Maximum triangle path sum
You are encouraged to solve this task according to the task description, using any language you may know.
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.
11l
<lang 11l>F solve(&tri)
L tri.len > 1
V t0 = tri.pop()
V t1 = tri.pop()
tri.append(enumerate(t1).map((i, t) -> max(@t0[i], @t0[i + 1]) + t))
R tri[0][0]
V 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(&data.split("\n").map(row -> row.split(‘ ’, group_delimiters' 1B).map(Int))))</lang>
- Output:
1320
Action!
<lang Action!>INT FUNC Max(INT a,b)
IF a>b THEN RETURN (a) FI
RETURN (b)
PROC Main()
DEFINE ROWCOUNT="18"
INT i,row,len,a,b
INT ARRAY rows(ROWCOUNT)
INT ARRAY 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]
row=0 len=1 FOR i=0 TO ROWCOUNT-1 DO rows(i)=row row==+len len==+1 OD
row=ROWCOUNT-2
WHILE row>=0
DO
len=row+1
FOR i=0 TO len-1
DO
a=data(rows(row+1)+i)
b=data(rows(row+1)+i+1)
data(rows(row)+i)==+Max(a,b)
OD
row==-1
OD
PrintI(data(0))
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
1320
Ada
<lang ada>with Ada.Text_Io; use Ada.Text_Io;
procedure Max_Sum is
Triangle : array (Positive range <>) of integer :=
(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);
Last : Integer := Triangle'Length; Tn : Integer := 1;
begin
while (Tn * (Tn + 1) / 2) < Last loop
Tn := Tn + 1;
end loop;
for N in reverse 2 .. Tn loop
for I in 2 .. N loop
Triangle (Last - N) := Triangle (Last - N) + Integer'Max(Triangle (Last - 1), Triangle (Last)); Last := Last - 1;
end loop;
Last := Last - 1;
end loop;
Put_Line(Integer'Image(Triangle(1)));
end Max_Sum;</lang>
- Output:
1320
ALGOL 68
Basically the same algorithm as Ada and C++ but using a triangular matrix. <lang algol68># create a triangular array of the required values #
[ 1]INT row 1 := ( 55 ); [ 2]INT row 2 := ( 94, 48 ); [ 3]INT row 3 := ( 95, 30, 96 ); [ 4]INT row 4 := ( 77, 71, 26, 67 ); [ 5]INT row 5 := ( 97, 13, 76, 38, 45 ); [ 6]INT row 6 := ( 07, 36, 79, 16, 37, 68 ); [ 7]INT row 7 := ( 48, 07, 09, 18, 70, 26, 06 ); [ 8]INT row 8 := ( 18, 72, 79, 46, 59, 79, 29, 90 ); [ 9]INT row 9 := ( 20, 76, 87, 11, 32, 07, 07, 49, 18 ); [10]INT row 10 := ( 27, 83, 58, 35, 71, 11, 25, 57, 29, 85 ); [11]INT row 11 := ( 14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55 ); [12]INT row 12 := ( 02, 90, 03, 60, 48, 49, 41, 46, 33, 36, 47, 23 ); [13]INT row 13 := ( 92, 50, 48, 02, 36, 59, 42, 79, 72, 20, 82, 77, 42 ); [14]INT row 14 := ( 56, 78, 38, 80, 39, 75, 02, 71, 66, 66, 01, 03, 55, 72 ); [15]INT row 15 := ( 44, 25, 67, 84, 71, 67, 11, 61, 40, 57, 58, 89, 40, 56, 36 ); [16]INT row 16 := ( 85, 32, 25, 85, 57, 48, 84, 35, 47, 62, 17, 01, 01, 99, 89, 52 ); [17]INT row 17 := ( 06, 71, 28, 75, 94, 48, 37, 10, 23, 51, 06, 48, 53, 18, 74, 98, 15 ); [18]INT row 18 := ( 27, 02, 92, 23, 08, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93 );
[18]REF[]INT triangle := ( row 1, row 2, row 3, row 4, row 5, row 6
, row 7, row 8, row 9, row 10, row 11, row 12
, row 13, row 14, row 15, row 16, row 17, row 18
);
PROC max = ( INT a, INT b )INT: IF a > b THEN a ELSE b FI;
- working backwards, we replace the elements of each row with the sum of that #
- element and the maximum of the two elements below it. #
- That destroys the triangle but leaves element [1][1] equal to the required #
- maximum #
FOR row FROM UPB triangle - 1 BY -1 TO 1
DO
FOR element FROM 1 TO UPB triangle[row]
DO
# the elements "under" triangle[row][element] are #
# triangle[row+1][element] and triangle[row+1][element+1] #
triangle[row][element]
+:= max( triangle[row+1][element], triangle[row+1][element+1] )
OD
OD;
print( ( triangle[1][1], newline ) ) </lang>
- Output:
+1320
AppleScript
<lang AppleScript>---------------- MAXIMUM TRIANGLE PATH SUM ---------------
-- Working from the bottom of the triangle upwards, -- summing each number with the larger of the two below -- until the maximum emerges at the top.
-- maxPathSum :: Int -> Int on maxPathSum(xss)
-- With the last row as the initial accumulator,
-- folding from the penultimate line,
-- towards the top of the triangle:
-- sumWithRowBelow :: [Int] -> [Int] -> [Int]
script sumWithRowBelow
on |λ|(row, accum)
-- plusGreaterOfTwoBelow :: Int -> Int -> Int -> Int
script plusGreaterOfTwoBelow
on |λ|(x, intLeft, intRight)
x + max(intLeft, intRight)
end |λ|
end script
-- The accumulator, zipped with the tail of the
-- accumulator, yields pairs of adjacent sums so far.
zipWith3(plusGreaterOfTwoBelow, row, accum, tail(accum))
end |λ|
end script
-- A list of lists folded down to a list of just one remaining integer.
-- Head returns that integer from the list.
head(foldr1(sumWithRowBelow, xss))
end maxPathSum
TEST -------------------------
on run
maxPathSum({¬
{55}, ¬
{94, 48}, ¬
{95, 30, 96}, ¬
{77, 71, 26, 67}, ¬
{97, 13, 76, 38, 45}, ¬
{7, 36, 79, 16, 37, 68}, ¬
{48, 7, 9, 18, 70, 26, 6}, ¬
{18, 72, 79, 46, 59, 79, 29, 90}, ¬
{20, 76, 87, 11, 32, 7, 7, 49, 18}, ¬
{27, 83, 58, 35, 71, 11, 25, 57, 29, 85}, ¬
{14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55}, ¬
{2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23}, ¬
{92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42}, ¬
{56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52}, ¬
{6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15}, ¬
{27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93} ¬
})
--> 1320
end run
GENERIC FUNCTIONS -------------------
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- foldr1 :: (a -> a -> a) -> [a] -> a
on foldr1(f, xs)
if length of xs > 1 then
tell mReturn(f)
set v to item -1 of xs
set lng to length of xs
repeat with i from lng - 1 to 1 by -1
set v to |λ|(item i of xs, v, i, xs)
end repeat
return v
end tell
else
xs
end if
end foldr1
-- head :: [a] -> a
on head(xs)
if length of xs > 0 then
item 1 of xs
else
missing value
end if
end head
-- max :: Ord a => a -> a -> a
on max(x, y)
if x > y then
x
else
y
end if
end max
-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then
y
else
x
end if
end min
-- minimum :: [a] -> a
on minimum(xs)
script min
on |λ|(a, x)
if x < a or a is missing value then
x
else
a
end if
end |λ|
end script
foldl(min, missing value, xs)
end minimum
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- tail :: [a] -> [a]
on tail(xs)
if length of xs > 1 then
items 2 thru -1 of xs
else
{}
end if
end tail
-- zipWith3 :: (a -> b -> c -> d) -> [a] -> [b] -> [c] -> [d]
on zipWith3(f, xs, ys, zs)
set lng to minimum({length of xs, length of ys, length of zs})
set lst to {}
tell mReturn(f)
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, item i of ys, item i of zs)
end repeat
return lst
end tell
end zipWith3</lang>
- Output:
1320
Astro
<lang python>fun maxpathsum(t): #: Array{Array{I}}
let a = val t
for i in a.length-1..-1..1, c in linearindices a[r]:
a[r, c] += max(a[r+1, c], a[r=1, c+1])
return a[1, 1]
let test = [
[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 maxpathsum test </lang>
AutoHotkey
<lang AutoHotkey></lang> Examples:<lang AutoHotkey>data :=[ (join ltrim
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 )]
i := data.MaxIndex() row := Ceil((Sqrt(8*i+1) - 1) / 2) path:=[]
loop % row { path[i] := data[i] i-- }
while i { row := Ceil((Sqrt(8*i+1) - 1) / 2) path[i] := data[i] "+" (data[i+row] > data[i+row+1] ? path[i+row] : path[i+row+1]) data[i] += data[i+row] > data[i+row+1] ? data[i+row] : data[i+row+1] i -- }
MsgBox % data[1] "`n" path[1]</lang>
Outputs:
1320 55+94+95+77+97+7+48+72+76+83+64+90+48+80+84+85+94+71
AWK
<lang AWK>
- syntax: GAWK -f MAXIMUM_TRIANGLE_PATH_SUM.AWK filename(s)
{ printf("%s\n",$0)
cols[FNR] = NF
for (i=1; i<=NF; i++) {
arr[FNR][i] = $i
}
} ENDFILE {
for (row=FNR-1; row>0; row--) {
for (col=1; col<=cols[row]; col++) {
arr[row][col] += max(arr[row+1][col],arr[row+1][col+1])
}
}
printf("%d using %s\n\n",arr[1][1],FILENAME)
delete arr
delete cols
} END {
exit(0)
} function max(x,y) { return((x > y) ? x : y) } </lang>
- Output:
55 94 48 95 30 96 77 71 26 67 321 using MAXIMUM_TRIANGLE_PATH_SUM_4.TXT 55 94 48 95 30 96 77 71 26 67 97 13 76 38 45 7 36 79 16 37 68 48 7 9 18 70 26 6 18 72 79 46 59 79 29 90 20 76 87 11 32 7 7 49 18 27 83 58 35 71 11 25 57 29 85 14 64 36 96 27 11 58 56 92 18 55 2 90 3 60 48 49 41 46 33 36 47 23 92 50 48 2 36 59 42 79 72 20 82 77 42 56 78 38 80 39 75 2 71 66 66 1 3 55 72 44 25 67 84 71 67 11 61 40 57 58 89 40 56 36 85 32 25 85 57 48 84 35 47 62 17 1 1 99 89 52 6 71 28 75 94 48 37 10 23 51 6 48 53 18 74 98 15 27 2 92 23 8 71 76 84 15 52 92 63 81 10 44 10 69 93 1320 using MAXIMUM_TRIANGLE_PATH_SUM_18.TXT
Bracmat
<lang bracmat>( "
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93 "
: ?triangle
& ( max
= a b . !arg:(?a.?b)&(!a:>!b|!b) )
& 0:?accumulator & whl
' ( @(!triangle:?row (\n|\r) ?triangle)
& :?newaccumulator
& 0:?first
& whl
' ( @(!row:? #%?n (" " ?row|:?row))
& !accumulator:#%?second ?accumulator
& !newaccumulator max$(!first.!second)+!n:?newaccumulator
& !second:?first
)
& !newaccumulator 0:?accumulator
)
& ( -1:?Max
& !accumulator
: ? (%@:>!Max:?Max&~) ?
| out$!Max
)
)</lang>
- Output:
1320
C
<lang C>
- include <stdio.h>
- include <math.h>
- define max(x,y) ((x) > (y) ? (x) : (y))
int tri[] = {
55,
94, 48, 95, 30, 96, 77, 71, 26, 67, 97, 13, 76, 38, 45, 7, 36, 79, 16, 37, 68, 48, 7, 9, 18, 70, 26, 6, 18, 72, 79, 46, 59, 79, 29, 90, 20, 76, 87, 11, 32, 7, 7, 49, 18, 27, 83, 58, 35, 71, 11, 25, 57, 29, 85, 14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55, 2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23, 92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42, 56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52, 6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15, 27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93 };
int main(void) {
const int len = sizeof(tri) / sizeof(tri[0]); const int base = (sqrt(8*len + 1) - 1) / 2; int step = base - 1; int stepc = 0;
int i;
for (i = len - base - 1; i >= 0; --i) {
tri[i] += max(tri[i + step], tri[i + step + 1]);
if (++stepc == step) {
step--;
stepc = 0;
}
}
printf("%d\n", tri[0]);
return 0;
} </lang>
- Output:
1320
C#
<lang csharp> using System;
namespace RosetaCode { class MainClass { public static void Main (string[] args) { int[,] list = new int[18,19]; string input = @"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"; var charArray = input.Split ('\n');
for (int i=0; i < charArray.Length; i++) { var numArr = charArray[i].Trim().Split(' ');
for (int j = 0; j<numArr.Length; j++) { int number = Convert.ToInt32 (numArr[j]); list [i, j] = number; } }
for (int i = 16; i >= 0; i--) { for (int j = 0; j < 18; j++) { list[i,j] = Math.Max(list[i, j] + list[i+1, j], list[i,j] + list[i+1, j+1]); } } Console.WriteLine (string.Format("Maximum total: {0}", list [0, 0])); } } }
</lang>
- Output:
Maximum total: 1320
C++
<lang cpp> /* Algorithm complexity: n*log(n) */
- include <iostream>
int main( int argc, char* argv[] ) {
int triangle[] =
{
55, 94, 48, 95, 30, 96, 77, 71, 26, 67, 97, 13, 76, 38, 45, 7, 36, 79, 16, 37, 68, 48, 7, 9, 18, 70, 26, 6, 18, 72, 79, 46, 59, 79, 29, 90, 20, 76, 87, 11, 32, 7, 7, 49, 18, 27, 83, 58, 35, 71, 11, 25, 57, 29, 85, 14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55, 2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23, 92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42, 56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52, 6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15, 27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93
};
const int size = sizeof( triangle ) / sizeof( int ); const int tn = static_cast<int>(sqrt(2.0 * size)); assert(tn * (tn + 1) == 2 * size); // size should be a triangular number
// walk backward by rows, replacing each element with max attainable therefrom
for (int n = tn - 1; n > 0; --n) // n is size of row, note we do not process last row
for (int k = (n * (n-1)) / 2; k < (n * (n+1)) / 2; ++k) // from the start to the end of row
triangle[k] += std::max(triangle[k + n], triangle[k + n + 1]);
std::cout << "Maximum total: " << triangle[0] << "\n\n";
} </lang>
- Output:
Maximum total: 1320
Clojure
<lang clojure> (ns clojure.examples.rosetta (:gen-class)
(:require [clojure.string :as string]))
(def rosetta "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")
- The technique is described here in more detail http://mishadoff.com/blog/clojure-euler-problem-018/
- Most of the code converts the string data to a nested array of integers.
- The code to calculate the max sum is then only a single line
- First convert string data to nested list
- with each inner list containing one row of the triangle
- [[55] [94 48] [95 30 96] ... [...10 69 93]
(defn parse-int [s]
" Convert digits to a number (finds digits when could be surrounded by non-digits" (Integer. (re-find #"\d+" s)))
(defn data-int-array [s]
" Convert string to integer array" (map parse-int (string/split (string/trim s) #"\s+")))
(defn nested-triangle [s]
" Convert triangle to nested vector, with each inner vector containing one triangle row"
(loop [lst s n 1 newlist nil]
(if (empty? lst) (reverse newlist)
(recur (drop n lst) (inc n) (cons (take n lst) newlist)))))
- Create nested list
(def nested-list (nested-triangle (data-int-array rosetta)))
- Function to compute maximum path sum
(defn max-sum [s]
" Compute maximum path sum using a technique described here: http://mishadoff.com/blog/clojure-euler-problem-018/" (reduce (fn [a b] (map + b (map max a (rest a)))) (reverse s)))
- Print result
(println (max-sum nested-list))
</lang>
- Output:
1320
Common Lisp
<lang lisp>(defun find-max-path-sum (s)
(let ((triangle (loop for line = (read-line s NIL NIL)
while line
collect (with-input-from-string (str line)
(loop for n = (read str NIL NIL)
while n
collect n)))))
(flet ((get-max-of-pairs (xs)
(maplist (lambda (ys)
(and (cdr ys) (max (car ys) (cadr ys))))
xs)))
(car (reduce (lambda (xs ys)
(mapcar #'+ (get-max-of-pairs xs) ys))
(reverse triangle))))))
(defparameter *small-triangle*
" 55
94 48
95 30 96
77 71 26 67")
(format T "~a~%" (with-input-from-string (s *small-triangle*)
(find-max-path-sum s)))
(format T "~a~%" (with-open-file (f "triangle.txt")
(find-max-path-sum f)))</lang>
- Output:
321 1320
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
Elena
ELENA 5.0 : <lang elena>import system'routines; import extensions; import extensions'math;
string input = "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";
public program() {
var list := IntMatrix.allocate(18,19);
int i := 0;
int j := 0;
input.splitBy(forward newLine).forEach:(string line)
{
j := 0;
line.trim().splitBy(" ").forEach:(string num)
{
list[i][j] := num.toInt();
j += 1
};
i += 1
};
for(int i := 16, i >= 0, i-=1)
{
for(int j := 0, j < 18, j += 1)
{
list[i][j] := max(list[i][j] + list[i+1][j], list[i][j] + list[i+1][j+1])
}
};
console.printLine("Maximum total: ", list[0][0])
}</lang>
- Output:
Maximum total: 1320
Elixir
<lang elixir>defmodule Maximum do
def triangle_path(text) do
text
|> String.split("\n", trim: true)
|> Enum.map(fn line ->
line
|> String.split()
|> Enum.map(&String.to_integer(&1))
end)
|> Enum.reduce([], fn x,total ->
[0]++total++[0]
|> Enum.chunk_every( 2, 1)
|> Enum.map(&Enum.max(&1))
|> Enum.zip(x)
|> Enum.map(fn{a,b} -> a+b end)
end)
|> Enum.max()
end
end
text = """
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 """
IO.puts Maximum.triangle_path(text) </lang>
- Output:
1320
Erlang
Reads the data from the file "triangle.txt" <lang Erlang> -mode(compile). -import(lists, [foldl/3]).
main(_) ->
{ok, Tmat} = file:open("triangle.txt", [read, raw, {read_ahead, 16384}]),
Max = max_sum(Tmat, []),
io:format("The maximum total is ~b~n", [Max]).
max_sum(FD, Last) ->
case file:read_line(FD) of
eof -> foldl(fun erlang:max/2, 0, Last);
{ok, Line} ->
Current = [binary_to_integer(B) || B <- re:split(Line, "[ \n]"), byte_size(B) > 0],
max_sum(FD, fold_row(Last, Current))
end.
% The first argument has one more element than the second, so compute % the initial sum so that both lists have identical length for fold_rest(). fold_row([], L) -> L; fold_row([A|_] = Last, [B|Bs]) ->
[A+B | fold_rest(Last, Bs)].
% Both lists must have same length fold_rest([A], [B]) -> [A+B]; fold_rest([A1 | [A2|_] = As], [B|Bs]) -> [B + max(A1,A2) | fold_rest(As, Bs)]. </lang>
- Output:
The maximum total is 1320
ERRE
<lang ERRE> PROGRAM TRIANGLE_PATH
CONST ROW=18
DIM TRI[200]
! ! for rosettacode,org !
FUNCTION MAX(X,Y)
MAX=-X*(X>=Y)-Y*(X<Y)
END FUNCTION
BEGIN
DATA(55)
DATA(94,48)
DATA(95,30,96)
DATA(77,71,26,67)
DATA(97,13,76,38,45)
DATA(7,36,79,16,37,68)
DATA(48,7,9,18,70,26,6)
DATA(18,72,79,46,59,79,29,90)
DATA(20,76,87,11,32,7,7,49,18)
DATA(27,83,58,35,71,11,25,57,29,85)
DATA(14,64,36,96,27,11,58,56,92,18,55)
DATA(2,90,3,60,48,49,41,46,33,36,47,23)
DATA(92,50,48,2,36,59,42,79,72,20,82,77,42)
DATA(56,78,38,80,39,75,2,71,66,66,1,3,55,72)
DATA(44,25,67,84,71,67,11,61,40,57,58,89,40,56,36)
DATA(85,32,25,85,57,48,84,35,47,62,17,1,1,99,89,52)
DATA(6,71,28,75,94,48,37,10,23,51,6,48,53,18,74,98,15)
DATA(27,2,92,23,8,71,76,84,15,52,92,63,81,10,44,10,69,93)
PRINT(CHR$(12);) !CLS
LUNG=ROW*(ROW+1)/2
FOR I%=0 TO LUNG-1 DO
READ(TRI[I%])
END FOR
BSE=(SQR(8*LUNG+1)-1)/2
STP=BSE-1
STEPC=0
FOR I%=LUNG-BSE-1 TO 0 STEP -1 DO
TRI[I%]=TRI[I%]+MAX(TRI[I%+STP],TRI[I%+STP+1])
STEPC=STEPC+1
IF STEPC=STP THEN
STP=STP-1
STEPC=0
END IF
END FOR
PRINT(TRI[0])
END PROGRAM </lang>
Factor
<lang factor>USING: grouping.extras io.encodings.utf8 io.files kernel math.order math.parser math.vectors prettyprint sequences splitting ; IN: rosetta-code.maximum-triangle-path-sum
- parse-triangle ( path -- seq )
utf8 file-lines [ " " split harvest ] map [ [ string>number ] map ] map ;
- max-triangle-path-sum ( seq -- n )
<reversed> unclip-slice [ swap [ max ] 2clump-map v+ ] reduce first ;
"triangle.txt" parse-triangle max-triangle-path-sum .</lang>
- Output:
1320
Forth
<lang forth> \ Triangle representation; words created by this defining word return the address of element \ specified by its row number and position within that row, both indexed from 0.
- TRIANGLE ( "name" -- |DOES: row pos -- addr )
CREATE DOES> ROT DUP 1+ * 2/ CELLS + SWAP CELLS +
18 CONSTANT #ROWS \ total number of rows in triangle TRIANGLE triang
55 ,
94 , 48 ,
95 , 30 , 96 ,
77 , 71 , 26 , 67 ,
97 , 13 , 76 , 38 , 45 ,
7 , 36 , 79 , 16 , 37 , 68 ,
48 , 7 , 9 , 18 , 70 , 26 , 6 ,
18 , 72 , 79 , 46 , 59 , 79 , 29 , 90 ,
20 , 76 , 87 , 11 , 32 , 7 , 7 , 49 , 18 ,
27 , 83 , 58 , 35 , 71 , 11 , 25 , 57 , 29 , 85 ,
14 , 64 , 36 , 96 , 27 , 11 , 58 , 56 , 92 , 18 , 55 ,
2 , 90 , 3 , 60 , 48 , 49 , 41 , 46 , 33 , 36 , 47 , 23 ,
92 , 50 , 48 , 2 , 36 , 59 , 42 , 79 , 72 , 20 , 82 , 77 , 42 ,
56 , 78 , 38 , 80 , 39 , 75 , 2 , 71 , 66 , 66 , 1 , 3 , 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 , 1 , 1 , 99 , 89 , 52 ,
6 , 71 , 28 , 75 , 94 , 48 , 37 , 10 , 23 , 51 , 6 , 48 , 53 , 18 , 74 , 98 , 15 ,
27 , 2 , 92 , 23 , 8 , 71 , 76 , 84 , 15 , 52 , 92 , 63 , 81 , 10 , 44 , 10 , 69 , 93 ,
\ Starting from the row above the bottom row and ending on the top, for every item in row \ find the bigger number from the two neighbours underneath and add it to this item. At \ the end, the result will be returned from the top element of the triangle.
- MAX-SUM ( -- n )
0 #ROWS 2 - DO
I 1+ 0 DO
J 1+ I triang @ J 1+ I 1+ triang @
MAX J I triang +!
LOOP
-1 +LOOP
0 0 triang @
MAX-SUM .</lang>
- Output:
1320
Fortran
This being Fortran, why not a brute-force scan of all possible paths? This is eased by noting that from a given position, only two numbers are accessible, and always two numbers. Just like binary digits. So, for three levels, the choices would be 000, 001, 010, 011, 100, 101, 110, 111 or somesuch. Since however the pinnacle of the pyramid is always chosen, there is no choice there so the digits would be 100, 101, 110, 111.
A triangular array can be defined in some languages, and in some circumstances a square array is used with a lower triangle and upper triangle partition, but here, a simple linear array is in order, with some attention to subscript usage. The first layer has one number, the second has two, the third has three, ... easy enough. The more refined method that determines the maximum sum without ascertaining the path through working upwards from the base employs a FOR ALL statement in adding the maximum of the two possible descendants to each brick in the current layer, employing array BEST that starts off with all the values of the bottom layer. As each layer is one value shorter than the one below and the expression computes BEST(i) = ... + MAX(BEST(i),BEST(i + 1)) the special feature of the FORALL statement, that all rhs expressions are evaluated before any results are placed on the lhs is not needed if a DO-loop were to be used instead.
For input, free-format is convenient. Bad input still is a problem, and can lead to puzzles. If say when N values are to be read but an input line is short of numbers, then additional lines will be read and confusion is likely. So, read the file's record into a text variable and then extract the expected N values from that. Should a problem arise, then the troublesome record can be shown.
<lang Fortran>
MODULE PYRAMIDS !Produces a pyramid of numbers in 1-D array.
INTEGER MANY !The usual storage issues.
PARAMETER (MANY = 666) !This should suffice.
INTEGER BRICK(MANY),IN,LAYERS !Defines a pyramid.
CONTAINS
SUBROUTINE IMHOTEP(PLAN)!The architect.
Counting is from the apex down, the Erich von Daniken construction.
CHARACTER*(*) PLAN !The instruction file.
INTEGER I,IT !Steppers.
CHARACTER*666 ALINE !A scratchpad for input.
IN = 0 !No bricks.
LAYERS = 0 !In no courses.
WRITE (6,*) "Reading from ",PLAN !Here we go.
OPEN(10,FILE=PLAN,FORM="FORMATTED",ACTION="READ",ERR=6) !I hope.
GO TO 10 !Why can't OPEN be a function?@*&%#^%!
6 STOP "Can't grab the file!"
Chew into the plan.
10 READ (10,11,END = 20) ALINE !Get the whole line in one piece.
11 FORMAT (A) !As plain text.
IF (ALINE .EQ. "") GO TO 10 !Ignoring any blank lines.
IF (ALINE(1:1).EQ."%") GO TO 10 !A comment opportunity.
LAYERS = LAYERS + 1 !Righto, this should be the next layer.
IF (IN + LAYERS.GT.MANY) STOP "Too many bricks!" !Perhaps not.
READ (ALINE,*,END = 15,ERR = 15) BRICK(IN + 1:IN + LAYERS) !Free format.
IN = IN + LAYERS !Insufficient numbers will provoke trouble.
GO TO 10 !Extra numbers/stuff will be ignored.
Caught a crab? A bad number, or too few numbers on a line? No read-next-record antics, thanks.
15 WRITE (6,16) LAYERS,ALINE !Just complain.
16 FORMAT ("Bad layer ",I0,": ",A)
Completed the plan.
20 WRITE (6,21) IN,LAYERS !Announce some details.
21 FORMAT (I0," bricks in ",I0," layers.")
CLOSE(10) !Finished with input.
Cast forth the numbers in a nice pyramid.
30 IT = 0 !For traversing the pyramid.
DO I = 1,LAYERS !Each course has one more number than the one before.
WRITE (6,31) BRICK(IT + 1:IT + I) !Sweep along the layer.
31 FORMAT (<LAYERS*2 - 2*I>X,666I4) !Leading spaces may be zero in number.
IT = IT + I !Thus finger the last of a layer.
END DO !On to the start of the next layer.
END SUBROUTINE IMHOTEP !The pyramid's plan is ready.
SUBROUTINE TRAVERSE !Clamber around the pyramid. Thoroughly.
C The idea is that a pyramid of numbers is provided, and then, starting at the peak, c work down to the base summing the numbers at each step to find the maximum value path. c The constraint is that from a particular brick, only the two numbers below left and below right c may be reached in stepping to that lower layer. c Since that is a 0/1 choice, recorded in MOVE, a base-two scan searches the possibilities.
INTEGER MOVE(LAYERS) !Choices are made at the various positions.
INTEGER STEP(LAYERS),WALK(LAYERS) !Thus determining the path.
INTEGER I,L,IT !Steppers.
INTEGER PS,WS !Scores.
WRITE (6,1) LAYERS !Announce the intention.
1 FORMAT (//,"Find the highest score path across a pyramid of ",
1 I0," layers."/) !I'm not worrying over singular/plural.
MOVE = 0 !All 0/1 values to zero.
MOVE(1) = 1 !Except the first.
STEP(1) = 1 !Every path starts here, without option.
WS = -666 !The best score so far.
Commence a multi-level loop, using the values of MOVE as the digits, one digit per level.
10 IT = 1 !All paths start with the first step.
PS = BRICK(1) !The starting score,.
c write (6,8) "Move",MOVE,WS
DO L = 2,LAYERS !Deal with the subsequent layers.
IT = IT + L - 1 + MOVE(L) !Choose a brick.
STEP(L) = IT !Remember this step.
PS = PS + BRICK(IT) !Count its score.
c WRITE (6,6) L,IT,BRICK(IT),PS
6 FORMAT ("Layer ",I0,",Brick(",I0,")=",I0,",Sum=",I0)
END DO !Thus is the path determined.
IF (PS .GT. WS) THEN !An improvement?
IF (WS.GT.0) WRITE (6,7) WS,PS !Yes! Announce.
7 FORMAT ("Improved path score: ",I0," to ",I0)
WRITE (6,8) "Moves",MOVE !Show the choices at each layer..
WRITE (6,8) "Steps",STEP !That resulted in this path.
WRITE (6,8) "Score",BRICK(STEP) !Whose steps were scored thus.
8 FORMAT (A8,666I4) !This should suffice.
WS = PS !Record the new best value.
WALK = STEP !And the path thereby.
END IF !So much for an improvement.
DO L = LAYERS,1,-1 !Now add one to the number in MOVE.
IF (MOVE(L).EQ.0) THEN !By finding the lowest order zero.
MOVE(L) = 1 !Making it one,
MOVE(L + 1:LAYERS) = 0 !And setting still lower orders back to zero.
GO TO 10 !And if we did, there's more to do!
END IF !But if that bit wasn't zero,
END DO !Perhaps the next one up will be.
WRITE (6,*) WS," is the highest score." !So much for that.
END SUBROUTINE TRAVERSE !All paths considered...
SUBROUTINE REFINE !Ascertain the highest score without searching.
INTEGER BEST(LAYERS) !A scratchpad.
INTEGER I,L !Steppers.
L = LAYERS*(LAYERS - 1)/2 + 1 !Finger the first brick of the lowest layer.
BEST = BRICK(L:L + LAYERS - 1)!Syncopation. Copy the lowest layer.
DO L = LAYERS - 1,1,-1 !Work towards the peak.
FORALL (I = 1:L) BEST(I) = BRICK(L*(L - 1)/2 + I) !Add to each brick's value
1 + MAXVAL(BEST(I:I + 1)) !The better of its two possibles.
END DO !On to the next layer.
WRITE (6,*) BEST(1)," is the highest score. By some path."
END SUBROUTINE REFINE !Who knows how we get there.
END MODULE PYRAMIDS
PROGRAM TRICKLE
USE PYRAMIDS
c CALL IMHOTEP("Sakkara.txt")
CALL IMHOTEP("Cheops.txt")
CALL TRAVERSE !Do this the definite way.
CALL REFINE !Only the result by more cunning.
END
</lang>
Output:
Reading from Cheops.txt
171 bricks in 18 layers.
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
7 36 79 16 37 68
48 7 9 18 70 26 6
18 72 79 46 59 79 29 90
20 76 87 11 32 7 7 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
2 90 3 60 48 49 41 46 33 36 47 23
92 50 48 2 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 2 71 66 66 1 3 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 1 1 99 89 52
6 71 28 75 94 48 37 10 23 51 6 48 53 18 74 98 15
27 2 92 23 8 71 76 84 15 52 92 63 81 10 44 10 69 93
Find the highest score path across a pyramid of 18 layers.
Moves 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Steps 1 2 4 7 11 16 22 29 37 46 56 67 79 92 106 121 137 154
Score 55 94 95 77 97 7 48 18 20 27 14 2 92 56 44 85 6 27
Improved path score: 864 to 904
Moves 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
Steps 1 2 4 7 11 16 22 29 37 46 56 67 79 92 106 121 138 155
Score 55 94 95 77 97 7 48 18 20 27 14 2 92 56 44 85 71 2
Improved path score: 904 to 994
Moves 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
Steps 1 2 4 7 11 16 22 29 37 46 56 67 79 92 106 121 138 156
Score 55 94 95 77 97 7 48 18 20 27 14 2 92 56 44 85 71 92
Improved path score: 994 to 1041
Moves 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1
Steps 1 2 4 7 11 16 22 29 37 46 56 67 79 93 108 124 141 159
Score 55 94 95 77 97 7 48 18 20 27 14 2 92 78 67 85 94 71
Improved path score: 1041 to 1087
Moves 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 1
Steps 1 2 4 7 11 16 22 29 37 46 56 68 80 93 108 124 141 159
Score 55 94 95 77 97 7 48 18 20 27 14 90 50 78 67 85 94 71
Improved path score: 1087 to 1104
Moves 1 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 1
Steps 1 2 4 7 11 16 22 29 37 46 56 68 81 95 109 124 141 159
Score 55 94 95 77 97 7 48 18 20 27 14 90 48 80 84 85 94 71
Improved path score: 1104 to 1137
Moves 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 1
Steps 1 2 4 7 11 16 22 29 37 46 57 68 80 93 108 124 141 159
Score 55 94 95 77 97 7 48 18 20 27 64 90 50 78 67 85 94 71
Improved path score: 1137 to 1154
Moves 1 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 1 1
Steps 1 2 4 7 11 16 22 29 37 46 57 68 81 95 109 124 141 159
Score 55 94 95 77 97 7 48 18 20 27 64 90 48 80 84 85 94 71
Improved path score: 1154 to 1193
Moves 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 1
Steps 1 2 4 7 11 16 22 29 37 47 57 68 80 93 108 124 141 159
Score 55 94 95 77 97 7 48 18 20 83 64 90 50 78 67 85 94 71
Improved path score: 1193 to 1210
Moves 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 1
Steps 1 2 4 7 11 16 22 29 37 47 57 68 81 95 109 124 141 159
Score 55 94 95 77 97 7 48 18 20 83 64 90 48 80 84 85 94 71
Improved path score: 1210 to 1249
Moves 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 1 1
Steps 1 2 4 7 11 16 22 29 38 47 57 68 80 93 108 124 141 159
Score 55 94 95 77 97 7 48 18 76 83 64 90 50 78 67 85 94 71
Improved path score: 1249 to 1266
Moves 1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 1 1
Steps 1 2 4 7 11 16 22 29 38 47 57 68 81 95 109 124 141 159
Score 55 94 95 77 97 7 48 18 76 83 64 90 48 80 84 85 94 71
Improved path score: 1266 to 1303
Moves 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 1 1
Steps 1 2 4 7 11 16 22 30 38 47 57 68 80 93 108 124 141 159
Score 55 94 95 77 97 7 48 72 76 83 64 90 50 78 67 85 94 71
Improved path score: 1303 to 1320
Moves 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 1
Steps 1 2 4 7 11 16 22 30 38 47 57 68 81 95 109 124 141 159
Score 55 94 95 77 97 7 48 72 76 83 64 90 48 80 84 85 94 71
1320 is the highest score.
1320 is the highest score. By some path.
FreeBASIC
<lang FreeBASIC>' version 21-06-2015 ' compile with: fbc -s console
Data " 55" Data " 94 48" Data " 95 30 96" Data " 77 71 26 67" Data " 97 13 76 38 45" Data " 07 36 79 16 37 68" Data " 48 07 09 18 70 26 06" Data " 18 72 79 46 59 79 29 90" Data " 20 76 87 11 32 07 07 49 18" Data " 27 83 58 35 71 11 25 57 29 85" Data " 14 64 36 96 27 11 58 56 92 18 55" Data " 02 90 03 60 48 49 41 46 33 36 47 23" Data " 92 50 48 02 36 59 42 79 72 20 82 77 42" Data " 56 78 38 80 39 75 02 71 66 66 01 03 55 72" Data " 44 25 67 84 71 67 11 61 40 57 58 89 40 56 36" Data " 85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52" Data " 06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15" Data " 27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93" Data "END" ' no more data
' ------=< MAIN >=------
Dim As String ln Dim As Integer matrix(1 To 20, 1 To 20) Dim As Integer x = 1, y, s1, s2, size
Do
Read ln
ln = Trim(ln)
If ln = "END" Then Exit Do
For y = 1 To x
matrix(x, y) = Val(Left(ln, 2))
ln = Mid(ln, 4)
Next
x += 1
size += 1
Loop
For x = size - 1 To 1 Step - 1
For y = 1 To x
s1 = matrix(x + 1, y)
s2 = matrix(x + 1, y + 1)
If s1 > s2 Then
matrix(x, y) += s1
Else
matrix(x, y) += s2
End If
Next
Next
Print Print " maximum triangle path sum ="; matrix(1, 1)
' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
maximum triangle path sum = 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
Or, inlining the data for quick testing, and using an applicative expression:
<lang haskell>---------------- MAXIMUM TRIANGLE PATH SUM ---------------
maxPathSum :: Int -> Int maxPathSum =
head
. foldr1
((<*> tail) . zipWith3 (\x y z -> x + max y z))
TEST -------------------------
main :: IO () main =
print $
maxPathSum
[ [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]
]</lang>
- Output:
1320
J
<lang j>padTri=: 0 ". ];._2 NB. parse triangle and (implicitly) pad with zeros maxSum=: [: {. (+ (0 ,~ 2 >./\ ]))/ NB. find max triangle path sum</lang>
Example Usage <lang j> maxSum padTri freads 'triangle.txt' 1320</lang>
Explanation:
First, we pad all short rows with trailing zeros so that all rows are the same length. This eliminates some ambiguity and simplifies the expression of both the data and the code.
Second, starting with the last row, for each pair of numbers we find the largest of the two (resulting in a list slightly shorter than before, so of course we pad it with a trailing zero) and add that row to the previous row. After repeating this through all the rows, the first value of the resulting row is the maximum we were looking for.
Instead of padding, we could instead trim the other argument to match the current reduced row length.
<lang J>maxsum=: ((] + #@] {. [)2 >./\ ])/</lang>
However, this turns out to be a slightly slower approach, because we are doing a little more work for each row.
(Note that the cost of padding every row to the same width averages out to an average 2x cost in space and time. So what we are saying here is that the interpreter overhead for changing the size of the memory region used in each operation with each row winds up being more than a 2x cost. You can probably beat that using compiled code, but of course the cost of compiling the program will itself be more than 2x - so not worth paying in a one-off experiment. You wind up with similar issues in any system involving one-off tests.)
Java
<lang java>import java.nio.file.*; import static java.util.Arrays.stream;
public class MaxPathSum {
public static void main(String[] args) throws Exception {
int[][] data = Files.lines(Paths.get("triangle.txt"))
.map(s -> stream(s.trim().split("\\s+"))
.mapToInt(Integer::parseInt)
.toArray())
.toArray(int[][]::new);
for (int r = data.length - 1; r > 0; r--)
for (int c = 0; c < data[r].length - 1; c++)
data[r - 1][c] += Math.max(data[r][c], data[r][c + 1]);
System.out.println(data[0][0]); }
}</lang>
1320
JavaScript
ES5
Imperative
<lang javascript> var arr = [ [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] ];
while (arr.length !== 1) {
var len = arr.length; var row = []; var current = arr[len-2]; var currentLen = current.length - 1; var end = arr[len-1];
for ( var i = 0; i <= currentLen; i++ ) {
row.push(Math.max(current[i] + end[i] || 0, current[i] + end[i+1] || 0) )
}
arr.pop(); arr.pop();
arr.push(row);
}
console.log(arr); </lang>
- Output:
<lang javascript> [ [ 1320 ] ] </lang>
Functional
<lang JavaScript>(function () {
// Right fold using final element as initial accumulator
// (a -> a -> a) -> t a -> a
function foldr1(f, lst) {
return lst.length > 1 ? (
f(lst[0], foldr1(f, lst.slice(1)))
) : lst[0];
}
// function of arity 3 mapped over nth items of each of 3 lists
// (a -> b -> c -> d) -> [a] -> [b] -> [c] -> [d]
function zipWith3(f, xs, ys, zs) {
return zs.length ? [f(xs[0], ys[0], zs[0])].concat(
zipWith3(f, xs.slice(1), ys.slice(1), zs.slice(1))) : [];
}
// Evaluating from bottom up (right fold)
// and with recursion left to right (head and first item of tail at each stage)
return foldr1(
function (xs, ys) {
return zipWith3(
function (x, y, z) {
return x + (y < z ? z : y);
},
xs, ys, ys.slice(1) // item above, and larger of two 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]
]
)[0];
})();</lang>
- Output:
<lang JavaScript>1320</lang>
ES6
Imperative
<lang javascript>function maximumTrianglePathSum(triangle) {
function distilLastLine() {
let lastLine = triangle.pop(),
aboveLine = triangle.pop();
for (let i = 0; i < aboveLine.length; i++)
aboveLine[i] = Math.max(
aboveLine[i] + lastLine[i],
aboveLine[i] + lastLine[i + 1]
);
triangle.push(aboveLine);
}
do {
distilLastLine();
} while (triangle.length > 1);
return triangle[0][0];
}
// testing let theTriangle = [ [55], [94, 48], [95, 30, 96], [77, 71, 26, 67], [97, 13, 76, 38, 45], [ 7, 36, 79, 16, 37, 68], [48, 7, 9, 18, 70, 26, 6], [18, 72, 79, 46, 59, 79, 29, 90], [20, 76, 87, 11, 32, 7, 7, 49, 18], [27, 83, 58, 35, 71, 11, 25, 57, 29, 85], [14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55], [ 2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23], [92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42], [56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52], [ 6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15], [27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93] ];
console.log(maximumTrianglePathSum(theTriangle));</lang>
- Output:
<lang javascript>1320</lang>
Functional
<lang JavaScript>(() => {
"use strict";
// ------------------ MAX PATH SUM -------------------
// Working from the bottom of the triangle upwards, // summing each number with the larger of the two below // until the maximum emerges at the top.
// maxPathSum ::Int -> Int const maxPathSum = xss => // A list of lists folded down to a list of just one // remaining integer. foldr1( // The accumulator, zipped with the tail of the // accumulator, yields pairs of adjacent sums. (ys, xs) => zipWith3(
// Plus greater of two below
(a, b, c) => a + Math.max(b, c)
)(xs)(ys)(ys.slice(1))
)(xss)[0];
// ---------------- GENERIC FUNCTIONS ----------------
// foldr1 :: (a -> a -> a) -> [a] -> a
const foldr1 = f =>
xs => 0 < xs.length ? (
xs.slice(0, -1).reduceRight(
f, xs.slice(-1)[0]
)
) : [];
// zipWith3 :: (a -> b -> c -> d) ->
// [a] -> [b] -> [c] -> [d]
const zipWith3 = f =>
xs => ys => zs => Array.from({
length: Math.min(
...[xs, ys, zs].map(x => x.length)
)
}, (_, i) => f(xs[i], ys[i], zs[i]));
// ---------------------- TEST -----------------------
return maxPathSum([
[55],
[94, 48],
[95, 30, 96],
[77, 71, 26, 67],
[97, 13, 76, 38, 45],
[7, 36, 79, 16, 37, 68],
[48, 7, 9, 18, 70, 26, 6],
[18, 72, 79, 46, 59, 79, 29, 90],
[20, 76, 87, 11, 32, 7, 7, 49, 18],
[27, 83, 58, 35, 71, 11, 25, 57, 29, 85],
[14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55],
[2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23],
[92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42],
[56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52],
[6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15],
[27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93]
]);
})();</lang>
- Output:
1320
jq
The following implementation illustrates the use of an inner function as a helper function, which is used here mainly for clarity. The inner function in effect implements the inner loop; the outer loop is implemented using reduce.
The input array is identical to that in the Javascript section and is therefore omitted here.<lang jq># Usage: TRIANGLE | solve def solve:
# update(next) updates the input row of maxima:
def update(next):
. as $maxima
| [ range(0; next|length)
| next[.] + ([$maxima[.], $maxima[. + 1]] | max) ];
. as $in
| reduce range(length -2; -1; -1) as $i
($in[-1]; update( $in[$i] ) ) ;</lang>
Julia
<lang julia># dynamic solution function maxpathsum(t::Array{Array{I, 1}, 1}) where I
T = deepcopy(t)
for r in length(T)-1:-1:1
for c in linearindices(T[r])
T[r][c] += max(T[r+1][c], T[r+1][c+1])
end
end
return T[1][1]
end
test = [[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]]
@show maxpathsum(test)</lang>
- Output:
maxpathsum(test) = 1320
Kotlin
<lang scala>// version 1.1.2
val tri = intArrayOf(
55,
94, 48,
95, 30, 96,
77, 71, 26, 67,
97, 13, 76, 38, 45,
7, 36, 79, 16, 37, 68,
48, 7, 9, 18, 70, 26, 6,
18, 72, 79, 46, 59, 79, 29, 90,
20, 76, 87, 11, 32, 7, 7, 49, 18,
27, 83, 58, 35, 71, 11, 25, 57, 29, 85,
14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55,
2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23,
92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42,
56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52,
6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15,
27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93
)
fun main(args: Array<String>) {
val triangles = arrayOf(tri.sliceArray(0..9), tri)
for (triangle in triangles) {
val size = triangle.size
val base = ((Math.sqrt(8.0 * size + 1.0) - 1.0)/ 2.0).toInt()
var step = base - 1
var stepc = 0
for (i in (size - base - 1) downTo 0) {
triangle[i] += maxOf(triangle[i + step], triangle[i + step + 1])
if (++stepc == step) {
step--
stepc = 0
}
}
println("Maximum total = ${triangle[0]}")
}
}</lang>
- Output:
Maximum total = 321 Maximum total = 1320
Lua
While the solutions here are clever, I found most of them to be hard to follow. In fact, none of them are very good for showing how the algorithm works. So I wrote this Lua version for maximum readability.
<lang lua>local triangleSmall = {
{ 55 },
{ 94, 48 },
{ 95, 30, 96 },
{ 77, 71, 26, 67 },
}
local triangleLarge = {
{ 55 },
{ 94, 48 },
{ 95, 30, 96 },
{ 77, 71, 26, 67 },
{ 97, 13, 76, 38, 45 },
{ 7, 36, 79, 16, 37, 68 },
{ 48, 7, 9, 18, 70, 26, 6 },
{ 18, 72, 79, 46, 59, 79, 29, 90 },
{ 20, 76, 87, 11, 32, 7, 7, 49, 18 },
{ 27, 83, 58, 35, 71, 11, 25, 57, 29, 85 },
{ 14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55 },
{ 2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23 },
{ 92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42 },
{ 56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52 },
{ 6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15 },
{ 27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93 },
};
function solve(triangle)
-- Get total number of rows in triangle. local nRows = table.getn(triangle)
-- Start at 2nd-to-last row and work up to the top. for row = nRows-1, 1, -1 do
-- For each value in row, add the max of the 2 children beneath it.
for i = 1, row do
local child1 = triangle[row+1][i]
local child2 = triangle[row+1][i+1]
triangle[row][i] = triangle[row][i] + math.max(child1, child2)
end
end
-- The top of the triangle now holds the answer. return triangle[1][1];
end
print(solve(triangleSmall)) print(solve(triangleLarge)) </lang>
- Output:
321 1320
Mathematica/Wolfram Language
<lang Mathematica>nums={{55},{94,48},{95,30,96},{77,71,26,67},{97,13,76,38,45},{7,36,79,16,37,68},{48,7,9,18,70,26,6},{18,72,79,46,59,79,29,90},{20,76,87,11,32,7,7,49,18},{27,83,58,35,71,11,25,57,29,85},{14,64,36,96,27,11,58,56,92,18,55},{2,90,3,60,48,49,41,46,33,36,47,23},{92,50,48,2,36,59,42,79,72,20,82,77,42},{56,78,38,80,39,75,2,71,66,66,1,3,55,72},{44,25,67,84,71,67,11,61,40,57,58,89,40,56,36},{85,32,25,85,57,48,84,35,47,62,17,1,1,99,89,52},{6,71,28,75,94,48,37,10,23,51,6,48,53,18,74,98,15},{27,2,92,23,8,71,76,84,15,52,92,63,81,10,44,10,69,93}}; ClearAll[DoStep,MaximumTrianglePathSum] DoStep[lst1_List,lst2_List]:=lst2+Join[{First[lst1]},Max/@Partition[lst1,2,1],{Last[lst1]}] MaximumTrianglePathSum[triangle_List]:=Max[Fold[DoStep,First[triangle],Rest[triangle]]]</lang>
- Output:
MaximumTrianglePathSum[nums] 1320
Nim
<lang nim>import sequtils, strutils, sugar
proc solve(tri: seq[seq[int]]): int =
var tri = tri while tri.len > 1: let t0 = tri.pop for i, t in tri[tri.high]: tri[tri.high][i] = max(t0[i], t0[i+1]) + t tri[0][0]
const 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"""
echo solve data.splitLines.map((x: string) => x.strip.split.map parseInt) </lang>
- Output:
1320
PARI/GP
<lang parigp>V=[[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]]; forstep(i=#V,2,-1,V[i-1]+=vector(i-1,j,max(V[i][j],V[i][j+1]))); V[1][1]</lang>
- Output:
%1 = 1320
Pascal
testet with freepascal, should run under Turbo Pascal, therefore using static array and val, and Delphi too. <lang pascal>program TriSum; {'triangle.txt'
- one element per line
55 94 48 95 30 96 ...} const
cMaxTriHeight = 18; cMaxTriElemCnt = (cMaxTriHeight+1)*cMaxTriHeight DIV 2 +1;
type
tElem = longint; tbaseRow = array[0..cMaxTriHeight] of tElem; tmyTri = array[0..cMaxTriElemCnt] of tElem;
function ReadTri( fname:string;
out t:tmyTri):integer;
{read triangle values into t and returns height} var
f : text; s : string; i : integer; ValCode : word;
begin
i := 0; fillchar(t,Sizeof(t),#0);
Assign(f,fname);
{$I-}
reset(f);
IF ioResult <> 0 then
begin
writeln('IO-Error ',ioResult);
close(f);
ReadTri := i;
EXIT;
end;
{$I+}
while NOT(EOF(f)) AND (i<cMaxTriElemCnt) do
begin
readln(f,s);
val(s,t[i],ValCode);
inc(i);
IF ValCode <> 0 then
begin
writeln(ValCode,' conversion error at line ',i);
fillchar(t,Sizeof(t),#0);
i := 0;
BREAK;
end;
end;
close(f);
ReadTri := round(sqrt(2*(i-1)));
end;
function TriMaxSum(var t: tmyTri;hei:integer):integer; {sums up higher values bottom to top} var
i,r,h,tmpMax : integer; idxN : integer; sumrow : tbaseRow;
begin
h := hei;
idxN := (h*(h+1)) div 2 -1;
{copy base row}
move(t[idxN-h+1],sumrow[0],SizeOf(tElem)*h);
dec(h);
{ for r := 0 to h do write(sumrow[r]:4);writeln;}
idxN := idxN-h;
while idxN >0 do
begin
i := idxN-h;
r := 0;
while r < h do
begin
tmpMax:= sumrow[r];
IF tmpMax<sumrow[r+1] then
tmpMax:=sumrow[r+1];
sumrow[r]:= tmpMax+t[i];
inc(i);
inc(r);
end;
idxN := idxN-h;
dec(h);
{ for r := 0 to h do write(sumrow[r]:4);writeln;}
end; TriMaxSum := sumrow[0];
end;
var
h : integer; triangle : tmyTri;
Begin { writeln(TriMaxSum(triangle,ReadTri('triangle.txt',triangle))); -> 1320}
h := ReadTri('triangle.txt',triangle);
writeln('height sum');
while h > 0 do
begin
writeln(h:4,TriMaxSum(triangle,h):7);
dec(h);
end;
end.</lang>
- Output:
height sum 18 1320 17 1249 .... 4 321 3 244 2 149 1 55
Perl
<lang perl>use 5.10.0; use List::Util 'max';
my @sum; while (<>) { my @x = split; @sum = ($x[0] + $sum[0], map($x[$_] + max(@sum[$_-1, $_]), 1 .. @x-2), $x[-1] + $sum[-1]); }
say max(@sum);</lang>
- Output:
% perl maxpath.pl triangle.txt 1320
Phix
with javascript_semantics sequence tri = {{55}, {94, 48}, {95, 30, 96}, {77, 71, 26, 67}, {97, 13, 76, 38, 45}, { 7, 36, 79, 16, 37, 68}, {48, 7, 9, 18, 70, 26, 6}, {18, 72, 79, 46, 59, 79, 29, 90}, {20, 76, 87, 11, 32, 7, 7, 49, 18}, {27, 83, 58, 35, 71, 11, 25, 57, 29, 85}, {14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55}, { 2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23}, {92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42}, {56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52}, { 6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15}, {27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93}} -- update each row from last but one upwards, with the larger -- child, so the first step is to replace 6 with 6+27 or 6+2. for r=length(tri)-1 to 1 by -1 do for c=1 to length(tri[r]) do tri[r][c] += max(tri[r+1][c..c+1]) end for end for ?tri[1][1]
- Output:
1320
PicoLisp
<lang PicoLisp>(de maxpath (Lst)
(let (Lst (reverse Lst) R (car Lst))
(for I (cdr Lst)
(setq R
(mapcar
+
(maplist
'((L)
(and (cdr L) (max (car L) (cadr L))) )
R )
I ) ) )
(car R) ) )</lang>
PL/I
<lang pli>*process source xref attributes or(!);
triang: Proc Options(Main); Dcl nn(18,18) Bin Fixed(31); Dcl (rows,i,j) Bin Fixed(31); Dcl (p,k,kn) Bin Fixed(31); Call f_r(1 ,' 55 '); Call f_r(2 ,' 94 48 '); Call f_r(3 ,' 95 30 96 '); Call f_r(4 ,' 77 71 26 67 '); Call f_r(5 ,' 97 13 76 38 45 '); Call f_r(6 ,' 07 36 79 16 37 68 '); Call f_r(7 ,' 48 07 09 18 70 26 06 '); Call f_r(8 ,' 18 72 79 46 59 79 29 90 '); Call f_r(9 ,' 20 76 87 11 32 07 07 49 18 '); Call f_r(10,' 27 83 58 35 71 11 25 57 29 85 '); Call f_r(11,' 14 64 36 96 27 11 58 56 92 18 55 '); Call f_r(12,' 02 90 03 60 48 49 41 46 33 36 47 23 '); Call f_r(13,' 92 50 48 02 36 59 42 79 72 20 82 77 42 '); Call f_r(14,' 56 78 38 80 39 75 02 71 66 66 01 03 55 72 '); Call f_r(15,' 44 25 67 84 71 67 11 61 40 57 58 89 40 56 36 '); Call f_r(16,' 85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52 '); Call f_r(17,' 06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15 '); Call f_r(18,' 27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93'); rows=hbound(nn,1);
do r=rows by -1 to 2;
p=r-1; /*traipse through triangle rows. */
do k=1 to p;
kn=k+1; /*re-calculate the previous row. */
nn(p,k)=max(nn(r,k),nn(r,kn))+nn(p,k); /*replace previous nn */
end;
end;
Put Edit('maximum path sum:',nn(1,1))(Skip,a,f(5)); /*display result*/
f_r: Proc(r,vl);
/* fill row r with r values */
Dcl r Bin Fixed(31);
Dcl vl Char(*);
Dcl vla Char(100) Var;
vla=' '!!trim(vl);
get string(vla) Edit((nn(r,j) Do j=1 To r))(f(3));
End;
End;</lang>
- Output:
maximum path sum: 1320
Prolog
<lang Prolog>max_path(N, V) :- data(N, T), path(0, T, V).
path(_N, [], 0) . path(N, [H | T], V) :- nth0(N, H, V0), N1 is N+1, path(N, T, V1), path(N1, T, V2), V is V0 + max(V1, V2).
data(1, P) :- P = [ [55], [94, 48], [95, 30, 96], [77, 71, 26, 67]].
data(2, P) :-
P =
[ [55],
[94, 48],
[95, 30, 96],
[77, 71, 26, 67],
[97, 13, 76, 38, 45],
[7, 36, 79, 16, 37, 68],
[48, 7, 9, 18, 70, 26, 6],
[18, 72, 79, 46, 59, 79, 29, 90],
[20, 76, 87, 11, 32, 7, 7, 49, 18],
[27, 83, 58, 35, 71, 11, 25, 57, 29, 85],
[14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55],
[2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23],
[92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42],
[56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52],
[6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15],
[27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93]].
</lang>
- Output:
?- max_path(1, V). V = 321 . ?- max_path(2, V). V = 1320 .
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]
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()])</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>
And, updating a little for Python 3 (in which itertools no longer defines imap, and reduce now has to be imported from functools), while inlining the data for ease of testing:
<lang python>Maximum triangle path sum
from functools import (reduce)
- maxPathSum :: Int -> Int
def maxPathSum(rows):
The maximum total among all possible
paths from the top to the bottom row.
return reduce(
lambda xs, ys: [
a + max(b, c) for (a, b, c)
in zip(ys, xs, xs[1:])
],
reversed(rows[:-1]), rows[-1]
)[0]
- ------------------------- TEST -------------------------
print(
maxPathSum([
[55],
[94, 48],
[95, 30, 96],
[77, 71, 26, 67],
[97, 13, 76, 38, 45],
[7, 36, 79, 16, 37, 68],
[48, 7, 9, 18, 70, 26, 6],
[18, 72, 79, 46, 59, 79, 29, 90],
[20, 76, 87, 11, 32, 7, 7, 49, 18],
[27, 83, 58, 35, 71, 11, 25, 57, 29, 85],
[14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55],
[2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23],
[92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42],
[56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 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, 1, 1, 99, 89, 52],
[6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15],
[27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93]
])
)</lang>
- Output:
1320
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
Raku
(formerly Perl 6)
The Z+ and Zmax are examples of the zipwith metaoperator. Note also we can use the Zmax metaoperator form because max is define as an infix in Perl 6. <lang perl6>my $triangle = q| 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|;
my @rows = $triangle.lines.map: { [.words] }
while @rows > 1 {
my @last := @rows.pop; @rows[*-1] = (@rows[*-1][] Z+ (@last Zmax @last[1..*])).List;
} put @rows;
- Here's a more FPish version. We define our own operator and the use it in the reduction metaoperator form, [op], which turns any infix into a list operator.
sub infix:<op>(@a,@b) { (@a Zmax @a[1..*]) Z+ @b } put [op] $triangle.lines.reverse.map: { [.words] }
- Or, instead of using reverse, one could also define the op as right-associative.
sub infix:<rop>(@a,@b) is assoc('right') { @a Z+ (@b Zmax @b[1..*]) } put [rop] $triangle.lines.map: { [.words] }</lang>
- Output:
1320 1320 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 program method's efficiency, see the discussion page.
<lang rexx>/*REXX program finds the maximum sum of a path of numbers in a pyramid of numbers. */
@.=.; @.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
- .=0
do r=1 while @.r\==. /*build another version of the pyramid.*/
do k=1 for r; #.r.k=word(@.r, k) /*assign a number to an array number. */
end /*k*/
end /*r*/
do r=r-1 by -1 to 2; p=r-1 /*traipse through the pyramid rows. */
do k=1 for p; _=k+1 /*re─calculate the previous pyramid row*/
#.p.k=max(#.r.k, #.r._) + #.p.k /*replace the previous number. */
end /*k*/
end /*r*/
/*stick a fork in it, we're all done. */
say 'maximum path sum: ' #.1.1 /*show the top (row 1) pyramid number. */</lang> output using the data within the REXX program:
maximum path sum: 1320
Ring
<lang ring>
- Project : Maximum triangle path sum
load "stdlib.ring" ln = list(19) ln[1] = " 55" ln[2] = " 94 48" ln[3] = " 95 30 96" ln[4] = " 77 71 26 67" ln[5] = " 97 13 76 38 45" ln[6] = " 07 36 79 16 37 68" ln[7] = " 48 07 09 18 70 26 06" ln[8] = " 18 72 79 46 59 79 29 90" ln[9] = " 20 76 87 11 32 07 07 49 18" ln[10] = " 27 83 58 35 71 11 25 57 29 85" ln[11] = " 14 64 36 96 27 11 58 56 92 18 55" ln[12] = " 02 90 03 60 48 49 41 46 33 36 47 23" ln[13] = " 92 50 48 02 36 59 42 79 72 20 82 77 42" ln[14] = " 56 78 38 80 39 75 02 71 66 66 01 03 55 72" ln[15] = " 44 25 67 84 71 67 11 61 40 57 58 89 40 56 36" ln[16] = " 85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52" ln[17] = " 06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15" ln[18] = " 27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93" ln[19] = "end"
matrix = newlist(20,20) x = 1 size = 0
for n = 1 to len(ln) - 1
ln2 = ln[n]
ln2 = trim(ln2)
for y = 1 to x
matrix[x][y] = number(left(ln2,2))
if len(ln2) > 4
ln2 = substr(ln2,4,len(ln2)-4)
ok
next
x = x + 1
size = size + 1
next
for x = size - 1 to 1 step - 1
for y = 1 to x
s1 = matrix[x+1][y]
s2 = matrix[x+1][y+1]
if s1 > s2
matrix[x][y] = matrix[x][y] + s1
else
matrix[x][y] = matrix[x][y] + s2
ok
next
next
see "maximum triangle path sum = " + matrix[1][1] </lang> Output:
maximum triangle path sum = 1320
Ruby
<lang ruby>triangle = " 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"
ar = triangle.each_line.map{|line| line.split.map(&:to_i)} puts ar.inject([]){|res,x|
maxes = [0, *res, 0].each_cons(2).map(&:max)
x.zip(maxes).map{|a,b| a+b}
}.max
- => 1320</lang>
Rust
<lang rust>use std::cmp::max;
fn max_path(vector: &mut Vec<Vec<u32>>) -> u32 {
while vector.len() > 1 {
let last = vector.pop().unwrap();
let ante = vector.pop().unwrap();
let mut new: Vec<u32> = Vec::new();
for (i, value) in ante.iter().enumerate() {
new.push(max(last[i], last[i+1]) + value);
};
vector.push(new);
};
vector[0][0]
}
fn main() {
let mut 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";
let mut vector = data.split("\n").map(|x| x.split(" ").map(|s: &str| s.parse::<u32>().unwrap())
.collect::<Vec<u32>>()).collect::<Vec<Vec<u32>>>();
let max_value = max_path(&mut vector);
println!("{}", max_value);
//=> 7273
}</lang>
Scala
<lang Scala>object MaximumTrianglePathSum extends App {
// Solution:
def sum(triangle: Array[Array[Int]]) =
triangle.reduceRight((upper, lower) =>
upper zip (lower zip lower.tail)
map {case (above, (left, right)) => above + Math.max(left, right)}
).head
// Tests:
def triangle = """
55
94 48
95 30 96
77 71 26 67
"""
def parse(s: String) = s.trim.split("\\s+").map(_.toInt)
def parseLines(s: String) = s.trim.split("\n").map(parse)
def parseFile(f: String) = scala.io.Source.fromFile(f).getLines.map(parse).toArray
println(sum(parseLines(triangle)))
println(sum(parseFile("triangle.txt")))
}</lang>
- Output:
321 1320
Sidef
Iterative solution: <lang ruby>var sum = [0]
ARGF.each { |line|
var x = line.words.map{.to_n}
sum = [
x.first + sum.first,
1 ..^ x.end -> map{|i| x[i] + [sum[i-1, i]].max}...,
x.last + sum.last,
]
}
say sum.max</lang>
Recursive solution: <lang ruby>var triangle = ARGF.slurp.lines.map{.words.map{.to_n}} func max_value(i=0, j=0) is cached {
i == triangle.len && return 0 triangle[i][j] + [max_value(i+1, j), max_value(i+1, j+1)].max
} say max_value()</lang>
- Output:
% sidef maxpath.sf triangle.txt 1320
Stata
<lang stata>import delimited triangle.txt, delim(" ") clear mata a = st_data(.,.) n = rows(a) for (i=n-1; i>=1; i--) { for (j=1; j<=i; j++) { a[i,j] = a[i,j]+max((a[i+1,j],a[i+1,j+1])) } } a[1,1] end</lang>
Output
1320
Tcl
<lang tcl>package require Tcl 8.6
proc maxTrianglePathSum {definition} {
# Parse the definition, stripping whitespace and leading zeroes.
set lines [lmap line [split [string trim $definition] "\n"] {
lmap val $line {scan $val %d}
}]
# Paths are bit strings (0 = go left, 1 = go right).
# Enumerate the possible paths.
set numPaths [expr {2 ** [llength $lines]}]
for {set path 0; set max -inf} {$path < $numPaths} {incr path} {
# Work out how much the current path costs. set sum [set idx [set row 0]] for {set bit 1} {$row < [llength $lines]} {incr row} { incr sum [lindex $lines $row $idx] if {$path & $bit} {incr idx} set bit [expr {$bit << 1}] } # Remember the max so far. if {$sum > $max} {set max $sum}
} return $max
}
puts [maxTrianglePathSum {
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 }]
- Reading from a file is left as an exercise…</lang>
- Output:
1320
VBScript
<lang vb> 'Solution derived from http://stackoverflow.com/questions/8002252/euler-project-18-approach.
Set objfso = CreateObject("Scripting.FileSystemObject") Set objinfile = objfso.OpenTextFile(objfso.GetParentFolderName(WScript.ScriptFullName) &_ "\triangle.txt",1,False)
row = Split(objinfile.ReadAll,vbCrLf)
For i = UBound(row) To 0 Step -1 row(i) = Split(row(i)," ") If i < UBound(row) Then For j = 0 To UBound(row(i)) If (row(i)(j) + row(i+1)(j)) > (row(i)(j) + row(i+1)(j+1)) Then row(i)(j) = CInt(row(i)(j)) + CInt(row(i+1)(j)) Else row(i)(j) = CInt(row(i)(j)) + CInt(row(i+1)(j+1)) End If Next End If Next
WScript.Echo row(0)(0)
objinfile.Close Set objfso = Nothing </lang>
- Input:
Input file
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
- Output:
1320
Wren
<lang ecmascript>var lines = [
" 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"
] var f = lines[-1].split(" ") var d = f.map { |s| Num.fromString(s) }.toList for (row in lines.count-2..0) {
var l = d[0]
var i = 0
for (s in lines[row].trimStart().split(" ")) {
var u = Num.fromString(s)
var r = d[i+1]
d[i] = (l > r) ? u + l : u + r
l = r
i = i + 1
}
} System.print(d[0])</lang>
- Output:
1320
zkl
The two Python solutions: <lang zkl>tri:=File("triangle.txt").pump(List,fcn(s){ s.strip().split(" ").apply("toInt") }).copy(); while(tri.len()>1){
t0:=tri.pop(); t1:=tri.pop(); tri.append( [[(it); t1.enumerate();
'wrap([(i,t)]){ t + t0[i].max(t0[i+1]) }]]) } tri[0][0].println();</lang> <lang zkl>data:=File("triangle.txt").pump(List,fcn(s){ s.strip().split(" ").apply("toInt") }); fcn f(x,y,z){ x + y.max(z) } fcn g(xs,ys){ Utils.zipWith(f,ys,xs,xs[1,*]); } data.reverse().reduce(g)[0].println();</lang>
<lang zkl>lines:=File("triangle.txt").pump(List,fcn(s){ s.strip().split(" ").apply("toInt") }); d:=lines[-1].copy(); foreach row in ([lines.len()-2..0,-1]){
d1:=d[1,*];
l :=d[0];
foreach i,u in (lines[row].enumerate()){
d[i] = u + l.max(r:=d1[i]);
l = r;
}
} println(d[0]);</lang>
- Output:
1320 1320 1320
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