# Pascal's triangle/Puzzle

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Pascal's triangle/Puzzle
You are encouraged to solve this task according to the task description, using any language you may know.

This puzzle involves a Pascals Triangle, also known as a Pyramid of Numbers.

```           [ 151]
[  ][  ]
[40][  ][  ]
[  ][  ][  ][  ]
[ X][11][ Y][ 4][ Z]
```

Each brick of the pyramid is the sum of the two bricks situated below it.
Of the three missing numbers at the base of the pyramid, the middle one is the sum of the other two (that is, Y = X + Z).

Write a program to find a solution to this puzzle.

## 11l

Translation of: D
```F e(&x, row, col) -> &
R x[row * (row + 1) I/ 2 + col]

F iterate(&v, &diff, do_print = 1B)
V tot = 0.0
L
e(&v, 0, 0) = 151
e(&v, 2, 0) = 40
e(&v, 4, 1) = 11
e(&v, 4, 3) = 4

L(i) 1..4
L(j) 0..i
e(&diff, i, j) = 0
I j < i
e(&diff, i, j) += e(&v, i - 1, j) - e(&v, i, j + 1) - e(&v, i, j)
I j != 0
e(&diff, i, j) += e(&v, i - 1, j - 1) - e(&v, i, j - 1) - e(&v, i, j)

L(i) 1..3
L(j) 0.<i
e(&diff, i, j) += e(&v, i + 1, j) + e(&v, i + 1, j + 1) - e(&v, i, j)

e(&diff, 4, 2) += e(&v, 4, 0) + e(&v, 4, 4) - e(&v, 4, 2)

L(i) 0 .< v.len
v[i] += diff[i] / 4

tot = sum(diff.map(a -> a * a))
I do_print
print(‘dev: ’tot)
I tot < 0.1
L.break

V v = [0.0] * 15
V diff = [0.0] * 15
iterate(&v, &diff)

V idx = 0
L(i) 5
L(j) 0..i
print(‘#4’.format(Int(0.5 + v[idx])), end' I j < i {‘ ’} E "\n")
idx++```
Output:
```dev: 73410
dev: 17968.6875
dev: 6388.46484375
dev: 2883.337402344
dev: 1446.593643188
...
dev: 0.136503592
dev: 0.125866452
dev: 0.116055273
dev: 0.107006115
dev: 0.098659977
151
81   70
40   41   29
16   24   17   12
5   11   13    4    8
```

The solution makes an upward run symbolically, though excluding Z. After that two blocks (1,1) and (3,1) being known yield a 2x2 linear system, from which X and Y are determined. Finally each block is revisited and printed.

```with Ada.Text_IO; use Ada.Text_IO;

procedure Pyramid_of_Numbers is

B_X, B_Y, B_Z : Integer := 0; -- Unknown variables

type Block_Value is record
Known   : Integer := 0;
X, Y, Z : Integer := 0;
end record;
X : constant Block_Value := (0, 1, 0, 0);
Y : constant Block_Value := (0, 0, 1, 0);
Z : constant Block_Value := (0, 0, 0, 1);
procedure Add (L : in out Block_Value; R : Block_Value) is
begin -- Symbolically adds one block to another
L.Known := L.Known + R.Known;
L.X := L.X + R.X - R.Z; -- Z is excluded as n(Y - X - Z) = 0
L.Y := L.Y + R.Y + R.Z;
procedure Add (L : in out Block_Value; R : Integer) is
begin -- Symbolically adds a value to the block
L.Known := L.Known + R;

function Image (N : Block_Value) return String is
begin -- The block value, when X,Y,Z are known
return Integer'Image (N.Known + N.X * B_X + N.Y * B_Y + N.Z * B_Z);
end Image;

procedure Solve_2x2 (A11, A12, B1, A21, A22, B2 : Integer) is
begin -- Don't care about things, supposing an integer solution exists
if A22 = 0 then
B_X := B2 / A21;
B_Y := (B1 - A11*B_X) / A12;
else
B_X := (B1*A22 - B2*A12) / (A11*A22 - A21*A12);
B_Y := (B1 - A11*B_X) / A12;
end if;
B_Z := B_Y - B_X;
end Solve_2x2;

B : array (1..5, 1..5) of Block_Value; -- The lower triangle contains blocks

begin
-- The bottom blocks

-- Upward run
for Row in reverse 1..4 loop
for Column in 1..Row loop
Add (B (Row, Column), B (Row + 1, Column));
Add (B (Row, Column), B (Row + 1, Column + 1));
end loop;
end loop;

-- Now have known blocks 40=(3,1), 151=(1,1) and Y=X+Z to determine X,Y,Z
Solve_2x2
(  B(1,1).X, B(1,1).Y, 151 - B(1,1).Known,
B(3,1).X, B(3,1).Y,  40 - B(3,1).Known
);

-- Print the results
for Row in 1..5 loop
New_Line;
for Column in 1..Row loop
Put (Image (B(Row,Column)));
end loop;
end loop;
end Pyramid_of_Numbers;
```
Output:
```
151
81 70
40 41 29
16 24 17 12
5 11 13 4 8
```

## ALGOL 68

Works with: ALGOL 68 version Standard - lu decomp and lu solve are from the ALGOL 68G/gsl library
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
```MODE
FIELD = REAL,
VEC = [0]REAL,
MAT = [0,0]REAL;
MODE BRICK = UNION(INT, CHAR);

FLEX[][]BRICK puzzle = (
( 151),
( " ", " "),
(  40, " ", " "),
( " ", " ", " ", " "),
( "x",  11, "y",  4, "z")
);

PROC mat col = (INT row, col)INT: row*(row-1)OVER 2 + col;
INT col x = mat col(5,1),
col y = mat col(5,3),
col z = mat col(5,5);

OP INIT = (REF VEC vec)VOID: FOR elem FROM LWB vec TO UPB vec DO vec[elem]:=0 OD;
OP INIT = (REF MAT mat)VOID: FOR row FROM LWB mat TO UPB mat DO INIT mat[row,] OD;

OP / = (MAT a, MAT b)MAT:( # matrix division #
[LWB b:UPB b]INT p ;
INT sign;
[,]FIELD lu = lu decomp(b, p, sign);
[LWB a:UPB a, 1 LWB a:2 UPB a]FIELD out;
FOR col FROM 2 LWB a TO 2 UPB a DO out[,col] := lu solve(b, lu, p, a[,col]) OD;
out
);

OP / = (VEC a, MAT b)VEC: ( # vector division #
[LWB a:UPB a,1]FIELD transpose a;
transpose a[,1]:=a;
(transpose a/b)[,LWB a]
);

INT upb mat = mat col(UPB puzzle, UPB puzzle);
[upb mat, upb mat] REAL mat; INIT mat;
[upb mat] REAL vec; INIT vec;

INT mat row := LWB mat;
INT known row := UPB mat - UPB puzzle + 1;

# build the simultaneous equation to solve #
FOR row FROM LWB puzzle TO UPB puzzle DO
FOR col FROM LWB puzzle[row] TO UPB puzzle[row] DO
IF row < UPB puzzle THEN
mat[mat row, mat col(row, col)] := 1;
mat[mat row, mat col(row+1, col)] := -1;
mat[mat row, mat col(row+1, col+1)] := -1;
mat row +:= 1
FI;
CASE puzzle[row][col] IN
(INT value):(
mat[known row, mat col(row, col)] := 1;
vec[known row] := value;
known row +:= 1
),
(CHAR variable):SKIP
ESAC
OD
OD;

# finally add x - y + z = 0 #
mat[known row, col x] := 1;
mat[known row, col y] := -1;
mat[known row, col z] := 1;

FORMAT real repr = \$g(-5,2)\$;

CO # print details of the simultaneous equation being solved #
FORMAT
vec repr = \$"("n(2 UPB mat-1)(f(real repr)", ")f(real repr)")"\$,
mat repr = \$"("n(1 UPB mat-1)(f(vec repr)", "lx)f(vec repr)")"\$;

printf((\$"Vec: "l\$,vec repr, vec, \$l\$));
printf((\$"Mat: "l\$,mat repr, mat, \$l\$));
END CO

# finally actually solve the equation #
VEC solution vec = vec/mat;

# and wrap up by printing the solution #
FLEX[UPB puzzle]FLEX[0]REAL solution;
FOR row FROM LWB puzzle TO UPB puzzle DO
solution[row] := LOC[row]REAL;
FOR col FROM LWB puzzle[row] TO UPB puzzle[row] DO
solution[row][col] := solution vec[mat col(row, col)]
OD;
printf((\$n(UPB puzzle-row)(4x)\$, \$x"("f(real repr)")"\$, solution[row], \$l\$))
OD;

FOR var FROM 1 BY 2 TO 5 DO
printf((\$5x\$,\$g\$,puzzle[UPB puzzle][var],"=", real repr, solution[UPB puzzle][var]))
OD```
Output:
```                 (151.0)
(81.00) (70.00)
(40.00) (41.00) (29.00)
(16.00) (24.00) (17.00) (12.00)
( 5.00) (11.00) (13.00) ( 4.00) ( 8.00)
x= 5.00     y=13.00     z= 8.00
```

## AutoHotkey

The main part is this:

```N1 := 11, N2 := 4, N3 := 40, N4 := 151
Z := (2*N4 - 7*N3 - 8*N2 + 6*N1) / 7
X := (N3 - 2*N1 - Z) / 2
MsgBox,, Pascal's Triangle, %X%`n%Z%
```

Message box shows:

```5.000000
8.000000```

The fun part is to create a GUI for entering different values for N1, N2, N3 and N4.

The GUI shows all values in the solved state.

```;---------------------------------------------------------------------------
; Pascal's triangle.ahk
; by wolf_II
;---------------------------------------------------------------------------
; http://rosettacode.org/wiki/Pascal's_triangle/Puzzle
;---------------------------------------------------------------------------

;---------------------------------------------------------------------------
AutoExecute: ; auto-execute section of the script
;---------------------------------------------------------------------------
#SingleInstance, Force          ; only one instance allowed
#NoEnv                          ; don't check empty variables
;-----------------------------------------------------------------------
AppName := "Pascal's triangle"
N1 := 11, N2 := 4, N3 := 40, N4 := 151

; monitor MouseMove events
OnMessage(0x0200, "WM_MOUSEMOVE")

; GUI
Gosub, GuiCreate
Gui, Show,, %AppName%

Return

;---------------------------------------------------------------------------
GuiCreate: ; create the GUI
;---------------------------------------------------------------------------
Gui, -MinimizeBox
Gui, Margin, 8, 8

; 15 edit controls
Loop, 5
Loop, % Row := A_Index {
xx := 208 + (A_Index - 5) * 50 - (Row - 5) * 25
yy := 8 + (Row - 1) * 22
vv := Row "_" A_Index
}
GuiControl, -WantReturn, Edit11
GuiControl, -WantReturn, Edit15

; buttons (2 hidden)
Gui, Add, Button, x8 w78, &Restart
Gui, Add, Button, x+8 wp, &Solve
Gui, Add, Button, x+8 wp, &Check
Gui, Add, Button, x8 wp, Cle&ar
Gui, Add, Button, xp wp Hidden, &Cancel
Gui, Add, Button, x+8 wp, &New
Gui, Add, Button, xp wp Hidden, &Apply
Gui, Add, Button, x+8 wp, E&xit

; status bar

; blue font
Gui, Font, bold cBlue
GuiControl, Font, Edit11
GuiControl, Font, Edit15
; falling through

;---------------------------------------------------------------------------
ButtonRestart: ; restart retaining the blue clues
;---------------------------------------------------------------------------
Controls(True) ; enable controls
Loop, 15
If A_Index Not In 1,4,11,12,14,15
GuiControl,, Edit%A_Index% ; clear
GuiControl,, Edit1, %N4%
GuiControl,, Edit4, %N3%
GuiControl,, Edit12, %N1%
GuiControl,, Edit14, %N2%
GuiControl,, Edit11, %X%
GuiControl,, Edit15, %Z%
GreenFont:
Gui, Font, bold cGreen
GuiControl, Font, Edit1
GuiControl, Font, Edit4
GuiControl, Font, Edit12
GuiControl, Font, Edit14

Return

;---------------------------------------------------------------------------
ButtonSolve: ; calculate solution
;---------------------------------------------------------------------------
; N1 := 11    N2 := 4    N3 := 40    N4 := 151
;-----------------------------------------------------------------------
; Y = X + Z
; 40  = (11+X) + (11+Y)
; A   = (11+Y) + (Y+4)
; B   =  (4+Y) + (4+Z)
; 151 = (40+A) + (A+B)
;-----------------------------------------------------------------------
Gosub, GreenFont
GuiControl,, Edit15, % Z := Round( (2*N4 - 7*N3 - 8*N2 + 6*N1) / 7 )
GuiControl,, Edit11, % X := Round( (N3 - 2*N1 - Z) / 2 )
; falling through

;---------------------------------------------------------------------------
ButtonCheck: ; check the [entry|solution] for errors
;---------------------------------------------------------------------------
Controls(False) ; disable controls
Gui, Submit, NoHide
X := 5_1, Z := 5_5
Loop, 5
Loop, % Row := A_Index
If (%Row%_%A_Index% = "")
%Row%_%A_Index% := 0
GuiControl,, Edit13, % 5_3 := 5_1 + 5_5
GuiControl,, Edit10, % 4_4 := 5_4 + 5_5
GuiControl,, Edit9,  % 4_3 := 5_3 + 5_4
GuiControl,, Edit8,  % 4_2 := 5_2 + 5_3
GuiControl,, Edit7,  % 4_1 := 5_1 + 5_2
GuiControl,, Edit6,  % 3_3 := 4_4 + 4_3
GuiControl,, Edit5,  % 3_2 := 4_3 + 4_2
GuiControl,, Edit4,  % 3_1 := 4_2 + 4_1
GuiControl,, Edit3,  % 2_2 := 3_3 + 3_2
GuiControl,, Edit2,  % 2_1 := 3_2 + 3_1
GuiControl,, Edit1,  % 1_1 := 2_2 + 2_1
Gui, Font, bold cRed
If Not 3_1 = N3
GuiControl, Font, Edit4
If Not 1_1 = N4
GuiControl, Font, Edit1

Return

;---------------------------------------------------------------------------
;---------------------------------------------------------------------------
X := Z := ""
Gosub, ButtonRestart

Return

;---------------------------------------------------------------------------
ButtonNew: ; enter new numbers for the puzzle
;---------------------------------------------------------------------------
Gosub, GreenFont
Loop, 15
If A_Index Not In 1,4,12,14
GuiControl,, Edit%A_Index% ; clear
Controls(False) ; disable controls
NewContr(True)  ; enable controls for new numbers

Return

;---------------------------------------------------------------------------
ButtonApply: ; remember the new numbers
;---------------------------------------------------------------------------
Gui, Submit, NoHide
N1 := 5_2, N2 := 5_4, N3 := 3_1, N4 := 1_1
NewContr(False) ; disable controls for new numbers
Controls(True)  ; enable controls

Return

;---------------------------------------------------------------------------
ButtonCancel: ; restore the old numbers
;---------------------------------------------------------------------------
GuiControl,, Edit1, %N4%
GuiControl,, Edit4, %N3%
GuiControl,, Edit12, %N1%
GuiControl,, Edit14, %N2%
NewContr(False) ; disable controls for new numbers
Controls(True)  ; enable controls

Return

;---------------------------------------------------------------------------
GuiClose:
;---------------------------------------------------------------------------
GuiEscape:
;---------------------------------------------------------------------------
ButtonExit:
;---------------------------------------------------------------------------
; common action
ExitApp

Return

;---------------------------------------------------------------------------
Controls(Bool) { ; [dis|re-en]able some controls
;---------------------------------------------------------------------------
Enable  := Bool ? "+" : "-"
Disable := Bool ? "-" : "+"

GuiControl, %Enable%TabStop, Edit11
GuiControl, %Enable%TabStop, Edit15

GuiControl, %Disable%Default, &Restart
GuiControl, %Enable%Default, &Check
GuiControl, %Disable%Disabled, &Check
GuiControl, %Enable%Disabled, &Restart
}

;---------------------------------------------------------------------------
NewContr(Bool) { ; [dis|re-en]able control for new numbers
;---------------------------------------------------------------------------
Enable  := Bool ? "+" : "-"
Disable := Bool ? "-" : "+"

GuiControl, %Enable%TabStop, Edit1
GuiControl, %Enable%TabStop, Edit4
GuiControl, %Enable%TabStop, Edit12
GuiControl, %Enable%TabStop, Edit14

GuiControl, %Enable%Hidden, Button1
GuiControl, %Enable%Hidden, Button2
GuiControl, %Enable%Hidden, Button3
GuiControl, %Enable%Hidden, Button4
GuiControl, %Disable%Hidden, Button5
GuiControl, %Enable%Hidden, Button6
GuiControl, %Disable%Hidden, Button7
GuiControl, %Enable%Hidden, Button8

}

;---------------------------------------------------------------------------
WM_MOUSEMOVE() { ; monitor MouseMove events
;---------------------------------------------------------------------------
; display quick help in StatusBar
;-----------------------------------------------------------------------
global AppName
CurrControl := A_GuiControl
IfEqual True,, MsgBox ; dummy

; mouse is over buttons
Else If (CurrControl = "&Restart")
SB_SetText("restart retaining the blue clues")
Else If (CurrControl = "&Solve")
SB_SetText("calculate solution")
Else If (CurrControl = "&Check")
SB_SetText("check if the entries are correct")
Else If (CurrControl = "Cle&ar")
Else If (CurrControl = "&New")
SB_SetText("enter new numbers for the puzzle")
Else If (CurrControl = "E&xit")
SB_SetText("exit " AppName)

; delete status bar text
Else SB_SetText("")
}
```

## BBC BASIC

```      INSTALL @lib\$ + "ARRAYLIB"

REM Describe the puzzle as a set of simultaneous equations:
REM  a + b = 151
REM  a - c = 40
REM  -b + c + d = 0
REM  e + f = 40
REM  -c + f + g = 0
REM  -d + g + h = 0
REM  e - x = 11
REM  f - y = 11
REM  g - y = 4
REM  h - z = 4
REM  x - y + z = 0
REM So we have 11 equations in 11 unknowns.

REM We can represent these equations as a matrix and a vector:
DIM matrix(10,10), vector(10)
matrix() = \ a, b, c, d, e, f, g, h, x, y, z
\            1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
\            1, 0,-1, 0, 0, 0, 0, 0, 0, 0, 0, \
\            0,-1, 1, 1, 0, 0, 0, 0, 0, 0, 0, \
\            0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, \
\            0, 0,-1, 0, 0, 1, 1, 0, 0, 0, 0, \
\            0, 0, 0,-1, 0, 0, 1, 1, 0, 0, 0, \
\            0, 0, 0, 0, 1, 0, 0, 0,-1, 0, 0, \
\            0, 0, 0, 0, 0, 1, 0, 0, 0,-1, 0, \
\            0, 0, 0, 0, 0, 0, 1, 0, 0,-1, 0, \
\            0, 0, 0, 0, 0, 0, 0, 1, 0, 0,-1, \
\            0, 0, 0, 0, 0, 0, 0, 0, 1,-1, 1
vector() = 151, 40, 0, 40, 0, 0, 11, 11, 4, 4, 0

REM Now solve the simultaneous equations:
PROC_invert(matrix())
vector() = matrix().vector()

PRINT "X = " ; vector(8)
PRINT "Y = " ; vector(9)
PRINT "Z = " ; vector(10)
```
Output:
```X = 5
Y = 13
Z = 8
```

## C

This solution is based upon algebraic necessities, namely that a solution exists when (top - 4(a+b))/7 is integral. It also highlights the type difference between floating point numbers and integers in C.

```/* Pascal's pyramid solver
*
*               [top]
*            [   ] [   ]
*         [mid] [   ] [   ]
*      [   ] [   ] [   ] [   ]
*   [ x ] [ a ] [ y ] [ b ] [ z ]
*             x + z = y
*
* This solution makes use of a little bit of mathematical observation,
* such as the fact that top = 4(a+b) + 7(x+z) and mid = 2x + 2a + z.
*/

#include <stdio.h>
#include <math.h>

void pascal(int a, int b, int mid, int top, int* x, int* y, int* z)
{
double ytemp = (top - 4 * (a + b)) / 7.;
if(fmod(ytemp, 1.) >= 0.0001)
{
x = 0;
return;
}
*y = ytemp;
*x = mid - 2 * a - *y;
*z = *y - *x;
}
int main()
{
int a = 11, b = 4, mid = 40, top = 151;
int x, y, z;
pascal(a, b, mid, top, &x, &y, &z);
if(x != 0)
printf("x: %d, y: %d, z: %d\n", x, y, z);
else printf("No solution\n");

return 0;
}
```
Output:
```x: 5, y: 13, z: 8
```

### Field equation solver

Treating relations between cells as if they were differential equations, and apply negative feedback to each cell at every iteration step. This is how field equations with boundary conditions are solved numerically. It is, of course, not the optimal solution for this particular task.

```#include <stdio.h>
#include <stdlib.h>

void show(int *x) {
int i, j;

for (i = 0; i < 5; i++)
for (j = 0; j <= i; j++)
printf("%4d%c", *(x++), j < i ? ' ' : '\n');
}

inline int sign(int i)
{
return i < 0 ? -1 : i > 0;
}

int iter(int *v, int *diff) {
int sum, i, j, e = 0;

#	define E(x, row, col) x[(row) * ((row) + 1) / 2 + (col)]
/* enforce boundary conditions */
E(v, 0, 0) = 151;
E(v, 2, 0) = 40;
E(v, 4, 1) = 11;
E(v, 4, 3) = 4;

/* calculate difference from equilibrium */
for (i = 1; i < 5; i++) {
for (j = 0; j <= i; j++) {
E(diff, i, j) = 0;
if (j < i)
E(diff, i, j) += E(v, i - 1, j) -
E(v, i, j + 1) -
E(v, i, j);
if (j)
E(diff, i, j) += E(v, i - 1, j - 1) -
E(v, i, j - 1) -
E(v, i, j);
}
}

for (i = 0; i < 4; i++)
for (j = 0; j < i; j++)
E(diff, i, j) += E(v, i + 1, j) +
E(v, i + 1, j + 1) -
E(v, i, j);

E(diff, 4, 2) += E(v, 4, 0) + E(v, 4, 4) - E(v, 4, 2);
#	undef E

/* Do feedback, check if we are done. */
for (i = sum = 0; i < 15; i++) {
sum += !!sign(e = diff[i]);

/* 1/5-ish feedback strength on average.  These numbers are highly
magical, depending on nodes' connectivities. */
if (e >= 4 || e <= -4) 		v[i] += e/5;
else if (rand() < RAND_MAX/4)	v[i] += sign(e);
}
return sum;
}

int main() {
int v[15] = { 0 }, diff[15] = { 0 }, i, s;

for (i = s = 1; s; i++) {
s = iter(v, diff);
printf("pass %d: %d\n", i, s);
}
show(v);

return 0;
}
```
Output:
```pass 1: 12
pass 2: 12
pass 3: 14
pass 4: 14
...
pass 113: 4
pass 114: 7
pass 115: 0

151
81   70
40   41   29
16   24   17   12

5   11   13    4    8```

## C#

```using System;

namespace Pyramid_of_Numbers
{
class Program
{
public static void Main(string[] args)
{
// Set console properties
Console.Title = " Pyramid of Numbers  /  Pascal's triangle Puzzle";
Console.SetBufferSize(80,1000);
Console.SetWindowSize(80,60);
Console.ForegroundColor = ConsoleColor.Green;

// Main Program Loop
ConsoleKeyInfo k = new ConsoleKeyInfo('Y', ConsoleKey.Y,true,true,true);
while (k.Key == ConsoleKey.Y)
{
Console.Clear();

Console.WriteLine("----------------------------------------------");
Console.WriteLine(" Pyramid of Numbers / Pascal's triangle Puzzle");
Console.WriteLine("----------------------------------------------");
Console.WriteLine();

//
// Declare new Pyramid array
//
int r = 5;// Number of rows
int [,] Pyramid = new int[r,r];

// Set initial Pyramid values
for (int i = 0; i < r; i++)
{
for(int j = 0; j < r; j++)
{
Pyramid[i,j] = 0;
}
}
// Show info on created array
Console.WriteLine(" Pyramid has " + r + " rows");
Console.WriteLine("--------------------------------------------");

// Enter Pyramid values
for(int i = 0; i <= r-1; i++)
{
Console.WriteLine(" Enter " + (i+1).ToString() + ". row values:");
Console.WriteLine("--------------------------------------------");

for(int j = 0; j < i+1; j++)
{
Console.Write(" " + (j+1).ToString() + ". value = ");

Pyramid[i,j] = v;
}
Console.WriteLine("--------------------------------------------");
}

//
// Show initial Pyramid values
//
Console.WriteLine();
Console.WriteLine(" Initial Pyramid Values ");
Console.WriteLine();

// Show Pyramid values
for(int i = 0; i <= r-1; i++)
{
for(int j = 0; j < i+1; j++)
{
Console.Write("{0,4}",Pyramid[i,j]);
}
Console.WriteLine();
}
Console.WriteLine("--------------------------------------------");

// Find solution
Solve_Pyramid(Pyramid);

Console.WriteLine();
Console.Write(" Start new calculation <Y/N>  . . . ");
}
}

//
// Solve Function
//
public static void Solve_Pyramid(int [,] Pyramid)
{
int r = 5; // Number of rows

// Calculate Y
int a = Pyramid[r-1,1];
int b = Pyramid[r-1,3];
int c = Pyramid[0,0];

int y =  (c - (4*a) - (4*b))/7;
Pyramid[r-1,2] = y;

// Create copy of Pyramid
int [,] Pyramid_Copy = new int[r,r];
Array.Copy(Pyramid,Pyramid_Copy,r*r);

int n = 0; // solution counter
for(int x = 0; x < y + 1; x++)
{
for(int z = 0; z < y + 1; z++)
{
if( (x+z) == y)
{
Pyramid[r-1,0]   = x;
Pyramid[r-1,r-1] = z;

// Recalculate Pyramid values
for(int i = r-1; i > 0; i--)
{
for(int j = 0; j < i; j++)
{
Pyramid[i-1,j] = Pyramid[i,j]+Pyramid[i,j+1];
}
}

// Compare Pyramid values
bool solved = true;
for(int i = 0; i < r-1; i++)
{
for(int j = 0; j < i+1; j++)
{
if(Pyramid_Copy[i,j]>0)
{
if(Pyramid[i,j] != Pyramid_Copy[i,j])
{
solved = false;
i = r;
break;
}
}
}
}

if(solved)
{
n++;
Console.WriteLine();
Console.WriteLine(" Solved Pyramid Values no." + n);
Console.WriteLine();

// Show Pyramid values
for(int i = 0; i <= r-1; i++)
{
for(int j = 0; j < i+1; j++)
{
Console.Write("{0,4}",Pyramid[i,j]);
}
Console.WriteLine();
}
Console.WriteLine();
Console.WriteLine(" X = " + Pyramid[r-1,0] + "   " +
" Y = " + Pyramid[r-1,2] + "   " +
" Z = " + Pyramid[r-1,4]);
Console.WriteLine();
Console.WriteLine("--------------------------------------------");
}

Array.Copy(Pyramid_Copy,Pyramid,r*r);
}
}
}

if(n == 0)
{
Console.WriteLine();
Console.WriteLine(" Pyramid has no solution ");
Console.WriteLine();
}
}

}
}```
Program Input and Output:
```
----------------------------------------------
Pyramid of Numbers / Pascal's triangle Puzzle
----------------------------------------------

Pyramid has 5 rows
--------------------------------------------
Enter 1. row values:
--------------------------------------------
1. value = 151
--------------------------------------------
Enter 2. row values:
--------------------------------------------
1. value = 0
2. value = 0
--------------------------------------------
Enter 3. row values:
--------------------------------------------
1. value = 40
2. value = 0
3. value = 0
--------------------------------------------
Enter 4. row values:
--------------------------------------------
1. value = 0
2. value = 0
3. value = 0
4. value = 0
--------------------------------------------
Enter 5. row values:
--------------------------------------------
1. value = 0
2. value = 11
3. value = 0
4. value = 4
5. value = 0
--------------------------------------------

Initial Pyramid Values

151
0   0
40   0   0
0   0   0   0
0  11   0   4   0
--------------------------------------------

Solved Pyramid Values no.1

151
81  70
40  41  29
16  24  17  12
5  11  13   4   8

X = 5    Y = 13    Z = 8

--------------------------------------------

Start new calculation <Y/N>  . . .

```

## C++

Translation of: C
```#include <iostream>
#include <iomanip>

inline int sign(int i) {
return i < 0 ? -1 : i > 0;
}

inline int& E(int *x, int row, int col) {
return x[row * (row + 1) / 2 + col];
}

int iter(int *v, int *diff) {
// enforce boundary conditions
E(v, 0, 0) = 151;
E(v, 2, 0) = 40;
E(v, 4, 1) = 11;
E(v, 4, 3) = 4;

// calculate difference from equilibrium
for (auto i = 1u; i < 5u; i++)
for (auto j = 0u; j <= i; j++) {
E(diff, i, j) = 0;
if (j < i)
E(diff, i, j) += E(v, i - 1, j) - E(v, i, j + 1) - E(v, i, j);
if (j)
E(diff, i, j) += E(v, i - 1, j - 1) - E(v, i, j - 1) - E(v, i, j);
}

for (auto i = 0u; i < 4u; i++)
for (auto j = 0u; j < i; j++)
E(diff, i, j) += E(v, i + 1, j) + E(v, i + 1, j + 1) - E(v, i, j);

E(diff, 4, 2) += E(v, 4, 0) + E(v, 4, 4) - E(v, 4, 2);

// do feedback, check if we are done
uint sum;
int e = 0;
for (auto i = sum = 0u; i < 15u; i++) {
sum += !!sign(e = diff[i]);

// 1/5-ish feedback strength on average.  These numbers are highly magical, depending on nodes' connectivities
if (e >= 4 || e <= -4)
v[i] += e / 5;
else if (rand() < RAND_MAX / 4)
v[i] += sign(e);
}
return sum;
}

void show(int *x) {
for (auto i = 0u; i < 5u; i++)
for (auto j = 0u; j <= i; j++)
std::cout << std::setw(4u) << *(x++) << (j < i ? ' ' : '\n');
}

int main() {
int v[15] = { 0 }, diff[15] = { 0 };
for (auto i = 1u, s = 1u; s; i++) {
s = iter(v, diff);
std::cout << "pass " << i << ": " << s << std::endl;
}
show(v);

return 0;
}
```

## Clojure

X and Z are the independent variables, so first work bottom up and determine the value of each cell in the form (n0 + n1*X + n2*Z). We'll use a vector [n0 n1 n2] to represent each cell.

```(def bottom [ [0 1 0], [11 0 0], [0 1 1], [4 0 0], [0 0 1] ])

(defn plus  [v1 v2] (vec (map + v1 v2)))
(defn minus [v1 v2] (vec (map - v1 v2)))
(defn scale [n v]   (vec (map #(* n %) v )))

(defn above [row] (map #(apply plus %) (partition 2 1 row)))

(def rows (reverse (take 5 (iterate above bottom))))
```

We know the integer value of cells c00 and c20 ( base-0 row then column numbers), so by subtracting these values we get two equations of the form 0=n0+n1*X+n2*Z.

```(def c00 (get-in rows [0 0]))
(def c20 (get-in rows [2 0]))

(def eqn0 (minus c00 [151 0 0]))
(def eqn1 (minus c20 [ 40 0 0]))
```

In this case, there are only two variables, so solving the system of linear equations is simple.

```(defn solve [m]
(assert (<= 1 m 2))
(let [n  (- 3 m)
v0 (scale (eqn1 n) eqn0)
v1 (scale (eqn0 n) eqn1)
vd (minus v0 v1)]
(assert (zero? (vd n)))
(/ (- (vd 0)) (vd m))))

(let [x (solve 1), z (solve 2), y (+ x z)]
(println "x =" x ", y =" y ", z =" z))
```

If you want to solve the whole pyramid, just add a call (show-pyramid x z) to the previous let form:

```(defn dot [v1 v2] (reduce + (map * v1 v2)))

(defn show-pyramid [x z]
(doseq [row rows]
(println (map #(dot [1 x z] %) row)))
```

## Craft Basic

```let x = -1

do

let x = x + 1
let z = 0

do

let e = x + 11
let f = 11 + (x + z)
let g = (x + z) + 4
let h = 4 + z

if e + f = 40 then

let c = f + g
let d = g + h
let a = 40 + c
let b = c + d
let q = 0

if a + b = 151 then

let q = 1

endif

endif

if q = 0 then

let z = z + 1

endif

wait

loopwhile z < 20 and q = 0

if q = 0 then

let z = -1

endif

wait

loopuntil z >= 0

print "x = ", x
print "y = ", x + z
print "z = ", z
```
Output:
```x = 5
y = 13

z = 8```

## Curry

Works with: PAKCS
```import CLPFD
import Constraint (allC, andC)
import Findall (findall)
import List (init, last)

solve :: [[Int]] -> Success
solve body@([n]:rest) =
domain (concat body) 1 n
& andC (zipWith atop body rest)
& labeling [] (concat body)
where
xs `atop` ys = andC \$ zipWith3 tri xs (init ys) (tail ys)

tri :: Int -> Int -> Int -> Success
tri x y z = x =# y +# z

test (x,y,z) | tri y x z =
[ [151]
, [ _,  _]
, [40,  _, _]
, [ _,  _, _, _]
, [ x, 11, y, 4, z]
]
main = findall \$ solve . test```
Output:
```Execution time: 0 msec. / elapsed: 0 msec.
[(5,13,8)]```

## D

Translation of: C
```import std.stdio, std.algorithm;

void iterate(bool doPrint=true)(double[] v, double[] diff) @safe {
static ref T E(T)(T[] x, in size_t row, in size_t col)
pure nothrow @safe @nogc {
return x[row * (row + 1) / 2 + col];
}

double tot = 0.0;
do {
// Enforce boundary conditions.
E(v, 0, 0) = 151;
E(v, 2, 0) = 40;
E(v, 4, 1) = 11;
E(v, 4, 3) = 4;

// Calculate difference from equilibrium.
foreach (immutable i; 1 .. 5) {
foreach (immutable j; 0 .. i + 1) {
E(diff, i, j) = 0;
if (j < i)
E(diff, i, j) += E(v, i - 1, j) - E(v, i, j + 1) - E(v, i, j);
if (j)
E(diff, i, j) += E(v, i - 1, j - 1) - E(v, i, j - 1) - E(v, i, j);
}
}

foreach (immutable i; 1 .. 4)
foreach (immutable j; 0 .. i)
E(diff, i, j) += E(v, i + 1, j) + E(v, i + 1, j + 1) - E(v, i, j);

E(diff, 4, 2) += E(v, 4, 0) + E(v, 4, 4) - E(v, 4, 2);

// Do feedback, check if we are close enough.
// 4: scale down the feedback to avoid oscillations.
v[] += diff[] / 4;
tot = diff.map!q{ a ^^ 2 }.sum;

static if (doPrint)
writeln("dev: ", tot);

// tot(dx^2) < 0.1 means each cell is no more than 0.5 away
// from equilibrium. It takes about 50 iterations. After
// 700 iterations tot is < 1e-25, but that's overkill.
} while (tot >= 0.1);
}

void main() {
static void show(in double[] x) nothrow @nogc {
int idx;
foreach (immutable i; 0 .. 5)
foreach (immutable j; 0 .. i+1) {
printf("%4d%c", cast(int)(0.5 + x[idx]), j < i ? ' ' : '\n');
idx++;
}
}

double[15] v = 0.0, diff = 0.0;
iterate(v, diff);
show(v);
}
```
Output:
```dev: 73410
dev: 17968.7
dev: 6388.46
dev: 2883.34
dev: 1446.59
dev: 892.753
dev: 564.678
[... several more iterations...]
dev: 0.136504
dev: 0.125866
dev: 0.116055
dev: 0.107006
dev: 0.0986599
151
81   70
40   41   29
16   24   17   12
5   11   13    4    8```

## F#

In a script, using the Math.NET Numerics library

```#load"Packages\MathNet.Numerics.FSharp\MathNet.Numerics.fsx"

open MathNet.Numerics.LinearAlgebra

let A = matrix [
[ 1.; 1.; 0.; 0.; 0.; 0.; 0.; 0.; 0.; 0.; 0. ]
[ -1.;  0.; 1.; 0.; 0.; 0.; 0.; 0.; 0.; 0.; 0. ]
[ 0.; -1.; 1.; 1.; 0.; 0.; 0.; 0.; 0.; 0.; 0. ]
[ 0.; 0.; 0.; 0.; 1.; 1.; 0.; 0.; 0.; 0.; 0. ]
[ 0.; 0.; -1.; 0.; 0.; 1.; 1.; 0.; 0.; 0.; 0. ]
[ 0.; 0.; 0.; -1.; 0.; 0.; 1.; 1.; 0.; 0.; 0. ]
[ 0.; 0.; 0.; 0.; -1.; 0.; 0.; 0.; 1.; 0.; 0. ]
[ 0.; 0.; 0.; 0.; 0.; -1.; 0.; 0.; 0.; 1.; 0. ]
[ 0.; 0.; 0.; 0.; 0.; 0.; -1.; 0.; 0.; 1.; 0. ]
[ 0.; 0.; 0.; 0.; 0.; 0.; 0.; -1.; 0.; 0.; 1. ]
[ 0.; 0.; 0.; 0.; 0.; 0.; 0.; 0.; 1.; -1.; 1. ]
]

let b = vector [151.; -40.; 0.; 40.; 0.; 0.; -11.; -11.; -4.; -4.; 0.]

let x = A.Solve(b)

printfn "x = %f, Y = %f, Z = %f" x.[8] x.[9] x.[10]
```
Output:
`x = 5.000000, Y = 13.000000, Z = 8.000000`

## Factor

Works with: Factor version 0.98
```USING: arrays backtrack combinators.extras fry grouping.extras
interpolate io kernel math math.ranges sequences ;

: base ( ?x ?z -- seq ) 2dup + swap '[ _ 11 _ 4 _ ] >array ;

: up ( seq -- seq' ) [ [ + ] 2clump-map ] twice ;

: find-solution ( -- x z )
10 [1,b] dup [ amb-lazy ] bi@ 2dup base
up dup first 40 = must-be-true
up first 151 = must-be-true ;

find-solution [I X = \${1}, Z = \${}I] nl
```
Output:
```X = 5, Z = 8
```

## FreeBASIC

Translation of: PureBasic
```Function SolveForZ(x As Integer) As Integer
Dim As Integer a, b, c, d, e, f, g, h, z
For z = 0 To 20
e = x + 11
f = 11 + (x+z)
g = (x+z) + 4
h = 4 + z
If e + f = 40 Then
c = f + g
d = g + h
a = 40 + c
b = c + d
If a + b = 151 Then Return z
End If
Next z
Return -1
End Function

Dim As Integer x = -1, z = 0
Do
x = x + 1
z = SolveForZ(x)
Loop Until z >= 0

Print "X ="; x
Print "Y ="; x + z
Print "Z ="; z
Sleep```
Output:
```X = 5
Y = 13
Z = 8```

## Go

This solution follows the way the problem might be solved with pencil and paper. It shows a possible data representation of the problem, uses the computer to do some arithmetic, and displays intermediate and final results.

```package main

import "fmt"

// representation of an expression in x, y, and z
type expr struct {
x, y, z float64 // coefficients
c       float64 // constant term
}

func addExpr(a, b expr) expr {
return expr{a.x + b.x, a.y + b.y, a.z + b.z, a.c + b.c}
}

// subtract two expressions
func subExpr(a, b expr) expr {
return expr{a.x - b.x, a.y - b.y, a.z - b.z, a.c - b.c}
}

// multiply expression by a constant
func mulExpr(a expr, c float64) expr {
return expr{a.x * c, a.y * c, a.z * c, a.c * c}
}

// given a row of expressions, produce the next row up, by the given
// sum relation between blocks
if len(l) == 0 {
panic("wrong")
}
r := make([]expr, len(l)-1)
for i := range r {
}
return r
}

// given expression b in a variable, and expression a,
// take b == 0 and substitute to remove that variable from a.
func substX(a, b expr) expr {
if b.x == 0 {
panic("wrong")
}
return subExpr(a, mulExpr(b, a.x/b.x))
}

func substY(a, b expr) expr {
if b.y == 0 {
panic("wrong")
}
return subExpr(a, mulExpr(b, a.y/b.y))
}

func substZ(a, b expr) expr {
if b.z == 0 {
panic("wrong")
}
return subExpr(a, mulExpr(b, a.z/b.z))
}

// given an expression in a single variable, return value of that variable
func solveX(a expr) float64 {
if a.x == 0 || a.y != 0 || a.z != 0 {
panic("wrong")
}
return -a.c / a.x
}

func solveY(a expr) float64 {
if a.x != 0 || a.y == 0 || a.z != 0 {
panic("wrong")
}
return -a.c / a.y
}

func solveZ(a expr) float64 {
if a.x != 0 || a.y != 0 || a.z == 0 {
panic("wrong")
}
return -a.c / a.z
}

func main() {
// representation of given information for bottom row
r5 := []expr{{x: 1}, {c: 11}, {y: 1}, {c: 4}, {z: 1}}
fmt.Println("bottom row:", r5)

// given definition of brick sum relation
fmt.Println("next row up:", r4)
fmt.Println("middle row:", r3)

// given relation y = x + z
xyz := subExpr(expr{y: 1}, expr{x: 1, z: 1})
fmt.Println("xyz relation:", xyz)
// remove z from third cell using xyz relation
r3[2] = substZ(r3[2], xyz)
fmt.Println("middle row after substituting for z:", r3)

// given cell = 40,
b := expr{c: 40}
// this gives an xy relation
xy := subExpr(r3[0], b)
fmt.Println("xy relation:", xy)
// substitute 40 for cell
r3[0] = b

// remove x from third cell using xy relation
r3[2] = substX(r3[2], xy)
fmt.Println("middle row after substituting for x:", r3)

// continue applying brick sum relation to get top cell
fmt.Println("next row up:", r2)
fmt.Println("top row:", r1)

// given top cell = 151, we have an equation in y
y := subExpr(r1[0], expr{c: 151})
fmt.Println("y relation:", y)
// using xy relation, we get an equation in x
x := substY(xy, y)
fmt.Println("x relation:", x)
// using xyz relation, we get an equation in z
z := substX(substY(xyz, y), x)
fmt.Println("z relation:", z)

fmt.Println("x =", solveX(x))
fmt.Println("y =", solveY(y))
fmt.Println("z =", solveZ(z))
}
```
Output:
```bottom row: [{1 0 0 0} {0 0 0 11} {0 1 0 0} {0 0 0 4} {0 0 1 0}]
next row up: [{1 0 0 11} {0 1 0 11} {0 1 0 4} {0 0 1 4}]
middle row: [{1 1 0 22} {0 2 0 15} {0 1 1 8}]
xyz relation: {-1 1 -1 0}
middle row after substituting for z: [{1 1 0 22} {0 2 0 15} {-1 2 0 8}]
xy relation: {1 1 0 -18}
middle row after substituting for x: [{0 0 0 40} {0 2 0 15} {0 3 0 -10}]
next row up: [{0 2 0 55} {0 5 0 5}]
top row: [{0 7 0 60}]
y relation: {0 7 0 -91}
x relation: {1 0 0 -5}
z relation: {0 0 -1 8}
x = 5
y = 13
z = 8
```

I assume the task is to solve any such puzzle, i.e. given some data

```puzzle = [["151"],["",""],["40","",""],["","","",""],["X","11","Y","4","Z"]]
```

one should calculate all possible values that fit. That just means solving a linear system of equations. We use the first three variables as placeholders for X, Y and Z. Then we can produce the matrix of equations:

```triangle n = n * (n+1) `div` 2

coeff xys x = maybe 0 id \$ lookup x xys

row n cs = [coeff cs k | k <- [1..n]]

eqXYZ n = [(0, 1:(-1):1:replicate n 0)]

eqPyramid n h = do
a <- [1..h-1]
x <- [triangle (a-1) + 1 .. triangle a]
let y = x+a
return \$ (0, 0:0:0:row n [(x,-1),(y,1),(y+1,1)])

eqConst n fields = do
(k,s) <- zip [1..] fields
guard \$ not \$ null s
return \$ case s of
"X" - (0, 1:0:0:row n [(k,-1)])
"Y" - (0, 0:1:0:row n [(k,-1)])
"Z" - (0, 0:0:1:row n [(k,-1)])
_   - (fromInteger \$ read s, 0:0:0:row n [(k,1)])

equations :: [[String]] - ([Rational], [[Rational]])
equations puzzle = unzip eqs where
fields = concat puzzle
eqs = eqXYZ n ++ eqPyramid n h ++ eqConst n fields
h = length puzzle
n = length fields
```

To solve the system, any linear algebra library will do (e.g hmatrix). For this example, we assume there are functions decompose for LR-decomposition, kernel to solve the homogenous system and solve to find a special solution for an imhomogenous system. Then

```normalize :: [Rational] - [Integer]
normalize xs = [numerator (x * v) | x <- xs] where
v = fromInteger \$ foldr1 lcm \$ map denominator \$ xs

run puzzle = map (normalize . drop 3) \$ answer where
(a, m) = equations puzzle
lr = decompose 0 m
answer = case solve 0 lr a of
Nothing - []
Just x  - x : kernel lr
```

will output one special solution and modifications that lead to more solutions, as in

```*Main run puzzle
[[151,81,70,40,41,29,16,24,17,12,5,11,13,4,8]]
*Main run [[""],["2",""],["X","Y","Z"]]
[[3,2,1,1,1,0],[3,0,3,-1,1,2]]
```

so for the second puzzle, not only X=1 Y=1 Z=0 is a solution, but also X=1-1=0, Y=1+1=2 Z=0+2=2 etc.

Note that the program doesn't attempt to verify that the puzzle is in correct form.

## J

Fixed points in the pyramid are 40 and 151, which I use to check a resulting pyramid for selection:

```chk=:40 151&-:@(2 4{{."1)
```

verb for the base of the pyramid:

```base=: [,11,+,4,]
```

the height of the pyramid:

```ord=:5
```

=> 'chk', 'base' and 'ord' are the knowledge rules abstracted from the problem definition.

The J-sentence that solves the puzzle is:

```    |."2(#~chk"2) 2(+/\)^:(<ord)"1 base/"1>,{ ;~i:28
```
``` 151  0  0  0 0
81 70  0  0 0
40 41 29  0 0
16 24 17 12 0
5 11 13  4 8```

Get rid of zeros:

```,.(1+i.5)<@{."0 1{.|."2(#~chk"2) 2(+/\)^:(<ord)"1 base/"1>,{ ;~i:28
```

or

```,.(<@{."0 1~1+i.@#){.|."2(#~chk"2) 2(+/\)^:(<ord)"1 base/"1>,{ ;~i:28
```
``` +-----------+
|151        |
+-----------+
|81 70      |
+-----------+
|40 41 29   |
+-----------+
|16 24 17 12|
+-----------+
|5 11 13 4 8|
+-----------+```

## Java

Generate 11 equations and 11 unknowns. Reuse code from Cramer's Rule.

```import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class PascalsTrianglePuzzle {

public static void main(String[] args) {
Matrix mat = new Matrix(Arrays.asList(1d, 0d, 0d, 0d, 0d, 0d, 0d, 0d, -1d, 0d, 0d),
Arrays.asList(0d, 1d, 0d, 0d, 0d, 0d, 0d, 0d, 0d, -1d, 0d),
Arrays.asList(0d, 0d, 0d, 0d, 0d, 0d, 0d, 0d, -1d, 1d, -1d),
Arrays.asList(0d, 0d, 1d, 0d, 0d, 0d, 0d, 0d, 0d, -1d, 0d),
Arrays.asList(0d, 0d, 0d, 1d, 0d, 0d, 0d, 0d, 0d, 0d, -1d),
Arrays.asList(1d, 1d, 0d, 0d, 0d, 0d, 0d, 0d, 0d, 0d, 0d),
Arrays.asList(0d, 1d, 1d, 0d, -1d, 0d, 0d, 0d, 0d, 0d, 0d),
Arrays.asList(0d, 0d, 1d, 1d, 0d, -1d, 0d, 0d, 0d, 0d, 0d),
Arrays.asList(0d, 0d, 0d, 0d, -1d, 0d, 1d, 0d, 0d, 0d, 0d),
Arrays.asList(0d, 0d, 0d, 0d, 1d, 1d, 0d, -1d, 0d, 0d, 0d),
Arrays.asList(0d, 0d, 0d, 0d, 0d, 0d, 1d, 1d, 0d, 0d, 0d));
List<Double> b = Arrays.asList(11d, 11d, 0d, 4d, 4d, 40d, 0d, 0d, 40d, 0d, 151d);
List<Double> solution = cramersRule(mat, b);
System.out.println("Solution = " + cramersRule(mat, b));
System.out.printf("X = %.2f%n", solution.get(8));
System.out.printf("Y = %.2f%n", solution.get(9));
System.out.printf("Z = %.2f%n", solution.get(10));
}

private static List<Double> cramersRule(Matrix matrix, List<Double> b) {
double denominator = matrix.determinant();
List<Double> result = new ArrayList<>();
for ( int i = 0 ; i < b.size() ; i++ ) {
}
return result;
}

private static class Matrix {

private List<List<Double>> matrix;

@Override
public String toString() {
return matrix.toString();
}

@SafeVarargs
public Matrix(List<Double> ... lists) {
matrix = new ArrayList<>();
for ( List<Double> list : lists) {
}
}

public Matrix(List<List<Double>> mat) {
matrix = mat;
}

public double determinant() {
if ( matrix.size() == 1 ) {
return get(0, 0);
}
if ( matrix.size() == 2 ) {
return get(0, 0) * get(1, 1) - get(0, 1) * get(1, 0);
}
double sum = 0;
double sign = 1;
for ( int i = 0 ; i < matrix.size() ; i++ ) {
sum += sign * get(0, i) * coFactor(0, i).determinant();
sign *= -1;
}
return sum;
}

private Matrix coFactor(int row, int col) {
List<List<Double>> mat = new ArrayList<>();
for ( int i = 0 ; i < matrix.size() ; i++ ) {
if ( i == row ) {
continue;
}
List<Double> list = new ArrayList<>();
for ( int j = 0 ; j < matrix.size() ; j++ ) {
if ( j == col ) {
continue;
}
}
}
return new Matrix(mat);
}

private Matrix replaceColumn(List<Double> b, int column) {
List<List<Double>> mat = new ArrayList<>();
for ( int row = 0 ; row < matrix.size() ; row++ ) {
List<Double> list = new ArrayList<>();
for ( int col = 0 ; col < matrix.size() ; col++ ) {
double value = get(row, col);
if ( col == column ) {
value = b.get(row);
}
}
}
return new Matrix(mat);
}

private double get(int row, int col) {
return matrix.get(row).get(col);
}

}

}
```
Output:
```
Solution = [16.0, 24.0, 17.0, 12.0, 41.0, 29.0, 81.0, 70.0, 5.0, 13.0, 8.0]
X = 5.00
Y = 13.00
Z = 8.00
```

## jq

Works with: jq

Also works with gojq, the Go implementation of jq, and with fq.

In the spirit of the simplicity of Pascal Triangle's definition, the first solution given in this entry works up from the bottom of the triangle using the basic Pascal Triangle equation for each brick that is above two others.

In the following, the value of the j-th brick in the i-th row is denoted by rij if initially known (so r11 refers to the apex), and by \$Rij if initially unknown, so the variable "X" in the puzzle could be denoted by \$R51.

It is assumed that:

• all values in the triangle must be positive integers;
• the task is to solve for any given values of r11, r31, r52, r54;
• all solutions for the triple [X, Y, Z] should be found.
```def solve(r11; r31; r52; r54):
range(1;r31 - 1) as \$X
| range(1; r31 - 1) as \$Y
| ((\$Y - \$X) | select(. > 0)) as \$Z
| (r52 + \$X) as \$R41
| (r52 + \$Y) as \$R42
| select(\$R41 + \$R42 == r31)
| (\$Y + r54) as \$R43
| (r54 + \$Z) as \$R44
| (\$R42 + \$R43) as \$R32
| (\$R43 + \$R44) as \$R33
| (r31 + \$R32) as \$R21
| (\$R32 + \$R33) as \$R22
| select(\$R21 + \$R22 == r11)
| [\$X, \$Y, \$Z];

solve(151; 40; 11; 4),```
Output:
```[5,13,8]
```

### Algebraic solution

As noted elsewhere on this page, elementary considerations show that the apex (top) value and the value in the third row (mid) can be written as:

```top = 4(a+b) + 7(x+z)
mid = 2x + 2a + z
```

where a and b are the known values at the base.

Using the jq program at Cramer's rule, with the unknown vector being [x, z]:

```include "rc-cramers-rule";
def solve(top; mid; a; b):
cramer(
[ [7, 7],
[2,  1]];
[top - 4*(a+b), mid-2*a]);

solve(151; 40; 11; 4)```

This gives the solution for [x,z] as:

Output:
```[5,8]
```

## Julia

Translation of: Kotlin
```function pascal(a::Integer, b::Integer, mid::Integer, top::Integer)
yd = round((top - 4 * (a + b)) / 7)
!isinteger(yd) && return 0, 0, 0
y  = Int(yd)
x  = mid - 2a - y
return x, y, y - x
end

x, y, z = pascal(11, 4, 40, 151)
if !iszero(x)
println("Solution: x = \$x, y = \$y, z = \$z.")
else
println("There is no solution.")
end
```
Output:
`Solution: x = 5, y = 13, z = 8.`

## Kotlin

Translation of: C
```// version 1.1.3

data class Solution(val x: Int, val y: Int, val z: Int)

fun Double.isIntegral(tolerance: Double = 0.0) =
(this - Math.floor(this)) <= tolerance || (Math.ceil(this) - this) <= tolerance

fun pascal(a: Int, b: Int, mid: Int, top: Int): Solution {
val yd = (top - 4 * (a + b)) / 7.0
if (!yd.isIntegral(0.0001)) return Solution(0, 0, 0)
val y = yd.toInt()
val x = mid - 2 * a - y
return Solution(x, y, y - x)
}

fun main(args: Array<String>) {
val (x, y, z) = pascal(11, 4, 40, 151)
if (x != 0)
println("Solution is: x = \$x, y = \$y, z = \$z")
else
println("There is no solutuon")
}
```
Output:
```Solution is: x = 5, y = 13, z = 8
```

## Maple

```sys := {22 + x + y = 40, 78 + 5*y + z = 151, x + z = y}:
solve(sys, {x,y,z});```
Output:
```
{x = 5, y = 13, z = 8}

```

## Mathematica/Wolfram Language

We assign a variable to each block starting on top with a, then on the second row b,c et cetera. k,m, and o are replaced by X, Y, and Z. We can write the following equations:

```b+c==a
d+e==b
e+f==c
g+h==d
h+i==e
i+j==f
l+X==g
l+Y==h
n+Y==i
n+Z==j
X+Z==Y
```

And we have the knowns

```a->151
d->40
l->11
n->4
```

Giving us 10 equations with 10 unknowns; i.e. solvable. So we can do so by:

```eqs={a==b+c,d+e==b,e+f==c,g+h==d,h+i==e,i+j==f,l+X==g,l+Y==h,n+Y==i,n+Z==j,Y==X+Z};
knowns={a->151,d->40,l->11,n->4};
Solve[eqs/.knowns,{b,c,e,f,g,h,i,j,X,Y,Z}]
```

gives back:

```{{b -> 81, c -> 70, e -> 41, f -> 29, g -> 16, h -> 24, i -> 17,  j -> 12, X -> 5, Y -> 13, Z -> 8}}
```

In pyramid form that would be:

```				151
81		70
40		41		29
16		24		17		12
5		11		13		4		8
```

An alternative solution in Mathematica 10, constructing the triangle:

```triangle[n_, m_] :=  Nest[MovingMap[Total, #, 1] &, {x, 11, y, 4, z}, n - 1][[m]]
Solve[{triangle[3, 1] == 40, triangle[5, 1] == 151, y == x + z}, {x, y, z}]
```

Three equations and three unknowns, which gives back:

```{{x -> 5, y -> 13, z -> 8}}
```

## MiniZinc

```%Pascal's Triangle Puzzle. Nigel Galloway, February 17th., 2020
int: N11=151; constraint N11=N21+N22;
var 1..N11: N21=N31+N32;
var 1..N11: N22=N32+N33;
int: N31=40; constraint N31=N41+N42;
var 1..N11: N32=N42+N43;
var 1..N11: N33=N43+N44;
var 1..N11: N41=X+11;
var 1..N11: N42=Y+11;
var 1..N11: N43=Y+4;
var 1..N11: N44=Z+4;
var 1..N11: X;
var 1..N11: Y=X+Z;
var 1..N11: Z;```
Output:
```Z = 8;
X = 5;
----------
```

## Nim

```import strutils

type

BlockValue = object
known: int
x, y, z: int

Variables = tuple[x, y, z: int]

func `+=`(left: var BlockValue; right: BlockValue) =
## Symbolically add one block to another.
left.known += right.known
left.x += right.x - right.z    # Z is excluded as n(Y - X - Z) = 0.
left.y += right.y + right.z

proc toString(n: BlockValue; vars: Variables): string =
## Return the representation of the block value, when X, Y, Z are known.
result = \$(n.known + n.x * vars.x + n.y * vars.y + n.z * vars.z)

proc Solve2x2(a11, a12, b1, a21, a22, b2: int): Variables =
## Solve a puzzle, supposing an integer solution exists.
if a22 == 0:
result.x = b2 div a21
result.y = (b1 - a11 * result.x) div a12
else:
result.x = (b1 * a22 - b2 * a12) div (a11 * a22 - a21 * a12)
result.y = (b1 - a11 * result.x) div a12
result.z = result.y - result.x

var blocks : array[1..5, array[1..5, BlockValue]]   # The lower triangle contains blocks.

# The bottom blocks.
blocks[5][1] = BlockValue(x: 1)
blocks[5][2] = BlockValue(known: 11)
blocks[5][3] = BlockValue(y: 1)
blocks[5][4] = BlockValue(known: 4)
blocks[5][5] = BlockValue(z: 1)

# Upward run.
for row in countdown(4, 1):
for column in 1..row:
blocks[row][column] += blocks[row + 1][column]
blocks[row][column] += blocks[row + 1][column + 1]

# Now have known blocks 40=[3][1], 151=[1][1] and Y=X+Z to determine X,Y,Z.
let vars = Solve2x2(blocks[1][1].x,
blocks[1][1].y,
151 - blocks[1][1].known,
blocks[3][1].x,
blocks[3][1].y,
40 - blocks[3][1].known)

# Print the results.
for row in 1..5:
var line = ""
for column in 1..row:
echo line
```
Output:
```151
81 70
40 41 29
16 24 17 12
5 11 13 4 8```

## Oz

```%% to compile : ozc -x <file.oz>
functor

import
System Application FD Search
define

proc{Quest Root Rules}

proc{Limit Rc Ls}
case Ls of nil then skip
[] X|Xs then
{Limit Rc Xs}
case X of N#V then
Rc.N =: V
[] N1#N2#N3 then
Rc.N1 =: Rc.N2 + Rc.N3
end
end
end

proc {Pyramid R}
{FD.tuple solution 15 0#FD.sup R}  %% non-negative integers domain
%%          01      , pyramid format
%%        02  03
%%      04  05  06
%%    07  08  09  10
%%  11  12  13  14  15
R.1 =: R.2 + R.3     %% constraints of Pyramid of numbers
R.2 =: R.4 + R.5
R.3 =: R.5 + R.6
R.4 =: R.7 + R.8
R.5 =: R.8 + R.9
R.6 =: R.9 + R.10
R.7 =: R.11 + R.12
R.8 =: R.12 + R.13
R.9 =: R.13 + R.14
R.10 =: R.14 + R.15

{Limit R Rules}      %% additional constraints

{FD.distribute ff R}
end
in
{Search.base.one Pyramid Root} %% search for solution
end

local
Root R
in
{Quest Root [1#151 4#40 12#11 14#4 13#11#15]} %% supply additional constraint rules
if {Length Root} >= 1 then
R = Root.1
{For 1 15 1
proc{\$ I}
if {Member I [1 3 6 10]} then
{System.printInfo R.I#'\n'}
else
{System.printInfo R.I#' '}
end
end
}
else
{System.showInfo 'No solution found.'}
end
end

{Application.exit 0}
end```

## PARI/GP

[ 6y+x+z+4a[2]+4a[4]= 7y +4a[2]+4a[4]] [3y+x+37 ][3y+z+23] [40=x+y+22][ 2y+15][ y+z+8 ] [ x+11 ][y+11 ][y+4 ][z+4 ] [ X][11][ Y][ 4][ Z]

this helped me...

```Pascals_triangle_puzzle(topvalue=151,leftsidevalue=40,bottomvalue1=11,bottomvalue2=4) = {
y=(topvalue-(4*(bottomvalue1+bottomvalue2)))/7;
x=leftsidevalue-(y+2*bottomvalue1);
z=y-x;
print(x","y","z); }```

I'm thinking of one to solve all puzzles regardless of size and positions. but the objective was to solve this puzzle.

## Perl

```# set up triangle
my \$rows = 5;
my @tri = map { [ map { {x=>0,z=>0,v=>0,rhs=>undef} } 1..\$_ ] } 1..\$rows;
\$tri[0][0]{rhs} = 151;
\$tri[2][0]{rhs} = 40;
\$tri[4][0]{x} = 1;
\$tri[4][1]{v} = 11;
\$tri[4][2]{x} = 1;
\$tri[4][2]{z} = 1;
\$tri[4][3]{v} = 4;
\$tri[4][4]{z} = 1;

# aggregate from bottom to top
for my \$row ( reverse 0..@tri-2 ) {
for my \$col ( 0..@{\$tri[\$row]}-1 ){
\$tri[\$row][\$col]{\$_} = \$tri[\$row+1][\$col]{\$_}+\$tri[\$row+1][\$col+1]{\$_} for 'x','z','v';
}
}
# find equations
my @eqn;
for my \$row ( @tri ) {
for my \$col ( @\$row ){
push @eqn, [ \$\$col{x}, \$\$col{z}, \$\$col{rhs}-\$\$col{v} ] if defined \$\$col{rhs};
}
}
# print equations
print "Equations:\n";
print "  x +   z = y\n";
printf "%d x + %d z = %d\n", @\$_ for @eqn;
# solve
my \$f = \$eqn[0][1] / \$eqn[1][1];
\$eqn[0][\$_] -=  \$f * \$eqn[1][\$_] for 0..2;
\$f = \$eqn[1][0] / \$eqn[0][0];
\$eqn[1][\$_] -=  \$f * \$eqn[0][\$_] for 0..2;
# print solution
print "Solution:\n";
my \$x = \$eqn[0][2]/\$eqn[0][0];
my \$z = \$eqn[1][2]/\$eqn[1][1];
my \$y = \$x+\$z;
printf "x=%d, y=%d, z=%d\n", \$x, \$y, \$z;
```
Output:
```Equations:
x +   z = y
7 x + 7 z = 91
2 x + 1 z = 18
Solution:
x=5, y=13, z=8
```

## Phix

```--
-- demo\rosetta\Pascal_triangle_Puzzle.exw
-- =======================================
--
-- I approached this with a view to solving general pyramid puzzles, not just the one given.
--
-- This little ditty converts the pyramid to rules quite nicely, then uses a modified copy
--  of solveN() from Solving_coin_problems#Phix to solve those simultaneous equations.
--
with javascript_semantics

sequence pyramid = {{151},
{"",""},
{40,"",""},
{"","","",""},
{"x",11,"y",4,"z"}}

sequence rules = {}

-- each cell in the pyramid is either an integer final value or an equation.
-- initially the equations are strings, we substitute all with triplets of
-- the form {k,x,z} ie k+l*x+m*z, and known values < last row become rules.

for r=5 to 1 by -1 do
for c=1 to length(pyramid[r]) do
object prc = pyramid[r][c], equ
if    prc="x" then  prc = {0,1,0}     -- ie 0 + one x
elsif prc="y" then  prc = {0,1,1}     -- ie 0 + one x plus one z
elsif prc="z" then  prc = {0,0,1}     -- ie 0 +            one z
else
if prc="" or r<=4 then
-- examples: x+11 is {0,1,0}+{11,0,0} -> {11,1,0},
--           11+y is {11,0,0}+{0,1,1} -> {11,1,1},
--       40=""+"" is {40,0,0}={22,2,1} ==> {18,2,1}
end if
if prc="" then  prc = equ
else            prc = {prc,0,0}
if r<=4 then
equ[1] = prc[1]-equ[1]
rules = append(rules,equ)
end if
end if
end if
pyramid[r][c] = prc
end for
end for

ppOpt({pp_Nest,1,pp_StrFmt,2,pp_IntCh,false})
?"equations"
pp(pyramid)
?"rules"
pp(rules)   -- {18,2,1} === 18=2x+z
-- {73,5,6} === 73=5x+6z
puts(1,"=====\n")

assert(length(rules)==2)    -- more work needed!?

-- modified copy of solveN() from Solving_coin_problems.exw as promised, a
-- bit of a sledgehammer to crack a peanut is the phrase you are looking for:
function solveN(sequence rules)
--
-- Based on https://mathcs.clarku.edu/~djoyce/ma105/simultaneous.html
--  aka the ancient Chinese Jiuzhang suanshu ~100 B.C. (!!)
--
-- Example (not related to the task problem):
--  rules = {{18,1,1},{38,1,5}}, ie 18==x+y, 38==x+5y
--  ==> {13,5}, ie x=13, y=5
--
--  In the elimination phase, both x have multipliers of 1, ie both rii and rij are 1,
--  so we can ignore the two sq_mul and just do [sq_sub] (38=x+5y)-(18=x+y)==>(20=4y).
--  Obviously therefore y is 5 and substituting backwards x is 13.
--
-- Example2 (taken from the task problem):
--  rules = {{18,2,1},{73,5,6}}, ie 18==2x+z, 73==5x+6z
--      ==> {{18,2,1},{56,0,7}}, ie rules[2]:=rules[2]*2-rules[1]*5     (eliminate)
--      ==> {{18,2,1},8},        ie rules[2]:=56/7, aka z:=8            (substitute)
--      ==> {{10,2,0},8},        ie rules[1]-=1z                        (substitute)
--      ==> {5,8},               ie rules[1]:=10/2, aka x:=5            (substitute)
--  ==> {5,8}, ie x=5, z=8
--
sequence ri, rj
integer l = length(rules), rii, rji
rules = deep_copy(rules)
for i=1 to l do
-- successively eliminate (grow lower left triangle of 0s)
ri = rules[i]
assert(length(ri)=l+1)
rii = ri[i+1]
assert(rii!=0)  -- (see note below)
for j=i+1 to l do
rj = rules[j]
rji = rj[i+1]
if rji!=0 then
rj = sq_sub(sq_mul(rj,rii),sq_mul(ri,rji))
assert(rj[i+1]==0) -- (job done)
rules[j] = rj
end if
end for
end for
for i=l to 1 by -1 do
-- then substitute each backwards
ri = rules[i]
rii = ri[1]/ri[i+1] -- (all else should be 0)
rules[i] = rii
for j=i-1 to 1 by -1 do
rj = rules[j]
rji = rj[i+1]
if rji!=0 then
rules[j] = 0
rj[1] -= rji*rii
rj[i+1] = 0
rules[j] = rj
end if
end for
end for
return rules
end function

-- Obviously these next two lines directly embody knowledge from the task, and
--  would need changing for an even slightly different version of the problem:
integer {x,z} = solveN(rules),
y = x+z -- (as per task desc)

printf(1,"x=%d, y=%d, z=%d\n",{x,y,z})

-- finally evaluate all the equations and print it.
for r=1 to length(pyramid) do
for c=1 to length(pyramid[r]) do
integer {k, l, m} = pyramid[r][c]
pyramid[r][c] = k+l*x+m*z
end for
end for

pp(pyramid)
```
Output:
```"equations"
{{{151,0,0}},
{{55,2,2}, {23,3,4}},
{{40,0,0}, {15,2,2}, {8,1,2}},
{{11,1,0}, {11,1,1}, {4,1,1}, {4,0,1}},
{{0,1,0}, {11,0,0}, {0,1,1}, {4,0,0}, {0,0,1}}}
"rules"
{{18,2,1},
{73,5,6}}
=====
x=5, y=13, z=8
{{151},
{81,70},
{40,41,29},
{16,24,17,12},
{5,11,13,4,8}}
```

Interestingly, this appears to match Python in that 40 is propagated up the tree, whereas Perl and Go appear to propagate 22+2x+z up, not that I can think of any case where that would make a difference.

A couple of other ways to use that solveN():

```?solveN({{18,1,1,0},{73,0,5,1},{0,1,-1,1}}) -- -- {5,13,8}

ppOpt({pp_Nest,0})
-- Or, hand-coding 11 simultaneous equations for 11 unknowns:
-- (be warned I had to reorder a bit to avoid rii==0 mishaps,
--  perhaps solveN() should find or even sort rules somehow)
--sequence pyramid = {{151},
--                  {"a","b"},
--                 {40,"c","d"},
--              {"e","f","g","h"},
--            {"x", 11,"y",  4,"z"}}
--               a  b  c  d  e  f  g  h  x  y  z
pp(solveN({{151, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0},  -- a+b=151
{ 40, 1, 0,-1, 0, 0, 0, 0, 0, 0, 0, 0},  -- 40+c=a
{  0, 0, 1,-1,-1, 0, 0, 0, 0, 0, 0, 0},  -- c+d=b
{  0, 0, 0, 0, 1, 0, 0,-1,-1, 0, 0, 0},  -- g+h=d
{ 40, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0},  -- e+f=40
{  0, 0, 0, 1, 0, 0,-1,-1, 0, 0, 0, 0},  -- f+g=c
{ 11, 0, 0, 0, 0, 1, 0, 0, 0,-1, 0, 0},  -- x+11=e
{  4, 0, 0, 0, 0, 0, 0, 1, 0, 0,-1, 0},  -- y+4=g
{ 11, 0, 0, 0, 0, 0, 1, 0, 0, 0,-1, 0},  -- 11+y=f
{  4, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,-1},  -- 4+z=h
{  0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1,-1}   -- x+z=y
--        })) -- {81,70,41,29,16,24,17,12,5,13,8}
})[-3..-1]) -- {5,13,8}
```

And finally, a cheeky little two-liner that does the whole job

```integer Y = (151-4*(11+4))/7, X = 40-2*11-Y, Z = Y-X
printf(1,"Two-liner: x=%d, y=%d, z=%d\n",{X,Y,Z})
```
Output:
```Two-liner: x=5, y=13, z=8
```

## Picat

Below are three different approaches using constraint modelling (cp solver).

### Using lists to represent the triangle

Translation of: Prolog
```import cp.

go =>
puzzle(T, X, Y,Z),
foreach(TT in T)
println(TT)
end,
println([x=X,y=Y,z=Z]),
nl,
fail, % are there any more solutions?
nl.

% Port of the Prolog solution
puzzle(Ts, X, Y, Z) :-
Ts =   [ [151],
[_, _],
[40, _, _],
[_, _, _, _],
[X, 11, Y, 4, Z]],
Y #= X + Z,
triangle(Ts),
Vs = vars(Ts),
Vs :: 0..10000,
solve(Vs).

triangle([T|Ts]) :-
( Ts = [N|_] -> triangle_(T, N), triangle(Ts) ; true ).

triangle_([], _).
triangle_([T|Ts],[A,B|Rest]) :-
T #= A + B, triangle_(Ts, [B|Rest]).```
Output:
```[151]
[81,70]
[40,41,29]
[16,24,17,12]
[5,11,13,4,8]
[x = 5,y = 13,z = 8]```

### Calculating the positions

Here is a constraint model which calculates the positions in the triangle that should be added together.

```import cp.

puzzle2 =>
N = 5, % number of rows
Len = (N*(N+1)) div 2, % number of entries

% The triangle numbers for 1..N
T = [I*(I+1) div 2 : I in 1..N],

% The index of first number to use in addition
% create the indices of the numbers to add,
foreach(I in 2..T[N-1])
% "jump" of 2 when i-1 is a triangle number
if membchk(I-1,T) then
else
end
end,

% the pyramid
MaxVal = 10_000,
L = new_list(Len),
L :: 1..MaxVal,

% The clues.
L =    [   151,
_, _,
40, _, _,
_, _,_ , _ ,
X, 11, Y, 4, Z
],

% The sums
foreach(I in 1..T[N-1])
end,

% The extra constraint
Y #= X + Z,

solve(L),
println([x=X,y=Y,z=Z]),
fail, % check if there is another solution
nl.```
Output:
`[x = 5,y = 13,z = 8]`

### Hard coded constraints

```import cp.

puzzle3 =>
%       1
%      2  3
%    4  5  6
%  7  8  9  10
%11 12 13 14 15

X = new_list(15),
X :: 0..10_000,
X[1] #= X[2]+X[3],
X[2] #= X[4]+X[5],
X[3] #= X[5]+X[6],
X[4] #= X[7]+X[8],
X[5] #= X[8]+X[9],
X[6] #= X[9]+X[10],
X[7] #= X[11]+X[12],
X[8] #= X[12]+X[13],
X[9] #= X[13]+X[14],
X[10] #= X[14]+X[15],
X[13] #= X[11] + X[15], % Y=X+Z,

% The hints
X[1] #= 151,
X[4] #= 40,
X[12] #= 11,
X[14] #= 4,

solve(X),
println([x=X[11],y=X[13],z=X[15]]),
fail,
nl.```
Output:
`[x = 5,y = 13,z = 8]`

## PicoLisp

```(be number (@N @Max)
(^ @C (box 0))
(repeat)
(or
((^ @ (>= (val (-> @C)) (-> @Max))) T (fail))
((^ @N (inc (-> @C)))) ) )

(be + (@A @B @Sum)
(^ @ (-> @A))
(^ @ (-> @B))
(^ @Sum (+ (-> @A) (-> @B))) )

(be + (@A @B @Sum)
(^ @ (-> @A))
(^ @ (-> @Sum))
(^ @B (- (-> @Sum) (-> @A)))
T
(^ @ (ge0 (-> @B))) )

(be + (@A @B @Sum)
(number @A @Sum)
(^ @B (- (-> @Sum) (-> @A))) )

#{
151
A   B
40  C   D
E  F  G    H
X  11  Y   4   Z
}#

(be puzzle (@X @Y @Z)
(+ @A @B 151)
(+ 40 @C @A)
(+ @C @D @B)
(+ @E @F 40)
(+ @F @G @C)
(+ @G @H @D)
(+ @X 11 @E)
(+ 11 @Y @F)
(+ @Y 4 @G)
(+ 4 @Z @H)
(+ @X @Z @Y)
T )```
Output:
```: (? (puzzle @X @Y @Z))
@X=5 @Y=13 @Z=8```

## Prolog

```:- use_module(library(clpfd)).

puzzle(Ts, X, Y, Z) :-
Ts =   [ [151],
[_, _],
[40, _, _],
[_, _, _, _],
[X, 11, Y, 4, Z]],
Y #= X + Z, triangle(Ts), append(Ts, Vs), Vs ins 0..sup, label(Vs).

triangle([T|Ts]) :- ( Ts = [N|_] -> triangle_(T, N), triangle(Ts) ; true ).

triangle_([], _).
triangle_([T|Ts], [A,B|Rest]) :- T #= A + B, triangle_(Ts, [B|Rest]).

% ?- puzzle(_,X,Y,Z).
% X = 5,
% Y = 13,
% Z = 8 ;
```

## PureBasic

Brute force solution.

```; Known;
; A.
;         [ 151]
;        [a ][b ]
;      [40][c ][d ]
;    [e ][f ][g ][h ]
;  [ X][11][ Y][ 4][ Z]
;
; B.
;  Y = X + Z

Procedure.i SolveForZ(x)
Protected a,b,c,d,e,f,g,h,z
For z=0 To 20
e=x+11: f=11+(x+z): g=(x+z)+4: h=4+z
If e+f=40
c=f+g : d=g+h: a=40+c: b=c+d
If a+b=151
ProcedureReturn z
EndIf
EndIf
Next z
ProcedureReturn -1
EndProcedure

Define x=-1, z=0, title\$="Pascal's triangle/Puzzle in PureBasic"
Repeat
x+1
z=SolveForZ(x)
Until z>=0
MessageRequester(title\$,"X="+Str(x)+#CRLF\$+"Y="+Str(x+z)+#CRLF\$+"Z="+Str(z))```

## Python

Works with: Python version 2.4+
```# Pyramid solver
#            [151]
#         [   ] [   ]
#      [ 40] [   ] [   ]
#   [   ] [   ] [   ] [   ]
#[ X ] [ 11] [ Y ] [ 4 ] [ Z ]
#  X -Y + Z = 0

def combine( snl, snr ):

cl = {}
if isinstance(snl, int):
cl['1'] = snl
elif isinstance(snl, string):
cl[snl] = 1
else:
cl.update( snl)

if isinstance(snr, int):
n = cl.get('1', 0)
cl['1'] = n + snr
elif isinstance(snr, string):
n = cl.get(snr, 0)
cl[snr] = n + 1
else:
for k,v in snr.items():
n = cl.get(k, 0)
cl[k] = n+v
return cl

def constrain(nsum, vn ):
nn = {}
nn.update(vn)
n = nn.get('1', 0)
nn['1'] = n - nsum
return nn

def makeMatrix( constraints ):
vmap = set()
for c in constraints:
vmap.update( c.keys())
vmap.remove('1')
nvars = len(vmap)
vmap = sorted(vmap)		# sort here so output is in sorted order
mtx = []
for c in constraints:
row = []
for vv in vmap:
row.append(float(c.get(vv, 0)))
row.append(-float(c.get('1',0)))
mtx.append(row)

if len(constraints) == nvars:
print 'System appears solvable'
elif len(constraints) < nvars:
print 'System is not solvable - needs more constraints.'
return mtx, vmap

def SolvePyramid( vl, cnstr ):

vl.reverse()
constraints = [cnstr]
lvls = len(vl)
for lvln in range(1,lvls):
lvd = vl[lvln]
for k in range(lvls - lvln):
sn = lvd[k]
ll = vl[lvln-1]
vn = combine(ll[k], ll[k+1])
if sn is None:
lvd[k] = vn
else:
constraints.append(constrain( sn, vn ))

print 'Constraint Equations:'
for cstr in constraints:
fset = ('%d*%s'%(v,k) for k,v in cstr.items() )
print ' + '.join(fset), ' = 0'

mtx,vmap = makeMatrix(constraints)

MtxSolve(mtx)

d = len(vmap)
for j in range(d):
print vmap[j],'=', mtx[j][d]

def MtxSolve(mtx):
# Simple Matrix solver...

mDim = len(mtx)			# dimension---
for j in range(mDim):
rw0= mtx[j]
f = 1.0/rw0[j]
for k in range(j, mDim+1):
rw0[k] *= f

for l in range(1+j,mDim):
rwl = mtx[l]
f = -rwl[j]
for k in range(j, mDim+1):
rwl[k] += f * rw0[k]

# backsolve part ---
for j1 in range(1,mDim):
j = mDim - j1
rw0= mtx[j]
for l in range(0, j):
rwl = mtx[l]
f = -rwl[j]
rwl[j]    += f * rw0[j]
rwl[mDim] += f * rw0[mDim]

return mtx

p = [ [151], [None,None], [40,None,None], [None,None,None,None], ['X', 11, 'Y', 4, 'Z'] ]
addlConstraint = { 'X':1, 'Y':-1, 'Z':1, '1':0 }
```
Output:
```Constraint Equations:
-1*Y + 1*X + 0*1 + 1*Z  = 0
-18*1 + 1*X + 1*Y  = 0
-73*1 + 5*Y + 1*Z  = 0
System appears solvable
X = 5.0
Y = 13.0
Z = 8.0
```

The Pyramid solver is not restricted to solving for 3 variables, or just this particular pyramid.

Alternative solution using the csp module (based on code by Gustavo Niemeyerby): http://www.fantascienza.net/leonardo/so/csp.zip

```from csp import Problem

p = Problem()
pvars = "R2 R3 R5 R6 R7 R8 R9 R10 X Y Z".split()
# 0-151 is the possible finite range of the variables
for sol in p.xsolutions():
print [sol[k] for k in "XYZ"]
```
Output:
`[5, 13, 8]`

## Racket

(Based on the clojure version)

Only X and Z are independent variables. We'll use a struct (cell v x z) to represent each cell, where the value is (v + x*X + z*Z).

```#lang racket/base
(require racket/list)

(struct cell (v x z) #:transparent)

(cell (+ (cell-v cx) (cell-v cy))
(+ (cell-x cx) (cell-x cy))
(+ (cell-z cx) (cell-z cy))))

(define (cell-sub cx cy)
(cell (- (cell-v cx) (cell-v cy))
(- (cell-x cx) (cell-x cy))
(- (cell-z cx) (cell-z cy))))
```

We first work bottom up and determine the value of each cell, starting from the bottom row.

```(define (row-above row) (map cell-add (drop row 1) (drop-right row 1)))

(define row0 (list (cell 0 1 0) (cell 11 0 0) (cell 0 1 1) (cell 4 0 0) (cell 0 0 1)))
(define row1 (row-above row0))
(define row2 (row-above row1))
(define row3 (row-above row2))
(define row4 (row-above row3))
```

We know the value of two additional cells, so by subtracting these values we get two equations of the form 0=v+x*X+z*Z. In the usual notation we get x*X+z*Z=-v, so v has the wrong sign.

```(define eqn40 (cell-sub (car row4) (cell 151 0 0)))
(define eqn20 (cell-sub (car row2) (cell 40 0 0)))
```

To solve the 2 equation system, we will use the Cramer's rule.

```(define (det2 eqnx eqny get-one get-oth)
(- (* (get-one eqnx) (get-oth eqny)) (* (get-one eqny) (get-oth eqnx))))

(define (cramer2 eqnx eqny get-val get-unk get-oth)
(/ (det2 eqnx eqny get-val get-oth)
(det2 eqnx eqny get-unk get-oth)))
```

To get the correct values of X, Y and Z we must change their signs.

```(define x (- (cramer2 eqn20 eqn40 cell-v cell-x cell-z)))
(define z (- (cramer2 eqn20 eqn40 cell-v cell-z cell-x)))

(displayln (list "X" x))
(displayln (list "Y" (+ x z)))
(displayln (list "Z" z))
```
Output:
```(X 5)
(Y 13)
(Z 8)
```

## Raku

(formerly Perl 6)

Translation of: Perl
```# set up triangle
my \$rows = 5;
my @tri = (1..\$rows).map: { [ { x => 0, z => 0, v => 0, rhs => Nil } xx \$_ ] }
@tri[0][0]<rhs> = 151;
@tri[2][0]<rhs> = 40;
@tri[4][0]<x> = 1;
@tri[4][1]<v> = 11;
@tri[4][2]<x> = 1;
@tri[4][2]<z> = 1;
@tri[4][3]<v> = 4;
@tri[4][4]<z> = 1;

# aggregate from bottom to top
for @tri - 2 ... 0 -> \$row {
for 0 ..^ @tri[\$row] -> \$col {
@tri[\$row][\$col]{\$_} = @tri[\$row+1][\$col]{\$_} + @tri[\$row+1][\$col+1]{\$_} for 'x','z','v';
}
}

# find equations
my @eqn = gather for @tri -> \$row {
for @\$row -> \$cell {
take [ \$cell<x>, \$cell<z>, \$cell<rhs> - \$cell<v> ] if defined \$cell<rhs>;
}
}

# print equations
say "Equations:";
say "  x +   z = y";
for @eqn -> [\$x,\$z,\$y] { say "\$x x + \$z z = \$y" }

# solve
my \$f = @eqn[0][1] / @eqn[1][1];
@eqn[0][\$_] -=  \$f * @eqn[1][\$_] for 0..2;
\$f = @eqn[1][0] / @eqn[0][0];
@eqn[1][\$_] -=  \$f * @eqn[0][\$_] for 0..2;

# print solution
say "Solution:";
my \$x = @eqn[0][2] / @eqn[0][0];
my \$z = @eqn[1][2] / @eqn[1][1];
my \$y = \$x + \$z;
say "x=\$x, y=\$y, z=\$z";
```
Output:
```Equations:
x +   z = y
7 x + 7 z = 91
2 x + 1 z = 18
Solution:
x=5, y=13, z=8```

## REXX

This REXX version also displays a diagram of the puzzle.

```/*REXX program solves a   (Pascal's)   "Pyramid of Numbers"   puzzle given four values. */
/* ╔══════════════════════════════════════════════════╗
║                            /                     ║
║              mid          /                      ║
║                 \       151                      ║
║                  \   ααα   ααα                   ║
║                   40    ααα   ααα                ║
║               ααα   ααα   ααα   ααα              ║
║              x    11     y     4     z           ║
║                  /              \                ║
║ find:           /                \               ║
║ x y z          b                  d              ║
╚══════════════════════════════════════════════════╝ */
do #=2;     _= sourceLine(#);  n= pos('_', _) /* [↓]  this DO loop shows (above) box.*/
if n\==0  then leave;             say _       /*only display  up to  the above line. */
end   /*#*/;                      say         /* [↑]  this is a way for in─line doc. */
parse arg  b  d  mid  answer  .                  /*obtain optional variables from the CL*/
if     b=='' |      b==","  then      b=  11     /*Not specified?  Then use the default.*/
if     d=='' |      d==","  then      d=   4     /* "      "         "   "   "     "    */
if    mid='' |    mid==","  then    mid=  40     /* "      "         "   "   "     "    */
big= answer - 4*b - 4*d    /*calculate  BIG  number less constants*/
do      x=-big  to big
do    y=-big  to big
if x+y\==mid - 2*b  then iterate            /*40 = x+2B+Y   ──or──   40-2*11 = x+y */
do z=-big  to big
if z \== y - x   then iterate            /*Z  has to equal   Y-X       (Y= X+Z) */
if x+y*6+z==big  then say right('x =', n)  x  right("y =",n)  y  right('z =',n)  z
end   /*z*/
end      /*y*/
end        /*x*/                              /*stick a fork in it,  we're all done. */
```
output   when using the default inputs:
```      /* ╔══════════════════════════════════════════════════╗
║                            /                     ║
║              mid          /                      ║
║                 \       151                      ║
║                  \   ααα   ααα                   ║
║                   40    ααα   ααα                ║
║               ααα   ααα   ααα   ααα              ║
║              x    11     y     4     z           ║
║                  /              \                ║
║ find:           /                \               ║
║ x y z          b                  d              ║
╚══════════════════════════════════════════════════╝ */

x = 5              y = 13              z = 8
```

## Ruby

```require 'rref'

pyramid = [
[ 151],
[nil,nil],
[40,nil,nil],
[nil,nil,nil,nil],
["x", 11,"y", 4,"z"]
]
pyramid.each{|row| p row}

equations = [[1,-1,1,0]]   # y = x + z

def parse_equation(str)
eqn = [0] * 4
lhs, rhs = str.split("=")
eqn[3] = rhs.to_i
for term in lhs.split("+")
case term
when "x" then eqn[0] += 1
when "y" then eqn[1] += 1
when "z" then eqn[2] += 1
else          eqn[3] -= term.to_i
end
end
eqn
end

-2.downto(-5) do |row|
pyramid[row].each_index do |col|
val = pyramid[row][col]
sum = "%s+%s" % [pyramid[row+1][col], pyramid[row+1][col+1]]
if val.nil?
pyramid[row][col] = sum
else
equations << parse_equation(sum + "=#{val}")
end
end
end

reduced = convert_to(reduced_row_echelon_form(equations), :to_i)

for eqn in reduced
if eqn[0] + eqn[1] + eqn[2] != 1
fail "no unique solution! #{equations.inspect} ==> #{reduced.inspect}"
elsif eqn[0] == 1 then x = eqn[3]
elsif eqn[1] == 1 then y = eqn[3]
elsif eqn[2] == 1 then z = eqn[3]
end
end

puts
puts "x == #{x}"
puts "y == #{y}"
puts "z == #{z}"

for row in pyramid
answer << row.collect {|cell| eval cell.to_s}
end
puts
```
Output:
```[151]
[nil, nil]
[40, nil, nil]
[nil, nil, nil, nil]
["x", 11, "y", 4, "z"]

x == 5
y == 13
z == 8

[151]
[81, 70]
[40, 41, 29]
[16, 24, 17, 12]
[5, 11, 13, 4, 8]
```

## Scala

```object PascalTriangle extends App {

val (x, y, z) = pascal(11, 4, 40, 151)

def pascal(a: Int, b: Int, mid: Int, top: Int): (Int, Int, Int) = {
val y = (top - 4 * (a + b)) / 7
val x = mid - 2 * a - y
(x, y, y - x)
}

println(if (x != 0) s"Solution is: x = \$x, y = \$y, z = \$z" else "There is no solution.")
}
```
Output:

See it in running in your browser by (JavaScript)

or by Scastie (JVM).

## Sidef

Translation of: Raku
```# set up triangle
var rows = 5
var tri = rows.of {|i| (i+1).of { Hash(x => 0, z => 0, v => 0, rhs => nil) } }
tri[0][0]{:rhs} = 151
tri[2][0]{:rhs} = 40
tri[4][0]{:x} = 1
tri[4][1]{:v} = 11
tri[4][2]{:x} = 1
tri[4][2]{:z} = 1
tri[4][3]{:v} = 4
tri[4][4]{:z} = 1

# aggregate from bottom to top
for row in (tri.len ^.. 1) {
for col in (^tri[row-1]) {
[:x, :z, :v].each { |key|
tri[row-1][col]{key} = (tri[row][col]{key} + tri[row][col+1]{key})
}
}
}

# find equations
var eqn = gather {
for r in tri {
for c in r {
take([c{:x}, c{:z}, c{:rhs} - c{:v}]) if defined(c{:rhs})
}
}
}

# print equations
say "Equations:"
say " x +  z = y"
for x,z,y in eqn { say "#{x}x + #{z}z = #{y}" }

# solve
var f = (eqn[0][1] / eqn[1][1])
{|i| eqn[0][i] -= (f * eqn[1][i]) } << ^3
f = (eqn[1][0] / eqn[0][0])
{|i| eqn[1][i] -= (f * eqn[0][i]) } << ^3

# print solution
say "Solution:"
var x = (eqn[0][2] / eqn[0][0])
var z = (eqn[1][2] / eqn[1][1])
var y = (x + z)
say "x=#{x}, y=#{y}, z=#{z}"
```
Output:
```Equations:
x +  z = y
7x + 7z = 91
2x + 1z = 18
Solution:
x=5, y=13, z=8
```

## SystemVerilog

We can view this as a problem of generating a set of random numbers that satisfy the constraints. Because there is only one solution, the result isn't very random...

```program main;

class Triangle;
rand bit [7:0] a,b,c,d,e,f,g,h,X,Y,Z;

function new();
randomize;
\$display("     [%0d]",                151);
\$display("    [%0d][%0d]",            a, b);
\$display("   [%0d][%0d][%0d]",        40,c,d);
\$display("  [%0d][%0d][%0d][%0d]",    e,f,g,h);
\$display(" [%0d][%0d][%0d][%0d][%0d]",X,11,Y,4,Z);
endfunction

constraint structure {
151 == a + b;

a == 40 + c;
b == c + d;

40 == e + f;
c == f + g;
d == g + h;

e == X + 11;
f == 11 + Y;
g == Y + 4;
h == 4 + Z;
};

constraint extra {
Y == X + Z;
};

endclass

endprogram
```
```     [151]
[81][70]
[40][41][29]
[16][24][17][12]
[5][11][13][4][8]
```

## Tcl

using code from Reduced row echelon form#Tcl

```package require Tcl 8.5
namespace path ::tcl::mathop

set pyramid {
{151.0 "" "" "" ""}
{"" "" "" "" ""}
{40.0 "" "" "" ""}
{"" "" "" "" ""}
{x 11.0 y 4.0 z}
}

set equations {{1 -1 1 0}}

proc simplify {terms val} {
set vars {0 0 0}
set x 0
set y 1
set z 2
foreach term \$terms {
switch -exact -- \$term {
x - y - z {
lset vars [set \$term] [+ 1 [lindex \$vars [set \$term]]]
}
default {
set val [- \$val \$term]
}
}
}
return [concat \$vars \$val]
}

for {set row [+ [llength \$pyramid] -2]} {\$row >= 0} {incr row -1} {
for {set cell 0} {\$cell <= \$row} {incr cell } {
set sum [concat [lindex \$pyramid [+ 1 \$row] \$cell] [lindex \$pyramid [+ 1 \$row] [+ 1 \$cell]]]
if {[set val [lindex \$pyramid \$row \$cell]] ne ""} {
lappend equations [simplify \$sum \$val]
} else {
lset pyramid \$row \$cell  \$sum
}
}
}

set solution [toRREF \$equations]
foreach row \$solution {
lassign \$row a b c d
if {\$a + \$b + \$c > 1} {
error "problem does not have a unique solution"
}
if {\$a} {set x \$d}
if {\$b} {set y \$d}
if {\$c} {set z \$d}
}
puts "x=\$x"
puts "y=\$y"
puts "z=\$z"

foreach row \$pyramid {
set newrow {}
foreach cell \$row {
if {\$cell eq ""} {
lappend newrow ""
} else {
lappend newrow [expr [join [string map [list x \$x y \$y z \$z] \$cell] +]]
}
}
lappend solved \$newrow
}
print_matrix \$solved
```
```x=5.0
y=13.0
z=8.0
151.0
81.0 70.0
40.0 41.0 29.0
16.0 24.0 17.0 12.0
5.0 11.0 13.0  4.0 8.0```

## Wren

Translation of: Kotlin
Library: Wren-fmt
```import "./fmt" for Fmt

var isIntegral = Fn.new { |x, tol| x.fraction.abs <= tol }

var pascal = Fn.new { |a, b, mid, top|
var yd = (top - 4 * (a + b)) / 7
if (!isIntegral.call(yd, 0.0001)) return [0, 0, 0]
var y = yd.truncate
var x = mid - 2*a - y
return [x, y, y - x]
}

var sol = pascal.call(11, 4, 40, 151)
if (sol[0] != 0) {
Fmt.print("Solution is: x = \$d, y = \$d, z = \$d", sol[0], sol[1], sol[2])
} else {
System.print("There is no solution")
}
```
Output:
```Solution is: x = 5, y = 13, z = 8
```

## XPL0

```proc Print; int N, A, B, C, D, E;
int  I, P;
def  Tab = \$09;
[P:= @A;        \point to first number
for I:= N to 5-1 do ChOut(0, Tab);
for I:= 0 to N-1 do
[IntOut(0, P(I));  ChOut(0, Tab);  ChOut(0, Tab)];
CrLf(0);
];

int N, P, Q, R, S, T, U, V, W, X, Y, Z; \       151
[for X:= 0 to 40-11 do                  \      N   P
for Z:= 0 to 151-4 do               \    Q   R   S
[Y:= X+Z;                       \  T   U   V   W
T:= X+11;                       \X   11  Y   4   Z
U:= 11+Y;
V:= Y+4;
W:= 4+Z;
if T+U = 40 then
[R:= U+V;
S:= V+W;
N:= 40+R;
P:= R+S;
if N+P = 151 then
[Print(1, 151);
Print(2, N, P);
Print(3, 40, R, S);
Print(4, T, U, V, W);
Print(5, X, 11, Y, 4, Z);
exit;
];
];
];
]```
Output:
```                                151
81              70
40              41              29
16              24              17              12
5               11              13              4               8
```

## zkl

Translation of: Python
```# Pyramid solver
#            [151]
#         [   ] [   ]
#      [ 40] [   ] [   ]
#   [   ] [   ] [   ] [   ]
#[ X ] [ 11] [ Y ] [ 4 ] [ Z ]
# Known: X - Y + Z = 0

p:=T( L(151), L(Void,Void), L(40,Void,Void), L(Void,Void,Void,Void),
L("X", 11, "Y", 4, "Z") );
addlConstraint:=Dictionary( "X",1, "Y",-1, "Z",1, "1",0 );
```fcn solvePyramid([List]vl,[Dictionary]cnstr){  //ListOfLists,Hash-->zip
vl=vl.reverse();
constraints:=L(cnstr);
lvls:=vl.len();
foreach lvln in ([1..lvls-1]){
lvd:=vl[lvln];
foreach k in (lvls-lvln){
sn:=lvd[k];
ll:=vl[lvln-1];
vn:=combine(ll[k], ll[k+1]);
if(Void==sn) lvd[k]=vn;
else constraints.append(constrainK(sn,vn));
}
}
println("Constraint Equations:");
constraints.pump(Console.println,fcn(hash){
hash.pump(List,fcn([(k,v)]){"%d*%s".fmt(v,k)}).concat(" + ") + " = 0"
});

mtx,vmap:=makeMatrix(constraints);
mtxSolve(mtx);

d:=vmap.len();
foreach j in (d){ println(vmap[j]," = ", mtx[j][d]); }
}

fcn [mixin=Dictionary] constrainK([Int]nsum,[Dictionary]vn){ //-->new hash of old hash, sum K
nn:=vn.copy(); nn["1"]=nn.find("1",0) - nsum;
}

fcn combine(snl,snr){ //Int|String|Hash *2 --> new Hash
cl:=Dictionary();
if(snl.isInstanceOf(Int))         cl["1"]=snl;
else if(snl.isInstanceOf(String)) cl[snl]=1;
else				     cl     =snl.copy();

if(snr.isInstanceOf(Int))         cl["1"]=cl.find("1",0) + snr;
else if(snr.isInstanceOf(String)) cl[snr]=cl.find(snr,0) + 1;
else{ foreach k,v in (snr){ 	     cl[k]  =cl.find(k,0)   + v; } }
}

//-->(listMatrix(row(X,Y,Z,c),row...),List("X","Y","Z"))
fcn makeMatrix([Dictionary]constraints){
vmap:=Dictionary();// create a sorted list of the variable names in constraints
foreach c in (constraints){ vmap.extend(c) }  // no duplicate names
vmap.del("1"); vmap=vmap.keys.sort();  # sort here so output is in sorted order

mtx:=constraints.pump(List,'wrap(c){ // create list of [writeable] rows
vmap.pump(List, c.find.fp1(0),"toFloat").copy()
.append(-c.find("1",0).toFloat())
}).copy();

nvars:=vmap.len();
if(constraints.len()==nvars) println("System appears solvable");
else if(constraints.len()<nvars)
println("System is not solvable - needs more constraints.");
return(mtx,vmap);
}

fcn mtxSolve([List]mtx){ //munge mtx	# Simple Matrix solver...
mDim:=mtx.len();			# num rows
foreach j in (mDim){
rw0:=mtx[j];
f:=1.0/rw0[j];
foreach k in ([j..mDim]){ rw0[k]=rw0[k]*f }
foreach l in ([j+1..mDim-1]){
rwl:=mtx[l]; f:=-rwl[j];
foreach k in ([j..mDim]){ rwl[k]+=f*rw0[k] }
}
}
# backsolve part ---
foreach j1 in ([1..mDim-1]){
j:=mDim - j1; rw0:=mtx[j];
foreach l in (j){
rwl:=mtx[l]; f:=-rwl[j];
rwl[j]   +=f*rw0[j];
rwl[mDim]+=f*rw0[mDim];
}
}
return(mtx);
}```
Output:
```Constraint Equations:
0*1 + 1*X + -1*Y + 1*Z = 0
-18*1 + 1*X + 1*Y = 0
-73*1 + 5*Y + 1*Z = 0
System appears solvable
X = 5
Y = 13
Z = 8
```