A Triangular Slice of Squared Pi

Question

Inspired by Bake a slice of Pi

Challenge

Given input 3 <= n <= 100 and 3 <= y <= n, construct an n x n matrix of the decimal portion of pi (14159...), starting in the top left. Then, take the upper-right triangle of size y x y and concatenate it together. Output the resulting number.

For example, for input n = 5, y = 3, the following matrix is constructed

14159
26535
89793
23846
26433

Then, the upper-right 3 x 3 triangle would be

159
 35
  3

so 159353 is the output.

Input

Two integers -- n representing the size of the square matrix of the digits of pi, and y representing the upper-right triangle -- in any convenient format.

Output

• The resulting sliced and concatenated number, either printed/displayed to the screen, returned as a string, etc.
• Trailing/leading whitespace is optional, so long as there's no whitespace in the output (i.e., 159 35 3 or the like would be invalid).
• Note that since we're explicitly looking for the digits of pi, and not an approximation or mathematical calculation, answers should not round the final digit of the matrix.

Rules

• This is code-golf so all the usual rules for golfing apply, and the shortest code (in bytes) wins.
• Either a full program or function are acceptable.
• Standard loopholes are forbidden.

Examples

 n  y  output
-------------
 3  3  141923
 5  3  159353
 6  4  1592589383
 6  6  141592535893238643794
20 12  358979323846950288419715820974944628620899211706792306647223172745025559196615

Answer 1

05AB1E, 19 bytes

Uses CP-1252 encoding.

nžs¦¦¹ôI£íRvyN>£J}R

Try it online!

Explanation

n=5, y=3 used for example

nžs                  # push pi to n^2 digits
                     # STACK: 3.1415926535897932384626433
   ¦¦                # remove the first 2 chars
                     # STACK: 1415926535897932384626433
     ¹ô              # split into n*n matrix
                     # STACK: ['14159', '26535', '89793', '23846', '26433']
       I£            # keep the first y rows
                     # STACK: ['14159', '26535', '89793']
         íR          # reverse the list of rows and each individual row
                     # STACK: ['39798', '53562', '95141']
           v     }   # for each y,N (row, index) in the list
            yN>£J    # keep the first index+1 digits of the row and join to string
                     # STACK: 353951
                  R  # reverse the string
                     # STACK: 159353
                     # implicit print

Answer 2

Python 2 (with sympy), 100 bytes

from sympy import*
lambda n,y:''.join(c for i,c in enumerate(`pi.round(n*n+1)`[2:])if i%n-i/n>n-y-1)

---

No sympy, 260 246 244 233 231 218 bytes

p=lambda n,y,a=-30,b=10,c=3,d=2,e=0,f=5,s='',i=0:i<n*n and p(n,y,*[((2*b+a)*f,b*d,c*f,d+1,(b*(7*d)+2+(a*f))/(c*f),f+2,s,i),(10*(a-e*c),10*b,c,d,((10*(3*b+a))/c)-10*e,f,s+(str(e)[:i%n-i/n>n-y-1]),i+1)][4*b+a-c<e*c])or s

This employs "The Spigot Algorithm For Pi" of Stanley Rabinowitz and Stan Wagon.

The standard arguments would be a,b,c,d,e,f=0,1,1,1,3,3 to yield the first digit of pi, 3, since that is not required the algorithm is initialised to the point before 1 is yielded, which saves two bytes even though a and b are longer as the result does not require slicing and i can start at 0 rather than -1.

Hits default recursion limit for last test case
Uses // for the first of the divisions so that str(v) may be replaced by `v` (otherwise it would end in L for a long).
repl.it

---

A non-recursive version for 232 bytes which evaluates the last test case too:

def p(n,y):
 a,b,c,d,e,f,i,s=-30,10,3,2,0,5,0,''
 while i<n*n:
    if 4*b+a-c<e*c:s+=`e`[:i%n-i/n>n-y-1];g=10*(a-e*c);e=((10*(3*b+a))//c)-10*e;b*=10;i+=1
    else:g=(2*b+a)*f;h=(b*(7*d)+2+(a*f))/(c*f);b*=d;c*=f;f+=2;d+=1;e=h
    a=g
 print s

repl.it (first indent is one space, second indent is one tab)

Answer 3

Mathematica, 82 bytes

Print@@Join@@Partition[RealDigits[Pi-3,10,#^2][[1]],#][[i,i-#2-1;;]]~Table~{i,#2}&

Answer 4

MATL, 23 22 27 bytes

1 Byte saved thanks to @Luis

UtEYPwY$IbH+&:)GetGi-&R!g)!

Try it Online

Explanation

        % Implicitly grab input (n)
Ut      % Square n and duplicate
E       % Multiply n^2 by 2
YP      % Pi literal
w       % Flip the stack
Y$      % Compute the first 2 * (n^2) digits of pi (accounts for rounding)
IbH+&:) % Grab the first n^2 digits after the decimal
Ge      % Reshape it into an n x n matrix in row-major ordering
t       % Duplicate this matrix
Gi-     % Grab the second input (y) and compute the difference between n and y
&R!     % Get the upper diagonal part and transpose to convert to lower diagonal
g)      % Convert it to a logical array and use it to select the digits of interest
!       % Transpose the result and implicitly display

Answer 5

Perl, 67 bytes

s/ /bpi++$_**2/e;$%=$';$%-=print/(.{$%})$/ for/\d{$`}/g

Requires the command line option -nMbignum=bpi, counted as 12. Input is taken from stdin.

Sample Usage

$ echo 3 3 | perl -nMbignum=bpi primo-square-pi.pl
141923

$ echo 5 3 | perl -nMbignum=bpi primo-square-pi.pl
159353

$ echo 6 4 | perl -nMbignum=bpi primo-square-pi.pl
1592589383

$ echo 6 6 | perl -nMbignum=bpi primo-square-pi.pl
141592535893238643794

$ echo 20 12 | perl -nMbignum=bpi primo-square-pi.pl
358979323846950288419715820974944628620899211706792306647223172745025559196615

Answer 6

C#, 232 bytes 268 bytes

Edit:

I originally used a constant string for Pi outside of the method, but it seems this was cheating. I've had to use the C# Math.PI value, which only has 14 decimal places, so the highest m value I can use is 3. Back to the drawing board...

Golfed:

IEnumerable<string>f(int m,int t){var a=new string[m, m];var b=Math.PI.ToString().Replace("3.","").Substring(0,m*m).ToArray();var c=0;for(int i=0;i<m;i++){for(int j=0;j<m;j++){a[i, j]=b[c]+"";c++;}}c=0;while(t>0){for(int i=t;i>0;i--){yield return a[c,m-i];}t--;c++;}}}

Ungolfed:

  class ATriangularSliceOfSquaredPi
  {
    //http://www.piday.org/million/
    //const string p = "1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831";

    public IEnumerable<string> f(int m, int t)
        {
          var a = new string[m, m];

          //var b = p.Substring(0, m * m).ToArray();
          var b = Math.PI.ToString().Replace("3.", "").Substring(0, m * m).ToArray();

          var c = 0;

          for (int i = 0; i < m; i++)
          {
            for (int j = 0; j < m; j++)
            {
              a[i, j] = b[c] + "";
              c++;
            }
          }

          c = 0;

          while (t > 0)
          {
            for (int i = t; i > 0; i--)
            {
              yield return a[c, m - i];
            }
            t--;
            c++;
          }
        }
      }

Not the shortest answer, but I was just happy I solved this one...

Test Output:

m   t   output
3   3   141923

5 3 159353
6 4 1592589383
6 6 141592535893238643794
20 12 358979323846950288419715820974944628620899211706792306647223172745025559196615