Maximum triangle path sum: Difference between revisions

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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.

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

Output:

Screenshot from Atari 8-bit computer

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.6.win32

Basically the same algorithm as Ada and C++ but using a triangular matrix.

# 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 ) )

Output:

      +1320

Astro

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

AutoHotkey

Examples:

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]

Outputs:

1320
55+94+95+77+97+7+48+72+76+83+64+90+48+80+84+85+94+71

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) }

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

BASIC

BASIC256

arraybase 1
dim ln(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"

dim matrix(20,20)
x = 1
tam = 0

for n = 1 to length(ln) - 1
	ln2 = trim(ln[n])
	for y = 1 to x
		matrix[x, y] = fromradix((left(ln2, 2)), 10)
		if length(ln2) > 4 then ln2 = mid(ln2, 4, length(ln2)-4)
	next y
	x += 1
	tam += 1
next n

for x = tam - 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] = matrix[x, y] + s1
		else
			matrix[x, y] = matrix[x, y] + s2
		end if
	next y
next x

print "  maximum triangle path sum = " + matrix[1, 1]

Output:

  maximum triangle path sum = 1320

Run BASIC

Works with: Just BASIC

Works with: Liberty BASIC

dim ln$(19)
ln$(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"

dim matrix(20,20)
x = 1
tam = 0

for n = 1 to 19 'ubound(ln$) - 1
    ln2$ = trim$(ln$(n))
    for y = 1 to x
        matrix(x, y) = val(left$(ln2$, 2))
        if len(ln2$) > 4 then ln2$ = mid$(ln2$, 4, len(ln2$)-4)
    next y
    x = x +1
    tam = tam +1
next n

for z = tam-1 to 1 step -1
    for y = 1 to z
        s1 = matrix(z+1, y)
        s2 = matrix(z+1, y+1)
        if s1 > s2 then
            matrix(z, y) = matrix(z, y) +s1
        else
            matrix(z, y) = matrix(z, y) +s2
        end if
    next y
next z

print "  maximum triangle path sum = "; matrix(1, 1)

Output:

  maximum triangle path sum = 1320

True BASIC

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

DIM matrix(1 TO 20, 1 TO 20)
LET x = 1
DO
   READ ln$
   LET ln$ = LTRIM$(RTRIM$(ln$))
   IF ln$ = "END" THEN EXIT DO
   FOR y = 1 TO x
       LET matrix(x, y) = VAL((ln$)[1:2])
       LET ln$ = (ln$)[4:maxnum]
   NEXT y
   LET x = x+1
   LET tam = tam+1
LOOP
FOR x = tam-1 TO 1 STEP -1
    FOR y = 1 TO x
        LET s1 = matrix(x+1, y)
        LET s2 = matrix(x+1, y+1)
        IF s1 > s2 THEN LET matrix(x, y) = matrix(x, y)+s1 ELSE LET matrix(x, y) = matrix(x, y)+s2
    NEXT y
NEXT x

PRINT "  maximum triangle path sum ="; matrix(1, 1)
END

Output:

  maximum triangle path sum = 1320

Yabasic

dim ln$(19)
ln$(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"

dim matrix(20,20)
x = 1
tam = 0

for n = 1 to arraysize(ln$(),1) - 1
	ln2$ = trim$(ln$(n))
	for y = 1 to x
		matrix(x, y) = val(left$(ln2$, 2))
		if len(ln2$) > 4  ln2$ = mid$(ln2$, 4, len(ln2$)-4)
	next y
	x = x + 1
	tam = tam + 1
next n

for x = tam - 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) = matrix(x, y) + s1
		else
			matrix(x, y) = matrix(x, y) + s2
		end if
	next y
next x

print "\n  maximum triangle path sum = ", matrix(1, 1)

Output:

  maximum triangle path sum = 1320

C#

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]));
		}
	}
}

Output:

Maximum total: 1320

C++

Translation of: Ada

/* 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";
}

Output:

Maximum total: 1320

Common 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)))

Output:

321
1320

Elena

Translation of: C#

ELENA 5.0 :

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])
}

Output:

Maximum total: 1320

Erlang

Reads the data from the file "triangle.txt"

-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)].

Output:

The maximum total is 1320

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

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.

      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

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

' 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

Output:

  maximum triangle path sum = 1320

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])
}

Output:

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

Output:

Or, inlining the data for quick testing, and using an applicative expression:

---------------- 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]
      ]

Output:

J

padTri=: 0 ". ];._2                  NB. parse triangle and (implicitly) pad with zeros
maxSum=: [: {. (+ (0 ,~ 2 >./\ ]))/  NB. find max triangle path sum

Example Usage

   maxSum padTri freads 'triangle.txt'
1320

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.

maxsum=: ((] + #@] {. [)2 >./\ ])/

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

Works with: Java version 8

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]);
    }
}

JavaScript

ES5

Imperative

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);

Output:

[ [ 1320 ] ]

Functional

Translation of: Haskell

(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];

})();

Output:

ES6

Imperative

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));

Output:

Functional

Translation of: Haskell

(() => {
    "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]
    ]);
})();

Output:

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.

# 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] ) ) ;

Julia

Works with: Julia version 0.6

# 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)

Output:

maxpathsum(test) = 1320

Kotlin

Translation of: C

// 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]}")
    }
}

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.

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))

Output:

321
1320

Mathematica/Wolfram Language

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]]]

Output:

MaximumTrianglePathSum[nums]
1320

Nim

Translation of: Python

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)

Output:

PARI/GP

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]

Output:

%1 = 1320

Pascal

testet with freepascal, should run under Turbo Pascal, therefore using static array and val, and Delphi too.

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.

Output:

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);

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:

Picat

Mode directed tabling

table (+,+,+,max)  
pp(Row,_Column,Tri,Sum),Row>Tri.length => Sum=0.
pp(Row,Column,Tri,Sum) ?=> 
  pp(Row+1,Column,Tri,Sum1),
  Sum = Sum1+Tri[Row,Column].
pp(Row,Column,Tri,Sum) => 
  pp(Row+1,Column+1,Tri,Sum1),
  Sum = Sum1+Tri[Row,Column].

Loop based approach

pp2(Row, Column, Sum, Tri, M) =>
  if Sum > M.get(max_val,0) then
     M.put(max_val,Sum)
  end,
  Row := Row + 1,
  if Row <= Tri.length then
    foreach(I in 0..1) 
       pp2(Row,Column+I, Sum+Tri[Row,Column+I], Tri, M)
     end
  end.

Recursion

Translation of: Prolog

max_path(N, V) :-
	data(N, T),
	path(1, T, V).
 
path(_N, [], 0) .
path(N, [H | T], V) :-
	nth(N, H, V0),
	N1 is N+1,
	path(N, T, V1),
	path(N1, T, V2),
	V = V0 +  max(V1, V2).
  
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]].

Test

import util.

go =>
  tri(Tri),

  println("Mode directed tabling:"),
  pp(1,1,Tri,Sum),
  writeln(max_val=Sum),
  nl,

  println("Loop based:"),
  M = new_map([max_val=0]),
  pp2(1,1, Tri[1,1], Tri, M),
  writeln(max_val=M.get(max_val)),
  nl,

  println("Adapted the Prolog solution:"),
  max_path(2, V2),
  println(max_val=V2),
  nl.

tri(Tri) =>
 Tri =
{
{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:

Mode directed tabling:
max_val = 1320

Loop based:
max_val = 1320

Adapted the Prolog solution:
max_val = 1320

PicoLisp

Translation of: Common Lisp

(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) ) )

PL/I

Translation of: REXX

*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;

Output:

maximum path sum: 1320

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]].

Output:

 ?- max_path(1, V).
V = 321 .

 ?- max_path(2, V).
V = 1320 .

Python

A simple mostly imperative solution:

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()])

Output:

A more functional version, similar to the Haskell entry (same output):

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]

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:

Translation of: Haskell

Translation of: JavaScript

Works with: 3 version 7

'''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]
    ])
)

Output:

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)
  )

Output:

Raku

(formerly Perl 6)

Works with: Rakudo version 2018.03

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.

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] }

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.

/*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. */

output using the data within the REXX program:

maximum path sum:  1320

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]

Output:

maximum triangle path sum = 1320

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

Rust

Works with: Rust version 1.3

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
}

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")))
}

Output:

321
1320

Sidef

Translation of: Perl

Iterative solution:

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

Recursive solution:

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()

Output:

% sidef maxpath.sf triangle.txt
1320

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

Output

Tcl

Works with: Tcl version 8.6

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…

Output:

VBScript

'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

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:

Wren

Translation of: Go

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])

Output:

zkl

Translation of: Python

The two Python solutions:

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();

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();

Translation of: Go

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]);

Output:

1320
1320
1320