Pascal's triangle/Puzzle: Difference between revisions
m
→{{header|Wren}}: Minor tidy
m (→{{header|Wren}}: Minor tidy) |
|||
| (7 intermediate revisions by 2 users not shown) | |||
Line 1,111:
(doseq [row rows]
(println (map #(dot [1 x z] %) row)))</syntaxhighlight>
=={{header|Craft Basic}}==
<syntaxhighlight lang="basic">let x = -1
do
let x = x + 1
let z = 0
do
let e = x + 11
let f = 11 + (x + z)
let g = (x + z) + 4
let h = 4 + z
if e + f = 40 then
let c = f + g
let d = g + h
let a = 40 + c
let b = c + d
let q = 0
if a + b = 151 then
let q = 1
endif
endif
if q = 0 then
let z = z + 1
endif
wait
loopwhile z < 20 and q = 0
if q = 0 then
let z = -1
endif
wait
loopuntil z >= 0
print "x = ", x
print "y = ", x + z
print "z = ", z</syntaxhighlight>
{{out| Output}}<pre>x = 5
y = 13
z = 8</pre>
=={{header|Curry}}==
Line 1,705 ⟶ 1,763:
In the spirit of the simplicity of Pascal Triangle's definition, the
basic Pascal Triangle equation for each brick that is above two
others.
Line 1,740 ⟶ 1,798:
<pre>
[5,13,8]
</pre>
===Algebraic solution===
As noted elsewhere on this page, elementary considerations show that
the apex (top) value and the value in the third row (mid) can be written as:
<pre>
top = 4(a+b) + 7(x+z)
mid = 2x + 2a + z
</pre>
where a and b are the known values at the base.
Using the jq program at [[Cramer%27s_rule#jq|Cramer's rule]], with the unknown vector being [x, z]:
<syntaxhighlight lang=jq>
include "rc-cramers-rule";
def solve(top; mid; a; b):
cramer(
[ [7, 7],
[2, 1]];
[top - 4*(a+b), mid-2*a]);
solve(151; 40; 11; 4)
</syntaxhighlight>
This gives the solution for [x,z] as:
{{output}}
<pre>
[5,8]
</pre>
Line 3,135 ⟶ 3,218:
{{trans|Kotlin}}
{{libheader|Wren-fmt}}
<syntaxhighlight lang="
var isIntegral = Fn.new { |x, tol| x.fraction.abs <= tol }
| |||