# Optimize my wings order

This tweet lists the possible orders for Wings of a Chinese restaurant1:

When ordering Pizza I usually calculate what size gives me the best Pizza-price ratio which is a simple calculation. However minimizing the price of an order at this restaurant isn't such a simple task, so I'd like to be prepared for my next order there.

# Challenge

Given an integer greater or equal to $$\4\$$, your task is to return one possible order which minimizes the price (overall cheapest) and the number of deals.

# Example

If I were to order $$\100\$$ Wings, it turns out the best bargain will cost $$\111.20\$$. However there are multiple orders which will cost that amount, namely:

[50,50],[25,25,50],[25,25,25,25]


Since the first order will use the least amount of deals ($$\2\$$) the result will be [50,50].

# Rules

• Input will be some integer $$\n \geq 4\$$
• Output will be a list/array/... of order sizes that sum up to $$\n\$$ and minimize the order's price
• you may choose to return all possible orders

# Testcases

4 -> [4]  (4.55)
23 -> [23]  (26.10)
24 -> [6,18],[9,15],[12,12]  (27.20)
31 -> [6,25]  (34.60)
32 -> [4,28],[6,26],[7,25]  (35.75)
33 -> [4,29],[5,28],[6,27],[7,26],[8,25]  (36.90)
34 -> [6,28],[9,25]  (38.00)
35 -> [35]  (39.15)
125 -> [125]  (139.00)
200 -> [25,50,125]  (222.40)
201 -> [26,50,125]  (223.55)
250 -> [125,125]  (278.00)
251 -> [26,50,50,125]  (279.15)
418 -> [15,28,125,125,125],[18,25,125,125,125]  (465.20)
1001 -> [26,50,50,125,125,125,125,125,125,125]  (1113.15)
12345 -> [15,80,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125],[25,70,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125],[45,50,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125,125]  (13728.10)


Note: These testcases list all possible outputs including the price, you're only required to output one and you're not required to output the price!

1: You can find the data as a CSV here.

• The real question is, who orders 200 or even 100 wings? ... Oct 26, 2018 at 21:43
• @Quintec: Why, do you need more testcases? Oct 26, 2018 at 23:24
• Two answers have misinterpreted the requirements, as only needing to minimise the price. Since minimising the price and number of deals is ambiguous (lowest price available from the ways with the lowest number of deals, or lowest number of deals available from the ways with the lowest price) it would be worth editing the requirement to be more explicit Oct 26, 2018 at 23:49
• Oct 27, 2018 at 4:12
• I notice that for $n \leq 23$ the price is given by $\frac 1 {20} \lceil \left ( \frac {68 n} 3 \right ) \rceil$, while the prices for $25 < n <= 50$ can be split into $25$ and $n - 25$ (for $n < 29$ you can't buy them separately but the formula still gives the appropriate value). $70$, $80$ and $125$ can also be derived from earlier values, so if you can use them they will reduce the number of deals. The other values are uneconomic.
– Neil
Oct 27, 2018 at 11:15

# JavaScript (ES6), 123 bytes

Returns the order as a space-separated string.

f=n=>n?(x=n>128|n==125?125:n>50?n<54?25:n-70?302256705>>n-80&n>79&n<109?80:50:n:n-24&&n-49?n<31|n%5<1?n:25:9)+' '+f(n-x):''


Try it online!

## How?

Given a number $$\n\$$ of wings, we recursively apply the following strategy.

### High values: $$\n>128\$$ or $$\n=125\$$

For high values, our best option is to buy $$\125\$$ wings. So that's what we're doing as long as $$\n\$$ is greater than or equal to $$\129\$$ or when $$\n\$$ is exactly equal to $$\125\$$.

We can't do that for $$\125 < n < 129\$$ because we'd be left with less than $$\4\$$ wings to buy, which is not possible.

### Low values: $$\n<31\$$

For $$\n<31\$$, the best deal is to buy all remaining wings at once, except for $$\n=24\$$ which must be purchased as either $$\2\times 12\$$, $$\18+6\$$ or $$\15+9\$$. We arbitrarily buy $$\9\$$ wings in that case.

### Range $$\31 \le n \le 50\$$

In this range, buying $$\25\$$ wings usually leads to an optimal order. But the following exceptions apply:

• If $$\n\$$ is a multiple of $$\5\$$, we must buy all remaining wings at once.
• If $$\n=49\$$, the best deals are $$\40+9\$$ and $$\28+21\$$. We arbitrarily buy $$\9\$$ wings in that case.

### Range $$\51 \le n \le 53\$$

We can't buy $$\50\$$ wings in that range or we'll be left with less than $$\4\$$ wings. Our best option is to buy $$\25\$$ of them instead. (We could buy $$\2\times26\$$ wings for $$\n=52\$$, but buying $$\25+27\$$ is just as good.)

### Range $$\54 \le n \le 128\$$ with $$\n \neq 125\$$

In this range, buying $$\50\$$ wings usually leads to an optimal order. But the following exceptions apply:

• If $$\n=70\$$, we must buy all of them at once.

• If $$\n\$$ is in $$\\{80,86,89,92,98,105,108\}\$$, we must buy $$\80\$$ wings. These values are encoded with the following bitmask, where the least significant bit is mapped to $$\80\$$ and the most significant one to $$\108\$$:

$$10010000001000001001001000001_{(2)} = 302256705_{(10)}$$

# JavaScript (Node.js), 112108106 105 bytes

f=n=>n?(x=n>128|n==125?125:n>53&n!=70?1629>>n/3+6&n<99==n%3/2?80:50:~n%25?n>30&&n%5?25:n:9)+' '+f(n-x):''


Try it online!

# Differences

• 51≤n≤53 is merged into 31≤n≤50 (8 bytes save)
• Rewrite the bitmap (3 bytes save)
• Rearrange some logic (4 6 7 bytes saved)

# Retina 0.8.2, 160 155 bytes

.+
$* {\b(1{80}(?=((111){2,6}|1{25}|1{28})?$)|1{70}$|1{9}(?=.{15}$|.{40}$)|(1{5}){6,9}$|1{26,29}$|1{4,23}$|1{125}|1{50}|1{25})+1,$& (1+),\1(1*)$
$.1,$2


Try it online! Link includes header that tests all values from $$\ n \$$ down to 1, remove it if you only want to test $$\ n \$$ itself. Uses @Arnauld's algorithm. Edit: Saved 5 bytes by purchasing 95 wings as 80 + 15 instead of 50 + 45. Explanation:

.+
$*  Convert to unary. {  Repeat until no more deals can be purchased. {\b(1{80}(?=((111){2,6}|1{25}|1{28})?$)|1{70}$|1{9}(?=.{15}$|.{40}$)|(1{5}){6,9}$|1{26,29}$|1{4,23}$|1{125}|1{50}|1{25})+1,$&  Find a way of purchasing deals and capture and duplicate one of the deals. (1+),\1(1*)$
$.1,$2


Convert the captured deal to decimal and subtract it from $$\ n \$$.

Deals are purchased under the following conditions:

1{80}(?=((111){2,6}|1{25}|1{28})?$)  Purchase 80 wings if that leaves 0, 6, 9, 12, 15, 18, 25 or 28 wings. 1{70}$


Purchase 70 wings if that's all we need.

1{9}(?=.{15}$|.{40}$)


Purchase 9 wings if that leaves 15 or 40 wings.

(1{5}){6,9}$ Purchase 30, 35, 40 or 45 wings if that's all we need. 1{26,29}$


Purchase 26, 27, 28 or 29 wings if that's all we need.

1{4,23}\$


Purchase 4 to 23 wings if that's all we need.

1{125}|1{50}|1{25}
`

Purchase 125, 50 or 25 wings if we can and if we can still purchase more wings exactly. Note that we have these options at the end of the alternation so that the exact purchases are tested first.