Fix #697: Quote NP in Typst math mode

docs/paper/reductions.typ

@@ -2305,7 +2305,7 @@ NP-completeness was established by Garey, Johnson, and Stockmeyer @gareyJohnsonS
 #problem-def("SumOfSquaresPartition")[
   Given a finite set $A = {a_0, dots, a_(n-1)}$ with sizes $s(a_i) in ZZ^+$, a positive integer $K lt.eq |A|$ (number of groups), and a positive integer $J$ (bound), determine whether $A$ can be partitioned into $K$ disjoint sets $A_1, dots, A_K$ such that $sum_(i=1)^K (sum_(a in A_i) s(a))^2 lt.eq J$.
 ][
-  Problem SP19 in Garey and Johnson @garey1979. NP-complete in the strong sense, so no pseudo-polynomial time algorithm exists unless $P = NP$. For fixed $K$, a dynamic-programming algorithm runs in $O(n S^(K-1))$ pseudo-polynomial time, where $S = sum s(a)$. The problem remains NP-complete when the exponent 2 is replaced by any fixed rational $alpha > 1$. #footnote[No algorithm improving on brute-force $O(K^n)$ enumeration is known for the general case.] The squared objective penalizes imbalanced partitions, connecting it to variance minimization, load balancing, and $k$-means clustering. Sum of Squares Partition generalizes Partition ($K = 2$, $J = S^2 slash 2$).
+  Problem SP19 in Garey and Johnson @garey1979. NP-complete in the strong sense, so no pseudo-polynomial time algorithm exists unless $P = "NP"$. For fixed $K$, a dynamic-programming algorithm runs in $O(n S^(K-1))$ pseudo-polynomial time, where $S = sum s(a)$. The problem remains NP-complete when the exponent 2 is replaced by any fixed rational $alpha > 1$. #footnote[No algorithm improving on brute-force $O(K^n)$ enumeration is known for the general case.] The squared objective penalizes imbalanced partitions, connecting it to variance minimization, load balancing, and $k$-means clustering. Sum of Squares Partition generalizes Partition ($K = 2$, $J = S^2 slash 2$).
 
   *Example.* Let $A = {5, 3, 8, 2, 7, 1}$ ($n = 6$), $K = 3$ groups, and bound $J = 240$. The partition $A_1 = {8, 1}$, $A_2 = {5, 2}$, $A_3 = {3, 7}$ gives group sums $9, 7, 10$ and sum of squares $81 + 49 + 100 = 230 lt.eq 240 = J$. With a tighter bound $J = 225$, the best achievable partition has group sums ${9, 9, 8}$ yielding $81 + 81 + 64 = 226 > 225$, so the answer is NO.
 ]