOTES:**

* Vowels in this context refers to: a e i o u y (including upper case)
* This is indexed from `[1..n]` (not zero indexed!)

Answer: {"fn_name": "vowel_indices", "inputs": [["mmm"], ["apple"], ["super"], ["orange"], ["grapes"], ["supercalifragilisticexpialidocious"], ["123456"], ["crIssUm"], ["Implied"], ["rIc"], ["UNDISARMED"], ["bialy"], ["stumpknocker"], ["narboonnee"], ["carlstadt"], ["ephodee"], ["spicery"], ["oftenness"], ["bewept"], ["capsized"]], "outputs": [[[]], [[1, 5]], [[2, 4]], [[1, 3, 6]], [[3, 5]], [[2, 4, 7, 9, 12, 14, 16, 19, 21, 24, 25, 27, 29, 31, 32, 33]], [[]], [[3, 6]], [[1, 5, 6]], [[2]], [[1, 4, 6, 9]], [[2, 3, 5]], [[3, 8, 11]], [[2, 5, 6, 9, 10]], [[2, 7]], [[1, 4, 6, 7]], [[3, 5, 7]], [[1, 4, 7]], [[2, 4]], [[2, 5, 7]]]}
user: ## Enough is enough!

Alice and Bob were on a holiday. Both of them took many pictures of the places they've been, and now they want to show Charlie their entire collection. However, Charlie doesn't like these sessions, since the motive usually repeats. He isn't fond of seeing the Eiffel tower 40 times. He tells them that he will only sit during the session if they show the same motive at most N times. Luckily, Alice and Bob are able to encode the motive as a number. Can you help them to remove numbers such that their list contains each number only up to N times, without changing the order?

## Task

Given a list lst and a number N, create a new list that contains each number of lst at most N times without reordering. For example if N = 2, and the input is [1,2,3,1,2,1,2,3], you take [1,2,3,1,2], drop the next [1,2] since this would lead to 1 and 2 being in the result 3 times, and then take 3, which leads to [1,2,3,1,2,3].

~~~if:nasm
## NASM notes

Write the output numbers into the `out` parameter, and return its length.

The input array will contain only integers between 1 and 50 inclusive. Use it to your advantage.
~~~

~~~if:c
For C:
* Assign the return array length to the pointer parameter `*szout`.
* Do not mutate the input array.
~~~

## Example
```python
  delete_nth ([1,1,1,1],2) # return [1,1]
  
  delete_nth ([20,37,20,21],1) # return [20,37,21]
```

Answer: {"fn_name": "delete_nth", "inputs": [[[20, 37, 20, 21], 1], [[1, 1, 3, 3, 7, 2, 2, 2, 2], 3], [[1, 2, 3, 1, 1, 2, 1, 2, 3, 3, 2, 4, 5, 3, 1], 3], [[1, 1, 1, 1, 1], 5], [[], 5]], "outputs": [[[20, 37, 21]], [[1, 1, 3, 3, 7, 2, 2, 2]], [[1, 2, 3, 1, 1, 2, 2, 3, 3, 4, 5]], [[1, 1, 1, 1, 1]], [[]]]}
user: ## Task

Given a positive integer `n`, calculate the following sum: 

```
n + n/2 + n/4 + n/8 + ...
``` 

All elements of the sum are the results of integer division.

## Example

```
25  =>  25 + 12 + 6 + 3 + 1 = 47
```

Answer: {"fn_name": "halving_sum", "inputs": [[25], [127], [38], [1], [320], [13], [15], [47], [101], [257]], "outputs": [[47], [247], [73], [1], [638], [23], [26], [89], [198], [512]]}
user: Scientists working internationally use metric units almost exclusively. Unless that is, they wish to crash multimillion dollars worth of equipment on Mars.

Your task is to write a simple function that takes a number of meters, and outputs it using metric prefixes.

In practice, meters are only measured in "mm" (thousandths of a meter), "cm" (hundredths of a meter), "m" (meters) and "km" (kilometers, or clicks for the US military).

For this exercise we just want units bigger than a meter, from meters up to yottameters, excluding decameters and hectometers.

All values passed in will be positive integers.
e.g.

```python
meters(5);
// returns "5m"

meters(51500);
// returns "51.5km"

meters(5000000);
// returns "5Mm"
```

See http://en.wikipedia.org/wiki/SI_prefix for a full list of prefixes

Answer: {"fn_name": "meters", "inputs": [[1], [999], [123456], [12300000], [9000000000.0], [9000000000000.0], [9000000000000000.0], [9e+18], [9e+21], [9e+24]], "outputs": [["1m"], ["999m"], ["123.456km"], ["12.3Mm"], ["9Gm"], ["9Tm"], ["9Pm"], ["9Em"], ["9Zm"], ["9Ym"]]}
user: Two red beads are placed between every two blue beads. There are N blue beads. After looking at the arrangement below work out the number of red beads.

@
@@
@
@@
@
@@
@
@@
@
@@
@

Implement count_red_beads(n) (in PHP count_red_beads($n); in Java, Javascript, TypeScript, C, C++ countRedBeads(n)) so that it returns the number of red beads.
If there are less than 2 blue beads return 0.

Answer: {"fn_name": "count_red_beads", "inputs": [[0], [1], [3], [5]], "outputs": [[0], [0], [4], [8]]}
user: Hector the hacker has stolen some information, but it is encrypted. In order to decrypt it, he needs to write a function that will generate a decryption key from the encryption key which he stole (it is in hexadecimal). To do this, he has to determine the two prime factors `P` and `Q` of the encyption key, and return the product `(P-1) * (Q-1)`.

**Note:** the primes used are < 10^(5)

## Examples

For example if the encryption key is `"47b"`, it is 1147 in decimal. This factors to 31\*37, so the key Hector needs is 1080 (= 30\*36).


More examples:
* input: `"2533"`, result: 9328 (primes: 89, 107)
* input: `"1ba9"`, result: 6912 (primes: 73, 97)

Answer: {"fn_name": "find_key", "inputs": [["47b"], ["2533"], ["1ba9"]], "outputs": [[1080], [9328], [6912]]}
user: Just like in the ["father" kata](http://www.codewars.com/kata/find-fibonacci-last-digit/), you will have to return the last digit of the nth element in the Fibonacci sequence (starting with 1,1, to be extra clear, not with 0,1 or other numbers).

You will just get much bigger numbers, so good luck bruteforcing your way through it ;)
```python
last_fib_digit(1) == 1
last_fib_digit(2) == 1
last_fib_digit(3) == 2
last_fib_digit(1000) == 5
last_fib_digit(1000000) == 5
```
``` haskell
lastFibDigit       1 == 1
lastFibDigit       2 == 1
lastFibDigit       3 == 2
lastFibDigit    1000 == 5
lastFibDigit 1000000 == 5
```

Answer: {"fn_name": "last_fib_digit", "inputs": [[1], [21], [302], [4003], [50004], [600005], [7000006], [80000007], [900000008], [1000000009]], "outputs": [[1], [6], [1], [7], [8], [5], [3], [8], [1], [9]]}
user: # Task
John was in math class and got bored, so he decided to fold some origami from a rectangular `a × b` sheet of paper (`a > b`). His first step is to make a square piece of paper from the initial rectangular piece of paper by folding the sheet along the bisector of the right angle and cutting off the excess part.



After moving the square piece of paper aside, John wanted to make even more squares! He took the remaining (`a-b`) × `b` strip of paper and went on with the process until he was left with a square piece of paper.

 Your task is to determine how many square pieces of paper John can make.

# Example:

 For: `a = 2, b = 1`, the output should be `2`.
 
 Given `a = 2` and `b = 1`, John can fold a `1 × 1` then another `1 × 1`. 
 
 So the answer is `2`.
 
 For: `a = 10, b = 7`, the output should be `6`.
 
 We are given `a = 10` and `b = 7`. The following is the order of squares John folds: `7 × 7, 3 × 3, 3 × 3, 1 × 1, 1 × 1, and 1 × 1`.
 
 Here are pictures for the example cases.



# Input/Output


 - `[input]` integer `a`

  `2 ≤ a ≤ 1000`


 - `[input]` integer `b`

  `1 ≤ b < a ≤ 1000`


 - `[output]` an integer

  The maximum number of squares.

Answer: {"fn_name": "folding", "inputs": [[2, 1], [10, 7], [3, 1], [4, 1], [3, 2], [4, 2], [1000, 700], [1000, 999]], "outputs": [[2], [6], [3], [4], [3], [2], [6], [1000]]}
user: Create a function that will return true if all numbers in the sequence follow the same counting pattern. If the sequence of numbers does not follow the same pattern, the function should return false.

Sequences will be presented in an array of varied length. Each array will have a minimum of 3 numbers in it.

The sequences are all simple and will not step up in varying increments such as a Fibonacci sequence.

JavaScript examples:

Answer: {"fn_name": "validate_sequence", "inputs": [[[1, 2, 3, 4, 5, 8, 7, 8, 9]], [[2, 8, 6, 7, 4, 3, 1, 5, 9]], [[1, 2, 3, 4, 5, 6, 7, 8, 9]], [[0, 1, 2, 3, 4, 5, 6, 7, 8]], [[1, 3, 5, 7, 9, 11, 13, 15]], [[1, 3, 5, 7, 8, 12, 14, 16]], [[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]], [[0, 2, 4, 6, 8]], [[2, 4, 6, 8, 10]], [[2, 4, 6, 8, 10, 12, 14, 16, 18, 20]], [[2, 4, 6, 8, 10, 12, 14, 16, 18, 22]], [[2, 4, 6, 8, 10, 12, 13, 16, 18, 20]], [[3, 7, 9]], [[3, 6, 9]], [[3, 6, 8]], [[50, 100, 200, 400, 800]], [[50, 100, 150, 200, 250]], [[100, 200, 300, 400, 500, 600]]], "outputs": [[false], [false], [true], [true], [true], [false], [false], [true], [true], [true], [false], [false], [false], [true], [false], [false], [true], [true]]}
user: # Introduction 

The ragbaby cipher is a substitution cipher that encodes/decodes a text using a keyed alphabet and their position in the plaintext word they are a part of.

To encrypt the text `This is an example.` with the key `cipher`, first construct a keyed alphabet:
```
c       i       p       h       e       r       a       b       d       f       g       j       k       l       m       n       o       q       s       t       u       v       w       x       y       z
```

Then, number the letters in the text as follows:
```
T       h       i       s               i       s               a       n               e       x       a       m       p       l       e       .
1       2       3       4               1       2               1       2               1       2       3       4       5       6       7        
```

To obtain the encoded text, replace each character of the word with the letter in the keyed alphabet the corresponding number of places to the right of it (wrapping if necessary). 
Non-alphabetic characters are preserved to mark word boundaries.

Our ciphertext is then `Urew pu bq rzfsbtj.`

# Task

Wirate functions `encode` and `decode` which accept 2 parameters:
- `text` - string - a text to encode/decode
- `key` -  string - a key

# Notes

- handle lower and upper case in `text` string
- `key` consists of only lowercase characters

Answer: {"fn_name": "encode", "inputs": [["cipher", "cipher"], ["cipher", "cccciiiiippphheeeeerrrrr"]], "outputs": [["ihrbfj"], ["ihrbfj"]]}
user: Your colleagues have been good enough(?) to buy you a birthday gift. Even though it is your birthday and not theirs, they have decided to play pass the parcel with it so that everyone has an even chance of winning. There are multiple presents, and you will receive one, but not all are nice... One even explodes and covers you in soil... strange office. To make up for this one present is a dog! Happy days! (do not buy dogs as presents, and if you do, never wrap them).

Depending on the number of passes in the game (y), and the present you unwrap (x), return as follows:

x == goodpresent --> return x with num of passes added to each charCode (turn to charCode, add y to each, turn back)
x == crap || x == empty --> return string sorted alphabetically
x == bang --> return string turned to char codes, each code reduced by number of passes and summed to a single figure
x == badpresent --> return 'Take this back!'
x == dog, return 'pass out from excitement y times' (where y is the value given for y).

Answer: {"fn_name": "present", "inputs": [["badpresent", 3], ["goodpresent", 9], ["crap", 10], ["bang", 27], ["dog", 23]], "outputs": [["Take this back!"], ["pxxmy{n|nw}"], ["acpr"], ["300"], ["pass out from excitement 23 times"]]}