Electrical Engineering Crossword (Issue 292)

The answers to Circuit Cellar’s November electronics engineering crossword puzzle are now available.292-crossword-(key)


  1. BITS—A nibble is 4 of these
  3. MICRO—Metric Prefix for 0.000001
  4. PATCH—To re-route a signal to a different circuit
  5. TOKEN—Used for authentication
  6. RELAY—A switch that is actuated by another electrical signal
  7. JITTER—The deviation of some aspect of a digital signal’s pulses
  8. RHEOSTAT—A variable resistor
  9. LUX—Lx
  10. VOLTA—Italian physicist who invented the first batteries
  12. CONDUIT—Wire piping
  13. PIGTAIL—Short wire connecting components
  14. RECTIFIER—A diode, used for converting AC into DC


  1. SCHOTTKY—High-speed diode that has very little junction capacitance
  3. INRUSH—A sudden input current surge
  4. CRESTFACTOR—The ratio of the peak value to the RMS value (two words)
  5. MICROFARAD—1,000,000 pF
  6. ANALOG—Constant signal processing



Electrical Engineering Crossword (Issue 291)

The answers to Circuit Cellar’s October electronics engineering crossword puzzle are now available.291crossword (key)


4.     PARAMETRON—Phase-locked oscillator

8.     UNBALANCED—Single-ended

9.     STRAY—Unwanted capacitance

10.   LEYDENJAR—Early capacitor [two words]

12.   ELECTROLYTE—Conducting fluid

14.   CROSSTALK—Caused when one circuit’s signal creates an unwanted effect on another

16.   ANECHOIC—Absorbs sound or electromagnetic wave reflections

17.   BIFILAR—Used in bipolar power-supply transformers to improve output voltage symmetry

18.   CRYSTALRECTIFIER—Semiconductor diode [two words]

19.   DOPPLEREFFECT—Frequency change that occurs when emitter and receiver move in unison [two words]


1.     BLEEDER—A resistor that draws the critical amount of load current

2.     GAUSSMETER—Detects magnetic anomalies

3.     HETERODYNE—Two frequencies combine to produce new ones

5.     SURFACEMOUNT—Place components directly on PCBs [two words]

6.     HASH—Garbage or gibberish

7.     GALVANOMETER—Measures small voltages

11.   RECTIFIER—Passes current in only one direction

13.   CATWHISKER—Sharp, flexible wire that connects to a semiconductor crystal’s surface [two words]

15.   ANOTRON—Cold-cathode-glow discharge diode


Electrical Engineering Crossword (Issue 290)

The answers to Circuit Cellar’s September electronics engineering crossword puzzle are now available.CrosswordEmptyGrid (key)


2.     THREE—Trivalent valence

3.     PHYSICS—Kilby’s Noble Prize in 2000

5.     INVERTER—Converts DC to AC

8.     BATCH—BAT file

9.     MAXIM—Founded ARRL in 1914

10.   KEYBOARD—If you are AFK, what are you away from?

11.   UPENN—University that housed the ENIAC in a 30’ × 40’ room

12.   HERTZ—1 cycle per second

14.   NIBBLE—4 bits

17.   EXPLAINER—Asimov was the great what?


1.     TRACK—PCB path

3.     PATCH—Quick fix

4.     SNIFFER—Used to monitor network traffic

6.     MAXWELL—A Gauss is one of these per square centimeter

7.     IBM—”Big Blue”

8.     BOOLE—“An Investigation of the Laws of Thought” (1854)

13.   TOGGLE—Move from setting A to B

15.   BLUE—Screen of death

16.   NINE—A nonet is a group of what?

17.   EW—Exawatt

Electrical Engineering Crossword (Issue 289)

The answers to Circuit Cellar’s August electronics engineering crossword puzzle are now available.289PuzzleGrid (key)


1.     FAST—Ethernet at 100 Mbps

3.     FAB—IC factory

6.     XOR—Logic gate

7.     VERILOG—HDL created in the early 1980s by Goel and Moorby

9.     MIL—0.001 inches = 25.4 what?

10.   AMPHOUR—Current flow over time [two words]

13.   SANTOS—Greek national soccer team manager with a degree in electrical engineering

14.   NOLEAD—Quad, flat, … [two words]

16.   BUCK—Step-down

17.   FEMTO—0.000000000000001

18.   GND—Ground pin

19.   NULL—Zero


2.     SLICE—Wafer or substrate

3.     FILO—Antonym for FIFO

4.     QUINARY—Base-5

5.     CODERDECODER—CODEC [two words]

7.     VERSORIUM—Gilbert’s static-detection device

8.     DISSIPATION—Release heat

11.   HAPTIC—Relates to touch

12.   JOULE—1 watt second

15.   DOPING—Process of purposely adding impurities

Electrical Engineering Crossword (Issue 288)

The answers to Circuit Cellar’s July electronics engineering crossword puzzle are now available.288Crossword (key)


2.     QUIESCE—Inactive but still available

4.     GLUELOGIC—Used for circuitry interfacing [two words]

7.     AMAYA—Open-source web tool developed by members of the World Wide Web Consortium (W3C)

8.     ROUNDROBIN—A continuous sequence [two words]

9.     FATCLIENT—A tower PC, for example [two words]

11.   LOGICBOMB—Explosive code [two words]

15.   HEISENBUG—A software glitch that changes its conduct when analyzed

16.   STROBOSCOPE—Makes things appear to move slowly or not at all

17.   STATAMPERE—Approximately 0.333 nanoampere

18.   KORNSHELL—Unix command-line interpreter developed by and named after a Bell Labs employee [two words]

19.   VOXEL—Defines a point in 3-D


1.     BEAMFORMING—Signal processing for sensor arrays

3.     SPIBUS—Works in double-duplex mode [two words]

4.     GREP—UNIX-based command-line utility

5.     SUPERHETERODYNE—Used to convert to intermediate frequencies

6.     ENDIAN—Creates data words

10.   PHOTOVOLTAICS—Uses solar power to create energy

12.   BITTORRENT—File sharing protocol

13.   BINARYPREFIX—E.g., gibi [two words]

14.   AUSTRUMI—Linux distribution based on Slackware


Issue 286: EQ Answers

Question 1—A divider is a logic module that takes two binary numbers and produces their numerical quotient (and optionally, the remainder). The basic structure is a series of subtractions and multiplexers, where the multiplexer uses the result of the subtraciton to select the value that gets passed to the next step. The quotient is formed from the bits used to control the multiplexers, and the remainder is the result of the last subtraction.

If it is implemented purely combinatorially, then the critical path through all of this logic is quite long (even with carry-lookahead in the subtractors) and the clock cycle must be very slow. What could be done to shorten the clock period without losing the ability to get a result on every clock?

Answer 1—Pretty much any large chunk of combinatorial logic can be pipelined in order to reduce the clock period. This allows it to produce more results in a given amount of time, at the expense of increasing the latency for any particular result.

Divider logic is very easy to pipeline, and the number of pipeline stages you can use is fairly arbitrary. You could insert a pipeline register after each subtract-mux pair, or you might choose to do two or more subtract-mux stages per pipeline register You could even go so far as to pipeline the subtracts and the muxes separately (or even pipeline *within* each subtract) in order to get the fastest possible clock speed, but this would be rather extreme.

The more pipeline registers you use, the shorter the critical path (and the clock period) can be, but you use more resources (the registers). Also, the overall latency goes up, since you need to account for the setup and propagation times of the pipeline registers in the clock period (in addition to the subtract-mux logic delays). This gets multiplied by the number of pipeline stages in order to compute the total latency.

Question 2—On the other hand, what could be done to reduce the amount of logic required for the divider, giving up the ability to have a result on every clock?


Answer 2—If you don’t need the level of performance provided by a pipelined divider, you can computes the quotient serially, one bit at a time. You would just need one subtractor and one multiplexer, along with registers to hold the input values, quotient bits and the intermediate result.

You could potentially compute more than one bit per clock period using additional subtract-mux stages. This gives you the flexibility to trade off space and time as needed for a particular application.

Question 3—An engineer wanted to build an 8-MHz filter that had a very narrow bandwidth, so he used a crystal lattice filter like this:


However, when he built and tested his filter, he discovered that while it worked fine around 8 MHz, the attenuation at very high frequencies (e.g., >80 MHz) was very much reduced. What caused this?

Answer 3—The equivalent circuit for a quartz crystal is something like this:EQ-fig2-CC287-June14

The components across the bottom represent the mechanical resonance of the crystal itself, while the capacitor at the top represents the capacitance of the electrodes and holder. Typical values are:

  • Cser: 10s of fF (yes, femtofarads, 10-15F)
  • L: 10s of mH
  • R: 10s of ohms
  • Cpar: 10s of pF

The crystal has a series-resonant frequency based on just Cser and L. It has a relatively low impedance (basically just R) at this frequency.

It also has a parallel-resonant (sometimes called “antiresonant”) frequency when you consider the entire loop, including Cpar. Since Cser and Cpar are essentially in series, together they have a slightly lower capacitance than Cser alone, so the parallel-resonant frequency is slightly higher. The crystal’s impedance is very high at this frequency.

But at frequencies much higher than either of the resonant frequencies, you can see that the impedance of Cparalone dominates, and this just keeps decreasing with increasing frequency. This reduces the crystal lattice filter to a simple capacitive divider, which passes high freqeuncies with little attenuation.

Question 4—Suppose you know that a nominal 10.000 MHz crystal has a series-resonant frequency of 9.996490 MHz and a parallel-resonant frequency of 10.017730 MHz. You also know that its equivalent series capacitance is 27.1 fF. How can you calculate the value of its parallel capacitance?

Answer 4—First, calculate the crystal’s equivalent inductance, based on the series-resonant frequency:EQ-equation1-CC287-June14

Next, calculate the capacitance required to resonate with that inductance at the parallel-resonant frequency:EQ-equation2-CC287-June14

Finally, calculate the value of Cpar required to give that value of capacitance when in series with Cser:EQ-equation3-CC287-June14

Note that all three equations can be combined into one, and this reduces to:EQ-equation4-CC287-June14

Electrical Engineering Crossword (Issue 287)

The answers to Circuit Cellar’s April electronics engineering crossword puzzle are now available.287 crossword (key)


1. BEAMFORMING—Signal processing technique

7. HETERODYNERECEIVER—Converts a signal to an intermediate frequency [two words]

8. AMIGA—A high-resolution PC based on Motorola’s 6800 microprocessor family

9. NAGLING—Creates “Russian doll”-type packets to improve a TCP/IP network’s performance

10. SERVERBLADE—A thin circuit board designed for one specific application [two words]

15. FUZZING—Tests for coding and security errors

17. PHASECHANGE—Nonvolatile RAM [two words]

18. WALLEDGARDEN—Restricts access to Web content and services [two words]

19. SERIALBACKPACK—A PCB interface that goes between a parallel LCD and a microcontroller [two words]

20. SLACKWARE—Open-source, Linux-based OS


2. ROOTMEANSQUARE—Determines an AC wave’s voltage [three words]

3. ROENTGEN—IBM’s active matrix LCD

4. KERNELPANIC—Happens when a fatal error is detected [two words]

5. BIPHASEENCODING—Requires a state transition at the end of every data bit [two words]

6. PERMITTIVITY—Ability to be polarized

11. VOODOO—Helps create realistic 3-D graphics

12. FLANGING—An audio process that combines signals to create a comb filter effect

13. BREADBOARDING—Used for circuit design experimentation

14. STATOHM—Five of these equal approximately 4.5 × 1012

16. TELEDACTYL—Utilizes human speech to code


Electrical Engineering Crossword (Issue 286)

The answers to Circuit Cellar’s April electronics engineering crossword puzzle are now available.



2. SAMBA—This networking protocol enables you to write to an embedded file system from a Windows PC

7. ELECTROMAGNETICFIELD—Used for data transfer [two words]

8. CAPACITOR—These types of microphones were commonly called “condensers” until about 1970

9. HOMODYNE—A receiver with direct amplification

13. BLENDER—Contains several modeling features and an integrated game engine

15. PULSESHAPING—Used to improve wired or wireless communication link performance [two words]

17. BRILLOUINSCATTERING—Occurs when certain types of light change their frequency and route [two words]

18. CLOCKSIGNAL—Used to coordinate circuits’ actions [two words]

19. RASPBERRYPI—Designed to encourage scholastic computer science lessons [two words]


1. NONRETURNTOZERO—Typically 1s are a positive voltage and 0s are a negative voltage [four words]

3. BITARRAY—Provides compact storage for computing and digital communications [two words]

4. DOUBLEDATARATE—Coordinates the rising and falling edges of an [18 Across] to transfer data [three words]

5. ANECHOIC—Absent of sound

6. FIRSTINFIRSTOUT—Oldest requests receive priority [four words]


11. CROSSTALK—Occurs when accidental coupling causes unwanted signals

12. BINISTOR—An electronic oscillator component

14. NANCY—Receiver that intercepts or demodulates IR radiation

16. BEAGLEA good breed of analyzer for I2C and SPI designs


Issue 284: EQ Answers

Can you name all of the signals in the original 25-pin RS-232 connector?

Pins 9, 10, 11, 18, and 25 are unassigned/reserved. The rest are:

Pin Abbreviation Source Description
1 PG - Protective ground
2 TD DTE Transmitted data
3 RD DCE Received data
4 RTS DTE Request to send
5 CTS DCE Clear to send
6 DSR DCE Data Set Ready
7 SG - Signal ground
8 CD DCE Carrier detect
12 SCD DCE Secondary carrier detect
13 SCTS DCE Secondary clear to send
14 STD DTE Secondary transmitted data
15 TC DCE Transmitter clock
16 SRD DCE Secondary received data
17 RC DCE Receiver clock
19 SRTS DTE Secondary request to send
20 DTR DTE Data terminal ready
21 SQ DCE Signal quality
22 RI DCE Ring indicator
23 - DTE Data rate selector
24 ETC DTE External transmitter clock


What is the key difference between a Moore state machine and a Mealy state machine?

The key difference between Moore and Mealy is that in a Moore state machine, the outputs depend only on the current state, while in a Mealy state machine, the outputs can also be affected directly by the inputs.


What are some practical reasons you might choose one state machine over the other?

In practice, the difference between Moore and Mealy in most situations is not very important. However, when you’re trying to optimize the design in certain ways, it sometimes is.

Generally speaking, a Mealy machine can have fewer state variables than the corresponding Moore machine, which will save physical resources on a chip. This can be important in low-power designs.

On the other hand, a Moore machine will typically have shorter logic paths between flip-flops (total combinatorial gate delays), which will enable it to run at a higher clock speed than the corresponding Mealy machine.


What is the key feature that distinguishes a DSP from any other general-purpose CPU?

Usually, the key distinguishing feature of a DSP when compared with a general-purpose CPU is that the DSP can execute certain signal-processing operations with few, if any, CPU cycles wasted on instructions that do not compute results.

One of the most basic operations in many key DSP algorithms is the MAC (multiply-accumulate) operation, which is the fundamental step used in matrix dot and cross products, FIR and IIR filters, and fast Fourier transforms (FFTs). A DSP will typically have a register and/or memory organization and a data path that enables it to do at least 64 MAC operations (and often many more) on unique data pairs in a row without any clocks wasted on loop overhead or data movement. General-purpose CPUs do not generally have enough registers to accomplish this without using additional instructions to move data between registers and memory.

Electrical Engineering Crossword (Issue 285)

The answers to Circuit Cellar’s April electronics engineering crossword puzzle are now available.


2.    STOKESSHIFT—Can reduce photon energy [two words]
8.    HYSTERESISLOOP—Its area measures the energy dispersed during a magnetization cycle [two words]
11.    NANDGATE—A shoe in when playing “true or false?” [two words]
13.    YOCTOPROJECT—An open-source alliance designed to help Linux aficionados [two words]
15.    RANKINE—°R
17.    INTERNALNET—A network that resides in and around you
18.    SEQUENTIALCIRCUIT—Dependent on past input [two words]
19.    NANOHENRY—Its abbreviation is the same as the state bordered by Massachusetts, Maine, and Vermont
20.    BINARYCODEDDECIMAL—Makes good use of a 4- or 8-bit nibble [three words]


1.    BIREFRINGENCE—Divides light into ordinary and extraordinary rays
3.    SQUIRREL—An object-oriented programming language
4.    SMARTMETER—Records and shares energy usage information [two words]
5.    MESHANALYSIS—A circuit evaluation method [two words]
6.    LYOTFILTER—Uses [1. Down] to produce a narrow frequency range of wavelengths [two words]
7.    LINEARREGULATOR—Keeps things steady [two words]
9.    BRAGGDIFFRACTION—Occurs when electromagnetic radiation disperses [two words]
10.    AUTODYNE—An amplifying vacuum tube-based circuit
12.    FEMTOWATT—10–15 W
14.    UNIJUCTION—Can be used to measure magnetic flux
16.    PEAKER—Increases gain at higher frequencies

Electrical Engineering Crossword (Issue 284)

The answers to Circuit Cellar’s March electronics engineering crossword puzzle are now available.



1.    CROSSEDFIELDAMPLIFIER—This vacuum tube is capable of high output power [three words]
3.    HYPERVISOR—Produces and runs virtual machines
5.    DYNATRON—Uses negative resistance to keep a tuned circuit oscillating
8.    ULTRAVIOLETLIGHT—Gives some substances “a healthy glow” [two words]
13.    ZEROMOMENTPOINT—A moment of respite for robots [three words]
14.    THERMOSONIC—Connects to silicon ICs
17.    CATSWHISKER—An outdated electronic component mainly used in antique radios [two words]
18.    FLEMINGVALVE—Invented in the early 1900s, this was known as the first vacuum tube [two words]
19.    BACKBONE—Makes LANs connect


2.    DEMODULATOR—Recovers information from a regulated waveform
4.    SQUEGGING—This type of circuit oscillates erratically
5.    DOWNMIXING—Audio manipulation process
6.    REYNOLDSNUMBER—Used for flow pattern predictions [two words]
7.    LATENCY—Used with bandwidth to ascertain network connection speed
9.    THICKFILM—This type of chip resistor is commonly used in electronic and electrical devices [two words]
10.    DYNAMIC—Its memory is volatile
11.    CRYOTRON—Operates via superconductivity
12.    NETMASK—Creates neighborhoods of IP addresses
15.    HOROLOGY—E.g., clepsydras, chronometers, and sundials
16.    SEEBECK—An effect that creates electricity



Issue 282: EQ Answers

Construct an electrical circuit to find the values of Xa, Xb, and Xc in this system of equations:

21Xa – 10Xb – 10Xc = 1
–10Xa + 22Xb – 10Xc = –2
–10Xa – 10Xb + 20Xc = 10

Your circuit should include only the following elements:

one 1-Ω resistor
one 2-Ω resistor
three 10-Ω resistors
three ideal constant voltage sources
three ideal ammeters

The circuit should be designed so that each ammeter displays one of the values Xa, Xb, or Xc. Given that the Xa, Xb, and Xc values represent currents, what kind of circuit analysis yields equations in this form?

You get equations in this form when you do mesh analysis of a circuit. Each equation represents the sum of the voltages around one loop in the mesh.

What do the coefficients on the left side of the equations represent? What about the constants on the right side?

The coefficients on the left side of each equation represent resistances. Resistance multiplied by current (the unknown Xa, Xb, and Xc values) yields voltage.
The “bare” numbers on the right side of each equation represent voltages directly (i.e., independent voltage sources).

What is the numerical solution for the equations?

To solve the equations directly, start by solving the third equation for Xc and substituting it into the other two equations:

Xc = 1/2 Xa + 1/2 Xb + 1/2

21Xa – 10Xb – 5Xa – 5Xb – 5 = 1
–10Xa + 22Xb – 5Xa – 5Xb – 5 = –2

16Xa – 15Xb = 6
–15Xa + 17Xb = 3

Solve for Xa by multiplying the first equation by 17 and the second equation by 15 and then adding them:

272Xa – 255Xb = 102
–225Xa + 255Xb = 45

47Xa = 147 → Xa = 147/47

Solve for Xb by multiplying the first equation by 15 and the second equation by 16 and then adding them:

240Xa – 225Xb = 90
–240Xa + 272Xb = 48

47Xb = 138 → Xb = 138/47

Finally, substitute those two results into the equation for Xc:

Xc = 147/94 + 138/94 + 47/94 = 332/94 = 166/47

Finally, what is the actual circuit? Draw a diagram of the circuit and indicate the required value of each voltage source.

The circuit is a mesh comprising three loops, each with a voltage source. The common elements of the three loops are the three 10-Ω resistors, connected in a Y configuration (see the figure below).

cc281_eq_fig1The values of the voltage sources in each loop are given directly by the equations, as shown. To verify the numeric solution calculated previously, you can calculate all of the node voltages around the outer loop, plus the voltage at the center of the Y, and ensure they’re self-consistent.

We’ll start by naming Va as ground, or 0 V:

Vb = Va + 2 V = 2 V

Vc = Vb + 2 Ω × Xb = 2V + 2 Ω × 138/47 A = 370/47 V = 7.87234 V

Vd = Vc + 1 Ω × Xa = 370/47 V + 1 Ω × 147/47A = 517/47 V = 11.000 V

Ve = Vd – 1 V = 11.000 V – 1.000 V = 10.000 V

Va = Ve – 10 V = 0 V

which is where we started.

The center node, Vf, should be at the average of the three voltages Va, Vc, and Ve:

0 V + 370/47 V + 10 V/3 = 840/141 V = 5.95745 V

We should also be able to get this value by calculating the voltage drops across each of the three 10-Ω resistors:

Va + (Xc – Xb) × 10 Ω = 0 V + (166 – 138)/47A × 10 Ω = 280/47 V = 5.95745 V

Vc + (Xb – Xa) × 10 Ω = 370/47V + (138-147)/47A × 10 Ω = 280/47 V = 5.95745 V

Ve + (Xa – Xc) × 10 Ω = 10 V + (147-166)/47 A × 10 Ω = 280/47 V = 5.95745 V

Electrical Engineering Crossword (Issue 283)

The answers to Circuit Cellar’s February electronics engineering crossword puzzle are now available.



2. LITZWIRE—Separately insulated strands woven together [two words]
4. LINKFIELD—First in a message buffer’s line [two words]
6. PETAFLOPS—Measures a processor’s floating point unit performance
8. ANION—negatively charged atom
9. LISP—Used to manipulate mathematical logic
11. STATCOULOMB—i.e., franklin (Fr)
12. AMBISONICS—Typically requires a soundfield microphone
15. BROUTER—This device can send data between networks and it can forward data to individual systems in a network
16. TRINITRON—This CRT technology was originally introduced the 1960s
17. OXIDE—The “O” in CMOS
18. DETENT—Used to prevent or stop something from spinning


3. RADECHON—A barrier-grid storage tube
5. COLPITTS—This oscillator uses two-terminal electrical components to create a specific oscillation frequency
7. BEAGLEBOARD—TI’s open-source SBC
10. PICONET—A network that is created using a wireless Bluetooth connection
13. BETATRON—Designed to accelerate electrons
14. TEBIBYTE—More than 1,000,000,000,000 bytes

Peter Baston Wins the CC Code Challenge (Week 31)

We have a winner of last week’s CC Weekly Code Challenge, sponsored by IAR Systems! We posted a code snippet with an error and challenged the engineering community to find the mistake!

Congratulations to Peter Baston of Flintshire, United Kingdom for winning the CC Weekly Code Challenge for Week 31! Peter will receive a Circuit Cellar 2012 & 2011 Archive CD.

Peter’s correct answer was randomly selected from the pool of responses that correctly identified an error in the code. Peter answered:

Line 35: Should not end with semi-colon


You can see the complete list of weekly winners and code challenges here.

What is the CC Weekly Code Challenge?
Each week, Circuit Cellar’s technical editors purposely insert an error in a snippet of code. It could be a semantic error, a syntax error, a design error, a spelling error, or another bug the editors slip in. You are challenged to find the error. Once the submission deadline passes, Circuit Cellar will randomly select one winner from the group of respondents who submit the correct answer.

The CC Weekly Code Challenge ran from June 3rd through December 30th, 2013. Subscribe to our CC.Post newsletter to stay informed of other contests and challenges, as well as recent news, new issue availability, and more!

Gait Boxman Wins the CC Code Challenge (Week 30)

We have a winner of last week’s CC Weekly Code Challenge, sponsored by IAR Systems! We posted a code snippet with an error and challenged the engineering community to find the mistake!

Congratulations to Gait Boxman of Gelderland, Netherlands for winning the CC Weekly Code Challenge for Week 30! Gait will receive an IAR Kickstart: KSK-FM3-48PMC-USB.

Gait’s correct answer was randomly selected from the pool of responses that correctly identified an error in the code. Gait answered:

Line 31: should be digits % 3 instead of digits / 3


You can see the complete list of weekly winners and code challenges here.

What is the CC Weekly Code Challenge?
Each week, Circuit Cellar’s technical editors purposely insert an error in a snippet of code. It could be a semantic error, a syntax error, a design error, a spelling error, or another bug the editors slip in. You are challenged to find the error.Once the submission deadline passes, Circuit Cellar will randomly select one winner from the group of respondents who submit the correct answer.

Inspired? Want to try this week’s challenge? Get started!

Submission Deadline: The deadline for each week’s challenge is Sunday, 12 PM EST. Refer to the Rules, Terms & Conditions for information about eligibility and prizes.