to precisely tune this quantity to zero simultaneously for the proton and the neutron (see Fig. 6). However, an approximate realization of this condition is possible – this is the tuning that underlies large SD/small SI points in Figs. 3 and 4 and reported in the literature ( e.g. [62]). 4. One can tune the contribution from the proton against the contribution from the neutron. The cancellation would only hold for a specific element. Since all experiments do not use the same elements, we will not pursue this case further. In what follows, we minimize the SI cross section by tuning the contributions from the h and H against each other (point 3 above). From Eq. (20), this cancellation requires (in the decoupling/large t$_{β}$ limit) sgn( Z$_{H}$$_{u}$ ) = sgn( Z$_{H}$$_{d}$ ). This condition for cancellations to be possible was first noted in [63]. Using DarkSUSY we have confirmed that this is a necessary condition, not just in this limit, but for any values of the pseudo-scalar Higgs mass ( m$_{A}$ ) and t$_{β}$ . This condition only occurs for certain signs of M$_{1}$ , M$_{2}$ and µ . If large SD/small SI were observed for neutralino DM, this would constrain the signs in the neutralino mass matrix. Let us estimate the maximum allowed suppression. To good approximation $^{2}$, the best one can do is to tune away the coupling to (for example) the proton: In order for Eq. (36) to have a guaranteed solution requires independent control of α and m$_{H}$ . Since there is a non-trivial relationship between α and m$_{H}$ (both are determined by m$_{A}$ ), our lower bound provides a conservative estimate. Using Eq. (9) to estimate c$_{q}$ and plugging in the relationship between c$_{u}$ and c$_{d}$ from Eq. (36) gives σ p SI = 0 and This gives an estimate for how small SI can be, absent taking some of M$_{1}$, M$_{2}$, µ → ∞ . The effects of the current uncertainties on the hadronic matrix elements described in Sec. II A 2 From Fig. 6 the absolute minimum of the total SI cross section occurs between the region where the coupling to the proton and neutron vanish. Therefore, the following analytic estimate will be off by a factor of a few. 23 ≈ 8 × 10 − $^{13}$pb$^{(}$ 115 GeV m$_{h}$ ) 4 ( y$_{χ}$ 0 . 1$^{)}$$^{2}$ (SI with cancellations) . (38) σ$_{SI}$ = σ n SI = 4 π m 2 n ( A − Z ) 2 A 2 m 2 $_{r}$y 2 χ 1 m 4 h (( f ( n ) Tu + 2 2 27 f ( n ) TG ) f ( p ) d/u −$^{(}$ f ( n ) Td + f ( n ) Ts + 2 27 f ( n ) TG )) 2 (37) c$_{u}$ m$_{u}$ = −$^{(}$ f ( p ) Td + f ( p ) Ts + 2 27 f ( p ) TG f ( p ) Tu + 2 2 27 f ( p ) TG ) c$_{d}$ m$_{d}$ ≡ − f ( p ) d/u c$_{d}$ m$_{d}$ ≈ − 1 . 64 c$_{d}$ m$_{d}$ . (36)