Pascal's triangle/Puzzle

From Rosetta Code
Task
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).


Task

Write a program to find a solution to this puzzle.

Ada[edit]

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;
end Add;
procedure Add (L : in out Block_Value; R : Integer) is
begin -- Symbolically adds a value to the block
L.Known := L.Known + R;
end Add;
 
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
Add (B(5,1),X); Add (B(5,2),11); Add (B(5,3),Y); Add (B(5,4),4); Add (B(5,5),Z);
 
-- 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[edit]

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[edit]

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
Gui, Add, Edit, x%xx% y%yy% w50 v%vv% Center ReadOnly -TabStop
}
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
Gui, Add, StatusBar
 
; 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
 
 
 
;---------------------------------------------------------------------------
ButtonClear: ; restart without the blue clues
;---------------------------------------------------------------------------
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, %Disable%ReadOnly, Edit11
GuiControl, %Disable%ReadOnly, Edit15
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, %Disable%ReadOnly, Edit1
GuiControl, %Disable%ReadOnly, Edit4
GuiControl, %Disable%ReadOnly, Edit12
GuiControl, %Disable%ReadOnly, Edit14
 
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")
SB_SetText("restart without the blue clues")
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[edit]

      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[edit]

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[edit]

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

Clojure[edit]

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)))

Curry[edit]

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[edit]

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#[edit]

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

Go[edit]

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
}
 
// add two expressions
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
func addRow(l []expr) []expr {
if len(l) == 0 {
panic("wrong")
}
r := make([]expr, len(l)-1)
for i := range r {
r[i] = addExpr(l[i], l[i+1])
}
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
r4 := addRow(r5)
fmt.Println("next row up:", r4)
r3 := addRow(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
r2 := addRow(r3)
fmt.Println("next row up:", r2)
r1 := addRow(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)
 
// show final answers
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

Haskell[edit]

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[edit]

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|
 +-----------+

Kotlin[edit]

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

Mathematica[edit]

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}}

Nim[edit]

Translation of Ada solution:

import math, strutils
 
var B_X, B_Y, B_Z : int = 0
 
type
Block_Value = object
Known  : int
X, Y, Z : int
 
let
X: Block_Value = Block_Value(Known:0, X:1, Y:0, Z:0)
Y: Block_Value = Block_Value(Known:0, X:0, Y:1, Z:0)
Z: Block_Value = Block_Value(Known:0, X:0, Y:0, Z:1)
 
proc Add (L : var Block_Value, R : Block_Value) =
# 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
 
proc Add (L: var Block_Value, R: int) =
# Symbolically adds a value to the block
L.Known = L.Known + R
 
proc Image (N : Block_Value): string =
# The block value, when X,Y,Z are known
result = $(N.Known + N.X * B_X + N.Y * B_Y + N.Z * B_Z)
 
proc Solve_2x2 (A11: int, A12:int, B1:int, A21:int, A22:int, B2: int) =
# Don't care about things, supposing an integer solution exists
if A22 == 0:
B_X = toInt(B2 / A21)
B_Y = toInt((B1 - (A11*B_X)) / A12)
else:
B_X = toInt((B1*A22 - B2*A12) / (A11*A22 - A21*A12))
B_Y = toInt((B1 - A11*B_X) / A12)
B_Z = B_Y - B_X
 
var B : array [1..5, array[1..5, Block_Value]] # The lower triangle contains blocks
 
# The bottom blocks
Add(B[5][1],X)
Add(B[5][2],11)
Add(B[5][3],Y)
Add(B[5][4],4)
Add(B[5][5],Z)
 
# Upward run
for Row in countdown(4,1):
for Column in 1 .. Row:
Add (B[Row][Column], B[Row + 1][Column])
Add (B[Row][Column], B[Row + 1][Column + 1])
 
# 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:
writeln(stdout,"")
for Column in 1..Row:
write(stdout, Image(B[Row][Column]), " ")
Output:
151 
81 70 
40 41 29 
16 24 17 12 
5 11 13 4 8

Oz[edit]

%% 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[edit]

[ 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[edit]

# 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

Perl 6[edit]

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

PicoLisp[edit]

(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) )
Output:
: (? (puzzle @X @Y @Z))
 @X=5 @Y=13 @Z=8

Prolog[edit]

:- 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[edit]

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[edit]

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 }
SolvePyramid( p, addlConstraint)
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
p.addvars(pvars, xrange(152))
p.addrule("R7 == X + 11")
p.addrule("R8 == Y + 11")
p.addrule("R9 == Y + 4")
p.addrule("R10 == Z + 4")
p.addrule("R7 + R8 == 40")
p.addrule("R5 == R8 + R9")
p.addrule("R6 == R9 + R10")
p.addrule("R2 == 40 + R5")
p.addrule("R3 == R5 + R6")
p.addrule("R2 + R3 == 151")
p.addrule("Y == X + Z")
for sol in p.xsolutions():
print [sol[k] for k in "XYZ"]
Output:
[5, 13, 8]

Racket[edit]

(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)
 
(define (cell-add cx cy)
(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)

REXX[edit]

/*REXX program solves a  "Pyramid of Numbers"  puzzle given four values.*/
/*┌──────────────────────────────────────────────┐
┌─┘ └─┐
│ answer │
│ mid / │
│ \ / │
│ \ 151 │
│ \ ααα ααα │
│ 40 ααα ααα │
│ ααα ααα ααα ααα │
│ x 11 y 4 z │
│ / \ │
│ / \ │
│ / \ │
│ B D │
└─┐ ┌─┘
└──────────────────────────────────────────────┘*/

parse arg x b y d z mid answer . /*get some values, others, just X*/
pad=left('',15) /*for inserting spaces in output.*/
top=answer - 4*b - 4*d /*calculate the top # - constants*/
middle=mid - 2*b /*calculate the mod # - constants*/
 
do x =-top to top
do y=-top to top
if x+y\==middle then iterate /*40 = x+2B+Y -or- 40-2*11 =x+y*/
y6=y*6 /*calculate a short cut. */
do z=-top to top
if z\==y-x then iterate /*z has to equal y-x (y=x+z) */
if x+y6+z==top then say pad 'x = ' x pad "y = " y pad 'z = ' z
end /*z*/
end /*y*/
end /*x*/
/*stick a fork in it, we're done.*/
Output:
using the following input: x 11 y 4 z 40 151
                x =  5                 y =  13                 z =  8

Ruby[edit]

uses Reduced row echelon form#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}"
 
answer = []
for row in pyramid
answer << row.collect {|cell| eval cell.to_s}
end
puts
answer.each{|row| p row}
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]

Sidef[edit]

Translation of: Perl 6
# 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[edit]

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
 
Triangle answer = new;
endprogram
     [151]
    [81][70]
   [40][41][29]
  [16][24][17][12]
 [5][11][13][4][8]

Tcl[edit]

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

zkl[edit]

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 );
solvePyramid(p, addlConstraint);
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;
return(nn.makeReadOnly());
}
 
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; } }
return(cl.makeReadOnly())
}
 
//-->(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