Beckman 792 *** 661-1-R1K

It's a resistor array, each contains 1k resistor. The connection is:

Pin 16: common
Pin 1-15: individual resistor, each 1 kilo ohm.

Roku 2 XS Internals

List of the chips inside:

• BCM 59002 From Broadcom: Mobile Power Management Unit
• SMC LAN 9512: Hi Speed USB 2.0 Hub and Ethernet controller
• AKM 4430 ET: 3.3volt 2 bit Stereo Audio digital audio converter
• BCM 4336 (inside the metal shield): Wireless LAN controller/transceiver
• BCM 20702 (inside the metal shield): Bluetooth controller
• TI 2051B:  Power distribution switch
• Samsung K4P2G324EC: LPDDR2 RAM 2-Gigabit (512 MByte)
• Hynix HY27UF082G2B NAND Flash (2Gbit - 256MB)

The CPU inside the SoC chip (BCM-4336), according to Broadcom website, is based on ARM Cortex-M3 architecture.  The chip integrates 802.11 a/b/g/n Wifi transceiver.

Another interesting site, but for Roku 2 XS: http://www.techrepublic.com/photos/cracking-open-the-roku-2-xs/6274959?seq=20&tag=thumbnail-view-selector;get-photo-roto

A Google interview

Question:
There are some data represented by(x,y,z). Now we want to find the K^th least data. We say (x1, y1, z1) > (x2, y2, z2) when value(x1, y1, z1) > value(x2, y2, z2) where value(x,y,z) = (2^x)*(3^y)*(5^z).

Now we cannot get it by calculating value(x,y,z) or through other indirect calculations as lg(value(x,y,z)). How to solve it?

Answer:

Say we have a set, H, which has these 3-tuple data.

Value(a) = 2^x3^y5^z
if Value(H[m]) > Value(H[n]):
      H[m] > H(n)

Test:
Assume h1 = (x1,y1,z1) = (0,0,0)
     Value(h1) = 2⁰ * 3⁰ * 5⁰ = 1*1*1 = 1

(x2,y2,z2) = (-1,-1,-1)
    Value(h2) = 2^-1* 3^-1* 5^-1 = 0.5 * 0.33 * 0.2, a value greater than 0 but less than 1

From the above, Value(h1) > Value(h2), so according to the statement above h1 should be > h2, or (0,0,0) > (-1,-1,-1).

One quick guess is that if z1 > z2,  the Value(x1,y1,z1) would be greater than Value(x2,y2,z2), for z=[some number to infinite], because the 5^z would be more significant than the contribution of 2^x*3^y alone.