I won't comment on your conclusion, but your analysis is *completely* wrong. Study the diagram carefully! The answer is evident by inspection.
Sorry I may still be *completely* wrong but I meant D4 as my first diode and D1 as my last. Or from most to least current : D4, (D5,D6)(same), D3, D2, D1 is my analysis still wrong?
I dont understand? current in D3 > D2 > D1... since the resistors are there its restricting the current going to D2 and then again to D1
You seem to be saying that the current that goes through D1 (the "last" diode) has to first go through all of the other diodes and resistors. Is that really what you are saying?
It has to go thru the first resistor and D4 D5 and D6 then the resistors around D3 D2 oh would that make D3 and D2 current the same as well? the resistor current flows around them straight to D1?
When the resistance is infinite, it's as though the resistor isn't there. Under those conditions, how much current is flowing in the diode? As the resistance decreases, what happens to the diode current? So if I have two of these circuits in series (so that both have Io going through them), and the resistance of the first is smaller than the resistance of the second, which diode has more current flowing in it? With this in mind, and thinking only of the branch with D1, D2, and D3 in it, which of these three diodes has the most current? Call that Dx. Now do a similar analysis thinking only of the branch with D4, D5, and D6 to find which of those three will have the most current in it. Call that Dy. Now the only question is which diode, Dx or Dy, has more current in it. To explore this, first ask what the answer would be if the two resistors in the first branch were make open circuits (i.e., made infinite resistance). Now ask what the answer would be if those two resistors were made shorts (i.e., zero resistance). If you get the same answer in both cases, the you have your answer. But if you don't get the same answer, then that means that there is some non-zero, non-infinite value of those resistors that would result in the current in Dx and Dy being the same and if you increase those resistances then one diode has more current and if you decrease them then the other diode has more current.
If the current flows out of the positive terminal of the battery, then over and through D4, D5, and D6, and then up through the branch that has D3, and D2, how does that current ever get to the negative terminal of the battery? Remember, the current in any component in a circuit must be part of a complete, closed loop that only goes through any given component in that loop once.
But if the current goes through D4,D5,D6 and then through D1,D2,D3 and then back through D4,D5,D6 it will have gone through come of those components more than once, right? Why, after going through D4, D5, and D6, would it not simply go straight to the negative terminal of the battery? Ask yourself which of those diodes, if any, are going to be forward biased and which of those diodes, if any, are going to be reverse-biased.