Your question is muddled and unclear (in my professional opinion - you may think otherwise and that's OK), but I think I can make one thing clearer.
- If you have two entangled systems, $A$ and $B$, and perform independent experiments on either side, then nothing about the choice of experiment you perform on $B$ will have any effect on the local results from $A$.
Note the keywords independent and local. If you look for correlations between the two experiments, then the situation can change. Thus, if you post-select the results from $A$ to include only the runs when $B$ performed experiment $B_1$ and obtained outcome $B_1(+)$, then the post-selected results from $A$ will depend, in general, on both the measurement and the outcome.
Having said this, I will continue to hammer on on what's been said before: your question is essentially undecidable unless you specify exactly what type of entanglement you imagine both photons to share. You seem to have a magical view of entanglement (which is alarmingly common) in which touching system $B$ will immediately and irrevocably change everything about system $A$, which is very, very far from the truth.
Allow me to elaborate by presenting two scenarios, which I believe are consistent with how you phrased your question:
Scenario 1
The photons are entangled in polarization, sharing the state $|HH\rangle+|VV\rangle$, but are otherwise indistinguishable.
In this case, nothing you do on $B$ will have any effect on the interference pattern (or lack thereof) shown in $A$, because the entanglement simply does not couple to the spatial modes.
Scenario 2
The photons are spatially entangled, in such a way that when photon $A$ hits its left slit then photon $B$ hits its right slit, and vice versa, i.e. they share the state $|LR\rangle+|RL\rangle$.
In this case, your actions on $B$, both your choice of measurement and the corresponding experimental outcome, can affect the results of $A$ if you postselect on the appropriate results on $B$, at least in principle. Other choices of measurement in $B$ will not have any effect on $A$. In particular:
If you perform a path measurement on $B$, where you detect which-way information for it, then (in this scenario) you also make available which-way information on $A$, which precludes it from showing an interference pattern. This is independent of the temporal sequence and spacing between the two measurements, and on whether the measurement in $B$ was randomized or not.
Suppose, on the other hand, that you do nothing with the photon in $B$ and allow it to run through the slits undisturbed. In this case (for this scenario), you will not observe any interference pattern on either photon. The reason for this is that each photon contains (in principle) which-way information on the other, so that the other one cannot interfere. The reduced state for both is $$\rho=\frac12\left(|L\rangle\langle L|+|R\rangle\langle R|\right),$$ and no coherent dynamics can be extracted from it. This is independent of the temporal sequence and spacing between the two measurements, and on whether the measurement in $B$ was randomized or not.
On the other hand, within this scenario there is indeed one measurement choice on $B$ which can restore the interference pattern on $A$, and it is of course the quantum eraser scheme. If you measure in the basis $\{|+\rangle,|-\rangle\}=\{|L\rangle+|R\rangle,|L\rangle-|R\rangle\}$, then the detector on $A$ will not turn up any interference pattern, but if you post-select and separate the counts on $A$ which coincided with $|+\rangle$ and $|-\rangle$ detections on $B$, then the blob will separate into two complementary interference patterns.
This behaviour is independent of the temporal sequence between the two measurements, their temporal separation, and whether the measurement in $B$ was randomized or not. If the measurement in $A$ is performed before the randomized decision to measure in $B$, then you have the delayed-choice quantum eraser experiment, which has been analyzed in detail elsewhere and which I will not attempt to clarify.
Note also that in the latter case, there is no measurement at $A$ that will enable you to predict the outcome in $B$. This is, again, exactly the same as with the delayed-choice quantum eraser, regardless of however many photons you send.
These two scenarios demonstrate that your question is ill-defined - the precise answer depends on information not contained in the question.
Some additional notes:
The measurement outcome is independent of however many photons you send through. Even in case 3 of scenario 2, if you try to send a million photons and observe the screen before you measure (or decide the measurement basis, or toss your coin, or whatever) on $B$, what you'll see in the screen in $A$ is... nothing, just an interferenceless blob. This is because each photon is an independent run of the experiment, and will have an independent result on the $|\pm\rangle$ basis, so if you add them up you are already performing a decision of what you'll do with the results from $B$.
The need for many photons to detect an interference pattern is a red herring; the "objection of many" is only raised by people with a hazy understanding of the quantum eraser experiment.
The goal is to detect coherence between $|L\rangle$ and $|R\rangle$ on the $A$ side, and this can be done without the need for a large number of photons, by being more clever about it. More specifically, you should collect the light behind each slit with an optical fiber and then shine them onto a beam splitter with detectors $M$ and $N$ on the output ports, calibrated so that $|L\rangle+|R\rangle$ will go exclusively into $M$ and $|L\rangle-|R\rangle$ will go exclusively into $N$. Thus a single count in $M$ rules out $|-\rangle$ and a single count in $N$ rules out $|+\rangle$.
This scheme is (provably) optimal. It will, of course, not give you enough information to completely determine the state from a single run of the experiment; this, however, is a fundamental limitation and the only way to get around it is via quantum state tomography.
This answer represents my best attempt in good faith to make sense of your question, which is a very tall proposition. If you feel I have not understood any aspect of the question, I would encourage you to think long and hard about how you are phrasing your post, instead of chucking more rep at it. Science depends crucially on its communication, and this includes making sure that one's audience can understand the material. If the entire audience says it is too obscure, blaming the audience will not benefit you at all; instead, try to clarify the material, and politely ask your audience what else you could do to make the point more evident. (If you want to, of course. Blaming the audience is lots of fun too.)
I tried answering this by going to the XCOM database where you can get a calculation of the stopping power of elements and compounds.
First - pick a few likely candidates. I found a table of elements with density which is a good place to start. The highest density elements are also among the highest Z ones:
proton
density name Z m density
-------+----------+---+-----+---------
15.4 Thorium 90 232 5.97
16.65 Tantalum 73 181 6.72
18.95 Neptunium 93 237 7.44
19.32 Gold 79 197 7.75
19.35 Tungsten 74 183 7.82
19.84 Americium 95 243 7.76
20.2 Uranium 92 235 7.91
21.04 Rhenium 75 185 8.53
21.45 Platinum 78 195 8.58
22.4 Iridium 77 191 9.03
22.6 Osmium 76 187 9.19
I got the first three columns from the above link, then added the atomic mass of the most stable (or one of the stable) isotopes, and used these to estimate proton density (from which electron density follows). Units of density and proton density are g/cm3.
I decided to plot the XCOM output for four elements: Pb, Os, Ir and U. You could repeat for others - but this gives a general direction. The result (log-log scale):
As you can see, lead is much worse than the other three - and the other three are almost indistinguishable at high energies, but have a little bit of difference at lower energies (where the binding energy of the electrons is comparable with the energy of the incoming radiation, and photoelectric effects dominate).
Whether Uranium or Osmium/Iridium "win" depends on the energy distribution of your "all energies" input beam - but this approach should give you an idea on how to tackle the question. Clearly, lead is worse than either of these three for most energies (it's better than Uranium just below Uranium's K edge...).
Best Answer
Each photon will be at rest relative to the other, since they both travel in the same direction at the same speed. A photon at rest has zero frequency, hence zero energy. So, at least to first order, there should be no gravitational interaction between the two photons within their mutual rest frame.
I'm not sure what an outside observer would see. It seems to be accepted that gravitational waves interact with each other very very weakly, which suggests that the same should be true of electromagnetic waves.