[euler] further whitespace, formatting cleanups

categories/euler/prob065-andreoss.pl

@@ -1,4 +1,3 @@
-#!/usr/bin/env perl6
 use v6;
 
 =begin pod
@@ -9,8 +8,6 @@
 
 L<https://projecteuler.net/problem=65>
 
-
-
 The square root of 2 can be written as an infinite continued fraction.
 
 √2 = 1 +  1
@@ -23,24 +20,25 @@
   	  	 ______
 	         2 + ...
 
-The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that
-2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)].
+The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates
+that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)].
+
+It turns out that the sequence of partial values of continued fractions for
+square roots provide the best rational approximations. Let us consider the
+convergents for √2.
 
-It turns out that the sequence of partial values of continued fractions for square
-roots provide the best rational approximations. Let us consider the convergents for √2.
-	 
 1 + 1
    ___              = 3/2
-    2 
+    2
 
 1 +    1
     _________       = 7/5
     2 +  1 / 2
-           
+
 ....
 
-Hence the sequence of the first ten convergents for √2 are:
-1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...
+Hence the sequence of the first ten convergents for √2 are: 1, 3/2, 7/5,
+17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...
 
 What is most surprising is that the important mathematical constant,
 e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
@@ -50,12 +48,12 @@
 
 The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.
 
-Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.
-
+Find the sum of digits in the numerator of the 100th convergent of the
+continued fraction for e.
 
 Expected result:  272
 
-=end pod		      
+=end pod
 
 sub continued-fraction(@sequence, :$depth)  {
     my $x = @sequence.shift;
@@ -67,3 +65,5 @@
 my @e := gather { take 2;  take (1; $_; 1) for 2,4 ... * };
 
 say [+] continued-fraction(@e, depth => 100).numerator.comb;
+
+# vim: expandtab shiftwidth=4 ft=perl6

categories/euler/prob066-andreoss.pl

@@ -1,4 +1,3 @@
-#!/usr/bin/perl6
 use v6;
 
 =begin pod
@@ -7,28 +6,31 @@
 
 =AUTHOR Andrei Osipov
 
-L<https://projecteuler.net/problem=66> 
+L<https://projecteuler.net/problem=66>
 
 Consider quadratic Diophantine equations of the form:
 
 x² – D×y² = 1
 
 For example, when D=13, the minimal solution in x is 649² – 13×180² = 1.
 
-It can be assumed that there are no solutions in positive integers when D is square.
+It can be assumed that there are no solutions in positive integers when D is
+square.
 
-By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
+By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the
+following:
 
 3²– 2×2²= 1
 2²– 3×1²= 1
 9²– 5×4²= 1
 5²– 6×2²= 1
 8²– 7×3²= 1
 
-Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.
-
-Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.
+Hence, by considering minimal solutions in x for D ≤ 7, the largest x is
+obtained when D=5.
 
+Find the value of D ≤ 1000 in minimal solutions of x for which the largest
+value of x is obtained.
 
 Expected result:  661
 
@@ -37,7 +39,7 @@
 
 =end pod
 
-subset NonSquarable where *.sqrt !%% 1; 
+subset NonSquarable where *.sqrt !%% 1;
 
 sub next-triplet([\a,\b,\k], \N) {
 
@@ -59,18 +61,18 @@
     my $a = N.sqrt.floor;
     my $b = 1;
     my $k = $a ** 2 - N;
-    
+
     $a, $b, $k;
 }
 
 sub chakravala(NonSquarable \N) {
-    # Start with a solution for a² - N b² = k 
+    # Start with a solution for a² - N b² = k
 
     my ($a, $b, $k) = simple-solution N;
-    
+
     ($a,$b,$k) = next-triplet [$a,$b,$k], N
 	while $k != 1;
-    
+
     $a, $b, $k;
 }
 
@@ -81,3 +83,5 @@ (NonSquarable \N)
     ==> sort *[2] ==> my @x;
 
 say @x.pop[0];
+
+# vim: expandtab shiftwidth=4 ft=perl6

categories/euler/prob080-andreoss.pl

@@ -6,21 +6,23 @@
 
 =AUTHOR Andrei Osipov
 
-L<https://projecteuler.net/problem=80> 
+L<https://projecteuler.net/problem=80>
 
-It is well known that if the square root of a natural number is not an integer, then it is irrational.
-The decimal expansion of such square roots is infinite without any repeating pattern at all.
-The square root of two is 1.41421356237309504880..., and the digital sum of the first one
-hundred decimal digits is 475.
+It is well known that if the square root of a natural number is not an
+integer, then it is irrational.  The decimal expansion of such square roots
+is infinite without any repeating pattern at all.  The square root of two is
+1.41421356237309504880..., and the digital sum of the first one hundred
+decimal digits is 475.
 
-For the first one hundred natural numbers, find the total of the digital sums of the first
-one hundred decimal digits for all the irrational square roots.
+For the first one hundred natural numbers, find the total of the digital
+sums of the first one hundred decimal digits for all the irrational square
+roots.
 
 The following algoritm was used for the solution:
 L<http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec02.pdf>
 
 Expected result:  40886
- 
+
 =end pod
 
 use v6;
@@ -39,7 +41,7 @@
 	    }
 	    when Less {
 		# add two zeros to a
-		$a *= 100; 
+		$a *= 100;
 		# add a zero to b just before the final digit (which will always be ‘5’).
 		$b = ($b - (my $x = $b % 10)) * 10 + $x;
 	    }
@@ -50,13 +52,12 @@
 
 sub MAIN(Bool :$verbose = False) {
     say [+] do for 1 ... 100 -> $n {
-	next if $n.sqrt.floor ** 2 == $n; 
+	next if $n.sqrt.floor ** 2 == $n;
 	my $x = [+] $n.&sqrt-subtraction.comb[^$limit];
 	say "$n $x"  if $verbose;
 	$x;
     }
     say "Done in {now - INIT now}" if $verbose;
 }
 
-
-
+# vim: expandtab shiftwidth=4 ft=perl6