### Project Euler: Problems 18 and 67

__Problem 18__By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3

7 4

2 4 6

8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75

95 64

17 47 82

18 35 87 10

20 04 82 47 65

19 01 23 75 03 34

88 02 77 73 07 63 67

99 65 04 28 06 16 70 92

41 41 26 56 83 40 80 70 33

41 48 72 33 47 32 37 16 94 29

53 71 44 65 25 43 91 52 97 51 14

70 11 33 28 77 73 17 78 39 68 17 57

91 71 52 38 17 14 91 43 58 50 27 29 48

63 66 04 68 89 53 67 30 73 16 69 87 40 31

04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

__Solution__*By extension, this solves problem 67 as well.*

This uses a tail-recursive function requiring the pyramid to be flattened into an array and reversed. Its 'intelligence' comes from calculating sums from the bottom-up, rather than top-down, and by flattening the pyramid into a binary heap. Since the recursion only keeps the higher sum between two (2) node comparisons, it reduces the calculations for a 100 hundred row pyramid, from the potential 2^99 routes, to fairly small number, 9950.

//calculates running sums up to the top node, which it then returns

let rec StripArray row (remainingList:array

if row = 0 then

//remainingList

runningSums

else

//create next array up the pyramid

let newList = [|for counterRow = row to (Array.length remainingList)-1 do

yield remainingList.[counterRow]|]

//if running sum is empty, create the first array

if Array.sum runningSums = 0 then

let baseNums = [|for counterNew = 0 to row - 1 do

yield remainingList.[counterNew]|]

StripArray (row-1) newList baseNums

else

//compares sum of each node pair with its node value, and retains greater of the two

let newSums = [|for counterNew = 0 to row - 1 do

let first = remainingList.[counterNew] + runningSums.[counterNew]

let second = remainingList.[counterNew] + runningSums.[counterNew + 1]

if first >= second then

yield first

else

yield second|]

StripArray (row - 1) newList newSums

let rec ItemsPerRowsOfArray row maxRows (arr:array

if row > maxRows then

arr

else

let lastValue = arr.[((Array.length arr) - 1)]

let newValue = [|lastValue+row|]

let newArr = Array.append arr newValue

ItemsPerRowsOfArray (row+1) maxRows newArr

//for a [pyramid converted to a binary heap, the number of rows

//this will throw an error if fed an array that is not a pyramid/binary heap

let FindNumberOfRows num =

let rowArray = ItemsPerRowsOfArray 1 (num/2) [|0|]

let rowNum = rowArray |> Array.findIndex (fun x -> x = num)

rowNum

let FindMaxValuePath (flattenedArray:array

let rows = FindNumberOfRows (Array.length flattenedArray)

StripArray rows (Array.rev flattenedArray) [||]

//general proof to verify the code works, in this case a small pyramid

let generalProof = FindMaxValuePath [|1;2;3;4;5;6;7;8;9;10|]

//the array for problem #18

let flattenedArray =

[|75;95;64;17;47;82;18;35;87;10;20;04;82;47;65;19;01;23;75;03;34;88;02;77;73;07;63;67;99;65;04;28;06;16;70;92;41;41;26;56;83;40;80;70;33;41;48;72;33;47;32;37;16;94;29;53;71;44;65;25;43;91;52;97;51;14;70;11;33;28;77;73;17;78;39;68;17;57;91;71;52;38;17;14;91;43;58;50;27;29;48;63;66;04;68;89;53;67;30;73;16;69;87;40;31;04;62;98;27;23;09;70;98;73;93;38;53;60;04;23|]

let Prob18Answer = FindMaxValuePath flattenedArray

//I created the array for problem #67 manually, but it is an intellectually trivial exercise to load the file provided into an array

let prob67FlattenedArray = [|59;73;41;...removed values for display purposes...26;15;59;63;35|]

let Prob67Answer = FindMaxValuePath prob67FlattenedArray