Maximum triangle path sum
Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26. You can compute the total of the numbers you have seen in such walk, in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task: find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
D
<lang d>void main() {
import std.stdio, std.algorithm, std.range, std.file, std.conv;
"triangle.txt".File.byLine.map!split.map!(to!(int[])).array.retro
.reduce!((x, y) => zip(y, x, x.dropOne)
.map!(t => t[0] + t[1 .. $].max)
.array)[0]
.writeln;
}</lang>
- Output:
1320
Go
<lang go>package main
import (
"fmt" "strconv" "strings"
)
const t = ` 55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93`
func main() {
lines := strings.Split(t, "\n")
f := strings.Fields(lines[len(lines)-1])
d := make([]int, len(f))
var err error
for i, s := range f {
if d[i], err = strconv.Atoi(s); err != nil {
panic(err)
}
}
d1 := d[1:]
var l, r, u int
for row := len(lines) - 2; row >= 0; row-- {
l = d[0]
for i, s := range strings.Fields(lines[row]) {
if u, err = strconv.Atoi(s); err != nil {
panic(err)
}
if r = d1[i]; l > r {
d[i] = u + l
} else {
d[i] = u + r
}
l = r
}
}
fmt.Println(d[0])
}</lang>
- Output:
1320
Haskell
<lang haskell>parse = map (map read . words) . lines f x y z = x + max y z g xs ys = zipWith3 f xs ys $ tail ys solve = head . foldr1 g main = readFile "triangle.txt" >>= print . solve . parse</lang>
- Output:
1320
Perl 6
The Z+ and Zmax are examples of the zipwith metaoperator. We ought to be able to use [Z+]= as an assignment operator here, but rakudo has a bug. Note also we can use the Zmax metaoperator form because max is define as an infix in Perl 6. <lang perl6>my @rows = slurp("triangle.txt").lines.map: { [.words] }
while @rows > 1 {
my @last := @rows.pop; @rows[*-1] = @rows[*-1] Z+ (@last Zmax @last[1..*]);
}
say @rows;</lang>
- Output:
1320
Here's a more FPish version with the same output. We define our own operator and the use it in the reduction metaoperator form, [op], which turns any infix into a list operator. <lang perl6>sub infix:<op>(@a,@b) { (@a Zmax @a[1..*]) Z+ @b }
say [op] slurp("triangle.txt").lines.reverse.map: { [.words] }</lang> Instead of using reverse, one could also define the op as right-associative. <lang perl6>sub infix:<op>(@a,@b) is assoc('right') { @a Z+ (@b Zmax @b[1..*]) }
say [op] slurp("triangle.txt").lines.map: { [.words] }</lang>
Python
A simple mostly imperative solution: <lang python>def solve(tri):
while len(tri) > 1:
t0 = tri.pop()
t1 = tri.pop()
tri.append([max(t0[i], t0[i+1]) + t for i,t in enumerate(t1)])
return tri[0][0]
def main():
data = """\
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93"""
print solve([map(int, row.split()) for row in data.splitlines()])
main()</lang>
- Output:
1320
A more functional version, similar to the Haskell entry (same output): <lang python>from itertools import imap
f = lambda x, y, z: x + max(y, z) g = lambda xs, ys: list(imap(f, ys, xs, xs[1:])) data = [map(int, row.split()) for row in open("triangle.txt")][::-1] print reduce(g, data)[0]</lang>
Racket
<lang racket>#lang racket (require math/number-theory)
(define (trinv n) ; OEIS A002024
(exact-floor (/ (+ 1 (sqrt (* 1 (* 8 n)))) 2)))
(define (triangle-neighbour-bl n)
(define row (trinv n)) (+ n (- (triangle-number row) (triangle-number (- row 1)))))
(define (maximum-triangle-path-sum T)
(define n-rows (trinv (vector-length T)))
(define memo# (make-hash))
(define (inner i)
(hash-ref!
memo# i
(λ ()
(+ (vector-ref T (sub1 i)) ; index is 1-based (so vector-refs need -1'ing)
(cond [(= (trinv i) n-rows) 0]
[else
(define bl (triangle-neighbour-bl i))
(max (inner bl) (inner (add1 bl)))])))))
(inner 1))
(module+ main
(maximum-triangle-path-sum
#(55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93)))
(module+ test
(require rackunit) (check-equal? (for/list ((n (in-range 1 (add1 10)))) (trinv n)) '(1 2 2 3 3 3 4 4 4 4)) ; 1 ; 2 3 ; 4 5 6 ; 7 8 9 10 (check-eq? (triangle-neighbour-bl 1) 2) (check-eq? (triangle-neighbour-bl 3) 5) (check-eq? (triangle-neighbour-bl 5) 8) (define test-triangle #(55 94 48 95 30 96 77 71 26 67)) (check-equal? (maximum-triangle-path-sum test-triangle) 321) )</lang>
- Output:
1320
REXX
The method used is very efficient and performs very well for triangles that have thousands of rows (lines).
For an expanded discussion of the method's efficiency, see the discussion page.
<lang rexx>/*REXX program finds the max sum of a "column" of numbers in a triangle.*/
@. =
@.1 = 55
@.2 = 94 48
@.3 = 95 30 96
@.4 = 77 71 26 67
@.5 = 97 13 76 38 45
@.6 = 07 36 79 16 37 68
@.7 = 48 07 09 18 70 26 06
@.8 = 18 72 79 46 59 79 29 90
@.9 = 20 76 87 11 32 07 07 49 18
@.10 = 27 83 58 35 71 11 25 57 29 85
@.11 = 14 64 36 96 27 11 58 56 92 18 55
@.12 = 02 90 03 60 48 49 41 46 33 36 47 23
@.13 = 92 50 48 02 36 59 42 79 72 20 82 77 42
@.14 = 56 78 38 80 39 75 02 71 66 66 01 03 55 72
@.15 = 44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
@.16 = 85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
@.17 = 06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
@.18 = 27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
do r=1 while @.r\== /*build a version of the triangle*/
do k=1 for words(@.r) /*build a row, number by number. */
#.r.k=word(@.r,k) /*assign a number to an array num*/
end /*k*/
end /*r*/
rows=r-1 /*compute the number of rows. */
do r=rows by -1 to 2; p=r-1 /*traipse through triangle rows. */
do k=1 for p; kn=k+1 /*re-calculate the previous row. */
#.p.k=max(#.r.k, #.r.kn) + #.p.k /*replace previous #. */
end /*k*/
end /*r*/
say 'maximum path sum:' #.1.1 /*display the top (row 1) number.*/
/*stick a fork in it, we're done.*/</lang>
output using the data within the REXX program:
maximum path sum: 1320