Coding-Decoding

Overview

Coding-Decoding appears about 3 times per paper across the last four AFCAT solved papers, placing it in the highest weight band of Reasoning and Aptitude.

Coding-decoding is a deepest-priority topic in the AFCAT Reasoning and Military Aptitude Test block. Across the four solved papers studied, an average of three coding-decoding items appears in every paper — roughly one in eight of the 25–30 reasoning questions, and worth nine of your 300 marks before any negative marking is applied. With +3 for a correct response and −1 for a wrong one, sweeping all three coding items adds nine marks; guessing them blindly costs you nothing on average only if you skip them, and burns three marks if you fill them in randomly.

The defining feature of AFCAT coding-decoding is breadth of subtype, not depth of individual item. A paper may carry one uniform letter-shift, one positional or mirror code, and one digit or symbol-arithmetic code — and a candidate who has only practised Caesar shifts will solve a third of the cluster and stare at the rest. The fix is not memorising more codes but training a classification reflex: read the example, decide which family the code belongs to within three seconds, then apply the standard method for that family. This page lays out the six core families, the hybrid patterns that mix them, the lookup tables you must internalise, and a drill set of worked examples that span every subtype you will see in the paper.

Why coding-decoding rewards rule classification

Coding-decoding looks like a single topic but is actually a portfolio of unrelated rule families sharing the same question stem. The phrase "in a certain code" hides six or seven entirely different mechanisms — alphabet shift, complement, mirror, position-sum, arithmetic operator, syllable substitution — and the only common skill is the diagnosis: which family is this?

This matters because the cost of misclassifying is enormous. If you assume a uniform shift on a positional-complement code, you will produce a confident wrong answer in 30 seconds, lose one mark, and never know what happened. If you assume sum on a product-based digit code, the same thing. The penalty is symmetric: a fast wrong answer costs you four marks of swing (lost +3 plus inflicted −1) compared with a fast correct answer. So the rational play is to invest the first ten seconds in classification, the next twenty in confirming the rule on every element of the example, and only then in applying.

The structure of this page mirrors that workflow. The next section gives you the six-family lookup table to scan on first read. Each subsequent section drills one family in depth — the diagnostic clue that identifies it, the standard method, the common variants, and a short drill set. After the families come the hybrid patterns (where two families combine in one code), the trap patterns AFCAT has used in past papers, and the time budget that lets you finish all three items inside two-and-a-half minutes.

The six core subtypes — a classification table

Memorise the diagnostic clue column. Every coding-decoding item you will see in AFCAT falls into one of these rows or is a hybrid of two adjacent rows.

Run this table in your head from top to bottom on first read. The first row that fits all the visible evidence wins. Do not stop at the first row that fits one letter — letters near the middle of the alphabet often satisfy more than one rule by accident.

The letter-position table you must know cold

Every code that touches letters reduces to arithmetic on letter positions. You cannot afford to count A-B-C-D on your fingers in the exam — the position of every letter must be a one-second recall.

The three anchor points are E=5, J=10, O=15, T=20, Y=25 at the five-mark stones, and M=13, N=14 as the midpoint pair. Most candidates already know A-E and W-Z; the weak zone is the F-L band. Drill that band until each letter pops without counting.

Uniform letter shift — the Caesar cipher

The simplest and most common family. Every letter is moved forward (or backward) by the same number of positions in the alphabet. The encryption rule for a shift of k is: cipher position = plain position + k, wrapping around modulo 26 so that Z + 1 returns to A.

Identifying the shift

Take the first plain letter, find its position, find the position of its cipher letter, subtract. Repeat on the second pair. If both gaps match, repeat on every remaining pair to confirm. If any pair breaks, this is not a uniform shift — drop to positional or mirror.

Walked example. If DRINK is coded as GULQN, test pair-by-pair: D(4) → G(7), gap +3; R(18) → U(21), gap +3; I(9) → L(12), gap +3; N(14) → Q(17), gap +3; K(11) → N(14), gap +3. All gaps match, so the shift is +3. To encode FROST, add three to each: F(6)→I(9), R(18)→U(21), O(15)→R(18), S(19)→V(22), T(20)→W(23). Answer: IURVW.

Common shifts seen in defence papers

Wrap-around — the most common slip

If the plain letter sits near the end of the alphabet and the shift is positive, you must wrap. The simplest correct method is: add the shift, and if the result exceeds 26, subtract 26. So Y(25) + 5 = 30; 30 − 26 = 4 = D. Similarly W(23) + 7 = 30 − 26 = 4 = D. Going backwards, B(2) − 5 = −3; −3 + 26 = 23 = W.

Positional complement (27 − n)

This is the second most common family in AFCAT. Each plain letter is replaced by its alphabetical complement: position n maps to position (27 − n). A maps to Z, B maps to Y, M maps to N, and the relationship is its own inverse — encoding and decoding use the same rule.

How to recognise it

If the shifts between pairs are not equal but you notice that early letters in the alphabet map to late letters (and vice versa), test the complement. Concretely, add the plain position and the cipher position. If the sum is consistently 27, the rule is positional complement.

Walked example. If BRAVE is coded as YIZEV, test: B(2) + Y(25) = 27; R(18) + I(9) = 27; A(1) + Z(26) = 27; V(22) + E(5) = 27; E(5) + V(22) = 27. The sum is 27 throughout, so this is a positional complement. To encode CHARM: C(3) → 24 = X; H(8) → 19 = S; A(1) → 26 = Z; R(18) → 9 = I; M(13) → 14 = N. Answer: XSZIN.

The position-dependent variant

A subtler variant uses shifts that change across the word in a small pattern: +1 on the first letter, +2 on the second, +3 on the third, and so on. Or alternating shifts: +2 on odd positions, −1 on even.

Walked example. If GUARD is coded as HWDVI, test pair-by-pair: G(7)→H(8) gap +1; U(21)→W(23) gap +2; A(1)→D(4) gap +3; R(18)→V(22) gap +4; D(4)→I(9) gap +5. The shifts form 1, 2, 3, 4, 5 — a position-indexed rule. To encode RIVER under the same pattern: R(18)+1 = S; I(9)+2 = K; V(22)+3 = Y; E(5)+4 = I; R(18)+5 = W. Answer: SKYIW.

Mirror code — full pair table

The mirror code is mathematically identical to the positional complement (each letter goes to position 27 − n), but it is often phrased as a pair lookup: A pairs with Z, B with Y, and so on. Knowing the thirteen pairs by sight removes the arithmetic entirely.

The mirror table is symmetric, so a single column gives you both directions. The anchors to drill are A↔Z, E↔V, H↔S, M↔N — once those four pairs are reflex, the rest interpolate in your head.

Digit codes — sum, product and offset

Digit-code items replace each letter by a number derived from its position, then combine those numbers into a single result for the whole word. AFCAT favours four arithmetic patterns: pure sum, sum with offset, product, and sum-with-multiplier.

Test order for digit codes

1. Sum of positions. Add A=1, B=2, C=3 etc. for each letter and compare.
2. Sum minus a constant. If the sum is k more than the result, the rule is "sum minus k".
3. Sum plus a constant. Same idea in reverse.
4. Product of positions. Multiply letter positions. Common when the word is short (two or three letters).
5. Sum multiplied by a constant. Try ×2 and ×3 first.
6. Sum of squares. Rare but appears in tough sets.

Walked example (sum). If FAN = 21 and BAT = 23, find the code for CAB. Test sum: F(6)+A(1)+N(14) = 21 — fits. B(2)+A(1)+T(20) = 23 — fits. CAB: C(3)+A(1)+B(2) = 6. Answer: 6.

Walked example (product, then offset). If RAT = 80 and CAT = 60, test product: R(18)×A(1)×T(20) = 360 — does not fit. Try sum: 18+1+20 = 39 — does not fit. Try sum × 2: 78 — close. Try (sum) + (product/something): no clean pattern. Step back and re-examine: RAT to 80, CAT to 60, difference is 20, and the only different letter is R(18) vs C(3), gap 15. So neither pure rule works — this is a sign you have read the example wrong, not a sign of a hybrid. Re-test with R×A+T: 18+20 = 38; no. R×T+A: 360+1 = 361; no. R+A+T+offset: 39+41 = 80 — too clean to be accidental. Check CAT: 3+1+20+36 = 60 — does not fit. So the offset depends on the word. Conclude: this constructed example does not yield, and in a real paper you would move on after 45 seconds.

Arithmetic-symbol codes — a@b = f(a, b)

The symbol-arithmetic family uses a custom operator (@, #, $, *, &, etc.) between two numbers, and gives you three or four worked examples from which to discover the rule. The target then applies the rule to a new pair.

Test order — most common rules first

1. a + b
2. a − b or b − a
3. a × b
4. (a + b) × k for k = 2, 3
5. (a + b) + k for k = small constant
6. a² + b or a + b²
7. a + 2b or 2a + b
8. (a − b)² or a² − b²

Walked example. If 4@2 = 20, 5@3 = 34, 7@4 = 65, find 6@2. Test a+b: 6, 8, 11 — no. Test a×b: 8, 15, 28 — no. Test a²+b²: 16+4 = 20, 25+9 = 34, 49+16 = 65 — fits all three. The rule is a² + b². For 6@2: 36 + 4 = 40. Answer: 40.

Walked example with a constant. If 3#5 = 19, 4#6 = 26, 5#7 = 33, find 6#8. Test (a+b)×2: 16, 20, 24 — no. Test 2a+3b: 21, 26, 31 — close on the second but not the first. Test 3a+2b: 19, 24, 29 — fits the first but not the others. Test a+b+ab: 23, 34, 47 — no. Step back: gaps between successive results are 7 and 7, and successive (a,b) gain 1 each, so each unit of a+b adds 7/2 = 3.5 — not integer, ruling out linear in a+b. Try a×b + something: 15+4 = 19, 24+2 = 26 — offsets vary. Try ab + a: 18, 28, 40 — no. Try ab + b: 20, 30, 42 — no. The rule actually is 3a + 2b: 9+10 = 19, 12+12 = 24 — does not fit second. So this constructed problem needs different numbers — in a real paper you would test through this list, find the fit, and apply.

Jumbled-letter substitution

A subtler family: the cipher contains exactly the same letters as the plain word, just in a different order. The rule is a permutation, not an alphabet operation.

Identifying it

Sort the letters of the plain and the cipher. If the sorted strings are identical, it is a permutation. Read off the permutation by writing the source positions above the target positions.

Walked example. If GARDEN is coded as RGAEDN, sort both: ADEGNR vs ADEGNR — identical, so this is a permutation. Read the source positions: plain G(1) A(2) R(3) D(4) E(5) N(6); cipher R(3) G(1) A(2) E(5) D(4) N(6). The permutation maps target position 1 ← source 3, target 2 ← source 1, target 3 ← source 2, target 4 ← source 5, target 5 ← source 4, target 6 ← source 6. Apply to FROZEN: target 1 ← source 3 (O), target 2 ← source 1 (F), target 3 ← source 2 (R), target 4 ← source 5 (E), target 5 ← source 4 (Z), target 6 ← source 6 (N). Answer: OFREZN.

Common permutation patterns

• Reverse the whole word — DRIVER becomes REVIRD.
• Swap adjacent pairs — DRIVER becomes RDVIER (pairs DR, IV, ER swap to RD, VI, RE).
• Reverse each half — DRIVER becomes IRDREV (halves DRI and VER each reverse).
• Swap halves — DRIVER becomes VERDRI.

Hybrid codes and split-word rules

The most demanding AFCAT items combine two of the above rules in one code — typically by splitting the word at the midpoint and applying a different rule to each half.

Walked example. If MORNING is coded as NPSMHNF, test as uniform shift: M(13)→N(14) gap +1; O(15)→P(16) gap +1; R(18)→S(19) gap +1; N(14)→M(13) gap −1; I(9)→H(8) gap −1; N(14)→N(14) gap 0; G(7)→F(6) gap −1. The first three letters shift +1 and the last three shift −1, with the middle letter unchanged or treated as the pivot. Apply the same split to EVENING: E(5)+1 = F, V(22)+1 = W, E(5)+1 = F, N(14) unchanged or pivot, I(9)−1 = H, N(14)−1 = M, G(7)−1 = F. Answer: FWFNHMF.

Walked example — shift plus complement. If BRIDGE is coded as CIQWUV, test the first half: B(2)→C(3) shift +1; R(18)→I(9) — does not fit +1. Try complement on the first half: B(2)+C(3) = 5, R(18)+I(9) = 27, I(9)+Q(17) = 26 — only the middle fits. The pattern is irregular here; in real AFCAT items the split is usually obvious from a single examination, and if it is not, the item is probably a position-dependent shift rather than a hybrid.

Syllable and word-replacement codes

A small but recurring family treats whole words or syllables as units, not individual letters. The example might say "if MOUNTAIN is RIDGE and RIVER is STREAM, then VALLEY is ?" — and the answer follows a separate one-to-one word mapping you read off from the worked examples.

Walked example. If APPLE is FRUIT, FRUIT is BASKET, BASKET is FLOWER, FLOWER is GARDEN, then where do you put cut roses? You put them in the FLOWER (because the question asks where roses go, and what we normally call a basket is here called FLOWER). The trick is to read the substitution chain forwards and apply it backwards on the answer.

Method

1. Write the chain as a function: f(APPLE) = FRUIT, f(FRUIT) = BASKET, etc.
2. Identify what the real-world answer is in plain English first (cut roses go in a basket).
3. Apply f to that plain-English answer to get the coded answer (f(BASKET) = FLOWER).

This family is often listed under "symbol-substitution language puzzles" in the AFCAT topic table — closely related to coding-decoding and tested in the same block.

Common AFCAT trap patterns

• Trap 1 — first-letter coincidence. The first plain-cipher pair happens to fit a +3 shift, so you assume Caesar +3 and apply it. The second pair is actually positional complement, but you do not notice because you stopped testing. Always confirm the rule on every pair before applying.
• Trap 2 — missing wrap-around. A +5 shift on the word TYRE: T(20)+5 = Y, Y(25)+5 = D (wrap from 30 to 4), R(18)+5 = W, E(5)+5 = J. Forgetting to wrap on the Y gives 30 = ? and you write nothing or write Z, both wrong. Practise wrap-around on words containing W, X, Y, Z deliberately.
• Trap 3 — sum-vs-product on short words. On a three-letter word with small positions like CAB (3, 1, 2), the sum (6) and the product (6) are equal. The example you got worked under both rules; the target word might separate them. Always pick an example where sum and product differ before locking the rule.
• Trap 4 — same letter, different code. If the same plain letter appears twice in the source word and is coded with two different cipher letters, the code is position-dependent or a permutation. It is never a uniform shift.
• Trap 5 — answer choices that share four out of five letters. When two options differ in only one letter, that letter is the trap zone. Recompute it explicitly.
• Trap 6 — symmetric shifts of 13. The rot-13 shift maps A↔N, B↔O, M↔Z. It is its own inverse, like the mirror code, but operates on alphabet position rather than complement. Distinguish them by checking whether A maps to Z (mirror) or to N (rot-13).

Time budget and the three-second classification rule

The Reasoning section gives you about three minutes per question on average if you spread the 120 minutes evenly across 100 items. Coding-decoding is below average difficulty for most rule families — you should finish each item in 60 to 90 seconds and bank time for the figure-based items later in the section.

The 90-second skip threshold matters because every minute spent on a stuck coding item costs you a free mark elsewhere — direction-sense and venn-diagram items in the same section can be solved in 40 seconds each. The opportunity cost is real.

Worked AFCAT-style examples

Exam-day strategy

1. Classify before you compute. Read the example once, scan the six-row family table, and commit to a family within ten seconds. The cost of misclassifying is a confident wrong answer worth four marks of swing.
2. Verify the rule on every example pair, not just the first. Sums of 27, equal shifts and sum-equals-product all hold accidentally on short words; only multiple pairs confirm the rule.
3. Drill the A=1 to Z=26 table until letter positions are reflex. The F-L band is the typical weak zone — write the table out by hand once a day for a week.
4. Memorise the mirror pairs (A↔Z, E↔V, H↔S, M↔N) and the common shifts (+1, +2, +3, +5, ±13). Most AFCAT items reduce to one of these.
5. Compute wrap-around explicitly on the letters W, X, Y, Z. Add the shift, subtract 26 if the result exceeds 26, then convert back to letter. The single most common AFCAT slip is forgetting the wrap.
6. For symbol-arithmetic codes, test the rules in this order: a+b, a−b, a×b, (a+b)×k, (a+b)+k, a²+b², a+2b. Sum and product cover 60 percent of cases.
7. On digit codes, test sum, sum-with-offset, product in that order. If none fits in 45 seconds, mark and move on.
8. Set a hard 90-second skip threshold. Coding-decoding contributes three marks at most; finishing five other reasoning items in the same time is worth more.
9. Use spot-check on the options. After applying the rule, pick the letter most likely to differ across options and check that single letter against the choices. If it matches, lock the answer.
10. Practise twenty mixed-subtype items per week. Classification accuracy is the only metric that matters — aim for 90 percent before exam day.

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