Universality Classes of Interaction Structures for NK Fitness Landscapes

Abstract

Kauffman’s NK-model is a paradigmatic example of a class of stochastic models of genotypic fitness landscapes that aim to capture generic features of epistatic interactions in multilocus systems. Genotypes are represented as sequences of L binary loci. The fitness assigned to a genotype is a sum of contributions, each of which is a random function defined on a subset of $k \le L$ loci. These subsets or neighborhoods determine the genetic interactions of the model. Whereas earlier work on the NK model suggested that most of its properties are robust with regard to the choice of neighborhoods, recent work has revealed an important and sometimes counter-intuitive influence of the interaction structure on the properties of NK fitness landscapes. Here we review these developments and present new results concerning the number of local fitness maxima and the statistics of selectively accessible (that is, fitness-monotonic) mutational pathways. In particular, we develop a unified framework for computing the exponential growth rate of the expected number of local fitness maxima as a function of L, and identify two different universality classes of interaction structures that display different asymptotics of this quantity for large k. Moreover, we show that the probability that the fitness landscape can be traversed along an accessible path decreases exponentially in L for a large class of interaction structures that we characterize as locally bounded. Finally, we discuss the impact of the NK interaction structures on the dynamics of evolution using adaptive walk models.

Appendices

A Asymptotics of $\pi _\mathrm {max}^{\mathrm {MF}}$ in the Joint Limit $k, L \rightarrow \infty$

We start from Eq. (34). Rescaling $y \rightarrow \frac{\eta y}{\sqrt{2}}$, we rewrite the equation in terms of the CDF of a standard Gaussian distribution $\varPhi (y)$ as

$$\begin{aligned} \pi _\mathrm {max}^{\mathrm {MF}}&= \sqrt{\frac{L \eta ^2}{4\pi } } \int dy e^{-L \eta ^2 y^2/4} \left[ \frac{1}{2} \left( \text {erf}\left( \frac{y}{\sqrt{2}}\right) +1\right) \right] ^L \nonumber \\&= \sqrt{\frac{\mu }{2\pi } } \int dy e^{-\mu y^2/2} \varPhi (y)^L, \end{aligned}$$

(99)

where $\mu \equiv \frac{L \eta ^2}{2}$ which converges to $\frac{(2-\alpha )}{\alpha }$ in the joint limit as can be seen from Eq. (32).

Interestingly, the only L-dependence shown in the above equation appears as an L-th power of the CDF $\varPhi (y)$, which converges monotonically to unity as $y \rightarrow \infty$. This implies that the conventional saddle point method cannot be applied here due to the absence of a maximum. Instead, we can rely on the extreme value theory by interpreting the term $\varPhi (y)^L$ as the probability that L randomly sampled standard Gaussian random variables are less than y. This leads immediately to the limit relation [16]

$$\begin{aligned} \varPhi \left( \frac{x}{a_L} + b_L\right) ^L \rightarrow G(x)( 1 + o(1)), \end{aligned}$$

(100)

where G(x) is the Gumbel CDF defined by $G(x) = e^{- e^{-x}}$, and the two scaling factors are given by $a_L = \sqrt{2 \ln L}$ and

$$\begin{aligned} b_L = \sqrt{2 \ln L} - \frac{\ln \ln L + \ln 4\pi }{2 \sqrt{2 \ln L}}. \end{aligned}$$

(101)

After making the change of variable $y = \frac{x}{a_L} + b_L$, the integral is now of the form

$$\begin{aligned} \pi _\mathrm {max}^{\mathrm {MF}}&= \frac{1}{a_L} \sqrt{\frac{\mu }{2\pi } } \int dx e^{-\mu \left( \frac{x}{a_L} + b_L \right) ^2/2} \varPhi \left( \frac{x}{a_L} + b_L\right) ^L \nonumber \\&= \frac{1}{a_L} \sqrt{\frac{\mu }{2\pi } } \int dx e^{- \mu \left( \frac{x}{a_L} + b_L \right) ^2/2} G(x) \left( 1 + o(1)\right) . \end{aligned}$$

(102)

The evaluation of the integral with respect to x is greatly simplified once one notices that the term $\frac{x^2}{a_L^2}$ in the exponent is sub-leading in L. Ignoring this term gives

$$\begin{aligned} \pi _\mathrm {max}^{\mathrm {MF}}&=\frac{1}{a_L} \sqrt{\frac{\mu }{2\pi } } \int dx e^{- \mu \left( b_L^2 + 2 \frac{b_L x}{a_L} \right) /2} G(x) \left( 1 + o(1)\right) \nonumber \\&= \frac{1}{a_L} \sqrt{\frac{\mu }{2\pi } } e^{-\mu b_L^2/2} \varGamma ( \mu )\left( 1 + o(1)\right) , \end{aligned}$$

(103)

where we have used the identity

$$\begin{aligned} \int _{-\infty }^{\infty } G(x) \exp (-M x) \, dx = \varGamma (M) \end{aligned}$$

(104)

for positive M. Next, expanding $a_L$ and $b_L$ and rearranging the terms gives

$$\begin{aligned} \pi _\mathrm {max}^{\mathrm {MF}}&= \frac{1}{a_L} \sqrt{\frac{ \mu }{2\pi } } e^{- \frac{\mu }{2} \left[ 2 \ln L - \left( \ln \ln L + \ln 4 \pi + o(1)\right) \right] } \varGamma (\mu )\left( 1 + o(1)\right) \nonumber \\&= \sqrt{\mu } {\frac{ \left( 4\pi \ln L\right) ^{\mu /2} }{\left( 4\pi \ln L\right) ^{1/2}}} \varGamma ( \mu ) L^{-\mu } \left( 1 + o(1)\right) . \end{aligned}$$

(105)

As expected from the formal analysis in Sect. 3.2.2, the leading order behavior is given by a power law with exponent $\mu = (2-\alpha )/\alpha$. By contrast, the existence of a non-trivial logarithmic correction is unexpected, in particular since such a correction does not appear in the exact result $\pi _\mathrm {max}^{\mathrm {HoC}} = (L+1)^{-1}$ for the HoC model ($\alpha = \mu = 1$). Remarkably, the logarithmic factors precisely cancel in this particular case.

B Variational Analysis at the Maximum of $\lambda _k^\mathrm {AN}$

In Fig. 4, we observed that $\lambda _2^\mathrm {AN}$ for the negative gamma distribution with shape parameter s is maximized at $s=1/2$. Furthermore, we claimed that this can be naturally generalized to arbitrary values of k if we replace the shape parameter by 1 / k. As a next question, one might further ask if $\lambda _k^\mathrm {AN}$ is an extremum also with respect to arbitrary variations in the space of base fitness distributions $p_f$. Here, we prove that this is indeed the case for distributions with support limited to the negative real axis.

Let us first evaluate the k-fold convolution of the gamma distribution needed to compute Eq. (42). This is easily achieved using the property that the gamma distribution is closed under the convolution operation, i.e., the k-fold convolution of the gamma distribution with shape parameter s is the gamma distribution with shape parameter sk. If we choose as our base distribution the negative gamma distribution with shape parameter $s=1/k$,

$$\begin{aligned} p_f(x) = p_{1/k}(x) \equiv g_{1/k}(-x), \end{aligned}$$

(106)

the k-fold convolution yields the gamma distribution with unit shape parameter a.k.a. a (negative) exponential distribution, characterized by the CDF $\tilde{F}_{1/k}^{(k)}(z) = e^z$ for $z<0$. Since $\tilde{F}_{1/k}^{(k)}(y_1 + y_2 + \cdots ) = e^{y_1} e^{y_2} \cdots$, Eq. (42) is fully factorized as

$$\begin{aligned} \pi _\mathrm {max}^{\mathrm {AN}} = \left( \int dy \, g_{1/k}(-y) e^{ky} \right) ^L = (k+1)^{-L/k}, \end{aligned}$$

(107)

which is exactly the result for the block model obtained in Eq. (26).

Next, let us derive a useful general formula for $\tilde{F}^{(k)}(z)$. Using the convolution theorem, it satisfies

$$\begin{aligned} \tilde{F}^{(k)}(z)&= \int _{-\infty }^{z} dz' \int _{z'}^{\infty } dy\, p_f(y) \, p^{(k-1)}_f(z'-y) \end{aligned}$$

(108)

where $p^{(k-1)}_f(z)$ is the PDF of the $k-1$ fold convolution of $p_f(z)$. It will later be convenient to exchange the order of integrals:

$$\begin{aligned} \tilde{F}^{(k)}(z)&= \int _{-\infty }^{z} dy\, p_f(y) \int _{-\infty }^{y} dz' \, p^{(k-1)}_f(z'-y) + \int _{z}^{\infty } dy\, p_f(y) \int _{-\infty }^{z} dz' \, p^{(k-1)}_f(z'-y) \nonumber \\&= \int _{-\infty }^{z} dy\, p_f(y) + \int _{z}^{\infty } dy\, p_f(y) \tilde{F}_s^{(k-1)}(z-y). \end{aligned}$$

(109)

In the first equality, we split the integral into two pieces to accommodate the condition $p^{(k-1)}_f(z) =0$ for positive z. In the next equality, we have used the fact that $\tilde{F}^{(k-1)}(0) =1$.

Now, we want to show that $\pi _\mathrm {max}^{\mathrm {AN}}$ is maximized when the base fitness distribution is given by Eq. (106). To this end, let us introduce a small perturbation $p_f(y) = p_{1/k}(y) + \epsilon \eta (y)$, with the properties that $\int dy \, \eta (y) = 0$ and $\eta (y) = 0$ for $y > 0$. Since the probability Eq. (42) is given by the product of 2L terms, there will be 2L linear terms in $O(\epsilon )$, i.e. $\pi _\mathrm {max}^{\mathrm {AN}}$ changes by

$$\begin{aligned} \delta \pi _\mathrm {max}^{\mathrm {AN}}&= \epsilon L \int dy\, \eta (y) \int \left( \prod _{r=2}^{L} dy_r p_{1/k}(y_r) \right) \prod _{l=0}^{L-1} \tilde{F}_{1/k}^{(k)}\left( \sum _{m=1 }^{k} y_{(l + m) \, \text {mod} \,L} \right) \nonumber \\&\quad \ + L \int \left( \prod _{r=1}^{L} dy_r p_{1/k}(y_r) \right) \delta \tilde{F}^{(k)}\left( \sum _{m=1 }^{k}y_{m} \right) \prod _{l=1}^{L-1} \tilde{F}_{1/k}^{(k)}\left( \sum _{m=1 }^{k} y_{(l + m) \, \text {mod} \,L} \right) \nonumber \\&\equiv L (J_1 + J_2). \end{aligned}$$

(110)

The first term is straightforward to evaluate. Since $\tilde{F}^{(k)}_{1/k}\left( \sum _{m=1}^{k} y_{(l + m) \, \text {mod} \,L} \right)$ is factorized, it readily follows that

$$\begin{aligned} J_1&= \epsilon \int dy\, \eta (y) \int \left( \prod _{r=1}^{L-1} dy_r p_{1/k}(y_r) \right) \prod _{l=0}^{L-1} \tilde{F}^{(k)}_{1/k}\left( \sum _{m=1 }^{k} y_{(l + m) \, \text {mod} \,L} \right) \nonumber \\&= \epsilon \int dy\, \eta (y) e^{ky} (k+1)^{-(L-1)/k}. \end{aligned}$$

(111)

To evaluate $J_2$, let us rewrite it in the following way:

$$\begin{aligned} J_2&= \int \left( \prod _{r=1}^{L} dy_r p_{1/k}(y_r) \right) \delta \tilde{F}^{(k)}\left( \sum _{m=1 }^{k}y_{m} \right) \prod _{l=1}^{L-1} \tilde{F}^{(k)}_{1/k}\left( \sum _{m=1 }^{k} y_{(l + m) \, \text {mod} \,L} \right) \nonumber \\&=(k+1)^{-(L-k)/k} \int \left( \prod _{r=1}^{k} dy_r p_{1/k}(y_r) e^{(k-1) y_r}\right) \delta \tilde{F}^{(k)}\left( \sum _{m=1 }^{k}y_{m} \right) . \end{aligned}$$

(112)

The argument of $\delta \tilde{F}^{(k)}$ is the sum of the variables $y_r$ that remain to be integrated over. To make them independent, let us introduce a delta function through the identity

$$\begin{aligned} 1 = \int dY \delta \left( \sum _{m=1 }^{k}y_{m} -Y\right) \varTheta (-Y) \end{aligned}$$

(113)

or, in the Fourier representation,

$$\begin{aligned} 1 = \int \frac{dY dZ}{2\pi } e^{ -i Z (\sum _{m=1 }^{k}y_{m} - Y) } \varTheta (-Y), \end{aligned}$$

(114)

where we impose the negativity of Y by inserting an additional theta function. Using the property $\int dx \delta (x-a) f(x) = \int dx \delta (x-a) f(a)$, we may now complete the integrations over the $y_r$ as

$$\begin{aligned}&\int \frac{dY dZ}{2\pi } \varTheta (-Y) \int \left( \prod _{r=1}^{k} dy_r g_{1/k}(y_r) e^{(k-1) y_r}\right) e^{ iZ (Y - \sum _{m}^{k} y_m )} \delta \tilde{F}^{(k)}(Y) \nonumber \\&\quad = \int \frac{dY dZ}{2\pi } \varTheta (-Y) (k-i Z)^{-1} e^{ iZ Y} \delta \tilde{F}^{(k)}(Y) = \int dY \varTheta (-Y) e^{ k Y} \delta \tilde{F}^{(k)}(Y), \end{aligned}$$

(115)

where we used Jordan’s lemma to evaluate the integral with respect to Z. With this result, $J_2$ is of the relatively simple form

$$\begin{aligned} J_2 = (k+1)^{-(L-k)/k} \int dY \varTheta (-Y) e^{ k Y} \delta \tilde{F}^{(k)}\left( Y \right) . \end{aligned}$$

(116)

Next, let us evaluate $\delta \tilde{F}^{(k)}(z)$. Using Eq. (109), we find that

$$\begin{aligned} \delta \tilde{F}^{(k)}(z)&= \epsilon k \left[ \int _{-\infty }^{z} dy\, \eta (y) + \int _{z}^{\infty } dy\, \eta (y) \tilde{F}^{(k-1)}_{1/k}(z-y) \right] \nonumber \\&= \epsilon k \left[ \int _{-\infty }^{\infty } dy\, \eta (y) + \int _{z}^{\infty } dy\, \eta (y) \left( \tilde{F}^{(k-1)}_{1/k}(z-y) -1\right) \right] \nonumber \\&= \epsilon k \int _{z}^{\infty } dy\, \eta (y) \left( \tilde{F}^{(k-1)}_{1/k}(z-y) -1\right) \nonumber \\&= \epsilon k \int _{-\infty }^{\infty } dy\, \eta (y) \left( \tilde{F}^{(k-1)}_{1/k}(z-y) -1\right) \varTheta (y-z), \end{aligned}$$

(117)

where the factor k comes from the k different choices of $p_f(y)$ in the variation of $\tilde{F}^{(k)}$ and the fact that $\int dy \, \eta (y) =0$ is used to eliminate the first term in the second equality. As expected, this implies that any perturbation made in the range $(-\infty ,z)$ does not change the behavior of $\tilde{F}^{(k)}(z)$. Inserting this result into $J_2$ gives

$$\begin{aligned} J_2&= (k+1)^{-(L-k)/k} \int dY \varTheta (-Y) e^{ k Y} \int dy\, \eta (y) \nonumber \\&\quad \ \times \epsilon k \left( \tilde{F}_{1/k}^{(k-1)}(Y-y) -1\right) \varTheta (y-Y). \end{aligned}$$

(118)

Now, the only technical point left is the integration with respect to Y. The integral domain is determined by two theta functions $\varTheta (-Y)$ and $\varTheta (y- Y)$, but since $\eta (y)$ is assumed to be supported only on the negative real axis, the condition imposed by $\varTheta (-Y)$ is irrelevant. Finally, using the identity

$$\begin{aligned} \int _{-\infty }^{0}dY\, k e^{k Y} \left( 1-\frac{\varGamma \left( \frac{k-1}{k},-Y\right) }{\varGamma \left( \frac{k-1}{k}\right) }\right) = (k+1)^{\frac{1}{k}-1}, \end{aligned}$$

(119)

we find

$$\begin{aligned} J_2 = - \epsilon \int dy\, \eta (y) e^{ky}(k+1)^{-(L-1)/k}. \end{aligned}$$

(120)

Thus, the two terms in Eq. (110) perfectly cancel, which completes the proof that $\delta \pi _\mathrm {max}^{\mathrm {AN}} = 0$.

C General Bounds on $\beta$ for Uniform and Regular Structures with Gaussian Fitness

In this appendix we derive some general upper and lower bounds on the coefficient $\beta$, defined in Eq. (86), for NK structures that are both uniform and regular. For this purpose we write the probability of $\sigma$ being a local optimum as

$$\begin{aligned} \pi _\text {max} = {\mathbb {E}\left[ {\prod _{l=1}^{L} \varTheta \left( -\varDelta _lF(\sigma )\right) }\right] } = {\mathbb {E}\left[ {\prod _{l=1}^L \varTheta \left( -\sum _{r=1}^{|{\mathcal {B}}|} \left( f_r\left( {\downarrow _{B_r}}\varDelta _l \sigma \right) -f_r\left( {\downarrow _{B_r}}\sigma \right) \right) \right) }\right] }.\nonumber \\ \end{aligned}$$

(121)

All fitness values of the partial landscapes $f_r$ are i.i.d. random variables. If $l\in B_r$, then $f_r\left( {\downarrow _{B_r}}\varDelta _l\sigma \right)$ and $f_r\left( {\downarrow _{B_r}}\sigma \right)$ are independent. Otherwise they are identical. Thus effectively only the sum over r with $l\in B_r$ remains. Due to regularity there are $\tilde{k} = \frac{Nk}{L}$ such elements for each l. For different r, the terms are always independent. The left-hand terms are also independent for different l. However the right-hand terms are correlated for different l but the same r, resulting in a non-trivial problem. Using these observations we can directly integrate out all terms $f_r\left( {\downarrow _{B_r}}\varDelta _l\sigma \right)$ and arrive at

$$\begin{aligned} \pi _\text {max} = {\mathbb {E}\left[ {\prod _{l=1}^{L} \varPhi _{\tilde{k}}\left( \sum _{r\;|\;l\in B_r} f_r\left( {\downarrow _{B_r}}\sigma \right) \right) }\right] }, \end{aligned}$$

(122)

where $\varPhi _{\tilde{k}}$ is the cumulative distribution function of the sum of $\tilde{k}$ i.i.d. fitness values. Introducing the short-hand notation $x_r = f_r\left( {\downarrow _{B_r}}\sigma \right)$, we can write the sum as a matrix product

$$\begin{aligned} \pi _\text {max} = {\mathbb {E}\left[ {\prod _{l=1}^{L} \varPhi _{\tilde{k}}\left( ({\mathbf {B}}x)_l\right) }\right] } \end{aligned}$$

(123)

where ${\mathbf {B}}$ is the incidence matrix of the NK structure, i.e. ${\mathbf {B}}_{lr} = b_{l,r} = 1$ if $l\in B_r$ and 0 otherwise.

If the base fitness distribution is a standard normal distribution, then the sum of $\tilde{k}$ i.i.d. fitness values is also normal distributed with variance $\tilde{k}$. Consequently we can simplify as

$$\begin{aligned} \pi _\text {max} = {\mathbb {E}\left[ {\prod _{l=1}^{L} \varPhi \left( \frac{1}{\sqrt{\tilde{k}}}({\mathbf {B}}x)_l\right) }\right] }. \end{aligned}$$

(124)

The random vector $y = \frac{1}{\sqrt{\tilde{k}}}{\mathbf {B}}x$ is then jointly normal distributed with zero mean and covariance matrix $\mathbf {C} = \frac{1}{\tilde{k}}{\mathbf {B}}{\mathbf {B}}^T$. This matrix is positive-semidefinite, and therefore

$$\begin{aligned} \pi _\text {max} = \int _{\mathbb {R}^L} \frac{\mathrm {d}y}{\sqrt{(2\pi )^L\det \mathbf {C}}}\exp \left( -\frac{1}{2}y^T\mathbf {C}^{-1}y + \sum _{l=1}^L \ln \varPhi (y_l)\right) . \end{aligned}$$

(125)

We can shift the integrand by a yet to be specified vector z, which yields

$$\begin{aligned} \pi _\text {max}&= \int _{\mathbb {R}^L} \frac{\mathrm {d}y}{\sqrt{(2\pi )^L\det \mathbf {C}}} \nonumber \\&\quad \ \times \exp \left( -\frac{1}{2}y^T\mathbf {C}^{-1}y -\frac{1}{2}z^T\mathbf {C}^{-1}z -z^T\mathbf {C}^{-1}y + \sum _{l=1}^L\ln \varPhi (y_l+z_l)\right) . \end{aligned}$$

(126)

Absorbing the first term in the exponent into a probability measure, we have again

$$\begin{aligned} \pi _\text {max} = e^{-\frac{1}{2}z^T\mathbf {C}^{-1}z}{\mathbb {E}\left[ {\exp \left( -z^T\mathbf {C}^{-1}y +\sum _{l=1}^L \ln \varPhi (y_l+z_l)\right) }\right] } \end{aligned}$$

(127)

where y is still jointly normal distributed with covariance matrix $\mathbf {C}$.

Notice that the all-ones vector $\bar{1}$ is an eigenvector of $\mathbf {C}$ with the eigenvalue k. This can be seen through the relations ${\mathbf {B}}\bar{1} = \tilde{k}\bar{1}$ and ${\mathbf {B}}^T\bar{1} = k\bar{1}$, as there are exactly $\tilde{k}$ ones in each row of ${\mathbf {B}}$ and k ones in each column. Thus let the $z_l = \bar{z}$ be equal for all l. Then

$$\begin{aligned} \pi _\text {max} = e^{-L\frac{\bar{z}^2}{2k}}\prod _{l=1}^L{\mathbb {E}\left[ {\exp \left( \sum _{l=1}^L\left( \ln \varPhi (y_l+\bar{z}) - \frac{\bar{z}}{k} y_l\right) \right) }\right] }. \end{aligned}$$

(128)

1.1 C. 1 Lower Bound

By Jensen’s inequality we have

$$\begin{aligned} \pi _\text {max} \ge e^{-L\frac{\bar{z}^2}{2k}}\prod _{l=1}^L\exp \left( {\mathbb {E}\left[ {\ln \varPhi (y_l+\bar{z}) - \frac{\bar{z}}{k} y_l}\right] }\right) . \end{aligned}$$

(129)

Because $y_l$ has a symmetric distribution, the mean of $\bar{z} y_l$ vanishes. The variance of $y_l$ is always 1, because by regularity and uniformity the diagonal elements of ${\mathbf {B}}{\mathbf {B}}^T$ are $\tilde{k}$, which is canceled to 1 by the pre-factor in $\mathbf {C}$. If we then assume $\bar{z}$ to be increasing in our limit of interest and noting that the Gaussian has a tail falling much quicker to zero than the tail of $\ln \varPhi$ falls to $-\infty$ at $x\rightarrow -\infty$, we can establish the bound

$$\begin{aligned} \pi _\text {max} \ge e^{-L\frac{\bar{z}^2}{2k}}\prod _{l=1}^L\exp \left( {\mathbb {E}\left[ {\varPhi (y_l+\bar{z})-1}\right] }(1+o(1))\right) \end{aligned}$$

(130)

which can be evaluated to

$$\begin{aligned} \pi _\text {max} \ge \exp \left( -L\frac{\bar{z}^2}{2k} + L\left( \varPhi \left( \frac{\bar{z}}{\sqrt{2}}\right) -1\right) (1+o(1))\right) . \end{aligned}$$

(131)

If we choose $\bar{z} = 2\sqrt{\ln k}$, then asymptotically for large k

$$\begin{aligned} \pi _\text {max} \ge \exp \left( -L\left( \frac{2\ln k}{k} + \mathcal {O}\left( \frac{1}{k \sqrt{\ln k}}\right) \right) \right) . \end{aligned}$$

(132)

Note that choosing $\bar{z} = \tilde{z} \sqrt{\ln k}$ with $\tilde{z} < 2$ will not give a better bound, as the right-hand term in the exponent in Eq. (131) would then dominate and approach zero more slowly than $\frac{\ln k}{k}$. This shows that $\beta \le 2$ for uniform and regular structures. With the MF model, which is uniform and regular, we have an example of a realization of $\beta = 2$. This shows that the bound is tight.

1.2 C. 2 Upper Bound

Starting from Eq. (128) we can find an upper bound by simply optimizing each term in the sum. The resulting sum is then an upper bound on the integrand, and because the expectation is taken with respect to a probability measure, it is bounded by the same value as well. If $0< \frac{\bar{z}}{k} < \frac{1}{\sqrt{2\pi }}$, the optimum must be at $y_l^\star +\bar{z} > 0$. Then by using the simplification $\ln \varPhi (y_l+z_l) \le \varPhi (y_l+z_l) -1$, the optimum is found to be at

$$\begin{aligned} y_l^\star = \sqrt{2\ln \left( \frac{k}{\sqrt{2\pi }\bar{z}}\right) }-\bar{z}. \end{aligned}$$

(133)

Inserting $y_l^\star$ back into the simplified argument of the expectation and assuming $\bar{z}\rightarrow \infty$ in the limit of interest we find

$$\begin{aligned} \pi _\text {max} \le \exp \left( -L\frac{\bar{z}^2}{2k} -L\left( \frac{\bar{z}}{k\sqrt{2\ln \left( \frac{k}{\sqrt{2\pi }\bar{z}}\right) }}(1+o(1)) + \frac{\bar{z}}{k}\sqrt{2\ln \left( \frac{k}{\sqrt{2\pi }\bar{z}}\right) } - \frac{\bar{z}^2}{k}\right) \right) . \end{aligned}$$

(134)

The left-most and right-most terms are of equal order, but the second one from the left is always of less significant order than the second from the right, as long as $\bar{z} = o(k)$.

The second term from the right becomes equal in order to the other two if $\bar{z} = \tilde{z}\sqrt{2\ln k}$ with a positive constant $\tilde{z}$. This satisfies the condition $\bar{z} = o(k)$ while still $\bar{z} \rightarrow \infty$, as required by previous assumptions (given that $k\rightarrow \infty$ in the limit of interest). With this we have

$$\begin{aligned} \pi _\text {max} \le \exp \left( -L\left( \frac{\ln k}{k}(2\tilde{z} - \tilde{z}^2) + \mathcal {O}\left( \frac{\ln \ln k}{k}\right) \right) \right) . \end{aligned}$$

(135)

The bound is best for $\tilde{z} = 1$, and so:

$$\begin{aligned} \pi _\text {max} \le \exp \left( -L\left( \frac{\ln k}{k} + \mathcal {O}\left( \frac{\ln \ln k}{k}\right) \right) \right) \end{aligned}$$

(136)

showing that $\beta \ge 1$ for regular and uniform NK structures with Gaussian fitness. This bound is realized by the AN and BN structures, for example, and thus it is tight.

D Simulation of the Number of Local Maxima

As first realized in [6], the choice of a Gaussian base fitness distribution greatly simplifies the computation of $\pi _\mathrm {max}$ through the numerical evaluation of Eq. (25), as it allows us to take advantage of an efficient algorithm. With this choice, the integrals over $\mathbf {q}$ and $\mathbf {y}$ can be cast into the form of multi-dimensional Gaussian integrals which may be evaluated for generally defined NK structures. Once these integrals are evaluated, we may construct a covariance matrix $\varSigma$ that satisfies the relation

$$\begin{aligned} \pi _\mathrm {max}= \int \mathcal {D} \mathbf {u} \exp \left( -\frac{1}{2}\sum _{j l} u_j \varSigma ^{-1}_{j l} u_l \right) , \end{aligned}$$

(137)

where $\int \mathcal {D} \mathbf {u} = \frac{1}{\sqrt{(2\pi )^L \det \varSigma }} \int _{0}^{\infty } \prod _j du_j$ and the matrix elements of $\varSigma$ are given by

$$\begin{aligned} \varSigma _{j l} = {\left\{ \begin{array}{ll} 2 \sum _{r} b_{l,r} &{} j = l \\ \sum _{r} b_{j,r} b_{l,r} &{} j \ne l. \end{array}\right. } \end{aligned}$$

(138)

Thus, the problem reduces to determining the probability that all the entries of the Gaussian random vector realized by the covariance matrix $\varSigma$ are positive. Since finding the probability for rectangular domains of multivariate Gaussian distribution is a well-known problem, an efficient algorithm has been known for a long time [28] and its implementation has been provided by the original authors as an R library [27].

Roughly speaking, this algorithm consists of two steps: i) transforming to an integral over a unit rectangular domain such that a rejection-free Monte-Carlo simulation is possible and ii) finding an ordering of loci that minimizes the variance of the Monte-Carlo step. However, since the loci in the NK models we consider in this review are statistically identical, the second step is irrelevant in this particular case. Thus, here we describe briefly how the transformation can be achieved from Eq. (137).

Since $\varSigma$ is positive-definite, the Cholesky decomposition ensures that there exists a triangular matrix C such that $\varSigma = C C^T$. The substitution $\mathbf {u} = C \mathbf {x}$ then diagonalizes the integral at the cost of nontrivial integral domain,

$$\begin{aligned} \pi _\mathrm {max}= \frac{1}{(2\pi )^{L/2}}\int _{ \mathbf {x} \in \mathcal {R}}\prod _{j=1}^{L} dx_j \exp \left( -\frac{1}{2}\sum _{j =1} ^L x_j^2 \right) , \end{aligned}$$

(139)

where the domain $\mathcal {R} = (a_1, \infty ) \times (a_2, \infty ) \times \cdots (a_L, \infty )$ and $a_j = - \sum _{l=1}^{j-1} x_l C_{j l } / C_{j j }$. Next, performing the canonical transformation to a standard uniform distribution $z_i = \varPhi (x_i)$, where $\varPhi (x)$ is the CDF of the standard Gaussian distribution, the integral becomes

$$\begin{aligned} \pi _\mathrm {max}= \int _{ \mathbf {z} \in \mathcal {R'}} \prod _{j=1}^{L} dz_j, \end{aligned}$$

(140)

where $\mathcal {R'} = (d_1, 1) \times (d_2, 1) \times \cdots (d_L, 1)$ and $d_j = \varPhi ( - \sum _{l=1}^{j-1} \varPhi ^{-1}(z_l) C_{j l} / C_{j j })$. Finally, another linear transformation $z_j = d_j + w_j (1- d_j)$ brings the integral into the form

$$\begin{aligned} \pi _\mathrm {max}= \int _{\mathbf {w} \in \mathcal {R''} } \prod _{j=1}^{L} (1- d_j) dw_j, \end{aligned}$$

(141)

where $\mathcal {R''} = (0,1)^{L}$. Now that the integral domain is the L-dimensional unit rectangle, this integral can be evaluated by sampling L random variables from a uniform distribution on (0, 1) and subsequently estimating the weight factors $d_j$.