- Added solutions by Mark Senn.

4 files changed, 64 insertions, 0 deletions

diff --git a/challenge-155/mark-senn/blog.txt b/challenge-155/mark-senn/blog.txt
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+https://engineering.purdue.edu/~mark/twc-155-1.pdf
diff --git a/challenge-155/mark-senn/blog1.txt b/challenge-155/mark-senn/blog1.txt
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+https://engineering.purdue.edu/~mark/twc-155-2.pdf
diff --git a/challenge-155/mark-senn/raku/ch-1.raku b/challenge-155/mark-senn/raku/ch-1.raku
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+++ b/challenge-155/mark-senn/raku/ch-1.raku
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+use v6.d;
+
+# Print the first 8 unique and sorted Fortunate Numbers.
+my $n = 8;
+
+# Guess that first 50 primes and primorials will give
+# the first 8 unique and sorted Fortunate Numbers.
+my $nn = 50;
+
+# Get the first nn primes.
+my @prime = (2..*).grep(&is-prime).head($nn);
+
+# Compute the first nn primorials.
+my @primorial;
+@primorial[0]= @prime[0];
+for (1..^$nn)
+{
+    @primorial[$_] = @primorial[$_-1] * @prime[$_];
+}
+
+# Save the first $nn m's.
+my @m = ();
+# Loop through the first nn primorials.
+for (0..^$nn) -> $i
+{
+    # For each primorial check if it plus m is prime, for m = 2 to whatever.
+    for (2..*) -> $m
+    {
+    	(@primorial[$i] + $m).is-prime and @m.push($m), last;
+    }
+}
+
+# Print the first 8 unique and sorted Fortunate Numbers.
+@m.unique.sort.head(8).join(', ').say;
diff --git a/challenge-155/mark-senn/raku/ch-2.raku b/challenge-155/mark-senn/raku/ch-2.raku
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+# Use Raku version 6.d.
+use v6.d;
+
+# Find the period of the 3rd Pisano period.
+my $modulo = 3;
+
+# Define a Fibonacci series (called sequence in Raku).
+# The last number in the series is the first number greater than 10,000.
+my $fibonacci := 0, 1, -> $a, $b {$a + $b} ... -> $a {$a > 10_000};
+
+# Compute the Fibonacci series modulo $modulo.
+my @modulo = $fibonacci >>%>> $modulo;
+
+# Find the period.
+# Test periods from 2 to whatever.
+for (2..*) -> $period
+{
+    # Divide @modulo into groups of period using ".rotor($period)"---groups
+    # with fewer than period elements are ignored.
+    # See
+    # https://docs.raku.org/language/operators#infix_eqv
+    # for information about "eqv".
+    # "[eqv] (1,2,3,4)" is equivalent to "1 eqv 2 eqv 3 eqv 4".
+    # "[eqv] ((1,2), (3,4))" is equivalent to "(1,2) eqv (3,4)".
+    # If all groups are equal that is the period.
+
+    [eqv] @modulo.rotor($period) and $period.say, last;
+}2