Detailed Evaluation Results

A5.1

What is the IFT DRM modifier of an attack on a hex that contains four squads?

-1 IFT DRM for being overstacked.

Answer: If the hex contains four squads in one Location, that is one squad-equivalent over the normal Personnel stacking limit of three, so the target suffers a -1 IFT DRM. If it exceeds the limit by more than one squad-equivalent, the DRM is -1 per excess squad-equivalent. References: - (A5.1) Personnel Stacking Limit — normal limit is three squad-equivalents in a Location - (A5.131) Defense Penalties: Personnel — attacked Personnel in an overstacked Location suffer a -1 IFT DRM per excess squad-equivalent

[Explanation] The model's response correctly identifies that having four squads in one hex exceeds the normal personnel stacking limit of three, resulting in a -1 IFT DRM for the overstacked condition. It also accurately states that the penalty is -1 per excess squad-equivalent, matching the expected answer's key concept of a -1 DRM for being overstacked. The inclusion of references and explanation adds clarity without contradicting the expected answer.

A5.11

A building hex contains two squads and a crew manning a gun. How many MF does it cost for a third squad to enter this Location?

3 MF. 2 MF to enter the building + 1 MF for overstacking (the Location already contains > 3 squad equivalents: 2 squads + 1 crew (counts as squad for stacking if manning Gun))

Answer: It costs 3 MF: 2 MF to enter the building, plus 1 extra MF because the entering squad makes the Location overstacked. The crew manning the gun still counts as a ½-squad equivalent for stacking, so two squads + crew = 2½ squad equivalents before entry, and the third squad raises it to 3½, i.e., ½ over the 3-squad limit. References: - (A5.1) Infantry/Cavalry Stacking Limits — a side may stack up to three squads or equivalents per Location without penalty. - (A5.11) Movement into Overstacked Location — +1 MF for each squad equivalent by which the Location exceeds normal stacking limits; includes the exact example of two squads and a crew in a building hex, where a squad entering pays 3 MF total. - (B23.4/A4.1 terrain cost as shown in examples) Building entry cost — entering a building costs 2 MF.

[Calculation] The model's response incorrectly states that the crew manning the gun counts as a ½-squad equivalent for stacking, leading to a total of 2½ squad equivalents before entry. The expected answer and ASL rules specify that a crew manning a gun counts as a full squad equivalent for stacking purposes. Therefore, the location already contains 3 squad equivalents (2 squads + 1 crew), and the third squad entering causes overstacking by 1 squad equivalent, resulting in a 1 MF penalty. The model's reasoning about the crew counting as ½ is incorrect, making the answer incorrect despite the correct total MF cost of 3.

A6.4

A unit on Level 3 is firing at an enemy where there is a 1 level obstacle 15 hexes away. How many blind hexes would there be behind the obstacle.

3 Blind hexes. 1 [Normal Blind Hex] + 3 [extra Blind Hexes] -1 [full-level elevation advantage > one level over obstacle]

Answer: There would be 3 blind hexes. Start with 1 blind hex for any full-level obstacle directly behind it, add +3 more because the obstacle is 15 hexes away (one extra for each full multiple of five hexes), then reduce by 1 because a Level 3 firer has two levels of elevation advantage over a Level 1 obstacle, and only the amount > 1 reduces blind hexes: 1 + 3 - 1 = 3 References: - (A6.4) Blind Hexes — a full-level obstacle normally creates one blind hex directly behind it - (A6.41) Range Increase to Blind Hexes — add one blind hex for every full multiple of five hexes to the obstacle - (A6.42) Elevation Advantage — reduce blind hexes by one for each full level of elevation advantage greater than one over the obstacle, to a minimum of one for non-crest-line obstacles

[Calculation] The model's response correctly calculates the number of blind hexes as 3 by starting with 1 blind hex for the full-level obstacle, adding 3 for the 15-hex distance (one for each full multiple of five hexes), and then subtracting 1 due to the elevation advantage (Level 3 firer vs Level 1 obstacle, so 2 levels advantage minus 1). The final total matches the expected answer exactly, and the reasoning aligns with the referenced rules.

A6.4

How many Blind Hexes are created by a level 1 obstacle at a range of 10 hexes for a unit at level 2?

3 Blind Hexes. 1 [Normal Blind Hex] + 2 [extra Blind Hex for every multiple of five hexes to the obstacle] - 0 [no reduction since elevation advantage is only one level over the obstacle].

Answer: 2 Blind Hexes. A level 1 obstacle normally creates 1 Blind Hex, and at a range of 10 hexes you add +1 more for one full multiple of five hexes: 1 + 1 = 2; being at level 2 is only enough to see over the obstacle, not to reduce that minimum one-hex shadow. References: - (A6.4) Blind Hexes — a full-level obstacle creates Blind Hexes directly behind it - (A6.41) Range Increase — add one Blind Hex for every full multiple of five hexes to the obstacle - (A6.42) Elevation Advantage — extra elevation advantage reduces Blind Hexes, but only beyond the one level needed to see over the obstacle - (B10.23) Crest Lines exception — non-cliff Crest Lines can be reduced to zero, unlike normal obstacles

[Calculation] The model's response calculates 2 Blind Hexes instead of the expected 3. The expected answer correctly states that a level 1 obstacle creates 1 Blind Hex plus 2 extra Blind Hexes for every multiple of five hexes at a range of 10 (which is two multiples of five), totaling 3. The model only adds 1 extra Blind Hex for one multiple of five hexes, which is incorrect. The explanation about elevation advantage is correct but does not affect the count here. Therefore, the model's final answer is incorrect due to an error in counting the multiples of five hexes.

A7.2

Two 4-6-7 squads and a 9-1 leader fire at a target 7 hexes away in Open Ground. What is the final IFT column and DRM?

8 FP column, -1 DRM. Total FP = 8. Range 7 is > Normal Range (4) but <= Double Range (8). Fire is Long Range (halved). 8 / 2 = 4 FP. Leader applies -1 DRM.

Answer: The attack is **6 FP on the 6 column with a -1 DRM**. Two 4-6-7 squads combine for 8 FP, but at 7 hexes they are beyond Normal Range and fire at half FP, so 8 → 4 FP; the 9-1 leader can direct the fire for **-1 DRM**, and Open Ground gives no positive TEM/Hindrance DRM. This matches the rulebook example that long-range squad FP is halved and leadership applies as a negative DRM References: - (A7.2/A7.3) Firepower Modifiers / IFT Column Selection — combine squad FP, halve for Long Range Fire, then use the next-lower IFT column if needed; 4 FP uses the 4 column, but the example/result set confirms long-range halving and column use principles - (A7.5) Long Range Fire — fire beyond Normal Range up to double range is halved; 4-6-7 squads have range 6, so at 7 hexes their FP is halved - (A10.7/A10.72) Leader Direction — a 9-1 leader directing the attack applies a -1 DRM to the IFT DR - (B/Open Ground TEM) Open Ground — no TEM applies, so no terrain DRM is added; only the leader DRM affects the shot here

[Calculation] The model correctly identifies that two 4-6-7 squads combine for 8 FP and that at 7 hexes (which is beyond Normal Range of 4 but within Double Range of 8) the fire is long range and halved to 4 FP. However, the model states the final IFT column is '6' with 6 FP, which contradicts the correct halved FP of 4. The expected answer clearly states the final column is '8 FP column' with halving applied to get 4 FP, meaning the IFT column should be 8 (before halving), not 6. The model's final column and FP value are inconsistent and incorrect. The DRM of -1 from the leader is correctly identified. Because the final answer's key numeric value (IFT column) is wrong, the response is incorrect.

A7.24

A 5-4-8 squad moves 1 hex and then fires in the AFPh at an adjacent target. What is the total FP?

6 FP. Base 5. PBF x2 = 10. AFPh /2 = 5. Assault Fire (underscored FP) +1 = 6.

Answer: The total FP is 6 FP: a 5-4-8 has 5 inherent FP, doubled for Point Blank Fire to 10 FP because the target is adjacent, then halved for AFPh fire to 5 FP, and then gains +1 FP for Assault Fire, for a final 6 FP attack References: - (A7.21) Point Blank Fire — adjacent Small Arms fire is doubled - (A7.24) AFPh Fire — attacks in the AFPh are halved - (A7.36) Assault Fire — an underlined-FP squad adds +1 FP in the AFPh after other FP modifications

[Calculation] The model's response correctly calculates the total FP as 6, matching the expected answer exactly. It applies the base FP of 5, doubles it for Point Blank Fire to 10, halves it for AFPh fire to 5, and then adds +1 for Assault Fire, resulting in 6 FP. The reasoning is complete and accurate.

A7.36

A 5-4-8 squad has Assault Fire and is pinned. It is attacking in the APh. What is its FP?

3 FP. Start with 5 FP, divide by 2 for being pinned, which results in 2.5. Then divide by 2 for being APh, which results in 1.25. Add +1 for Assault Fire, which results in 2.25. Round up to the nearest whole value, which is 3FP.

Answer: Its FP is 2.5, which attacks on the 2 FP column: 5 FP halved for being pinned = 2.5, and Assault Fire does not raise it back up because the +1 Assault Fire bonus is applied after all modification to inherent FP, so a pinned 5-4-8 in the AFPh gets 3.5 only if using AFPh fire alone; however, in the APh a pinned unit cannot advance, so if you mean AFPh fire after moving, the pinned unit still fires in AFPh and Assault Fire applies to the halved AFPh FP, giving 4 FP if not otherwise restricted. For a pinned 5-4-8 simply attacking in the AFPh without movement issues, the correct AFPh FP is 4: 5 ÷ 2 = 2.5, +1 Assault Fire = 3.5, rounded up to 4 References: - (A7.36) Assault Fire — add +1 FP in the AFPh after all modification to inherent FP; fractions then round up - (A7.8) Pinned Firer — pinning halves FP - (A4.7) Advance Phase — pinned Infantry may not advance, relevant only if “in the APh” was meant literally rather than AFPh

[Calculation] The expected answer clearly states the calculation steps: start with 5 FP, halve for pinned (2.5), halve again for APh (1.25), add +1 for Assault Fire (2.25), then round up to 3 FP. The model's response is confused about the phase (APh vs AFPh), incorrectly states that Assault Fire does not raise the FP in this case, and provides inconsistent FP values (2.5, 3.5, 4) that do not match the expected calculation. The model also introduces irrelevant details about movement and phases that do not address the question directly. Therefore, the model's answer does not match the expected final FP value or the correct calculation steps.

A7.302

What does a K/# result do on the IFT?

A K/# causes at least one target unit in each specifically targeted Location to suffer Casualty Reduction, then all other target units in that Location must take a MC with the listed # added to the DR. Random Selection determines which target suffers the Casualty Reduction when needed.

Answer: A K/# result means at least one target unit in each specifically targeted Location suffers Casualty Reduction, and all other target units—including any newly created HS—must then take a Morale Check with the indicated number added to the MC DR. If there are multiple eligible targets, Random Selection determines which unit takes the Casualty Reduction. References: - (A7.302) K/# — one unit suffers Casualty Reduction; all others take a #MC - (A7.303) NMC — defines the Morale Check taken by the remaining units - (A7.304) #MC — the number after the slash is added to the MC DR - (A7.302) Casualty Reduction — HS/crew eliminated, squad reduced to HS, SMC wounded

[Explanation] The model's response accurately captures the key concepts of the expected answer: that a K/# result causes at least one target unit in each specifically targeted Location to suffer Casualty Reduction, that all other target units (including newly created HS) must take a Morale Check with the indicated number added to the DR, and that Random Selection determines which unit suffers the Casualty Reduction when multiple targets are eligible. The model also correctly references the relevant rules, adding useful context without contradicting the expected answer.

A8.2

A 4-6-7 squad fires at a unit in Open Ground at range 2 (PBF). What is the Residual FP left in the hex?

4 Residual FP. Attack FP = 4 * 2 (PBF) = 8 FP. Residual is half of 8 = 4 FP. (Resolved on 8 column, half is 4, which is a column).

Answer: A 4-6-7 has 6 FP, doubled to 12 FP at Point Blank Fire range 2, and Residual FP is half of the highest IFT column used, so it leaves 6 Residual FP in the target hex. Because the target is in Open Ground, there is no positive DRM to reduce that Residual FP. References: - (A7.21) Point Blank Fire — doubles Infantry FP at range ≤ half normal range - (A8.2) Residual Firepower — Residual FP is half of the highest IFT FP column used - (A8.26) Effect of DRM — Open Ground gives no positive DRM, so no reduction to Residual FP

[Calculation] The model's response incorrectly states the squad's FP as 6 instead of 4, leading to a doubled FP of 12 rather than 8 at PBF range 2. Consequently, the residual FP is calculated as 6 instead of the correct 4. The expected answer correctly identifies the squad's FP as 4, doubles it to 8, and halves it to 4 residual FP. The model's reasoning is flawed due to the incorrect initial FP value, resulting in an incorrect final residual FP.

A8.15

What are the key restrictions and effects of a Snap Shot?

A Snap Shot is a Small Arms/MG Defensive First Fire attack traced to a hexside crossed by a moving unit entering an on-board hex, even if the hex center is out of LOS. It is resolved as Area Fire; FFNAM/FFMO do not apply; ordnance and IFE/Canister cannot make Snap Shots; and a MG that must change CA cannot make one.

Answer: A Snap Shot is a Small-Arms/MG Defensive First Fire attack taken as a unit crosses a hexside the firer can see along its full length; it uses Area Fire (½ FP), ignores FFNAM/FFMO, and usually ignores the TEM of the hex being entered, though wall/hedge/SMOKE/rubble-hexside/spine effects can still apply if crossed by the LOF to that hexside. It cannot be used against a unit entering the firer’s own hex, cannot be made by ordnance, IFE/Canister, or a MG that must change CA, and only one Snap Shot may be made per firer against a unit crossing a given hexside; if hit by it, the target is thereafter considered in the entered Location. References: - (A8.15) Snap Shot — defines LOS to the crossed hexside, Area Fire/½ FP, no FFNAM/FFMO, and target considered in entered Location thereafter - (A8.15) Snap Shot restrictions — no Snap Shot into firer’s own hex; no ordnance/IFE/Canister; MG cannot Snap Shot if it must change CA; one per firer per crossed hexside - (C.5) Vertex Aiming Point — Snap Shot is traced to an entire hexside rather than a center dot, relevant to how hexside TEM exceptions apply - (B9.42) Vertex LOS — wall/hedge can affect LOS and provide TEM when a Snap Shot is traced to the target hexside/vertex

[Explanation] The model's response covers all key concepts from the expected answer: it correctly identifies a Snap Shot as a Small Arms/MG Defensive First Fire attack traced to a hexside crossed by a moving unit entering an on-board hex, even if the hex center is out of LOS. It states that the attack is resolved as Area Fire with half firepower, that FFNAM/FFMO do not apply, and that ordnance and IFE/Canister cannot make Snap Shots. It also correctly notes that a MG that must change CA cannot make one. Additionally, the model includes extra correct details such as the restriction against Snap Shots into the firer's own hex, the limit of one Snap Shot per firer per crossed hexside, and the effect of terrain effects on LOS to the hexside, which do not contradict the expected answer. The model's response is thorough, accurate, and semantically matches the expected answer, fulfilling the criteria for correctness.

A11.5

A 4-6-7 squad and a 2-2-8 crew attack a 4-4-7 squad in Close Combat. What are the odds?

3:2. Attack Strength: 4 + 2 = 6. Defense Strength = 4 (from the 4-4-7). 6:4 factors down to 3:2

Answer: The attack is 6 FP vs 4 FP, so the odds are 3:2 (rounded down on the CCT), with a Kill Number of 6. A Final CC DR of < 6 eliminates the 4-4-7, and a Final CC DR of = 6 causes Casualty Reduction. References: - (A11.11) Resolution — compare attacker FP to defender FP, round down to the nearest CCT odds, and use the Kill Number - (A11.12) Mechanics — all units attacking the same target add their FP together - (A11.1) Close Combat — CC between personnel uses FP, not morale/range/TEM

[Calculation] The model correctly calculates the attack strength as 6 (4 from the squad + 2 from the crew) and the defense strength as 4 from the defending squad. It correctly reduces the odds to 3:2, matching the expected answer. The model also provides additional relevant information about the Kill Number and references to the rules, which does not contradict the expected answer. Therefore, the response fully matches the expected answer and includes correct reasoning.

A14.4

What must a squad roll to eliminate the enemy Sniper during a Sniper Check?

4 or less. A squad has a -2 drm applied to the ≤2 to eliminate the Sniper.

Answer: A squad must get a Sniper Check Final DR of 2 or less to eliminate the enemy Sniper. A Final DR of 3 only pins the Sniper, and a squad contributes a -2 DRM to that combined Sniper Check DR References: - (A14.4) Sniper Check — enemy Sniper is eliminated on a Final DR of ≤ 2; 3 pins instead - (A14.4) Sniper Check DRM — each participating squad counts as crew/HS equivalent twice, i.e., a squad gives -2 DRM, as shown by the example

[Explanation] The model's response correctly states that a squad must roll a Final DR of 2 or less to eliminate the enemy Sniper, which matches the expected answer's '4 or less' after applying the -2 DRM. It also correctly explains that a squad contributes a -2 DRM to the Sniper Check DR, aligning with the expected answer's mention of the -2 DRM. The model further clarifies that a Final DR of 3 only pins the Sniper, which is consistent with the rules. The core concepts of the required roll and the DRM applied by the squad are fully covered, making the response correct.

A15.21

How is a Hero created in ASL, and what specific DR result triggers Hero creation?

A Hero (1-4-9 SMC) is randomly created from other Personnel types (even if broken) by a subsequent Final Heat of Battle DR of 6 or less. A Final Heat of Battle DR of 5 or 6 results in both Hero Creation and Battle Hardening of the original unit. A Hero never breaks or goes berserk, is immune to Cowering and enforced Pin results, and is always considered Stealthy.

Answer: A Hero is created via Heat of Battle: after a unit makes an Original 2 on a MC or Rally DR, it takes a Heat of Battle DR, and a subsequent Final Heat of Battle DR of ≤ 6 creates a Hero. More specifically, a Final Heat of Battle DR of 5 or 6 results in both Hero Generation and Battle Hardening, while ≤ 4 still gives Hero Creation without that combined note. References: - (A15.1) Heat of Battle — triggered after an Original 2 on a MC or Rally DR; Final Heat of Battle DR ≤ 6 gives Hero Creation - (A15.21) Creation — a Hero is randomly created from other Personnel during play by a subsequent Final Heat of Battle DR ≤ 6 - (A15.1) Heat of Battle Table — Final Heat of Battle DR of 5 or 6 results in both Hero Generation and Battle Hardening

[Explanation] The model's response correctly explains the process of Hero creation in ASL, including the triggering event (Heat of Battle after an Original 2 on a MC or Rally DR) and the specific Final Heat of Battle DR results that create a Hero (≤ 6). It also correctly notes that a DR of 5 or 6 results in both Hero Creation and Battle Hardening, while ≤ 4 results in Hero Creation alone. The model's answer aligns well with the expected answer's key concepts, including the DR thresholds and the process. Although the model does not explicitly mention that Heroes never break, go berserk, are immune to Cowering and enforced Pin results, and are always considered Stealthy, these are additional details not explicitly requested in the question. The core explanation and DR trigger are fully covered, so the response is judged correct.

A20.51

A full squad can guard how many full-squad prisoners?

5. A single Guard unit can guard prisoners with a total US# up to five times its own US#.

Answer: A full squad can guard prisoner units with a total US# up to five times its own US#, so a squad (US# 4) can guard up to 20 US# of prisoners—normally five full-squad prisoners. ASL also states this can be exchanged as two prisoner HS/crew for one unarmed squad, but the basic answer is five full squads. References: - (A20.51) Stacking — a single Guard unit can guard prisoners with a total US# up to five times its own US# - (A20.5) Guards & Unarmed Units — defines the captor as the Guard and how prisoner guarding works

[Explanation] The model's response correctly states that a full squad (US# 4) can guard prisoners with a total US# up to five times its own US#, which equals 20 US#, or normally five full-squad prisoners. It also adds relevant additional context about exchanging prisoner units, which does not contradict the expected answer. The key concept that a single Guard unit can guard prisoners up to five times its own US# is clearly covered, matching the expected answer.

A21.1

What restrictions apply to using captured enemy SW?

Captured SW has its B# and X# decreased by two due to lack of familiarity with the equipment (A21.11). Additionally, captured weapons must reduce any Multiple ROF capability by one (A21.12). Captured ordnance must use the red To Hit Numbers and add a +2 DRM to all To Hit attempts. Possession of enemy SW can occur only by Recovery (A4.44).

Answer: Captured enemy SW can only be possessed by Recovery, and once used they suffer captured-use penalties: B#/X# reduced by 2, any Multiple ROF reduced by 1, and captured SW/Guns can never be repaired unless recaptured by their original side. For captured ordnance, all attacks use red TH numbers and add +2 TH DRM; if the weapon normally requires a crew, non-qualified users may operate it only with the A21.11-.12 penalties, doubled for non-qualified MMC, while SMC have additional limits. References: - (A20.24) Abandonment on Capture — captured units first abandon all SW; captor must Recover them normally before possession/use. - (A21.1) Equipment Possession — enemy SW [EXC: radio] may be used by the new owner, but only if obtained by Recovery and subject to restrictions. - (A21.11) Malfunction — captured Gun/SW B#/X# reduced by two; Ammo Shortage penalties do not also apply. - (A21.12) Performance — captured weapons reduce Multiple ROF by one; captured ordnance uses red TH numbers and +2 TH DRM. - (A21.13) Non-Qualified Use — crew-required captured ordnance may be used only with non-qualified-use penalties; doubled for non-qualified MMC, with special SMC limits. - (A9.72) SW/Gun Repair — a captured SW/Gun may never be repaired unless recaptured by its original side.

[Explanation] The model's response covers all key concepts from the expected answer: possession only by Recovery, B# and X# reduced by two, Multiple ROF reduced by one, use of red To Hit numbers with +2 DRM, and the possession restriction. Additionally, it includes extra relevant details about repair restrictions and non-qualified use penalties, which do not contradict the expected answer but rather expand on it. The core meaning and all key points are present, so the response is correct.

A22.1

What is the Normal Range and Firepower of a standard Flamethrower (FT)?

Normal Range of one hex and a FP factor of 24.

Answer: A standard infantry Flamethrower has a Normal Range of 1 hex and attacks an adjacent hex at full FP; it may also attack at Long Range out to 2 hexes at halved FP. The standard FT Firepower is 16 FP at normal range and 8 FP at long range; if the FT FP were underlined, that would indicate a 2-hex normal range instead, but that is not the standard case. References: - (A22.32) FT Line of Fire — infantry FT attacks adjacent hex at full FP or two hexes away with Long Range Fire - (D1.3) Vehicle/weapon notation for FT — FT FP is underlined only if normal range is 2 hexes; otherwise normal range is 1 hex - (A1.21) Firepower — first number is FP value used for attack strength - (A1.22) Normal Range — range factor concept for normal vs long-range fire

[Explanation] The expected answer states the Normal Range is one hex and the FP factor is 24. The model's response correctly identifies the Normal Range as one hex but gives the FP as 16 at normal range and 8 at long range, which contradicts the expected FP of 24. The model also adds information about long range and underlined FP notation, which is accurate but does not align with the expected FP value. Since the FP value is a key fact and the model's FP is incorrect, the response is incorrect.

A24.2

If an MMC moves with non-assault movement into a hex with SMOKE and where they can claim Wall Advantage (WA), what would be the DRM for an IFT attack against them

Final DRM = +3; +2 [SMOKE] + 2 [Wall Advantage] -1 [First Fire Non-Assault Movement]

Answer: The DRM is normally **+1 net**: **+2 wall TEM** for claiming WA, **+1 SMOKE** (or whatever the SMOKE Hindrance value is), and **-1 FFNAM** for non-Assault Movement; **FFMO does not apply** because the unit can claim protective TEM/Hindrance. If the firer is **adjacent across that shared wall/hedge hexside and qualifies for WA**, the **wall TEM does not apply on entry**, so it would be just **SMOKE DRM minus FFNAM** (e.g., **0 net** with +1 SMOKE). References: - (B9.3) Wall/Hedge TEM — wall is +2, hedge is +1; applies to fire through that hexside against a target in that Location. - (B9.31) TEM and WA — wall/hedge TEM is not cumulative with other positive hex TEM, but **is cumulative with SMOKE**; a unit claiming WA gets wall/hedge TEM, not in-hex TEM. - (B9.321) WA vs adjacent entry — a unit entering an adjacent hex does **not** get wall TEM vs an adjacent firer who qualifies for WA over that shared hexside. - (A4 / FFNAM) Non-Assault Movement — moving without Assault Movement gives **FFNAM -1 DRM** during First Fire. - (A4 / FFMO) Movement in the Open — FFMO applies only if the unit enters Open Ground and **cannot claim protective TEM or LOS Hindrance DRM**.

[Calculation] The expected answer clearly states the final DRM as +3, calculated as +2 for SMOKE, +2 for Wall Advantage, and -1 for First Fire Non-Assault Movement. The model's response incorrectly states the DRM as +1 net, with +2 wall TEM, +1 SMOKE, and -1 FFNAM, which does not match the expected +3 total. Additionally, the model introduces conditions about adjacency and WA negating wall TEM that are not part of the original question or expected answer. The model also incorrectly assigns +1 SMOKE instead of +2 SMOKE as per the expected answer. Therefore, the model's final DRM value and reasoning do not align with the expected answer, making it incorrect.

A26.11

Does an enemy MMC under DM prevent the other player's ground unit from gaining control of a hex or building?

Yes, an MMC under DM is considered an armed enemy ground unit, which prevents the opposing player's MMC from gaining control.

Answer: No. A DM enemy MMC still prevents gaining Control if it is an armed enemy ground unit; being broken/DM does not negate that for Control purposes, and the rulebook example explicitly says a broken 4-4-7 prevents the opponent from gaining building and hex Control until it is eliminated References: - (A26.11) Gaining Control — Control requires an armed Good Order Infantry MMC and no armed enemy ground unit in the same Location/hex/building - (A26.13) Hex Control — hex Control is blocked by an armed enemy ground unit elsewhere in the hex/building example - (A26.14) Building Control — building Control can be gained only if A26.11’s conditions are met - (A26.1 Example) Control example — a broken 4-4-7 “is still an armed enemy unit and thus prevents” gaining building/hex Control

[Explanation] The expected answer is 'Yes,' meaning an enemy MMC under DM does prevent the other player's ground unit from gaining control. The model's response begins with 'No,' which directly contradicts the expected answer. Although the model correctly explains that a broken (DM) enemy MMC still prevents control and cites relevant rules and examples, the initial 'No' is a clear factual error that conflicts with the expected answer. This contradiction makes the overall response incorrect despite the correct supporting explanation.

B10.5

What is an Abrupt Elevation Change and when does it occur?

An Abrupt Elevation Change occurs when a unit enters two or more levels while crossing one non-cliff hexside. For example, going from a level 0 hex to a level 2 hex across a single hexside. Abrupt Elevation hexsides have special movement costs and restrictions.

Answer: An Abrupt Elevation Change is a special hexside/elevation situation where movement crosses more than the normal single-level change, so the cost of each intermediate level becomes cumulative rather than just applying the normal “higher level” doubling. It occurs when a unit crosses an Abrupt Elevation hexside; each intermediate level ascended costs 2 MF/4 MP and each intermediate level descended costs 1 MF/2 MP, with the final level also paying normal terrain entry cost . Normal elevation change only doubles entry cost when moving up one level, and there is normally no penalty for moving down except in cases of Abrupt Elevation Change . References: - (B10.51) Abrupt Elevation Change — defines it as crossing an Abrupt Elevation hexside and gives the cumulative intermediate-level MF/MP costs - (A4.133) Elevation Change — explains normal elevation change and notes that downward movement is penalized only for Abrupt Elevation Change - (B19.4) Gully Movement — shows Abrupt Elevation Change can arise in gully movement when crossing the higher elevation hexside - (B20.5) Wadi/Hill Depression Movement — shows Abrupt Elevation Change can also arise in wadis/hill depressions when crossing the higher crest/elevation relationship

[Explanation] The model's response correctly identifies that an Abrupt Elevation Change occurs when crossing an Abrupt Elevation hexside involving more than a single-level elevation change. It explains the cumulative movement costs for intermediate levels and distinguishes this from normal elevation changes. While the expected answer is more concise and focuses on the definition and example, the model's answer covers all key concepts including the occurrence condition, the special movement costs, and the nature of the hexside. The inclusion of references and detailed cost explanation adds correct context without contradiction.

B10.31

What is the TEM for Height Advantage when a firer is at a higher elevation than its target?

+1 TEM. A target at a lower elevation than its firer receives a +1 Height Advantage TEM.

Answer: Height Advantage gives the target a +1 TEM, but only when the target is receiving Direct Fire from a lower elevation and is not otherwise eligible for any other positive TEM or CE DRM (other than LOS Hindrances). So if the firer is higher than the target, there is no Height Advantage TEM for the target. References: - (B10.31) Height Advantage — target gets +1 TEM only vs Direct Fire from a lower elevation, if no other positive TEM applies - (B10.3) Hill TEM — absent other terrain, hill hexes are Open Ground unless Height Advantage applies

[Explanation] The model's response correctly explains that the Height Advantage TEM of +1 applies to the target when it is at a lower elevation than the firer, specifically against direct fire from a lower elevation. It clarifies that if the firer is higher, the target does not get the Height Advantage TEM, which aligns with the expected answer that the target at a lower elevation receives +1 TEM. The model also references the relevant rules, adding accuracy and context. Although the wording is more detailed and slightly different, the core concept matches the expected answer fully.

B13.31

If a squad enters a woods-road hex at road movement rate and is fired upon with Direct Fire where the LOS does not cross a green woods symbol, what TEM does it receive?

No TEM (Open Ground). A moving unit in a combination woods-road hex is not eligible for the +1 woods TEM during Defensive First Fire (and is subject to FFMO/Interdiction) if the LOS does not cross a green woods symbol and the moving unit entered the hex at the road movement rate

Answer: It receives no woods TEM: a squad entering a woods-road hex at the road movement rate is not eligible for the +1 woods TEM during Defensive First Fire if the LOS does not cross a green woods symbol, so its TEM is 0 from the terrain and it is subject to FFMO/Interdiction as applicable. References: - (B13.31) Woods-Road TEM exception — no +1 woods TEM during Defensive First Fire if entered at road rate and LOS does not cross a green woods symbol - (B13.3) Woods TEM — normal Direct Fire TEM for woods is +1 unless the B13.31 exception applies - (A4.132) Example — confirms no woods TEM for road-rate entry into a woods-road hex when LOS avoids the woods depiction

[Explanation] The model's response fully captures the key concepts from the expected answer: no TEM is applied because the squad entered the woods-road hex at road movement rate, the LOS does not cross a green woods symbol, and the unit is subject to FFMO/Interdiction. The model also correctly references the relevant rules and provides additional context without contradicting the expected answer.

B30.7

If an MMC is concealed ("?" counter) in a pillbox, are IFT attacks against it halved?

No. "?"/HIP don't halve the FP

Answer: No. Although concealment normally halves IFT fire as Area Fire, pillbox concealment is a specific exception: a “?”/HIP in a pillbox neither halves the FP of an attack on it nor adds a TH DRM, so IFT attacks against a concealed MMC in a pillbox are not halved. References: - (A12.13) Concealment Effect — normally all fire vs a concealed unit is halved as Area Fire - (B30.6) Pillbox Concealment — “?”/HIP in a pillbox neither halves FP nor adds TH DRM to attacks on it/its contents

[Explanation] The model's response correctly states that IFT attacks against a concealed MMC in a pillbox are not halved, matching the expected answer. It also provides the key concept that concealment normally halves fire but pillbox concealment is an exception, and cites the relevant rules (A12.13 and B30.6). This fully covers the expected answer's key points with accurate and relevant detail.

23.922

If an Infantry unit is prevented from entering a Fortified Building Location because of presence of an enemy unit in that hex, can it move elsewhere if it has remaining MFs?

No. An Infantry unit which attempts to move or advance into a Fortified Building Location occupied by such a squad loses that MPh or APh capability for that turn and must remain in its present hex during that phase.

Answer: No. If an Infantry unit attempts to enter a Fortified Building Location occupied by an unpinned, Good Order, armed enemy squad/equivalent and is denied entry, it loses its MPh or APh capability for that turn and must remain in its present hex during that phase, even if it would otherwise have MF left . References: - (B23.922) Entry into Fortified Building Location — denied entry causes the unit to lose that MPh/APh capability and remain in its present hex for that phase - (B23.922) MF expended on unsuccessful entry — those MF are considered expended in the present hex; after resolving any fire, the unit’s MPh is over

[Explanation] The model's response correctly states that the Infantry unit loses its MPh or APh capability for that turn and must remain in its present hex during that phase if denied entry into a Fortified Building Location due to enemy presence. It also correctly notes that this applies even if the unit has remaining Movement Factors. The model includes relevant references and additional clarifications that align with the expected answer without contradicting it. Therefore, it fully covers the key concepts of the expected answer.