Number of grid points satisfying the triangle inequality

Question

Background: The following questions arise from the Wigner $3j$ symbol, see here. It is well known that the angular momenta $(j_1,j_2,j_3)$ in the Wigner $3j$ symbol must satisfy the triangle inequality.

Q1: Assume three nonnegative integer numbers $J_1,J_2,J_3\in \mathbb{N}$. I would like to know how to calculate the total number of the triplet $(j_1,j_2,j_3)$ satisfying the triangle inequality, see Eq. (34.2.1) here $$ |j_1-j_2| \leq j_3 \leq j_1+j_2, $$ where $j_i=0,1,...,J_i,i=1,2,3$. The total number is denoted by $N(J_1,J_2,J_3)$.

It is clear that the total number of the triplet $(j_1,j_2,j_3)$ without satisfying the triangle inequality is $$ M(J_1,J_2,J_3) = (J_1+1)(J_2+1)(J_3+1). $$ To make the question clear, here lists the results for some values. It is observed that $N\approx M/2$.

$J_1$	$J_2$	$J_3$	$M$	$N$
0	0	0	1	1
1	0	0	2	1
1	1	0	4	2
1	1	1	8	5
2	0	0	3	1
2	1	0	6	2
2	1	1	12	6
2	2	1	18	9
2	2	2	27	15

Q2: Assume the set $A$ contains all the triplets $(j_1,j_2,j_3)$ satisfying the triangle inequality for given numbers $J_1,J_2,J_3$. The total number of elements is $N$ as mentioned above. How to effectively index the element of $A$? It means that we need to find a relation between the index $j = 0, 1, ..., N$ and the triplet $(j_1,j_2,j_3)$ satisfying the triangle inequality.

@JeanMarie Hi, $J_i$ specifies the end point of $j_i$, where $i=1,2,3$. I have mentioned this just below the first equation. — Jiaxin Zhong, Dec 29 '22 at 15:53
I don't think this will be much useful, anyway apparently: $N(k,k,k)=(k+1)((k+1)^2+1)/2$, $N(k,k+1,k)=(k+1)^2(k+2)/2$, $N(k,k+1,k+1)=(k+1)(k+2)^2/2$, $N(k,2k,3k)=(k+1)(5k^2+7k+3)/3$, $N(k,2k,2k)=(3k^3 + 7k^2 + 6k + 2)/2$ and more particular cases. — Fabius Wiesner, Dec 30 '22 at 21:51
@BillyJoe Interesting! How to obtain these results? I think they might bring insights into the derivation of the solution to the general cases. — Jiaxin Zhong, Dec 31 '22 at 12:45
@Jiaxin Zhong I have just written a program and then searched the output sequences at OEIS. This is just a numerical evidence without proof. — Fabius Wiesner, Dec 31 '22 at 15:20

score 3 · Accepted Answer · answered Jan 05 '23 at 12:50

3

Suppose $J_1 \le J_2 \le J_3$ and first suppose $J_3 \ge J_1+J_2$. Then whatever the values of $j_1$ and $j_2$, any value of $j_3$ completing the triangle inequality is possible. So the number of triples is $$\sum_{j_1=0}^{J_1}\sum_{j_2=j_1}^{J_2} (2j_1+1)+ \sum_{j_1=1}^{J_1}\sum_{j_2=0}^{j_1-1}(2j_2+1) $$ which gives $N(J_1, J_2,J_3)=\frac{1}{3}(J_1+1)\big(3(J_1J_2+J_2+1)+J_1-J_1^2\big)$.

If, on the other hand, $J_3<J_1+J_2$, you need to take away from this the triangles that become impossible because you can't select a sufficiently large $j_3$. There are $$\sum_{j_1=J_3-J_2+1}^{J_1}\sum_{j_2=J_3-j_1+1}^{J_2} (J_2-J_3+j_1)$$ of these which is $\frac{1}{6}(J_1+J_2-J_3)(J_1+J_2-J_3+1)(J_1+J_2-J_3+2)$.

I think these agree with the special cases in the comments.

answered Jan 05 '23 at 12:50

mcd

3,448

With my test program this fails for $N(k,2k,k)$, $k \le 2$. If my program is correct, your formula gives a lower result, e.g. $17$ instead of $19$ for $N(2,4,2)$. I confirm that it gives the same results for the cases in the question comments. – Fabius Wiesner Jan 06 '23 at 22:25
I don't think so @BillyJoe . Take $J_1=J_2=2$ and you get 19. Note the ordering in my result - you need to calculate $N(k, k, 2k)$ and there is no subtraction, as $2k \ge k + k$. – mcd Jan 06 '23 at 23:53
Why can you exchange $J_2$ and $J_3$? Exchanging $J_1$ and $J_2$ is clear. Shouldn't you consider the case $J_1 \le J_3 \le J_2$ in your answer? – Fabius Wiesner Jan 07 '23 at 09:30
No, I don't think so @BillyJoe . $N(a, b, c)$ does not change when you permute $a$, $b$ and $c$, so you can always calculate $N(a, b, c) = N(A, B, C)$, where $[A, B, C]$ is the permutation of $[a, b, c]$ with $A \le B \le C$ (or one of the permutations, if not all of $a$, $b$, $c$ are distinct). Thus you only need a formula for the case $J_1 \le J_2 \le J_3$. – mcd Jan 07 '23 at 10:46
@BillyJoe: I verified $N(J_1,J_2,J_3)=N(J_1,J_3,J_2)$ using $(j_1,j_2,j_3)\mapsto(j_1,j_3,j_2)$ and iterating through the four cases $j_3\ge j_1$ or lower, $j_2\ge j_1$ or lower. – Matija Jan 07 '23 at 11:39

leonbloy · Answer 2 · 2023-01-07T14:44:11.480

Not an answer but some asymptotics.

For large $J_k$, we can assume real coordinates, and expect that $p=N/M$ tends to the probability that the real random variables $x,y,z$, chosen uniformly inside a box of lengths $(a,b,c)$, satisfy the triangle inequality.

The case $a=b=c=1$ is simple and indeed gives $p=1/2$.

Because of symmetry and scaling invariance, we can assume WLOG $a\le\ b \le c$, and $a=1$. (In the original setting, this corresponds to assume $J_3 \ge J_2 \ge J_1 \gg 1$, and set $b=J_2/J_1$, $c= J_3/J_1$)

To compute $p(b,c)$ in general, we need to compute $p(b,c)=\frac{1}{bc}I(b,c)$ where $$I(b,c)= \int_{0}^1 \int_{0}^b [\min(c,x+y) - |x-y|] \, dy \,dx $$

There are several ranges to consider, I'll compute the two extremes.

For the range $c\le 2$ (not too dissimilar sides) (which includes the $b=c=1$ case) we get

$$I(b,c)=\frac12(c+b-1) -\frac12(c-b)^2 +\frac16(c-b)^3$$

For the range $c \ge b+1$ we get

$$I(b,c)= b-\frac13$$

and $p(b,c)=\frac{1}{c}-\frac{1}{3bc}$. This agrees (asympotically) with the first answer by @mcd.

Number of grid points satisfying the triangle inequality

2 Answers2