All Figures

$\begin{figure} \par\includegraphics[width=8.5cm,clip]{peebles.ps}\end{figure}$ Figure 1: A schematic demonstration of Peebles' reconstruction of the trajectories of the members of the local neighbourhood using a variational approach based on the minimization of the Euler-Lagrange action. In most cases there is more than one allowed trajectory due to orbit crossing (closely related to the multistreaming of the underlying dark matter fluid). The pink (darker) orbits correspond to taking the minimum of the action whereas the yellow (brighter) orbits were obtained by taking the saddle-point solution. Of particular interest is the orbit of N6822 which in the former solution is on its first approach towards us and in the second solution is in its passing orbit. A better agreement between the evaluated and observed velocities was shown to correspond to the saddle point solution.
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In the text

$\begin{figure} \par\includegraphics[width=8.3cm,clip]{piza-marseille.ps} \end{figure}$ Figure 2: Solving the reconstruction problem as an assignment problem. An example of a system of N=9 particles is sketched. The cost matrix is shown. For example the entry C₃₂ is the cost of getting from the Lagrangian position $\vec{q}_2$ to the Eulerian position $\vec{x}_3$ . The cost of this transport is $C_{32}=(\vec{q}_2-\vec{x}_3)^2$ . The total cost $C_{\rm T}$ of a given permutation is the sum of all such costs which are chosen for that permutation. That is $C_{\rm T}=\sum _i C_{ij(i)}$ . Generally, there are N! possible permutations (or $C_{\rm T}s$ ). Only one of these N! permutations is the optimal assignment: ${C_{\rm optimal}}= \min_{j(\cdot)}(C_{\rm T})$ . For this system there are 362 889 different costs possible, each obtained by a different permutation. Algorithms with factorial complexity are clearly impractical even for small systems. However, assignment algorithms have complexities of polynomial degrees.
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In the text

$\begin{figure} \par\includegraphics[width=8.8cm,clip]{exactpizaseed1-exactpizaseed2-2.ps} \end{figure}$ Figure 3: The lack of uniqueness in the results of two runs using a stochastic algorithm to solve the assignment problems. In the stochastic algorithm, based on pair-interchange, one searches for, what is frequently referred to in pure mathematics as, a monotonic map instead of a true cyclic monotone map. The red and blue points (stars and boxes respectively) are the perfectly-reconstructed Lagrangian positions using the same stochastic code with two different random seeds. The outputs of the two runs do not coincide, meaning that the reconstructed Lagrangian positions (and hence peculiar velocities) depend on the initial random assignment and on the random pair interchange. The lack of uniqueness in this case is superficial and a shortcoming of the chosen numerical method. Such difficulties do not arise when deterministic algorithms are used to solve the assignment problem.
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In the text

$\begin{figure} \par\includegraphics[width=7.5cm,clip]{henonmarseille-1.ps} \end{figure}$ Figure 4: Hénon's mechanical device which solves the assignment problem. The device acts as an analog computer. The rows A_i remain parallel to the y axis, are constrained to move vertically and have positive weights. The columns B_j remain parallel to the x axis, can only move vertically and have negative weights. Vertical studs are placed on the columns, in such a way that each stud enforces a minimal distance between row A_i and column B_j. Initially all rods are kept at fixed positions by stops Pand Q. Then the rods are released by removing the stops Q and P and the system starts evolving. Rows go down and columns go up and aggregates of rows and columns are made. Thus a complex evolution takes place. The rules for the formation of aggregates is demonstrated by a simple example in the figure that follows.
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In the text

$\begin{figure} \par\includegraphics[width=8.5cm,clip]{henonmarseille-2.ps} \end{figure}$ Figure 5: A step-by-step ((a) to (l)) progress of Hénon's algorithm on a simple example with three initial and final positions (columns and rows) is shown. The table on the top right shows the values of the costs (distances between rows and columns). When executing the algorithm by hand, it is convenient to keep track of the distances between the rows and the studs, i.e. the quantities $\gamma _{ij}=\alpha _i-\beta _j-c_{ij}$ where $\alpha _i$ is the variable height of the row A_i and $\beta _j$ is the variable height of the column B_j. A graph of the initial state is made with the first contact being made between the second row and second column which have the largest cost (note that the code originally written by Hénon, in fact, finds the maximum cost; for our purpose of finding the minimum cost, the matrix elements c_ij should be subtracted from a large number). Thus, at this point the entries in the $\gamma$ matrix change since now the second row and column are in contact and hence c₂₂=0. Obviously the distances between all other rows and columns should also be modified. The second matrix shows the new distances and automatically a contact is made between third row and third column whose separation is now also zero. Since the second and third rows cannot move now, the next contact is made between the first column and the second row and the break occurs where the total force on the branch is weakest. Since this is where the second row meets the support Q, part of the Q tree is captured by the P tree, as demonstrated. The next contact can now only be made between the first row and the first column and the break occurs at the weakest branch. The equilibrium position is now reached where each row is supported by one column. In the final optimal assignment Col. 1 is attached to row 1, Col. 2 goes to row 2 and Col. 3 is associated to row 3. For this simple exercise one can easily see that this procedure achieves the maximum cost (which in this example is 21).
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In the text

$\begin{figure} \par\includegraphics[width=8.5cm,clip]{fig2-periodic-1-10to4.ps} \end{figure}$ Figure 6: Test of MAK reconstruction of the Lagrangian positions, using a $\Lambda$ CDM simulation of 128³ particles in a box of size $200 ~{\rm Mpc}^3/h^3$ . In the scatter plot, the dots near the diagonal are a scatter plot of reconstructed initial points versus simulation initial points for a grid of size 6 Mpc/h with about 20 000 points. The scatter diagram uses a quasi-periodic projection coordinate $\tilde{\vec{q}}\equiv (q_x+\sqrt 2 q_y+{\sqrt 3} q_z)/(1+ \sqrt 2+\sqrt 3)$ which guarantees a one-to-one correspondence between $\tilde{\vec{q}}$ values and points on the regular Lagrangian grid. The upper left inset is a histogram (by percentage) of distances in reconstruction mesh units between such points; the first bin corresponds to perfect reconstruction; the lower-inset is a similar histogram for reconstruction on a finer grid of about 3 Mpc/h using 100 000 points. With the 6 Mpc/h grid 62% of the points, and with 3 Mpc/h grid more than 50% of the points, are assigned perfectly.
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In the text

$\begin{figure} \par\includegraphics[width=8.8cm,clip]{mak1-piza_nonmonotonic_seed1.ps} \end{figure}$ Figure 7: A comparison of a PIZA-type, stochastic method, versus our MAK, deterministic method of solving the assignment problem is shown. The outputs of the simulation used for Fig. 6 have also been used here. In the stochastic method, almost one billion pair interchanges are made, at which point the cost-reduction cannot be reduced anymore. However, even with such a large number of interchanges, the number of exactly reconstructed points is well below that achieved by MAK. The upper sets are the scatter plots of reconstructed versus simulation Lagrangian positions for PIZA (left top set) and for MAK (right top set). The lower sets are histograms of the distances between the reconstructed Lagrangian positions and those given by the simulation. The bin sizes are taken to be slightly less than one mesh. Hence, all the points in the first (darker) bin correspond to those which have been exactly reconstructed. Using PIZA, we achieve a maximum of about 5000 out of the 17 178 points exactly-reconstructed positions whereas with MAK this number reaches almost 8000 out of 17 178 points. Note that, for the sole purpose of comparing MAK with PIZA it is not necessary to account for periodicity corrections, which would improve both performances equally. Accounting for the periodicity improves exactly-reconstructed MAK positions to almost 11 000 points out of 17 178 points used for reconstruction, as shown in the upper inset of Fig. 6. Starting with an initial random cost of about 200 million (Mpc/h)² (5000 in our box unit which runs from zero to one), after one billion pair interchange, a minimum cost of about 1 960 000 (Mpc/h)² (49 in our box unit) is achieved. Continuing to run the code on a 2 GHz DEC alpha workstation, consuming almost a week of CPU time, does not reduce the cost any further. With the MAK algorithm, the minimum cost is achieved, on the same machine, in a few minutes. The cost for MAK is 1 836 000 (Mpc/h)² (45.9 in box units) which is significantly lower than the minimum cost of PIZA. Considering that the average particle displacement in one Hubble time is about 10 Mpc/h (about 1/20 of the box size) this discrepancy between MAK and PIZA costs is rather significant.
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In the text

$\begin{figure} \par\includegraphics[width=8.8cm,clip]{redshift-mak-sim.ps} \end{figure}$ Figure 8: Reconstruction test in redshift space with the same data as that used for real-space reconstruction tested in the upper left histogram of Fig. 6. The velocities are smoothed over a sphere of radius 2 Mpc/h. The observer is taken to be at the centre of the simulation box. The Lagrangian reconstructed points are plotted against the simulation Lagrangian positions using the quasi-periodic projection coordinates used for the scatter plot of Fig. 6. The histogram corresponds to the distances between the reconstructed and the simulation Lagrangian positions for each Eulerian position. The bins of the histogram are smaller than one mesh and the first one corresponds to exactly-reconstructed Lagrangian positions. We obtain $43\%$ of perfect reconstruction.
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In the text

$\begin{figure} \par\includegraphics[width=8.5cm,clip]{redshiftmak-realmak.ps} \end{figure}$ Figure 9: Comparison of redshift versus real space reconstruction allows a test of the robustness of MAK reconstruction against systematic errors. The same data as those used for Fig. 8 has been used to obtain the above result. In both real and redshift space, large number (almost $50\%$ ) of the exactly reconstructed points, i.e. those which fall in the first bin of the histogram, correspond to the same particles.
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In the text

$\begin{figure} \par\includegraphics[width=8.5cm,clip]{failed.ps} \end{figure}$ Figure 10: A thin slice (26 Mpc/h) of the box (of sides 200 Mpc/h) of the $128^3, \Lambda$ CDM simulation the same as that used for Fig. 6 is shown. Out of the 100 000 points which have been used for reconstruction, more than half are exactly reconstructed. The points which fail exact reconstruction are highlighted (yellow, brighter points). The plot verifies that these points lie mainly in the overdense regions where we do not expect our reconstruction hypothesis to hold well.
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In the text