This circuit can be redrawn into a simpler version. The circuit so formed will exactly resemble as the "Wheatstone's bridge" with a resistor in place of a galvanometer(which is commonly used for checking slightest of the current passing through it).
First mark the points at the junctions as "1","2","3","4" respectively from left to right. Then you can observe that point 1 to point 2 has resistor no.1.so draw it first. Then we can observe that point 1 to point 3 has resistor no. 4, so draw Resistor no. 4 from the point1 (marked before) to point3. Now we see that point 2 to point 3 has resistor no. 2 in between, so draw it between the points 2 and 3 (both points are marked in the earlier steps so just draw a resistor between them). Now draw resistor 3 and resistor 5 from points 3 and 4 respectively to point 5. Now point 1 resembles point "A" and point 5 resembles point "B"(as no resistor is placed between point "A" and point 1 and between point "B" and point 5). Hint: Draw the straight i.e. without any curve between two points.
You'll observe that a Rhombus resembling Quadrilateral is formed with a resistor on all its side and a diagonal. Prefer to draw same size of the wire. For solving and knowing more about the Wheatstone's bridge thus formed.