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
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. <lang ada>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;</lang>
- Output:
151 81 70 40 41 29 16 24 17 12 5 11 13 4 8
ALGOL 68
<lang algol68>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</lang>
- 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: <lang autohotkey>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%</lang> 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. <lang autohotkey>;---------------------------------------------------------------------------
- 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("")
}</lang>
BBC BASIC
<lang bbcbasic> 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)</lang>
- Output:
X = 5 Y = 13 Z = 8
C#
<lang c sharp> 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 = "); int v = int.Parse(Console.ReadLine());
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> . . . "); k = Console.ReadKey(true); } }
//
// 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(); } }
} } </lang>
- 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
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.
<lang 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;
} </lang>
- 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. <lang c>#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; }</lang>
- Output:
pass 1: 12pass 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 17 125 11 13 4 8
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. <lang clojure>(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))))</lang> 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. <lang clojure>(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]))</lang> In this case, there are only two variables, so solving the system of linear equations is simple. <lang clojure>(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))</lang>
If you want to solve the whole pyramid, just add a call (show-pyramid x z) to the previous let form: <lang clojure>(defn dot [v1 v2] (reduce + (map * v1 v2)))
(defn show-pyramid [x z]
(doseq [row rows] (println (map #(dot [1 x z] %) row)))</lang>
Curry
<lang curry>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</lang>
- Output:
Execution time: 0 msec. / elapsed: 0 msec. [(5,13,8)]
D
<lang d>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);
}</lang>
- 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
<lang fsharp>
- 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]</lang>
- Output:
x = 5.000000, Y = 13.000000, Z = 8.000000
Factor
<lang factor>USING: arrays backtrack combinators.extras fry grouping.extras interpolate io kernel math math.ranges sequences ; IN: rosetta-code.pascals-triangle-puzzle
- 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</lang>
- Output:
X = 5, 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. <lang go>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))
}</lang>
- 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
I assume the task is to solve any such puzzle, i.e. given some data
<lang haskell>puzzle = [["151"],["",""],["40","",""],["","","",""],["X","11","Y","4","Z"]]</lang>
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:
<lang haskell>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</lang>
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
<lang haskell>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</lang>
will output one special solution and modifications that lead to more solutions, as in
<lang haskell>*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]]</lang>
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:
<lang j>chk=:40 151&-:@(2 4{{."1)</lang>
verb for the base of the pyramid:
<lang j>base=: [,11,+,4,]</lang>
the height of the pyramid:
<lang j>ord=:5</lang>
=> 'chk', 'base' and 'ord' are the knowledge rules abstracted from the problem definition.
The J-sentence that solves the puzzle is:
<lang j> |."2(#~chk"2) 2(+/\)^:(<ord)"1 base/"1>,{ ;~i:28</lang>
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:
<lang j>,.(1+i.5)<@{."0 1{.|."2(#~chk"2) 2(+/\)^:(<ord)"1 base/"1>,{ ;~i:28</lang> or <lang j>,.(<@{."0 1~1+i.@#){.|."2(#~chk"2) 2(+/\)^:(<ord)"1 base/"1>,{ ;~i:28</lang>
+-----------+ |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.
<lang java> 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++ ) {
result.add(matrix.replaceColumn(b, i).determinant() / denominator);
}
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) {
matrix.add(list);
}
}
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;
}
list.add(get(i, j));
}
mat.add(list);
}
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);
}
list.add(value);
}
mat.add(list);
}
return new Matrix(mat);
}
private double get(int row, int col) {
return matrix.get(row).get(col);
}
}
} </lang>
- 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
Julia
<lang julia>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</lang>
- Output:
Solution: x = 5, y = 13, z = 8.
Kotlin
<lang scala>// 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")
}</lang>
- Output:
Solution is: x = 5, y = 13, z = 8
Maple
<lang Maple> sys := {22 + x + y = 40, 78 + 5*y + z = 151, x + z = y}: solve(sys, {x,y,z}); </lang>
- Output:
{x = 5, y = 13, z = 8}
Mathematica
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: <lang Mathematica>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</lang> And we have the knowns <lang Mathematica>a->151 d->40 l->11 n->4</lang> Giving us 10 equations with 10 unknowns; i.e. solvable. So we can do so by: <lang Mathematica>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}]</lang> gives back: <lang Mathematica>{{b -> 81, c -> 70, e -> 41, f -> 29, g -> 16, h -> 24, i -> 17, j -> 12, X -> 5, Y -> 13, Z -> 8}}</lang> In pyramid form that would be: <lang Mathematica> 151 81 70 40 41 29 16 24 17 12 5 11 13 4 8</lang>
An alternative solution in Mathematica 10, constructing the triangle: <lang Mathematica>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}]</lang> Three equations and three unknowns, which gives back: <lang Mathematica>{{x -> 5, y -> 13, z -> 8}}</lang>
Nim
Translation of Ada solution: <lang nim>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 = $(N.Known + N.X * B_X + N.Y * B_Y + N.Z * B_Z)
proc Solve_2x2 (A11: int, A12:int, B1:int, A21:int, A22:int, B2: int) =
# Don't care about things, supposing an integer solution exists
if A22 == 0:
B_X = toInt(B2 / A21)
B_Y = toInt((B1 - (A11*B_X)) / A12)
else:
B_X = toInt((B1*A22 - B2*A12) / (A11*A22 - A21*A12))
B_Y = toInt((B1 - A11*B_X) / A12)
B_Z = B_Y - B_X
var B : array [1..5, array[1..5, Block_Value]] # The lower triangle contains blocks
- The bottom blocks
Add(B[5][1],X) Add(B[5][2],11) Add(B[5][3],Y) Add(B[5][4],4) Add(B[5][5],Z)
- Upward run
for Row in countdown(4,1):
for Column in 1 .. Row:
Add (B[Row][Column], B[Row + 1][Column])
Add (B[Row][Column], B[Row + 1][Column + 1])
- Now have known blocks 40=[3][1], 151=[1][1] and Y=X+Z to determine X,Y,Z
Solve_2x2( B[1][1].X,
B[1][1].Y,
151 - B[1][1].Known,
B[3][1].X,
B[3][1].Y,
40 - B[3][1].Known)
- Print the results
for Row in 1..5:
writeln(stdout,"")
for Column in 1..Row:
write(stdout, Image(B[Row][Column]), " ")</lang>
- Output:
151 81 70 40 41 29 16 24 17 12 5 11 13 4 8
Oz
<lang oz>%% to compile : ozc -x <file.oz> functor
import
System Application FD Search
define
proc{Quest Root Rules}
proc{Limit Rc Ls}
case Ls of nil then skip
[] X|Xs then
{Limit Rc Xs}
case X of N#V then
Rc.N =: V
[] N1#N2#N3 then
Rc.N1 =: Rc.N2 + Rc.N3
end
end
end
proc {Pyramid R}
{FD.tuple solution 15 0#FD.sup R} %% non-negative integers domain
%% 01 , pyramid format
%% 02 03
%% 04 05 06
%% 07 08 09 10
%% 11 12 13 14 15
R.1 =: R.2 + R.3 %% constraints of Pyramid of numbers
R.2 =: R.4 + R.5
R.3 =: R.5 + R.6
R.4 =: R.7 + R.8
R.5 =: R.8 + R.9
R.6 =: R.9 + R.10
R.7 =: R.11 + R.12
R.8 =: R.12 + R.13
R.9 =: R.13 + R.14
R.10 =: R.14 + R.15
{Limit R Rules} %% additional constraints
{FD.distribute ff R}
end
in
{Search.base.one Pyramid Root} %% search for solution
end
local
Root R
in
{Quest Root [1#151 4#40 12#11 14#4 13#11#15]} %% supply additional constraint rules
if {Length Root} >= 1 then
R = Root.1
{For 1 15 1
proc{$ 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</lang>
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... <lang parigp> 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); } </lang>
I'm thinking of one to solve all puzzles regardless of size and positions. but the objective was to solve this puzzle.
Perl
<lang 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; </lang>
- Output:
Equations: x + z = y 7 x + 7 z = 91 2 x + 1 z = 18 Solution: x=5, y=13, z=8
Perl 6
<lang perl6># 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";</lang>
- Output:
Equations: x + z = y 7 x + 7 z = 91 2 x + 1 z = 18 Solution: x=5, y=13, z=8
Phix
Converts the pyramid to rules quite nicely, however I will concede
that solving those two rules (18=2x+z and 73=5x+6z) and specifically converting them
into xrule(35=7x) and zrule(56=7z) is somewhat amateurish - suggestions welcome.
UPDATE: See Solving_coin_problems#Phix, routine solveN()
<lang Phix>-- demo\rosetta\Pascal_triangle_Puzzle.exw
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 one x
elsif prc="y" then prc = {0,1,1} -- ie one x plus one z
elsif prc="z" then prc = {0,0,1} -- ie 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}
equ = sq_add(pyramid[r+1][c],pyramid[r+1][c+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,1}) ?"equations" pp(pyramid) ?"rules" pp(rules) puts(1,"=====\n")
if length(rules)!=2 then ?9/0 end if -- more work needed!?
-- admittedly this bit is rather amateurish, and maybe problem-specific: -- UPDATE: as above, see that solveN() routine. sequence xrule = sq_sub(sq_mul(rules[1],rules[2][3]),sq_mul(rules[2],rules[1][3])),
zrule = sq_sub(sq_mul(rules[2],rules[1][2]),sq_mul(rules[1],rules[2][2]))
?{"xrule",xrule} ?{"zrule",zrule}
integer x = xrule[1]/xrule[2],
z = zrule[1]/zrule[3],
y = x+z
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)</lang>
- 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}}
=====
{"xrule",{35,7,0}}
{"zrule",{56,0,7}}
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.
PicoLisp
<lang 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 )</lang>
- Output:
: (? (puzzle @X @Y @Z)) @X=5 @Y=13 @Z=8
Prolog
<lang 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 ;</lang>
PureBasic
Brute force solution. <lang PureBasic>; 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))</lang>
Python
<lang python># 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)</lang>
- 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
<lang python>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"]</lang>
- 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>
- 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))))
</lang>
We first work bottom up and determine the value of each cell, starting from the bottom row. <lang Racket> (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)) </lang>
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.
<lang Racket> (define eqn40 (cell-sub (car row4) (cell 151 0 0))) (define eqn20 (cell-sub (car row2) (cell 40 0 0))) </lang>
To solve the 2 equation system, we will use the Cramer's rule. <lang Racket> (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)))
</lang>
To get the correct values of X, Y and Z we must change their signs. <lang Racket> (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)) </lang>
- Output:
(X 5) (Y 13) (Z 8)
REXX
This REXX version also displays a diagram of the puzzle. <lang rexx>/*REXX program solves a (Pascal's) "Pyramid of Numbers" puzzle given four values. */
/* ╔══════════════════════════════════════════════════╗
║ answer ║
║ / ║
║ 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 /* " " " " " " */ if answer= | answer=="," then answer= 151 /* " " " " " " */
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. */</lang>
- output when using the default inputs:
/* ╔══════════════════════════════════════════════════╗
║ answer ║
║ / ║
║ 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
uses Reduced row echelon form#Ruby <lang 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}</lang>
- 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
<lang 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.")
}</lang>
- Output:
See it in running in your browser by (JavaScript)
or by Scastie (JVM).
Sidef
<lang ruby># 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])
f = (eqn[1][0] / eqn[0][0])- print solution
- 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... <lang SystemVerilog>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</lang>
[151]
[81][70]
[40][41][29]
[16][24][17][12]
[5][11][13][4][8]
Tcl
using code from Reduced row echelon form#Tcl <lang 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 Template: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</lang>
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
<lang zkl># 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);</lang> <lang zkl>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);
}</lang>
- 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