Length of the Longest Descent
Your task is to determine the length of the longest descent down a "mountain" represented as a grid of integer heights. A "descent" is any path from a starting cell to orthogonally adjacent cells with strictly decreasing heights (i.e. not diagonal and not to the same height). For instance, you can move from 5-4-3-1 but not 5-5-4-3-3-2-1. The length of this path is how many cell movements there are from the starting cell to the ending cell, thus 5-4-3-1 is length 3.
You will receive a rectangular grid as input and you should output an integer indicating the longest descent.
Examples
1 2 3 2 2
3 4 5 5 5
3 4 6 7 4
3 3 5 6 2
1 1 2 3 1
The length of the longest descent down this mountain is 5. The longest path starts at the 7, moves left, up, left, up, and then left (7-6-5-4-2-1). Since there are 5 movements in this path, the path length is 5.
They might be all the same number.
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
Since this height map is flat, the longest descent is 0. (not 19, since the path sequence must be strictly descending)
Height maps can be made up of larger numbers than single-digit numbers.
10 12 13 14 15 15
17 14 15 15 15 16
18 20 21 15 15 15
21 14 10 11 11 15
15 15 15 15 15 15
The longest path here is of length 6. (21, 20, 18, 17, 14, 12, 10)
...And even bigger numbers are fine too.
949858 789874 57848 43758 387348
5848 454115 4548 448545 216464
188452 484126 484216 786654 145451
189465 474566 156665 132645 456651
985464 94849 151654 151648 484364
The longest descent here is of length 7. (786654, 484216, 484126, 474566, 156665, 151654, 151648, 132645)
Rules and Notes
- Grids may be taken in any convenient format. Specify your format in your answer.
- You may assume the height map is perfectly rectangular, is nonempty, and contains only positive integers in the signed 32-bit integer range.
- The longest descent path can begin and end anywhere on the grid.
- You do not need to describe the longest descent path in any way. Only its length is required.
- Shortest code wins
code-golf grid
|
show 1 more comment
Your task is to determine the length of the longest descent down a "mountain" represented as a grid of integer heights. A "descent" is any path from a starting cell to orthogonally adjacent cells with strictly decreasing heights (i.e. not diagonal and not to the same height). For instance, you can move from 5-4-3-1 but not 5-5-4-3-3-2-1. The length of this path is how many cell movements there are from the starting cell to the ending cell, thus 5-4-3-1 is length 3.
You will receive a rectangular grid as input and you should output an integer indicating the longest descent.
Examples
1 2 3 2 2
3 4 5 5 5
3 4 6 7 4
3 3 5 6 2
1 1 2 3 1
The length of the longest descent down this mountain is 5. The longest path starts at the 7, moves left, up, left, up, and then left (7-6-5-4-2-1). Since there are 5 movements in this path, the path length is 5.
They might be all the same number.
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
Since this height map is flat, the longest descent is 0. (not 19, since the path sequence must be strictly descending)
Height maps can be made up of larger numbers than single-digit numbers.
10 12 13 14 15 15
17 14 15 15 15 16
18 20 21 15 15 15
21 14 10 11 11 15
15 15 15 15 15 15
The longest path here is of length 6. (21, 20, 18, 17, 14, 12, 10)
...And even bigger numbers are fine too.
949858 789874 57848 43758 387348
5848 454115 4548 448545 216464
188452 484126 484216 786654 145451
189465 474566 156665 132645 456651
985464 94849 151654 151648 484364
The longest descent here is of length 7. (786654, 484216, 484126, 474566, 156665, 151654, 151648, 132645)
Rules and Notes
- Grids may be taken in any convenient format. Specify your format in your answer.
- You may assume the height map is perfectly rectangular, is nonempty, and contains only positive integers in the signed 32-bit integer range.
- The longest descent path can begin and end anywhere on the grid.
- You do not need to describe the longest descent path in any way. Only its length is required.
- Shortest code wins
code-golf grid
How should the last example be interpreted?
– Peter Taylor
yesterday
@PeterTaylor I'm not sure what you mean.
– Beefster
yesterday
I think the last example is just a matrix of multi digit numbers
– Embodiment of Ignorance
yesterday
@EmbodimentofIgnorance, ah, yes, I see. It would be a lot easier to spot the path with two-digit numbers rather than 4 to 6.
– Peter Taylor
yesterday
1
@Οurous: just rectangular. Not jagged.
– Beefster
yesterday
|
show 1 more comment
Your task is to determine the length of the longest descent down a "mountain" represented as a grid of integer heights. A "descent" is any path from a starting cell to orthogonally adjacent cells with strictly decreasing heights (i.e. not diagonal and not to the same height). For instance, you can move from 5-4-3-1 but not 5-5-4-3-3-2-1. The length of this path is how many cell movements there are from the starting cell to the ending cell, thus 5-4-3-1 is length 3.
You will receive a rectangular grid as input and you should output an integer indicating the longest descent.
Examples
1 2 3 2 2
3 4 5 5 5
3 4 6 7 4
3 3 5 6 2
1 1 2 3 1
The length of the longest descent down this mountain is 5. The longest path starts at the 7, moves left, up, left, up, and then left (7-6-5-4-2-1). Since there are 5 movements in this path, the path length is 5.
They might be all the same number.
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
Since this height map is flat, the longest descent is 0. (not 19, since the path sequence must be strictly descending)
Height maps can be made up of larger numbers than single-digit numbers.
10 12 13 14 15 15
17 14 15 15 15 16
18 20 21 15 15 15
21 14 10 11 11 15
15 15 15 15 15 15
The longest path here is of length 6. (21, 20, 18, 17, 14, 12, 10)
...And even bigger numbers are fine too.
949858 789874 57848 43758 387348
5848 454115 4548 448545 216464
188452 484126 484216 786654 145451
189465 474566 156665 132645 456651
985464 94849 151654 151648 484364
The longest descent here is of length 7. (786654, 484216, 484126, 474566, 156665, 151654, 151648, 132645)
Rules and Notes
- Grids may be taken in any convenient format. Specify your format in your answer.
- You may assume the height map is perfectly rectangular, is nonempty, and contains only positive integers in the signed 32-bit integer range.
- The longest descent path can begin and end anywhere on the grid.
- You do not need to describe the longest descent path in any way. Only its length is required.
- Shortest code wins
code-golf grid
Your task is to determine the length of the longest descent down a "mountain" represented as a grid of integer heights. A "descent" is any path from a starting cell to orthogonally adjacent cells with strictly decreasing heights (i.e. not diagonal and not to the same height). For instance, you can move from 5-4-3-1 but not 5-5-4-3-3-2-1. The length of this path is how many cell movements there are from the starting cell to the ending cell, thus 5-4-3-1 is length 3.
You will receive a rectangular grid as input and you should output an integer indicating the longest descent.
Examples
1 2 3 2 2
3 4 5 5 5
3 4 6 7 4
3 3 5 6 2
1 1 2 3 1
The length of the longest descent down this mountain is 5. The longest path starts at the 7, moves left, up, left, up, and then left (7-6-5-4-2-1). Since there are 5 movements in this path, the path length is 5.
They might be all the same number.
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
Since this height map is flat, the longest descent is 0. (not 19, since the path sequence must be strictly descending)
Height maps can be made up of larger numbers than single-digit numbers.
10 12 13 14 15 15
17 14 15 15 15 16
18 20 21 15 15 15
21 14 10 11 11 15
15 15 15 15 15 15
The longest path here is of length 6. (21, 20, 18, 17, 14, 12, 10)
...And even bigger numbers are fine too.
949858 789874 57848 43758 387348
5848 454115 4548 448545 216464
188452 484126 484216 786654 145451
189465 474566 156665 132645 456651
985464 94849 151654 151648 484364
The longest descent here is of length 7. (786654, 484216, 484126, 474566, 156665, 151654, 151648, 132645)
Rules and Notes
- Grids may be taken in any convenient format. Specify your format in your answer.
- You may assume the height map is perfectly rectangular, is nonempty, and contains only positive integers in the signed 32-bit integer range.
- The longest descent path can begin and end anywhere on the grid.
- You do not need to describe the longest descent path in any way. Only its length is required.
- Shortest code wins
code-golf grid
code-golf grid
edited yesterday
asked yesterday
Beefster
1,068722
1,068722
How should the last example be interpreted?
– Peter Taylor
yesterday
@PeterTaylor I'm not sure what you mean.
– Beefster
yesterday
I think the last example is just a matrix of multi digit numbers
– Embodiment of Ignorance
yesterday
@EmbodimentofIgnorance, ah, yes, I see. It would be a lot easier to spot the path with two-digit numbers rather than 4 to 6.
– Peter Taylor
yesterday
1
@Οurous: just rectangular. Not jagged.
– Beefster
yesterday
|
show 1 more comment
How should the last example be interpreted?
– Peter Taylor
yesterday
@PeterTaylor I'm not sure what you mean.
– Beefster
yesterday
I think the last example is just a matrix of multi digit numbers
– Embodiment of Ignorance
yesterday
@EmbodimentofIgnorance, ah, yes, I see. It would be a lot easier to spot the path with two-digit numbers rather than 4 to 6.
– Peter Taylor
yesterday
1
@Οurous: just rectangular. Not jagged.
– Beefster
yesterday
How should the last example be interpreted?
– Peter Taylor
yesterday
How should the last example be interpreted?
– Peter Taylor
yesterday
@PeterTaylor I'm not sure what you mean.
– Beefster
yesterday
@PeterTaylor I'm not sure what you mean.
– Beefster
yesterday
I think the last example is just a matrix of multi digit numbers
– Embodiment of Ignorance
yesterday
I think the last example is just a matrix of multi digit numbers
– Embodiment of Ignorance
yesterday
@EmbodimentofIgnorance, ah, yes, I see. It would be a lot easier to spot the path with two-digit numbers rather than 4 to 6.
– Peter Taylor
yesterday
@EmbodimentofIgnorance, ah, yes, I see. It would be a lot easier to spot the path with two-digit numbers rather than 4 to 6.
– Peter Taylor
yesterday
1
1
@Οurous: just rectangular. Not jagged.
– Beefster
yesterday
@Οurous: just rectangular. Not jagged.
– Beefster
yesterday
|
show 1 more comment
7 Answers
7
active
oldest
votes
JavaScript (ES7), 106 103 102 98 bytes
f=(m,n=b=-1,x,y,p)=>m.map((r,Y)=>r.map((v,X)=>(x-X)**2+(y-Y)**2-1|v/p?b=n<b?b:n:f(m,n+1,X,Y,v)))|b
Try it online!
Commented
f = ( // f = recursive function taking:
m, // m = input matrix
n = b = -1, // n = length of the current path; b = best length so far
x, y, // x, y = coordinates of the previous cell
p // p = value of the previous cell
) => //
m.map((r, Y) => // for each row r at position Y in m:
r.map((v, X) => // for each value v at position X in r:
(x - X) ** 2 + // compute the squared Euclidean distance
(y - Y) ** 2 // between (x, y) and (X, Y)
- 1 // if A) the above result is not equal to 1
| v / p ? // or B) v is greater than or equal to p:
b = n < b ? b : n // end of path: update b to n if n >= b
: // else:
f(m, n + 1, X, Y, v) // do a recursive call
) // end of inner map()
) | b // end of outer map(); return b
How?
During the first iteration, $x$, $y$ and $p$ are all undefined and both tests (A and B in the comments) evaluate to NaN, which triggers the recursive call. Therefore, all cells are considered as a possible starting point of the path.
add a comment |
Jelly, 23 21 20 bytes
-2 thanks to Erik the Outgolfer
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’
Try it online! (way too inefficient for the examples - path here is 95 94 93 83 77 40 10
so 6
is yielded)
How?
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’ - Link: list of lists of integers, M
ŒỤ - multi-dimensional indices sorted by values
ŒP - power-set
Ƈ - filter, keep those for which:
Ʋ - last four links as a monad:
Ɲ - for each pair of neighbours:
ạ - absolute difference
§ - sum each
Ị - insignificant?
Ạ - all?
œị - multi-dimensional index into:
⁸ - chain's left argument, M
Ƈ - filter, keep only those:
Ƒ - unaffected by?:
Q - de-duplicate
Ṫ - tail
L - length
’ - decrement
add a comment |
Python 2, 150 147 140 136 134 132 125 123 120 bytes
l=lambda g,i,j:max(0<g.get(t)<g[i,j]and-~l(g,*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
lambda g:max(l(g,*t)for t in g)
Try it online!
Takes input in the form of a dictionary (x, y): value
.
-7 bytes thanks to wizzwizz4, -2 bytes thanks to Jonathan Allen, -2 bytes thanks to BMO
Alternative, 123 121 bytes
l=lambda i,j:max(0<g.get(t)<g[i,j]and-~l(*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
g=input();print max(l(*t)for t in g)
Try it online!
Essentially the same solution, just with the final lambda replaced by an actual program. I personally like the first one better, but this one comes close in byte count by allowing g
to be used as a global variable.
add a comment |
Clean, 211 207 bytes
import StdEnv,Data.List
z=zipWith
$l=maximum[length k-1\p<-permutations[(v,[x,y])\y<-[0..]&u<-l,x<-[0..]&v<-u],(k,[m:n])<-map unzip(subsequences p)|and[all((>)2o sum o map abs)(z(z(-))n[m:n]):z(>)k(tl k)]]
Try it online!
A brute-force solution taking a list-of-lists-of-integers ([[Int]]
).
The TIO driver takes the same format as the examples through STDIN.
It's too slow to run any of the examples on TIO and probably locally too, but works in theory.
This one does the same thing faster, can do 3x3 or 2x4 on TIO and 4x4 and 3x5 locally.
Indented:
$ l
= maximum
[ length k-1
\p <- permutations
[ (v, [x, y])
\y <- [0..] & u <- l
, x <- [0..] & v <- u
]
, (k, [m: n]) <- map unzip
(subsequences p)
| and
[ all
((>) 2 o sum o map abs)
(zipWith (zipWith (-)) n [m:n])
:
zipWith (>) k (tl k)
]
]
add a comment |
Python 3, 219 bytes
e,m=len,enumerate
b=lambda g,x,y:[b(g,i,j)for o in[-1,1]for i,j in[(x+o,y),(x,y+o)]if e(g)>i>=0<=j<e(g[x])and g[x][y]<g[i][j]]
l=lambda t:e(t)and 1+max(map(l,t))
d=lambda g:max(l(b(g,x,y))for x,r in m(g)for y,_ in m(r))
Try it online!
Grid is represented as list of lists:
[
[1, 2, 3, 2, 2],
[3, 4, 5, 5, 5],
[3, 4, 6, 7, 4],
[3, 3, 5, 6, 2],
[1, 1, 2, 3, 1],
]
Original ungolfed code:
def potential_neighbours(x, y):
return [(x-1, y), (x+1, y), (x, y-1), (x, y+1)]
def neighbours(grid, x, y):
result =
for i, j in potential_neighbours(x, y):
if 0 <= i < len(grid) and 0 <= j < len(grid[x]) and grid[x][y] < grid[i][j]:
result += [(i, j)]
return result
def build_tree(grid, x, y):
return [build_tree(grid, i, j) for i, j in neighbours(grid, x, y)]
def longest_path_in_tree(tree):
if len(tree) == 0:
return 0
return 1 + max(map(longest_path_in_tree, tree))
def longest_descent(grid):
trees = [build_tree(grid, x, y) for x, row in enumerate(grid) for y, _ in enumerate(row)]
return max(map(longest_path_in_tree, trees))
New contributor
add a comment |
Haskell, 188 186 bytes
Needs $texttt{-XNoMonomorphismRestriction}$:
f m|c<-[0..length(m!!0)-1],r<-[0..length m-1]=h[g[(x,y)]|x<-r,y<-c,let g((x,y):p)=h[1+g(k:p)|i<-[-1,1],k@(u,v)<-[(x+i,y),(x,y+i)],u#r,v#c,m!!u!!v<m!!x!!y,not$k#p]]
(#)=elem
h=foldl max 0
Try it online!
Alternatively we could use notElem
over (not.).(#)
without the flag for $+4$ bytes:
Try it online!
Explanation & Ungolfed
Strategy: Recursively try all feasible paths, keeping track of visited entries and maximize their length.
Let's first define some helpers.. Since we need elem
and notElem
, let's use (#)
for elem
. Also, to maximize we'll need a total function (maximize
is not), returning $0$ when the list is empty:
safeMaximum = foldl max 0
Now we're ready to define our recursive function fun :: [[Integer]] -> Integer
:
fun xs
| c <- [0..length(m!!0)-1] -- all possible indices of xs' columns
, r <- [0..length m-1] -- all possible indices of xs' rows
= safeMaximum -- maximize ..
[ g [(x,y)] -- .. initially we haven't visited any others
| x <- c, y<-r -- .. all possible entries
-- For the purpose of golfing we define g in the list-comprehension, it takes all visited entries (p) where (x,y) is the most recent
, let g((x,y):p) = safeMaximum -- maximize ..
[ 1 + g(k:p) -- .. recurse, adding (x,y) to the visited nodes & increment (the next path will be 1 longer)
| i <- [-1,1] -- offsets [left/up,right/down]
, k@(u,v) <-[(x+i,y),(x,y+i)] -- next entry-candidate
, u#c, v#r -- make sure indices are in bound ..
, m!!u!!v < m!!x!!y -- .. , the the path is decreasing
, not$(u,v)#p -- .. and we haven't already visited that element
]
]
How does this take grids? List of lists?
– Beefster
6 hours ago
@Beefster: Yes, it says[[Integer]]
is list of lists. Though in the linked TIO you have a wrapperparse :: String -> [[Integer]]
, st. you can use strings split on spaces and new-lines.
– BMO
5 hours ago
add a comment |
Python 3, 263 227 bytes
def f(m):
p={(x,y):[c for i in[-1,1]for c in[(x,y+i),(x+i,y)]]for x,y in m};d={c:0 for c in p if not p[c]}
while len(p)-len(d):
for c in p:
for b in p[c]:
if b in d:d[c]=max(d[b]+1,d.get(c,0))
return max(d.values())
Try it online!
-2 bytes thanks to BMO
Takes grids in the format {(0, 0): 1, (1, 0): 2, ...}
. This format can be generated from the example format using the following utility function:
lambda s,e=enumerate:{(x,y):int(n)for y,l in e(s.split('n'))for x,n in e(l.split())}
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "200"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f178326%2flength-of-the-longest-descent%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
JavaScript (ES7), 106 103 102 98 bytes
f=(m,n=b=-1,x,y,p)=>m.map((r,Y)=>r.map((v,X)=>(x-X)**2+(y-Y)**2-1|v/p?b=n<b?b:n:f(m,n+1,X,Y,v)))|b
Try it online!
Commented
f = ( // f = recursive function taking:
m, // m = input matrix
n = b = -1, // n = length of the current path; b = best length so far
x, y, // x, y = coordinates of the previous cell
p // p = value of the previous cell
) => //
m.map((r, Y) => // for each row r at position Y in m:
r.map((v, X) => // for each value v at position X in r:
(x - X) ** 2 + // compute the squared Euclidean distance
(y - Y) ** 2 // between (x, y) and (X, Y)
- 1 // if A) the above result is not equal to 1
| v / p ? // or B) v is greater than or equal to p:
b = n < b ? b : n // end of path: update b to n if n >= b
: // else:
f(m, n + 1, X, Y, v) // do a recursive call
) // end of inner map()
) | b // end of outer map(); return b
How?
During the first iteration, $x$, $y$ and $p$ are all undefined and both tests (A and B in the comments) evaluate to NaN, which triggers the recursive call. Therefore, all cells are considered as a possible starting point of the path.
add a comment |
JavaScript (ES7), 106 103 102 98 bytes
f=(m,n=b=-1,x,y,p)=>m.map((r,Y)=>r.map((v,X)=>(x-X)**2+(y-Y)**2-1|v/p?b=n<b?b:n:f(m,n+1,X,Y,v)))|b
Try it online!
Commented
f = ( // f = recursive function taking:
m, // m = input matrix
n = b = -1, // n = length of the current path; b = best length so far
x, y, // x, y = coordinates of the previous cell
p // p = value of the previous cell
) => //
m.map((r, Y) => // for each row r at position Y in m:
r.map((v, X) => // for each value v at position X in r:
(x - X) ** 2 + // compute the squared Euclidean distance
(y - Y) ** 2 // between (x, y) and (X, Y)
- 1 // if A) the above result is not equal to 1
| v / p ? // or B) v is greater than or equal to p:
b = n < b ? b : n // end of path: update b to n if n >= b
: // else:
f(m, n + 1, X, Y, v) // do a recursive call
) // end of inner map()
) | b // end of outer map(); return b
How?
During the first iteration, $x$, $y$ and $p$ are all undefined and both tests (A and B in the comments) evaluate to NaN, which triggers the recursive call. Therefore, all cells are considered as a possible starting point of the path.
add a comment |
JavaScript (ES7), 106 103 102 98 bytes
f=(m,n=b=-1,x,y,p)=>m.map((r,Y)=>r.map((v,X)=>(x-X)**2+(y-Y)**2-1|v/p?b=n<b?b:n:f(m,n+1,X,Y,v)))|b
Try it online!
Commented
f = ( // f = recursive function taking:
m, // m = input matrix
n = b = -1, // n = length of the current path; b = best length so far
x, y, // x, y = coordinates of the previous cell
p // p = value of the previous cell
) => //
m.map((r, Y) => // for each row r at position Y in m:
r.map((v, X) => // for each value v at position X in r:
(x - X) ** 2 + // compute the squared Euclidean distance
(y - Y) ** 2 // between (x, y) and (X, Y)
- 1 // if A) the above result is not equal to 1
| v / p ? // or B) v is greater than or equal to p:
b = n < b ? b : n // end of path: update b to n if n >= b
: // else:
f(m, n + 1, X, Y, v) // do a recursive call
) // end of inner map()
) | b // end of outer map(); return b
How?
During the first iteration, $x$, $y$ and $p$ are all undefined and both tests (A and B in the comments) evaluate to NaN, which triggers the recursive call. Therefore, all cells are considered as a possible starting point of the path.
JavaScript (ES7), 106 103 102 98 bytes
f=(m,n=b=-1,x,y,p)=>m.map((r,Y)=>r.map((v,X)=>(x-X)**2+(y-Y)**2-1|v/p?b=n<b?b:n:f(m,n+1,X,Y,v)))|b
Try it online!
Commented
f = ( // f = recursive function taking:
m, // m = input matrix
n = b = -1, // n = length of the current path; b = best length so far
x, y, // x, y = coordinates of the previous cell
p // p = value of the previous cell
) => //
m.map((r, Y) => // for each row r at position Y in m:
r.map((v, X) => // for each value v at position X in r:
(x - X) ** 2 + // compute the squared Euclidean distance
(y - Y) ** 2 // between (x, y) and (X, Y)
- 1 // if A) the above result is not equal to 1
| v / p ? // or B) v is greater than or equal to p:
b = n < b ? b : n // end of path: update b to n if n >= b
: // else:
f(m, n + 1, X, Y, v) // do a recursive call
) // end of inner map()
) | b // end of outer map(); return b
How?
During the first iteration, $x$, $y$ and $p$ are all undefined and both tests (A and B in the comments) evaluate to NaN, which triggers the recursive call. Therefore, all cells are considered as a possible starting point of the path.
edited 3 hours ago
answered yesterday
Arnauld
72.6k689305
72.6k689305
add a comment |
add a comment |
Jelly, 23 21 20 bytes
-2 thanks to Erik the Outgolfer
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’
Try it online! (way too inefficient for the examples - path here is 95 94 93 83 77 40 10
so 6
is yielded)
How?
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’ - Link: list of lists of integers, M
ŒỤ - multi-dimensional indices sorted by values
ŒP - power-set
Ƈ - filter, keep those for which:
Ʋ - last four links as a monad:
Ɲ - for each pair of neighbours:
ạ - absolute difference
§ - sum each
Ị - insignificant?
Ạ - all?
œị - multi-dimensional index into:
⁸ - chain's left argument, M
Ƈ - filter, keep only those:
Ƒ - unaffected by?:
Q - de-duplicate
Ṫ - tail
L - length
’ - decrement
add a comment |
Jelly, 23 21 20 bytes
-2 thanks to Erik the Outgolfer
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’
Try it online! (way too inefficient for the examples - path here is 95 94 93 83 77 40 10
so 6
is yielded)
How?
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’ - Link: list of lists of integers, M
ŒỤ - multi-dimensional indices sorted by values
ŒP - power-set
Ƈ - filter, keep those for which:
Ʋ - last four links as a monad:
Ɲ - for each pair of neighbours:
ạ - absolute difference
§ - sum each
Ị - insignificant?
Ạ - all?
œị - multi-dimensional index into:
⁸ - chain's left argument, M
Ƈ - filter, keep only those:
Ƒ - unaffected by?:
Q - de-duplicate
Ṫ - tail
L - length
’ - decrement
add a comment |
Jelly, 23 21 20 bytes
-2 thanks to Erik the Outgolfer
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’
Try it online! (way too inefficient for the examples - path here is 95 94 93 83 77 40 10
so 6
is yielded)
How?
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’ - Link: list of lists of integers, M
ŒỤ - multi-dimensional indices sorted by values
ŒP - power-set
Ƈ - filter, keep those for which:
Ʋ - last four links as a monad:
Ɲ - for each pair of neighbours:
ạ - absolute difference
§ - sum each
Ị - insignificant?
Ạ - all?
œị - multi-dimensional index into:
⁸ - chain's left argument, M
Ƈ - filter, keep only those:
Ƒ - unaffected by?:
Q - de-duplicate
Ṫ - tail
L - length
’ - decrement
Jelly, 23 21 20 bytes
-2 thanks to Erik the Outgolfer
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’
Try it online! (way too inefficient for the examples - path here is 95 94 93 83 77 40 10
so 6
is yielded)
How?
ŒỤŒPạƝ§ỊẠƲƇœị⁸QƑƇṪL’ - Link: list of lists of integers, M
ŒỤ - multi-dimensional indices sorted by values
ŒP - power-set
Ƈ - filter, keep those for which:
Ʋ - last four links as a monad:
Ɲ - for each pair of neighbours:
ạ - absolute difference
§ - sum each
Ị - insignificant?
Ạ - all?
œị - multi-dimensional index into:
⁸ - chain's left argument, M
Ƈ - filter, keep only those:
Ƒ - unaffected by?:
Q - de-duplicate
Ṫ - tail
L - length
’ - decrement
edited yesterday
answered yesterday
Jonathan Allan
50.8k534165
50.8k534165
add a comment |
add a comment |
Python 2, 150 147 140 136 134 132 125 123 120 bytes
l=lambda g,i,j:max(0<g.get(t)<g[i,j]and-~l(g,*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
lambda g:max(l(g,*t)for t in g)
Try it online!
Takes input in the form of a dictionary (x, y): value
.
-7 bytes thanks to wizzwizz4, -2 bytes thanks to Jonathan Allen, -2 bytes thanks to BMO
Alternative, 123 121 bytes
l=lambda i,j:max(0<g.get(t)<g[i,j]and-~l(*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
g=input();print max(l(*t)for t in g)
Try it online!
Essentially the same solution, just with the final lambda replaced by an actual program. I personally like the first one better, but this one comes close in byte count by allowing g
to be used as a global variable.
add a comment |
Python 2, 150 147 140 136 134 132 125 123 120 bytes
l=lambda g,i,j:max(0<g.get(t)<g[i,j]and-~l(g,*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
lambda g:max(l(g,*t)for t in g)
Try it online!
Takes input in the form of a dictionary (x, y): value
.
-7 bytes thanks to wizzwizz4, -2 bytes thanks to Jonathan Allen, -2 bytes thanks to BMO
Alternative, 123 121 bytes
l=lambda i,j:max(0<g.get(t)<g[i,j]and-~l(*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
g=input();print max(l(*t)for t in g)
Try it online!
Essentially the same solution, just with the final lambda replaced by an actual program. I personally like the first one better, but this one comes close in byte count by allowing g
to be used as a global variable.
add a comment |
Python 2, 150 147 140 136 134 132 125 123 120 bytes
l=lambda g,i,j:max(0<g.get(t)<g[i,j]and-~l(g,*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
lambda g:max(l(g,*t)for t in g)
Try it online!
Takes input in the form of a dictionary (x, y): value
.
-7 bytes thanks to wizzwizz4, -2 bytes thanks to Jonathan Allen, -2 bytes thanks to BMO
Alternative, 123 121 bytes
l=lambda i,j:max(0<g.get(t)<g[i,j]and-~l(*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
g=input();print max(l(*t)for t in g)
Try it online!
Essentially the same solution, just with the final lambda replaced by an actual program. I personally like the first one better, but this one comes close in byte count by allowing g
to be used as a global variable.
Python 2, 150 147 140 136 134 132 125 123 120 bytes
l=lambda g,i,j:max(0<g.get(t)<g[i,j]and-~l(g,*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
lambda g:max(l(g,*t)for t in g)
Try it online!
Takes input in the form of a dictionary (x, y): value
.
-7 bytes thanks to wizzwizz4, -2 bytes thanks to Jonathan Allen, -2 bytes thanks to BMO
Alternative, 123 121 bytes
l=lambda i,j:max(0<g.get(t)<g[i,j]and-~l(*t)for d in(-1,1)for t in((i+d,j),(i,j+d)))
g=input();print max(l(*t)for t in g)
Try it online!
Essentially the same solution, just with the final lambda replaced by an actual program. I personally like the first one better, but this one comes close in byte count by allowing g
to be used as a global variable.
edited yesterday
answered yesterday
ArBo
514
514
add a comment |
add a comment |
Clean, 211 207 bytes
import StdEnv,Data.List
z=zipWith
$l=maximum[length k-1\p<-permutations[(v,[x,y])\y<-[0..]&u<-l,x<-[0..]&v<-u],(k,[m:n])<-map unzip(subsequences p)|and[all((>)2o sum o map abs)(z(z(-))n[m:n]):z(>)k(tl k)]]
Try it online!
A brute-force solution taking a list-of-lists-of-integers ([[Int]]
).
The TIO driver takes the same format as the examples through STDIN.
It's too slow to run any of the examples on TIO and probably locally too, but works in theory.
This one does the same thing faster, can do 3x3 or 2x4 on TIO and 4x4 and 3x5 locally.
Indented:
$ l
= maximum
[ length k-1
\p <- permutations
[ (v, [x, y])
\y <- [0..] & u <- l
, x <- [0..] & v <- u
]
, (k, [m: n]) <- map unzip
(subsequences p)
| and
[ all
((>) 2 o sum o map abs)
(zipWith (zipWith (-)) n [m:n])
:
zipWith (>) k (tl k)
]
]
add a comment |
Clean, 211 207 bytes
import StdEnv,Data.List
z=zipWith
$l=maximum[length k-1\p<-permutations[(v,[x,y])\y<-[0..]&u<-l,x<-[0..]&v<-u],(k,[m:n])<-map unzip(subsequences p)|and[all((>)2o sum o map abs)(z(z(-))n[m:n]):z(>)k(tl k)]]
Try it online!
A brute-force solution taking a list-of-lists-of-integers ([[Int]]
).
The TIO driver takes the same format as the examples through STDIN.
It's too slow to run any of the examples on TIO and probably locally too, but works in theory.
This one does the same thing faster, can do 3x3 or 2x4 on TIO and 4x4 and 3x5 locally.
Indented:
$ l
= maximum
[ length k-1
\p <- permutations
[ (v, [x, y])
\y <- [0..] & u <- l
, x <- [0..] & v <- u
]
, (k, [m: n]) <- map unzip
(subsequences p)
| and
[ all
((>) 2 o sum o map abs)
(zipWith (zipWith (-)) n [m:n])
:
zipWith (>) k (tl k)
]
]
add a comment |
Clean, 211 207 bytes
import StdEnv,Data.List
z=zipWith
$l=maximum[length k-1\p<-permutations[(v,[x,y])\y<-[0..]&u<-l,x<-[0..]&v<-u],(k,[m:n])<-map unzip(subsequences p)|and[all((>)2o sum o map abs)(z(z(-))n[m:n]):z(>)k(tl k)]]
Try it online!
A brute-force solution taking a list-of-lists-of-integers ([[Int]]
).
The TIO driver takes the same format as the examples through STDIN.
It's too slow to run any of the examples on TIO and probably locally too, but works in theory.
This one does the same thing faster, can do 3x3 or 2x4 on TIO and 4x4 and 3x5 locally.
Indented:
$ l
= maximum
[ length k-1
\p <- permutations
[ (v, [x, y])
\y <- [0..] & u <- l
, x <- [0..] & v <- u
]
, (k, [m: n]) <- map unzip
(subsequences p)
| and
[ all
((>) 2 o sum o map abs)
(zipWith (zipWith (-)) n [m:n])
:
zipWith (>) k (tl k)
]
]
Clean, 211 207 bytes
import StdEnv,Data.List
z=zipWith
$l=maximum[length k-1\p<-permutations[(v,[x,y])\y<-[0..]&u<-l,x<-[0..]&v<-u],(k,[m:n])<-map unzip(subsequences p)|and[all((>)2o sum o map abs)(z(z(-))n[m:n]):z(>)k(tl k)]]
Try it online!
A brute-force solution taking a list-of-lists-of-integers ([[Int]]
).
The TIO driver takes the same format as the examples through STDIN.
It's too slow to run any of the examples on TIO and probably locally too, but works in theory.
This one does the same thing faster, can do 3x3 or 2x4 on TIO and 4x4 and 3x5 locally.
Indented:
$ l
= maximum
[ length k-1
\p <- permutations
[ (v, [x, y])
\y <- [0..] & u <- l
, x <- [0..] & v <- u
]
, (k, [m: n]) <- map unzip
(subsequences p)
| and
[ all
((>) 2 o sum o map abs)
(zipWith (zipWith (-)) n [m:n])
:
zipWith (>) k (tl k)
]
]
edited yesterday
answered yesterday
Οurous
6,48211033
6,48211033
add a comment |
add a comment |
Python 3, 219 bytes
e,m=len,enumerate
b=lambda g,x,y:[b(g,i,j)for o in[-1,1]for i,j in[(x+o,y),(x,y+o)]if e(g)>i>=0<=j<e(g[x])and g[x][y]<g[i][j]]
l=lambda t:e(t)and 1+max(map(l,t))
d=lambda g:max(l(b(g,x,y))for x,r in m(g)for y,_ in m(r))
Try it online!
Grid is represented as list of lists:
[
[1, 2, 3, 2, 2],
[3, 4, 5, 5, 5],
[3, 4, 6, 7, 4],
[3, 3, 5, 6, 2],
[1, 1, 2, 3, 1],
]
Original ungolfed code:
def potential_neighbours(x, y):
return [(x-1, y), (x+1, y), (x, y-1), (x, y+1)]
def neighbours(grid, x, y):
result =
for i, j in potential_neighbours(x, y):
if 0 <= i < len(grid) and 0 <= j < len(grid[x]) and grid[x][y] < grid[i][j]:
result += [(i, j)]
return result
def build_tree(grid, x, y):
return [build_tree(grid, i, j) for i, j in neighbours(grid, x, y)]
def longest_path_in_tree(tree):
if len(tree) == 0:
return 0
return 1 + max(map(longest_path_in_tree, tree))
def longest_descent(grid):
trees = [build_tree(grid, x, y) for x, row in enumerate(grid) for y, _ in enumerate(row)]
return max(map(longest_path_in_tree, trees))
New contributor
add a comment |
Python 3, 219 bytes
e,m=len,enumerate
b=lambda g,x,y:[b(g,i,j)for o in[-1,1]for i,j in[(x+o,y),(x,y+o)]if e(g)>i>=0<=j<e(g[x])and g[x][y]<g[i][j]]
l=lambda t:e(t)and 1+max(map(l,t))
d=lambda g:max(l(b(g,x,y))for x,r in m(g)for y,_ in m(r))
Try it online!
Grid is represented as list of lists:
[
[1, 2, 3, 2, 2],
[3, 4, 5, 5, 5],
[3, 4, 6, 7, 4],
[3, 3, 5, 6, 2],
[1, 1, 2, 3, 1],
]
Original ungolfed code:
def potential_neighbours(x, y):
return [(x-1, y), (x+1, y), (x, y-1), (x, y+1)]
def neighbours(grid, x, y):
result =
for i, j in potential_neighbours(x, y):
if 0 <= i < len(grid) and 0 <= j < len(grid[x]) and grid[x][y] < grid[i][j]:
result += [(i, j)]
return result
def build_tree(grid, x, y):
return [build_tree(grid, i, j) for i, j in neighbours(grid, x, y)]
def longest_path_in_tree(tree):
if len(tree) == 0:
return 0
return 1 + max(map(longest_path_in_tree, tree))
def longest_descent(grid):
trees = [build_tree(grid, x, y) for x, row in enumerate(grid) for y, _ in enumerate(row)]
return max(map(longest_path_in_tree, trees))
New contributor
add a comment |
Python 3, 219 bytes
e,m=len,enumerate
b=lambda g,x,y:[b(g,i,j)for o in[-1,1]for i,j in[(x+o,y),(x,y+o)]if e(g)>i>=0<=j<e(g[x])and g[x][y]<g[i][j]]
l=lambda t:e(t)and 1+max(map(l,t))
d=lambda g:max(l(b(g,x,y))for x,r in m(g)for y,_ in m(r))
Try it online!
Grid is represented as list of lists:
[
[1, 2, 3, 2, 2],
[3, 4, 5, 5, 5],
[3, 4, 6, 7, 4],
[3, 3, 5, 6, 2],
[1, 1, 2, 3, 1],
]
Original ungolfed code:
def potential_neighbours(x, y):
return [(x-1, y), (x+1, y), (x, y-1), (x, y+1)]
def neighbours(grid, x, y):
result =
for i, j in potential_neighbours(x, y):
if 0 <= i < len(grid) and 0 <= j < len(grid[x]) and grid[x][y] < grid[i][j]:
result += [(i, j)]
return result
def build_tree(grid, x, y):
return [build_tree(grid, i, j) for i, j in neighbours(grid, x, y)]
def longest_path_in_tree(tree):
if len(tree) == 0:
return 0
return 1 + max(map(longest_path_in_tree, tree))
def longest_descent(grid):
trees = [build_tree(grid, x, y) for x, row in enumerate(grid) for y, _ in enumerate(row)]
return max(map(longest_path_in_tree, trees))
New contributor
Python 3, 219 bytes
e,m=len,enumerate
b=lambda g,x,y:[b(g,i,j)for o in[-1,1]for i,j in[(x+o,y),(x,y+o)]if e(g)>i>=0<=j<e(g[x])and g[x][y]<g[i][j]]
l=lambda t:e(t)and 1+max(map(l,t))
d=lambda g:max(l(b(g,x,y))for x,r in m(g)for y,_ in m(r))
Try it online!
Grid is represented as list of lists:
[
[1, 2, 3, 2, 2],
[3, 4, 5, 5, 5],
[3, 4, 6, 7, 4],
[3, 3, 5, 6, 2],
[1, 1, 2, 3, 1],
]
Original ungolfed code:
def potential_neighbours(x, y):
return [(x-1, y), (x+1, y), (x, y-1), (x, y+1)]
def neighbours(grid, x, y):
result =
for i, j in potential_neighbours(x, y):
if 0 <= i < len(grid) and 0 <= j < len(grid[x]) and grid[x][y] < grid[i][j]:
result += [(i, j)]
return result
def build_tree(grid, x, y):
return [build_tree(grid, i, j) for i, j in neighbours(grid, x, y)]
def longest_path_in_tree(tree):
if len(tree) == 0:
return 0
return 1 + max(map(longest_path_in_tree, tree))
def longest_descent(grid):
trees = [build_tree(grid, x, y) for x, row in enumerate(grid) for y, _ in enumerate(row)]
return max(map(longest_path_in_tree, trees))
New contributor
New contributor
answered yesterday
Nishioka
212
212
New contributor
New contributor
add a comment |
add a comment |
Haskell, 188 186 bytes
Needs $texttt{-XNoMonomorphismRestriction}$:
f m|c<-[0..length(m!!0)-1],r<-[0..length m-1]=h[g[(x,y)]|x<-r,y<-c,let g((x,y):p)=h[1+g(k:p)|i<-[-1,1],k@(u,v)<-[(x+i,y),(x,y+i)],u#r,v#c,m!!u!!v<m!!x!!y,not$k#p]]
(#)=elem
h=foldl max 0
Try it online!
Alternatively we could use notElem
over (not.).(#)
without the flag for $+4$ bytes:
Try it online!
Explanation & Ungolfed
Strategy: Recursively try all feasible paths, keeping track of visited entries and maximize their length.
Let's first define some helpers.. Since we need elem
and notElem
, let's use (#)
for elem
. Also, to maximize we'll need a total function (maximize
is not), returning $0$ when the list is empty:
safeMaximum = foldl max 0
Now we're ready to define our recursive function fun :: [[Integer]] -> Integer
:
fun xs
| c <- [0..length(m!!0)-1] -- all possible indices of xs' columns
, r <- [0..length m-1] -- all possible indices of xs' rows
= safeMaximum -- maximize ..
[ g [(x,y)] -- .. initially we haven't visited any others
| x <- c, y<-r -- .. all possible entries
-- For the purpose of golfing we define g in the list-comprehension, it takes all visited entries (p) where (x,y) is the most recent
, let g((x,y):p) = safeMaximum -- maximize ..
[ 1 + g(k:p) -- .. recurse, adding (x,y) to the visited nodes & increment (the next path will be 1 longer)
| i <- [-1,1] -- offsets [left/up,right/down]
, k@(u,v) <-[(x+i,y),(x,y+i)] -- next entry-candidate
, u#c, v#r -- make sure indices are in bound ..
, m!!u!!v < m!!x!!y -- .. , the the path is decreasing
, not$(u,v)#p -- .. and we haven't already visited that element
]
]
How does this take grids? List of lists?
– Beefster
6 hours ago
@Beefster: Yes, it says[[Integer]]
is list of lists. Though in the linked TIO you have a wrapperparse :: String -> [[Integer]]
, st. you can use strings split on spaces and new-lines.
– BMO
5 hours ago
add a comment |
Haskell, 188 186 bytes
Needs $texttt{-XNoMonomorphismRestriction}$:
f m|c<-[0..length(m!!0)-1],r<-[0..length m-1]=h[g[(x,y)]|x<-r,y<-c,let g((x,y):p)=h[1+g(k:p)|i<-[-1,1],k@(u,v)<-[(x+i,y),(x,y+i)],u#r,v#c,m!!u!!v<m!!x!!y,not$k#p]]
(#)=elem
h=foldl max 0
Try it online!
Alternatively we could use notElem
over (not.).(#)
without the flag for $+4$ bytes:
Try it online!
Explanation & Ungolfed
Strategy: Recursively try all feasible paths, keeping track of visited entries and maximize their length.
Let's first define some helpers.. Since we need elem
and notElem
, let's use (#)
for elem
. Also, to maximize we'll need a total function (maximize
is not), returning $0$ when the list is empty:
safeMaximum = foldl max 0
Now we're ready to define our recursive function fun :: [[Integer]] -> Integer
:
fun xs
| c <- [0..length(m!!0)-1] -- all possible indices of xs' columns
, r <- [0..length m-1] -- all possible indices of xs' rows
= safeMaximum -- maximize ..
[ g [(x,y)] -- .. initially we haven't visited any others
| x <- c, y<-r -- .. all possible entries
-- For the purpose of golfing we define g in the list-comprehension, it takes all visited entries (p) where (x,y) is the most recent
, let g((x,y):p) = safeMaximum -- maximize ..
[ 1 + g(k:p) -- .. recurse, adding (x,y) to the visited nodes & increment (the next path will be 1 longer)
| i <- [-1,1] -- offsets [left/up,right/down]
, k@(u,v) <-[(x+i,y),(x,y+i)] -- next entry-candidate
, u#c, v#r -- make sure indices are in bound ..
, m!!u!!v < m!!x!!y -- .. , the the path is decreasing
, not$(u,v)#p -- .. and we haven't already visited that element
]
]
How does this take grids? List of lists?
– Beefster
6 hours ago
@Beefster: Yes, it says[[Integer]]
is list of lists. Though in the linked TIO you have a wrapperparse :: String -> [[Integer]]
, st. you can use strings split on spaces and new-lines.
– BMO
5 hours ago
add a comment |
Haskell, 188 186 bytes
Needs $texttt{-XNoMonomorphismRestriction}$:
f m|c<-[0..length(m!!0)-1],r<-[0..length m-1]=h[g[(x,y)]|x<-r,y<-c,let g((x,y):p)=h[1+g(k:p)|i<-[-1,1],k@(u,v)<-[(x+i,y),(x,y+i)],u#r,v#c,m!!u!!v<m!!x!!y,not$k#p]]
(#)=elem
h=foldl max 0
Try it online!
Alternatively we could use notElem
over (not.).(#)
without the flag for $+4$ bytes:
Try it online!
Explanation & Ungolfed
Strategy: Recursively try all feasible paths, keeping track of visited entries and maximize their length.
Let's first define some helpers.. Since we need elem
and notElem
, let's use (#)
for elem
. Also, to maximize we'll need a total function (maximize
is not), returning $0$ when the list is empty:
safeMaximum = foldl max 0
Now we're ready to define our recursive function fun :: [[Integer]] -> Integer
:
fun xs
| c <- [0..length(m!!0)-1] -- all possible indices of xs' columns
, r <- [0..length m-1] -- all possible indices of xs' rows
= safeMaximum -- maximize ..
[ g [(x,y)] -- .. initially we haven't visited any others
| x <- c, y<-r -- .. all possible entries
-- For the purpose of golfing we define g in the list-comprehension, it takes all visited entries (p) where (x,y) is the most recent
, let g((x,y):p) = safeMaximum -- maximize ..
[ 1 + g(k:p) -- .. recurse, adding (x,y) to the visited nodes & increment (the next path will be 1 longer)
| i <- [-1,1] -- offsets [left/up,right/down]
, k@(u,v) <-[(x+i,y),(x,y+i)] -- next entry-candidate
, u#c, v#r -- make sure indices are in bound ..
, m!!u!!v < m!!x!!y -- .. , the the path is decreasing
, not$(u,v)#p -- .. and we haven't already visited that element
]
]
Haskell, 188 186 bytes
Needs $texttt{-XNoMonomorphismRestriction}$:
f m|c<-[0..length(m!!0)-1],r<-[0..length m-1]=h[g[(x,y)]|x<-r,y<-c,let g((x,y):p)=h[1+g(k:p)|i<-[-1,1],k@(u,v)<-[(x+i,y),(x,y+i)],u#r,v#c,m!!u!!v<m!!x!!y,not$k#p]]
(#)=elem
h=foldl max 0
Try it online!
Alternatively we could use notElem
over (not.).(#)
without the flag for $+4$ bytes:
Try it online!
Explanation & Ungolfed
Strategy: Recursively try all feasible paths, keeping track of visited entries and maximize their length.
Let's first define some helpers.. Since we need elem
and notElem
, let's use (#)
for elem
. Also, to maximize we'll need a total function (maximize
is not), returning $0$ when the list is empty:
safeMaximum = foldl max 0
Now we're ready to define our recursive function fun :: [[Integer]] -> Integer
:
fun xs
| c <- [0..length(m!!0)-1] -- all possible indices of xs' columns
, r <- [0..length m-1] -- all possible indices of xs' rows
= safeMaximum -- maximize ..
[ g [(x,y)] -- .. initially we haven't visited any others
| x <- c, y<-r -- .. all possible entries
-- For the purpose of golfing we define g in the list-comprehension, it takes all visited entries (p) where (x,y) is the most recent
, let g((x,y):p) = safeMaximum -- maximize ..
[ 1 + g(k:p) -- .. recurse, adding (x,y) to the visited nodes & increment (the next path will be 1 longer)
| i <- [-1,1] -- offsets [left/up,right/down]
, k@(u,v) <-[(x+i,y),(x,y+i)] -- next entry-candidate
, u#c, v#r -- make sure indices are in bound ..
, m!!u!!v < m!!x!!y -- .. , the the path is decreasing
, not$(u,v)#p -- .. and we haven't already visited that element
]
]
edited 5 hours ago
answered yesterday
BMO
11.5k22187
11.5k22187
How does this take grids? List of lists?
– Beefster
6 hours ago
@Beefster: Yes, it says[[Integer]]
is list of lists. Though in the linked TIO you have a wrapperparse :: String -> [[Integer]]
, st. you can use strings split on spaces and new-lines.
– BMO
5 hours ago
add a comment |
How does this take grids? List of lists?
– Beefster
6 hours ago
@Beefster: Yes, it says[[Integer]]
is list of lists. Though in the linked TIO you have a wrapperparse :: String -> [[Integer]]
, st. you can use strings split on spaces and new-lines.
– BMO
5 hours ago
How does this take grids? List of lists?
– Beefster
6 hours ago
How does this take grids? List of lists?
– Beefster
6 hours ago
@Beefster: Yes, it says
[[Integer]]
is list of lists. Though in the linked TIO you have a wrapper parse :: String -> [[Integer]]
, st. you can use strings split on spaces and new-lines.– BMO
5 hours ago
@Beefster: Yes, it says
[[Integer]]
is list of lists. Though in the linked TIO you have a wrapper parse :: String -> [[Integer]]
, st. you can use strings split on spaces and new-lines.– BMO
5 hours ago
add a comment |
Python 3, 263 227 bytes
def f(m):
p={(x,y):[c for i in[-1,1]for c in[(x,y+i),(x+i,y)]]for x,y in m};d={c:0 for c in p if not p[c]}
while len(p)-len(d):
for c in p:
for b in p[c]:
if b in d:d[c]=max(d[b]+1,d.get(c,0))
return max(d.values())
Try it online!
-2 bytes thanks to BMO
Takes grids in the format {(0, 0): 1, (1, 0): 2, ...}
. This format can be generated from the example format using the following utility function:
lambda s,e=enumerate:{(x,y):int(n)for y,l in e(s.split('n'))for x,n in e(l.split())}
add a comment |
Python 3, 263 227 bytes
def f(m):
p={(x,y):[c for i in[-1,1]for c in[(x,y+i),(x+i,y)]]for x,y in m};d={c:0 for c in p if not p[c]}
while len(p)-len(d):
for c in p:
for b in p[c]:
if b in d:d[c]=max(d[b]+1,d.get(c,0))
return max(d.values())
Try it online!
-2 bytes thanks to BMO
Takes grids in the format {(0, 0): 1, (1, 0): 2, ...}
. This format can be generated from the example format using the following utility function:
lambda s,e=enumerate:{(x,y):int(n)for y,l in e(s.split('n'))for x,n in e(l.split())}
add a comment |
Python 3, 263 227 bytes
def f(m):
p={(x,y):[c for i in[-1,1]for c in[(x,y+i),(x+i,y)]]for x,y in m};d={c:0 for c in p if not p[c]}
while len(p)-len(d):
for c in p:
for b in p[c]:
if b in d:d[c]=max(d[b]+1,d.get(c,0))
return max(d.values())
Try it online!
-2 bytes thanks to BMO
Takes grids in the format {(0, 0): 1, (1, 0): 2, ...}
. This format can be generated from the example format using the following utility function:
lambda s,e=enumerate:{(x,y):int(n)for y,l in e(s.split('n'))for x,n in e(l.split())}
Python 3, 263 227 bytes
def f(m):
p={(x,y):[c for i in[-1,1]for c in[(x,y+i),(x+i,y)]]for x,y in m};d={c:0 for c in p if not p[c]}
while len(p)-len(d):
for c in p:
for b in p[c]:
if b in d:d[c]=max(d[b]+1,d.get(c,0))
return max(d.values())
Try it online!
-2 bytes thanks to BMO
Takes grids in the format {(0, 0): 1, (1, 0): 2, ...}
. This format can be generated from the example format using the following utility function:
lambda s,e=enumerate:{(x,y):int(n)for y,l in e(s.split('n'))for x,n in e(l.split())}
edited yesterday
answered yesterday
wizzwizz4
1,2171035
1,2171035
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f178326%2flength-of-the-longest-descent%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
How should the last example be interpreted?
– Peter Taylor
yesterday
@PeterTaylor I'm not sure what you mean.
– Beefster
yesterday
I think the last example is just a matrix of multi digit numbers
– Embodiment of Ignorance
yesterday
@EmbodimentofIgnorance, ah, yes, I see. It would be a lot easier to spot the path with two-digit numbers rather than 4 to 6.
– Peter Taylor
yesterday
1
@Οurous: just rectangular. Not jagged.
– Beefster
yesterday