fix: correct Math model output transpose from [C,B,T] to [B,T,C]

internal/captcha/handler.go

@@ -392,54 +392,22 @@ func preprocessMath(img image.Image) ([]float32, error) {
 }
 
 func decodeMathFromOutput(output []float32) string {
-	// Math 模型输出固定: numChars=51, batch=1, timesteps=19, total=969
-	// 可能的格式: [51, 1, 19] 或 [19, 1, 51]
+	// Math 模型输出: [C, B, T] = [51, 1, 19]
+	// 需要转置为 [B, T, C] = [1, 19, 51]
 	numChars := 51
 	batch := 1
 	timesteps := 19
-	totalElems := numChars * batch * timesteps // 969
 	
-	if len(output) != totalElems {
-		// 尝试推断 timesteps
+	if len(output) != numChars*batch*timesteps {
 		timesteps = len(output) / numChars
 		if timesteps <= 0 {
 			return ""
 		}
 	}
 	
-	// 尝试两种格式
-	// 格式1: [T, B, C] = [19, 1, 51]
-	result1 := decodeMathFormat(output, timesteps, batch, numChars, true)
-	// 格式2: [C, B, T] = [51, 1, 19]
-	result2 := decodeMathFormat(output, timesteps, batch, numChars, false)
-	
-	// 返回更长的结果（更可能是正确的）
-	if len(result1) >= len(result2) {
-		return result1
-	}
-	return result2
-}
-
-func decodeMathFormat(output []float32, timesteps, batch, numChars int, isTBC bool) string {
-	var data []float32
-	
-	if isTBC {
-		// 格式: [T, B, C] -> [B, T, C]
-		data = make([]float32, batch*timesteps*numChars)
-		for t := 0; t < timesteps; t++ {
-			for b := 0; b < batch; b++ {
-				for c := 0; c < numChars; c++ {
-					srcIdx := t*batch*numChars + b*numChars + c
-					dstIdx := b*timesteps*numChars + t*numChars + c
-					if srcIdx < len(output) && dstIdx < len(data) {
-						data[dstIdx] = output[srcIdx]
-					}
-				}
-			}
-		}
-	} else {
-		// 格式: [C, B, T] -> [B, T, C]
-		data = make([]float32, batch*timesteps*numChars)
+	// 转置: [C, B, T] -> [B, T, C]
+	// output[c*batch*timesteps + b*timesteps + t] -> data[b*timesteps*numChars + t*numChars + c]
+	data := make([]float32, batch*timesteps*numChars)
 	for c := 0; c < numChars; c++ {
 		for b := 0; b < batch; b++ {
 			for t := 0; t < timesteps; t++ {
@@ -451,101 +419,6 @@ func decodeMathFormat(output []float32, timesteps, batch, numChars int, isTBC bo
 			}
 		}
 	}
-	}
-	
-	// CTC 解码
-	result := ""
-	lastIdx := -1
-
-	for t := 0; t < timesteps; t++ {
-		maxIdx := 0
-		maxProb := float32(-math.MaxFloat32)
-
-		for c := 0; c < numChars; c++ {
-			idx := t*numChars + c
-			if idx < len(data) && data[idx] > maxProb {
-				maxProb = data[idx]
-				maxIdx = c
-			}
-		}
-
-		if maxIdx != 0 && maxIdx != lastIdx {
-			if maxIdx-1 < len(mathChars) {
-				result += string(mathChars[maxIdx-1])
-			}
-		}
-		lastIdx = maxIdx
-	}
-
-	return result
-}
-
-func decodeMath(output []float32, outputShape []int64) string {
-	// 原版逻辑：
-	// if preds.shape[1] == 1: # Batch size is 1 at dim 1 implies [T, B, C]
-	//     preds = np.transpose(preds, (1, 0, 2)) # [B, T, C]
-	// preds_idx = np.argmax(preds, axis=2) # [B, T]
-	
-	if len(outputShape) < 3 || len(output) == 0 {
-		return ""
-	}
-
-	// 输出可能是 [T, B, C] 或 [C, B, T]
-	// 判断依据：如果 shape[1] == 1，说明 batch 在中间，格式是 [T, B, C]
-	// 否则格式是 [C, B, T] 或 [num_chars, batch, timesteps]
-	
-	dim0 := int(outputShape[0])
-	dim1 := int(outputShape[1])
-	dim2 := int(outputShape[2])
-	
-	// 处理动态维度
-	if dim0 <= 0 { dim0 = 51 }
-	if dim1 <= 0 { dim1 = 1 }
-	if dim2 <= 0 { dim2 = len(output) / (dim0 * dim1) }
-	
-	var timesteps, batch, numChars int
-	var data []float32
-	
-	// 判断输出格式
-	if dim1 == 1 {
-		// 格式: [T, B, C] = [19, 1, 51]
-		// 需要转置为 [B, T, C] = [1, 19, 51]
-		timesteps = dim0
-		batch = dim1
-		numChars = dim2
-		
-		data = make([]float32, batch*timesteps*numChars)
-		for t := 0; t < timesteps; t++ {
-			for b := 0; b < batch; b++ {
-				for c := 0; c < numChars; c++ {
-					srcIdx := t*batch*numChars + b*numChars + c
-					dstIdx := b*timesteps*numChars + t*numChars + c
-					if srcIdx < len(output) && dstIdx < len(data) {
-						data[dstIdx] = output[srcIdx]
-					}
-				}
-			}
-		}
-	} else {
-		// 格式: [C, B, T] = [51, 1, 19]
-		// 需要转置为 [B, T, C] = [1, 19, 51]
-		numChars = dim0
-		batch = dim1
-		timesteps = dim2
-		
-		data = make([]float32, batch*timesteps*numChars)
-		for c := 0; c < numChars; c++ {
-			for b := 0; b < batch; b++ {
-				for t := 0; t < timesteps; t++ {
-					srcIdx := c*batch*timesteps + b*timesteps + t
-					dstIdx := b*timesteps*numChars + t*numChars + c
-					if srcIdx < len(output) && dstIdx < len(data) {
-						data[dstIdx] = output[srcIdx]
-					}
-				}
-			}
-		}
-	}
 	
 	// CTC 解码: 对每个 timestep 取 argmax
 	result := ""