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A number spiral is an infinite grid whose upper-left square has number 1. Here are the first five layers of the spiral:
Your task is to find out the number in row y and column x.
Example:
Input: 2 3
Out : 8
Input: 1 1
Out : 1
Input: 4 2
Out : 15
Note:
(Originally) Adapted from Doorknob's solution over a few beers.
wV
nU²ÒNra
wV\nnU²ÒNra :Implicit input of integers N=[U=y, V=X]
wV :Maximum of U & V
\n :Reassign to U, leaving the value in N unchanged
n :Subtract U from
U² : U squared
Ò : Subtract the bitwise NOT of
Nr : N reduced by
a : Absolute difference
f(x,y,z){z=x>y?x:y;z=z*z-~(z%2?y-x:x-y)-z;}
The spiral has several "arms":
12345
22345
33345
44445
55555
The position \$(x, y)\$ is located on arm \$\max(x, y)\$ (assigned to variable z). Then, the largest number on arm \$n\$ is \$n^2\$, which alternates between being in the bottom left and top right position on the arm. Subtracting \$x\$ from \$y\$ gives the sequence \$-n+1, -n+2, \ldots, -1, 0, 1, \ldots, n-1, n-2\$ moving along arm \$n\$, so we choose the appropriate sign based on the parity of \$n\$, adjust by \$n-1\$ to get a sequence starting at 0, and subtract this value from \$n^2\$.
Thanks to Mr. Xcoder for saving a byte.
f(x,y,z){z=x>y?x:y;z=z*z-~(z%2?x-y:y-x)-z;} saves 1 byte.
\$\endgroup\$
– Mr. Xcoder
Aug 17 '18 at 15:52
f(1, 1) returns the value 1. The Footer section loops through x=1 through 5 and y=1 through 5, calls the function for all such values, and prints its output in a grid, to demonstrate that the function is correct for all inputs shown in the question.
\$\endgroup\$
– Doorknob♦
Aug 17 '18 at 19:38
def f(a,b):M=max(a,b);return(a-b)*(-1)**M+M*M-M+1
-4 bytes thanks to @ChasBrown
-1 bytes thanks to @Shaggy
First time golfing! I'm more than aware this is not optimal, but whatever.
Essentially runs on the same principle as @Doorknob C code.
def f(a,b): approach, see here.
\$\endgroup\$
– Chas Brown
Aug 17 '18 at 21:21
M**2 can be replaced with M*M.
\$\endgroup\$
– Shaggy
Aug 17 '18 at 22:36
X>ttq*QwoEqGd*+
Try it online!
Collect and print as a matrix
Edit: Same technique as @Doorknob's answer, just arrived at differently.
The difference between the diagonal elements of the spiral is the arithmetic sequence \$ 0, 2, 4, 6, 8, \ldots \$. Sum of \$ n \$ terms of this is \$ n(n - 1) \$ (by the usual AP formula). This sum, incremented by 1, gives the diagonal element at position \$ (n, n) \$.
Given \$ (x, y) \$, we find the maximum of these two, which is the "layer" of the spiral that this point belongs to. Then, we find the diagonal value of that layer as \$ v = n(n-1) + 1 \$. For even layers, the value at \$ (x, y) \$ is then \$ v + x - y \$, for odd layers \$ v - x + y \$.
X> % Get the maximum of the input coordinates, say n
ttq* % Duplicate that and multiply by n-1
Q % Add 1 to that. This is the diagonal value v at layer n
wo % Bring the original n on top and check if it's odd (1 or 0)
Eq % Change 1 or 0 to 1 or -1
Gd % Push input (x, y) again, get y - x
* % Multiply by 1 or -1
% For odd layers, no change. For even layers, y-x becomes x-y
+ % Add that to the diagonal value v
% Implicit output
Alternate 21 byte solution:
Pdt|Gs+ttqq*4/QJb^b*+
Try it online!
Collect and print as a matrix
From the above, we know that the function we want is
$$ f = m * (m - 1) + 1 + (-1)^m * (x - y) $$
where \$ m = max(x, y) \$.
Some basic calculation will show that one expression for max of two numbers is
$$ m = max(x, y) = \frac{x + y + abs(x - y)}{2} $$
Plugging one into another, we find that one alternate form for \$ f \$ is:
$$ f = (x-y)\cdot i^{k} + \frac{1}{4}((k-2)\cdot k) + 1 $$
where \$ k = abs(x-y) + x + y \$.
This is the function the solution implements.
A~Qh.MZQh-+*-GH^_1Q*
An almost literal translation of Rushabh Mehta's answer.
Explanation:A~Qh.MZQh-+*-GH^_1Q* | Full code
A~Qh.MZQh-+*-GH^_1Q*QQQ | Code with implicit variables filled
| Assign Q as the evaluated input (implicit)
A | Assign [G,H] as
~Q | Q, then assign Q as
h.MZQ | Q's maximal value.
| Print (implicit)
h-+*-GH^_1Q*QQQ | (G-H)*(-1)^Q+Q*Q-Q+1
ṀḂḤ’×I+²_’ṀƲ
Computes the diagonal term with ²_’Ṁ and adds/subtracts to the correct index value with ṀḂḤ’×I.
((({}<>))<>[(({}))]<{({}[()])<>}>)<>{}((){({}[()])({})<><([{}])><>}{}<>{}<>)
ZÐ<*>ŠGR}¥+
-1 byte thanks to @Emigna changing Èi to G.
Port of @sundar's MATL answer, so make sure to upvote him!
Try it online or verify all test cases.
Explanation:
Z # Get the maximum of the (implicit) input-coordinate
# i.e. [4,5] → 5
Ð # Triplicate this maximum
< # Decrease it by 1
# i.e. 5 - 1 → 4
* # Multiply it
# i.e. 5 * 4 → 20
> # Increase it by 1
# i.e. 20 + 1 → 21
Š # Triple swap the top threes values on the stack (a,b,c to c,a,b)
# i.e. [4,5], 5, 21 → 21, [4,5], 5
G } # Loop n amount of times
R # Reverse the input-coordinate each iteration
# i.e. 5 and [4,5] → [5,4]→[4,5]→[5,4]→[4,5] → [5,4]
¥ # Calculate the delta of the coordinate
# [5,4] → [1]
+ # And add it to the earlier calculate value (output the result implicitly)
# 21 + [1] → [22]
uses math;var x,y,z:word;begin read(x,y);z:=max(x,y);write(z*z-z+1+(1and z*2-1)*(y-x))end.
Port of Doorknob's answer, but sundar's answer gave me idea for z mod 2*2-1 which I transformed into 1and z*2-1 to remove space.
x = {5, 8};
so:
m = Max[x];
Subtract @@ x (-1)^m + m^2 - m + 1
(*
54
*)
f=(r,c,x)=>r<c?f(c,r,1):r%2-!x?r*r-c+1:--r*r+c
Also adapted from Doorknob's solution.
listen to Y
listen to X
let M be Y is less than X and X-0 or Y-0
let R be M*M-M
let R be-M-Y and Y-X or X-Y
say R+1
Try it here (Code will need to be pasted in, with each input integer on an individual line)
