Circular Limited Sums

Question

Challenge

Let's imagine an N-tuple of integers between 0 and M inclusive, and let's call it F.

There are (M + 1) ** N possible Fs in total.

How many such Fs satisfy all of the following inequalities (index is one-based)?

• F[n] + F[n+1] <= M for 1 <= n < N
• F[N] + F[1] <= M

Write a program or function that takes two positive integers N and M and outputs the answer in any convenient form.

Test Cases

(N,M) => Answer

(1,1) => 1
(2,1) => 3
(3,1) => 4
(4,1) => 7

(1,2) => 2
(2,2) => 6
(3,2) => 11
(4,2) => 26

(10,3) => 39175
(10,4) => 286555
(10,5) => 1508401

(25,3) => 303734663372
(25,4) => 43953707972058
(25,5) => 2794276977562073

(100,3) => 8510938110502117856062697655362747468175263710
(100,4) => 3732347514675901732382391725971022481763004479674972370
(100,5) => 60964611448369808046336702581873778457326750953325742021695001

Explanation

M (max value of element) = 1

F[1] + F[1] <= 1; F = [0]
(1,1) => 1

F[1] + F[2] <= 1; F = [0,0], [0,1], [1,0]
(2,1) => 3

F = [0,0,0], [0,0,1], [0,1,0], [1,0,0]
(3,1) => 4

F = [0,0,0,0], [0,0,0,1], [0,0,1,0], [0,1,0,0], [0,1,0,1], [1,0,0,0], [1,0,1,0]
(4,1) => 7

---

M = 2

F[1] + F[1] <= 2; F = [0], [1]
(1,2) => 2

F = [0,0], [0,1], [0,2], [1,0], [1,1], [2,0]
(2,2) => 6

F = [0,0,0], [0,0,1], [0,0,2], [0,1,0], [0,1,1], [0,2,0], [1,0,0], [1,0,1],
[1,1,0], [1,1,1], [2,0,0]
(3,2) => 11

(4,2) => 26 (left as exercise for you)

Rules

• This is a restricted-complexity challenge. The time complexity of your code should be polynomial in M and N (e.g. you can't generate all (M + 1) ** N tuples and then check for the condition). Please explain your approach in your submission.
• Standard code-golf rules apply. The shortest answer in bytes wins.

Answer 1

Python with numpy, 59 bytes

lambda M,N:trace(mat(tri(M+1)[::-1])**N)
from numpy import*

Try it online!

Uses matrix multiplication to count paths. If float precision is an issue, the mat could specify mat(...,int).

Answer 2

Pyth, 27 bytes

.N?Ys:RTtYh-QNgQ+NTs:Rdtszh

Demonstration

Expects input in the format:

M
N

This is classic dynamic programming, over the left end of the values set so far, the right end, and the current size of the gap.

How it works, in pseudocode/Python:

.N          | define memoized fill(left, right, gap):
?           | if cap > 0 then
s:RTtY      | sum(fill(i, right, gap - 1)
h-QN        |     for i in range(M - left + 1))
gQ+NT       | else M >= left + right
            | output:
s:Rdtsz     | sum(fill(i, i, N - 1)
h           |     for i in range(M + 1))

Q is used for M, z is used for N, : is fill, N is left, T is right, Y is gap.

Answer 3

MATL, 13 12 bytes

Q:&>~PiY^Xds

Try it online! This is a direct translation of xnor's Python answer and my first MATL answer, so it's most likely not optimal. E.g. there is likely a shorter way to get an upper-left triangular matrix of ones than t&lYRP. Edit: And it turns out there is, namely :&>~P. Thanks to Luis Mendo for -1 byte!

               M is the first input and N the second
Q:             increment M and generate range from 1 to M+1
  &>           compare vector element wise with itself with greater-than function
               results in a upper-right triangular matrix
    ~          inverse to get lower-left triangular matrix
     P         flip rows to get upper-left triangular matrix
      i        input N
       Y^      take the matrix to the power of N
         Xds   compute the sum of the main diagonal

Answer 4

Stax, 17 bytes

°(√&╒íƽ╨⌂'├╖▼1_Z

Run and debug it

Unpacked, ungolfed, and commented, it looks like this.

^1](    [1, 0, ... 0] with M zeroes
:)      get all rotations of the array
{       begin block
  {:+rm map each array to reverse(prefixSums(arr))
},v*    execute preceding block N-1 times
F       for each array, execute the rest of the program
  iT    remove the last i elements from the array, where i is the iteration index
  F+    add the remaining elements to the running total
        implicitly print output

Run this one

Answer 5

R, 72 bytes

function(M,N)sum(diag(Reduce(`%*%`,rep(list(outer(0:M,0:M,"+")<=M),N))))

Try it online!

Ports xnor's approach.

Fails for larger test cases as R only has 32-bit integer support (they get cast to double once the max int value is reached), so using gmp or another arbitrary precision arithmetic library would be required.

Strangely, R lacks a matrix power operator, as ^ always applies elementwise.

Answer 6

APL (Dyalog Unicode), ³¹ 25 bytes

+/1 1⍉+.×⍣(⎕-1)⍨⌽∘.≤⍨⍳1+⎕

Try it online!

A full program which takes N, then M as inputs.

Follows the same algorithm as xnor's Python solution.

-6 bytes from Bubbler.