Bloons TD 6 Upgrade Paths

Question

Background

The monkeys need help organizing their defense and have asked you, Benjamin the code monkey, to create a program that will list all tower upgrade options. Each tower has three unique upgrade "paths", each having a tier represented by a number between 0 and 5 inclusive, 0 meaning no upgrade. Up to two paths may be chosen for upgrading, that is contain an upgrade tier 1 or greater. Additionally, only one path can can contain a tier 3 or greater.

Task

Output in some reasonable format all valid upgrade path triples in any order (the triples themselves are ordered). Triples can be represented in any reasonable way, such as 025 or 0-2-5. The triples must be distinguishable from each other in some way, so a flat list of numbers without triple delimiters is not allowed.

Here is an example list of all 64 possible triples, as they appear in-game for your insta-monkey collection:

5-2-0
5-0-2
5-1-0
5-0-1
5-0-0
2-5-0
0-5-2
1-5-0
0-5-1
0-5-0
2-0-5
0-2-5
1-0-5
0-1-5
0-0-5
4-2-0
4-0-2
4-1-0
4-0-1
4-0-0
2-4-0
0-4-2
1-4-0
0-4-1
0-4-0
2-0-4
0-2-4
1-0-4
0-1-4
0-0-4
3-2-0
3-0-2
3-1-0
3-0-1
3-0-0
2-3-0
0-3-2
1-3-0
0-3-1
0-3-0
2-0-3
0-2-3
1-0-3
0-1-3
0-0-3
2-2-0
2-0-2
2-1-0
2-0-1
2-0-0
0-2-2
1-2-0
0-2-1
0-2-0
1-0-2
0-1-2
0-0-2
1-1-0
1-0-1
1-0-0
0-1-1
0-1-0
0-0-1
0-0-0

\$\begingroup\$ Are we allowed to output duplicates? (assuming no) \$\endgroup\$
– thejonymyster
Commented Jun 30, 2022 at 23:06 — thejonymyster, Commented Jun 30, 2022 at 23:06
\$\begingroup\$ @thejonymyster no, there should be exactly 64 triples \$\endgroup\$
– qwr
Commented Jun 30, 2022 at 23:21 — qwr, Commented Jun 30, 2022 at 23:21

ovs · Accepted Answer · 2022-07-01 09:48:38Z

7

K (ngn/k), 20 18 bytes

?,/+'!'1|3*<'+!3#3

Try it online!

+!3#3 All triples with values in 0 1 2.
<' Grade up each triple. This results in all permutations of 0 1 2 with some duplicates.
1|3* Multiply by 3 and take maximum with 1 to generate all permutations of 6 3 1.
+'!' For each permutation, generate all triples where the each entry is a non-negative integer less than the number in the permutation at the same index.
?,/ Flatten into a matrix with three columns and take the unique rows.

edited Jul 1, 2022 at 9:48

answered Jun 30, 2022 at 18:04

ovs

60.1k3 gold badges47 silver badges159 bronze badges

Add a comment |

chunes · Accepted Answer · 2022-06-30 22:42:39Z

5

Factor + `math.combinatorics`, 59 58 bytes

6 iota 3 selections [ natural-sort "\0"before? ] filter .

Try it online!

-1 byte from a tip by @ovs.

Note the string "\0" has literal control character 3 embedded as well as the 0; you can see it on TIO. This string is equivalent to the longer (but clearer) sequence { 0 3 }.

6 iota 3 selections

Generate all 3-selections of \$[0..5]\$: { { 0 0 0 } { 0 0 1 } ... { 5 5 5 } }
[ natural-sort { 0 3 } before? ] filter

Select those that when sorted are less than { 0 3 }.
.

Print them.

edited Jun 30, 2022 at 22:42

answered Jun 30, 2022 at 18:19

chunes

26.2k3 gold badges23 silver badges53 bronze badges

Add a comment |

Kevin Cruijssen · Accepted Answer · 2022-07-02 11:17:24Z

4

Python 2, 82 80 bytes

from itertools import*
print[k for k in product(*[range(6)]*3)if[0,3]>sorted(k)]

-2 bytes thanks to @ovs.

Try it online.

Explanation:

from itertools import*: Import itertools for the cartesian product builtin
range(6): Push list [0,1,2,3,4,5]
[^]*3: Repeat it three times: [[0,1,2,3,4,5],[0,1,2,3,4,5],[0,1,2,3,4,5]]
product(*^): Use the cartesian product builtin to get all triplets
k for k in ^: Loop over these triplets
^if: And filter to only keep the triple-tuplets k that are:
- sorted(k): When sorted and converted to an array
- [0,3]>^: are smaller than [0,3]
[^]: Wrap all these tuplets into an array
print^: And print it as result

edited Jul 2, 2022 at 11:17

answered Jun 30, 2022 at 19:04

Kevin Cruijssen

129k12 gold badges142 silver badges381 bronze badges

3

\$\begingroup\$ I think [0,3]>sorted(k) is sufficient \$\endgroup\$
– ovs
Commented Jun 30, 2022 at 21:06
\$\begingroup\$ Surely [range(6] is roughly equivalent to [[0,1,2,3,4,5]], so that [range[6]]*3 is [[0,1,2,3,4,5],[0,1,2,3,4,5],[0,1,2,3,4,5]]? \$\endgroup\$
– Neil
Commented Jul 2, 2022 at 9:42
\$\begingroup\$ @Neil You're right. I've changed the explanation accordingly. \$\endgroup\$
– Kevin Cruijssen
Commented Jul 2, 2022 at 11:17

Add a comment |

lynn · Accepted Answer · 2022-07-10 21:51:23Z

4

Python, 71 bytes

for x in range(521):max(s:='%03d'%x)<'6'!=['0','3']>sorted(s)==print(s)

Attempt This Online!

I took the ['0','3']>sorted(s) idea from ovs.

answered Jul 10, 2022 at 21:51

lynn

69k11 gold badges133 silver badges282 bronze badges

Add a comment |

ovs · Accepted Answer · 2022-06-30 21:10:49Z

3

Husk, 10 bytes

uΣmPΠmŀ∫ḣ3

Try it online!

JavaScript (V8), 75 67 bytes

Saved 8 bytes thanks to @tsh

Prints one comma-separated triplet per line.

for(i=216;i--;)[...a=[i/36%6|0,i/6%6|0,i%6]].sort()<'0,3'&&print(a)

Try it online!

edited Jul 1, 2022 at 16:36

answered Jun 30, 2022 at 17:48

Arnauld

195k20 gold badges178 silver badges644 bronze badges

1

\$\begingroup\$ Maybe for(i=216;i--;)[...a=[i/36%6|0,i/6%6|0,i%6]].sort()<'0,3'&&print(a)? \$\endgroup\$
– tsh
Commented Jul 1, 2022 at 3:31

Add a comment |

Kevin Cruijssen · Accepted Answer · 2022-06-30 18:30:41Z

05AB1E, 13 11 bytes

5Ý3ãʒ{₆1š‹P

-2 bytes thanks to a tip of @ovs using 136.

Outputs as a list of triplets.

Try it online.

Explanation:

5Ý           # Push a list in the range [0,5]
  3ã         # Cartesian power of 3: get all triplets using these [0,1,2,3,4,5]
    ʒ        # Filter this list by:
     {       #  Sort the triplet from lowest to highest
      ƵZ     #  Push compressed integer 136
        S    #  Convert it to a list of digits: [1,3,6]
         ‹   #  Check for sorted triplet [a,b,c] whether [a<1,b<3,c<6]
          P  #  Check if all three are truthy
             # (after which the filtered list of 64 triplets is output implicitly)

ƵZS could alternatively be ₆1š for the same byte-count: try it online.

      ₆      #  Push 36
       1š    #  Convert it to a list of digits, and prepend 1

See this 05AB1E tip of mine (section How to compress large integers?) to understand why ƵZ is 136.

There is a 11-byter that uses a single filter. ƵZ might be a hint ;) — ovs, Commented Jun 30, 2022 at 18:19

Neil · Accepted Answer · 2022-07-01 00:02:03Z

2

Retina 0.8.2, 50 bytes


520$*¶

00$.`
.+(...)
$1
G`0
A`[3-5].*[3-5]|[6-9]

Try it online! Link includes footer that prettifies the output. Explanation:


520$*¶

00$.`
.+(...)
$1

List all the integers from 0 to 520 inclusive, padded to 3 digits.

G`0

Only keep those integers with at least one 0 digit.

A`[3-5].*[3-5]|[6-9]

Discard those with more than one digit greater than 2 or with a digit greater than 5.

answered Jul 1, 2022 at 0:02

Neil

174k12 gold badges73 silver badges277 bronze badges

\$\begingroup\$ I was looking forward to a regex solution! \$\endgroup\$
– qwr
Commented Jul 1, 2022 at 16:28

Add a comment |

the default. · Accepted Answer · 2022-07-04 06:22:13Z

2

Python 3, 75 bytes

print([r for r in[(x//36,x//6%6,x%6)for x in range(216)]if[0,3]>sorted(r)])

Try it online!

answered Jul 4, 2022 at 6:22

the default.

8,4251 gold badge35 silver badges62 bronze badges

Add a comment |

Neil · Accepted Answer · 2022-07-01 00:13:36Z

Charcoal, 26 bytes

ΦＥφ﹪%03dι∧№ι0∧›6⌈ι›²ΣＥι‹2λ

Try it online! Link is to verbose version of code. Explanation:

  φ                         Predefined variable `1000`
 Ｅ                          Map over implicit range
        ι                   Current value
   ﹪%03d                    Formatted to 3 0-filled digits
Φ                           Filtered where
           ι                Current value
          №                 Contains
            0               Literal string `0`
         ∧                  Logical And
               6            Literal string `6`
              ›             Is greater than
                 ι          Current value
                ⌈           Maximum character
             ∧              Logical And
                   ²        Literal integer `2`
                  ›         Is greater than
                      ι     Current value
                     Ｅ      Map over digits
                        2   Literal string `2`
                       ‹    Is less than
                         λ  Current digit
                    Σ       Take the sum
                            Implicitly print

28 bytes for the prettier version:

ΦＥφ⪫﹪%03dι-∧№ι0∧›6⌈ι›²ΣＥι‹2λ

Try it online! Link is to verbose version of code.

G B · Accepted Answer · 2022-07-01 06:01:29Z

1

Ruby, 62 bytes

p (z=*0..5).product(z,z).select{|x|x.min<1&&(x-[3,4,5])[1]}|[]

Try it online!

edited Jul 1, 2022 at 6:01

answered Jul 1, 2022 at 5:56

G B

22.5k1 gold badge20 silver badges53 bronze badges

Add a comment |

mousetail · Accepted Answer · 2022-07-01 09:21:35Z

1

Rust, 106 bytes

for i in 0..6{for j in 0..6{for k in 0..6{let mut t=[i,j,k];t.sort();if t<[0,3,6]{print!("{i}{j}{k}
")}}}}

Actually using nested loops is barely shorter than my attempt using itertoools.

Attempt This Online!

answered Jul 1, 2022 at 9:21

mousetail

13k1 gold badge36 silver badges83 bronze badges

\$\begingroup\$ if[0,3,6]>t saves 1 byte \$\endgroup\$
– corvus_192
Commented Jan 20 at 14:26

Add a comment |

math junkie · Accepted Answer · 2022-07-02 14:09:07Z

1

Pyth, 12 bytes

f>,Z3ST^U6 3

Try it online!

Port of Kevin Cruijssen's Python answer

f>,Z3ST^U6 3
        U6    Range up to 6 = [0,1,2,3,4,5]
       ^   3  Cartesian product with itself 3 times = [0,1,2,3,4,5] * [0,1,2,3,4,5] * [0,1,2,3,4,5]
f             Filter for elements T such that:
 >,Z3ST       [0,3] > sorted(T)

edited Jul 2, 2022 at 14:09

answered Jul 1, 2022 at 15:34

math junkie

7,2471 gold badge22 silver badges46 bronze badges

Add a comment |

thejonymyster · Accepted Answer · 2022-07-11 02:47:38Z

1

Regenerate `-a`, 73 64 bytes

[345]([12]0|0[012])|[12]([1-5]0|0[0-5])|0([012][0-5]|[345][012])

Attempt This Online!

Surprised nopony did this yet. Probably golfable!

EDIT: -9 thanks to math junkie for reminding me of the shorter output format. Now it's 1 byte per combination :P I still think it can be golfed further, possibly with a new approach :-)c

edited Jul 11, 2022 at 2:47

answered Jul 11, 2022 at 1:07

thejonymyster

3,33713 silver badges44 bronze badges

1

\$\begingroup\$ I think you can drop the - separators since that's allowed by the challenge rules \$\endgroup\$
– math junkie
Commented Jul 11, 2022 at 1:23
\$\begingroup\$ @mathjunkie omg i forgot thank u \$\endgroup\$
– thejonymyster
Commented Jul 11, 2022 at 2:40

Add a comment |

Fmbalbuena · Accepted Answer · 2022-12-16 16:35:18Z

1

Python 3, 161 146 141 135 bytes

from itertools import*
for u in product(*[range(6)]*3):
 i,j,k=u
 if 0not in u or i>2and j>2or j>2and k>2or k>2and i>2:1
 else:print(u)

Try it online!

-5 thanks to qwr

edited Dec 16, 2022 at 16:35

answered Jun 30, 2022 at 19:33

Fmbalbuena

4,2051 gold badge9 silver badges52 bronze badges

\$\begingroup\$ You can probably use itertools cartesian product and filter lambda to save some bytes over triple for loop \$\endgroup\$
– qwr
Commented Jun 30, 2022 at 20:55
\$\begingroup\$ @qwr I don't know what is itertools, Can you upload a TIO link? \$\endgroup\$
– Fmbalbuena
Commented Jun 30, 2022 at 21:02
\$\begingroup\$ See Kevin Cruijssen's answer for an example \$\endgroup\$
– qwr
Commented Jun 30, 2022 at 21:04
\$\begingroup\$ @qwr Thanks, -5 bytes. \$\endgroup\$
– Fmbalbuena
Commented Jul 1, 2022 at 0:14

Add a comment |

Nick Kennedy · Accepted Answer · 2023-12-30 08:16:41Z

1

Jelly, 11 bytes

6p3’Ż€Œ!€ẎQ

Try it online!

A niladic link that returns the 64 triples. This is my answer to the follow-up question but with the sorting code removed.

answered Dec 30, 2023 at 8:16

Nick Kennedy

21.2k3 gold badges17 silver badges43 bronze badges

Add a comment |

pacman256 · Accepted Answer · 2024-09-05 14:42:53Z

1

Vyxal, 66 bits^v2, 8.25 bytes

⁺ŀfʀΠvṖÞfU

Try it Online!

Bitstring:

111111011001001101001110100101101000111101101011000011001000110100

answered 2 days ago

pacman256

4,2251 gold badge9 silver badges32 bronze badges

Add a comment |

Yousername · Accepted Answer · 2022-07-10 21:00:52Z

0

Batch, 271 bytes

@!! 2>nul||cmd/q/v/c%0&&exit/b&for /l %%i in (0,1,5)do @(for /l %%j in (0,1,5)do @(for /l %%k in (0,1,5)do @(set s=%%i%%j%%k&set p=!s:0=!&if !p:~!==~ set p=0
if !p!. neq !s!. set/af=!p!/10&set/al=!p!%%10&if !f! lss !l! (if !f! leq 2 echo !s!)else if !l! leq 2 echo !s!)))

answered Jul 10, 2022 at 21:00

Yousername

4,05010 silver badges24 bronze badges

Add a comment |

Stack Exchange Network

Bloons TD 6 Upgrade Paths

Background

Task

18 Answers 18

K (ngn/k), 20 18 bytes

Factor + `math.combinatorics`, 59 58 bytes

Python 2, 82 80 bytes

Python, 71 bytes

Husk, 10 bytes

JavaScript (V8), 75 67 bytes

05AB1E, 13 11 bytes

Retina 0.8.2, 50 bytes

Python 3, 75 bytes

Charcoal, 26 bytes

Ruby, 62 bytes

Rust, 106 bytes

Pyth, 12 bytes

Regenerate `-a`, 73 64 bytes

Python 3, 161 146 141 135 bytes

Jelly, 11 bytes

Vyxal, 66 bits^v2, 8.25 bytes

Batch, 271 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
number
kolmogorov-complexity
or ask your own question.

Linked

Hot Network Questions

Bloons TD 6 Upgrade Paths

Background

Task

18 Answers 18

K (ngn/k), 20 18 bytes

Factor + math.combinatorics, 59 58 bytes

Python 2, 82 80 bytes

Python, 71 bytes

Husk, 10 bytes

JavaScript (V8), 75 67 bytes

05AB1E, 13 11 bytes

Retina 0.8.2, 50 bytes

Python 3, 75 bytes

Charcoal, 26 bytes

Ruby, 62 bytes

Rust, 106 bytes

Pyth, 12 bytes

Regenerate -a, 73 64 bytes

Python 3, 161 146 141 135 bytes

Jelly, 11 bytes

Vyxal, 66 bitsv2, 8.25 bytes

Batch, 271 bytes

Your Answer

Sign up or log in

Post as a guest

Not the answer you're looking for? Browse other questions tagged code-golfnumberkolmogorov-complexity or ask your own question.

Linked

Related

Hot Network Questions

Factor + `math.combinatorics`, 59 58 bytes

Regenerate `-a`, 73 64 bytes

Vyxal, 66 bits^v2, 8.25 bytes

Not the answer you're looking for? Browse other questions tagged
code-golf
number
kolmogorov-complexity
or ask your own question.